Exercise-11.4, Class 9th, Maths, Chapter 11, NCERT

Q1. Find the volume of a sphere whose radius is

(i) $7\text{ cm}$
(ii) $0.63\text{ m}$

Volume formula: $V={\displaystyle \frac{4}{3}}\pi r^3$

(i) $r=7$

$$V=\frac{4}{3}\times \frac{22}{7}\times 7^3=\frac{4}{3}\times \frac{22}{7}\times 343=\frac{88}{3}\times 49=\frac{4312}{3}{\text{cm}}^3\approx 1437.33{\text{cm}}^3\mathrm{.}$$

(ii) $r=\mathrm{0.63, }$ $r^3={0.63}^3=0.250047{\text{m}}^3\mathrm{}$

$$V=\frac{4}{3}\times \frac{22}{7}\times 0.250047=\frac{88}{21}\times 0.250047\approx 1.047815{\text{m}}^3\mathrm{}$$

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Q2. Find the amount of water displaced by a solid spherical ball of diameter

(i) $28\text{ cm}$ $\Rightarrow r=14\text{ cm}$
(ii) $0.21\text{ m}$ $\Rightarrow r=0.105\text{ m}$

(Amount displaced = volume of the sphere.)

(i) $r=14$ cm.

$$V=\frac{4}{3}\pi (14{)}^3=\frac{4}{3}\times \frac{22}{7}\times 2744=\frac{34496}{3}{\text{cm}}^3\approx 11498.67{\text{cm}}^3\mathrm{.}$$

In litres: $11498.67{\text{cm}}^3\approx 11.499\text{ L}\mathrm{}$

(ii) $r=0.105$ m. $r^3={0.105}^3=0.001157625{\text{m}}^3\mathrm{}$

$$V=\frac{4}{3}\times \frac{22}{7}\times 0.001157625\approx 0.00485195{\text{m}}^3\approx 4.85195\text{ L}\mathrm{.}$$

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Q3. The diameter of a metallic ball is $4.2\text{ cm}$. What is the mass of the ball, if the density of the metal is $8.9{\text{g/cm}}^3$?

Radius $r=2.1$ cm. Volume:

$$V=\frac{4}{3}\pi (2.1{)}^3=\frac{88}{21}\times 9.261\approx 38.808{\text{cm}}^3\mathrm{.}$$

Mass $=\text{density}\times V=8.9\times 38.808\approx 345.51\text{ g}\mathrm{}$

Answer: $\approx 345.5\text{ g}\mathrm{}$

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Q4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

If linear scale factor $=\frac{1}{4}$, volume scales as cube ⇒ fraction $={\left(\frac{1}{4}\right)}^3=\frac{1}{64}\mathrm{}$

Answer: ${\displaystyle \frac{1}{64}}\mathrm{}$

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Q5. How many litres of milk can a hemispherical bowl of diameter $10.5\text{ cm}$ hold?

Radius $r=5.25$ cm. Volume of hemisphere = ${\displaystyle \frac{2}{3}}\pi r^3$

Compute $r^3={5.25}^3=144.703125{\text{cm}}^3\mathrm{}$

$$V=\frac{2}{3}\times \frac{22}{7}\times 144.703125=\frac{44}{21}\times 144.703125\approx 303.1875{\text{cm}}^3\mathrm{.}$$

Convert to litres: $303.1875{\text{cm}}^3=0.3031875\text{ L}\approx 303.19\text{ mL}\mathrm{}$

Answer: $\approx 0.3032\text{ L}$

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Q6. A hemispherical tank is made up of an iron sheet $1\text{ cm}$ thick. If the inner radius is $1\text{ m}$, find the volume of the iron used.

Inner radius $r=1.00\text{ m}$. Thickness $=1\text{ cm}=0.01\text{ m}$. Outer radius $R=1.01\text{ m}$.

Volume of iron = volume(outer hemisphere) – volume(inner hemisphere)

$$V=\frac{2}{3}\pi (R^3-r^3)=\frac{44}{21}({1.01}^3-1^3){\text{m}}^3\mathrm{}$$

${1.01}^3-1=0.030301$. Thus

$$V\approx \frac{44}{21}\times 0.030301\approx 0.06349{\text{m}}^3\mathrm{.}$$

In litres: $\approx 63.49\text{ L}\mathrm{}$

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Q7. Find the volume of a sphere whose surface area is $154{\text{cm}}^2$.

Surface area $=4\pi r^2=154$. So

$$r^2=\frac{154}{4\pi }=\frac{154}{88\mathrm{/}7}=\frac{154\times 7}{88}=12.25,$$

so $r=3.5\text{ cm}\mathrm{}$

Volume $V={\displaystyle \frac{4}{3}}\pi r^3={\displaystyle \frac{4}{3}}\times \frac{22}{7}\times (3.5{)}^3={\displaystyle \frac{4}{3}}\times \frac{22}{7}\times 42.875$

$=\frac{4}{3}\times \frac{22}{7}\times 42.875\approx 179.594{\text{cm}}^3\mathrm{.}$

(You can also simply answer $r=3.5$ if only radius is required — question asked volume, so value above.)

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Q8. A dome is a hemisphere. From inside it was white-washed at total cost $\text{₹}4989.60$. If white-washing costs $\text{₹}20$ per m${}^2$, find (i) inside surface area of the dome, (ii) volume of air inside the dome.

(i) Area $={\displaystyle \frac{\text{cost}}{\text{rate}}}={\displaystyle \frac{4989.60}{20}}=249.48{\text{m}}^2\mathrm{<span\; class="base"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="mord">.</span></span>}$

This is the curved surface area of the hemisphere: $2\pi r^2=249.48$So

$$r^2=\frac{249.48}{2\pi }=\frac{249.48}{44\mathrm{/}7}=39.69\Rightarrow r=6.3\text{ m}\mathrm{.}$$

(ii) Volume of air (hemisphere) $={\displaystyle \frac{2}{3}}\pi r^3$
$r^3={6.3}^3=250.047$. Hence

$$V=\frac{44}{21}\times 250.047\approx 523.91{\text{m}}^3\mathrm{.}$$

Answers: (i) $249.48{\text{m}}^2$ (given), (ii) $r=6.3\text{ m}$ and $V\approx 523.91{\text{m}}^3\mathrm{}$

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Q9. Twenty seven solid iron spheres, each of radius $r$ and surface area $S$, are melted to form a single sphere with surface area $S^{\mathrm{\prime }}$. Find (i) radius $r^{\mathrm{\prime }}$of the new sphere, (ii) ratio $S:S^{\mathrm{\prime }}$.

Each small sphere volume = ${\displaystyle \frac{4}{3}}\pi r^3$. Total volume of 27 spheres $=27\cdot \frac{4}{3}\pi r^3$.

Let new radius be $r^{\mathrm{\prime }}$. Then

$$\frac{4}{3}\pi r^{\mathrm{\prime }3}=27\cdot \frac{4}{3}\pi r^3\Rightarrow r^{\mathrm{\prime }3}=27r^3\Rightarrow r^{\mathrm{\prime }}=3r\mathrm{}$$

Surface area of one small sphere $S=4\pi r^2$. New sphere $S^{\mathrm{\prime }}=4\pi r^{\mathrm{\prime }2}=4\pi (3r{)}^2=9\cdot 4\pi r^2=9S\mathrm{}$

So $S:S^{\mathrm{\prime }}=1:9$

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Q10. A capsule of medicine is a sphere of diameter $3.5\text{ mm}$. How much medicine (in ${\text{mm}}^3$) is needed to fill it?

Diameter $=3.5$ mm ⇒ $r=1.75$ mm $=\frac{7}{4}$ mm.

Volume:

$$V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi {\left(\frac{7}{4}\right)}^3=\frac{4}{3}\pi \cdot \frac{343}{64}=\frac{343}{48}\pi \mathrm{}$$

With $\pi ={\displaystyle \frac{22}{7}}$:

$$V=\frac{343}{48}\times \frac{22}{7}=\frac{539}{24}{\text{mm}}^3\approx 22.4583{\text{mm}}^3\mathrm{.}$$