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Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-2
Class 12th Class 12th Maths

Question 2

The two equal sides of an isosceles triangle with fixed base $b$ are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

Solution

Let the triangle be isosceles with:
- Base $= b$ (constant)
- Two equal sides each $= x$ (changing with time)
Given:

$\frac{d x}{d t} = - 3 cm/sec$

We want:

$\frac{d A}{d t} when x = b$

Step 1: Find the height of the triangle

For an isosceles triangle, dropping a perpendicular from the vertex to the base divides it into two equal parts:

$h = \sqrt{x^{2} - {(\frac{b}{2})}^{2}}$

Step 2: Write the area formula

$A = \frac{1}{2} \cdot b \cdot h = \frac{1}{2} b \sqrt{x^{2} - \frac{b^{2}}{4}}$

Step 3: Differentiate with respect to time $t$

$A = \frac{b}{2} \sqrt{x^{2} - \frac{b^{2}}{4}}$ $\frac{d A}{d t} = \frac{b}{2} \cdot \frac{1}{2 \sqrt{x^{2} - \frac{b^{2}}{4}}} \cdot (2 x) \frac{d x}{d t}$ $\frac{d A}{d t} = \frac{b x}{2 \sqrt{x^{2} - \frac{b^{2}}{4}}} \cdot \frac{d x}{d t}$

Step 4: Substitute $x = b$

$\frac{d A}{d t} = \frac{b \cdot b}{2 \sqrt{b^{2} - \frac{b^{2}}{4}}} \cdot (- 3)$ $= \frac{b^{2}}{2 \sqrt{\frac{3 b^{2}}{4}}} \cdot (- 3)$ $= \frac{b^{2}}{2 \cdot \frac{b \sqrt{3}}{2}} \cdot (- 3)$ $= \frac{b}{\sqrt{3}} \cdot (- 3)$ $\frac{d A}{d t} = - b \sqrt{3} {cm}^{2} / sec$

Final Answer

$The area is decreasing at b \sqrt{3} {cm}^{2} / sec when the equal sides equal the base.$
November 30, 2025
Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-1

Class 12th Class 12th Maths

Miscellaneous Exercise on Chapter 6

Question 1

Show that the function given by

$f (x) = \frac{\log x}{x}$

has maximum at $x = e$ .

Solution

Given:

$f (x) = \frac{\log x}{x}, x > 0$

Differentiate using quotient rule:

$f^{'} (x) = \frac{(1 / x) \cdot x - (\log x) \cdot 1}{x^{2}}$

Simplify numerator:

$f^{'} (x) = \frac{1 - \log x}{x^{2}}$

To find critical points, set $f^{'} (x) = 0$

$\frac{1 - \log x}{x^{2}} = 0$

Since $x^{2} > 0$ always,

$1 - \log x = 0$ $\log x = 1$ $x = e$

So the critical point is $x = e$ .

Check whether it is maximum (Second Derivative Test)

Find $f^{''} (x)$ :

$f^{'} (x) = \frac{1 - \log x}{x^{2}}$

Differentiate again:

$f^{''} (x) = \frac{- 1 / x \cdot x^{2} - (1 - \log x) \cdot 2 x}{x^{4}}$

Simplify numerator:

$f^{''} (x) = \frac{- x - 2 x (1 - \log x)}{x^{4}}$ $f^{''} (x) = \frac{- x - 2 x + 2 x \log x}{x^{4}}$ $f^{''} (x) = \frac{x (2 \log x - 3)}{x^{4}}$ $f^{''} (x) = \frac{2 \log x - 3}{x^{3}}$

Now check at $x = e$ :

$f^{''} (e) = \frac{2 \cdot 1 - 3}{e^{3}} = \frac{- 1}{e^{3}} < 0$

Since $f^{''} (e) < 0$ , the function is concave down at $x = e$ , therefore:

$x = e is a point of maximum$

Final Answer

$The function f (x) = \frac{\log x}{x} has a maximum at x = e .$

November 30, 2025
Class 10th Science Chapter 10 – The Human Eye and the Colourful World – Exercises
Q1. The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to

(a) presbyopia
(b) accommodation
(c) near-sightedness
(d) far-sightedness

Answer: (b) accommodation

Explanation:

Accommodation is the ability of the eye lens to change its focal length by changing its curvature with the help of ciliary muscles so that objects at different distances can be seen clearly.

Q2. The human eye forms the image of an object at its

(a) cornea
(b) iris
(c) pupil
(d) retina

Answer: (d) retina

Explanation:

The retina is a light-sensitive screen inside the eye where a real and inverted image is formed. It contains cells that convert light into electrical signals sent to the brain.

Q3. The least distance of distinct vision for a young adult with normal vision is about

(a) 25 m
(b) 2.5 cm
(c) 25 cm
(d) 2.5 m

Answer: (c) 25 cm

Explanation:

The minimum distance at which an object can be seen clearly without strain is 25 cm, known as the near point.

Q4. The change in focal length of an eye lens is caused by the action of the

(a) pupil
(b) retina
(c) ciliary muscles
(d) iris

Answer: (c) ciliary muscles

Explanation:

Ciliary muscles adjust the curvature of the eye lens, changing its focal length to focus objects clearly.

Q5.

Question

A person needs a lens of power –5.5 D for correcting his distant vision. For correcting his near vision, he needs a lens of power +1.5 D. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?

Formula

$P = \frac{1}{f} \Rightarrow f = \frac{1}{P}$

(i) For distant vision

Given:

$P = - 5.5 D$ $f = \frac{1}{- 5.5} = - 0.1818 m$

Convert to cm:

$f = - 0.1818 \times 100 = - 18.18 c m$

(ii) For near vision

Given:

$P = + 1.5 D$ $f = \frac{1}{1.5} = 0.67 m$ Convert to cm:

$f = 0.67 \times 100 = 67 c m$

Q6.

Question

The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?

Answer:

Given

For a myopic (near-sighted) person:
- Far point = 80 cm
- For normal vision, the far point should be infinity.
To correct myopia, a concave (diverging) lens is used to form the image of a distant object at the person’s far point.

So the image distance (v) should be:

$v = - 80 cm$

For distant objects, u = –∞.

Lens power calculation

Using lens formula for distant vision:

$P = \frac{1}{f}$

When the object distance is infinity:

$f = v = - 80 cm$

Convert focal length to metres:

$f = - 80 cm = - 0.8 m$

So: $P = \frac{1}{- 0.8} = - 1.25 D$

Q7.

Question

Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of a normal eye is 25 cm.

Answer:

(Second Image is the corrected Hypermetropic eye using a lens)

Given
- Near point of hypermetropic eye: 1 m = 100 cm
- Near point of normal eye: 25 cm
- Object distance (u) = –25 cm (normal reading distance)
- Image must form at v = –100 cm (to be seen clearly by the hypermetropic eye)
- Lens formula:
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

Substituting values

$\frac{1}{f} = \frac{1}{- 100} - \frac{1}{- 25}$

$\frac{1}{f} = - \frac{1}{100} + \frac{1}{25}$

$\frac{1}{f} = - 0.01 + 0.04 = 0.03$

$f = \frac{1}{0.03} = 33.33 cm = 0.33 m$

Power

$P = \frac{1}{f} = \frac{1}{0.33} = + 3.0 D$

Q8. Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Answer:

A normal human eye cannot see objects clearly if they are placed closer than 25 cm because:
- The ciliary muscles of the eye cannot contract any further to increase the curvature of the eye lens.
- Therefore, the lens cannot shorten its focal length sufficiently.
- As a result, light rays from a very close object are not focused on the retina, and the image appears blurred.
Final Point

The minimum distance at which the eye can see clearly without strain is 25 cm, which is called the least distance of distinct vision.

Q9. What happens to the image distance in the eye when we increase the distance of an object from the eye?

Answer:

When we increase the distance of an object from the eye, the image distance in the eye remains the same.

Explanation:
- In the human eye, the image is always formed on the retina, regardless of how far the object is.
- When the object distance increases, the ciliary muscles adjust the curvature and focal length of the eye lens so that the image continues to form at the same position — the retina.
- Therefore, the image distance (distance between lens and retina) does not change; instead, the focal length of the lens changes to maintain focus.
Q10. Why do stars twinkle?

Answer:

Stars twinkle because of the atmospheric refraction of starlight.

Explanation:
- The light from a star travels through the different layers of Earth’s atmosphere.
- These layers have varying densities, so the light bends (refracts) irregularly.
- Because of this, the apparent brightness and position of the star keep changing continuously.
- Therefore, a star appears to twinkle.
Q11. Explain why the planets do not twinkle.

Answer:

Planets do not twinkle because they appear much closer to Earth and act as extended sources of light.

Explanation:
- A planet’s light is considered as coming from a large area, not a single point.
- The refraction effects from the atmosphere average out over the larger disk of light.
- Therefore, the brightness does not fluctuate like stars.
- So, planets do not twinkle, they appear steady.
Q12. Why does the sky appear dark instead of blue to an astronaut?

Answer:

The sky appears dark to an astronaut because there is no atmosphere in space to scatter sunlight.

Explanation:
- On Earth, the atmosphere scatters shorter wavelengths of light (blue).
- This scattered blue light makes the sky appear blue.
- In space, there are no air molecules to scatter sunlight.
- Therefore, no light reaches the eye from any direction, and the sky appears black or dark.
November 18, 2025
Class 10th Science Chapter 10 – The Human Eye and the Colourful World – In-text Questions
Page 164 – Intext Questions & Answers

Q1. What is meant by power of accommodation of the eye?

Answer:
The power of accommodation of the eye is the ability of the eye lens to change its focal length so that it can clearly focus both near and distant objects on the retina by adjusting its curvature using ciliary muscles.

Q2. A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of corrective lens used to restore proper vision?

Answer:
A concave lens (diverging lens) should be used.

Reason:
In myopia, images of distant objects are formed in front of the retina. A concave lens diverges the incoming light rays so that the image forms on the retina, restoring clear vision.

Q3. What is the far point and near point of the human eye with normal vision?

Answer:
- Near point (least distance of distinct vision) = 25 cm
- Far point = Infinity (∞)
Meaning:
A normal human eye can see objects clearly from 25 cm to infinity.

Q4. A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?

Answer:
The child is likely suffering from Myopia (near-sightedness).

Correction:
It can be corrected by using concave (diverging) lenses, which help focus distant images clearly onto the retina.
November 18, 2025
Class 10th Science Chapter 9 – Light – Reflection and Refraction – Exercises
Q1. Which one of the following materials cannot be used to make a lens?

(a) Water
(b) Glass
(c) Plastic
(d) Clay

Answer: (d) Clay

Explanation:

A material used to make a lens must be transparent to allow light to pass and refract through it.
- Water, glass and plastic are transparent materials and can be shaped into lenses.
- Clay is opaque, so it cannot be used to make a lens.
Q2. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?

(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.

Answer: (d) Between the pole of the mirror and its principal focus

Explanation:

A concave mirror forms a virtual, erect, and magnified image only when the object is placed between P and F.
In all other positions, the image formed is real and inverted.

Q3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?

(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre and its principal focus

Answer: (b) At twice the focal length

Explanation:

For a convex lens:
- When the object is at 2F, the image is real, inverted, and same size formed at 2F on the other side.
Q4. A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be

(a) both concave
(b) both convex
(c) the mirror is concave and the lens is convex
(d) the mirror is convex, but the lens is concave

Answer: (d) the mirror is convex, but the lens is concave

Explanation:

According to the sign conventions:
- Focal length of a concave mirror is negative
- Focal length of a convex mirror is positive
- Focal length of a convex lens is positive
- Focal length of a concave lens is negative
Q5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be

(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.

Answer: (d) either plane or convex

Explanation:
- A plane mirror always forms an erect and same-sized image at all distances.
- A convex mirror always forms an erect and diminished virtual image no matter where the object is placed.
- A concave mirror can form inverted images when the object is far away.
Q6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.

Answer: (c) A convex lens of focal length 5 cm

Explanation:

To magnify tiny letters, we use a magnifying glass, which is a convex lens with short focal length so that it produces a large, virtual, and erect image.
- Convex lens helps in magnification.
- Short focal length = greater magnifying power.
Q 7 We wish to obtain an erect image of an object using a concave mirror of focal length 15 cm.
- What should be the range of distance of the object from the mirror?
- What is the nature of the image?
- Is the image larger or smaller than the object?
- Draw a ray diagram.
Answer:

Range of distance

To get an erect image using a concave mirror, the object must be placed:

$Between the pole (P) and the principal focus (F)$

Given:

$f = 15 cm$

So, the object distance must be:

$Less than 15 cm from the mirror$

Nature of the Image
- Virtual
- Erect
- Magnified (larger than the object)
Size of the Image

The image is larger (magnified) when the object is placed between P and F.

Q8. Name the type of mirror used in the following situations.

(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Support your answer with reason.

Answer:

(a) Headlights of a car

Mirror used: Concave mirror

Reason:
A concave mirror, when the bulb is placed at its focus, reflects light as a powerful parallel beam of light. This helps the driver to see long distances clearly at night.

(b) Side/rear-view mirror of a vehicle

Mirror used: Convex mirror

Reason:
A convex mirror:
- Always forms an erect image
- Forms a diminished image, allowing a wide field of view
  Thus, it helps the driver see more traffic area behind the vehicle.
(c) Solar furnace

Mirror used: Concave mirror

Reason:
A concave mirror converges parallel rays of sunlight to its focus, concentrating solar energy at a single point, producing very high temperature for heating or melting materials.

Q9. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Answer:

Yes, the lens will still produce a complete image of the object, even if one-half of the convex lens is covered with black paper.

Explanation:

Every part of the lens refracts light rays from all parts of the object.
So even when half of the lens is covered, the remaining half still allows light rays from the whole object to pass and form the complete image.

However:
- The image becomes dimmer (less bright),
- Because lesser light passes through the uncovered portion.
Experimental Verification:

Materials needed:

Convex lens, candle (object), screen, black paper.

Procedure:
1. Place a convex lens on a stand.
2. Place a burning candle in front of the lens and place a screen behind the lens.
3. Adjust until you get a clear sharp image.
4. Now cover the upper half of the lens with black paper.
5. Observe the image on the screen.
Observation:
- The complete image of the candle is still formed.
- The image becomes less bright compared to before covering the lens.
Conclusion:

Even a small portion of a lens contains the ability to refract light from all parts of an object, forming a complete imagebut with reduced brightness.

Q10. Solution

Question

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Given Data
- Height of object, h = 5 cm
- Object distance, u = –25 cm
- Focal length, f = +10 cm (positive because converging/convex lens)
Using Lens Formula

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $\frac{1}{v} - \frac{1}{- 25} = \frac{1}{10}$ $\frac{1}{v} + \frac{1}{25} = \frac{1}{10}$ $\frac{1}{v} = \frac{1}{10} - \frac{1}{25} = \frac{5 - 2}{50} = \frac{3}{50}$ $v = \frac{50}{3} = 16.67 cm (approx)$

Magnification

$m = \frac{v}{u} = \frac{16.67}{- 25} = - 0.67$ $h^{'} = m \times h = - 0.67 \times 5 = - 3.33 cm$

Q11. Solution

Question

A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Given
- Focal length of concave lens, f = –15 cm
- Image distance, v = –10 cm (negative for virtual image on same side as object)
Using Lens Formula

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $\frac{1}{- 10} - \frac{1}{u} = \frac{1}{- 15}$ $- \frac{1}{10} + \frac{1}{15} = \frac{1}{u}$ $\frac{- 3 + 2}{30} = \frac{1}{u}$ $\frac{- 1}{30} = \frac{1}{u}$ $u = - 30 cm$

Final Answer Summary

Quantity Result

Object distance (u) 30 cm in front of the lens

Nature of image Virtual and erect

Size Diminished

Location

10 cm from the lens on the same side

Q12.

Question

An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Solution:

Given
- Object distance: u = –10 cm (negative as per sign convention, object placed in front)
- Focal length of convex mirror: f = +15 cm
  (Convex mirror focal length is positive because its focus lies behind the mirror)
Using Mirror Formula

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ $\frac{1}{v} - \frac{1}{10} = \frac{1}{15}$ $\frac{1}{v} = \frac{1}{15} + \frac{1}{10}$ $\frac{1}{v} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$ $v = + 6 cm$

Interpretation

The image distance is positive, meaning the image forms behind the mirror.

Final Result

Feature Result

Position of image 6 cm behind the mirror

Nature of image Virtual

Orientation Erect

Size Diminished

Final Statement

For an object placed 10 cm in front of a convex mirror of focal length 15 cm, a virtual, erect and smaller image is formed 6 cm behind the mirror.

Q13.

Question

The magnification produced by a plane mirror is +1. What does this mean?

Answer

Magnification +1 means that:

1. The image formed is of the same size as the object

$m = \frac{height of image}{height of object} = + 1$

So, image height = object height.

2. The positive sign indicates the image is erect

Positive magnification → Erect image
Negative magnification → Inverted image

3. The mirror forms a virtual image

A plane mirror always produces a virtual, erect, and same-sized image.

Final Statement

Magnification +1 means the image is virtual, erect, and equal in size to the object, which is a characteristic of a plane mirror.

Q14.

Question

An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

Answer:

Given
- Object height, h = +5.0 cm
- Object distance, u = –20 cm
- Radius of curvature, R = 30 cm
- For spherical mirrors:
$f = \frac{R}{2} = \frac{30}{2} = 15 cm$

For a convex mirror, focal length is positive

$f = + 15 cm$

Mirror Formula

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ $\frac{1}{v} - \frac{1}{20} = \frac{1}{15}$ $\frac{1}{v} = \frac{1}{15} + \frac{1}{20}$

$\frac{1}{v} = \frac{4 + 3}{60} = \frac{7}{60}$ $v = \frac{60}{7} = + 8.57 cm (approx)$

Interpretation
- Positive v indicates the image is formed behind the mirror
- Convex mirrors always form virtual, erect and diminished images
Magnification

$m = \frac{v}{u} = \frac{8.57}{- 20} = - 0.4285$ $h^{'} = m \times h = - 0.4285 \times 5 = - 2.14 cm$

Magnitude of height = 2.14 cm
Negative sign means image is upright (virtual & erect).

Q15.

Question

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained? Find the size and the nature of the image.

Answer:

Given
- Object height: h = +7.0 cm
- Object distance: u = –27 cm (object in front of mirror → negative)
- Focal length for concave mirror: f = –18 cm
Using Mirror Formula

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ $\frac{1}{v} - \frac{1}{27} = \frac{1}{18}$ $\frac{1}{v} = \frac{1}{18} + \frac{1}{27}$

$\frac{1}{v} = \frac{3}{54} + \frac{2}{54} = \frac{5}{54}$ $v = \frac{54}{5} = 10.8 cm$

Position of Image

Screen should be placed at:

v = 10.8 cm in front of the mirror

(Because real images form in front of concave mirrors)

Magnification

$m = \frac{v}{u} = \frac{10.8}{- 27} = - 0.4$ $h^{'} = m \times h = - 0.4 \times 7 = - 2.8 cm$

Q16. Solution

Question

Find the focal length of a lens of power –2.0 D. What type of lens is this?

Answer:

Given

Power of lens,

$P = - 2.0 D$

Formula

$P = \frac{1}{f}$

So,

$f = \frac{1}{P}$

Substitution

$f = \frac{1}{- 2.0} = - 0.5 m$

Convert into cm:

$f = - 0.5 \times 100 = - 50 c m$

Q17.

Question

A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Answer:

Given

Power of lens:

$P = + 1.5 D$

Formula

$P = \frac{1}{f}$

Therefore,

$f = \frac{1}{P}$

Substitution

$f = \frac{1}{1.5} = 0.67 m$

Convert into cm:

$0.67 \times 100 = 67 c m$

Final Answer

Quantity Result

Focal length +0.67 m or +67 cm

Type of lens Converging (Convex) lens

Explanation
- A positive power always represents a convex (converging) lens
- Convex lenses correct hypermetropia (far-sightedness)
November 18, 2025
Class 10th Science Chapter 9 – Light – Reflection and Refraction – In-text Questions
Page 142 – Questions

Q1. Define the principal focus of a concave mirror.

Answer:
The principal focus of a concave mirror is the point on the principal axis at which all the light rays parallel to the principal axis converge after reflection from the mirror.

Q2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Answer:
We know: R = 2f
So, f = R/2 = 20/2 = 10 cm

Q3. Name a mirror that can give an erect and enlarged image of an object.

Answer:
A concave mirror can give an erect and enlarged image when the object is placed between the pole (P) and the focus (F).

Q4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Answer:
Convex mirrors are used as rear-view mirrors because:
- They always form an erect image.
- The image is diminished, so a larger area can be seen.
- They provide a wide field of view, helping the driver see more traffic behind.
Page 145 – Questions

Q1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Answer:
Using R = 2f
So, f = R/2 = 32/2 = 16 cm

Q2. A concave mirror produces three times magnified real image of an object placed at 10 cm in front of it. Where is the image located?

Answer:
Magnification m = –v/u (negative because image is real and inverted)

Given: m = 3, u = –10 cm

$3 = - v / (- 10) \Rightarrow v = 30 cm$

So, the image is formed 30 cm in front of the mirror.

Page 150 – Questions

Q1. A ray of light travelling in air enters obliquely into water. Does it bend towards the normal or away from the normal? Why?

Answer:
It bends towards the normal, because light travels slower in water (denser medium) than in air (rarer medium).

Q2. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass?

Answer:

$n = \frac{c}{v} \Rightarrow v = \frac{c}{n} = \frac{3 \times 10^{8}}{1.50}$

$v = 2 \times 10^{8} m/s$

Q3. Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density.

Answer:
- Highest optical density: Diamond (n = 2.42)
- Lowest optical density: Air (n = 1.0003)
Q4. You are given kerosene, turpentine and water. In which of these does the light travel fastest?

Answer:
Light travels fastest in the medium with lowest refractive index.

Refractive indices:
Water = 1.33, Kerosene = 1.44, Turpentine = 1.47

So, light travels fastest in water.

Q5. The refractive index of diamond is 2.42. What is the meaning of this statement?

Answer:
It means the speed of light in air is 2.42 times more than the speed of light in diamond, or light slows down by 2.42 times when entering diamond.

Page 158 – Questions

Q1. Define 1 dioptre of power of a lens.

Answer:
1 dioptre is the power of a lens whose focal length is 1 metre.

$1 D = 1 m^{- 1}$

Q2. A convex lens forms a real and inverted image of a needle at 50 cm from it. Where is the needle placed if the image is equal in size to the object? Also find the power of the lens.

Answer:
For a convex lens, an image equal in size to the object is formed when object is at 2F and image is also at 2F.

So, 2F = 50 cm → F = 25 cm

Thus, object is placed 25 cm from lens.

Power:

$P = 1 / f = 1 / 0.25 = 4 D$

Power = +4 dioptre

Q3. Find the power of a concave lens of focal length 2 m.

Answer:

$P = 1 / f = 1 / (- 2) = – 0.5 D$

Power = –0.5 dioptre (concave lens)
November 18, 2025
Class 10th Science Chapter 8 – Heredity – Exercises
Q1. A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as:

(a) TTWW
(b) TTww
(c) TtWW
(d) TtWw

Correct Answer: (c) TtWW

Detailed Explanation:
- The tall plant crossed with short plant produced progeny where all flowers were violet.
  → This means violet (W) is dominant over white (w), and the tall parent must have two dominant alleles W W, since no white appeared in F₁.
- In the same cross, almost half of the plants were short.
  → For short plants to appear, the tall parent must carry the recessive gene t along with dominant T, so when crossed with short plant (tt), some offspring receive t and become short.
Thus, tall parent genotype =

$T t W W$

Q2. A study found that children with light-coloured eyes are likely to have parents with light-coloured eyes. Can we say light-eye colour is dominant or recessive? Why or why not?

Answer:

No, we cannot conclude whether light-eye colour is dominant or recessive based on the information given.

Detailed Explanation:
- The fact that parents and children both show the same trait only proves that the trait is inherited, not whether it is dominant or recessive.
- To determine dominance, we need to study trait expression over multiple generations, including:
  - Crosses between mixed-trait parents
  - Occurrence of traits in grandchildren or siblings
For example, only if light-eyed children appear even when one parent has dark eyes, we could infer dominance.
Here, we don’t have enough evidence.

Q3. Outline a project which aims to find the dominant coat colour in dogs.

Expanded Project Outline:
1. Observe and Record
  - Visit households or shelters where dogs have different coat colours.
  - Record the colours of parents and their puppies in a chart.
2. Collect Data
  - Note which colour is more frequent in the offspring from mixed-colour parents.
3. Create Possible Crosses
  - Example: Black × Brown → Puppies mostly black?
  - Repeat with several families for reliability.
4. Analyze Results
  - If one colour consistently appears more often in puppies, even when parents are mixed, that colour is dominant.
Conclusion
- The colour that masks the other in offspring is dominant.
Q4. How is the equal genetic contribution of male and female parents ensured in the progeny?

Detailed Answer:
- Humans have 23 pairs (46) chromosomes in body cells.
- During gamete formation (sperms and eggs), the number becomes 23 (half) due to meiosis.
- During fertilisation:
23 (father) + 23 (mother) = 46 (zygote)
]
- Thus both parents contribute equally to the genetic makeup of the child.
- Each parent provides one set of chromosomes, ensuring equal hereditary information.
November 17, 2025
Class 10th Science Chapter 8 – Heredity – In-text Questions
PAGE 129 – In-text Questions (Answered)

Q1. If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier?

Answer:

Trait B is likely to have arisen earlier because:
- In asexually reproducing organisms, traits spread only by DNA copying
- A trait present in more individuals (60%) has had more generations to spread
- So trait B must have appeared before trait A
Q2. How does the creation of variations in a species promote survival?

Answer:

Variations help species survive by:
- Allowing some individuals to withstand environmental changes
- Helping organisms adapt to climate, food availability, and diseases
- Increasing chances of survival in natural selection
Example:

Bacteria resistant to heat survive a heat wave, others die → resistant type survives and continues the species.

PAGE 133 – In-text Questions (Answered)

Q1. How do Mendel’s experiments show that traits may be dominant or recessive?

Answer:

Mendel crossed tall pea plants with short pea plants:
- F₁ generation → all plants were tall
- F₂ generation → 3 tall : 1 short
This shows:
- Tall trait (T) is dominant
- Short trait (t) is recessive
Q2. How do Mendel’s experiments show that traits are inherited independently?

Answer:

When Mendel crossed:
- Tall round (T R) × short wrinkled (t r)
He found in F₂ generation 4 types of plants:
- Tall round
- Tall wrinkled
- Short round
- Short wrinkled
This shows seed shape and plant height are inherited independently.

Q3. A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits – blood group A or O – is dominant? Why or why not?

Answer:

No, this is not enough information because:
- Blood group A can be AA or AO
- For the child to have O blood group, both parents must pass O allele
- So the father must be AO
- We cannot conclude dominance because we only know inheritance pattern, not relative strength
Q4. How is the sex of the child determined in human beings?

Answer:

Sex is determined by father’s chromosome:
- Mother always contributes X chromosome
- Father contributes X or Y
Combination Child

X from father + X from mother Girl (XX)

Y from father + X from mother Boy (XY)

So the father determines the sex of the child.
November 17, 2025

Quantity	Result
Object distance (u)	30 cm in front of the lens
Nature of image	Virtual and erect
Size	Diminished
Location	10 cm from the lens on the same side

Feature	Result
Position of image	6 cm behind the mirror
Nature of image	Virtual
Orientation	Erect
Size	Diminished

Quantity	Result
Focal length	+0.67 m or +67 cm
Type of lens	Converging (Convex) lens

Combination	Child
X from father + X from mother	Girl (XX)
Y from father + X from mother	Boy (XY)

Class 10th Science Chapter 7 – How do Organisms Reproduce? – Exercises

Q1. Asexual reproduction takes place through budding in

(a) Amoeba
(b) Yeast
(c) Plasmodium
(d) Leishmania

Answer: (b) Yeast

Explanation:

Yeast reproduces asexually by budding, where a small outgrowth (bud) forms and grows into a new individual.
Amoeba → binary fission
Plasmodium → multiple fission
Leishmania → binary fission (longitudinal)

Q2. Which of the following is not a part of the female reproductive system in human beings?

(a) Ovary
(b) Uterus
(c) Vas deferens
(d) Fallopian tube

Answer: (c) Vas deferens

Explanation:

Vas deferens is a part of the male reproductive system, which carries sperm.
Ovary, uterus and fallopian tube are parts of the female reproductive system.

Q3. The anther contains

(a) sepals
(b) ovules
(c) pistil
(d) pollen grains

Answer: (d) Pollen grains

Explanation:

Anther is a part of the male reproductive organ (stamen) in flowers.
It produces pollen grains, which contain male gametes.
Ovules are found in ovary (female part).

Q4. What are the advantages of sexual reproduction over asexual reproduction?

Answer:

Sexual reproduction has several advantages:

Produces variations in offspring
Helps in evolution and development of better-adapted organisms
Ensures greater survival in changing environmental conditions
Results in offspring that are genetically different from parents

Conclusion

Sexual reproduction improves the survival and stability of a species.

Q5. What are the functions performed by the testis in human beings?

Answer:

Testis performs two main functions:

Production of sperms (male reproductive cells)
Production of testosterone hormone, which:
- controls the development of male sexual features
- regulates sperm production

Q6. Why does menstruation occur?

Answer:

If fertilisation does not occur, the thick lining of the uterus breaks down and comes out through the vagina along with blood and tissue.
This process is called menstruation.

Occurs once every 28 days approximately.

Q7. Draw a labelled diagram of the longitudinal section of a flower.

Answer (description for notebook):

A labelled diagram must include:

Petal
Sepal
Stamen (Anther + Filament)
Carpel / Pistil (Stigma, Style, Ovary)
Ovule
Receptacle

Q8. What are the different methods of contraception?

Answer:

Different methods of contraception include:

1. Barrier methods

Condom, diaphragm
Prevent sperm from reaching egg

2. Chemical / Hormonal methods

Oral pills
Injections
Prevent ovulation

3. Intrauterine devices (IUDs)

Copper-T, Loop
Placed inside uterus to block sperm entry

4. Surgical methods

Vasectomy (males – cutting vas deferens)
Tubectomy (females – cutting fallopian tubes)

5. Natural methods

Withdrawal method
Rhythm / safe period method

Q9. How are the modes for reproduction different in unicellular and multicellular organisms?

Answer:

Unicellular organisms	Multicellular organisms
Reproduce by asexual methods such as binary fission, budding, multiple fission	Reproduce mostly by sexual reproduction
One parent is involved	Two parents (male & female) are usually involved
Offspring are genetically identical to parent	Offspring show variations
Simple and fast process	Complex and slower process
Example: Amoeba, Yeast	Example: Humans, Animals, Flowering plants

Q10. How does reproduction help in providing stability to populations of species?

Answer:

Reproduction helps maintain the population of a species by:

Replacing individuals that die
Producing new individuals to continue species existence
Providing variations which help organisms adapt and survive environmental changes

Conclusion:

Without reproduction, species would disappear over time.

Q11. What could be the reasons for adopting contraceptive methods?

Answer:

Contraceptive methods are adopted to:

Prevent unwanted or unplanned pregnancies
Control population growth
Maintain the health of the mother
Allow proper spacing between children
Prevent transmission of sexually transmitted diseases (STDs) (e.g., HIV, AIDS, Syphilis) – by using condoms
Improve economic and social conditions of the family

November 17, 2025

Class 10th Science Chapter 7 – How do Organisms Reproduce? – In-text Questions

PAGE 114 – QUESTIONS & ANSWERS

Q1. What is the importance of DNA copying in reproduction?

Answer:

DNA copying is important because:

It ensures that genetic information is passed from parents to offspring.
It maintains similar body design and characteristics in the next generation.
It also introduces variations, which help in the survival and evolution of species.

$Parent Cell DNA ⟶ Copied DNA ⟶ Daughter Cells$

Q2. Why is variation beneficial to the species but not necessarily for the individual?

Answer:

Variation helps a species survive changes in the environment.
For example, if climate or temperature changes, some individuals with useful variations survive.

But variation may not help the same individual because:

It might not provide immediate advantage.
Some variations may even be harmful.

Example: If temperatures rise suddenly, heat-resistant bacteria survive, others die.

PAGE 119 – QUESTIONS & ANSWERS

Q1. How does binary fission differ from multiple fission?

Binary Fission	Multiple Fission
One parent cell divides into two equal daughter cells	One parent cell divides into many daughter cells
Example: Amoeba, Leishmania	Example: Plasmodium

Q2. How will an organism be benefited if it reproduces through spores?

Answer:

Spores can survive harsh conditions because they have thick protective walls.
They spread easily by wind or water.
They reproduce quickly to form new individuals.

Example: Bread mould (Rhizopus)

Q3. Can you think of reasons why more complex organisms cannot give rise to new individuals through regeneration?

Answer:

Complex organisms have:

Highly specialised tissues and organs arranged in a fixed pattern.
Cannot be divided into pieces without damaging vital organs.
Therefore, regeneration cannot form a complete individual.

Example: Humans cannot regrow full bodies.

Q4. Why is vegetative propagation practised for growing some types of plants?

Answer:

Vegetative propagation is used because:

Plants grow faster and bear fruits earlier.
New plants are genetically identical to parents (same qualities preserved).
Useful for plants not producing seeds naturally.

Q5. Why is DNA copying an essential part of the process of reproduction?

Answer:

Because it ensures:

Transfer of hereditary information to offspring.
Development of new individuals similar to parents.
Introduction of variations that support evolution.

PAGE 126 – QUESTIONS & ANSWERS

Q1. How is the process of pollination different from fertilisation?

Pollination	Fertilisation
Transfer of pollen from anther to stigma	Fusion of male gamete & female gamete
Occurs by wind, water, insects	Occurs inside ovule
No zygote formation	Zygote is formed

Q2. What is the role of the seminal vesicles and the prostate gland?

Answer:

They produce fluid that mixes with sperms to form semen, which:

Provides nutrition
Helps transport the sperms easily

Q3. What are the changes seen in girls at the time of puberty?

Answer:

Development of breasts
Menstruation begins
Growth of hairs in armpits & pubic region
Widening of hips
Oily skin and pimples

Q4. How does the embryo get nourishment inside the mother’s body?

Answer:

Through the placenta, which:

Transfers oxygen & nutrients from mother’s blood to embryo
Removes waste products from embryo’s blood

Q5. If a woman is using a copper-T, will it help in protecting her from sexually transmitted diseases?

Answer:

No.
Copper-T prevents pregnancy by blocking the movement of sperms,
but it does not prevent STDs like AIDS or syphilis.

Condoms help prevent STDs.

November 17, 2025

Class 10th Science Chapter 6 – Control and Coordination – Exercises

Q1. Which of the following is a plant hormone?

(a) Insulin
(b) Thyroxin
(c) Oestrogen
(d) Cytokinin

Answer: (d) Cytokinin

Explanation:

Insulin, thyroxin and oestrogen are human hormones.
Cytokinin is a plant hormone that promotes cell division and growth.

Q2. The gap between two neurons is called a

(a) dendrite
(b) synapse
(c) axon
(d) impulse

Answer: (b) synapse

Explanation:

A synapse is a microscopic gap between two neurons through which nerve impulses pass using chemical neurotransmitters.

Q3. The brain is responsible for

(a) thinking
(b) regulating the heartbeat
(c) balancing the body
(d) all of the above

Answer: (d) all of the above

Explanation:

Thinking is controlled by cerebrum
Heartbeat is regulated by medulla
Balance and posture are controlled by cerebellum
Therefore, the brain performs all of these functions.

Q4. What is the function of receptors in our body? Think of situations where receptors do not work properly. What problems are likely to arise?

Answer:

Receptors are special cells present in our sense organs that receive stimuli such as light, sound, smell, taste, pressure, pain, and temperature.
They send this information to the brain through nerve impulses for interpretation and response.

If receptors do not work properly:

We may not feel pain or danger signals (e.g., touching a hot object)
We may lose senses such as hearing, vision, smell, taste, or touch
The body may fail to respond correctly to the environment, causing accidents or injuries

Example:

A person whose pain receptors are damaged may touch fire without realising and get burnt.

Q5. Draw the structure of a neuron and explain its function.

Answer:

Diagram of a Neuron (rough sketch format for exam)

Function:

A neuron is the structural and functional unit of the nervous system.
Its function is to receive and transmit nerve impulses from one part of the body to another.

Q6. How does phototropism occur in plants?

Answer:

Phototropism is the growth movement of plants in response to light.

How it occurs:

When light falls on one side of the plant shoot, the auxin hormone moves to the shaded side.
Cells on the shaded side elongate more than cells on the lighted side.
This causes the shoot to bend towards the light.

Example:

Bending of a plant stem towards sunlight.

Q7. Which signals will get disrupted in case of a spinal cord injury?

Answer:

If the spinal cord is damaged, the following signals will be disrupted:

Reflex actions
Communication between brain and body
Movement and muscle control
Sensory signals from body parts
Control of organs below the injury

Result:

Paralysis, loss of sensation, and inability to coordinate body movements.

Q8. How does chemical coordination occur in plants?

Answer:

Chemical coordination in plants occurs through plant hormones.
These hormones are produced in one part of the plant and transported to other parts where they regulate growth and responses.

Important plant hormones:

Hormone	Function
Auxin	Cell elongation, bending towards light
Gibberellin	Stem growth and seed germination
Cytokinin	Cell division
Abscisic acid	Closes stomata, inhibits growth, stress response

Q9. What is the need for a system of control and coordination in an organism?

Answer:

An organism needs a system of control and coordination to:

Receive information from the surroundings
Process and respond to stimuli correctly
Coordinate functions of different organs
Maintain balance and homeostasis
Ensure survival in changing environments

Explanation:

Control and coordination help different body parts work together smoothly, for example:

Running away from danger
Adjusting body temperature
Regulating heartbeat and breathing

Q10. How are involuntary actions and reflex actions different from each other?

Involuntary Actions	Reflex Actions
Occur without conscious control	Quick, automatic response to a stimulus
Controlled by brain (medulla)	Controlled by spinal cord
Slow compared to reflex	Very fast
Example: heartbeat, breathing	Example: removing hand from hot object

Q11. Compare and contrast nervous and hormonal mechanisms for control and coordination in animals.

Feature	Nervous System	Hormonal (Endocrine) System
Mode of transmission	Electrical impulses	Chemical hormones in blood
Speed	Very fast	Slow
Duration	Short-lived	Long-lasting
Target	Specific organs or tissues	Many organs at once
Example	Touch, reflex action	Growth, puberty, sugar control

Q12. What is the difference between the manner in which movement takes place in a sensitive plant and movement in our legs?

Sensitive Plant Movement	Movement in Human Legs
Response to stimuli (touch)	Controlled by brain and spinal cord
Chemical changes in cells cause water to move	Muscle contraction causes movement
Not growth-based	Requires coordinated muscles & nervous system
Fast and temporary	Slower and under voluntary control
Example: Touch-me-not leaf closes	Walking, running, jumping

November 17, 2025

Class 10th Science Chapter 6 – Control and Coordination – In-text Questions

PAGE 105 – QUESTIONS & ANSWERS

Q1. What is the difference between a reflex action and walking?

Answer:

Reflex Action	Walking
Automatic, rapid, involuntary action	Slow, voluntary, controlled action
Does not involve thinking	Requires thinking and decision
Controlled by spinal cord	Controlled by brain (cerebellum)
Example: Pulling hand away from flame	Example: Walking, running

Q2. What happens at the synapse between two neurons?

Answer:

At the synapse, the electrical impulse arriving at the end of one neuron triggers the release of chemical neurotransmitters into the gap.
These chemicals cross the synapse and start a new electrical impulse in the next neuron.

Q3. Which part of the brain maintains posture and equilibrium of the body?

Answer:

The cerebellum, located in the hindbrain, maintains balance, posture and precision of voluntary movements.

Q4. How do we detect the smell of an agarbatti (incense stick)?

Answer:

Odour molecules from agarbatti are detected by olfactory receptors in the nose.
These receptors send information to the forebrain, which interprets it as smell.

Q5. What is the role of the brain in reflex action?

Answer:

The reflex action is coordinated by the spinal cord, but signals also reach the brain, which interprets the situation and allows us to become aware of what happened after the action has already occurred.

Q1. What are plant hormones?

Answer:

Plant hormones are chemical substances produced in plants that regulate growth and responses to stimuli.
Example: Auxin, Cytokinin, Gibberellin, Abscisic acid

Q2. How is the movement of leaves of the sensitive plant different from the movement of a shoot towards light?

Answer:

Sensitive plant movement	Shoot bending towards light
Very quick response	Slow response
Not dependent on growth	Depends on growth
Due to change in water content in cells	Due to uneven auxin distribution

Q3. Give an example of a plant hormone that promotes growth.

Answer:

Auxin – promotes cell elongation.

Q4. How do auxins promote the growth of a tendril around a support?

Answer:

When a tendril touches support, auxin collects on the opposite side, causing cells on that side to grow longer.
This unequal growth causes the tendril to curve around the support.

Q5. Design an experiment to demonstrate hydrotropism.

Answer:

Experiment:

Take a pot divided into two sections.
Fill both sides with soil.
Place seeds on both sides.
Pour water only on one side.
Cover pot for a few days.

Observation:

Roots bend towards the side containing water.

Conclusion:

Roots show positive hydrotropism.

PAGE 111 – QUESTIONS & ANSWERS

Q1. How does chemical coordination take place in animals?

Answer:

Animals use hormones, secreted by endocrine glands, which are released into the bloodstream and reach target organs, regulating growth, metabolism, reproduction, and emergency responses.

Q2. Why is the use of iodised salt advisable?

Answer:

Because iodine is necessary for the thyroid gland to produce thyroxin hormone.
Its deficiency causes goitre (swelling of the neck).

Q3. How does our body respond when adrenaline is secreted into the blood?

Answer:

Adrenaline prepares the body for fight or flight:

Faster heartbeat
Increased breathing rate
Increased blood flow to muscles
More oxygen supply
Energy release increases

Q4. Why are some patients of diabetes treated by giving injections of insulin?

Answer:

Diabetes occurs because the pancreas does not produce enough insulin, so blood sugar rises.
Injected insulin helps to control blood sugar levels.

November 17, 2025

Class 10th Science Chapter 5 – Life Processes – Exercises

Q1. The kidneys in human beings are a part of the system for:

(a) nutrition
(b) respiration
(c) excretion
(d) transportation

Answer: (c) excretion

Explanation:

Kidneys remove nitrogenous wastes such as urea, uric acid and excess salts from the blood.
They filter blood and form urine, which is then expelled from the body.
This process is called excretion, handled by the excretory system.

Therefore, kidneys belong to the excretory system.

Q2. The xylem in plants are responsible for:

(a) transport of water
(b) transport of food
(c) transport of amino acids
(d) transport of oxygen

Answer: (a) transport of water

Explanation:

Xylem is a conducting tissue in plants responsible for the upward transport of water and minerals from the roots to different parts of the plant.
Phloem transports food (such as sugars) prepared by leaves during photosynthesis to other parts of the plant.

Q3. The autotrophic mode of nutrition requires

(a) carbon dioxide and water
(b) chlorophyll
(c) sunlight
(d) all of the above

Answer: (d) all of the above

Explanation:

Autotrophic nutrition is the process by which green plants prepare their own food through photosynthesis.

Q4. The breakdown of pyruvate to give carbon dioxide, water and energy takes place in

(a) cytoplasm
(b) mitochondria
(c) chloroplast
(d) nucleus

Answer: (b) mitochondria

Explanation:

After glycolysis in the cytoplasm, pyruvate enters the mitochondria, where it is broken down in the presence of oxygen (aerobic respiration).

$Pyruvate + O_{2} \to C O_{2} + H_{2} O + Energy (ATP)$

Q5. How are fats digested in our bodies? Where does this process take place?

Answer:

Fats are digested in the small intestine.
The liver secretes bile, which is stored in the gall bladder and released into the small intestine.
Bile breaks large fat globules into tiny droplets (emulsification), making digestion easier.
Pancreatic lipase enzyme then converts emulsified fats into fatty acids and glycerol.

Reaction summary:

$Fat \overset{bile}{\to} small droplets \overset{lipase}{\to} fatty acids + glycerol$

Q6. What is the role of saliva in the digestion of food?

Answer:

Saliva plays two important roles:

Moistens and softens food to make it easy to swallow.
Contains an enzyme called salivary amylase which breaks down starch (carbohydrate) into simple sugar (maltose).

Equation:

$Starch \overset{amylase}{\to} maltose$

Q7. What are the necessary conditions for autotrophic nutrition and what are its by-products?

Answer:

Necessary conditions for autotrophic nutrition:

Sunlight
Chlorophyll
Carbon dioxide
Water

By-products:

$Glucose (food) and Oxygen$

Photosynthesis reaction:

$6 C O_{2} + 6 H_{2} O \overset{sunlight, chlorophyll}{\to} C_{6} H_{12} O_{6} + 6 O_{2}$

Q8. What are the differences between aerobic and anaerobic respiration? Name some organisms that use the anaerobic mode of respiration.

Answer:

Aerobic Respiration	Anaerobic Respiration
Occurs in presence of oxygen	Occurs in absence of oxygen
Produces CO₂ + H₂O	Produces alcohol + CO₂ or lactic acid
Occurs in mitochondria	Occurs in cytoplasm
Releases large amount of energy	Releases less energy

Examples of organisms using anaerobic respiration:

Yeast
Certain bacteria
Muscle cells in humans (during vigorous exercise)

Equation examples:

$Glucose \to Ethanol + C O_{2} + Energy (yeast)$

Q9. How are the alveoli designed to maximise the exchange of gases?

Answer:

Alveoli are designed to maximise gas exchange because:

They are very small and numerous (millions in number), providing a very large surface area.
They have thin walls (one cell thick) which allow faster diffusion of gases.
They are surrounded by a network of capillaries with rich blood supply.
They remain moist, helping gases dissolve and diffuse easily.

Result:

Oxygen diffuses into the blood, and carbon dioxide diffuses out rapidly.

Q10. What would be the consequences of a deficiency of haemoglobin in our bodies?

Answer:

Haemoglobin carries oxygen in the blood.
If haemoglobin is deficient:

Less oxygen will be transported to body cells.
The person becomes tired, weak and pale.
Severe deficiency causes anaemia.
In extreme cases, tissues can be permanently damaged due to lack of oxygen.

Q11. Describe double circulation of blood in human beings. Why is it necessary?

Answer:

Double circulation means that blood flows through the heart twice during one complete cycle.

Two circuits involved:

Pulmonary circulation
- From heart → lungs → heart
- Carries deoxygenated blood to lungs and returns oxygenated blood.
Systemic circulation
- From heart → body → heart
- Carries oxygenated blood to body and returns deoxygenated blood.

Why is it necessary?

To separate oxygenated and deoxygenated blood
For maintaining efficient supply of oxygen
To release large amounts of energy needed by warm-blooded organisms like humans
For proper functioning of organs

Q12. What are the differences between the transport of materials in xylem and phloem?

Feature	Xylem	Phloem
Transport material	Water & minerals	Food (sucrose)
Direction	Upward only	Both upward & downward
Requires energy	No	Yes (active transport)
Tissue type	Dead cells	Living cells
Mechanism	Transpiration pull, root pressure, capillary action	Translocation

Q13. Compare the functioning of alveoli in the lungs and nephrons in the kidneys with respect to their structure and functioning.

Feature	Alveoli	Nephrons
Function	Gas exchange	Urine formation / filtration
Structure	Thin-walled sacs with capillaries	Bowman’s capsule + tubule + ducts
Purpose	Oxygen enters blood & CO₂ removed	Waste removed & useful substances reabsorbed
Surface area	Very large (millions of alveoli)	Very large (millions of nephrons)
Mechanism	Diffusion	Filtration and reabsorption

Conclusion

Both alveoli and nephrons:

Provide large surface area
Have thin membranes
Are highly vascular (rich blood supply)

November 17, 2025

Class 10th Science Chapter 5 – Life Processes – In-text Questions

PAGE 81 – QUESTIONS & ANSWERS

Q1. Why is diffusion insufficient to meet the oxygen requirements of multicellular organisms like humans?

Answer:

Diffusion is insufficient in multicellular organisms because:

Their bodies are large and complex, made up of millions of cells.
All cells are not directly in contact with the environment.
Diffusion is a slow process, and oxygen cannot reach each cell quickly or efficiently through diffusion alone.
Therefore, they need a specialised respiratory and circulatory system to transport oxygen to all parts of the body.

Q2. What criteria do we use to decide whether something is alive?

Answer:

We decide something is alive if it shows life processes such as:

Nutrition
Respiration
Growth
Excretion
Movement (visible or molecular)
Reproduction
Even if motion is not visible, molecular movements inside the body show the organism is alive.

Q3. What are outside raw materials used for by an organism?

Answer:

Outside raw materials are used for:

Energy production (e.g., glucose breakdown)
Growth and repair of body tissues
Maintaining life processes
Examples: oxygen, water, minerals, and food.

Q4. What processes would you consider essential for maintaining life?

Answer:

Essential processes for maintaining life are:

Nutrition
Respiration
Transportation
Excretion

These processes supply energy, distribute useful materials, and remove wastes to maintain life.

PAGE 87 – QUESTIONS & ANSWERS

Q1. What are the differences between autotrophic nutrition and heterotrophic nutrition?

Autotrophic nutrition	Heterotrophic nutrition
Organisms prepare their own food	Organisms depend on others for food
Uses CO₂ and H₂O	Uses complex organic food
Occurs in green plants & some bacteria	Occurs in animals, fungi & some bacteria
Requires sunlight and chlorophyll	No need for sunlight

Q2. Where do plants get each of the raw materials required for photosynthesis?

Raw material	Source
Carbon dioxide (CO₂)	From air through stomata
Water (H₂O)	From soil through roots
Sunlight	From sun (trapped by chlorophyll)
Chlorophyll	Present in chloroplasts of leaves

Q3. What is the role of the acid in our stomach?

Answer:

Creates an acidic medium for enzymes like pepsin to work.
Kills harmful bacteria present in food.
Helps in digestion of proteins.

Q4. What is the function of digestive enzymes?

Answer:

Digestive enzymes break down complex food molecules into simpler, absorbable forms:

Proteins → amino acids
Carbohydrates → glucose
Fats → fatty acids & glycerol

Q5. How is the small intestine designed to absorb digested food?

Answer:

It has villi – finger-like projections increasing surface area for absorption.
Villi contain blood vessels, carrying absorbed food across the body.
Walls are one cell thick, enabling easy diffusion.
Long and coiled tube provides maximum time for food absorption.

PAGE NO. 91 – QUESTIONS & ANSWERS

Q1. What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration?

Answer:

Terrestrial organisms obtain oxygen directly from air, which contains more oxygen than water.
Aquatic organisms obtain dissolved oxygen from water, which is present in much lower concentration, so they need to take in large amounts of water to get sufficient oxygen.

Therefore, terrestrial organisms require less energy for breathing than aquatic organisms.

Q2. What are the different ways in which glucose is oxidised to provide energy in various organisms?

Answer:

Condition	Process	End products	Energy
With oxygen (aerobic respiration)	In mitochondria	CO₂ + H₂O	Large amount
Without oxygen (anaerobic respiration in yeast)	Fermentation	Ethanol + CO₂	Less
Lack of oxygen (in muscle cells)	Anaerobic respiration	Lactic acid	Less

Main reaction:

$Glucose \to Pyruvate$

Q3. How is oxygen and carbon dioxide transported in human beings?

Answer:

Gas	Mode of transport
Oxygen	Carried by haemoglobin in red blood cells as oxyhaemoglobin
Carbon dioxide	Mostly transported as bicarbonates in blood plasma, and partly dissolved in blood

Q4. How are the lungs designed in human beings to maximise the area for exchange of gases?

Answer:

Lungs contain millions of tiny air sacs called alveoli, which:

Greatly increase surface area for gas exchange
Have thin walls and rich blood supply
Provide large surface for diffusion of gases

PAGE NO. 96 – QUESTIONS & ANSWERS

Q1. What are the components of the transport system in human beings?

Answer:

Components of the transport system:

Heart
Blood (plasma, RBC, WBC, platelets)
Blood vessels (arteries, veins, capillaries)

Q2. What are the functions of the components of the transport system in human beings?

Answer:

Heart – pumps blood
Arteries – carry blood away from the heart
Veins – carry blood toward the heart
Capillaries – allow exchange of gases and nutrients
Blood – transports oxygen, CO₂, nutrients, hormones, waste

Q3. Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds?

Answer:

Separation allows efficient oxygen supply to body tissues and maintains high energy level needed for a warm-bloodedlifestyle.
Mixing would reduce the oxygen-carrying efficiency.

Q4. What are the components of the transport system in highly organised plants?

Answer:

Xylem – transports water and minerals
Phloem – transports food (sugar)

Q5. How are water and minerals transported in plants?

Answer:

Absorbed by root hairs
Move upward through xylem
Driven by transpiration pull, root pressure and capillary action

Q6. How is food transported in plants?

Answer:

Food is transported by phloem through a process called translocation, using energy from ATP, from leaves (source) to storage organs (sink).

PAGE NO. 98 – QUESTIONS & ANSWERS

Q1. Describe the structure and function of nephrons.

Answer:

Nephron is the functional unit of the kidney.

Structure Components

Bowman’s capsule
Glomerulus
Tubules
Collecting duct

Functions

Filtration of blood
Removal of nitrogenous wastes (urea, uric acid)
Reabsorption of useful substances
Formation of urine

Q2. What are the methods used by plants to get rid of excretory products?

Answer:

Plants remove waste by:

Diffusion through stomata and lenticels
Transpiration
Storing waste in leaves, bark, fruits
Gums, resins, latex production

Q3. How is the amount of urine produced regulated?

Answer:

Urine production depends on:

Amount of water in the body
ADH (Antidiuretic hormone) regulating water reabsorption
Salts present in the blood
Body’s hydration level and temperature

If more water in body → more urine
If less water → concentrated urine

November 17, 2025

Class 10th Science Chapter 4 – Carbon and Its Compounds – Exercises

Q1. Ethane, with the molecular formula C₂H₆ has

(a) 6 covalent bonds
(b) 7 covalent bonds
(c) 8 covalent bonds
(d) 9 covalent bonds

Answer: (b) 7 covalent bonds

Explanation:

Structure of ethane (C₂H₆):

$C H_{3} - C H_{3}$

C–C bond → 1 covalent bond
Each carbon forms 3 C–H bonds, so 3 + 3 = 6 covalent bonds

Total covalent bonds = 1 + 6 = 7

Q2. Butanone is a four-carbon compound with the functional group

(a) carboxylic acid
(b) aldehyde
(c) ketone
(d) alcohol

Answer: (c) ketone

Explanation:

Butanone (C₄H₈O) has the structure:

$C H_{3} - C O - C H_{2} - C H_{3}$

Contains the CO (carbonyl) group in the middle, which is a ketone functional group.

Functional group	Example
–COOH	Carboxylic acid
–CHO	Aldehyde
–CO–	Ketone
–OH	Alcohol

Q3. While cooking, if the bottom of the vessel is getting blackened on the outside, it means that

(a) the food is not cooked completely
(b) the fuel is not burning completely
(c) the fuel is wet
(d) the fuel is burning completely

Answer: (b) the fuel is not burning completely

Explanation:

Incomplete burning of fuel produces soot (carbon particles).
This soot deposits as a black layer on the vessel’s bottom.

Burning type	Flame	Effect
Complete combustion	Blue flame	No soot
Incomplete combustion	Yellow flame	Black soot forms

Q4. Explain the nature of the covalent bond using the bond formation in CH₃Cl.

Answer:

CH₃Cl (chloromethane) is formed by covalent bonding, in which atoms share electrons to achieve stable electronic configuration.

Bond Formation Explanation:

Carbon (C) has 4 valence electrons → needs 4 more to complete its octet.
Hydrogen (H) has 1 valence electron → needs 1 more.
Chlorine (Cl) has 7 valence electrons → needs 1 more.

Sharing of electrons:

Carbon shares one electron each with three hydrogen atoms → forms three C–H single covalent bonds.
Carbon shares one electron with chlorine → forms one C–Cl single covalent bond.
All atoms achieve stable octet/duplet configuration.

Electron dot (Lewis) structure:

$H : C : H$

(with Cl attached forming shared pair)

Detailed representation:

$\begin{array}{c} H \\ ∣ \\ H - C - C l \\ ∣ \\ H \end{array}$

Nature of bonds: All bonds (C–H and C–Cl) are covalent because they are formed by sharing of electrons, not by electron transfer.

Q5. Draw the electron dot structures for:

(a) Ethanoic acid (CH₃COOH)

(b) Hydrogen sulphide (H₂S)

S has 6 valence electrons and forms two single covalent bonds with two H atoms.

$H : S : H$

(two lone pairs left on S)

(c) Propanone (C₃H₆O)

Structure:

$C H_{3} - C O - C H_{3}$

Electron dot structure overview:

(O joins central carbon using double bond)

(d) Fluorine (F₂)

Each fluorine atom has 7 valence electrons and shares 1 pair forming a single covalent bond.

$: F — F :$

(Each F has 3 lone pairs outside the shared pair)

Q6. What is a homologous series? Explain with an example.

Answer:

A homologous series is a series of organic compounds having the same functional group and similar chemical properties, in which the molecular formula of each successive member differs by –CH₂– (14 u).

Example: Alkanes

Member	Formula
Methane	CH₄
Ethane	C₂H₆
Propane	C₃H₈
Butane	C₄H₁₀

All have the functional group — single covalent bonds (C–C, C–H), and differ by CH₂.

Q7. How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties?

Answer:

Property	Ethanol	Ethanoic acid
Smell	Wine-like smell	Vinegar-like smell
Effect on litmus	No change	Turns blue litmus red
Reaction with NaHCO₃	No reaction	Brisk effervescence due to CO₂ gas
Melting point	Liquid at room temperature	Freezes into solid ice-like mass at 16.6°C

$C H_{3} C O O H + N a H C O_{3} \to C H_{3} C O O N a + H_{2} O + C O_{2} ↑$

Q8. Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?

Answer:

Soap molecules have two ends:

Hydrophobic tail (repels water, attracts dirt/oil)
Hydrophilic head (attracts water)

When soap is added to water, hydrophobic ends surround the dirt and hydrophilic ends remain in water, forming micelles.

Micelle diagram explanation:

Dirt is trapped inside a spherical micelle.

Micelles in ethanol?

No micelles will not form in ethanol because ethanol is not polar enough.
Micelle formation requires polar water molecules to keep hydrophilic ends outward.

Q9. Why are carbon and its compounds used as fuels for most applications?

Answer:

Carbon and its compounds are used as fuels because:

They have high calorific value (release large amount of heat on burning)
Their combustion is controlled and clean
They produce fewer pollutants when burnt completely

Example:

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O + heat$

Q10. Explain the formation of scum when hard water is treated with soap.

Answer:

Hard water contains calcium (Ca²⁺) and magnesium (Mg²⁺) ions.
Soap reacts with these ions to form insoluble salts called scum.

$2 C_{17} H_{35} C O O N a + C a^{2 +} \to (C_{17} H_{35} C O O)_{2} C a + 2 N a^{+}$

Scum wastes soap and prevents lather formation.

Q11. What change will you observe if you test soap with litmus paper (red and blue)?

Answer:

Soap solution is basic in nature.

Litmus paper	Result
Red litmus	Turns blue
Blue litmus	No change

Q12. What is hydrogenation? What is its industrial application?

Answer:

Hydrogenation is a process in which hydrogen is added to unsaturated compounds in the presence of a palladium, nickel or platinum catalyst.

Example:

$C_{2} H_{4} + H_{2} \overset{N i}{\to} C_{2} H_{6}$

Industrial application:

Used for converting vegetable oils into solid fats like margarine / vanaspati ghee.

Q13. Which of the following hydrocarbons undergo addition reactions:

C₂H₆, C₃H₈, C₃H₆, C₂H₂ and CH₄?

Answer:

Unsaturated hydrocarbons undergo addition reaction.
Unsaturated compounds contain double or triple bonds.

Correct answer:

$C_{3} H_{6} (propene), C_{2} H_{2} (ethyne)$

Q14. Give a test that can be used to differentiate between saturated and unsaturated hydrocarbons.

Answer:

Bromine water test

Experiment:

Add bromine water (orange) to the sample.

Observation	Hydrocarbon Type
Colour disappears (decolourisation)	Unsaturated
Colour remains	Saturated

Q15. Explain the mechanism of cleaning action of soaps.

Answer:

Soap molecules contain:

Hydrophobic tail (binds with grease and dirt)
Hydrophilic head (binds with water)

Steps:

Hydrophobic tails attach to oil/dirt particles.
Hydrophilic heads stay in water.
Micelles are formed trapping dirt inside.
Agitation lifts micelles away and washes them off.

Diagram:

(Dirt particle in the centre surrounded by soap molecules in a spherical micelle)

November 17, 2025

Class 10th Science Chapter 4 – Carbon and Its Compounds – In-text Questions

PAGE NO. 61 – QUESTIONS & ANSWERS

Q1. What would be the electron dot structure of carbon dioxide which has the formula CO₂?

Answer:

Carbon dioxide has one carbon atom and two oxygen atoms.
Carbon has 4 valence electrons and each oxygen has 6 valence electrons.

Carbon shares two electrons with each oxygen, forming two double bonds.

Electron dot structure:

$\cdot \cdot O : : C : : O \cdot \cdot$

O = C = O

Explanation:

Carbon needs 4 electrons to complete its octet, and oxygen needs 2 electrons.
By sharing electrons, all atoms achieve stable configuration.

Q2. What would be the electron dot structure of a molecule of sulphur which is made up of eight atoms of sulphur?

(Hint: The eight atoms of sulphur are joined together in the form of a ring.)

Answer:

Sulphur (atomic number 16) has 6 valence electrons.
In S₈, eight sulphur atoms form a ring structure, each sharing 2 electrons with adjacent sulphur atoms.

Electron dot structure (simplified ring form):

$S - S - S - S - S - S - S - S$

Each S atom has 6 electrons around it (2 shared, 4 unshared).

Explanation:

Sulphur does not form double or triple bonds like oxygen; instead it forms a ring structure by single covalent bonds.

PAGE NO. 68 – QUESTIONS

Q1. How many structural isomers can you draw for pentane?

Answer:

Pentane (C₅H₁₂) has three structural isomers:

Isomer	Structure
Normal pentane	CH₃–CH₂–CH₂–CH₂–CH₃
Iso-pentane	CH₃–CH(CH₃)–CH₂–CH₃
Neo-pentane	C(CH₃)₄

Q2. What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?

Answer:

Catenation – Ability of carbon atoms to bond with other carbon atoms forming long chains, branched chains or rings.
Tetravalency – Carbon forms 4 covalent bonds with other atoms (H, O, N, Cl etc.)

Q3. What will be the formula and electron dot structure of cyclopentane?

Answer:

Formula of cyclopentane = C₅H₁₀

Electron dot structure:

Five carbon atoms form a closed ring, each carbon bound to two other carbons and two hydrogens.

Ring form (simplified):

$C - C - C - C - C (closed ring)$

Q4. Draw the structures for the following compounds

(i) Ethanoic acid
(ii) Bromopentane
(iii) Butanone
(iv) Hexanal

Answer:

Compound	Structure
Ethanoic acid (CH₃COOH)	CH₃–COOH
Bromopentane (C₅H₁₁Br)	CH₃–CH₂–CH₂–CH₂–CH₂–Br (one example; structural isomers possible)
Butanone (C₄H₈O)	CH₃–CO–CH₂–CH₃
Hexanal (C₆H₁₂O)	CH₃–CH₂–CH₂–CH₂–CH₂–CHO

Q5. How would you name the following compounds?

Answer:

(i) Bromoethane
(ii) Propanone
(iii) Hexanol

PAGE 71 – QUESTIONS & ANSWERS

1. Why is the conversion of ethanol to ethanoic acid an oxidation reaction?

Answer:

The conversion of ethanol to ethanoic acid is an oxidation reaction because oxygen is added to ethanol during the process.

Reaction:

$C H_{3} C H_{2} O H \overset{(O)}{\to} C H_{3} C O O H$

Explanation:

Ethanol (CH₃CH₂OH) gets oxidised to ethanoic acid (CH₃COOH) when oxidising agents such as alkaline potassium permanganate (KMnO₄) or acidified potassium dichromate (K₂Cr₂O₇) are used.
In this reaction, the ethanol molecule loses hydrogen atoms and gains oxygen, which is the definition of oxidation.

$Oxidation = addition of oxygen or removal of hydrogen$

Therefore, the conversion is considered an oxidation reaction.

2. A mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used?

Answer:

A mixture of ethyne and air is not used for welding because it does not produce sufficient heat.

Explanation:

Burning ethyne in air gives a yellow, smoky flame and produces a temperature of about 1500°C, which is not hot enough to melt metals for welding.
However, when ethyne is burnt in oxygen, it gives a hot, clean blue flame with a temperature of about 3000°C, which is suitable for welding metals.

Reactions:

$C_{2} H_{2} + O_{2} \to C O_{2} + H_{2} O + heat (lower)$ $C_{2} H_{2} + 2.5 O_{2} \to 2 C O_{2} + H_{2} O + very high heat$

Q3. How would you distinguish experimentally between ethanol and ethanoic acid?

Answer:

Ethanol and ethanoic acid can be distinguished by the sodium bicarbonate test (baking soda test).

Experiment:

Take a small amount of ethanol in one test tube and ethanoic acid in another.
Add a pinch of sodium bicarbonate (NaHCO₃) or baking soda solution to both test tubes.

Observation:

Substance	Observation
Ethanoic acid	Brisk effervescence (bubbles due to CO₂ gas)
Ethanol	No effervescence

Reason / Reaction:

Ethanoic acid reacts with sodium bicarbonate to produce carbon dioxide gas:

$C H_{3} C O O H + N a H C O_{3} \to C H_{3} C O O N a + H_{2} O + C O_{2} ↑$

Ethanol does not react with NaHCO₃, therefore no gas is evolved.

Q4. What are oxidising agents?

Answer:

Oxidising agents are chemical substances that add oxygen or remove hydrogen from other substances in a chemical reaction.

Example:

Alkaline potassium permanganate (KMnO₄)
Acidified potassium dichromate (K₂Cr₂O₇)

Explanation:

Oxidising agents help in converting:

$Ethanol \to Ethanoic acid$

by supplying oxygen.

PAGE 74 – QUESTIONS & ANSWERS

1. How would you distinguish experimentally between an alcohol and a carboxylic acid?

Answer:

An alcohol and a carboxylic acid can be distinguished using the sodium bicarbonate (baking soda) test.

Experiment:

Take a small amount of the unknown liquid in two different test tubes.
Add a pinch of sodium bicarbonate (NaHCO₃) or baking soda solution to each.

Observation:

Substance	Result / Observation
Carboxylic acid (e.g., ethanoic acid)	Effervescence (bubbles of CO₂ gas)
Alcohol (e.g., ethanol)	No effervescence

Reason / Reaction:

Carboxylic acids react with NaHCO₃ to produce carbon dioxide gas (CO₂), which causes bubbling:

$C H_{3} C O O H + N a H C O_{3} \to C H_{3} C O O N a + H_{2} O + C O_{2} ↑$

Alcohols do not react with sodium bicarbonate, so no gas is evolved.

2. What are oxidising agents?

Answer:

Oxidising agents are substances that add oxygen or remove hydrogen from other substances during a chemical reaction.

Examples of oxidising agents:

Alkaline potassium permanganate (KMnO₄)
Acidified potassium dichromate (K₂Cr₂O₇)

Explanation:

Oxidising agents are used in reactions such as:

$C H_{3} C H_{2} O H \overset{[O]}{\to} C H_{3} C O O H$

Here, ethanol is oxidised to ethanoic acid because oxygen is supplied by the oxidising agent.

PAGE 76 – QUESTIONS & ANSWERS

Q1. Would you be able to check if water is hard by using a detergent?

Answer:

No, we would not be able to check if water is hard by using a detergent.

Explanation:

Detergents form lather easily in both soft and hard water because they do not react with calcium (Ca²⁺) and magnesium (Mg²⁺) ions present in hard water.
Therefore, detergents cannot indicate whether water is hard or soft.

Q2. People use a variety of methods to wash clothes. Usually, after adding the soap, they ‘beat’ the clothes on a stone, or beat them with a paddle, scrub with a brush or the mixture is agitated in a washing machine. Why is agitation necessary to get clean clothes?

Answer:

Agitation is necessary because it helps soap micelles remove dirt from the fabric.

Explanation:

Soap molecules form micelles around dirt particles.
The hydrophobic ends of soap molecules attach to dirt and oil.
The hydrophilic ends remain in water.
Agitation (scrubbing, beating or washing machine movement) helps pull the dirt-filled micelles away from the fabric surface so the dirt can be washed away with water.

Conclusion:

Without agitation, micelles cannot detach dirt particles, and clothes will not become completely clean.

November 17, 2025

Class 10th Science Chapter 3 – Metals and Non-metals – Exercises

Q1. Which of the following pairs will give displacement reactions?

(a) NaCl solution and copper metal
(b) MgCl₂ solution and aluminium metal
(c) FeSO₄ solution and silver metal
(d) AgNO₃ solution and copper metal

Answer:

Correct option: (d) AgNO₃ solution and copper metal

Explanation:

A displacement reaction occurs when a more reactive metal displaces a less reactive metal from its salt solution.

Reactivity series (part):

$K > N a > C a > M g > A l > Z n > F e > P b > H > C u > A g > A u$

Q2. Which of the following methods is suitable for preventing an iron frying pan from rusting?

(a) Applying grease
(b) Applying paint
(c) Applying a coating of zinc
(d) All of the above

Answer:

Correct option: (c) Applying a coating of zinc

Explanation:

Rusting of iron occurs when iron reacts with oxygen and moisture.
To prevent rusting, the iron surface must be protected from air and water.

Q3. An element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be

(a) calcium
(b) carbon
(c) silicon
(d) iron

Answer:

(a) calcium

Explanation:

Calcium reacts with oxygen to form calcium oxide (CaO):

$2 C a + O_{2} \to 2 C a O$

Properties of calcium oxide (CaO):

High melting point
Soluble in water, forming calcium hydroxide (lime water):

$C a O + H_{2} O \to C a (O H)_{2}$

Q4. Food cans are coated with tin and not with zinc because:

(a) zinc is costlier than tin.
(b) zinc has a higher melting point than tin.
(c) zinc is more reactive than tin.
(d) zinc is less reactive than tin.

Answer:

(c) zinc is more reactive than tin

Explanation:

Tin (Sn) is less reactive than zinc (Zn).
If zinc were used to coat food containers, it could react with acids present in food items, forming harmful compounds.
Tin does not react easily with food substances and therefore is safe for coating food cans.

Q5. You are given a hammer, a battery, a bulb, wires and a switch.

(a) How could you use them to distinguish between samples of metals and non-metals?

Answer:

Test 1 – Hammer Test (Malleability Test)

Place each sample on a hard surface.
Strike each sample gently with the hammer.

Observation:

Test 2 – Electrical Conductivity Test

Set up an electrical circuit using the battery, bulb, wires and switch.
Connect each sample between the open ends of the circuit.
Switch on the circuit.

Observation:

(b) Assess the usefulness of these tests in distinguishing between metals and non-metals.

Answer:

These tests are very useful because:

Metals are malleable whereas non-metals are brittle
Metals conduct electricity well whereas non-metals do not

Therefore, by using a hammer and conductivity test, we can easily identify whether an element is a metal or non-metal.

Q6. What are amphoteric oxides? Give two examples of amphoteric oxides.

Answer:

Amphoteric oxides are metal oxides that react with both acids and bases to form salt and water.

Examples:

Aluminium oxide (Al₂O₃)
Zinc oxide (ZnO)

Explanation with equations:

Aluminium oxide reacting with acid:

$A l_{2} O_{3} + 6 H C l \to 2 A l C l_{3} + 3 H_{2} O$

Aluminium oxide reacting with base:

$A l_{2} O_{3} + 2 N a O H \to 2 N a A l O_{2} + H_{2} O$

Zinc oxide reacting with acid:

$Z n O + 2 H C l \to Z n C l_{2} + H_{2} O$

Zinc oxide reacting with base:

$Z n O + 2 N a O H \to N a_{2} Z n O_{2} + H_{2} O$

Q7. Name two metals which will displace hydrogen from dilute acids, and two metals which will not.

Answer:

Metals that will displace hydrogen from dilute acids:

Magnesium (Mg)
Zinc (Zn)

Reaction example:

$Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2} ↑$

These metals are more reactive than hydrogen.

Metals that will not displace hydrogen from dilute acids:

Copper (Cu)
Silver (Ag)

These metals are less reactive than hydrogen, so no reaction occurs.

Q8. In the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte?

Answer:

In the electrolytic refining of metal M:

Component	Used Material
Anode	Impure metal M
Cathode	Pure metal M plate
Electrolyte	Solution of a salt of metal M (e.g., MNO₃ or MCl₂)

Explanation:

During electrolysis, impure metal dissolves from the anode and pure metal gets deposited on the cathode.
Impurities settle down at the bottom as anode mud.

Q9. Pratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test tube over it.

(a) What will be the action of gas on:

(i) Dry litmus paper?
→ No change

(ii) Moist litmus paper?
→ Turns blue litmus red (acidic gas)

Reason:

The gas formed is sulphur dioxide (SO₂).
SO₂ dissolves in water forming sulphurous acid (H₂SO₃), which is acidic:

$S O_{2} + H_{2} O \to H_{2} S O_{3}$

Hence only moist litmus changes colour, dry litmus does not.

(b) Write the balanced chemical equation for the reaction taking place.

$S + O_{2} \to S O_{2}$

Q10. State two ways to prevent the rusting of iron.

Answer:

Two methods to prevent rusting are:

Galvanisation – coating iron with zinc
Painting / Greasing / Oiling – prevents air and moisture from coming in contact with iron

Other acceptable methods (if needed):

Alloying (e.g., making stainless steel)
Electroplating
Plastic coating

Q11. What type of oxides are formed when non-metals combine with oxygen?

Answer:

When non-metals react with oxygen, they form acidic oxides.

Example:

$C + O_{2} \to C O_{2} (acidic oxide)$

Explanation:

Acidic oxides dissolve in water to form acids:

$C O_{2} + H_{2} O \to H_{2} C O_{3}$

Q12. Give reasons:

(a) Platinum, gold and silver are used to make jewellery.

Answer:

Platinum, gold and silver are used to make jewellery because:

They are lustrous (shiny and attractive appearance)
They are malleable and ductile (can be beaten into sheets and drawn into wires)
They are least reactive, so they do not corrode or tarnish easily
They maintain their shine for a long time

(b) Sodium, potassium and lithium are stored under oil.

Answer:

Sodium, potassium and lithium are stored under oil because:

They are highly reactive metals
They react vigorously with oxygen and moisture present in air
They can catch fire or even explode when exposed to water or air

Storage under oil prevents their contact with air and moisture.

(c) Aluminium is a highly reactive metal, yet it is used to make utensils for cooking.

Answer:

Although aluminium is highly reactive, it is used to make utensils because:

It forms a thin protective layer of aluminium oxide (Al₂O₃) on its surface
This oxide layer prevents further reaction with air and water
It is lightweight, strong and good conductor of heat

Therefore, it becomes safe and suitable for utensils.

(d) Carbonate and sulphide ores are usually converted into oxides during the process of extraction.

Answer:

Carbonate and sulphide ores are converted into oxides because:

It is easier to reduce metal oxides to pure metal than carbonates or sulphides
Sulphides are first roasted to convert into oxides:

$2 Z n S + 3 O_{2} \to 2 Z n O + 2 S O_{2}$

Carbonates are calcined to form oxides:

$Z n C O_{3} \to Z n O + C O_{2}$

Once converted into oxides, metals can be extracted more easily by reduction.

Q13. You must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.

Answer:

Copper vessels become tarnished due to the formation of basic copper carbonate on their surface when copper reacts with moist air.

$2 C u + H_{2} O + C O_{2} + O_{2} \to C u C O_{3} \cdot C u (O H)_{2}$

Lemon and tamarind contain acids (citric acid and tartaric acid).
These acids react with the basic copper carbonate and dissolve it, bringing back the shine.

$C u C O_{3} \cdot C u (O H)_{2} + 2 H^{+} \to 2 C u^{2 +} + C O_{2} + 3 H_{2} O$

So, tarnish is removed and the vessel becomes shiny.

Q14. Differentiate between metals and non-metals on the basis of their chemical properties.

Chemical Property	Metals	Non-metals
Reaction with oxygen	Form basic oxides (e.g., MgO, CaO)	Form acidic or neutral oxides (e.g., CO₂, SO₂)
Reaction with water	Some react to release H₂ gas (e.g., Na, Mg)	Generally do not react
Reaction with acids	React to produce H₂ gas	Do not produce hydrogen gas
Reaction with salt solutions	More reactive metal displaces less reactive one	No displacement reaction
Conductivity	Conductors	Poor conductors (except graphite)

Q15. A man posed as a goldsmith and dipped gold bangles in a solution, which made them shine but reduced their weight. What was the solution likely to be?

Answer:

The solution used was aqua regia, a mixture of concentrated hydrochloric acid and concentrated nitric acid in the ratio 3:1.

Explanation:

Aqua regia dissolves a thin layer of gold from the surface.
This makes the jewellery look shiny, but at the same time reduces its weight, because some gold dissolves.

$A u + H N O_{3} + 4 H C l \to H A u C l_{4} + N O_{2} + H_{2} O$

So, the man fraudulently removed real gold to make money.

Q16. Give reasons why copper is used to make hot water tanks and not steel (an alloy of iron).

Answer:

Copper is used because:

It does not react with water, even at high temperature
It does not rust or corrode easily
It is a good conductor of heat

Steel (alloy of iron), however:

Rusts easily when exposed to water and air
Would get damaged or leak over time

Therefore, copper is a better and safer option.

November 17, 2025

Class 10th Science Chapter 3 – Metals and Non-metals – In-text Questions

TEXTBOOK QUESTION–ANSWERS WITH EXPLANATION

SECTION – PAGE QUESTIONS

Page 40 – Questions

Q1. Give an example of a metal which

(i) is a liquid at room temperature
(ii) can be easily cut with a knife
(iii) is the best conductor of heat
(iv) is a poor conductor of heat

Answer:

Property	Metal	Explanation
(i) Liquid at room temperature	Mercury (Hg)	Only metal existing as liquid at RT
(ii) Can be cut with knife	Sodium / Potassium	Very soft metals
(iii) Best conductor of heat	Silver (Ag)	Highest thermal conductivity
(iv) Poor conductor of heat	Lead (Pb)	Conducts heat very poorly

Q2. Explain the meanings of malleable and ductile.

Answer

Malleable → metals can be beaten into thin sheets (e.g., aluminium foil).
Ductile → metals can be drawn into thin wires (e.g., copper wires).

SECTION – CHEMICAL PROPERTIES OF METALS (Page 46)

Q1. Why is sodium kept immersed in kerosene oil?

Answer:
Sodium reacts violently with oxygen and water, producing heat and hydrogen gas.
To prevent accidental fire, it is stored under kerosene.

Q2. Write equations for reactions of

(i) Iron with steam
(ii) Calcium and potassium with water

Answer

(i)

$3 F e + 4 H_{2} O \overset{s t e a m}{\to} F e_{3} O_{4} + 4 H_{2} ↑$

(ii)

$C a + 2 H_{2} O \to C a (O H)_{2} + H_{2} ↑$ $2 K + 2 H_{2} O \to 2 K O H + H_{2} ↑ + heat$

Q3. Table of metals A, B, C, D

Metal	FeSO₄	CuSO₄	ZnSO₄	AgNO₃
A	No reaction	Displacement
B	Displacement		No reaction
C	No reaction	No reaction	No reaction	Displacement
D	No reaction	No reaction	No reaction	No reaction

Answers

(i) Most reactive metal = B (displaces Fe)
(ii) When B added to CuSO₄ → reddish brown copper is deposited
(iii) Decreasing reactivity order: B > A > C > D

Q4. Which gas is produced when dilute HCl is added to a reactive metal? Write reaction for Fe + H₂SO₄.

Answer

Gas released = Hydrogen (H₂)

$F e + H_{2} S O_{4} \to F e S O_{4} + H_{2} ↑$

Q5. What would you observe when zinc is added to iron(II) sulphate solution?

Answer

Zinc is more reactive than iron, so it displaces iron.

$Z n + F e S O_{4} \to Z n S O_{4} + F e$

Appearance: brown iron deposits, solution fades from green

PAGE NO. 49 – QUESTIONS & ANSWERS

Q1. (i) Write the electron-dot structures for sodium, oxygen and magnesium.

(ii) Show the formation of Na₂O and MgO by the transfer of electrons.
(iii) What are the ions present in these compounds?**

Answer:

(i) Electron-dot structures

Element	Atomic No.	Electronic Configuration	Valence Electrons	Electron-dot Structure
Sodium (Na)	11	2, 8, 1	1	Na·
Oxygen (O)	8	2, 6	6	·O: (6 dots around O)
Magnesium (Mg)	12	2, 8, 2	2	Mg:··

(ii) Formation of Na₂O and MgO

Formation of Sodium Oxide (Na₂O)

Sodium loses one electron each to become Na⁺

$2 N a \to 2 N a^{+} + 2 e^{-}$

Oxygen gains two electrons to become O²⁻

$O + 2 e^{-} \to O^{2 -}$

Combination

$2 N a^{+} + O^{2 -} \to N a_{2} O$

Formation of Magnesium Oxide (MgO)

Magnesium loses two electrons

$M g \to M g^{2 +} + 2 e^{-}$

Oxygen gains two electrons

$O + 2 e^{-} \to O^{2 -}$

Combination

$M g^{2 +} + O^{2 -} \to M g O$

(iii) Ions present in these compounds

Compound	Cation	Anion
Na₂O	Na⁺	O²⁻
MgO	Mg²⁺	O²⁻

Q2. Why do ionic compounds have high melting points?

Answer:

Ionic compounds have high melting points because the strong electrostatic forces of attraction between the oppositely charged ions require a large amount of energy to break.
Therefore, ionic compounds are usually solid at room temperature and have very high melting and boiling points.

PAGE NO. 53 – QUESTIONS & ANSWERS

Q1. Define the following terms:

(i) Mineral

A mineral is a naturally occurring substance found in the earth’s crust that contains one or more metals or their compounds.

(ii) Ore

An ore is a mineral from which a metal can be profitably extracted.

(iii) Gangue

Gangue refers to the impurities such as sand, soil, and rocky materials that are present in the ore and must be removed before metal extraction.

Q2. Name two metals which are found in nature in the free state.

Answer:

Gold and Silver

Explanation:
These metals are least reactive and do not combine with other elements, so they are found in free or native form.

Q3. What chemical process is used for obtaining a metal from its oxide?

Answer:

The process used is called reduction.

Explanation:

In reduction, oxygen is removed from metal oxide to obtain the pure metal.
Example:

$Z n O + C \to Z n + C O$

PAGE NO. 55 – Question & Answer

Q1. Metallic oxides of zinc, magnesium and copper were heated with the following metals.

Metal	Zinc oxide	Magnesium oxide	Copper oxide
Zinc	?	?	?
Magnesium	?	?	?
Copper	?	?	?

In which cases will you find displacement reactions taking place?

Answer:

To determine displacement reactions, we use the reactivity series:

$Most reactive: M g > Z n > C u : Least reactive$

Rule:

A more reactive metal can displace a less reactive metal from its oxide.

Case-by-case outcomes

1. Zinc heated with Zinc oxide

No reaction — a metal cannot displace itself.

2. Zinc heated with Magnesium oxide

No reaction — Zn is less reactive than Mg.

3. Zinc heated with Copper oxide

Reaction occurs because Zn is more reactive than Cu.

$Z n + C u O \to Z n O + C u$

4. Magnesium heated with Zinc oxide

Reaction occurs because Mg is more reactive than Zn.

$M g + Z n O \to M g O + Z n$

5. Magnesium heated with Magnesium oxide

No reaction — metal cannot displace itself.

6. Magnesium heated with Copper oxide

Reaction occurs because Mg is more reactive than Cu.

$M g + C u O \to M g O + C u$

7. Copper heated with any oxide

No reaction — Cu is least reactive, cannot displace any metal.

Final Result Summary Table

Metal Added	Zinc oxide (ZnO)	Magnesium oxide (MgO)	Copper oxide (CuO)
Zinc	❌ No reaction	❌ No reaction	✔ Reaction occurs
Magnesium	✔ Reaction occurs	❌ No reaction	✔ Reaction occurs
Copper	❌ No reaction	❌ No reaction	❌ No reaction

Therefore, displacement reactions take place in:

✔ Magnesium + Zinc oxide

✔ Magnesium + Copper oxide

✔ Zinc + Copper oxide

Q2. Which metals do not corrode easily?

Gold, silver and platinum
They are least reactive (do not react with air or moisture).

Q3. What are alloys?

An alloy is a homogeneous mixture of two or more metals, or a metal and a non-metal, made to improve strength, hardness or corrosion resistance.

Examples: Brass (Cu + Zn), Bronze (Cu + Sn), Stainless steel (Fe + Cr + Ni)

November 17, 2025

Class 10th Science Chapter 2 – Acid, Bases and Salt – Exercises
1. A solution turns red litmus blue, its pH is likely to be

(a) 1 (b) 4 (c) 5 (d) 10

Answer: (d) 10

Explanation:
A solution turning red litmus blue is a base, and basic solutions have pH greater than 7. Among the given options, pH 10 is strongly basic.

2. A solution reacts with crushed egg-shells to give a gas that turns lime-water milky. The solution contains

(a) NaCl (b) HCl (c) LiCl (d) KCl

Answer: (b) HCl

Explanation:
Egg-shells contain calcium carbonate (CaCO₃). Acids like HCl react with carbonates to produce CO₂ gas, which turns lime water milky.

${CaCO}_{3} + 2 HCl \to {CaCl}_{2} + H_{2} O + {CO}_{2} ↑$

3. 10 mL NaOH solution is neutralised by 8 mL HCl solution. How much HCl will be needed for 20 mL of NaOH?

(a) 4 mL (b) 8 mL (c) 12 mL (d) 16 mL

Answer: (d) 16 mL

Explanation:
Neutralisation is directly proportional:

$10 mL NaOH \to 8 mL HCl$

So,

$20 mL NaOH \to 8 \times 2 = 16 mL HCl$

4. Which type of medicine is used for treating indigestion?

(a) Antibiotic (b) Analgesic (c) Antacid (d) Antiseptic

Answer: (c) Antacid

Explanation:
Indigestion is caused by excess acid in the stomach. Antacids like magnesium hydroxide or baking soda neutralise the acid.

5. Write word equations and balanced equations for the following reactions:

(a) Dilute sulphuric acid reacts with zinc granules

Word Equation:
Sulphuric acid + Zinc → Zinc sulphate + Hydrogen gas

Balanced Equation:

$H_{2} {SO}_{4} + Zn \to {ZnSO}_{4} + H_{2} ↑$

(b) Dilute hydrochloric acid reacts with magnesium ribbon

Word Equation:
Hydrochloric acid + Magnesium → Magnesium chloride + Hydrogen gas

Balanced Equation:

$2 HCl + Mg \to {MgCl}_{2} + H_{2} ↑$

(c) Dilute sulphuric acid reacts with aluminium powder

Word Equation:
Sulphuric acid + Aluminium → Aluminium sulphate + Hydrogen gas

Balanced Equation:

$3 H_{2} {SO}_{4} + 2 Al \to {Al}_{2} {(SO}_{4})_{3} + 3 H_{2} ↑$

(d) Dilute hydrochloric acid reacts with iron filings

Word Equation:
Hydrochloric acid + Iron → Iron chloride + Hydrogen gas

Balanced Equation:

$2 HCl + Fe \to {FeCl}_{2} + H_{2} ↑$

6. Compounds such as alcohol and glucose contain hydrogen but are not acids. Describe an activity to prove it.

Activity:
- Take solutions of alcohol and glucose in two beakers.
- Fix two carbon/platinum electrodes and connect them to a bulb and battery.
- Switch on the circuit.
Observation:

The bulb does not glow in either solution.

Conclusion:

Although alcohol and glucose contain hydrogen, they do not produce H⁺ ions in water, so they do not conduct electricity and are not acids.

7. Why does distilled water not conduct electricity, whereas rain water does?

Answer:
- Distilled water is completely pure and contains no ions, so it cannot conduct electricity.
- Rain water contains dissolved acids and salts (like carbonic acid formed by dissolving CO₂), which provide ions, so it conducts electricity.
8. Why do acids not show acidic behaviour in the absence of water?

Answer:
Acids do not show acidic behaviour in the absence of water because they do not produce hydrogen ions (H⁺) without dissolving in water.
Acids ionise only in aqueous solution to release H⁺ or H₃O⁺ ions, which are responsible for acidic properties.

$HCl + H_{2} O \to H_{3} O^{+} + {Cl}^{-}$

Without water → no ionisation → no acidic behaviour.

9. Five solutions A,B,C,D,E show pH values 4, 1, 11, 7 and 9

(a) Neutral: pH 7 → Solution D

(b) Strongly alkaline: pH 11 → Solution C

(c) Strongly acidic: pH 1 → Solution B

(d) Weakly acidic: pH 4 → Solution A

(e) Weakly alkaline: pH 9 → Solution E

Arrange pH in increasing order of hydrogen-ion concentration

Greater H⁺ concentration = stronger acidity = lower pH
Order:

$pH 11 < pH 9 < pH 7 < pH 4 < pH 1$

10. Equal lengths of magnesium ribbons placed in test tubes A & B with equal concentrations of acids HCl and CH₃COOH. Which shows more fizzing and why?

Answer:
Fizzing will be more vigorous in test tube A (HCl).

Reason:
Hydrochloric acid is a strong acid and ionises completely in water to release more H⁺ ions.
Acetic acid is a weak acid and ionises partially.

$Mg + 2 HCl \to {MgCl}_{2} + H_{2} ↑$

More H⁺ ions → faster reaction → more hydrogen gas bubbles (fizzing).

11. Fresh milk has pH 6. How will pH change as it turns into curd? Explain.

Answer:
As milk turns into curd, its pH decreases below 6.

Reason:
Bacteria convert lactose (milk sugar) into lactic acid, increasing acidity and lowering pH.

12. A milkman adds a very small amount of baking soda to fresh milk.

(a) Why does he shift the pH from 6 to slightly alkaline?

To prevent milk from turning sour quickly, as acids formed will first neutralise the baking soda.

(b) Why does the milk take a longer time to set as curd?

Because curd formation requires acidic conditions, and the added baking soda neutralises acids initially, slowing down the process.

13. Plaster of Paris should be stored in a moisture-proof container. Explain why.

Answer:
Plaster of Paris reacts with moisture to convert back into gypsum, forming a hard solid mass.

${CaSO}_{4} \cdot \frac{1}{2} H_{2} O + \frac{3}{2} H_{2} O \to {CaSO}_{4} \cdot 2 H_{2} O$

So, it becomes useless if exposed to moisture.

14. What is a neutralisation reaction? Give two examples.

Answer:
A neutralisation reaction is a reaction in which an acid reacts with a base to form salt and water.

$Acid + Base \to Salt + Water$

Examples:
1. HCl + NaOH → NaCl + H₂O
2. $H_{2} {SO}_{4} + {Ca(OH)}_{2} \to {CaSO}_{4} + 2 H_{2} O$

15. Give two important uses of washing soda and baking soda.

Washing Soda (Na₂CO₃·10H₂O)
- Used in softening hard water
- Used in the manufacture of glass, soap and paper
Baking Soda (NaHCO₃)
- Used in baking to make cakes and bread soft and fluffy
- Used as an antacid to remove acidity in the stomach
November 17, 2025
Class 10th Science Chapter 2 – Acid, Bases and Salt – In-text Questions
PAGE 18 – Activity Question

Q 1. You have been provided with three test tubes… identify the contents using only red litmus paper.

Answer:
- Dip red litmus paper into each solution.
- Solution which turns red litmus blue is a base.
- Solution which does not change colour could be water or acid.
- Now dip the same strip (already blue) into the two remaining test tubes:
  - If blue litmus turns red → acidic solution
  - If no change → distilled water
Explanation:
Acids turn blue litmus red, bases turn red litmus blue, and neutral water shows no change.

PAGE 22 – Short Questions

Q1. Why should curd and sour substances not be kept in brass or copper vessels?

Answer:
Curd and sour foods contain acids which react with copper or brass (metals) forming toxic salts that may cause food poisoning.
This is due to metal reacting with acid to form salt + hydrogen gas.

Q2. Which gas is usually liberated when acid reacts with metal? How do you test it?

Answer:
Hydrogen gas is liberated.
Test: Bring a burning splinter near the gas bubbles. It burns with a ‘pop’ sound.
Reaction Example:

${Zn + H}_{2} {SO}_{4} \to {ZnSO}_{4} + H_{2} ↑$

Q3. Metal compound A reacts with HCl producing a gas that extinguishes flame; CO₂ is produced. Write balanced equation.

Answer:

${CaCO}_{3} + 2 HCl \to {CaCl}_{2} + {CO}_{2} + H_{2} O$

Explanation: Gas formed turns lime water milky → indicates CO₂.

PAGE 25 – pH Section Questions

Q1. Why do HCl, HNO₃ show acidic character in aqueous solution but glucose & alcohol do not?

Answer:

HCl and HNO₃ show acidic character in aqueous solution because they ionise in water and produce hydrogen ions (H⁺ / H₃O⁺).
These free ions are responsible for acidic properties.

Example:

$HCl + H_{2} O \to H_{3} O^{+} + {Cl}^{-}$ ${HNO}_{3} + H_{2} O \to H_{3} O^{+} + {NO}_{3}^{-}$

Q2. Why does an aqueous solution of an acid conduct electricity?

Answer:
An aqueous acid conducts electricity because acids dissociate in water to produce ions, particularly hydrogen ions (H⁺ / H₃O⁺). These ions are charged particles that carry electric current through the solution.

Example:

$HCl (aq) \to H^{+} (aq) + {Cl}^{-} (aq)$

Explanation:
Electric current in a solution is carried by free-moving ions. When an acid dissolves in water, it breaks into positive and negative ions, allowing electricity to flow.
That is why the bulb glows in acid solution during the conductivity test.

Q3. Why does dry HCl gas not change the colour of dry litmus paper?

Answer:
Dry HCl gas does not change the colour of dry litmus paper because it does not produce hydrogen ions (H⁺) in the absence of water.
Acids show acidic behaviour only when they ionise in water.

$HCl (gas) does not ionise \Rightarrow {No H}^{+} ions \Rightarrow No acidic property$

When moisture (water) is present:

$HCl + H_{2} O \to H_{3} O^{+} + {Cl}^{-}$

These H⁺ (H₃O⁺) ions are responsible for colour change of litmus.

Conclusion

Condition Behaviour of HCl Effect on litmus

Dry HCl gas No ionisation No colour change

Aqueous HCl Produces H⁺ ions Turns blue litmus red

Q4. While diluting an acid, why is it recommended that the acid should be added to water and not water to the acid?

Answer:
While diluting an acid, acid must always be added slowly to water because the process of mixing acid with water is highly exothermic (releases a lot of heat).
If water is added directly to a concentrated acid, the heat produced may cause the mixture to splash out suddenly, which can cause severe burns or break the container.

However, when acid is added gradually to a larger volume of water, the heat gets absorbed safely by water and no splashing occurs.

Explanation

$Acid + Water → Heat released (exothermic)$

If water is added to acid If acid is added to water

Sudden boiling and splashing Heat safely absorbed

Dangerous, may cause burns
Safe dilution

Q5. How is the concentration of hydronium ions (H₃O⁺) affected when a solution of an acid is diluted?

Answer:
When a solution of an acid is diluted by adding water, the concentration of hydronium ions (H₃O⁺) decreases.
This happens because the number of H₃O⁺ ions per unit volume reduces as more water molecules are added, spreading the ions over a larger volume.

Explanation

$Acid + H_{2} O \to H_{3} O^{+} + ions$

When water is added:
- The total volume increases
- The number of H₃O⁺ ions remains the same
- Therefore, ionic concentration (H₃O⁺ per unit volume) decreases
Q6. How is the concentration of hydroxide ions (OH⁻) affected when excess base is dissolved in a solution of sodium hydroxide?

Answer:
When excess base is added to a solution of sodium hydroxide (NaOH), the concentration of hydroxide ions (OH⁻) increases because more OH⁻ ions are released into the solution.

Explanation

Sodium hydroxide is a strong base, and it dissociates completely in water:

$NaOH (aq) \to {Na}^{+} + {OH}^{-}$

If more base (e.g., NaOH, KOH) is added, the solution gets additional OH⁻ ions, increasing the strength of alkalinity.

Page No. 28 – Answers

Q1. You have two solutions, A and B. The pH of solution A is 6 and pH of solution B is 8.
Which solution has more hydrogen ion concentration? Which of these is acidic and which one is basic?

Answer:
- Solution A (pH 6) has more hydrogen ion (H⁺) concentration than solution B (pH 8).
- Solution A is acidic (pH < 7).
- Solution B is basic (pH > 7).
Explanation

The pH scale indicates acidity or alkalinity of a solution.
- Lower pH value → higher H⁺ ion concentration → stronger acidity
- Higher pH value → lower H⁺ ion concentration → stronger basicity
Therefore:

$pH 6 (A) < pH 8 (B) \Rightarrow [H^{+}] in A > [H^{+}] in B$

Q2. What effect does the concentration of H⁺(aq) ions have on the nature of the solution?

Answer:
The nature of a solution depends on the concentration of H⁺(aq) ions:
- Higher concentration of H⁺ ions makes the solution more acidic.
- Lower concentration of H⁺ ions makes the solution less acidic (or more basic).
Explanation

On the pH scale:

$Higher [H^{+}] \Rightarrow Lower pH \Rightarrow Stronger acid$ $Lower [H^{+}] \Rightarrow Higher pH \Rightarrow Weaker acid / Basic$

Q3. Do basic solutions also have H⁺(aq) ions? If yes, then why are these basic?

Answer:

Yes, basic solutions also contain H⁺(aq) ions, but they are basic because the concentration of hydroxide ions (OH⁻)is much higher than the concentration of hydrogen ions (H⁺).
The excess OH⁻ ions make the solution alkaline.

Explanation with Equations

Water undergoes slight ionisation:

$H_{2} O ⇌ H^{+} + {OH}^{-}$

When a base like sodium hydroxide dissolves in water:

$NaOH (s) \overset{H_{2} O}{\to} {Na}^{+} + {OH}^{-}$

This produces a large amount of OH⁻ ions.
So, although a small amount of H⁺ is always present (from water ionisation), the solution remains basic because:

$[{OH}^{-}] > [H^{+}]$

Q4. Under what soil condition do you think a farmer would treat the soil of his fields with quick lime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate)?

Answer:
A farmer would treat the soil with quick lime (CaO), slaked lime (Ca(OH)₂), or chalk (CaCO₃) when the soil is too acidic (low pH).

These substances are basic in nature and neutralise excess acids present in the soil, helping to maintain the optimal pH required for plant growth.

Explanation

When the soil becomes acidic due to excessive use of chemical fertilisers or acid rain, plant growth is hampered.
Adding lime compounds increases the pH of the soil, making it less acidic and more suitable for agriculture.

Neutralisation reaction example:

${Ca(OH)}_{2} + 2 HCl \to {CaCl}_{2} + 2 H_{2} O$

PAGE 33 – Answers

Q1. What is the common name of the compound Ca(ClO)₂?

Answer:
The common name of Ca(ClO)₂ is Bleaching Powder.

Q2. Name the substance which on treatment with chlorine yields bleaching powder.

Answer:
Slaked lime (Calcium hydroxide, Ca(OH)₂)

Reaction:

$2 {Ca(OH)}_{2} + 2 {Cl}_{2} \to {Ca(ClO)}_{2} + {CaCl}_{2} + 2 H_{2} O$

Q3. Name the sodium compound which is used for softening hard water.

Answer:
Washing soda (Sodium carbonate – Na₂CO₃ · 10H₂O)

Q4. What will happen if a solution of sodium hydrogencarbonate is heated? Give the equation of the reaction involved.

Answer:
On heating, sodium hydrogencarbonate decomposes to form sodium carbonate, carbon dioxide and water.

$2 {NaHCO}_{3} \overset{Heat}{\to} {Na}_{2} {CO}_{3} + H_{2} O + {CO}_{2} ↑$

Q5. Write an equation to show the reaction between Plaster of Paris and water.

Answer:

${CaSO}_{4} \cdot \frac{1}{2} H_{2} O + \frac{3}{2} H_{2} O \to {CaSO}_{4} \cdot 2 H_{2} O$
November 17, 2025

Condition	Behaviour of HCl	Effect on litmus
Dry HCl gas	No ionisation	No colour change
Aqueous HCl	Produces H⁺ ions	Turns blue litmus red

Tag: NCERT Question Answers

Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-2

Class 12th Class 12th Maths

Question 2

Solution

Step 1: Find the height of the triangle

Step 2: Write the area formula

Step 3: Differentiate with respect to time t

Step 4: Substitute x=b

Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-1

Class 12th Class 12th Maths

Miscellaneous Exercise on Chapter 6

Question 1

Solution

To find critical points, set f′(x)=0

Check whether it is maximum (Second Derivative Test)

Final Answer

Class 10th Science Chapter 10 – The Human Eye and the Colourful World – Exercises

Q1. The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to

Answer: (b) accommodation

Explanation:

Q2. The human eye forms the image of an object at its

Answer: (d) retina

Explanation:

Q3. The least distance of distinct vision for a young adult with normal vision is about

Answer: (c) 25 cm

Explanation:

Q4. The change in focal length of an eye lens is caused by the action of the

Answer: (c) ciliary muscles

Explanation:

Q5.

Question

Formula

(i) For distant vision

(ii) For near vision

Q6.

Question

Answer:

Given

Lens power calculation

Q7.

Question

Answer:

(Second Image is the corrected Hypermetropic eye using a lens)

Given

Substituting values

Power

Q8. Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Answer:

Final Point

Q9. What happens to the image distance in the eye when we increase the distance of an object from the eye?

Answer:

Explanation:

Q10. Why do stars twinkle?

Answer:

Explanation:

Q11. Explain why the planets do not twinkle.

Answer:

Explanation:

Q12. Why does the sky appear dark instead of blue to an astronaut?

Answer:

Explanation:

Class 10th Science Chapter 10 – The Human Eye and the Colourful World – In-text Questions

Page 164 – Intext Questions & Answers

Q1. What is meant by power of accommodation of the eye?

Q2. A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of corrective lens used to restore proper vision?

Q3. What is the far point and near point of the human eye with normal vision?

Q4. A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?

Class 10th Science Chapter 9 – Light – Reflection and Refraction – Exercises

Q1. Which one of the following materials cannot be used to make a lens?

Answer: (d) Clay

Explanation:

Q2. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?

Answer: (d) Between the pole of the mirror and its principal focus

Explanation:

Q3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?

Answer: (b) At twice the focal length

Explanation:

Q4. A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be

Answer: (d) the mirror is convex, but the lens is concave

Step 3: Differentiate with respect to time $t$

Step 4: Substitute $x = b$

To find critical points, set $f^{'} (x) = 0$