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Class 11th Physics Chapter-8 Solutions
Mechanical Properties of Solids

Question 8.1

A steel wire of length 4.7 m and cross-sectional area 3.0 × 10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area 4.0 × 10⁻⁵ m² when both are subjected to the same load. Find the ratio of the Young’s modulus of steel to that of copper.

Solution

For a wire under tension, the extension is given by:

$Δ L = \frac{F L}{A Y}$

Since both wires experience the same load and same extension,

$\frac{F L_{s}}{A_{s} Y_{s}} = \frac{F L_{c}}{A_{c} Y_{c}}$

Cancelling $F$ and rearranging,

$\frac{Y_{s}}{Y_{c}} = \frac{L_{s} A_{c}}{L_{c} A_{s}}$

Substituting values:

$\frac{Y_{s}}{Y_{c}} = \frac{(4.7) (4.0 \times 10^{- 5})}{(3.5) (3.0 \times 10^{- 5})}$

$= \frac{18.8}{10.5} = 1.79 \approx 1.8$

Question 8.2

Figure 8.9 shows the stress–strain curve for a given material.
Find:
1. Young’s modulus of the material, and
2. Approximate yield strength.
Understanding the Graph (Figure 8.9)
- X-axis: Strain (dimensionless)
- Y-axis: Stress (in MPa)
- The curve is linear initially and then bends after a point.
The linear portion obeys Hooke’s law and its slope gives Young’s modulus.
The point where the curve deviates from linearity gives the yield strength.

(a) Young’s Modulus

From the straight-line portion of the graph (approximate reading):
- Stress ≈ 150 MPa
- Corresponding strain ≈ 0.002
$Young’s modulus (Y) = \frac{Stress}{Strain}$ $Y = \frac{150 MPa}{0.002} = 75,000 MPa = 7.5 \times 10^{10} {N m}^{- 2}$

Young’s Modulus

$Y \approx 7.5 \times 10^{10} {N m}^{- 2}$

(b) Approximate Yield Strength
- Yield point is where the curve starts deviating from linearity
- From the graph, this occurs at stress ≈ 250 MPa
Yield Strength

$Yield strength \approx 250 MPa$

Question 8.3

The stress–strain graphs for materials A and B are shown in Fig. 8.10.
The graphs are drawn to the same scale.

(a) Which of the materials has the greater Young’s modulus?
(b) Which of the two is the stronger material?

Solution

Key Concepts Used
1. Young’s Modulus (Y)
  - Equal to the slope of the linear (elastic) part of the stress–strain curve.
  - Steeper slope ⇒ larger Young’s modulus.
2. Strength of a Material
  - Determined by the maximum stress the material can withstand before breaking.
  - Higher maximum stress ⇒ stronger material.
(a) Greater Young’s Modulus
- From Fig. 8.10, material B has a steeper initial (linear) portion of the stress–strain curve than material A.
- Since:
  
  $Y = \frac{Stress}{Strain}$
  
  a steeper slope means larger Young’s modulus.
Answer (a):

$Material B has the greater Young’s modulus$

(b) Stronger Material
- Strength depends on the maximum stress reached before fracture.
- From the graph, material A can withstand a higher stress before breaking than material B.
Answer (b):

$Material A is the stronger material$

Question 8.4

Read the following statements carefully and state, with reasons, whether they are true or false:

(a) The Young’s modulus of rubber is greater than that of steel.
(b) The stretching of a coil is determined by its shear modulus.

Solution

(a) The Young’s modulus of rubber is greater than that of steel.

❌ False

Reason:
- Young’s modulus is a measure of stiffness (resistance to deformation).
- Rubber stretches much more than steel for the same applied force.
- Hence, rubber has a much smaller Young’s modulus compared to steel.
Conclusion:
Steel is more elastic (stiffer) than rubber, even though rubber stretches more.

(b) The stretching of a coil is determined by its shear modulus.

✅ True

Reason:
- When a coil (spring) is stretched, the wire forming the coil undergoes shearing deformation, not longitudinal stretching.
- Therefore, the restoring force and extension of a coil depend on the shear modulus (modulus of rigidity) of the material.
Question 8.5

Two wires of equal diameter 0.25 cm, one made of steel and the other of brass, are loaded as shown in Fig. 8.11. The unloaded length of the steel wire is 1.5 m and that of the brass wire is 1.0 m.

Calculate the elongations of the steel wire and the brass wire.

(Young’s modulus: Steel = $2.0 \times 10^{11} {N m}^{- 2}$ ,
Brass = $9.0 \times 10^{10} {N m}^{- 2}$ )

Solution

Given

Diameter of each wire:

$d = 0.25 cm = 2.5 \times 10^{- 3} m$

Radius:

$r = 1.25 \times 10^{- 3} m$

Cross-sectional area:

$A = π r^{2} = π (1.25 \times 10^{- 3})^{2} = 4.9 \times 10^{- 6} m^{2}$

Lengths:
- Steel wire: $L_{s} = 1.5 m$
- Brass wire: $L_{b} = 1.0 m$
Young’s modulus:
- Steel: $Y_{s} = 2.0 \times 10^{11} {N m}^{- 2}$
- Brass: $Y_{b} = 9.0 \times 10^{10} {N m}^{- 2}$
Load on each wire (from Fig. 8.11):

$F = 100 N$

Formula Used

$Δ L = \frac{F L}{A Y}$

(a) Elongation of Steel Wire

$Δ L_{s} = \frac{(100) (1.5)}{(4.9 \times 10^{- 6}) (2.0 \times 10^{11})}$

$Δ L_{s} = 1.53 \times 10^{- 4} m = 0.153 mm$

(b) Elongation of Brass Wire

$Δ L_{b} = \frac{(100) (1.0)}{(4.9 \times 10^{- 6}) (9.0 \times 10^{10})}$

$Δ L_{b} = 2.27 \times 10^{- 4} m = 0.227 mm$

Question 8.7

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner radius of each column is 30 cm and the outer radius is 60 cm. Assuming that the load is uniformly distributed, calculate the compressional strain of each column.

(Young’s modulus of mild steel = $2.0 \times 10^{11} {N m}^{- 2}$ ;
take $g = 9.8 {m s}^{- 2}$ )

Solution

Step 1: Total load on the structure

$W = m g = 50,000 \times 9.8 = 4.9 \times 10^{5} N$

Since there are four identical columns, load on each column:

$F = \frac{4.9 \times 10^{5}}{4} = 1.225 \times 10^{5} N$

Step 2: Cross-sectional area of one hollow column

Outer radius:

$R = 60 cm = 0.6 m$

Inner radius:

$r = 30 cm = 0.3 m$

$A = π (R^{2} - r^{2}) = π ({0.6}^{2} - {0.3}^{2})$

$A = π (0.36 - 0.09) = π (0.27) = 0.848 m^{2}$

Step 3: Stress on each column

$Stress = \frac{F}{A} = \frac{1.225 \times 10^{5}}{0.848}$

$Stress \approx 1.44 \times 10^{5} {N m}^{- 2}$

Step 4: Compressional strain

$Strain = \frac{Stress}{Y}$

$Strain = \frac{1.44 \times 10^{5}}{2.0 \times 10^{11}}$

$Strain = 7.2 \times 10^{- 7}$

Question 8.8

A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension by a force of 44,500 N, producing only elastic deformation.
Calculate the resulting strain.

(Young’s modulus of copper = $1.1 \times 10^{11} {N m}^{- 2}$ )

Solution

Step 1: Given data

Force applied:

$F = 44,500 N$

Cross-sectional dimensions:

$15.2 mm = 15.2 \times 10^{- 3} m$

$19.1 mm = 19.1 \times 10^{- 3} m$

Cross-sectional area:

$A = (15.2 \times 10^{- 3}) (19.1 \times 10^{- 3}) = 2.9032 \times 10^{- 4} m^{2}$

Young’s modulus of copper:

$Y = 1.1 \times 10^{11} {N m}^{- 2}$

Step 2: Calculate stress

$Stress = \frac{F}{A} = \frac{44,500}{2.9032 \times 10^{- 4}}$

$Stress \approx 1.53 \times 10^{8} {N m}^{- 2}$

Step 3: Calculate strain

$Strain = \frac{Stress}{Y} = \frac{1.53 \times 10^{8}}{1.1 \times 10^{11}}$

$Strain \approx 1.39 \times 10^{- 3}$

Question 8.9

A steel cable of radius 1.5 cm supports a chairlift at a ski area.
If the maximum stress in the cable is not to exceed

$1.0 \times 10^{8} {N m}^{- 2},$

calculate the maximum load that the cable can support.

Solution

Step 1: Given data

Radius of cable:

$r = 1.5 cm = 1.5 \times 10^{- 2} m$

Maximum allowable stress:

$σ_{\max} = 1.0 \times 10^{8} {N m}^{- 2}$

Step 2: Cross-sectional area of the cable

$A = π r^{2} = π (1.5 \times 10^{- 2})^{2}$

$A = π (2.25 \times 10^{- 4}) \approx 7.07 \times 10^{- 4} m^{2}$

Step 3: Maximum load supported

Stress is given by:

$σ = \frac{F}{A}$

$F_{\max} = σ_{\max} \times A$

$F_{\max} = (1.0 \times 10^{8}) (7.07 \times 10^{- 4})$

$F_{\max} \approx 7.07 \times 10^{4} N$

Question 8.10

A rigid bar of mass 15 kg is supported symmetrically by three vertical wires, each of length 2.0 m. The two end wires are of copper, while the middle wire is of iron. Determine the ratio of the diameters of the copper and iron wires if each wire is to have the same tension.

(Young’s modulus:
Copper $= 1.1 \times 10^{11} {N m}^{- 2}$ ,
Iron $= 2.0 \times 10^{11} {N m}^{- 2}$ )

Solution

Key Physical Idea
- The bar is rigid and supported symmetrically, so for the bar to remain horizontal:
  - All three wires must undergo the same extension.
- The tension in each wire is the same (given condition).
Step 1: Use the extension formula

For a stretched wire:

$Δ L = \frac{F L}{A Y}$

Here:
- $F$ = tension (same for all wires)
- $L$ = length (same for all wires)
- $A$ = cross-sectional area
- $Y$ = Young’s modulus
Step 2: Condition for equal extension

Since $Δ L$ is same for copper and iron wires:

$\frac{F L}{A_{c} Y_{c}} = \frac{F L}{A_{i} Y_{i}}$

Cancelling $F$ and $L$ :

$A_{c} Y_{c} = A_{i} Y_{i}$

Step 3: Express area in terms of diameter

$A = \frac{π d^{2}}{4}$

So, $d_{c}^{2} Y_{c} = d_{i}^{2} Y_{i}$ $\frac{d_{c}}{d_{i}} = \sqrt{\frac{Y_{i}}{Y_{c}}}$

Step 4: Substitute values

$\frac{d_{c}}{d_{i}} = \sqrt{\frac{2.0 \times 10^{11}}{1.1 \times 10^{11}}} = \sqrt{1.82} \approx 1.35$

Question 8.11

A mass of 14.5 kg, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle. At the lowest point of the circle, the angular speed is 2 revolutions per second. The cross-sectional area of the wire is 0.065 cm².

Calculate the elongation of the wire when the mass is at the lowest point of its path.

(Young’s modulus of steel = $2.0 \times 10^{11} {N m}^{- 2}$ )

Solution

Step 1: Given data

Mass:

$m = 14.5 kg$

Length of wire: $L = 1.0 m$

Angular speed:

$f = 2 rev/s$

$ω = 2 π f = 4 π rad/s$

Cross-sectional area:

$A = 0.065 {cm}^{2} = 0.065 \times 10^{- 4} m^{2} = 6.5 \times 10^{- 6} m^{2}$

Young’s modulus of steel:

$Y = 2.0 \times 10^{11} {N m}^{- 2}$

Step 2: Forces acting at the lowest point

At the lowest point, tension $T$ provides:
- Centripetal force
- Weight support
$T - m g = m ω^{2} r$ Here $r \approx L = 1.0 m$

$T = m ω^{2} L + m g$

Substitute values:

$T = 14.5 \times (4 π)^{2} \times 1.0 + 14.5 \times 9.8$

$T = 14.5 \times 16 π^{2} + 142.1$

$T \approx 14.5 \times 157.9 + 142.1$

$T \approx 2290 + 142$

$T \approx 2432 N$

Step 3: Formula for elongation

$Δ L = \frac{T L}{A Y}$

Step 4: Substitute values

$Δ L = \frac{2432 \times 1.0}{(6.5 \times 10^{- 6}) (2.0 \times 10^{11})}$

$Δ L = \frac{2432}{1.3 \times 10^{6}}$

$Δ L \approx 1.87 \times 10^{- 3} m$

Question 8.12

Compute the bulk modulus of water from the following data:
- Initial volume of water = 100.0 litre
- Final volume of water = 100.5 litre
- Increase in pressure = 100.0 atm
  
  $(1 atm = 1.013 \times 10^{5} Pa)$
Also, compare the bulk modulus of water with that of air (at constant temperature) and explain in simple terms why the ratio is so large.

Solution

Step 1: Convert given quantities into SI units

Initial volume:

$V = 100.0 litre = 0.100 m^{3}$

Final volume:

$V_{f} = 100.5 litre$

Change in volume:

$Δ V = 100.5 - 100.0 = 0.5 litre = 5.0 \times 10^{- 4} m^{3}$

Increase in pressure:

$Δ P = 100.0 \times 1.013 \times 10^{5} = 1.013 \times 10^{7} Pa$

Step 2: Formula for bulk modulus

$B = \frac{Δ P}{Δ V / V}$

Step 3: Substitute values

$\frac{Δ V}{V} = \frac{5.0 \times 10^{- 4}}{0.100} = 5.0 \times 10^{- 3}$

$B = \frac{1.013 \times 10^{7}}{5.0 \times 10^{- 3}}$

$B \approx 2.03 \times 10^{9} Pa$

(a) Bulk Modulus of Water

$B_{water} \approx 2.0 \times 10^{9} {N m}^{- 2}$

(This agrees well with the standard value.)

(b) Comparison with Bulk Modulus of Air
- Bulk modulus of air (at constant temperature):
  
  $B_{air} \approx 1.0 \times 10^{5} {N m}^{- 2}$
Ratio

$\frac{B_{water}}{B_{air}} = \frac{2.0 \times 10^{9}}{1.0 \times 10^{5}} = 2.0 \times 10^{4}$

$B_{water} \approx 20,000 \times B_{air}$

(c) Explanation (Why is the ratio so large?)
- In liquids (water), molecules are closely packed with very little empty space.
- Hence, liquids are very difficult to compress, giving a very large bulk modulus.
- In gases (air), molecules are far apart, so gases compress easily.
- Therefore, air has a very small bulk modulus compared to water.
Question 8.13

What is the density of water at a depth where the pressure is 80.0 atm, given that the density of water at the surface is

$1.03 \times 10^{3} {kg m}^{- 3} ?$

(Bulk modulus of water $= 2.2 \times 10^{9} {N m}^{- 2}$ ;
$1 atm = 1.013 \times 10^{5} Pa$ )

Solution

Step 1: Given data

Density at surface:

$ρ_{0} = 1.03 \times 10^{3} {kg m}^{- 3}$ Pressure at depth:

$Δ P = 80.0 atm = 80 \times 1.013 \times 10^{5} = 8.10 \times 10^{6} Pa$

Bulk modulus of water:

$B = 2.2 \times 10^{9} {N m}^{- 2}$

Step 2: Relation between density and bulk modulus

$B = \frac{Δ P}{Δ ρ / ρ}$

$\Rightarrow \frac{Δ ρ}{ρ} = \frac{Δ P}{B}$

Step 3: Calculate fractional change in density

$\frac{Δ ρ}{ρ} = \frac{8.10 \times 10^{6}}{2.2 \times 10^{9}} \approx 3.68 \times 10^{- 3}$

Step 4: Calculate change in density

$Δ ρ = ρ_{0} \times \frac{Δ ρ}{ρ}$

$Δ ρ = (1.03 \times 10^{3}) \times (3.68 \times 10^{- 3}) \approx 3.8 {kg m}^{- 3}$

Step 5: Density at depth

$ρ = ρ_{0} + Δ ρ = 1030 + 3.8$

$ρ \approx 1.034 \times 10^{3} {kg m}^{- 3}$

Question 8.14

Compute the fractional change in volume of a glass slab when it is subjected to a hydraulic pressure of 10 atm.

(Bulk modulus of glass $= 3.7 \times 10^{10} {N m}^{- 2}$ ;
$1 atm = 1.013 \times 10^{5} Pa$ )

Solution

Step 1: Given data

Pressure applied:

$Δ P = 10 atm = 10 \times 1.013 \times 10^{5} = 1.013 \times 10^{6} Pa$

Bulk modulus of glass:

$B = 3.7 \times 10^{10} {N m}^{- 2}$

Step 2: Formula for bulk modulus

$B = \frac{Δ P}{Δ V / V}$

$\Rightarrow \frac{Δ V}{V} = \frac{Δ P}{B}$

Step 3: Substitute values

$\frac{Δ V}{V} = \frac{1.013 \times 10^{6}}{3.7 \times 10^{10}}$

$$ $\frac{Δ V}{V} \approx 2.74 \times 10^{- 5}$

Question 8.15

Determine the volume contraction of a solid copper cube, each edge of which is 10 cm, when it is subjected to a hydraulic pressure of $7.0 \times 10^{6} Pa$ .

(Bulk modulus of copper $= 1.4 \times 10^{11} {N m}^{- 2}$ )

Solution

Step 1: Given data

Edge of the cube:

$a = 10 cm = 0.10 m$ Initial volume:

$V = a^{3} = (0.10)^{3} = 1.0 \times 10^{- 3} m^{3}$

Applied pressure:

$Δ P = 7.0 \times 10^{6} Pa$

Bulk modulus of copper:

$B = 1.4 \times 10^{11} {N m}^{- 2}$

Step 2: Formula used

Bulk modulus is given by:

$B = \frac{Δ P}{Δ V / V}$ Rearranging,

$Δ V = \frac{Δ P}{B} \times V$

Step 3: Substitute values

$Δ V = \frac{7.0 \times 10^{6}}{1.4 \times 10^{11}} \times 1.0 \times 10^{- 3}$

$Δ V = 5.0 \times 10^{- 8} m^{3}$

Question 8.16

How much should the pressure on one litre of water be changed so as to compress it by 0.10%?

(Bulk modulus of water $= 2.2 \times 10^{9} {N m}^{- 2}$ )

Solution

Step 1: Given data

Volume of water:

$V = 1 litre$

(Note: actual volume value is not required because we are given fractional change.)

Fractional change in volume:

$\frac{Δ V}{V} = 0.10 % = \frac{0.10}{100} = 1.0 \times 10^{- 3}$

Bulk modulus of water:

$B = 2.2 \times 10^{9} {N m}^{- 2}$

Step 2: Formula for bulk modulus

$B = \frac{Δ P}{Δ V / V}$

Rearranging,

$Δ P = B \times \frac{Δ V}{V}$

Step 3: Substitute values

$Δ P = (2.2 \times 10^{9}) \times (1.0 \times 10^{- 3})$

$Δ P = 2.2 \times 10^{6} Pa$
December 19, 2025
Class 11th Physics Chapter-5 Solutions
Go back to CLASS 11th Physics Page

Question 5.1

The sign of work done by a force on a body is important to understand. State carefully whether the following quantities are positive or negative:

Answer:

(a) Work done by a man in lifting a bucket

Positive
The force applied by the man is in the same direction as the displacement of the bucket (upward), so the work done is positive.

(b) Work done by gravitational force

Negative
Gravity acts downward while the bucket moves upward. Since force and displacement are in opposite directions, the work done by gravity is negative.

(c) Work done by friction on a body sliding down an inclined plane

Negative
Friction always acts opposite to the direction of motion. As the body slides down, friction acts upward along the plane, so its work is negative.

(d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity

Positive
The applied force acts in the direction of motion to balance friction. Hence, the work done by the applied force is positive
(while friction does equal negative work).

(e) Work done by the resistive force of air on a vibrating pendulum

Negative
Air resistance opposes the motion of the pendulum and removes energy from it, bringing it to rest. Therefore, the work done by air resistance is negative.

Question 5.2

A body of mass 2 kg, initially at rest, moves under the action of an applied horizontal force of 7 N on a table. The coefficient of kinetic friction between the body and the table is 0.1.

Compute:
(a) work done by the applied force in 10 s,
(b) work done by friction in 10 s,
(c) work done by the net force on the body in 10 s,
(d) change in kinetic energy of the body in 10 s,

and interpret the results.

Solution

Given:

$m = 2 kg, F = 7 N, μ_{k} = 0.1, t = 10 s$

Take $g = 10 {m s}^{- 2}$

Step 1: Forces acting on the body

Normal reaction:

$N = m g = 2 \times 10 = 20 N$

Frictional force:

$f_{k} = μ_{k} N = 0.1 \times 20 = 2 N$

Net force:

$F_{net} = 7 - 2 = 5 N$

Step 2: Acceleration of the body

$a = \frac{F_{net}}{m} = \frac{5}{2} = 2.5 {m s}^{- 2}$

Step 3: Displacement in 10 s

Initial velocity $u = 0$

$s = u t + \frac{1}{2} a t^{2}$

$s = 0 + \frac{1}{2} \times 2.5 \times (10)^{2}$

$s = 125 m$

(a) Work done by the applied force

$W_{applied} = F \times s = 7 \times 125$

$W_{applied} = 875 J$

(b) Work done by friction

Friction acts opposite to motion:

$W_{friction} = - f_{k} \times s = - 2 \times 125$

$W_{friction} = - 250 J$

(c) Work done by the net force

$W_{net} = F_{net} \times s = 5 \times 125$

$W_{net} = 625 J$

(d) Change in kinetic energy

Final velocity:

$v = u + a t = 0 + 2.5 \times 10 = 25 {m s}^{- 1}$

Initial kinetic energy:

$K_{i} = 0$

Final kinetic energy:

$K_{f} = \frac{1}{2} m v^{2} = \frac{1}{2} \times 2 \times 25^{2} = 625 J$

$Δ K = 625 J$

Question 5.3

Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the particleis indicated by a cross on the ordinate axis.

In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

Answer:

For a particle moving in one dimension, the total energy $E$ is the sum of kinetic and potential energies:

$E = K + V (x)$

Since kinetic energy $K \geq 0$ , the particle can exist only in regions where

$E \geq V (x)$

Regions where $V (x) > E$ are classically forbidden.

Case (a): Constant potential energy

Regions where the particle cannot be found:
- None, provided $E$ is greater than or equal to the constant value of $V$ .
Minimum total energy:

$E_{\min} = V$

Physical context:
- A free particle moving on a smooth horizontal surface.
- An idealised electron moving freely inside a metal.
Case (b): Potential well (minimum in the middle)

Regions where the particle cannot be found:
- Regions on either side of the well where
$V (x) > E$
- The particle is confined between two turning points where $E = V (x)$ .
Minimum total energy:

$E_{\min} = V_{\min}$

(the lowest value of the potential energy)

Physical context:
- A ball in a valley.
- A mass–spring system oscillating about equilibrium.
- Vibrational motion of atoms in a molecule.
Case (c): Potential barrier or step potential

Regions where the particle cannot be found:
- If $E < V_{0}$ , the region beyond the barrier (where $V = V_{0}$ ) is forbidden.
- If $E \geq V_{0}$ , there is no forbidden region.
Minimum total energy:

$E_{\min} = lowest value of V (x)$

Physical context:
- A particle approaching a wall or step.
- A classical analogue of a particle facing a potential barrier.
Question 5.4

The potential energy function for a particle executing linear simple harmonic motion is given by

$V (x) = \frac{1}{2} k x^{2},$ where $k$ is the force constant of the oscillator. For

$k = 0.5 {N m}^{- 1},$

the graph of $V (x)$ versus $x$ is shown in Fig. 5.12. Show that a particle of total energy 1 J moving under this potential must turn back when it reaches

$x = \pm 2 m .$

Solution

For motion in one dimension, the total energy $E$ of the particle is

$E = K + V (x),$

where $K$ is kinetic energy and $V (x)$ is potential energy.

Since kinetic energy cannot be negative,

$K \geq 0 \Rightarrow E \geq V (x) .$

Step 1: Write the potential energy function

Given:

$V (x) = \frac{1}{2} k x^{2}$

Substitute $k = 0.5 {N m}^{- 1}$ :

$V (x) = \frac{1}{2} \times 0.5 x^{2} = 0.25 x^{2}$

Step 2: Find the turning points

At the turning points, the particle momentarily comes to rest, so

$K = 0 \Rightarrow E = V (x)$

Given total energy:

$E = 1 J$

So,

$0.25 x^{2} = 1$

$x^{2} = 4$

$x = \pm 2 m$

Question 5.5

Answer the following:

(a)

The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained — the rocket or the atmosphere?

Answer:
The heat energy required for burning is obtained at the expense of the rocket’s kinetic energy.
Due to air friction, part of the rocket’s mechanical energy is converted into heat. The atmosphere only acts as a medium; it does not supply energy.

(b)

Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?

Answer:
Gravitational force is a conservative force.
For conservative forces, the work done over a closed path is zero, irrespective of the shape of the path.
Since a complete orbit is a closed path, the total work done by gravity on the comet over one revolution is zero.

(c)

An artificial satellite orbiting the earth in a very thin atmosphere loses its energy gradually due to atmospheric resistance. Why then does its speed increase as it comes closer and closer to the earth?

Answer:
As the satellite loses energy, it moves to a lower orbit, closer to the earth.
In a lower orbit, the gravitational pull is stronger, and the satellite must move with a higher speed to remain in orbit.
Thus, although the total mechanical energy decreases, the kinetic energy (and speed) increases as the satellite approaches the earth.

(d)

In Fig. 5.13(i), a man walks 2 m carrying a mass of 15 kg in his hands. In Fig. 5.13(ii), he walks the same distance pulling a rope over a pulley with a 15 kg mass hanging at the other end. In which case is the work done greater?

Answer:
- Case (i): Carrying the mass
  The force applied by the man on the mass is vertical, while the displacement is horizontal.
  Since force and displacement are perpendicular,
  $W = F d \cos 90^{\circ} = 0$
  Hence, no work is done on the mass.
- Case (ii): Pulling the rope
  The applied force causes the hanging mass to move upward.
  Force and displacement are in the same direction, so positive work is done.
$More work is done in case (ii).$

Question 5.6

Underline the correct alternative:

(a) When a conservative force does positive work on a body, the potential energy of the body
increases / decreases / remains unaltered

✔ Correct: decreases
(Because $Δ V = - W$ )

(b) Work done by a body against friction always results in a loss of its
kinetic / potential energy

✔ Correct: kinetic
(Friction converts kinetic energy into heat)

(c) The rate of change of total momentum of a many-particle system is proportional to the
external force / sum of the internal forces on the system

✔ Correct: external force
(Internal forces cancel in pairs)

(d) In an inelastic collision of two bodies, the quantity which does not change after the collision is
total kinetic energy / total linear momentum / total energy of the system of two bodies

✔ Correct: total linear momentum

Question 5.7

State whether each of the following statements is true or false. Give reasons for your answer.

(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.

False.
In an elastic collision, the total linear momentum and total kinetic energy of the system are conserved, not those of each individual body. The momentum and kinetic energy of each body generally change due to interaction during the collision.

(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.

True.
The total energy of the universe is always conserved. Energy may transform from one form to another (mechanical, thermal, chemical, etc.), but it is never destroyed. External forces may change the mechanical energy of a system, but the total energy remains conserved when all forms are included.

(c) Work done in the motion of a body over a closed loop is zero for every force in nature.

False.
This is true only for conservative forces (like gravity or spring force).
For non-conservative forces such as friction, the work done over a closed path is not zero.

(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

True.
In an inelastic collision, part of the initial kinetic energy is converted into other forms of energy such as heat, sound, or deformation. Hence, the final kinetic energy is less than the initial kinetic energy (maximum loss occurs in a perfectly inelastic collision).

Question 5.8

Answer carefully, with reasons:

(a)

In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?

Answer: No.
During the short time when the two balls are in contact, they are deformed and part of their kinetic energy is temporarily stored as elastic potential energy. Hence, the total kinetic energy is not conserved at every instant during the collision.
However, once the collision is over and the balls separate, the total kinetic energy of the system is restored to its initial value.

(b)

Is the total linear momentum conserved during the short time of an elastic collision of two balls?

Answer: Yes.
The forces acting between the two balls during collision are internal forces, which are equal and opposite at every instant (Newton’s third law). Therefore, the total linear momentum of the system remains conserved at all times, including during the collision.

(c)

What are the answers to (a) and (b) for an inelastic collision?
- Kinetic energy: Not conserved, even after the collision.
  In an inelastic collision, a part of the initial kinetic energy is permanently converted into other forms of energy such as heat, sound, or energy of deformation.
- Linear momentum: Conserved during the collision.
  As in all collisions, the internal forces between the bodies are equal and opposite at every instant, so the total linear momentum of the system remains conserved (provided external forces are negligible).
(d)

If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic?

Answer: The collision is elastic.

Reason:
If the potential energy depends only on the separation between the centres of the balls, the force during collision is a conservative force. For conservative forces, energy stored during deformation is completely recovered when the bodies separate. Hence, no kinetic energy is lost, and the collision is elastic.

Question 5.9

A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time $t$ is proportional to:

(i) $t^{1 / 2}$
(ii) $t$
(iii) $t^{3 / 2}$
(iv) $t^{2}$

Solution

For constant acceleration $a$ :

Velocity at time $t$ :

$v = a t$ Force on the body: $F = m a = constant$ Instantaneous power:

$P = F v$

$P \propto v \propto t$

Correct answer (5.9):

$(i i) t$

Question 5.10

A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time $t$ is proportional to:

(i) $t^{1 / 2}$
(ii) $t$
(iii) $t^{3 / 2}$
(iv) $t^{2}$

Solution

Power is given by:

$P = F v$

Since power is constant:

$F \propto \frac{1}{v}$

Acceleration: $a = \frac{F}{m} \propto \frac{1}{v}$

But:

$a = \frac{d v}{d t}$

$\frac{d v}{d t} \propto \frac{1}{v}$ $v d v \propto d t$

Integrating:

$v^{2} \propto t \Rightarrow v \propto t^{1 / 2}$

Now, displacement:

$s = \int v d t \propto \int t^{1 / 2} d t$

$s \propto t^{3 / 2}$

Correct answer (5.10):

$(i i i) t^{3 / 2}$

Question 5.11

A body constrained to move along the z-axis of a coordinate system is subject to a constant force

$\vec{F} = - \hat{i} + 2 \hat{j} + 3 \hat{k} N$

where $\hat{i}, \hat{j}, \hat{k}$ are unit vectors along the x-, y- and z-axes respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?

Solution

Displacement vector:

$\vec{s} = 4 \hat{k} m$ Work done:

$W = \vec{F} \cdot \vec{s} = (- \hat{i} + 2 \hat{j} + 3 \hat{k}) \cdot (4 \hat{k})$ Using dot products:

$\hat{i} \cdot \hat{k} = 0, \hat{j} \cdot \hat{k} = 0, \hat{k} \cdot \hat{k} = 1$

$W = 3 \times 4 = 12 J$

Question 5.12

An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds.

$m_{e} = 9.11 \times 10^{- 31} kg, m_{p} = 1.67 \times 10^{- 27} kg, 1 eV = 1.60 \times 10^{- 19} J$

Solution

For non-relativistic speeds, kinetic energy:

$K = \frac{1}{2} m v^{2} \Rightarrow v = \sqrt{\frac{2 K}{m}}$

Convert energies to joules
- Electron:
$K_{e} = 10 keV = 10 \times 10^{3} \times 1.60 \times 10^{- 19} = 1.60 \times 10^{- 15} J$
- Proton:
$K_{p} = 100 keV = 100 \times 10^{3} \times 1.60 \times 10^{- 19} = 1.60 \times 10^{- 14} J$

Ratio of speeds

$\frac{v_{e}}{v_{p}} = \sqrt{\frac{K_{e} / m_{e}}{K_{p} / m_{p}}} = \sqrt{\frac{K_{e} m_{p}}{K_{p} m_{e}}}$ $= \sqrt{\frac{(1.60 \times 10^{- 15}) (1.67 \times 10^{- 27})}{(1.60 \times 10^{- 14}) (9.11 \times 10^{- 31})}}$ Canceling common factors:

$= \sqrt{0.1 \times \frac{1.67 \times 10^{- 27}}{9.11 \times 10^{- 31}}} = \sqrt{0.1 \times 1836} = \sqrt{183.6}$

$\frac{v_{e}}{v_{p}} \approx 13.5$

Question 5.13

A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of air) until, at half its original height, it attains its terminal speed, and moves with uniform speed thereafter.

What is the work done by the gravitational force on the drop in the first and second half of its journey?
What is the work done by the resistive force during the entire journey, if its speed on reaching the ground is 10 m s⁻¹?

Solution

Given
- Radius of drop:
$r = 2 mm = 2 \times 10^{- 3} m$
- Height fallen:
$h = 500 m$
- Final speed:
$v = 10 {m s}^{- 1}$
- Density of water:
$ρ = 1000 {kg m}^{- 3}$
- Take
$g = 10 {m s}^{- 2}$

Step 1: Mass of the rain drop

Volume of the drop:

$V = \frac{4}{3} π r^{3} = \frac{4}{3} π (2 \times 10^{- 3})^{3} = 3.35 \times 10^{- 8} m^{3}$ Mass:

$m = ρ V = 1000 \times 3.35 \times 10^{- 8} = 3.35 \times 10^{- 5} kg$

(a) Work done by gravity

Work done by gravity depends only on vertical displacement, not on speed.

First half (250 m):

$W_{g 1} = m g (250) = 3.35 \times 10^{- 5} \times 10 \times 250$

$W_{g 1} \approx 0.084 J$

Second half (250 m):

$W_{g 2} = m g (250)$

$W_{g 2} \approx 0.084 J$

(b) Work done by resistive force (entire journey)

Total work done by gravity

$W_{g} = m g (500) = 3.35 \times 10^{- 5} \times 10 \times 500$

$W_{g} = 0.168 J$

Change in kinetic energy

Initial speed = 0
Final speed = 10 m s⁻¹

$Δ K = \frac{1}{2} m v^{2} = \frac{1}{2} \times 3.35 \times 10^{- 5} \times (10)^{2}$

$Δ K = 1.675 \times 10^{- 3} J$

Using work–energy theorem

$Δ K = W_{g} + W_{r}$

$W_{r} = Δ K - W_{g}$

$W_{r} = (1.675 \times 10^{- 3}) - 0.168$

$W_{r} \approx - 0.166 J$

(The negative sign shows energy loss due to air resistance.)

Question 5.14

A molecule in a gas container hits a horizontal wall with speed 200 m s⁻¹ making an angle of 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision?
Is the collision elastic or inelastic?

Answer

Momentum conservation
- The momentum of the molecule alone is not conserved.
- During the collision, the wall exerts a force on the molecule, changing the component of momentum normal to the wall.
- Hence, the molecule’s momentum changes.
However,
- If we consider the combined system of molecule + wall (or Earth), then the total momentum of the system is conserved, because the force between the molecule and the wall is internal to this larger system.
$Momentum of the molecule alone is not conserved;$

$momentum of the system is conserved.$

Nature of collision (elastic or inelastic)
- The molecule rebounds with the same speed as before collision.
- Hence, its kinetic energy remains unchanged.
- No kinetic energy is lost in the collision.
$The collision is elastic.$

Question 5.15

A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m³ in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

Solution

Given
- Volume of water, $V = 30 m^{3}$
- Time, $t = 15 min = 900 s$
- Height, $h = 40 m$
- Efficiency, $η = 30 % = 0.30$
- Density of water, $ρ = 1000 {kg m}^{- 3}$
- $g = 10 {m s}^{- 2}$
Step 1: Mass of water

$m = ρ V = 1000 \times 30 = 3.0 \times 10^{4} kg$

Step 2: Work done in lifting the water

$W = m g h = 3.0 \times 10^{4} \times 10 \times 40$

$W = 1.2 \times 10^{7} J$

Step 3: Useful power output of the pump

$P_{out} = \frac{W}{t} = \frac{1.2 \times 10^{7}}{900}$

$P_{out} \approx 1.33 \times 10^{4} W$

Step 4: Electric power consumed

$η = \frac{P_{out}}{P_{in}} \Rightarrow P_{in} = \frac{P_{out}}{η}$

$P_{in} = \frac{1.33 \times 10^{4}}{0.30}$

$P_{in} \approx 4.4 \times 10^{4} W$

Question 5.16

Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass, moving initially with speed V. If the collision is elastic, which of the following (Fig. 5.14) is a possible result after collision?

Answer and Explanation

The correct result is the one in which:
- The incoming ball comes to rest,
- The middle ball remains at rest,
- The last ball moves forward with speed V.
Reason

All three balls have the same mass, and the collision is elastic. Therefore:
1. Total linear momentum is conserved
2. Total kinetic energy is conserved
This situation is identical to the well-known Newton’s cradle effect.
- Momentum before collision:
$m V$
- After collision, for momentum and kinetic energy to be conserved:
  - Only one ball can move with speed $V$ ,
  - That ball must be the last one,
  - The other two must remain at rest.
Any other outcome (e.g., two balls moving, or all three moving) would violate either momentum conservation or kinetic energy conservation.

Final Answer

$The incoming ball stops and only the farthest ball moves with speed V .$

Hence, the figure showing only the last ball moving with speed V is the correct one.

Question 5.17

The bob A of a pendulum, released from 30° to the vertical, hits another bob B of the same mass at rest on a table, as shown in Fig. 5.15. How high does bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Solution

Step 1: Speed of bob A just before collision

Bob A is released from an angle $30^{\circ}$ .
Loss in gravitational potential energy = gain in kinetic energy.

If $l$ is the length of the pendulum, the vertical drop is:

$h = l (1 - \cos 30^{\circ})$

Using energy conservation:

$\frac{1}{2} m v^{2} = m g l (1 - \cos 30^{\circ})$

This gives the speed $v$ of bob A just before collision.

Step 2: Nature of collision
- Collision is elastic
- Masses of A and B are equal
- Bob B is initially at rest
- Collision is head-on
For a head-on elastic collision between equal masses:
- The moving body (A) comes to rest
- The stationary body (B) moves off with the same speed
So, immediately after collision:

$v_{A} = 0$

Step 3: Rise of bob A after collision

Since bob A comes to rest immediately after collision, it has no kinetic energy left to convert into potential energy.

Hence, it does not rise at all.

Question 5.18

The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipates 5% of its initial energy against air resistance?

Solution

Step 1: Initial potential energy

When released from the horizontal position, the vertical drop of the bob equals the length of the pendulum:

$h = l = 1.5 m$

Initial potential energy:

$E_{i} = m g h$ Take:

$g = 10 {m s}^{- 2}$

$E_{i} = m \times 10 \times 1.5 = 15 m J$

Step 2: Energy lost due to air resistance

Given that 5% of initial energy is dissipated:

$E_{lost} = 0.05 \times 15 m = 0.75 m$

Remaining energy at the lowest point:

$E_{f} = 15 m - 0.75 m = 14.25 m$

Step 3: Kinetic energy at the lowest point

At the lowest point, all remaining energy is kinetic:

$\frac{1}{2} m v^{2} = 14.25 m$ Cancel $m$ :

$\frac{1}{2} v^{2} = 14.25$

$v^{2} = 28.5$

$v = \sqrt{28.5} \approx 5.34 {m s}^{- 1}$

Question 5.19

A trolley of mass 300 kg carrying a sandbag of mass 25 kg is moving uniformly with a speed of 27 km h⁻¹ on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s⁻¹.

What is the speed of the trolley after the entire sandbag is empty?

Solution

Key physical idea
- The track is frictionless, so there is no external horizontal force on the trolley–sand system.
- The sand leaks vertically downward, so it carries no horizontal momentum relative to the ground.
- Hence, the horizontal momentum of the trolley does not change.
Apply conservation of momentum

Initial total mass:

$M_{i} = 300 + 25 = 325 kg$

Initial speed: $u = 27 {km h}^{- 1} = 7.5 {m s}^{- 1}$ Initial momentum:

$p_{i} = 325 \times 7.5$

After the entire sandbag is empty:

Final mass of trolley:

$M_{f} = 300 kg$ Let final speed be $v$ .

Since no external horizontal force acts,

$p_{i} = p_{f}$

$325 \times 7.5 = 300 \times v$

Solve for $v$

$v = \frac{325 \times 7.5}{300}$

$v = 8.125 {m s}^{- 1}$ Convert back to km h⁻¹:

$v = 8.125 \times 3.6 \approx 29.3 {km h}^{- 1}$

Question 5.20

A body of mass 0.5 kg travels in a straight line with velocity

$v = a x^{3 / 2},$ where

$a = 5 m^{- 1 / 2} s^{- 1} .$

What is the work done by the net force during its displacement from

$x = 0 to x = 2 m ?$

Solution

The work done by the net force equals the change in kinetic energy (work–energy theorem):

$W = Δ K$

Step 1: Expression for kinetic energy

$K = \frac{1}{2} m v^{2}$ Given: $v = a x^{3 / 2} \Rightarrow v^{2} = a^{2} x^{3}$

$K = \frac{1}{2} m a^{2} x^{3}$

Step 2: Initial and final kinetic energies
- At $x = 0$ :
$K_{i} = \frac{1}{2} m a^{2} (0)^{3} = 0$
- At $x = 2$ :
$K_{f} = \frac{1}{2} \times 0.5 \times (5)^{2} \times (2)^{3}$

$K_{f} = \frac{1}{4} \times 25 \times 8$

$K_{f} = 50 J$

Step 3: Work done

$W = K_{f} - K_{i} = 50 - 0$

Question 5.21

The blades of a windmill sweep out a circle of area A.

(a) If the wind flows with velocity v perpendicular to the circle, what is the mass of air passing through it in time t?
(b) What is the kinetic energy of the air?
(c) Assume that the windmill converts 25% of the wind’s energy into electrical energy.
Given:

$A = 30 m^{2}, v = 36 {km h}^{- 1}, ρ = 1.2 {kg m}^{- 3}$

Find the electrical power produced.

Solution

(a) Mass of air passing in time $t$

In time $t$ , air travels a distance $v t$ .
Volume of air crossing the area:

$V = A v t$

Mass of air:

$m = ρ V = ρ A v t$

$m = ρ A v t$

(b) Kinetic energy of the air

$K = \frac{1}{2} m v^{2}$

Substitute $m = ρ A v t$ :

$K = \frac{1}{2} ρ A v t v^{2}$

$K = \frac{1}{2} ρ A v^{3} t$

(c) Electrical power produced

First convert speed:

$v = 36 {km h}^{- 1} = 10 {m s}^{- 1}$

Power available in the wind

$P_{wind} = \frac{K}{t} = \frac{1}{2} ρ A v^{3}$

$P_{wind} = \frac{1}{2} \times 1.2 \times 30 \times (10)^{3}$

$P_{wind} = 0.6 \times 30 \times 1000 = 18000 W$

Electrical power produced (25% efficiency)

$P_{electric} = 0.25 \times 18000$

$P_{electric} = 4500 W = 4.5 kW$

Question 5.22

A person trying to lose weight (dieter) lifts a 10 kg mass, 1000 times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.

(a) How much work does she do against the gravitational force?
(b) Fat supplies $3.8 \times 10^{7}$ J of energy, which is converted to mechanical energy with 20% efficiency. How much fat will the dieter use up?

Solution

Given
- Mass lifted, $m = 10$
- Height each time, $h = 0.5$
- Number of lifts, $n = 1000$
- $g = 10$
- Energy from fat = $3.8 \times 10^{7}$
- Efficiency, $η = 20 % = 0.20$
(a) Work done against gravity

Work done in one lift:

$W_{1} = m g h = 10 \times 10 \times 0.5 = 50 J$

Work done in 1000 lifts:

$W = 1000 \times 50 = 5.0 \times 10^{4} J$

(The energy lost while lowering is dissipated, so it does not cancel this work.)

(b) Fat consumed

Only 20% of the energy from fat is converted into useful mechanical work.

So, energy obtained from fat:

$E_{fat} = \frac{W}{η} = \frac{5.0 \times 10^{4}}{0.20} = 2.5 \times 10^{5} J$

Mass of fat used: $m_{fat} = \frac{E_{fat}}{3.8 \times 10^{7}}$

$m_{fat} = \frac{2.5 \times 10^{5}}{3.8 \times 10^{7}} \approx 6.6 \times 10^{- 3} kg$

$m_{fat} \approx 6.6 g$

Question 5.23

A family uses 8 kW of power.

(a) Direct solar energy is incident on a horizontal surface at an average rate of
200 W m⁻². If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?

(b) Compare this area to that of the roof of a typical house.

Solution

(a) Area required

Solar power incident per unit area:

$P_{incident} = 200 {W m}^{- 2}$

Useful electrical power obtained (20% efficiency):

$P_{useful} = 0.20 \times 200 = 40 {W m}^{- 2}$

Required power:

$P = 8 kW = 8000 W$

Area needed:

$A = \frac{P}{P_{useful}} = \frac{8000}{40} = 200 m^{2}$

$A = 200 m^{2}$

(b) Comparison with a typical house roof

The roof area of a typical house is about 100–150 m².
- Required area = 200 m²
- This is larger than the roof area of a typical house.
$Thus, the required area is comparable to,$

$but somewhat larger than, a typical house roof.$
December 18, 2025
Class 11th Physics Chapter 4 Solutions
Go Back to CLASS 11TH – PHYSICS PAGE

(For simplicity in numerical calculations, take $g = 10 {m s}^{- 2}$ )

Question 4.1

Give the magnitude and direction of the net force acting on:

(a) A drop of rain falling down with a constant speed

Since the drop is falling with constant speed, its acceleration is zero.
Hence, the net force acting on it is zero.

Answer:
- Magnitude of net force: 0 N
- Direction: None
(b) A cork of mass 10 g floating on water

The weight of the cork acting downward is balanced by the upthrust (buoyant force) of water acting upward.
Therefore, the resultant force is zero.

Answer:
- Magnitude of net force: 0 N
- Direction: None
(c) A kite skillfully held stationary in the sky

The forces acting on the kite (weight, lift, tension, air drag) balance each other.
Since the kite is stationary, its acceleration is zero.

Answer:
- Magnitude of net force: 0 N
- Direction: None
(d) A car moving with a constant velocity of 30 km h⁻¹ on a rough road

Constant velocity means zero acceleration.
The driving force balances the frictional force.

Answer:
- Magnitude of net force: 0 N
- Direction: None
(e) A high-speed electron in space far from all material objects and free of electric and magnetic fields

No external force acts on the electron.

Answer:
- Magnitude of net force: 0 N
- Direction: None
Question 4.2

A pebble of mass 0.05 kg is thrown vertically upward. Give the direction and magnitude of the net force on the pebble:
(Ignore air resistance)

Weight of the pebble:

$F = m g = 0.05 \times 10 = 0.5 N$

(a) During its upward motion

Only gravitational force acts, directed downward.

Answer:
- Magnitude of net force: 0.5 N
- Direction: Vertically downward
(b) During its downward motion

Gravity continues to act downward.

Answer:
- Magnitude of net force: 0.5 N
- Direction: Vertically downward
(c) At the highest point where it is momentarily at rest

Although velocity is zero, gravitational force still acts.

Answer:
- Magnitude of net force: 0.5 N
- Direction: Vertically downward
Question 4.3

Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg:
(Neglect air resistance)

Weight of the stone:

$F = m g = 0.1 \times 10 = 1 N$

(a) Just after it is dropped from the window of a stationary train

Only gravitational force acts.

Answer:
- Magnitude of net force: 1 N
- Direction: Vertically downward
(b) Just after it is dropped from the window of a train running at a constant velocity of 36 km h⁻¹

Horizontal motion does not affect the force acting on the stone.
Only gravity acts.

Answer:
- Magnitude of net force: 1 N
- Direction: Vertically downward
(c) Just after it is dropped from the window of a train accelerating at 1 m s⁻²

After release, the stone is no longer influenced by the train’s acceleration.
Only gravity acts.

Answer:
- Magnitude of net force: 1 N
- Direction: Vertically downward
(d) A stone lying on the floor of a train accelerating at 1 m s⁻², the stone being at rest relative to the train

The stone accelerates along with the train.

$F = m a = 0.1 \times 1 = 0.1 N$

Answer:
- Magnitude of net force: 0.1 N
- Direction: In the direction of the train’s acceleration
Question 4.4

One end of a string of length $l$ is connected to a particle of mass $m$ and the other end to a small peg on a smooth horizontal table. If the particle moves in a circle with speed $v$ , the net force on the particle (directed towards the centre) is:

(i) $T$
(ii) $T - \frac{m v^{2}}{l}$
(iii) $T + \frac{m v^{2}}{l}$
(iv) $0$

( $T$ is the tension in the string. Choose the correct alternative.)

Solution

The particle is moving in a horizontal circular path on a smooth table, so:
- There is no friction.
- The weight of the particle and the normal reaction act vertically and cancel each other.
- The only horizontal force acting on the particle is the tension $T$ in the string.
This tension provides the centripetal force required for circular motion.

$Net force towards the centre = \frac{m v^{2}}{l}$

But this centripetal force is provided entirely by the tension $T$ .

$∴ T = \frac{m v^{2}}{l}$

Hence, the net force acting on the particle towards the centre is equal to the tension $T$ .

Question 4.5

A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s⁻¹. How long does the body take to stop?

Solution

Given:
- Retarding force, $F = 50 N$
- Mass of the body, $m = 20 kg$
- Initial velocity, $u = 15 {m s}^{- 1}$
- Final velocity, $v = 0 {m s}^{- 1}$ (body stops)
Step 1: Find the acceleration

Using Newton’s second law,

$F = m a$

Since the force is retarding, acceleration will be negative.

$a = \frac{F}{m} = \frac{50}{20} = 2.5 {m s}^{- 2}$ $\Rightarrow a = - 2.5 {m s}^{- 2}$

Step 2: Use the equation of motion

$v = u + a t$

Substitute the values:

$0 = 15 + (- 2.5) t$ $2.5 t = 15$ $t = \frac{15}{2.5} = 6 s$

Answer:

$The body takes 6 seconds to stop.$

Question 4.6

A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s⁻¹ to 3.5 m s⁻¹ in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

Solution

Given:
- Mass of the body, $m = 3.0 kg$
- Initial speed, $u = 2.0 {m s}^{- 1}$
- Final speed, $v = 3.5 {m s}^{- 1}$
- Time taken, $t = 25 s$
Step 1: Find the acceleration

Using the first equation of motion,

$a = \frac{v - u}{t}$ $a = \frac{3.5 - 2.0}{25} = \frac{1.5}{25} = 0.06 {m s}^{- 2}$

Step 2: Find the force

Using Newton’s second law,

$F = m a$ $F = 3.0 \times 0.06 = 0.18 N$

Direction of the force

Since the speed increases and the direction of motion remains unchanged, the force acts in the direction of motion.

Question 4.7

A body of mass 5 kg is acted upon by two perpendicular forces of magnitudes 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

Solution

Given:
- Mass of the body, $m = 5 kg$
- Two perpendicular forces:
  $F_{1} = 8 N$ and $F_{2} = 6 N$
Step 1: Find the resultant force

Since the forces are perpendicular, the resultant force is given by:

$F = \sqrt{F_{1}^{2} + F_{2}^{2}}$ $F = \sqrt{8^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10 N$

Step 2: Find the acceleration

Using Newton’s second law,

$a = \frac{F}{m}$ $a = \frac{10}{5} = 2 {m s}^{- 2}$

Step 3: Find the direction of acceleration

The acceleration is in the direction of the resultant force.

Let the direction of the 8 N force be along the x-axis.

$\tan θ = \frac{F_{2}}{F_{1}} = \frac{6}{8} = \frac{3}{4}$ $θ = \tan^{- 1} (\frac{3}{4})$

Question 4.8

The driver of a three-wheeler moving with a speed of 36 km h⁻¹ sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.

Solution

Given:
- Speed of the three-wheeler,
  $u = 36 {km h}^{- 1} = 10 {m s}^{- 1}$
- Final velocity,
  $v = 0 {m s}^{- 1}$
- Time taken to stop,
  $t = 4.0 s$
- Mass of three-wheeler = 400 kg
- Mass of driver = 65 kg
$Total mass, m = 400 + 65 = 465 kg$

Step 1: Calculate the acceleration

Using the first equation of motion,

$a = \frac{v - u}{t}$

$a = \frac{0 - 10}{4} = - 2.5 {m s}^{- 2}$

(The negative sign indicates retardation.)

Step 2: Calculate the average retarding force

Using Newton’s second law,

$F = m a$ $F = 465 \times (- 2.5) = - 1162.5 N$

Question 4.9

A rocket with a lift-off mass of 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s⁻². Calculate the initial thrust (force) of the blast.
(Take $g = 10 {m s}^{- 2}$ )

Solution

Given:
- Mass of the rocket,
  $m = 20,000 kg$
- Upward acceleration,
  $a = 5.0 {m s}^{- 2}$
- Acceleration due to gravity,
  $g = 10 {m s}^{- 2}$
Step 1: Forces acting on the rocket
- Thrust $T$ acts upward
- Weight $m g$ acts downward
The net upward force on the rocket is:

$T - m g = m a$

Step 2: Calculate the thrust

$T = m (a + g)$ $T = 20,000 \times (5 + 10)$ $T = 20,000 \times 15$ $T = 3.0 \times 10^{5} N$

Question 4.10

A body of mass 0.40 kg moving initially with a constant speed of 10 m s⁻¹ to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be $t = 0$ , the position of the body at that time to be $x = 0$ , and predict its position at $t = - 5$ , $25$ and $100$ s.

Solution

Given:
- Mass of the body, $m = 0.40 kg$
- Initial speed, $u = 10 {m s}^{- 1}$ (towards north)
- Force applied, $F = 8.0 N$ (towards south)
- Time for which force acts = 30 s
- Initial position at $t = 0$ , $x = 0$
Step 1: Choose sign convention
- Take north as positive direction
- Hence, force towards south is negative
Step 2: Find acceleration

Using Newton’s second law:

$a = \frac{F}{m} = \frac{8.0}{0.40} = 20 {m s}^{- 2}$

Since force is towards south:

$a = - 20 {m s}^{- 2}$

Step 3: Equation of motion

Position as a function of time:

$x = u t + \frac{1}{2} a t^{2}$

$x = 10 t + \frac{1}{2} (- 20) t^{2}$

$x = 10 t - 10 t^{2}$

(a) Position at $t = - 5$

$x = 10 (- 5) - 10 (25)$

$x = - 50 - 250 = - 300 m$ Answer: $x = - 300 m (300 m towards south)$

(b) Position at $t = 25$

$x = 10 (25) - 10 (625)$

$x = 250 - 6250 = - 6000 m$ Answer: $x = - 6000 m (6000 m towards south)$

(c) Position at $t = 100$

$x = 10 (100) - 10 (10000)$

$x = 1000 - 100000 = - 99000 m$ Answer $x = - 99000 m (99 km towards south)$

Question 4.11

A truck starts from rest and accelerates uniformly at $2.0 {m s}^{- 2}$ . At $t = 10$ s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at $t = 11$ ?
(Neglect air resistance; take $g = 10 {m s}^{- 2}$ )

Solution (NCERT-style)

Given:
- Acceleration of the truck, $a_{truck} = 2.0 {m s}^{- 2}$
- Truck starts from rest
- Stone is released at $t = 10$
- Time at which quantities are required: $t = 11$
  ⇒ Time of motion of stone after release,
  $Δ t = 11 - 10 = 1 s$
Step 1: Velocity of the truck at $t = 10$ s

$v_{truck} = u + a t = 0 + 2.0 \times 10 = 20 {m s}^{- 1}$

When the stone is dropped, it has:
- Horizontal velocity = 20 m s⁻¹
- Vertical velocity = 0
(a) Velocity of the stone at $t = 11$ s

Horizontal component
- No horizontal force acts on the stone after release.
$v_{x} = 20 {m s}^{- 1}$

Vertical component

$v_{y} = u_{y} + g t = 0 + 10 \times 1 = 10 {m s}^{- 1}$

Resultant velocity

$v = \sqrt{v_{x}^{2} + v_{y}^{2}}$

$v = \sqrt{20^{2} + 10^{2}} = \sqrt{400 + 100} = \sqrt{500}$

$v = 10 \sqrt{5} {m s}^{- 1}$

Direction of velocity

$\tan θ = \frac{v_{y}}{v_{x}} = \frac{10}{20} = \frac{1}{2}$

$θ = \tan^{- 1} (\frac{1}{2})$

The velocity is directed below the horizontal.

Answer (a):
- Velocity:
$10 \sqrt{5} {m s}^{- 1}$
- Direction:
  At an angle $\tan^{- 1} (1 / 2)$ below the horizontal
(b) Acceleration of the stone at $t = 11$

After release, the only force acting on the stone is gravity.

$Acceleration = g = 10 {m s}^{- 2}$

Answer (b):
- Acceleration:
$10 {m s}^{- 2} vertically downward$

Question 4.12

A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s⁻¹. What is the trajectory of the bob if the string is cut when the bob is
(a) at one of its extreme positions,
(b) at its mean position?

Solution

The bob is executing simple pendulum motion. Its motion and hence its trajectory after cutting the string depend on the velocity of the bob at the instant the string is cut.

(a) String is cut at one of the extreme positions

At the extreme position of oscillation:
- The velocity of the bob is zero.
- Only gravitational force acts on the bob after the string is cut.
Since the bob has no horizontal velocity at that instant, it will simply fall under gravity.

Trajectory:

The bob will move vertically downward in a straight line.

Answer (a):
- Trajectory: Straight vertical line downward
- Reason: Velocity at extreme position is zero
(b) String is cut at the mean position

At the mean (lowest) position:
- The bob has maximum speed, given as
  $v = 1 {m s}^{- 1}$
- The direction of velocity at the mean position is horizontal.
- After the string is cut, the bob has:
  - Initial horizontal velocity = 1 m s⁻¹
  - Vertical acceleration due to gravity = $g$
Thus, the bob behaves like a horizontally projected projectile.

Trajectory:

The bob will follow a parabolic path.

Answer (b):
- Trajectory: Parabola
- Reason: Horizontal velocity with vertical acceleration due to gravity
Question 4.13

A man of mass 70 kg stands on a weighing scale in a lift which is moving:

(a) upwards with a uniform speed of 10 m s⁻¹,
(b) downwards with a uniform acceleration of 5 m s⁻²,
(c) upwards with a uniform acceleration of 5 m s⁻².

What would be the readings on the scale in each case?

(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

(Take $g = 10 {m s}^{- 2}$ )

Solution

Given:
Mass of the man, $m = 70 kg$
Acceleration due to gravity, $g = 10 {m s}^{- 2}$

The reading of the weighing scale is the normal reaction $N$ exerted by the scale on the man.

(a) Lift moving upwards with uniform speed

Uniform speed ⇒ acceleration $a = 0$

$N = m g = 70 \times 10 = 700 N$

Reading on the scale: $700 N$

(b) Lift moving downwards with uniform acceleration $5 {m s}^{- 2}$

Acceleration is downward.

$N = m (g - a)$

$N = 70 (10 - 5) = 70 \times 5 = 350 N$

Reading on the scale

$350 N$

(c) Lift moving upwards with uniform acceleration $5 {m s}^{- 2}$

Acceleration is upward.

$N = m (g + a)$

$N = 70 (10 + 5) = 70 \times 15 = 1050 N$

Reading on the scale:

$1050 N$

(d) Lift falls freely under gravity

In free fall:

$a = g$

$N = m (g - g) = 0$

Reading on the scale: $0$

(The man feels weightless.)

Question 4.14

Figure 4.16 shows the position–time graph of a particle of mass 4 kg. What is the
(a) force on the particle for $t < 0$ , $0 < t < 4 s$ , and $t > 4 s$ ?
(b) impulse at $t = 0$ and $t = 4 s$ ?
(Consider one-dimensional motion only.)

Given / From the position–time graph (Fig. 4.16)
- The graph consists of straight line segments with sharp corners (kinks) at
  $t = 0$ and $t = 4 s$ .
- Slope of a position–time graph = velocity.
From the graph:
- For $t < 0$ : slope is constant ⇒ velocity is constant
- For $0 < t < 4 s$ : slope is constant (different from earlier)
- For $t > 4 s$ : slope is constant again
Hence, velocity is constant in each interval, but changes suddenly at
$t = 0$ and $t = 4 s$ .

Mass of particle:

$m = 4 kg$

(a) Force on the particle

For $t < 0$

Velocity is constant ⇒ acceleration $a = 0$

$F = m a = 0$

For $0 < t < 4 s$

Velocity is again constant ⇒ acceleration $a = 0$

$F = 0$

Force = 0 N

For $t > 4 s$

Velocity remains constant ⇒ acceleration $a = 0$

$F = 0$

Answer (a):

$Force is zero for t < 0, 0 < t < 4 s and t > 4 s$

(b) Impulse at $t = 0$ and $t = 4 s$

Impulse is produced when velocity changes suddenly.

$Impulse = Δ p = m (v_{2} - v_{1})$

Impulse at $t = 0$

From the graph:
- Velocity just before $t = 0$ : $v_{1} = 2 {m s}^{- 1}$
- Velocity just after $t = 0$ : $v_{2} = 0$
$I_{0} = 4 (0 - 2) = - 8 N s$

Impulse at $t = 0$ : $- 8 N s$

(negative sign indicates direction opposite to initial motion)

Impulse at $t = 4 s$

From the graph:
- Velocity just before $t = 4$ : $v_{1} = 0$
- Velocity just after $t = 4$ : $v_{2} = - 2 {m s}^{- 1}$
$I_{4} = 4 (- 2 - 0) = - 8 N s$

Impulse at $t = 4 s$ :

$- 8 N s$

Question 4.15

Two bodies of masses 10 kg and 20 kg respectively, kept on a smooth horizontal surface, are tied to the ends of a light string. A horizontal force F = 600 N is applied along the direction of the string to
(i) the 10 kg body (A),
(ii) the 20 kg body (B).

Find the tension in the string in each case.

Solution

Let
Mass of A = 10 kg
Mass of B = 20 kg

Since the surface is smooth, there is no friction.

Step 1: Acceleration of the system

Total mass

$m = 10 + 20 = 30 kg$

$a = \frac{F}{m} = \frac{600}{30} = 20 {m s}^{- 2}$

(i) Force applied on the 10 kg body (A)

The 20 kg body (B) is pulled only by the tension T.

$T = m_{B} \times a = 20 \times 20 = 400 N$

Tension in the string = $400 N$

(ii) Force applied on the 20 kg body (B)

The 10 kg body (A) is pulled only by the tension T.

$T = m_{A} \times a = 10 \times 20 = 200 N$

Tension in the string = $200 N$

Question 4.16

Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string which passes over a frictionless pulley. Find
(i) the acceleration of the masses, and
(ii) the tension in the string, when the system is released.

Solution

Let

$m_{1} = 8 kg, m_{2} = 12 kg$

Since $m_{2} > m_{1}$ , the 12 kg mass moves downward and the 8 kg mass moves upward.

Take acceleration = $a$ , tension in string = $T$ , and $g = 9.8 {m s}^{- 2}$ .

Equations of motion

For 12 kg mass (downward):

$12 g - T = 12 a (1)$

For 8 kg mass (upward):

$T - 8 g = 8 a (2)$

Step 1: Find acceleration

Add equations (1) and (2):

$12 g - 8 g = 12 a + 8 a$

$4 g = 20 a$

$a = \frac{4 g}{20} = \frac{g}{5}$

$a = 1.96 {m s}^{- 2}$

Step 2: Find tension

Substitute $a = \frac{g}{5}$ in equation (2):

$T - 8 g = 8 (\frac{g}{5})$

$T = 8 g + \frac{8 g}{5} = \frac{48 g}{5}$

$T = 94.08 N$

Question 4.17

A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei, the products must move in opposite directions.

Solution

Let the initial nucleus be at rest in the laboratory frame.

$Initial momentum = 0$

Suppose the nucleus disintegrates into two smaller nuclei of masses $m_{1}$ and $m_{2}$ , moving with velocities ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ respectively.

Using law of conservation of momentum

Since no external force acts on the system,

$Total momentum before disintegration = Total momentum after disintegration$

$0 = m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2}$

Rearranging, $m_{1} {\vec{v}}_{1} = - m_{2} {\vec{v}}_{2}$

Question 4.18

Two billiard balls, each of mass 0.05 kg, moving in opposite directions with a speed of
6 m s⁻¹, collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?

Solution

Given:
Mass of each ball $m = 0.05 kg$ Initial speed $u = 6 {m s}^{- 1}$

Final speed (after rebound)

$v = 6 {m s}^{- 1}$

Impulse

Impulse $J$ is equal to the change in momentum:

$J = Δ p = m (v - u)$

For one billiard ball

Take the initial direction as positive.

Initial velocity

$u = + 6 {m s}^{- 1}$

After collision, the ball rebounds, so its velocity becomes

$v = - 6 {m s}^{- 1}$

Change in momentum

$Δ p = m (v - u)$

$Δ p = 0.05 (- 6 - 6)$

$Δ p = 0.05 (- 12) = - 0.6 {kg m s}^{- 1}$ Impulse

Magnitude of impulse imparted to each ball:

$J = 0.6 N s$

(The negative sign only indicates change of direction.)

Question 4.19

A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s⁻¹, find the recoil speed of the gun.

Solution

Given:
Mass of shell,

$m = 0.020 kg$ Mass of gun, $M = 100 kg$ Velocity of shell, $v = 80 {m s}^{- 1}$

Let recoil speed of the gun be $V$ .

Using law of conservation of momentum

Initially, the gun–shell system is at rest, so total momentum is zero.

$Initial momentum = 0$

After firing: $m v - M V = 0$

$M V = m v$

$V = \frac{m v}{M}$

Calculation

$V = \frac{0.020 \times 80}{100}$

$V = \frac{1.6}{100} = 0.016 {m s}^{- 1}$

Question 4.20

A batsman deflects a ball by an angle of 45° without changing its initial speed, which is
54 km h⁻¹. What is the impulse imparted to the ball?
(Mass of the ball = 0.15 kg)

Solution

Given:
Mass of ball, $m = 0.15 kg$

Initial speed = Final speed $v = 54 {km h}^{- 1} = 15 {m s}^{- 1}$

Angle between initial and final velocities,

$θ = 45^{\circ}$

Impulse

Impulse $J$ = change in momentum

$J = m ∣ Δ \vec{v} ∣$

Magnitude of change in velocity when direction changes by angle $θ$ :

$∣ Δ \vec{v} ∣ = \sqrt{v^{2} + v^{2} - 2 v^{2} \cos θ}$

$∣ Δ \vec{v} ∣ = v \sqrt{2 (1 - \cos 45^{\circ})}$ $∣ Δ \vec{v} ∣ = 15 \sqrt{2 (1 - 0.707)}$ $∣ Δ \vec{v} ∣ = 15 \times 0.765 \approx 11.48 {m s}^{- 1}$

Impulse imparted

$J = 0.15 \times 11.48$

$J \approx 1.72 N s$

Question 4.21

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev min⁻¹ in a horizontal plane.

(i) What is the tension in the string?
(ii) What is the maximum speed with which the stone can be whirled if the string can withstand a maximum tension of 200 N?

Solution

Given:

$m = 0.25 kg, r = 1.5 m$

Speed of rotation:

$f = 40 {rev min}^{- 1} = \frac{40}{60} = \frac{2}{3} {rev s}^{- 1}$

Angular speed:

$ω = 2 π f = 2 π \times \frac{2}{3} = \frac{4 π}{3} {rad s}^{- 1}$

(i) Tension in the string

Linear speed:

$v = ω r = \frac{4 π}{3} \times 1.5 = 6.28 {m s}^{- 1}$

Tension provides the centripetal force:

$T = \frac{m v^{2}}{r}$

$T = \frac{0.25 \times (6.28)^{2}}{1.5}$

$T \approx 6.6 N$

$T \approx 6.6 N$

(ii) Maximum speed when maximum tension = 200 N

$T_{\max} = \frac{m v_{\max}^{}}{r}$

$v_{\max} = \sqrt{\frac{T_{\max} r}{m}}$

$v_{\max} = \sqrt{\frac{200 \times 1.5}{0.25}}$

$v_{\max} = \sqrt{1200}$

$v_{\max} \approx 34.6 {m s}^{- 1}$

Question 4.22

If, in Exercise 4.21, the speed of the stone is increased beyond the maximum permissible value and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks?

(a) The stone moves radially outwards.
(b) The stone flies off tangentially from the instant the string breaks.
(c) The stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle.

Answer and Explanation

The correct option is (b).

Reason
- While the stone is moving in a circle, the tension in the string provides the centripetal force, continuously changing the direction of velocity.
- At the instant the string breaks, this centripetal force vanishes suddenly.
- According to Newton’s first law of motion, the stone will continue to move with the velocity it had at that instant.
- The instantaneous velocity of a particle in circular motion is always along the tangent to the circle.
Conclusion

$The stone flies off tangentially from the instant the string breaks.$

Hence, option (b) is correct.

Question 4.23

Explain why:

(a) a horse cannot pull a cart and run in empty space,
(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
(c) it is easier to pull a lawn mower than to push it,
(d) a cricketer moves his hands backwards while holding a catch.

Answer

(a) A horse cannot pull a cart and run in empty space

A horse pulls a cart by pushing the ground backward with its legs. The ground provides an equal and opposite frictional force on the horse, which helps it move forward. In empty space, there is no ground and hence no friction, so the horse cannot get a forward reaction force and cannot pull the cart.

(b) Passengers are thrown forward when a speeding bus stops suddenly

This happens due to inertia of motion. When the bus stops suddenly, the lower part of the passenger’s body in contact with the seat stops, but the upper part tends to continue moving forward, causing the passenger to be thrown forward.

(c) It is easier to pull a lawn mower than to push it

When a lawn mower is pulled, the applied force has an upward component which reduces the normal reaction and hence reduces friction. When it is pushed, the downward component of force increases the normal reaction and friction, making it harder to move.

(d) A cricketer moves his hands backwards while holding a catch

By moving his hands backward, the cricketer increases the time during which the ball’s momentum is brought to zero. Since force is given by

$F = \frac{Δ p}{Δ t},$

increasing the time reduces the force acting on the hands, preventing injury.
December 18, 2025

Class 11th Physics Chapter-3 Solutions

Question 3.1

State, for each of the following physical quantities, whether it is a scalar or a vector:

volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.

Answer

Physical Quantity	Scalar / Vector
Volume	Scalar
Mass	Scalar
Speed	Scalar
Acceleration	Vector
Density	Scalar
Number of moles	Scalar
Velocity	Vector
Angular frequency (ω)	Scalar
Displacement	Vector
Angular velocity	Vector

Exam-tip (1 line logic)

Scalar → has magnitude only
Vector → has magnitude + direction
Triangle Law can be applied

Question 3.2

Pick out the two scalar quantities in the following list :

force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.

Answer

The two scalar quantities are:

$Work, Current$

Question 3.3

Pick out the only vector quantity in the following list :

Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

Answer

$Impulse$

Impulse = change in momentum, and momentum is a vector, hence impulse is a vector.

Question 3.4

State with reasons whether the following algebraic operations with scalar and vector physical quantities are meaningful:

(a) Adding any two scalars

✅ Meaningful
Reason: Scalars have magnitude only and obey the rules of ordinary algebra (provided they have the same dimensions).
Example: 5 kg + 3 kg = 8 kg

(b) Adding a scalar to a vector of the same dimensions

❌ Not meaningful
Reason: Scalars and vectors are different types of physical quantities; a scalar has no direction while a vector has both magnitude and direction. They cannot be added.
Example: Mass (scalar) + displacement (vector) ❌

(c) Multiplying any vector by any scalar

✅ Meaningful
Reason: Multiplying a vector by a scalar changes its magnitude (and possibly its direction if the scalar is negative).
Example: 2 v, −3 F

(d) Multiplying any two scalars

✅ Meaningful
Reason: Scalars follow ordinary multiplication rules.
Example: Pressure × volume, mass × acceleration magnitude

(e) Adding any two vectors

❌ Not always meaningful
Reason: Two vectors can be added only if they represent the same physical quantity.
Example: Velocity + velocity ✔️, but velocity + force ❌

(f) Adding a component of a vector to the same vector

❌ Not meaningful
Reason: A component (e.g., $A_{x}$ ) is a scalar, while the vector $\vec{A}$ is a vector; scalar and vector cannot be added.
Example: $\vec{A} + A_{x}$ ❌

Question 3.5

Read each statement below carefully and state with reasons whether it is true or false :

(a) The magnitude of a vector is always a scalar

✅ True
Reason: The magnitude gives only the size of the vector and has no direction, hence it is a scalar.
Example: Speed is the magnitude of velocity and is scalar.

(b) Each component of a vector is always a scalar

❌ False
Reason: The numerical component (like $A_{x}$ ) is a scalar, but the component vector (like $A_{x} \hat{i}$ ) is a vector. Hence the statement is not always true.

(c) The total path length is always equal to the magnitude of the displacement vector of a particle

❌ False
Reason: Total path length depends on the actual path taken, whereas displacement depends only on initial and final positions.
Equality occurs only for straight-line motion without change in direction.

(d) The average speed of a particle is either greater than or equal to the magnitude of average velocity of the particle over the same time interval

✅ True
Reason:

$Average speed = \frac{total path length}{t}, ∣ average velocity ∣ = \frac{∣ displacement ∣}{t}$

Since total path length ≥ displacement, average speed ≥ magnitude of average velocity.

(e) Three vectors not lying in a plane can never add up to give a null vector

✅ True
Reason: For vectors to add up to zero, they must form a closed polygon, which is possible only if they are coplanar.

Question 3.6

Establish the following vector inequalities geometrically or otherwise :

$\begin{aligned} (a) & ∣ a + b ∣ < ∣ a ∣ + ∣ b ∣ \\ (b) & ∣ a + b ∣ > ∣ ∣ a ∣ - ∣ b ∣ ∣ \\ (c) & ∣ a - b ∣ < ∣ a ∣ + ∣ b ∣ \\ (d) & ∣ a - b ∣ > ∣ ∣ a ∣ - ∣ b ∣ ∣ \end{aligned}$

When does the equality sign above apply?

Answer:

Geometrical idea (triangle law)

If two vectors a and b are placed head-to-tail, they form two sides of a triangle; the resultant (a+b) is the third side. For any triangle:

the length of one side is less than the sum of the other two,
and greater than the difference of the other two.

(a) $∣ a + b ∣ < ∣ a ∣ + ∣ b ∣$

Reason: By the triangle law, the magnitude of the resultant side of a triangle is less than the sum of the magnitudes of the other two sides.
Equality holds only when a and b are in the same direction.

(b) $∣ a + b ∣ > ∣ ∣ a ∣ - ∣ b ∣ ∣$

Reason: In any triangle, one side is greater than the difference of the other two sides.
Equality holds when a and b are in opposite directions.

(c) $∣ a - b ∣ < ∣ a ∣ + ∣ b ∣$

Note that $a - b = a + (- b)$

Reason: Applying result (a) to vectors a and −b, whose magnitude is $∣ b ∣$ :

$∣ a - b ∣ = ∣ a + (- b) ∣ < ∣ a ∣ + ∣ b ∣ .$

(d) $∣ a - b ∣ > ∣ ∣ a ∣ - ∣ b ∣ ∣$

Reason: Applying result (b) to vectors a and −b:

$∣ a - b ∣ = ∣ a + (- b) ∣ > ∣ ∣ a ∣ - ∣ b ∣ ∣$

Equality holds when the vectors are collinear (parallel), either in the same or opposite direction as required.

Question 3.7

Given

$a + b + c + d = 0$

Which of the following statements are correct?

(a) $a, b, c, d$ must each be a null vector

❌ False
Reason: Non-zero vectors can add up to zero if they form a closed polygon (e.g., a quadrilateral). Each vector need not be zero individually.

(b) The magnitude of $(a + c)$ equals the magnitude of $(b + d)$

✅ True
Reason: From the given condition,

$a + c = - (b + d)$

Vectors equal in magnitude and opposite in direction have the same magnitude.

(c) The magnitude of $a$ can never be greater than the sum of the magnitudes of $b, c, d$

✅ True
Reason: From

$a = - (b + c + d)$

Taking magnitudes and using the triangle inequality,

$∣ a ∣ \leq ∣ b ∣ + ∣ c ∣ + ∣ d ∣$

(d) $b + c$ must lie in the plane of $a$ and $d$ if $a$ and $d$ are not collinear, and in the line of $a$ and $d$ if they are collinear

✅ True
Reason: From

$b + c = - (a + d)$

The vector $a + d$ lies:

in the plane of a and d if they are not collinear,
along the same line if they are collinear.

Hence $b + c$ must lie in the same plane or line respectively.

Question 3.8

Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 3.19. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skated?

Answer

Magnitude of displacement

Displacement depends only on initial and final positions, not on the path taken.

Points P and Q are diametrically opposite points on a circle of radius
$R = 200 m$
Distance PQ = diameter of the circle
$P Q = 2 R = 2 \times 200 = 400 m$

✅ Magnitude of displacement for each girl = 400 m

Path length vs displacement

Girl skating along the diameter (straight line PQ):
- Actual path length = 400 m
- Displacement = 400 m
  ✅ Path length equals displacement
Girls skating along curved paths (arc or other curved routes):
- Actual path length greater than 400 m
- Displacement = 400 m
  ❌ Path length ≠ displacement

Question 3.9

A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 3.20. If the round trip takes 10 min, find:

(a) net displacement
(b) average velocity
(c) average speed

Given

Radius of circular park, $R = 1 km$
Total time, $t = 10 min = \frac{1}{6} h$

(a) Net displacement

Initial position = O (centre)
Final position = O (centre)

$Net displacement = 0$

(b) Average velocity

$Average velocity = \frac{Net displacement}{Total time}$

Since net displacement = 0,

$Average velocity = 0$

(c) Average speed

First calculate the total distance travelled:

O to P (radius) = $1 km$
Along circumference from P to Q = semicircle
$= \frac{1}{2} (2 π R) = π R = π \times 1 = π km$
Q to O (radius) = $1 km$

Total distance:

$Distance = 1 + π + 1 = (2 + π) km$

Average speed:

$Average speed = \frac{2 + π}{1 / 6} = 6 (2 + π)$

$Average speed = 6 (2 + π) {km h}^{- 1} \approx 30.8 {km h}^{- 1}$

Question 3.10

On an open ground, a motorist follows a track that turns to his left by an angle of $60^{\circ}$ after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

Understanding the motion

Each straight segment = 500 m
Turn after every segment = $60^{\circ}$ to the left
This motion traces the sides of a regular hexagon (since exterior angle = $60^{\circ}$ ).

Let side length $s = 500 m$ .

(i) Displacement at the 3rd turn

After 3 sides of a regular hexagon, the motorist reaches the vertex opposite to the starting point.
Displacement = distance between opposite vertices of a hexagon

$= 2 s = 2 \times 500 = 1000 m$ Path length covered:

$3 \times 500 = 1500 m$ Comparison:

$Displacement (1000 m) < Path length (1500 m)$

(ii) Displacement at the 6th turn

After 6 sides, the motorist completes one full hexagon and returns to the starting point.

$Displacement = 0$ Path length covered:

$6 \times 500 = 3000 m$ Comparison:

$Displacement = 0 ≪ Path length$

(iii) Displacement at the 8th turn

8 turns = 6 turns + 2 more turns
Equivalent to displacement after 2 sides of the hexagon.
Two vectors of magnitude $s$ at an angle of $60^{\circ}$ .

Using vector addition:

$∣ \vec{R} ∣ = \sqrt{s^{2} + s^{2} + 2 s^{2} \cos 60^{\circ}} = \sqrt{3 s^{2}} = s \sqrt{3}$ $∣ \vec{R} ∣ = 500 \sqrt{3} \approx 866 m$

Path length covered:

$8 \times 500 = 4000 m$

Comparison:

$Displacement (\approx 866 m) < Path length (4000 m)$

Question 3.11

A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. Find:

(a) the average speed of the taxi
(b) the magnitude of average velocity
Are the two equal?

Given

Straight-line distance (displacement) = 10 km
Actual path length = 23 km
Time taken = 28 min = $\frac{28}{60} = \frac{7}{15}$

(a) Average speed

$Average speed = \frac{total distance}{time} = \frac{23}{7 / 15} = \frac{23 \times 15}{7}$

$Average speed \approx 49.3 {km h}^{- 1}$

(b) Magnitude of average velocity

$Magnitude of average velocity = \frac{displacement}{time} = \frac{10}{7 / 15} = \frac{150}{7}$

$∣ Average velocity ∣ \approx 21.4 {km h}^{- 1}$

Are the two equal?

❌ No

Reason:

Average speed depends on total path length
Average velocity depends on displacement only

Since the path taken is longer than the straight-line distance,

$Average speed > ∣ Average velocity ∣$

Question 3.12

The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s⁻¹ can go without hitting the ceiling of the hall?

Given

Speed of projection: $u = 40 {m s}^{- 1}$
Maximum allowed height: $H = 25 m$
Acceleration due to gravity: $g = 9.8 {m s}^{- 2}$

Step 1: Condition for not hitting the ceiling

Maximum height of a projectile:

$H = \frac{u^{2} \sin^{2} θ}{2 g}$

For maximum range without touching the ceiling, the ball must just reach the ceiling:

$25 = \frac{(40)^{2} \sin^{2} θ}{2 \times 9.8}$

$25 = \frac{1600 \sin^{2} θ}{19.6}$

$\sin^{2} θ = \frac{25 \times 19.6}{1600} = 0.30625$

$\sin θ = 0.553$

Step 2: Horizontal range

Horizontal range:

$R = \frac{u^{2} \sin 2 θ}{g}$ First find $\sin 2 θ$ :

$\sin 2 θ = 2 \sin θ \cos θ = 2 (0.553) \sqrt{1 - {0.553}^{2}}$

$\cos θ = \sqrt{1 - 0.30625} = 0.833$

$\sin 2 θ = 2 (0.553) (0.833) \approx 0.921$

Now range: $R = \frac{1600 \times 0.921}{9.8} \approx 150.4 m$

Question 3.13

A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?

Concept used

Maximum horizontal range:

$R_{\max} = \frac{u^{2}}{g}$

Maximum height (when thrown vertically upward with same speed):

$H_{\max} = \frac{u^{2}}{2 g}$

Step 1: Find the speed of projection

Given: $R_{\max} = 100 m$ $\frac{u^{2}}{g} = 100 \Rightarrow u^{2} = 100 g$

Step 2: Find the maximum height

$H_{\max} = \frac{u^{2}}{2 g} = \frac{100 g}{2 g} = 50 m$

Question 3.14

A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, find the magnitude and direction of the acceleration of the stone.

Given

Radius of circular path:
$R = 80 cm = 0.8 m$
Number of revolutions: $n = 14$
Time taken: $t = 25 s$

Step 1: Find angular speed

$ω = \frac{2 π n}{t} = \frac{2 π \times 14}{25} = \frac{28 π}{25} \approx 3.52 {rad s}^{- 1}$

Step 2: Find centripetal acceleration

For uniform circular motion,

$a = ω^{2} R$ $a = (3.52)^{2} \times 0.8 \approx 12.38 \times 0.8 \approx 9.9 {m s}^{- 2}$

Direction of acceleration

The acceleration is centripetal, i.e.
always directed towards the centre of the circular path, radially inward.

Question 3.15

An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km h⁻¹. Compare its centripetal acceleration with the acceleration due to gravity.

Given

Radius of loop:R
$= 1.00 km = 1000 m$
Speed of aircraft:
$v = 900 {km h}^{- 1} = \frac{900 \times 1000}{3600} = 250 {m s}^{- 1}$
Acceleration due to gravity:
$g = 9.8 {m s}^{- 2}$

Centripetal acceleration

$a_{c} = \frac{v^{2}}{R} = \frac{(250)^{2}}{1000} = \frac{62500}{1000} = 62.5 {m s}^{- 2}$

Comparison with gravity

$\frac{a_{c}}{g} = \frac{62.5}{9.8} \approx 6.4$

Question 3.16

Read each statement below carefully and state, with reasons, whether it is true or false :

(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre

❌ False

Reason:

In uniform circular motion, the net acceleration is centripetal, directed towards the centre.
But in non-uniform circular motion, there is also a tangential acceleration due to change in speed.
Hence the net acceleration is not always purely radial.

(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point

✅ True

Reason:
Velocity is the time rate of change of position, and in the limit $Δ t \to 0$ , the displacement is along the tangent to the path. Hence instantaneous velocity is tangential.

(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector

✅ True

Reason:
In uniform circular motion:

The magnitude of centripetal acceleration is constant,
But its direction continuously changes and covers all directions in one complete cycle.
The vector sum (and hence average) of acceleration over one full cycle is zero.

Question 3.17

The position of a particle is given by

$\vec{r} (t) = (3.0 t^{2}) \hat{i} - (2.0 t) \hat{j} + (4.0) \hat{k} m$

where $t$ is in seconds.

(a) Find the velocity $\vec{v}$ and acceleration $\vec{a}$ of the particle.
(b) Find the magnitude and direction of the velocity at $t = 2.0 s$ .

(a) Velocity and acceleration

Velocity is the time derivative of position:

$\vec{v} = \frac{d \vec{r}}{d t} = (6.0 t) \hat{i} - (2.0) \hat{j} + 0 \hat{k}$

$\vec{v} (t) = (6.0 t) \hat{i} - 2.0 \hat{j} {m s}^{- 1}$

Acceleration is the time derivative of velocity:

$\vec{a} = \frac{d \vec{v}}{d t} = 6.0 \hat{i}$

$\vec{a} = 6.0 \hat{i} {m s}^{- 2}$

(b) Velocity at $t = 2.0 s$

Substitute $t = 2.0$ s in $\vec{v} (t)$ :

$\vec{v} (2) = (12) \hat{i} - 2 \hat{j} {m s}^{- 1}$

Magnitude

$∣ \vec{v} ∣ = \sqrt{12^{2} + (- 2)^{2}} = \sqrt{148} \approx 12.2 {m s}^{- 1}$

Direction

Angle $θ$ with the positive $x$ -axis:

$\tan θ = \frac{∣ v_{y} ∣}{v_{x}} = \frac{2}{12} = \frac{1}{6} \Rightarrow θ \approx {9.5}^{\circ}$

Since $v_{x} > 0$ and $v_{y} < 0$ , the direction is below the +x-axis.

$Direction: {9.5}^{\circ} below the positive x -axis$

Question 3.18

A particle starts from the origin at $t = 0 s$ with a velocity

${\vec{v}}_{0} = 10.0 \hat{j} {m s}^{- 1}$

and moves in the x–y plane with a constant acceleration

$\vec{a} = (8.0 \hat{i} + 2.0 \hat{j}) {m s}^{- 2} .$

(a) At what time is the x-coordinate of the particle 16 m? What is the y-coordinate at that time?
(b) What is the speed of the particle at that time?

Given

Initial position: $x_{0} = 0, y_{0} = 0$
Initial velocity components:
$u_{x} = 0, u_{y} = 10.0 {m s}^{- 1}$
Acceleration components:
$a_{x} = 8.0, a_{y} = 2.0 {m s}^{- 2}$

(a) Time when $x = 16 m$ and corresponding $y$

x–motion:

$x = u_{x} t + \frac{1}{2} a_{x} t^{2} = 0 + \frac{1}{2} (8.0) t^{2} = 4 t^{2}$

Set $x = 16$ :

$4 t^{2} = 16 \Rightarrow t^{2} = 4 \Rightarrow t = 2.0 s$

y–motion at $t = 2.0 s$ :

$y = u_{y} t + \frac{1}{2} a_{y} t^{2} = 10 (2) + \frac{1}{2} (2) (2^{2}) = 20 + 4 = 24 m$

(b) Speed at $t = 2.0 s$

Velocity components at time $t$ :

$v_{x} = u_{x} + a_{x} t = 0 + 8 (2) = 16 {m s}^{- 1}$

$v_{y} = u_{y} + a_{y} t = 10 + 2 (2) = 14 {m s}^{- 1}$

Speed:

$v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{16^{2} + 14^{2}} = \sqrt{452} \approx 21.3 {m s}^{- 1}$

Question 3.19

$\hat{i}$ and $\hat{j}$ are unit vectors along the x-axis and y-axis respectively.

(a) What are the magnitude and direction of the vectors

$(\hat{i} + \hat{j}) and (\hat{i} - \hat{j}) ?$

(b) What are the components of a vector

$\vec{A} = 2 \hat{i} + 3 \overset{^}{j}$

along the directions of $(\hat{i} + \hat{j})$ and $(\hat{i} - \hat{j})$ ?
(Graphical method may be used.)

(a) Magnitude and direction

Vector $\hat{i} + \hat{j}$

Magnitude

$∣ \hat{i} + \hat{j} ∣ = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$

Direction

$\tan θ = \frac{1}{1} = 1 \Rightarrow θ = 45^{\circ}$

➡️ Direction: 45° above the +x-axis

Vector $\hat{i} - \hat{j}$

Magnitude

$∣ \hat{i} - \hat{j} ∣ = \sqrt{1^{2} + (- 1)^{2}} = \sqrt{2}$

Direction

$\tan θ = \frac{- 1}{1} = - 1 \Rightarrow θ = - 45^{\circ}$

➡️ Direction: 45° below the +x-axis

(b) Components of $\vec{A} = 2 \hat{i} + 3 \hat{j}$

First find unit vectors along the given directions.

Unit vector along $(\hat{i} + \hat{j})$

${\hat{n}}_{1} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$

Unit vector along $(\hat{i} - \hat{j})$

${\hat{n}}_{2} = \frac{\hat{i} - \hat{j}}{\sqrt{2}}$

Component of $\vec{A}$ along $(\hat{i} + \hat{j})$

$A_{1} = \vec{A} \cdot {\hat{n}}_{1} = \frac{1}{\sqrt{2}} (2 + 3) = \frac{5}{\sqrt{2}}$

Component of $\vec{A}$ along $(\hat{i} - \hat{j})$

$A_{2} = \vec{A} \cdot {\hat{n}}_{2} = \frac{1}{\sqrt{2}} (2 - 3) = - \frac{1}{\sqrt{2}}$

Question 3.20

For any arbitrary motion in space, which of the following relations are true? Give reasons.

(a)

${\vec{v}}_{average} = \frac{1}{2} [\vec{v} (t_{1}) + \vec{v} (t_{2})]$

❌ False

Reason:
This relation holds only for motion with constant acceleration.
For arbitrary (general) motion, velocity need not vary linearly with time.

(b)

${\vec{v}}_{average} = \frac{\vec{r} (t_{2}) - \vec{r} (t_{1})}{t_{2} - t_{1}}$

✅ True

Reason:
This is the definition of average velocity, valid for all kinds of motion.

(c)

$\vec{v} (t) = \vec{v} (0) + \vec{a} t$

❌ False

Reason:
This equation is valid only when acceleration is constant.
For arbitrary motion, acceleration may vary with time.

(d)

$\vec{r} (t) = \vec{r} (0) + \vec{v} (0) t + \frac{1}{2} \vec{a} t^{2}$

❌ False

Reason:
This is a kinematic equation applicable only for constant acceleration.
Not valid for arbitrary motion.

(e)

${\vec{a}}_{average} = \frac{\vec{v} (t_{2}) - \vec{v} (t_{1})}{t_{2} - t_{1}}$

✅ True

Reason:
This is the definition of average acceleration, valid for any motion.

Question 3.21

Read each statement below carefully and state, with reasons and examples, whether it is true or false:

A scalar quantity is one that—

(a) is conserved in a process

❌ False

Reason:
Being a scalar does not imply conservation. Conservation depends on physical laws, not on whether a quantity is scalar or vector.

Example:

Energy (scalar) is conserved (in isolated systems).
Temperature (scalar) is not conserved.

(b) can never take negative values

❌ False

Reason:
Scalars can be positive, negative, or zero.

Example:

Temperature on the Celsius scale can be negative.
Electric potential can be negative.

(c) must be dimensionless

❌ False

Reason:
Most scalars have dimensions and units.

Example:

Mass (kg), time (s), energy (J) are scalars with dimensions.
Only some scalars (e.g., coefficient of friction) are dimensionless.

(d) does not vary from one point to another in space

❌ False

Reason:
A scalar quantity may vary from point to point, forming a scalar field.

Example:

Temperature varies from place to place in a room.
Pressure varies with depth in a fluid.

(e) has the same value for observers with different orientations of axes

✅ True

Reason:
Scalars are independent of the choice of coordinate axes and their orientation.

Example:

Mass, time, temperature remain the same regardless of how axes are oriented.

Question 3.22

An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is $30^{\circ}$ , what is the speed of the aircraft?

Understanding the situation

The aircraft flies horizontally at a constant height.
From a fixed point on the ground, the lines of sight to the aircraft at two positions (separated by 10 s) make an angle of 30° with each other.
Height of aircraft:

$h = 3400 m$

For maximum clarity (and as done in NCERT), we assume the aircraft is observed before and after it passes the point of closest approach, so the two angles with the vertical are symmetric.

Hence, each angle with the vertical:

$θ = \frac{30^{\circ}}{2} = 15^{\circ}$

Horizontal distance covered in 10 s

Let the horizontal distance from the observer to each position be $x$ .

From right-angled triangle:

$\tan 15^{\circ} = \frac{x}{3400}$ $x = 3400 \tan 15^{\circ}$

Using $\tan 15^{\circ} \approx 0.268$ :

$x \approx 3400 \times 0.268 \approx 911 m$

So, total horizontal distance travelled in 10 s:

$Distance = 2 x \approx 2 \times 911 = 1822 m$

Speed of the aircraft

$v = \frac{distance}{time} = \frac{1822}{10} \approx 182 {m s}^{- 1}$

December 17, 2025

Class 11th Physics Chapter-2 Solutions (Question – 2.10 to 2.18
Go Back to CLASS 11TH PHYSICS Page

Question 2.10

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h⁻¹. Finding the market closed, he instantly turns back and walks home with a speed of 7.5 km h⁻¹.

Find:

(a) Magnitude of average velocity
(b) Average speed

for the time intervals:
(i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min

Given
- Distance to market = 2.5 km
- Speed while going = 5 km h⁻¹
- Speed while returning = 7.5 km h⁻¹
Step 1: Time taken in each part

Time to reach market

$t_{1} = \frac{2.5}{5} = 0.5 h = 30 min$

Time to return home

$t_{2} = \frac{2.5}{7.5} = \frac{1}{3} h = 20 min$

Total time for complete journey

$t_{total} = 30 + 20 = 50 min$

(i) Time interval: 0 to 30 min

During this time, the man only walks towards the market.

Displacement

$= 2.5 km$

Total path length

$= 2.5 km$

(a) Magnitude of average velocity

$= \frac{2.5}{0.5} = 5 {km h}^{- 1}$

(b) Average speed

$= \frac{2.5}{0.5} = 5 {km h}^{- 1}$

(ii) Time interval: 0 to 50 min

The man goes to the market and returns home.

Displacement

$= 0 (initial and final positions are same)$

Total path length

$= 2.5 + 2.5 = 5 km$

(a) Magnitude of average velocity

$= \frac{0}{50 / 60} = 0$

(b) Average speed

$= \frac{5}{50 / 60} = 6 {km h}^{- 1}$

(iii) Time interval: 0 to 40 min
- First 30 min: reaches market
- Next 10 min: returns home
Distance covered in 10 min return

Speed = 7.5 km h⁻¹
Time = 10 min = 1/6 h

$Distance = 7.5 \times \frac{1}{6} = 1.25 km$

Position after 40 min

$2.5 - 1.25 = 1.25 km from home$

Displacement

$= 1.25 km$

Total path length

$= 2.5 + 1.25 = 3.75 km$

(a) Magnitude of average velocity

$= \frac{1.25}{40 / 60} = 1.875 {km h}^{- 1}$

(b) Average speed

$= \frac{3.75}{40 / 60} = 5.625 {km h}^{- 1}$

Question 2.11

In Exercises 2.9 and 2.10, we distinguished between average speed and magnitude of average velocity. However, no such distinction is required for instantaneous speed and instantaneous velocity. Why? Explain.

Answer

Key Idea

The difference between speed and velocity arises because they are defined over a finite time interval. But instantaneous quantities are defined at a single instant of time.

Definitions

Instantaneous velocity

Instantaneous velocity at a given instant is defined as:

$\vec{v} = \lim_{Δ t \to 0} \frac{Δ \vec{x}}{Δ t}$

It has both magnitude and direction.

Instantaneous speed

Instantaneous speed is defined as:

$Instantaneous speed = \lim_{Δ t \to 0} \frac{distance travelled}{Δ t}$

It has only magnitude, no direction.

Why are they always equal in magnitude?
- Over an infinitesimally small time interval (Δt→0):
  - The distance travelled becomes equal to the magnitude of displacement.
  - There is no change of direction in that tiny interval.
- Hence,
$Instantaneous speed = ∣ Instantaneous velocity ∣$

Why the distinction disappears
- For average quantities, the particle may:
  - change direction,
  - retrace its path,
  - or return to the starting point.
- For instantaneous quantities, motion is examined at a single point in time, so:
  - path length = displacement (in magnitude),
  - speed = magnitude of velocity.
Question 2.12

Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Answer:

In one-dimensional motion:
1. At a given instant of time, a particle can have only one position.
2. Speed can never be negative.
3. A graph must satisfy the physical meaning of the quantities on the axes.
(a) – Possible
- This graph represents a particle moving forward and then backward.
- At every instant of time, there is only one position.
- ✔️ Physically possible in one-dimensional motion.
(b) – Not Possible ❌
- This graph shows more than one position for the same time.
- A vertical line cuts the graph at more than one point.
- This means the particle is at two positions at the same time, which is impossible.
Cannot represent one-dimensional motion.

(c) – Possible
- This graph represents motion where speed increases and then decreases.
- Speed remains positive throughout.
- ✔️ Physically possible.
(d) – Not Possible ❌
- This graph shows negative values of speed.
- Speed is a scalar quantity and cannot be negative.
- Negative values have no physical meaning for speed.
Cannot represent one-dimensional motion.

Question 2.13

Figure 2.11 shows the x–t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for $t < 0$ and on a parabolic path for $t > 0$ ? If not, suggest a suitable physical context for this graph.

Answer

No, this interpretation is NOT correct.

The statement is incorrect because the given figure is an x–t (position–time) graph, not a path diagram.

Correct Explanation
- The graph shows how position x
  
  of the particle changes with time t
- It does not represent the actual path in space.
- Therefore:
  - A straight line in an x–t graph does not mean straight-line motion in space.
  - A parabolic curve in an x–t graph does not mean a parabolic path.
In fact, the motion is one-dimensional throughout, i.e., the particle always moves along the same straight line (x-axis) for all values of time.

Question 2.14

A police van moving on a highway with a speed of 30 km h⁻¹ fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h⁻¹. If the muzzle speed of the bullet is 150 m s⁻¹, with what speed does the bullet hit the thief’s car?

(Relevant speed is the speed of the bullet relative to the car.)

Step 1: Convert all speeds into SI units

Police van speed

$30 {km h}^{- 1} = \frac{30 \times 1000}{3600} = 8.33 {m s}^{- 1}$

Thief’s car speed

$192 {km h}^{- 1} = \frac{192 \times 1000}{3600} = 53.33 {m s}^{- 1}$

Step 2: Speed of bullet relative to ground

The bullet is fired from the moving police van, so its speed relative to the ground is:

$v_{bullet, ground} = 150 + 8.33 = 158.33 {m s}^{- 1}$

Step 3: Speed of bullet relative to thief’s car

This is the speed that causes damage.

$v_{relative} = v_{bullet} - v_{car}$

$v_{relative} = 158.33 - 53.33 = 105 {m s}^{- 1}$

Question 2.15

Suggest a suitable physical situation for each of the graphs shown in Fig. 2.12.

Graph-wise Analysis

Graph (a): x–t graph (Position–Time) ✅ Possible
- The particle moves forward, reaches a maximum position, and then moves backward.
- At every instant of time, there is only one position.
- The sharp corner only indicates a sudden change in velocity, which is allowed in an idealised graph.
✔️ This CAN represent one-dimensional motion.

Graph (b): v–t graph (Velocity–Time) ❌ Not Possible
- At the same instant of time, the graph shows more than one velocity (multiple slanted lines overlapping the same time).
- In one-dimensional motion, a particle can have only one velocity at a given instant.
❌ This CANNOT represent one-dimensional motion.

Graph (c): a–t graph (Acceleration–Time) ✅ Possible
- Acceleration is zero most of the time and becomes non-zero for a short interval.
- This can happen when a force acts briefly, such as:
  - a bat hitting a ball,
  - a hammer striking a nail.
✔️ This CAN represent one-dimensional motion.

Question 2.16

Figure 2.13 gives the x–t plot of a particle executing one-dimensional simple harmonic motion.
(You will learn about this motion in more detail in Chapter 13).

Give the signs of position, velocity and acceleration variables of the particle at

$t = 0.3 s, 1.2 s, - 1.2 s .$

Answer

From the given x–t graph of SHM:
- Position (x): sign depends on whether the curve is above (+) or below (–) the time axis.
- Velocity (v): sign depends on the slope of the x–t graph
  - rising → v positive
  - falling → v negative
- Acceleration (a) in SHM:
  $a = - ω^{2} x$
  so acceleration is always opposite in sign to displacement.
At $t = 0.3 s$
- Position: negative (below x-axis)
- Velocity: negative (curve falling)
- Acceleration: positive (opposite to x)
$x < 0, v < 0, a > 0$

At $t = 1.2 s$
- Position: positive (above x-axis)
- Velocity: negative (curve falling)
- Acceleration: negative
$x > 0, v < 0, a < 0$

At $t = - 1.2 s$
- Position: positive
- Velocity: positive (curve rising)
- Acceleration: negative
$x > 0, v > 0, a < 0$

Final Summary Table

Time (s) Position (x) Velocity (v) Acceleration (a)

0.3 – – +

1.2 + – –

–1.2 + + –

Question 2.17

Figure 2.14 gives the x–t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown.

(a) In which interval is the average speed greatest, and in which is it the least?
(b) Give the sign of average velocity for each interval.

Answer

Key ideas to use:
- Average speed = total distance covered ÷ time
  → depends on how steep the x–t graph is (ignores direction).
- Average velocity = displacement ÷ time
  → sign depends on whether x increases (+) or decreases (–) with time.
(a) Average speed

From the graph (equal time intervals):
- Greatest average speed → Interval II
  (the graph is steepest → maximum distance covered in the same time)
- Least average speed → Interval I
  (the graph is least steep → minimum distance covered)
(b) Sign of average velocity
- Interval I: position increases with time
  → average velocity is positive (+)
- Interval II: position still increases with time
  → average velocity is positive (+)
- Interval III: position decreases with time
  → average velocity is negative (–)
Question 2.18

Figure 2.15 gives a speed–time graph of a particle in motion along a constant direction.
Three equal intervals of time are shown.

(a) In which interval is the average acceleration greatest in magnitude?
(b) In which interval is the average speed greatest?
(c) Choosing the positive direction as the constant direction of motion, give the signs of velocity (v) and acceleration (a) in the three intervals.
(d) What are the accelerations at the points A, B, C and D?

Answer

Key ideas from a speed–time (v–t) graph
- Average acceleration = slope of the v–t graph over an interval
- Average speed = average value of speed in that interval
- Instantaneous acceleration at a point = slope of the tangent at that point
- Motion is along a constant positive direction, so v is always positive
(a) Interval with greatest average acceleration (magnitude)
- The steepest change of speed occurs in Interval I
$Greatest average acceleration (magnitude): Interval I$

(b) Interval with greatest average speed
- The highest speeds overall are in Interval III
$Greatest average speed: Interval III$

(c) Signs of velocity and acceleration

Since the particle always moves in the chosen positive direction:

Interval Velocity (v) Acceleration (a) Reason

I + + Speed increasing

II + – Speed decreasing

III + + Speed increasing

(d) Accelerations at points A, B, C and D

Acceleration at a point = slope of the v–t curve at that point.

Point Acceleration

A Positive (speed increasing)

B

Zero (maximum speed → slope zero)

C

Negative (speed decreasing)

D Zero (minimum speed → slope zero)
December 17, 2025
Class 11th Physics Chapter-2 Solutions
From Question 2.1 to 2.9

Question 2.1
In which of the following examples of motion, can the body be considered approximately a point object?
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.

Answer:

A body can be treated as a point object when:
- its size is negligible compared to the distance travelled, and
- rotation or shape does not affect the description of motion.
(a) Railway carriage ✔
The distance between stations is very large compared to the size of the carriage. Its size does not affect the motion.

Can be treated as a point object.

(b) Monkey on a man cycling ✔
The motion of the system (man + monkey) is smooth and its size is small compared to the circular track.

Can be treated as a point object.

(c) Spinning cricket ball ✘
The spinning and turning of the ball are important. Size and rotation cannot be ignored.

Cannot be treated as a point object.

(d) Tumbling beaker ✘
The beaker rotates and changes orientation while falling. Its shape matters.
Cannot be treated as a point object.

Correct answer: (a) and (b)

Question 2.2
The position–time (x–t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown. Choose the correct entries in the brackets below:
(a) (A/B) lives closer to the school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) (A/B) walks faster than (B/A)
(d) A and B reach home at the (same/different) time
(e) (A/B) overtakes (B/A) on the road (once/twice)

Answer:

Important ideas used:
- Distance from school → final position on x-axis
- Who starts earlier → whose graph begins earlier on time axis
- Speed → slope of x–t graph
- Overtaking → point where two graphs intersect
(a) A lives closer to the school than B
A’s final position is less than B’s.
A lives closer.

(b) B starts earlier than A
B’s graph begins at an earlier time.
B starts earlier.

(c) A walks faster than B
Slope of A’s graph is steeper than that of B.
A walks faster.

(d) A and B reach home at the same time
Both graphs end at the same time on the time axis.
Same time.

(e) A overtakes B once
The two graphs intersect at one point.
A overtakes B once.

Question 2.3
A woman starts from her home at 9.00 am, walks with a speed of 5 km h⁻¹ on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h⁻¹. Choose suitable scales and plot the x–t graph of her motion.

Step 1: Choose origin and directions
- Take home as origin (x = 0).
- Direction from home to office is taken as positive x-direction.
- Motion is one-dimensional along a straight road.
Step 2: Motion from home to office
- Distance = 2.5 km
- Speed = 5 km h⁻¹
$Time taken = \frac{2.5}{5} = 0.5 h = 30 minutes$

She starts at 9:00 am, so she reaches office at 9:30 am.

This part of motion is uniform motion, so x–t graph is a straight line with positive slope.

Step 3: Stay at office
- She stays from 9:30 am to 5:00 pm.
- Position remains constant at x = 2.5 km.
- Duration of rest = 7.5 hours.
On x–t graph, this is shown by a horizontal straight line.

Step 4: Return journey (office to home)
- Speed of auto = 25 km h⁻¹
- Distance = 2.5 km
$Time taken = \frac{2.5}{25} = 0.1 h = 6 minutes$

She starts at 5:00 pm and reaches home at 5:06 pm.

Since she is moving towards home, displacement decreases.
The x–t graph is a straight line with negative slope, steeper than the walking line.

Step 5: Summary of key points for the graph

Time Position x (km) Nature of motion

9:00 am 0 Start from home

9:30 am 2.5 Walks uniformly

9:30 am – 5:00 pm 2.5 At rest

5:06 pm 0 Returns home uniformly

Step 6: Choosing suitable scales

You may choose, for example:
- Time axis (t-axis):
  1 cm = 1 hour
- Position axis (x-axis):
  1 cm = 0.5 km
(Any other convenient and proportional scale is acceptable.)

Step 7: Shape of the x–t graph
1. Straight rising line from (9:00 am, 0) to (9:30 am, 2.5 km)
2. Horizontal line from 9:30 am to 5:00 pm at x = 2.5 km
3. Steep straight falling line from (5:00 pm, 2.5 km) to (5:06 pm, 0)
Question 2.4
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x–t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

Step 1: Understand the motion
- Length of each step = 1 m
- Time for each step = 1 s
One complete cycle:
- Forward steps = 5 m in 5 s
- Backward steps = 3 m in 3 s
Net displacement in one cycle

$= 5 - 3 = 2 m$

Time for one cycle

$= 5 + 3 = 8 s$

Step 2: Motion in successive cycles

Cycle Time (s) Net position (m)

1 8 2

2 16 4

3 24 6

4 32 8

5 40 10

6 48 12

After 6 cycles, the drunkard reaches 12 m.

Step 3: Final motion to the pit

The pit is at 13 m from the start.

At the end of 6 cycles:
- Position = 12 m
- Time = 48 s
Now the drunkard moves forward:
- 1 forward step = 1 m in 1 s
So, to go from 12 m to 13 m:

$Time required = 1 s$

Step 4: Total time (by calculation)

$Total time = 48 + 1 = 49 s$

Step 5: x–t graph description (graphical method)

How to draw the x–t graph:
- x-axis → time (s)
- y-axis → position (m)
Nature of graph:
- Slanted rising straight lines → forward motion
- Slanted falling straight lines → backward motion
- Repeated zig-zag pattern
- The graph finally reaches x = 13 m at t = 49 s
From the graph, the point where the line first touches x = 13 m gives the same result:

$t = 49 s$

Question 2.5
A car moving along a straight highway with speed of 126 km h⁻¹ is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

Step 1: Write the given data
- Initial speed, $u = 126 {km h}^{- 1} = \frac{126 \times 1000}{3600} = 35 {m s}^{- 1}$
- Final speed, $v = 0 {m s}^{- 1} (car comes to rest)$
- Distance covered before stopping, $s = 200 m$
Step 2: Find the retardation

Use the equation of motion:

$v^{2} = u^{2} + 2 a s$

Substitute the values:

$0 = (35)^{2} + 2 a (200)$ $0 = 1225 + 400 a$ $a = - \frac{1225}{400} = - 3.06 {m s}^{- 2}$

Retardation

$3.06 {m s}^{- 2}$

(Negative sign indicates deceleration or retardation.)

Step 3: Find the time taken to stop

Use the equation:

$v = u + a t$ $0 = 35 + (- 3.06) t$ $t = \frac{35}{3.06} \approx 11.4 s$

Question 2.6
A player throws a ball upwards with an initial speed of 29.4 m s⁻¹.

(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c) Choose x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands?
(Take g = 9.8 m s⁻² and neglect air resistance.)

(a) Direction of acceleration during upward motion

The acceleration of the ball is due to gravity, which always acts vertically downward, irrespective of the direction of motion.

Answer:

The acceleration during upward motion is vertically downward.

(b) Velocity and acceleration at the highest point

At the highest point:
- The ball momentarily comes to rest, so velocity is zero.
- Gravity still acts downward, so acceleration is not zero.
Answer:
- Velocity = 0
- Acceleration = 9.8 m s⁻² downward
(c) Signs of position, velocity and acceleration

Given sign convention:
- Highest point → x = 0 m
- Time at highest point → t = 0 s
- Downward direction is positive
Upward motion (before reaching highest point)
- Position: Below the highest point → negative
- Velocity: Moving upward (opposite to positive direction) → negative
- Acceleration: Gravity acts downward → positive
Downward motion (after highest point)
- Position: Below the highest point → positive
- Velocity: Moving downward → positive
- Acceleration: Gravity acts downward → positive
(d) Maximum height and time of flight

Given:
- Initial speed, $u = 29.4 {m s}^{- 1}$
- Acceleration due to gravity, $a = - 9.8 {m s}^{- 2}$
- Velocity at highest point, $v = 0$
Maximum height reached

Use:

$v^{2} = u^{2} + 2 a s$

Time to reach highest point

Use:

$v = u + a t$

Total time to return to player

Time of ascent = Time of descent

$Total time = 2 \times 3 = 6 s$

Question 2.7
Read each statement below carefully and state with reasons and examples, if it is true or false. A particle in one-dimensional motion:

(a) with zero speed at an instant may have non-zero acceleration at that instant

Answer: True

Reason:
Speed can be zero at an instant while acceleration is non-zero if the velocity is changing at that instant.

Example:
A ball thrown vertically upward has zero speed at the highest point, but its acceleration due to gravity is 9.8 m s⁻² downward.

(b) with zero speed may have non-zero velocity

Answer: False

Reason:
Velocity is a vector quantity whose magnitude is speed.
If speed is zero, the magnitude of velocity is zero, hence velocity must also be zero.

Example:
An object at rest has zero speed and zero velocity.

(c) with constant speed must have zero acceleration

Answer: False

Reason:
Acceleration depends on change in velocity. Velocity can change due to change in direction even if speed remains constant.

Example:
Uniform circular motion has constant speed, but acceleration is present because direction of velocity changes continuously.

(Note: In one-dimensional straight-line motion, constant speed implies zero acceleration.)

(d) with positive value of acceleration must be speeding up

Answer: False

Reason:
Whether a particle speeds up or slows down depends on the relative directions of velocity and acceleration, not just the sign of acceleration.

Example:
A ball thrown vertically upward has positive acceleration downward, but its speed decreases during upward motion.

Question 2.8

A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one-tenth of its speed. Plot the speed–time graph of its motion between t = 0 to 12 s.

Given
- Height, $h = 90 m$
- Initial velocity, $u = 0$
- Acceleration due to gravity, $g = 9.8 {m s}^{- 2}$
- At each collision, speed reduces to $\frac{9}{10}$ of its value just before impact.
Step 1: First fall (0 to first impact)

Using

$h = \frac{1}{2} g t^{2}$

$90 = \frac{1}{2} \times 9.8 \times t^{2}$

$t^{2} = \frac{180}{9.8} \approx 18.37$

$t_{1} \approx 4.3 s$

Speed just before hitting the floor:

$v_{1} = g t_{1} = 9.8 \times 4.3 \approx 42.1 {m s}^{- 1}$

Step 2: Speed just after first collision

Loss of one-tenth speed:

$v_{1}^{'} = \frac{9}{10} \times 42.1 \approx 37.9 {m s}^{- 1}$

This is upward speed immediately after collision.

Step 3: Motion after first bounce
- The ball moves upward, speed decreases uniformly to zero.
- Time to reach maximum height:
$t_{up} = \frac{v_{1}^{'}}{g} = \frac{37.9}{9.8} \approx 3.9 s$
- Total time till top of first bounce:
$4.3 + 3.9 = 8.2 s$

Speed becomes zero at this point.

Step 4: Second fall (within 12 s)
- From 8.2 s onward, the ball starts falling again.
- Speed increases linearly from 0.
- Time available:
$12 - 8.2 = 3.8 s$

Speed at $t = 12 s$ :

$v = g t = 9.8 \times 3.8 \approx 37.2 {m s}^{- 1}$

(No second collision occurs within 12 s.)

Speed–Time Graph (Description)
0 to 4.3 s → Straight line rising from 0 to 42.1 m/s
Sudden drop at 4.3 s to 37.9 m/s (collision)
4.3 to 8.2 s → Straight line falling to zero (upward motion)
8.2 to 12 s → Straight line rising again (downward motion)
⚠️ Important NCERT Point:
Speed–time graph has sharp vertical drops at collision points because speed changes suddenly.

Question 2.9

Explain clearly, with examples, the distinction between:
(a) Magnitude of displacement and total path length
(b) Magnitude of average velocity and average speed

Show that in both cases the second quantity is greater than or equal to the first.
When does equality hold?
(Consider one-dimensional motion only.)

(a) Displacement (magnitude) vs Total Path Length

Displacement
- Displacement is the shortest straight-line distance between the initial and final positions of a particle.
- It depends only on the initial and final positions, not on the actual path followed.
- It can be positive, negative or zero.
- Its magnitude is always less than or equal to the path length.
Total Path Length (Distance)
- Total path length is the actual length of the path travelled by the particle.
- It depends on the entire motion.
- It is always positive.
- It is always greater than or equal to the magnitude of displacement.
Example

Suppose a particle moves:
- From $x = 0$ m to $x = 10$ ,
- Then back to $x = 4$ .
- Magnitude of displacement
  
  $= ∣ 4 - 0 ∣ = 4 m$
- Total path length
  
  $= 10 + 6 = 16 m$
$Path length > Magnitude of displacement$

Equality condition
- Equality holds only when the particle moves in a straight line without changing direction.
(b) Magnitude of Average Velocity vs Average Speed

Average Velocity
- Average velocity is defined as:
$Average velocity = \frac{Displacement}{Time interval}$
- Its magnitude depends only on displacement.
- It can be zero, even if the particle has moved.
Average Speed
- Average speed is defined as:
$Average speed = \frac{Total path length}{Time interval}$
- It depends on the entire distance travelled.
- It is always positive.
Why average speed ≥ magnitude of average velocity

Since:

$Total path length \geq Magnitude of displacement$

Dividing both sides by time:

$Average speed \geq Magnitude of average velocity$

Example

A person walks:
- 10 m forward in 10 s,
- then 10 m backward in 10 s.
- Displacement = 0
- Total path length = 20 m
- Time = 20 s
$Magnitude of average velocity = \frac{0}{20} = 0$ $Average speed = \frac{20}{20} = 1 {m s}^{- 1}$

$Average speed > Magnitude of average velocity$

Equality condition
- Equality holds only when motion is along a straight line without reversal of direction.
Final Conclusion (Very Important for Exams)

Quantity Comparison

Path length ≥ Magnitude of displacement

Average speed
≥ Magnitude of average velocity

Equality holds only for straight-line motion in one direction.
December 16, 2025
Class 11th Physics Exercises-1 Solutions
Exercise 1.1 – Fill in the blanks (Solved)

(a)

Question:
The volume of a cube of side 1 cm is equal to _____ m³.

Solution:
Side of cube = 1 cm

$Volume = (1 cm)^{3} = 1 {cm}^{3}$ Since,

$1 cm = 10^{- 2} m$ $1 {cm}^{3} = (10^{- 2})^{3} = 10^{- 6} m^{3}$

Answer:

$1 \times 10^{- 6} m^{3}$

(b)

Question:
The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to _____ (mm)².

Solution:
Radius $r = 2.0$
Height $h = 10.0$

Surface area of a solid cylinder:

$S = 2 π r (r + h)$ $S = 2 π \times 2 \times (2 + 10)$ $S = 48 π {cm}^{2}$

Convert cm² to mm²:

$1 {cm}^{2} = 100 {mm}^{2}$ $S = 48 π \times 100 \approx 1.51 \times 10^{4} {mm}^{2}$

Answer:

$1.51 \times 10^{4} (mm)^{2}$

(c)

Question:
A vehicle moving with a speed of 18 km h⁻¹ covers _____ m in 1 s.

Solution:

$18 {km h}^{- 1} = \frac{18 \times 1000}{3600}$

$= 5 {m s}^{- 1}$

Distance covered in 1 s:

$= 5 m$

Answer:

$5 m$

(d)

Question:
The relative density of lead is 11.3. Its density is _____ g cm⁻³ or _____ kg m⁻³.

Solution:
Relative density =

$\frac{Density of substance}{Density of water}$

Density of water = 1 g cm⁻³

$Density of lead = 11.3 \times 1 = 11.3 {g cm}^{- 3}$

Convert to kg m⁻³:

$1 {g cm}^{- 3} = 1000 {kg m}^{- 3}$ $11.3 \times 1000 = 1.13 \times 10^{4}$

Answer:

$11.3 {g cm}^{- 3} or 1.13 \times 10^{4} {kg m}^{- 3}$

Exercise 1.2 – Fill in the blanks by suitable conversion of units (Solved)

(a)

Question:
1 kg m² s⁻² = _____ g cm² s⁻².

Solution: $1 kg = 10^{3} g$ $1 m = 10^{2} cm \Rightarrow 1 m^{2} = 10^{4} {cm}^{2}$ $1 {kg m}^{2} s^{- 2} = 10^{3} \times 10^{4}$ $= 10^{7} {g cm}^{2} s^{- 2}$

Answer:

$10^{7} {g cm}^{2} s^{- 2}$

Question 1.3

A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1 J = 1 kg m² s⁻². Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude $4.2 α^{- 1} β^{- 2} γ^{2}$ in terms of the new units.

Solution

We are given:

$1 calorie = 4.2 J$

Also,

$1 J = 1 {kg m}^{2} s^{- 2}$

Step 1: Write the new fundamental units
- New unit of mass = $α kg$
- New unit of length = $β m$
- New unit of time = $γ s$
Step 2: Express SI units in terms of new units

$1 kg = α^{- 1} (new mass unit)$ $1 m = β^{- 1} (new length unit)$ $1 s = γ^{- 1} (new time unit)$

Step 3: Convert joule into the new system

$1 J = 1 {kg m}^{2} s^{- 2}$

Substitute the above relations:

$1 J = (α^{- 1}) (β^{- 1})^{2} (γ^{- 1})^{- 2}$ $= α^{- 1} β^{- 2} γ^{2}$

Step 4: Convert calorie

$1 calorie = 4.2 J$ $= 4.2 \times α^{- 1} β^{- 2} γ^{2}$

Final Answer

$1 calorie = 4.2 α^{- 1} β^{- 2} γ^{2}$

Question 1.4

Explain this statement clearly:

“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison.”

In view of this, reframe the following statements wherever necessary :
(

Explanation

A dimensional quantity has physical dimensions such as length, mass, time, etc.
Calling such a quantity large or small has no meaning unless it is compared with a suitable standard or reference quantity of the same kind.

For example, saying that a mass is “large” is meaningless unless we specify large compared to what—a human body, a planet, or an atom.

Hence, every statement about size, mass, speed, or number must include a reference or comparison to be scientifically meaningful.

Reframed Statements

(a) Atoms are very small objects.

Reframed:
Atoms are very small compared to ordinary macroscopic objects such as grains of sand or cells.

(b) A jet plane moves with great speed.

Reframed:
A jet plane moves with great speed compared to road vehicles such as cars or trains.

(c) The mass of Jupiter is very large.

Reframed:
The mass of Jupiter is very large compared to the mass of the Earth or other planets.

(d) The air inside this room contains a large number of molecules.

Reframed:
The air inside this room contains a large number of molecules compared to the number of molecules in a small volume of air, such as a test tube.

(e) A proton is much more massive than an electron.

Reframed:
A proton is much more massive than an electron (about 1836 times more massive).

✔️ (This statement is already meaningful because a clear comparison is given.)

(f) The speed of sound is much smaller than the speed of light.

Reframed:
The speed of sound is much smaller than the speed of light in vacuum.

Question 1.5

A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

Solution

Step 1: Given data

Speed of light in vacuum:

$c = 1 (in the new system of units)$

Time taken by light:

$8 min 20 s$

Convert time into seconds:

$8 min = 8 \times 60 = 480 s$ $Total time = 480 + 20 = 500 s$

Step 2: Use the formula for distance

$Distance = Speed \times Time$

Since $c = 1$ in the new unit system:

$Distance = 1 \times 500 = 500$

Final Answer

$Distance between the Sun and the Earth = 500 (new units of length)$

Question 1.6

Which of the following is the most precise device for measuring length :
(a) a vernier callipers with 20 divisions on the sliding scale
(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale
(c) an optical instrument that can measure length to within a wavelength of light ?

Solution

Precision of a measuring instrument depends on its least count.
The instrument with the smallest least count is the most precise.

(a) Vernier callipers

For a standard vernier callipers:
- 1 main scale division (MSD) = 1 mm
- 20 vernier divisions = 19 MSD
Least count:

$LC = 1 MSD - 1 VSD = 1 - \frac{19}{20} = 0.05 mm$

(b) Screw gauge

Given:
- Pitch = 1 mm
- Number of divisions = 100
Least count:

$LC = \frac{Pitch}{No. of divisions} = \frac{1}{100} = 0.01 mm$

(c) Optical instrument

An optical instrument can measure length up to one wavelength of light.

Typical wavelength of visible light:

$λ \approx 5 \times 10^{- 7} m = 0.0005 mm$

This is much smaller than the least count of both vernier callipers and screw gauge.

Answer

Since the optical instrument has the smallest least count, it is the most precise.

$(c) An optical instrument that can measure length to within a wavelength of light$

Question 1.7

A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ?

Solution

Step 1: Understand magnification

Magnification $M$ is given by:

$M = \frac{Apparent width}{Actual width}$

Given:
- Magnification, $M = 100$
- Apparent (observed) width of hair = 3.5 mm
Step 2: Find actual thickness of hair

$Actual thickness = \frac{Observed width}{Magnification}$

$= \frac{3.5 mm}{100}$ $= 0.035 mm$

Step 3: Convert into metre (optional)

$0.035 mm = 3.5 \times 10^{- 5} m$

Answer

$Thickness of human hair = 0.035 mm (= 3.5 \times 10^{- 5} m)$

Question 1.8

Answer the following :

(a)

You are given a thread and a metre scale. How will you estimate the diameter of the thread ?

Answer:
- Wind the thread closely and uniformly around a pencil or cylindrical rod for a large number of turns (say $n$ turns), without leaving gaps.
- Measure the total length $L$ occupied by these $n$ turns using the metre scale.
- The diameter of the thread is then given by:
$Diameter of thread = \frac{L}{n}$
- Taking a large number of turns reduces error and gives a more accurate result.
(b)

A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ?

Answer:

No, it is not possible to increase the accuracy arbitrarily by increasing the number of divisions.

Explanation:
- Least count of screw gauge:
$LC = \frac{Pitch}{Number of divisions}$
- Increasing the number of divisions reduces the least count, but:
  - Mechanical limitations
  - Backlash error
  - Wear and tear
  - Manufacturing precision
    limit further improvement.
- Beyond a certain point, increasing divisions does not improve accuracy.
(c)

The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only ?

Answer:
- Every measurement contains random errors.
- When a large number of measurements (like 100) are taken:
  - Random errors tend to cancel out.
  - The mean value becomes closer to the true value.
- With only 5 measurements:
  - Random errors have a larger effect.
  - The estimate is less reliable.
Hence, 100 measurements give a more accurate and reliable mean diameter than only 5 measurements.

Question 1.9

The photograph of a house occupies an area of 1.75 cm² on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m². What is the linear magnification of the projector–screen arrangement?

Solution

Step 1: Understand the relation between area and linear magnification

If linear magnification = $M$ , then:

$Areal magnification = M^{2}$ $M = \sqrt{\frac{Area on screen}{Area on slide}}$

Step 2: Convert all areas into the same unit

Given:
- Area on slide = $1.75 {cm}^{2}$
- Area on screen = $1.55 m^{2}$
Convert $m^{2}$ to ${cm}^{2}$ :

$1 m^{2} = 10^{4} {cm}^{2}$ $1.55 m^{2} = 1.55 \times 10^{4} {cm}^{2}$

Step 3: Calculate linear magnification

$M = \sqrt{\frac{1.55 \times 10^{4}}{1.75}}$ $= \sqrt{8857}$ $\approx 94$

Answer

$Linear magnification \approx 94$

Question 1.10

State the number of significant figures in the following :

(a) 0.007 m²

Leading zeros are not significant.
Only 7 is significant.

Answer:

$1 significant figure$

(b) 2.64 × 10²⁴ kg

All digits in scientific notation are significant.
Digits: 2, 6, 4

Answer:

$3 significant figures$

(c) 0.2370 g cm⁻³

Leading zero is not significant.
Trailing zero after decimal is significant.

Digits: 2, 3, 7, 0

Answer:

$4 significant figures$

(d) 6.320 J

Trailing zero after decimal point is significant.

Digits: 6, 3, 2, 0

Answer:

$4 significant figures$

(e) 6.032 N m⁻²

Zero between non-zero digits is significant.

Digits: 6, 0, 3, 2

Answer:

$4 significant figures$

(f) 0.0006032 m²

Leading zeros are not significant.
Zero between 6 and 3 is significant.

Digits: 6, 0, 3, 2

Answer:

$4 significant figures$

Question 1.11

The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Solution

Given data
- Length, $l = 4.234 m$ → 4 significant figures
- Breadth, $b = 1.005 m$ → 4 significant figures
- Thickness, $t = 2.01 cm = 0.0201 m$ → 3 significant figures
(a) Area of the sheet

$Area = l \times b$ $= 4.234 \times 1.005$ $= 4.25517 m^{2}$

Both length and breadth have 4 significant figures, so the result should be given to 4 significant figures.

Area (correct significant figures)

$Area = 4.255 m^{2}$

(b) Volume of the sheet

$Volume = l \times b \times t$ $= 4.234 \times 1.005 \times 0.0201$ $= 0.0855389 m^{3}$

The least number of significant figures among the quantities is 3 (thickness).

Volume (correct significant figures)

$Volume = 0.0855 m^{3}$

Question 1.12

The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is:
(a) the total mass of the box,
(b) the difference in the masses of the pieces, to correct significant figures?

Solution

Given data
- Mass of box = 2.30 kg → 3 significant figures
- Mass of first gold piece = 20.15 g
- Mass of second gold piece = 20.17 g
(a) Total mass of the box

Step 1: Convert gold masses into kg

$20.15 g = 0.02015 kg$ $20.17 g = 0.02017 kg$

Step 2: Add all masses

$Total mass = 2.30 + 0.02015 + 0.02017$ $= 2.34032 kg$

Step 3: Apply significant-figure rule (addition)

In addition, the result should have the least number of decimal places among the quantities.
- Mass of box = 2.30 kg → 2 decimal places
So, round the result to 2 decimal places:

$Total mass = 2.34 kg$

(b) Difference in the masses of the gold pieces

Step 1: Subtract the two masses

$Difference = 20.17 g - 20.15 g$ $= 0.02 g$

Step 2: Apply significant-figure rule (subtraction)

Both values are correct to two decimal places, so the result should also be given to two decimal places.

$Difference in masses = 0.02 g$

Question 1.13

A famous relation in physics relates ‘moving mass’ $m$ to the ‘rest mass’ $m_{0}$ of a particle in terms of its speed $v$ and the speed of light $c$ . (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant $c$ . He writes :

$m = \frac{m_{0}}{(1 - v^{2})^{1 / 2}}$

Guess where to put the missing $c$ .

Solution

Step 1: Use dimensional analysis
- Speed $v$ has dimensions:
  
  $[v] = L T^{- 1}$
- Therefore,
  
  $[v^{2}] = L^{2} T^{- 2}$
The quantity inside the square root must be dimensionless.
But in the given expression:

$(1 - v^{2})$

the term $v^{2}$ has dimensions, so the expression is dimensionally incorrect.

Step 2: Make the expression dimensionless

To cancel the dimensions of $v^{2}$ , it must be divided by a quantity having the same dimensions, i.e. $c^{2}$ , where $c$ is the speed of light.

Thus, the correct dimensionless term is:

$\frac{v^{2}}{c^{2}}$

Step 3: Write the correct formula

Substituting this into the expression:

$m = \frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

Answer

$m = \frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

Question 1.14

The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å:
$1 \overset{˚}{A} = 10^{- 10} m$ .
The size of a hydrogen atom is about $0.5 \overset{˚}{A}$ .
What is the total atomic volume in $m^{3}$ of a mole of hydrogen atoms?

Solution

Step 1: Given data
- Radius (size) of hydrogen atom:
$r = 0.5 \overset{˚}{A} = 0.5 \times 10^{- 10} m$
- Number of atoms in one mole:
$N = 6.02 \times 10^{23}$

Step 2: Volume of one hydrogen atom

Assuming the atom to be spherical,

$Volume of one atom = \frac{4}{3} π r^{3}$

$= \frac{4}{3} π (0.5 \times 10^{- 10})^{3}$ $= \frac{4}{3} π (0.125 \times 10^{- 30})$ $= 5.24 \times 10^{- 31} m^{3} (approximately)$

Step 3: Total atomic volume of one mole

$Total volume = N \times volume of one atom$ $= 6.02 \times 10^{23} \times 5.24 \times 10^{- 31}$ $= 3.15 \times 10^{- 7} m^{3}$

Answer

$Total atomic volume of one mole of hydrogen \approx 3.1 \times 10^{- 7} m^{3}$

Question. 1.15
One mole of an ideal gas at STP occupies 22.4 L. What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Size of hydrogen molecule ≈ 1 Å). Why is this ratio so large?

Step 1: Molar volume of hydrogen gas at STP

$V_{molar} = 22.4 L = 22.4 \times 10^{- 3} m^{3}$

Step 2: Volume of one hydrogen molecule

Size (diameter) of hydrogen molecule ≈ $1 \overset{˚}{A} = 10^{- 10} m$

Assume the molecule to be spherical:

$V_{one molecule} = \frac{4}{3} π r^{3}$

$r = \frac{1}{2} \times 10^{- 10} = 0.5 \times 10^{- 10} m$ $V_{one molecule} \approx \frac{4}{3} π (0.5 \times 10^{- 10})^{3} \approx 5.2 \times 10^{- 31} m^{3}$

Step 3: Atomic (molecular) volume of one mole of hydrogen

Number of molecules in 1 mole:

$N_{A} = 6.02 \times 10^{23}$ $V_{atomic (1 mole)} = 6.02 \times 10^{23} \times 5.2 \times 10^{- 31}$ $V_{atomic} \approx 3.1 \times 10^{- 7} m^{3}$

Step 4: Required ratio

$Ratio = \frac{V_{molar}}{V_{atomic}} = \frac{22.4 \times 10^{- 3}}{3.1 \times 10^{- 7}}$ $Ratio \approx 7.2 \times 10^{4}$

Question 1.16

Explanation of the Observation

When you look out of the window of a fast-moving train, the apparent motion of objects depends on their distance from you.

1. Nearby objects (trees, houses, poles)
- These objects are close to the observer.
- As the train moves, your position with respect to these objects changes rapidly.
- This causes a large change in their angular position in a short time.
- Hence, nearby objects appear to move rapidly in the opposite direction to the train’s motion.
This effect is called relative motion.

2. Distant objects (hills, mountains)
- These objects are very far away.
- Even though the train moves, the change in their angular position is very small.
- Therefore, they appear to move very slowly or seem almost stationary.
3. Very distant objects (Moon, stars)
- These objects are at extremely large distances.
- The change in their position relative to the observer is negligible.
- As a result, they appear to be completely stationary or appear to move along with you.
Question 1.17

The Sun is a hot plasma, so one might initially guess that its density should be like that of gases. Let us now check this guess by calculation using the given data.

Given data
- Mass of the Sun,
  
  $M = 2.0 \times 10^{30} kg$
- Radius of the Sun,
  
  $R = 7.0 \times 10^{8} m$
Step 1: Volume of the Sun

The Sun is approximately spherical.

$V = \frac{4}{3} π R^{3}$

$V = \frac{4}{3} π (7.0 \times 10^{8})^{3}$ $V \approx 1.44 \times 10^{27} m^{3}$

Step 2: Average density of the Sun

$ρ = \frac{M}{V}$ $ρ = \frac{2.0 \times 10^{30}}{1.44 \times 10^{27}}$

$ρ \approx 1.4 \times 10^{3} {kg m}^{- 3}$

Step 3: Interpretation
- Density of gases (at STP):
  
  $\sim 1 {kg m}^{- 3}$
- Density of liquids/solids (water):
  
  $\sim 10^{3} {kg m}^{- 3}$
The Sun’s average density ≈ 1400 kg m⁻³, which is:

much greater than gases
comparable to liquids and solids
December 16, 2025

Time (s)	Position (x)	Velocity (v)	Acceleration (a)
0.3	–	–	+
1.2	+	–	–
–1.2	+	+	–

Interval	Velocity (v)	Acceleration (a)	Reason
I	+	+	Speed increasing
II	+	–	Speed decreasing
III	+	+	Speed increasing

Point	Acceleration
A	Positive (speed increasing)
B	Zero (maximum speed → slope zero)
C	Negative (speed decreasing)
D	Zero (minimum speed → slope zero)

Time	Position x (km)	Nature of motion
9:00 am	0	Start from home
9:30 am	2.5	Walks uniformly
9:30 am – 5:00 pm	2.5	At rest
5:06 pm	0	Returns home uniformly

Cycle	Time (s)	Net position (m)
1	8	2
2	16	4
3	24	6
4	32	8
5	40	10
6	48	12

Quantity	Comparison
Path length	≥ Magnitude of displacement
Average speed	≥ Magnitude of average velocity

Class 9th Science Chapter-12 Exercises

Exercises – Chapter 12: Improvement in Food Resources

Questions with Answers

Question 1

Explain any one method of crop production which ensures high yield.

Answer:
One method of crop production that ensures high yield is crop variety improvement.
In this method, improved crop varieties are developed through selection, hybridisation or genetic manipulation. These varieties are high-yielding, disease-resistant and adaptable to different environmental conditions, which helps in increasing crop production.

Question 2

Why are manure and fertilizers used in fields?

Answer:
Manure and fertilizers are used in fields to replenish essential nutrients in the soil.
They improve soil fertility and provide nutrients required for healthy growth, development and higher yield of crops.

Question 3

What are the advantages of inter-cropping and crop rotation?

Answer:

Advantages of inter-cropping:

Ensures better utilisation of nutrients
Reduces the spread of pests and diseases
Gives higher overall yield

Advantages of crop rotation:

Maintains soil fertility
Prevents nutrient depletion
Reduces pest and weed growth

Question 4

What is genetic manipulation? How is it useful in agricultural practices?

Answer:
Genetic manipulation is the introduction of desirable genes into crop plants to obtain improved characteristics.

Usefulness:

Produces high-yielding varieties
Provides resistance to diseases and pests
Improves quality of crops
Helps crops tolerate abiotic stresses like drought and salinity

Question 5

How do storage grain losses occur?

Answer:
Storage grain losses occur due to:

Biotic factors: insects, rodents, fungi, mites and bacteria
Abiotic factors: improper moisture and unsuitable temperature

These factors cause loss in weight, quality, germination and market value of grains.

Question 6

How do good animal husbandry practices benefit farmers?

Answer:
Good animal husbandry practices:

Improve health and productivity of animals
Increase production of milk, eggs and meat
Reduce disease-related losses
Increase income and livelihood security of farmers

Question 7

What are the benefits of cattle farming?

Answer:
The benefits of cattle farming are:

Provides milk, an important food source
Supplies draught power for agricultural work
Produces dung, used as manure and fuel
Generates additional income for farmers

Question 8

For increasing production, what is common in poultry, fisheries and bee-keeping?

Answer:
In poultry, fisheries and bee-keeping, improved management practices are common, such as:

Selection of improved breeds
Proper feeding and nutrition
Disease control and hygiene
Scientific management techniques

These practices help in increasing production.

Question 9

How do you differentiate between capture fishing, mariculture and aquaculture?

Answer:

Capture fishing	Mariculture	Aquaculture
Fish are obtained from natural water bodies	Farming of marine organisms in seawater	Rearing of aquatic organisms in controlled water bodies
No breeding is involved	Marine fish are cultured	Includes freshwater and marine organisms
Yield depends on natural availability	Higher yield than capture fishing	High and controlled yield

December 15, 2025

Class 9th Science Chapter-12 In-Text Questions

Chapter 12: Improvement in Food Resources

Question with Answer

Page 141

Question 1.

What do we get from cereals, pulses, fruits and vegetables?

Answer:

Cereals provide carbohydrates, which give energy.
Pulses provide proteins, which are needed for growth and repair of the body.
Fruits and vegetables provide vitamins and minerals, which protect us from diseases and keep the body healthy.

Page 142

Question 1

How do biotic and abiotic factors affect crop production?

Answer:

Biotic factors such as insects, pests, diseases, weeds and pathogens damage crop plants and reduce their yield.
Abiotic factors such as drought, salinity, waterlogging, heat, cold and frost affect the growth and productivity of crops.

Thus, both biotic and abiotic factors reduce crop production by affecting plant health and growth.

Question 2

What are the desirable agronomic characteristics for crop improvements?

Answer:
Desirable agronomic characteristics for crop improvement include:

Tallness and profuse branching in fodder crops to obtain more biomass
Dwarfness in cereal crops so that less nutrients are used and lodging is prevented

These characteristics help in obtaining higher yield and better productivity.

Page 143

Question 1

What are macro-nutrients and why are they called macro-nutrients?

Answer:
Macro-nutrients are nutrients that are required by plants in large quantities for their growth and development.
They are called macro-nutrients because plants need them in comparatively large amounts.

Examples of macro-nutrients are nitrogen, phosphorus, potassium, calcium, magnesium and sulphur.

Question 2

How do plants get nutrients?

Answer:
Plants get nutrients in the following ways:

Carbon and oxygen are obtained from air.
Hydrogen is obtained from water.
Other nutrients are obtained from the soil in dissolved form through the roots.

These nutrients are then used by plants for growth and development.

Page 144

Question 1

Compare the use of manure and fertilizers in maintaining soil fertility.

Answer:

Manure	Fertilizers
Manure is a natural substance obtained by the decomposition of plant and animal wastes	Fertilizers are chemical substances manufactured in factories
It supplies nutrients in small quantities	It supplies nutrients in large quantities
It adds organic matter to the soil	It does not add organic matter to the soil
It improves soil structure, water-holding capacity and aeration	It does not improve soil structure
It increases soil fertility in the long term	Continuous use may reduce soil fertility

Page 145

Question 1

Which of the following conditions will give the most benefits? Why?

(a) Farmers use high-quality seeds, do not adopt irrigation or use fertilizers.
(b) Farmers use ordinary seeds, adopt irrigation and use fertilizer.
(c) Farmers use quality seeds, adopt irrigation, use fertilizer and use crop protection measures.

Answer:

Option (c) will give the most benefits.

Explanation:
Quality seeds ensure better yield potential.
Irrigation supplies adequate water at different stages of crop growth.
Fertilizers provide essential nutrients required for healthy plant growth.
Crop protection measures protect crops from weeds, pests and diseases.

Thus, the combined use of quality seeds, irrigation, fertilizers and crop protection results in maximum yield and better crop production.

Page 146

Question 1

Why should preventive measures and biological control methods be preferred for protecting crops?

Answer:
Preventive measures and biological control methods should be preferred because:

They do not cause environmental pollution.
They do not harm useful organisms present in the ecosystem.
They help in maintaining ecological balance.
They reduce the harmful effects caused by excessive use of chemical pesticides.

Question 2

What factors may be responsible for losses of grains during storage?

Answer:
Losses of grains during storage occur due to the following factors:

Biotic factors: insects, rodents, fungi, mites and bacteria
Abiotic factors: improper moisture and unsuitable temperature conditions

These factors lead to loss in weight, quality and germination capacity of stored grains.

Page 147

Question 1

Which method is commonly used for improving cattle breeds and why?

Answer:
The commonly used method for improving cattle breeds is cross-breeding.

Explanation:
Cross-breeding is done between indigenous breeds (which have good disease resistance) and exotic breeds (which have high milk yield and long lactation period).
This helps in obtaining cattle with both high milk production and disease resistance.

Page 148

Set – 1

Question 1

Discuss the implications of the following statement:
“It is interesting to note that poultry is India’s most efficient converter of low-fibre food stuff (which is unfit for human consumption) into highly nutritious animal protein food.”

Answer:

The statement highlights the importance and efficiency of poultry farming in food production.

Explanation:

Poultry birds can consume low-fibre agricultural by-products such as broken grains and crop residues, which are not suitable for direct human consumption.
These low-quality food materials are efficiently converted into high-quality animal protein in the form of eggs and meat.
Poultry farming requires less space, less time and lower investment compared to other animal husbandry practices.
It helps in reducing food wastage and provides a cheap and rich source of protein to humans.
Thus, poultry farming plays an important role in nutritional security and income generation.

Hence, poultry is an efficient system for converting low-value feed into nutritious food for humans.

Set – 2

Question 1

What management practices are common in dairy and poultry farming?

Answer:
The common management practices in both dairy and poultry farming are:

Proper housing and shelter
Balanced and nutritious feed
Cleanliness and sanitation
Prevention and control of diseases through vaccination

Question 2

What are the differences between broilers and layers and in their management?

Answer:

Broilers	Layers
Raised for meat production	Raised for egg production
Require protein-rich diet with more fat	Require balanced diet rich in vitamins and minerals
Short growth period	Longer rearing period
Management focuses on rapid growth	Management focuses on egg laying capacity

Page 150

Set – 1

Question 1

How are fish obtained?

Answer:
Fish are obtained in two ways:

Capture fishing: Fish are caught from natural water bodies such as seas, rivers, lakes and ponds.
Culture fishery (fish farming): Fish are bred and reared in controlled water bodies like ponds, tanks and reservoirs.

Question 2

What are the advantages of composite fish culture?

Answer:
The advantages of composite fish culture are:

Different species of fish having different food habits are grown together.
Fish use food from all levels of the pond (surface, middle and bottom).
There is no competition for food among fish species.
It results in higher fish yield from the same pond.

Set – 2

Question 1

What are the desirable characters of bee varieties suitable for honey production?

Answer:
Desirable characters of bee varieties suitable for honey production are:

High honey collection capacity
Less aggressive nature (sting less)
Ability to stay in the hive for long periods
Good breeding capacity
Resistance to diseases

These characteristics help in obtaining higher yield of honey.

Question 2

What is pasturage and how is it related to honey production?

Answer:
Pasturage refers to the availability of flowers from which bees collect nectar and pollen.

Relation to honey production:

The quantity of honey depends on the abundance of flowers available.
The quality and taste of honey depend on the type of flowers present in the pasturage.

Thus, good pasturage results in better quality and higher production of honey.

December 15, 2025

Class 9th Science Chapter-11 Exercises
Exercises – Chapter – 11: Sound

Questions with Answers

Question 1

What is sound and how is it produced?

Answer:
Sound is a form of energy that produces the sensation of hearing in our ears.
Sound is produced when an object vibrates. These vibrations set the particles of the surrounding medium into motion, producing sound.

Question 2

Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.

Answer:
When a vibrating object moves forward, it compresses the air in front of it, creating a region of high pressure called compression.
When it moves backward, it creates a region of low pressure called rarefaction.
As the object vibrates continuously, a series of compressions and rarefactions is formed, which travels through air as a sound wave.

Question 3

Why is sound wave called a longitudinal wave?

Answer:
Sound wave is called a longitudinal wave because the particles of the medium vibrate back and forth parallel to the direction of propagation of the wave.

Question 4

Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?

Answer:
The quality (or timbre) of sound helps us to identify a person by his or her voice.

Question 5

Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?

Answer:
Light travels much faster than sound.
Therefore, the flash of lightning reaches our eyes almost instantly, while the sound of thunder takes more time to reach our ears.

Question 6

A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies?
(Take speed of sound in air $v = 344 m s^{- 1}$ )

Answer:

Using the relation:

$λ = \frac{v}{ν}$

For $ν = 20 H z$ :

$λ = \frac{344}{20} = 17.2 m$

For $ν = 20 000 H z$ :

$λ = \frac{344}{20000} = 0.0172 m$

Question 7

Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.

Answer:

Speed of sound in air = 344 m s⁻¹
Speed of sound in aluminium = 6420 m s⁻¹

Time taken $t \propto \frac{1}{v}$

$Ratio of times = \frac{v_{aluminium}}{v_{air}} = \frac{6420}{344} \approx 18.7$

Ratio = 18.7 : 1

Question 8

The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?

Answer:
Frequency = number of vibrations per second

In 1 second = 100 vibrations
In 60 seconds:

$100 \times 60 = 6000$

Number of vibrations = 6000

Question 9

Does sound follow the same laws of reflection as light does? Explain.

Answer:
Yes, sound follows the same laws of reflection as light.

Explanation:
- The angle of incidence is equal to the angle of reflection.
- The incident sound, reflected sound and the normal lie in the same plane.
Question 10

When a sound is reflected from a distant object, an echo is produced. If the distance between the reflecting surface and the source remains the same, do you hear echo sound on a hotter day?

Answer:
Yes, the echo is heard earlier on a hotter day.

Explanation:
On a hotter day, the speed of sound increases, so the reflected sound returns faster.

Question 11

Give two practical applications of reflection of sound waves.

Answer:
1. Megaphones and loudhailers are designed using reflection of sound to direct sound forward.
2. Stethoscope uses multiple reflections of sound to transmit heartbeats clearly to the doctor’s ears.
Question 12

A stone is dropped from the top of a tower 500 m high into a pond at the base. When is the splash heard at the top?
(Given $g = 10 m s^{- 2}$ , speed of sound $= 340 m s^{- 1}$

Answer:

Time taken by stone to fall:

$s = \frac{1}{2} g t^{2} \Rightarrow 500 = 5 t^{2} \Rightarrow t = 10 s$

Time taken by sound to travel up:

$t = \frac{500}{340} \approx 1.47 s$

Total time:

$10 + 1.47 = 11.47 s$

The splash is heard after approximately 11.5 s.

Question 13

A sound wave travels at a speed of 339 m s⁻¹. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?

Answer:

Given:
Speed, $v = 339 m s^{- 1}$
Wavelength, $λ = 1.5 c m = 0.015 m$

$ν = \frac{v}{λ} = \frac{339}{0.015} = 22600 H z$

The audible range for humans is 20 Hz to 20 kHz.

Therefore, the sound will not be audible, as its frequency is greater than 20 kHz (ultrasound).

Question 14

What is reverberation? How can it be reduced?

Answer:
Reverberation is the persistence of sound in a large hall due to multiple reflections from walls, ceiling and other surfaces, even after the source has stopped producing sound.

It can be reduced by:
- Covering walls and ceilings with sound-absorbing materials such as fibreboard or rough plaster
- Using curtains, carpets and cushioned seats
Question 15

What is loudness of sound? What factors does it depend on?

Answer:
Loudness is the physiological response of the human ear to the sound.

It depends on:
- The amplitude of the sound wave
- The sensitivity of the human ear
A sound with larger amplitude is heard as louder.

Question 16

How is ultrasound used for cleaning?

Answer:
In ultrasonic cleaning, objects are placed in a liquid and ultrasonic waves are passed through it.
These high-frequency waves produce vibrations that dislodge dust, grease and dirt from even hard-to-reach places.
Thus, the objects get thoroughly cleaned.

Question 17

Explain how defects in a metal block can be detected using ultrasound.

Answer:
Ultrasonic waves are passed through the metal block.
If there is a defect or crack inside, the waves get reflected back from that place.
By detecting the reflected waves, the presence and location of defects in the metal block can be identified.
December 15, 2025

Class 9th Science Chapter-11 In-Text Questions

Chapter 11 – Sound

Page 129

Questions with Answers

Question 1

How does the sound produced by a vibrating object in a medium reach your ear?

Answer:
The vibrating object sets the particles of the medium around it into vibration. These vibrations are passed on from one particle to the next in the form of compressions and rarefactions. In this way, the disturbance travels through the medium and reaches the ear, producing the sensation of sound.

Question 2

Explain how sound is produced by your school bell.

Answer:
When the school bell is struck, it starts vibrating. These vibrations produce compressions and rarefactions in the surrounding air. The sound thus produced travels through air and reaches our ears.

Question 3

Why are sound waves called mechanical waves?

Answer:
Sound waves are called mechanical waves because they require a material medium for their propagation and are produced due to the vibrations of particles of the medium.

Question 4

Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?

Answer:
No, sound cannot be heard on the moon because there is no medium like air to transmit sound waves.

Page 132

Question 1

Which wave property determines (a) loudness, (b) pitch?

Answer:
(a) Loudness depends on the amplitude of the sound wave.
(b) Pitch depends on the frequency of the sound wave.

Question 2

Guess which sound has a higher pitch: guitar or car horn?

Answer:
The guitar produces a sound of higher pitch than a car horn.

Question 1

What are wavelength, frequency, time period and amplitude of a sound wave?

Answer:

Wavelength (λ): The distance between two consecutive compressions or two consecutive rarefactions.
Frequency (ν): The number of oscillations per second.
Time period (T): The time taken for one complete oscillation.
Amplitude (A): The maximum displacement of particles of the medium from their mean position.

Question 2

How are the wavelength and frequency of a sound wave related to its speed?

Answer:
The speed of sound is given by:

$v = λ ν$

Question 3

Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m s⁻¹ in a given medium.

Answer:

$λ = \frac{v}{ν} = \frac{440}{220} = 2 m$

Question 4

A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?

Answer:
Time interval between successive compressions equals the time period.

$T = \frac{1}{ν} = \frac{1}{500} = 0.002 s$

Page 133

Sound

Questions with Answers

Set – 1

Question 1

Distinguish between loudness and intensity of sound.

Answer:

Loudness	Intensity
Loudness is a physiological response of the human ear	Intensity is a physical quantity
It depends on the sensitivity of the ear	It depends on the amount of sound energy passing per second through unit area
It cannot be measured accurately	It can be measured
It is expressed in decibel (dB)	It is expressed in watt m⁻²

Set – 2

Question 1

In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?

Answer:
Sound travels fastest in iron, then in water, and slowest in air.

Explanation:
The speed of sound depends on the nature of the medium. In solids like iron, the particles are closely packed and have strong intermolecular forces. When sound is produced, vibrations are transferred more quickly from one particle to the next.

In liquids, particles are less closely packed than in solids, so sound travels slower than in solids but faster than in gases.
In gases like air, particles are far apart, so vibrations take more time to pass from one particle to another, making sound travel slowest.

Hence, sound travels fastest in iron, followed by water, and slowest in air.

Page 134

Question 1

An echo is heard in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m s⁻¹?

Answer:

Speed of sound, $v = 342 m s^{- 1}$
Time for echo, $t = 3 s$

Distance travelled by sound:

$Distance = v \times t = 342 \times 3 = 1026 m$

Since the sound travels to the reflecting surface and back, the distance of the reflecting surface is:

$\frac{1026}{2} = 513 m$

Distance of the reflecting surface = 513 m

Page – 135

Question 1

Why are the ceilings of concert halls curved?

Answer:
The ceilings of concert halls are curved so that sound waves, after reflection, spread uniformly in all directions. This helps the sound to reach all corners of the hall clearly, improving audibility for the audience.

Page 136

Question 1

What is the audible range of the average human ear?

Answer:
The audible range of the average human ear is from 20 Hz to 20 kHz.

Question 2

What is the range of frequencies associated with
(a) Infrasound?
(b) Ultrasound?

Answer:
(a) Infrasound: Frequencies below 20 Hz
(b) Ultrasound: Frequencies above 20 kHz

December 15, 2025

Class 9th Science Chapter-10 Exercises
Exercises – Chapter 10: Work and Energy

Question 1

Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.
(i) Suma is swimming in a pond.
(ii) A donkey is carrying a load on its back.
(iii) A wind-mill is lifting water from a well.
(iv) A green plant is carrying out photosynthesis.
(v) An engine is pulling a train.
(vi) Food grains are getting dried in the sun.
(vii) A sailboat is moving due to wind energy.

Answer:

(i) Yes, work is done (force causes displacement).
(ii) No, work is not done (no displacement in the direction of force).
(iii) Yes, work is done (water is lifted against gravity).
(iv) No, work is not done (no mechanical displacement).
(v) Yes, work is done (train is displaced by force).
(vi) No, work is not done (no displacement due to force).
(vii) Yes, work is done (wind causes displacement of boat).

Question 2

An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?

Answer:
The work done by the force of gravity is zero, because the vertical displacement of the object is zero.

Question 3

A battery lights a bulb. Describe the energy changes involved in the process.

Answer:
Chemical energy of the battery is converted into electrical energy, which is further converted into light energy and heat energy in the bulb.

Question 4

A certain force acting on a 20 kg mass changes its velocity from 5 m s⁻¹ to 2 m s⁻¹. Calculate the work done by the force.

Answer:

$W = \frac{1}{2} m (v^{2} - u^{2})$

$W = \frac{1}{2} \times 20 \times (2^{2} - 5^{2})$

$W = 10 \times (4 - 25) = - 210 J$

Work done = –210 J

Question 5

A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.

Answer:
The work done by the gravitational force is zero, because there is no vertical displacement of the object.

Question 6

The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

Answer:
No, this does not violate the law of conservation of energy.
As the object falls, its potential energy decreases and an equal amount of kinetic energy increases. The total energy remains constant.

Question 7

What are the various energy transformations that occur when you are riding a bicycle?

Answer:
Chemical energy of food → muscular energy → mechanical energy → kinetic energy of the bicycle.

Question 8

Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?

Answer:
No work is done on the rock as there is no displacement.
The energy spent is converted into heat energy in the muscles and surroundings.

Question 9

A certain household has consumed 250 units of energy during a month. How much energy is this in joules?

Answer:

1 unit = 1 kWh = $3.6 \times 10^{6} J$

$Energy = 250 \times 3.6 \times 10^{6}$

$= 9.0 \times 10^{8} J$

Question 10

An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.

Answer:

Potential energy at height 5 m:

$E_{p} = m g h = 40 \times 10 \times 5 = 2000 J$

At half-way down (height = 2.5 m):

Potential energy remaining:

$E_{p} = 40 \times 10 \times 2.5 = 1000 J$

Kinetic energy at that point:

$E_{k} = 2000 - 1000 = 1000 J$

Question 11

What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.

Answer:
The work done by the force of gravity is zero.
The force of gravity acts towards the centre of the earth, while the displacement of the satellite is along the circular path. Since the force is perpendicular to the displacement, no work is done.

Question 12

Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.

Answer:
Yes, an object can have displacement without any force acting on it if it is moving with uniform velocity.
In such a case, no net force is required to maintain motion.

Question 13

A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.

Answer:
No work is done on the bundle of hay because there is no displacement of the bundle in the direction of the force.
The person gets tired due to energy spent by the muscles.

Question 14

An electric heater is rated 1500 W. How much energy does it use in 10 hours?

Answer:

$E = P \times t$

$E = 1500 \times (10 \times 3600)$

$E = 5.4 \times 10^{7} J$

Question 15

Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?

Answer:
When the pendulum bob is raised, it has potential energy.
As it swings down, potential energy is converted into kinetic energy.
During oscillation, energy keeps changing between potential and kinetic forms.

The bob eventually comes to rest due to air resistance and friction, which convert mechanical energy into heat energy.
This does not violate the law of conservation of energy because energy is only transformed, not destroyed.

Question 16

An object of mass $m$ is moving with a constant velocity $v$ . How much work should be done on the object in order to bring the object to rest?

Answer:
Initial kinetic energy $= \frac{1}{2} m v^{2}$
Final kinetic energy $= 0$

Work done $=$ change in kinetic energy

$W = 0 - \frac{1}{2} m v^{2} = - \frac{1}{2} m v^{2}$

Question 17

Calculate the work required to be done to stop a car of mass 1500 kg moving at a velocity of 60 km h⁻¹.

Answer:

$u = 60 {km h}^{- 1} = \frac{50}{3} {m s}^{- 1}, v = 0$

$W = \frac{1}{2} m (v^{2} - u^{2})$

$W = \frac{1}{2} \times 1500 (0 - {(\frac{50}{3})}^{2})$

$W = - \frac{1}{2} \times 1500 \times \frac{2500}{9} = - 2.08 \times 10^{5} J (approximately)$

Question 18

In each of the following cases a force $F$ acts on an object of mass $m$ . The direction of displacement is from west to east. State whether the work done by the force is positive, negative or zero.

Answer:
- Work done is positive if the force acts from west to east.
- Work done is negative if the force acts from east to west.
- Work done is zero if the force acts perpendicular to the direction of displacement.
Question 19

Soni says that the acceleration of an object could be zero even when several forces are acting on it. Do you agree with her? Why?

Answer:
Yes, the acceleration can be zero if the net force acting on the object is zero, even though several forces may be acting on it.

Question 20

Find the energy in joules consumed in 10 hours by four devices of power 500 W each.

Answer:

Total power $= 4 \times 500 = 2000 W$

$E = P \times t = 2000 \times (10 \times 3600)$

$E = 7.2 \times 10^{7} J$

Question 21

A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?

Answer:
The kinetic energy of the object is converted into heat energy, sound energy, and energy used in deforming the object and the ground.
December 15, 2025
Class 9th Science Chapter-10 In-Text Questions
Chapter 10: Work and Energy

Page 115 – NCERT Class 9 Science

Questions with Answers

Question 1

A force of 7 N acts on an object. The displacement is 8 m in the direction of the force. Let us take it that the force acts on the object through the displacement. What is the work done in this case?

Answer

$Work done = Force \times Displacement$

$W = 7 \times 8 = 56 J$

Work done = 56 joules

Page 116 – NCERT Class 9 Science

Question 1

When do we say that work is done?

Answer:
Work is said to be done when a force acts on an object and the object gets displaced in the direction of the force.

Question 2

Write an expression for the work done when a force is acting on an object in the direction of its displacement.

Answer:

$W = F \times s$

where $F$ is force and $s$ is displacement.

Question 3

Define 1 J of work.

Answer:
1 joule is the work done when a force of 1 newton displaces an object by 1 metre in the direction of the force.

Question 4

A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?

Answer:

Given Force = 140 N, Distance covered = 15 m

$W = F \times s = 140 \times 15 = 2100 J$

Page 119

Question 1

What is the kinetic energy of an object?

Answer:
The energy possessed by an object due to its motion is called kinetic energy.

Question 2

Write an expression for the kinetic energy of an object.

Answer:
The kinetic energy of an object of mass $m$ moving with velocity $v$ is given by:

$E_{k} = \frac{1}{2} m v^{2}$

Question 3

The kinetic energy of an object of mass $m$ moving with a velocity of $5 m s^{- 1}$ is $25 J$ . What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?

Answer:

Since kinetic energy is proportional to the square of velocity:
- When velocity is doubled:
$E_{k} = 4 \times 25 = 100 J$
- When velocity is tripled:
$E_{k} = 9 \times 25 = 225 J$

Page 123

Question 1

What is power?

Answer:
Power is defined as the rate of doing work or the rate of transfer of energy.

Question 2

Define 1 watt of power.

Answer:
1 watt is the power when 1 joule of work is done in 1 second.

$1 W = 1 J s^{- 1}$

Question 3

A lamp consumes 1000 J of electrical energy in 10 s. What is its power?

Answer:

$P = \frac{W}{t} = \frac{1000}{10} = 100 W$

Question 4

Define average power.

Answer:
Average power is defined as the total work done divided by the total time taken.

$Average power = \frac{Total work done}{Total time taken}$
December 15, 2025
Class 9th Science Chapter-9 Exercises
NCERT Class 9 Science – Gravitation

Exercises | Complete Answers

Question 1. How does the force of gravitation between two objects change when the distance between them is reduced to half?

Answer:

Gravitational force is inversely proportional to the square of the distance between two objects.

$F \propto \frac{1}{d^{2}}$

If distance is reduced to half:

$d \to \frac{d}{2}$ $F \propto \frac{1}{(d / 2)^{2}} = \frac{4}{d^{2}}$

The gravitational force becomes 4 times greater.

Question 2. Gravitational force acts on all objects in proportion to their masses. Why then does a heavy object not fall faster than a light object?

Answer:

According to Newton’s second law:

$a = \frac{F}{m}$

Though a heavy object experiences a greater gravitational force, its mass is also greater.
These two effects cancel each other.

As a result, all objects fall with the same acceleration (g), irrespective of their mass (ignoring air resistance).

Hence, a heavy object does not fall faster than a light object.

Question 3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface?

Given:
- Mass of earth, $M = 6 \times 10^{24} k g$
- Mass of object, $m = 1 k g$
- Radius of earth, $R = 6.4 \times 10^{6} m$
- Gravitational constant,
  $G = 6.7 \times 10^{- 11} N m^{2} k g^{- 2}$
Formula:

$F = G \frac{M m}{R^{2}}$

Calculation:

$F = \frac{6.7 \times 10^{- 11} \times 6 \times 10^{24} \times 1}{(6.4 \times 10^{6})^{2}}$ $F = 9.8 N$

Answer: The gravitational force is 9.8 N.

Question 4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater, smaller, or the same as the force with which the moon attracts the earth? Why?

Answer:

The earth attracts the moon with the same force as the moon attracts the earth.

Reason:
According to Newton’s third law of motion, every action has an equal and opposite reaction.

Hence, the forces are equal in magnitude but opposite in direction.

Question 5. If the moon attracts the earth, why does the earth not move towards the moon?

Answer:

Although the moon attracts the earth, the mass of the earth is very large compared to the moon.

From Newton’s second law:

$a = \frac{F}{m}$

Due to its huge mass, the acceleration of the earth is extremely small, so its motion is not noticeable.

Therefore, the earth does not move noticeably towards the moon.

Question 6. What happens to the force between two objects if:

(i) the mass of one object is doubled?

$F \propto m$

Force becomes double.

(ii) the distance between the objects is doubled and tripled?

$F \propto \frac{1}{d^{2}}$
- Distance doubled →
  
  $F = \frac{1}{4} times$
- Distance tripled →
  
  $F = \frac{1}{9} times$
(iii) the masses of both objects are doubled?

$F \propto M \times m$ $(2 M) \times (2 m) = 4 M m$

Force becomes 4 times.

Question 7. What is the importance of universal law of gravitation?

Answer:
The universal law of gravitation explains many natural phenomena, such as:
- The force that binds objects to the earth
- The motion of the moon around the earth
- The motion of planets around the Sun
- The occurrence of tides due to the moon and the Sun
Thus, it shows that the same gravitational force acts everywhere in the universe.

Question 8. What is the acceleration of free fall?

Answer:
The acceleration of an object when it falls towards the earth under the influence of gravity alone is called acceleration of free fall.
It is denoted by g and its value near the earth’s surface is:

$g = 9.8 m s^{- 2}$

Question 9. What do we call the gravitational force between the earth and an object?

Answer:
The gravitational force between the earth and an object is called the weight of the object.

$Weight = m \times g$

Question 10. Amit buys a few grams of gold at the poles and gives it to his friend at the equator. Will the friend agree with the weight? If not, why?

Answer:
No, the friend will not agree with the weight.

Reason:
- The value of g is greater at the poles than at the equator.
- Since weight = m × g, the gold will weigh less at the equator.
- Although the mass remains the same, the weight decreases at the equator.
Question 11. Why will a sheet of paper fall slower than one that is crumpled into a ball?

Answer:
A sheet of paper has a larger surface area, so it experiences more air resistance.
A crumpled paper has a smaller surface area, so air resistance is less.

Therefore, the sheet of paper falls slower than the crumpled ball.

Question 12. Gravitational force on the moon is $\frac{1}{6}$ of that on the earth. Find the weight of a 10 kg object on the earth and on the moon.

Given:
- Mass, $m = 10 k g$
- $g_{e a r t h} = 9.8 m s^{- 2}$
Weight on Earth

$W_{e} = m \times g = 10 \times 9.8 = 98 N$

Weight on Moon

$W_{m} = \frac{1}{6} \times W_{e} = \frac{1}{6} \times 98$ $W_{m} = 16.3 N (approx.)$

Answer:
- Weight on Earth = 98 N
- Weight on Moon = ≈ 16.3 N
Question 13. A ball is thrown vertically upwards with a velocity of 49 m/s.

Find
(i) maximum height,
(ii) total time to return to the earth.

Given:
Initial velocity $u = 49 m s^{- 1}$
Acceleration $a = - 9.8 m s^{- 2}$ (upward motion)
Final velocity at top $v = 0$

(i) Maximum height

$v^{2} = u^{2} + 2 a s$ $0 = (49)^{2} + 2 (- 9.8) s$ $s = \frac{49^{2}}{2 \times 9.8} = \frac{2401}{19.6} = 122.5 m$

Answer: Maximum height = 122.5 m

(ii) Total time of flight

Time to reach top:

$v = u + a t \Rightarrow 0 = 49 - 9.8 t \Rightarrow t = 5 s$

Total time $= 2 \times 5 = 10 s$

Answer: Total time = 10 s

Question 14. A stone is released from the top of a tower of height 19.6 m.

Find the final velocity just before touching the ground.

Given:
$u = 0$ , $s = 19.6 m$ , $a = 9.8 m s^{- 2}$

$v^{2} = u^{2} + 2 a s = 0 + 2 \times 9.8 \times 19.6$ $v^{2} = 384.16 \Rightarrow v = 19.6 m s^{- 1}$

Answer: Final velocity = 19.6 m s⁻¹ (downwards)

Question 15. A stone is thrown vertically upward with $u = 40 m / s$ .

Take $g = 10 m / s^{2}$ . Find
(i) maximum height,
(ii) net displacement,
(iii) total distance covered.

Given: $u = 40 m / s$ , $a = - 10 m / s^{2}$ , $v = 0$ at top.

(i) Maximum height

$0 = 40^{2} + 2 (- 10) s \Rightarrow s = \frac{1600}{20} = 80 m$

Maximum height = 80 m

(ii) Net displacement

The stone returns to the starting point.

Net displacement = 0 m

(iii) Total distance covered

Upward distance $= 80 m$
Downward distance $= 80 m$

Total distance = 160 m

Question 16. Force of gravitation between the earth and the Sun

Given:
$M = 6 \times 10^{24} k g$
$m = 2 \times 10^{30} k g$
$d = 1.5 \times 10^{11} m$
$G = 6.7 \times 10^{- 11} N m^{2} k g^{- 2}$

Formula:

$F = G \frac{M m}{d^{2}}$

Calculation:

$F = \frac{6.7 \times 10^{- 11} \times 6 \times 10^{24} \times 2 \times 10^{30}}{(1.5 \times 10^{11})^{2}}$

$F = \frac{8.04 \times 10^{44}}{2.25 \times 10^{22}} = 3.57 \times 10^{22} N$

Answer:

$F = 3.57 \times 10^{22} N$

Question 17. Two stones are released simultaneously
- Stone A: dropped from top of a 100 m tower
- Stone B: projected upwards from ground with 25 m s⁻¹
  Find when and where they meet.
Take: $g = 10 m s^{- 2}$

For stone A (downward):

$s_{1} = \frac{1}{2} g t^{2} = 5 t^{2}$

For stone B (upward):

$s_{2} = u t - \frac{1}{2} g t^{2} = 25 t - 5 t^{2}$

They meet when:

$s_{1} + s_{2} = 100$

$5 t^{2} + (25 t - 5 t^{2}) = 100$

$25 t = 100 \Rightarrow t = 4 s$

Position:

$s_{1} = 5 (4)^{2} = 80 m$

Answer:
- Time = 4 s
- Meeting point = 80 m below the top (or 20 m above ground)
Question 18. A ball returns to the thrower after 6 s

Find
(a) initial velocity
(b) maximum height
(c) position after 4 s

Given: Total time = 6 s → time to reach top = 3 s
$g = 9.8 m s^{- 2}$

(a) Initial velocity

$u = g t = 9.8 \times 3 = 29.4 m s^{- 1}$

(b) Maximum height

$h = u t - \frac{1}{2} g t^{2}$

$h = (29.4 \times 3) - \frac{1}{2} (9.8) (3^{2})$

$h = 88.2 - 44.1 = 44.1 m$

(c) Position after 4 s

$s = u t - \frac{1}{2} g t^{2}$

$s = (29.4 \times 4) - \frac{1}{2} (9.8) (16)$

$s = 117.6 - 78.4 = 39.2 m$

Answers:
- (a) 29.4 m s⁻¹
- (b) 44.1 m
- (c) 39.2 m above the point of projection
19. In what direction does the buoyant force act?

Answer:
The buoyant force acts vertically upward, opposite to the direction of gravity.

20. Why does a block of plastic released under water come up to the surface?

Answer:
The buoyant force acting upward is greater than the weight of the plastic block.
Hence, the net force is upward and the block rises to the surface.

Question 21

The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g cm⁻³, will the substance float or sink?

Answer:

First, calculate the density of the substance.

$Density = \frac{Mass}{Volume}$

$= \frac{50}{20} = 2.5 g c m^{- 3}$

The density of the substance (2.5 g cm⁻³) is greater than the density of water (1 g cm⁻³).

Therefore, the substance will sink in water.

Question 22

The volume of a 500 g sealed packet is 350 cm³. Will the packet float or sink in water if the density of water is 1 g cm⁻³? What will be the mass of the water displaced by this packet?

Answer:

Step 1: Calculate the density of the packet

$Density = \frac{500}{350} = 1.43 g c m^{- 3}$

Since the density of the packet (1.43 g cm⁻³) is greater than the density of water (1 g cm⁻³):

The packet will sink in water.
December 15, 2025

Class 9th Science Chapter-9 In-Text Questions

Chapter-9 Gravitation

NCERT Class 9 Science

Question 1

State the universal law of gravitation.

Answer:

The universal law of gravitation states that:

Every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. The force acts along the line joining the centres of the two objects.

Mathematically,

$F = G \frac{M m}{d^{2}}$

where

$F$ = gravitational force
$M, m$ = masses of the two objects
$d$ = distance between their centres
$G$ = universal gravitational constant

Question 2

Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.

Answer:

The formula is:

$F = G \frac{M m}{R^{2}}$

where

$F$ = gravitational force between earth and object
$G$ = universal gravitational constant
$M$ = mass of the earth
$m$ = mass of the object
$R$ = radius of the earth

Page 104

Questions

Question 1

What do you mean by free fall?

Answer:

Free fall is the motion of an object when it falls towards the earth under the influence of gravitational force alone, without any other force acting on it.

Question 2

What do you mean by acceleration due to gravity?

Answer:

The acceleration produced in an object due to the gravitational force of the earth is called acceleration due to gravity.
It is denoted by g and its value near the earth’s surface is 9.8 m s⁻².

Page 106 –

Question 1

What are the differences between the mass of an object and its weight?

Answer:

Mass	Weight
Mass is the amount of matter in an object	Weight is the force of gravity acting on the object
It is constant everywhere	It changes from place to place
SI unit is kilogram (kg)	SI unit is newton (N)
It has only magnitude	It has both magnitude and direction
Measured by beam balance	Measured by spring balance

Question 2

Why is the weight of an object on the moon $\frac{1}{6}$ th its weight on the earth?

Answer:

The mass of the moon is much smaller than the mass of the earth.
Due to this, the gravitational force of the moon is weaker, and the value of acceleration due to gravity on the moon is about one-sixth of that on the earth.
Therefore, the weight of an object on the moon becomes one-sixth of its weight on the earth.

Questions (Page 109)

Question 1

Why is it difficult to hold a school bag having a strap made of a thin and strong string?

Answer:

A thin strap has a small area of contact with the shoulder.
Since pressure = thrust / area, a smaller area produces more pressure.
Therefore, the bag hurts the shoulder and becomes difficult to hold.

Question 2

What do you mean by buoyancy?

Answer:

Buoyancy is the upward force exerted by a fluid (liquid or gas) on an object when it is immersed in it.
This upward force is also called buoyant force.

Question 3

Why does an object float or sink when placed on the surface of water?

Answer:

An object floats or sinks depending on its density compared to water.

If the density of the object is less than the density of water, the buoyant force is greater than its weight → object floats
If the density of the object is greater than the density of water, the buoyant force is less than its weight → object sinks

Questions (Page 110)

Question 1

You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?

Answer:

Your mass is more than 42 kg.

Reason:
A weighing machine actually measures weight, not mass.
Due to the buoyant force of air, the apparent weight is slightly less than the true weight.
Hence, the actual mass is slightly more than 42 kg.

Question 2

You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than the other. Can you say which one is heavier and why?

Answer:

The bag of cotton is actually heavier.

Reason:

The cotton bag has a larger volume, so it displaces more air.
Greater air displacement causes a greater buoyant force, reducing the apparent weight more.
Therefore, although both show the same mass on the weighing machine, the true weight of the cotton bag is more than that of the iron bar.

December 15, 2025

Class 9th Science Chapter-8 Exercises
Exercises – Force and Laws of Motion (Class 9)

Question 1

An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions.

Answer:

Yes, it is possible.

Condition:

If the object is already moving and all external forces are balanced (net force = 0), then the object will continue to move with a constant (non-zero) velocity in a straight line.

Example:

A car moving on a straight road at constant speed.

Question 2

When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer:

When the carpet is beaten, the carpet moves suddenly, but the dust particles tend to remain at rest due to inertia of rest.
As a result, the dust gets detached and falls off.

Question 3

Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer:

When the bus starts, stops, or turns suddenly, the luggage tends to maintain its state of motion due to inertia and may fall off.
Tying the luggage prevents it from falling by providing the necessary force.

Question 4

A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because:

(a) the batsman did not hit the ball hard enough.

(b) velocity is proportional to the force exerted on the ball.

(c) there is a force on the ball opposing the motion.

(d) there is no unbalanced force on the ball, so the ball

would want to come to rest.

Correct Answer:

(c) there is a force on the ball opposing the motion.

Explanation:

Friction between the ball and the ground opposes motion and gradually reduces the speed to zero.

Question 5

A truck starts from rest and rolls down a hill with a constant acceleration. It travels 400 m in 20 s. Find its acceleration and the force acting on it if its mass is 7 tonnes.

Given:
- Initial velocity, $u = 0$
- Distance, $s = 400 m$
- Time, $t = 20 s$
- Mass, $m = 7 tonnes = 7000 kg$
Step 1: Find acceleration

Using:

$s = u t + \frac{1}{2} a t^{2}$ $400 = 0 + \frac{1}{2} a (20)^{2}$ $400 = 200 a$ $a = 2 {m s}^{- 2}$

Step 2: Find force

Using:

$F = m a$ $F = 7000 \times 2 = 14000 N$

Answer:
- Acceleration = 2 m s⁻²
- Force acting = 14,000 N
Question 6

A stone of 1 kg is thrown with a velocity of 20 m s⁻¹ on ice and comes to rest after 50 m. Find the force of friction.

Given:
- Mass, $m = 1 kg$
- Initial velocity, $u = 20 {m s}^{- 1}$
- Final velocity, $v = 0$
- Distance, $s = 50 m$
Step 1: Find acceleration

Using:

$v^{2} = u^{2} + 2 a s$ $0 = 400 + 100 a$ $a = - 4 {m s}^{- 2}$

Step 2: Find friction force

$F = m a = 1 \times (- 4) = - 4 N$

Answer:

Force of friction = 4 N (opposite to motion)

Question 7

An 8000 kg engine pulls 5 wagons of 2000 kg each. Engine force = 40000 N, friction = 5000 N. Find:
(a) net accelerating force
(b) acceleration

Step 1: Total mass

$m = 8000 + (5 \times 2000) = 18000 kg$

Step 2: Net force

$F_{net} = 40000 - 5000 = 35000 N$

Step 3: Acceleration

$a = \frac{F}{m} = \frac{35000}{18000}$

$a \approx 1.94 {m s}^{- 2}$

Answer:
- Net accelerating force = 35,000 N
- Acceleration = 1.94 m s⁻²
Question 8

An automobile of mass 1500 kg is stopped with a negative acceleration of 1.7 m s⁻². Find the force between the vehicle and the road.

Given:
- Mass, $m = 1500 kg$
- Acceleration, $a = - 1.7 {m s}^{- 2}$
Calculation:

$F = m a = 1500 \times (- 1.7) = - 2550 N$

Answer:

Force = 2550 N (opposite to direction of motion)

Question 9

What is the momentum of an object of mass m, moving with velocity v?

Correct answer:
(d) mv

Explanation:
Momentum $p = m \times v$

Question 10

Using a horizontal force of 200 N, a wooden cabinet is moved across the floor at constant velocity. What is the friction force acting on the cabinet?

Answer:

When an object moves with constant velocity, net force = 0.

$Applied force = Friction force$

$Friction force = 200 N$

Friction force = 200 N (opposite to motion)

Question 11

A student says that when we push a parked truck, the equal and opposite forces cancel each other, so the truck does not move. Comment on this logic.

Answer:

The student’s logic is incorrect.
- According to Newton’s third law, action and reaction forces act on different objects, not on the same object.
- The force applied by the person acts on the truck, while the reaction force acts on the person.
- The truck does not move because its large mass (high inertia) and friction with the road require a much larger force to produce noticeable acceleration.
Question 12

A hockey ball of mass 200 g moving at 10 m s⁻¹ returns with velocity 5 m s⁻¹. Find the magnitude of change in momentum.

Given:
- Mass, $m = 200 g = 0.2 kg$
- Initial velocity, $u = + 10 {m s}^{- 1}$
- Final velocity, $v = - 5 {m s}^{- 1}$ (opposite direction)
Calculation:

Initial momentum:

$p_{1} = m u = 0.2 \times 10 = 2 {kg m s}^{- 1}$

Final momentum:

$p_{2} = m v = 0.2 \times (- 5) = - 1 {kg m s}^{- 1}$

Change in momentum:

$Δ p = p_{2} - p_{1} = - 1 - 2 = - 3$

Magnitude of change in momentum = 3 kg m s⁻¹

Question 13

A bullet of mass 10 g moving at 150 m s⁻¹ stops in 0.03 s. Find the distance penetrated and the force exerted.

Given:
- Mass, $m = 10 g = 0.01 kg$
- Initial velocity, $u = 150 {m s}^{- 1}$
- Final velocity, $v = 0$
- Time, $t = 0.03 s$
Step 1: Acceleration

$a = \frac{v - u}{t} = \frac{0 - 150}{0.03} = - 5000 {m s}^{- 2}$

Step 2: Distance penetrated

$s = u t + \frac{1}{2} a t^{2}$

$s = 150 (0.03) + \frac{1}{2} (- 5000) (0.03)^{2}$ $s = 4.5 - 2.25 = 2.25 m$

Step 3: Force exerted

$F = m a = 0.01 \times (- 5000) = - 50 N$

Answers:
- Distance of penetration = 2.25 m
- Force exerted = 50 N (opposite to motion)
Question 14

A 1 kg object moving at 10 m s⁻¹ sticks to a stationary 5 kg block. Find momentum before and after collision and final velocity.

Given:
- $m_{1} = 1 kg, u_{1} = 10 {m s}^{- 1}$
- $m_{2} = 5 kg, u_{2} = 0$
Momentum before collision

$p = (1 \times 10) + (5 \times 0) = 10 {kg m s}^{- 1}$

Momentum after collision

Momentum is conserved:

$p = 10 {kg m s}^{- 1}$

Final velocity

$v = \frac{p}{m_{1} + m_{2}} = \frac{10}{6} = 1.67 {m s}^{- 1}$

Answers:
- Momentum before impact = 10 kg m s⁻¹
- Momentum after impact = 10 kg m s⁻¹
- Final velocity = 1.67 m s⁻¹
Question 15

A 100 kg object accelerates from 5 m s⁻¹ to 8 m s⁻¹ in 6 s. Find initial momentum, final momentum and force.

Given:
- $m = 100 kg$
- $u = 5 {m s}^{- 1}$
- $v = 8 {m s}^{- 1}$
- $t = 6 s$
Initial momentum

$p_{1} = m u = 100 \times 5 = 500 {kg m s}^{- 1}$

Final momentum

$p_{2} = m v = 100 \times 8 = 800 {kg m s}^{- 1}$

Force

$a = \frac{v - u}{t} = \frac{3}{6} = 0.5 {m s}^{- 2}$ $F = m a = 100 \times 0.5 = 50 N$

Answers:
- Initial momentum = 500 kg m s⁻¹
- Final momentum = 800 kg m s⁻¹
- Force = 50 N
Question 16

An insect hits a moving car’s windscreen. Comment on the views of Akhtar, Kiran and Rahul.

Correct Explanation:

Rahul is correct.

Reason:
- According to Newton’s third law, the insect and the car exert equal and opposite forces on each other.
- The change in momentum of both is equal in magnitude, but:
  - The insect has a very small mass, so it undergoes large acceleration and gets crushed.
  - The car has a very large mass, so its acceleration is negligible.
❌ Why others are wrong:
- Kiran: Change in momentum is equal, not greater for insect.
- Akhtar: Force on both bodies is equal, not larger on insect.
Question 17

How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm?
(Take downward acceleration = 10 m s⁻²)

Given:
- Mass of dumb-bell,
  
  $m = 10 kg$
- Height,
  
  $h = 80 cm = 0.8 m$
- Acceleration due to gravity,
  
  $g = 10 {m s}^{- 2}$
Step 1: Find velocity just before hitting the floor

Using the equation of motion:

$v^{2} = u^{2} + 2 g h$

Since the dumb-bell is dropped from rest:

$u = 0$ $v^{2} = 2 \times 10 \times 0.8$ $v^{2} = 16$ $v = 4 {m s}^{- 1}$

Step 2: Calculate momentum

Momentum transferred to the floor equals the momentum of the dumb-bell just before impact.

$p = m v$ $p = 10 \times 4$ $p = 40 {kg m s}^{- 1}$

Final Answer

Momentum transferred to the floor = 40 kg m s⁻¹
December 15, 2025

Class 9th Science Chapter-8 In-Text Questions

Chapter – 8 Force and the Laws of Motion

Page No. 91 – Questions & Answers

Question 1

Which of the following has more inertia?
(a) a rubber ball and a stone of the same size
(b) a bicycle and a train
(c) a five-rupees coin and a one-rupee coin

Answer:

(a) Stone
→ Because the stone has greater mass than the rubber ball.

(b) Train
→ A train has much larger mass than a bicycle.

Key idea:
Greater the mass, greater is the inertia.

Question 2

In the following example, try to identify the number of times the velocity of the ball changes and the agent supplying the force in each case:

“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team.”

Answer:

Velocity of the ball changes 4 times.

Event	Change in velocity	Agent supplying force
First player kicks the ball	At kick	First player’s foot
Second player kicks the ball	At kick	Second player’s foot
Goalkeeper stops the ball	While stopping	Goalkeeper’s hands
Goalkeeper kicks the ball back	At kick	Goalkeeper’s foot

Question 3

Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

Answer:
When the branch is shaken suddenly, the branch moves, but the leaves tend to remain at rest due to inertia. This causes the leaves to get detached from the branch and fall down.

Question 4

Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?

Answer:

When the bus stops suddenly:
Your feet stop with the bus, but the upper part of your body keeps moving forward due to inertia of motion, so you fall forward.
When the bus starts suddenly:
Your feet move with the bus, but the upper part of your body tends to remain at rest due to inertia of rest, so you fall backward.

EXTRA NOTES:

Newton’s Laws of Motion

Sir Isaac Newton gave three fundamental laws to explain the motion of objects. These are called Newton’s Laws of Motion.

1. First Law of Motion (Law of Inertia)

Statement

An object remains at rest or continues to move with uniform velocity in a straight line unless acted upon by an unbalanced external force.

Explanation (Easy Language)

A body at rest will remain at rest unless a force acts on it.
A moving body will keep moving at the same speed and in the same direction unless a force changes its motion.
Only an unbalanced force can change the state of rest or motion.

Examples

A book lying on a table does not move unless pushed.
Passengers fall forward when a moving bus stops suddenly.
Passengers fall backward when a bus starts suddenly.

Why it is called the Law of Inertia

This law explains the property called inertia.

Law of Inertia

Definition

Inertia is the natural tendency of an object to resist any change in its state of rest or motion.

Types of Inertia

Inertia of Rest – A body at rest resists motion
Example: Dust falls off a carpet when beaten.
Inertia of Motion – A moving body resists stopping
Example: A person falls forward when a bus stops suddenly.
Inertia of Direction – A body resists change in direction
Example: A person leans sideways when a car takes a sharp turn.

Relation between Mass and Inertia

Greater the mass → Greater the inertia
A train has more inertia than a bicycle.

2. Second Law of Motion

Statement

The rate of change of momentum of an object is directly proportional to the applied unbalanced force and takes place in the direction of the force.

Explanation

Force causes a change in momentum.
A larger force produces a greater acceleration.
The change occurs in the direction of the applied force.

Mathematical Form

$Force = mass \times acceleration$

$F = m a$

Examples

A cricket ball hurts more than a tennis ball because it has more mass.
A footballer pulls his hands backward while catching a fast ball to reduce force.
A loaded truck needs more force to move than an empty one.

SI Unit of Force

Newton (N)
1 N = force needed to give 1 kg mass an acceleration of 1 m s⁻²

3. Third Law of Motion

Statement

For every action, there is an equal and opposite reaction.

Explanation

Forces always act in pairs.
Action and reaction forces:
- Are equal in magnitude
- Act in opposite directions
- Act on different objects

Examples

While walking, we push the ground backward; the ground pushes us forward.
Recoil of a gun when a bullet is fired.
Swimming: pushing water backward moves the swimmer forward.

Important Note

Action and reaction do not cancel each other because they act on different bodies.

December 15, 2025

Class 9th Science Chapter-7 Exercises
Exercise – Motion (Answers)

Question 1

An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Given:
- Diameter of track = 200 m
- Radius, $r = 100$ m
- Time for 1 round = 40 s
- Total time = 2 min 20 s = 140 s
Step 1: Number of rounds

$Number of rounds = \frac{140}{40} = 3.5$

Step 2: Distance covered

Circumference of track:

$2 π r = 2 \times π \times 100 = 200 π m$

Distance in 3.5 rounds:

$= 3.5 \times 200 π = 700 π m$ $= 2200 m (approximately)$

Step 3: Displacement

After 3.5 rounds, the athlete reaches the diametrically opposite point.

$Displacement = Diameter = 200 m$

Answer:
- Distance covered = 2200 m
- Displacement = 200 m
Question 2

Joseph jogs from A to B (300 m) in 2 min 30 s and then jogs back 100 m to point C in 1 min. Find average speed and average velocity for:
(a) A to B
(b) A to C

(a) From A to B

Given:
- Distance = 300 m
- Time = 2 min 30 s = 150 s
Average speed

$= \frac{300}{150} = 2 {m s}^{- 1}$

Average velocity

$= \frac{Displacement}{Time} = \frac{300}{150} = 2 {m s}^{- 1}$

Answer (A to B):
- Average speed = 2 m s⁻¹
- Average velocity = 2 m s⁻¹
(b) From A to C

Total distance travelled

$= 300 + 100 = 400 m$

Total time

$= 150 + 60 = 210 s$

Displacement (A to C)

$= 300 - 100 = 200 m$

Average speed

$= \frac{400}{210} = 1.90 {m s}^{- 1}$

Average velocity

$= \frac{200}{210} = 0.95 {m s}^{- 1}$

Answer (A to C):
- Average speed = 1.90 m s⁻¹
- Average velocity = 0.95 m s⁻¹
Question 3

Abdul’s average speed to school is 20 km h⁻¹ and on return it is 30 km h⁻¹. Find the average speed for the whole trip.

Formula (important):

$Average speed = \frac{2 v_{1} v_{2}}{v_{1} + v_{2}}$

Calculation:

$= \frac{2 \times 20 \times 30}{20 + 30} = \frac{1200}{50} = 24 {km h}^{- 1}$

Answer:
Average speed for the whole trip = 24 km h⁻¹

Question 4

A motorboat starts from rest and accelerates uniformly at 3.0 m s⁻² for 8.0 s. Find the distance travelled.

Given:

$u = 0, a = 3.0 {m s}^{- 2}, t = 8 s$

Formula:

$s = u t + \frac{1}{2} a t^{2}$

Calculation:

$s = 0 + \frac{1}{2} \times 3 \times 8^{2} = 1.5 \times 64 = 96 m$

Answer:
Distance travelled = 96 m

Question 5

A driver of a car travelling at 52 km h⁻¹ applies the brakes.
(a) Shade the area on the speed–time graph that represents the distance travelled.
(b) Which part of the graph represents uniform motion?

(a) Answer:

The area under the speed–time graph represents the distance travelled by the car.
Shade the region between the graph line and the time axis.

(b) Answer:

The horizontal straight-line portion of the speed–time graph represents uniform motion, because speed remains constant there.

Exercise – Question 6

Fig. 7.10 shows the distance–time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest?

Answer:
Object B is travelling the fastest.

Explanation:
In a distance–time graph, the object with the steepest slope (greatest inclination) has the highest speed.
Object B has the steepest line.

(b) Are all three ever at the same point on the road?

Answer:
Yes, all three objects are at the same point at the same time.

Explanation:
All three distance–time graphs intersect at one point, which indicates that their distances from the origin are the same at that instant.

How far has C travelled when B passes A?

Answer:
When B passes A, both B and A are at the same distance from the origin.
From the graph, at this point, object C has travelled 6 km.

Distance travelled by C = 6 km

(d) How far has B travelled by the time it passes C?

Answer:
When B passes C, their graphs intersect again.
From the graph, the distance corresponding to this intersection point is 8 km.

Distance travelled by B = 8 km

Question 7

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s⁻², with what velocity will it strike the ground? After what time will it strike the ground?

Given:
- Initial velocity,
  
  $u = 0 {m s}^{- 1}$
  
  (Ball is gently dropped)
- Distance fallen,
  
  $s = 20 m$
- Acceleration due to gravity,
  
  $a = 10 {m s}^{- 2}$
(a) Velocity with which the ball strikes the ground

Using the equation of motion:

$v^{2} = u^{2} + 2 a s$

Substituting the values:

$v^{2} = 0 + 2 \times 10 \times 20$ $v^{2} = 400$ $v = 20 {m s}^{- 1}$

Velocity at the ground = 20 m s⁻¹ (downward)

(b) Time taken to strike the ground

Using the equation:

$v = u + a t$

Substituting the values:

$20 = 0 + 10 t$ $t = 2 s$

Time taken = 2 s

Question 8

The speed–time graph for a car is shown in Fig. 7.11.

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

(b) Which part of the graph represents uniform motion of the car?

(a) Distance travelled in the first 4 seconds

Concept Used

Distance travelled = Area under the speed–time graph

In the first 4 seconds, the graph is a straight sloping line starting from the origin, forming a triangle.

From the graph (Fig. 7.11):
- Time = 4 s
- Maximum speed at 4 s = 8 m s⁻¹
Area of triangle

$Distance = \frac{1}{2} \times base \times height$ $= \frac{1}{2} \times 4 \times 8$ $= 16 m$

Distance travelled in first 4 seconds = 16 m

Shading instruction:
Shade the triangular area under the graph from 0 to 4 s.

(b) Which part of the graph represents uniform motion?

Answer:
The horizontal straight-line portion of the speed–time graph represents uniform motion.

Explanation:
- In this part, speed remains constant
- Constant speed ⇒ zero acceleration
- Hence, motion is uniform
Question 9

State which of the following situations are possible and give an example for each:

(a) An object with a constant acceleration but with zero velocity

Answer:
Yes, this situation is possible.

Example:
A ball thrown vertically upward has zero velocity at the highest point, but it still has a constant acceleration due to gravity (10 m s⁻² downward).

(b) An object moving with an acceleration but with uniform speed

Answer:
Yes, this situation is possible.

Example:
An object moving in a circular path at constant speed, such as:
- A car moving on a circular track
- The moon revolving around the Earth
Explanation:
Although speed is constant, the direction of motion keeps changing, so the velocity changes, which means the object has acceleration.

(c) An object moving in a certain direction with an acceleration in the perpendicular direction

Answer:
Yes, this situation is possible.

Example:
An object in uniform circular motion, such as:
- A stone tied to a string and whirled in a circle
Explanation:
The acceleration (centripetal acceleration) is directed towards the centre of the circle, which is perpendicular to the direction of motion at every point.

Question 10

An artificial satellite is moving in a circular orbit of radius 42,250 km. Calculate its speed if it takes 24 hours to revolve around the Earth.

Given:

Radius of orbit,

$r = 42,250 km$

Time period,

$T = 24 h$

Formula Used (Uniform Circular Motion):

$Speed = \frac{Circumference of orbit}{Time period} = \frac{2 π r}{T}$

Calculation:

Circumference of orbit:

$2 π r = 2 \times \frac{22}{7} \times 42,250 = 265,714 km (approx.)$

Speed:

$v = \frac{265,714}{24} = 11,071 {km h}^{- 1}$

(Optional – in m s⁻¹)

$11,071 \times \frac{1000}{3600} = 3075 {m s}^{- 1} (approx.)$
December 15, 2025

Class 9th Science Chapter-7 In-Text Questions

Chapter – 7 Motion

Page No. 74 – Questions & Answers

Question 1

An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Answer:
Yes, an object can have zero displacement even after moving through a distance.

Explanation:
Displacement depends on the initial and final positions of an object. If the final position is the same as the initial position, displacement becomes zero.

Example:
If a person walks 10 m forward and then 10 m backward, the distance travelled is 20 m, but since the person returns to the starting point, the displacement is zero.

Question 2

A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Answer:

Side of square field = 10 m
Perimeter of square = 4 × 10 = 40 m
Time to complete 1 round = 40 s

Total time given = 2 min 20 s = 140 s

Number of rounds completed =
140 ÷ 40 = 3.5 rounds

After 3 full rounds, the farmer comes back to the starting point.
After half round, he reaches the opposite corner of the square.

Magnitude of displacement = diagonal of square

$Displacement = \sqrt{10^{2} + 10^{2}} = \sqrt{200} = 10 \sqrt{2} m$

Final Answer:
Magnitude of displacement = $10 \sqrt{2}$ m

Question 3

Which of the following is true for displacement?

(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.

Answer:

Neither (a) nor (b) is true.

Explanation:

Displacement can be zero when the initial and final positions are the same.
The magnitude of displacement is always less than or equal to the distance travelled, never greater.

Page No. 76 – Questions & Answers

Question 1

Distinguish between speed and velocity.

Speed	Velocity
Speed is the distance travelled per unit time.	Velocity is the displacement per unit time.
It is a scalar quantity (has magnitude only).	It is a vector quantity (has magnitude and direction).
Direction is not required.	Direction is compulsory.
Speed is always positive or zero.	Velocity can be positive, negative or zero.

Question 2

Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Answer:
The magnitude of average velocity is equal to average speed when the object moves in a straight line without changing direction.

Explanation:
In this case, distance = displacement, so both average speed and average velocity become equal.

Question 3

What does the odometer of an automobile measure?

Answer:
An odometer measures the total distance travelled by an automobile.

Question 4

What does the path of an object look like when it is in uniform motion?

Answer:
When an object is in uniform motion, it moves along a straight-line path.

Question 5

During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station?
(The signal travels at the speed of light = $3 \times 10^{8}$ m s⁻¹)

Given:
Speed of signal,

$v = 3 \times 10^{8} {m s}^{- 1}$

Time taken,

$t = 5 minutes = 5 \times 60 = 300 s$

Distance = Speed × Time

$s = v \times t$ $s = 3 \times 10^{8} \times 300$ $s = 9 \times 10^{10} m$

Final Answer:
Distance of the spaceship = $9 \times 10^{10}$ m

Page No. 77 – Questions & Answers

Question 1

When will you say a body is in
(i) uniform acceleration?
(ii) non-uniform acceleration?

Answer:

(i) Uniform acceleration:
A body is said to be in uniform acceleration when its velocity changes by equal amounts in equal intervals of time, no matter how small the time intervals are.

Example:
A freely falling body under gravity.

(ii) Non-uniform acceleration:
A body is said to be in non-uniform acceleration when its velocity changes by unequal amounts in equal intervals of time.

Example:
A car moving in heavy traffic where speed changes irregularly.

Question 2

A bus decreases its speed from 80 km h⁻¹ to 60 km h⁻¹ in 5 s. Find the acceleration of the bus.

Given:
Initial velocity,

$u = 80 {km h}^{- 1} = \frac{80 \times 1000}{3600} = 22.22 {m s}^{- 1}$

Final velocity,

$v = 60 {km h}^{- 1} = \frac{60 \times 1000}{3600} = 16.67 {m s}^{- 1}$

Time,

$t = 5 s$

Formula:

$a = \frac{v - u}{t}$

Calculation:

$a = \frac{16.67 - 22.22}{5} = \frac{- 5.55}{5} = - 1.11 {m s}^{- 2}$

Answer:
Acceleration of the bus = –1.11 m s⁻²
(Negative sign indicates deceleration)

Question 3

A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km h⁻¹ in 10 minutes. Find its acceleration.

Given:
Initial velocity,

$u = 0$

Final velocity,

$v = 40 {km h}^{- 1} = \frac{40 \times 1000}{3600} = 11.11 {m s}^{- 1}$

Time,

$t = 10 minutes = 600 s$

Formula:

$a = \frac{v - u}{t}$

Calculation:

$a = \frac{11.11 - 0}{600} = 0.0185 {m s}^{- 2}$

Answer:
Acceleration of the train = 0.0185 m s⁻²

Page No. 81 – Questions & Answers

Question 1

What is the nature of the distance–time graphs for uniform and non-uniform motion of an object?

Answer:

Uniform motion:
The distance–time graph is a straight line because the object covers equal distances in equal intervals of time.
Non-uniform motion:
The distance–time graph is a curved line because the object covers unequal distances in equal intervals of time.

Question 2

What can you say about the motion of an object whose distance–time graph is a straight line parallel to the time axis?

Answer:
The object is at rest.

Explanation:
A straight line parallel to the time axis means that the distance does not change with time, so the object is not moving.

Question 3

What can you say about the motion of an object if its speed–time graph is a straight line parallel to the time axis?

Answer:
The object is moving with uniform speed (constant speed).

Explanation:
A straight line parallel to the time axis in a speed–time graph indicates that speed remains constant with time, and acceleration is zero.

Question 4

What is the quantity which is measured by the area occupied below the velocity–time graph?

Answer:
The area under the velocity–time graph represents the displacement of the object.

Page No. 82–83 : Questions & Answers

Question 1

A bus starting from rest moves with a uniform acceleration of 0.1 m s⁻² for 2 minutes. Find
(a) the speed acquired
(b) the distance travelled

Given:

Initial velocity,

$u = 0$

Acceleration,

$a = 0.1 {m s}^{- 2}$

Time,

$t = 2 min = 120 s$

(a) Speed acquired

Using equation:

$v = u + a t$ $v = 0 + (0.1 \times 120)$ $v = 12 {m s}^{- 1}$

Speed acquired = 12 m s⁻¹

(b) Distance travelled

Using equation:

$s = u t + \frac{1}{2} a t^{2}$

$s = 0 + \frac{1}{2} \times 0.1 \times (120)^{2}$ $s = 0.05 \times 14400$ $s = 720 m$

Distance travelled = 720 m

Question 2

A train is travelling at a speed of 90 km h⁻¹. Brakes are applied to produce a uniform acceleration of –0.5 m s⁻². Find how far the train will go before it is brought to rest.

Given:

Initial velocity,

$u = 90 {km h}^{- 1} = 25 {m s}^{- 1}$

Final velocity,

$v = 0$

Acceleration,

$a = - 0.5 {m s}^{- 2}$

Using equation:

$v^{2} = u^{2} + 2 a s$ $0 = (25)^{2} + 2 (- 0.5) s$ $0 = 625 - s$ $s = 625 m$

Distance travelled before stopping = 625 m

Question 3

A trolley, while going down an inclined plane, has an acceleration of 2 cm s⁻². What will be its velocity 3 s after the start?

Given:

Initial velocity,

$u = 0$

Acceleration,

$a = 2 {cm s}^{- 2}$

Time,

$t = 3 s$

Using equation:

$v = u + a t$ $v = 0 + (2 \times 3)$ $v = 6 {cm s}^{- 1}$

Velocity after 3 s = 6 cm s⁻¹

Question 4

A racing car has a uniform acceleration of 4 m s⁻². What distance will it cover in 10 s after start?

Given:

$u = 0, a = 4 {m s}^{- 2}, t = 10 s$

Using equation:

$s = u t + \frac{1}{2} a t^{2}$ $s = 0 + \frac{1}{2} \times 4 \times (10)^{2}$ $s = 2 \times 100$ $s = 200 m$

Distance covered = 200 m

Question 5

A stone is thrown vertically upwards with a velocity of 5 m s⁻¹. If acceleration is 10 m s⁻² downward, find:
(a) maximum height attained
(b) time taken to reach maximum height

Given:

$u = 5 {m s}^{- 1}, v = 0, a = - 10 {m s}^{- 2}$

(a) Maximum height

Using equation:

$v^{2} = u^{2} + 2 a s$ $0 = 25 - 20 s$ $s = 1.25 m$

Maximum height = 1.25 m

(b) Time taken

Using equation:

$v = u + a t$ $0 = 5 - 10 t$ $t = 0.5 s$

Time taken = 0.5 s

December 15, 2025

Class 9th Science Chapter-6 Exercises

Exercises – Chapter 6: Tissues

Question 1

Define the term “tissue”.

Answer:
A tissue is a group of similar cells having a common origin that work together to perform a specific function.

Question 2

How many types of elements together make up the xylem tissue? Name them.

Answer:
Xylem tissue is made up of four types of elements:

Tracheids
Vessels
Xylem fibres
Xylem parenchyma

Question 3

How are simple tissues different from complex tissues in plants?

Answer:

Simple Tissues	Complex Tissues
Made of one type of cell	Made of more than one type of cell
Perform basic functions	Perform complex functions like transport
Example: Parenchyma, Collenchyma, Sclerenchyma	Example: Xylem, Phloem

Question 4

Differentiate between parenchyma, collenchyma and sclerenchyma on the basis of their cell wall.

Answer:

Tissue	Cell wall
Parenchyma	Thin cell wall
Collenchyma	Unevenly thickened at corners
Sclerenchyma	Very thick, lignified cell wall

Question 5

What are the functions of the stomata?

Answer:
Functions of stomata:

Exchange of gases (O₂ and CO₂)
Transpiration (loss of water vapour)
Regulation of water balance in plants

Question 6

Diagrammatically show the difference between the three types of muscle fibres.

Answer (what to draw in exam):

Striated muscle:
Long, cylindrical fibres with alternate light and dark bands, many nuclei.
Unstriated muscle:
Spindle-shaped fibres, no striations, single nucleus.
Cardiac muscle:
Branched fibres with striations, single nucleus, connected by intercalated discs.

Question 7

What is the specific function of the cardiac muscle?

Answer:
The cardiac muscle contracts rhythmically and continuously to pump blood throughout the body.

Question 8

Differentiate between striated, unstriated and cardiac muscles on the basis of their structure and site/location in the body.

Answer:

Feature	Striated	Unstriated	Cardiac
Striations	Present	Absent	Present
Control	Voluntary	Involuntary	Involuntary
Shape	Long, cylindrical	Spindle-shaped	Branched
Location	Limbs	Stomach, intestine	Heart

Question 9

Draw a labelled diagram of a neuron.

Answer (what to draw):

Dendrites
Cell body (cyton)
Nucleus
Axon
Axon terminal

(Draw a neat neuron and label all parts clearly.)

Question 10

Name the following:

(a) Tissue that forms the inner lining of our mouth
→ Squamous epithelium

(b) Tissue that connects muscle to bone in humans
→ Tendon

(c) Tissue that transports food in plants
→ Phloem

(d) Tissue that stores fat in our body
→ Adipose tissue

(e) Connective tissue with a fluid matrix
→ Blood

(f) Tissue present in the brain
→ Nervous tissue

Question 11

Identify the type of tissue in the following:

Structure	Tissue
Skin	Squamous epithelium
Bark of tree	Protective tissue
Bone	Connective tissue
Lining of kidney tubule	Cuboidal epithelium
Vascular bundle	Complex permanent tissue

Question 12

Name the regions in which parenchyma tissue is present.

Answer:
Parenchyma tissue is present in:

Cortex and pith of stems and roots
Mesophyll of leaves
Fleshy parts of fruits
Seeds and endosperm

Question 13

What is the role of epidermis in plants?

Answer:
The epidermis forms the outer protective layer of plants. Its main roles are:

Protection from mechanical injury and infection
Prevention of water loss by cuticle
Gas exchange through stomata
Absorption of water and minerals in roots (root hairs)

Question 14

How does the cork act as a protective tissue?

Answer:
Cork acts as a protective tissue because:

Its cells are dead and tightly packed
Cell walls are coated with suberin, which makes them impermeable to water and gases
It protects the plant from water loss, pathogens, and mechanical damage

Question 15

Complete the following chart:

Tissue	Location	Function
Areolar tissue	Between skin and muscles	Packing, binding and support
Adipose tissue	Below skin, around organs	Storage of fat, insulation
Bone	Skeleton	Support, protection, movement
Tendon	Between muscle and bone	Attaches muscle to bone
Blood	Blood vessels	Transport of gases, food, wastes

December 14, 2025

Tag: NCERT Question Answers

Class 11th Physics Chapter-8 Solutions

Mechanical Properties of Solids

Question 8.1

Solution

Question 8.2

Understanding the Graph (Figure 8.9)

(a) Young’s Modulus

Young’s Modulus

(b) Approximate Yield Strength

Yield Strength

Question 8.3

Solution

Key Concepts Used

(a) Greater Young’s Modulus

Answer (a):

(b) Stronger Material

Answer (b):

Question 8.4

Solution

(a) The Young’s modulus of rubber is greater than that of steel.

Reason:

(b) The stretching of a coil is determined by its shear modulus.

Reason:

Question 8.5

Solution

Given

Formula Used

(a) Elongation of Steel Wire

(b) Elongation of Brass Wire

Question 8.7

Solution

Step 1: Total load on the structure

Step 2: Cross-sectional area of one hollow column

Step 3: Stress on each column

Step 4: Compressional strain

Question 8.8

Solution

Step 1: Given data

Step 2: Calculate stress

Step 3: Calculate strain

Question 8.9

Solution

Step 1: Given data

Step 2: Cross-sectional area of the cable

Step 3: Maximum load supported

Question 8.10

Solution

Key Physical Idea

Step 1: Use the extension formula

Step 2: Condition for equal extension

Step 3: Express area in terms of diameter

Step 4: Substitute values

Question 8.11

Solution

Step 1: Given data

Step 2: Forces acting at the lowest point

Step 3: Formula for elongation

Step 4: Substitute values

Question 8.12

Solution

Step 1: Convert given quantities into SI units

Step 2: Formula for bulk modulus

Step 3: Substitute values

(a) Bulk Modulus of Water

(b) Comparison with Bulk Modulus of Air

Ratio

(c) Explanation (Why is the ratio so large?)

Question 8.13

Solution

Step 1: Given data

Step 2: Relation between density and bulk modulus

Step 3: Calculate fractional change in density

Step 4: Calculate change in density

Step 5: Density at depth

Question 8.14

Solution

Step 1: Given data

Step 2: Formula for bulk modulus

Step 3: Substitute values

$K_{i} = 0$

$More work is done in case (ii).$