NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes Exercise 11.4

Exercise 11.4

$\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

1. Find the volume of the sphere whose radius is

(i) $\text{7 cm}$

Ans: It is given the radius of sphere $\text{r = 7 cm}$

The volume of the sphere $\text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{V =}\left[ \text{ }\dfrac{\text{4}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{7} \right)}^{\text{3}}} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\dfrac{\text{4312}}{\text{3}}\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 1437}\text{.33 c}{{\text{m}}^{\text{3}}}$

Therefore, the volume of the sphere is $\text{1437}\text{.33 c}{{\text{m}}^{\text{3}}}$.

(ii) $\text{0}\text{.63 m}$ 

Ans: It is given the radius of sphere $\text{r = 0}\text{.63 m}$

The volume of the sphere $\text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{V =}\left[ \text{ }\dfrac{\text{4}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{0}\text{.63} \right)}^{\text{3}}} \right]\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 1}\text{.0478 }{{\text{m}}^{\text{3}}}$

Therefore, the volume of the sphere is $\text{1}\text{.0478 }{{\text{m}}^{\text{3}}}$.

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

(i) $\text{28 cm}$

Ans: It is given the diameter of ball $\text{= 28 cm}$

So, the radius of ball $\text{r = }\dfrac{\text{28}}{\text{2}}\text{ = 14 cm}$

The volume of the ball $\text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{V =}\left[ \text{ }\dfrac{\text{4}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{14} \right)}^{\text{3}}} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 11498 c}{{\text{m}}^{\text{3}}}$

Therefore, volume of the sphere is $\text{11498 c}{{\text{m}}^{\text{3}}}$.

(ii) $\text{0}\text{.21 m}$ 

Ans: It is given the diameter of ball $\text{= 0}\text{.21 m}$

So, the radius of ball $\text{r = }\dfrac{\text{0}\text{.21}}{\text{2}}\text{ = 0}\text{.105 m}$

The volume of the sphere $\text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{V =}\left[ \text{ }\dfrac{\text{4}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{0}\text{.105} \right)}^{\text{3}}} \right]\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 0}\text{.004851 }{{\text{m}}^{\text{3}}}$

Therefore, the volume of the sphere is $\text{0}\text{.004851 }{{\text{m}}^{\text{3}}}$.

3. The diameter of a metallic ball is $\text{4}\text{.2 cm}$. What is the mass of the ball, if the density of the metal is $\text{8}\text{.9 g per c}{{\text{m}}^{\text{3}}}$? Ans: It is given the diameter of metallic ball $\text{= 4}\text{.2 cm}$

So, the radius of ball $\text{r = }\dfrac{\text{4}\text{.2}}{\text{2}}\text{ = 2}\text{.1 cm}$

The volume of the sphere $\text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{V =}\left[ \text{ }\dfrac{\text{4}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{2}\text{.1} \right)}^{\text{3}}} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 38}\text{.808 c}{{\text{m}}^{\text{3}}}$

We know that $\text{Density = }\dfrac{\text{Mass}}{\text{Volume}}$

$\Rightarrow \text{Mass = Density  }\times\text{  Volume}$

$\Rightarrow \text{Mass = }\left( \text{8}\text{.9  }\times\text{  38}\text{.808} \right)\text{ g}$

$\Rightarrow \text{Mass = 345}\text{.39 g}$

Therefore, the mass of the metallic ball is $\text{345}\text{.39 g}$.

4. 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?

Ans: Let us assume the diameter of the earth is $\text{d}$.

So, the radius of earth will be $\text{R = }\dfrac{\text{d}}{\text{2}}$.

From the question, we can write the diameter of the moon as $\dfrac{\text{d}}{\text{4}}$.

So, the radius of moon will be $\text{r = }\dfrac{\text{d}}{\text{8}}$.

The volume of earth $\text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{R}}^{\text{3}}}$

$\Rightarrow \text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\left( \dfrac{\text{d}}{\text{2}} \right)}^{\text{3}}}$

$\Rightarrow \text{V = }\dfrac{\text{1}}{8}\text{  }\times\text{  }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{d}}^{\text{3}}}$

The volume of moon $\text{{V}' = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{{V}' = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\left( \dfrac{\text{d}}{\text{8}} \right)}^{\text{3}}}$

$\Rightarrow \text{{V}' = }\dfrac{\text{1}}{\text{512}}\text{  }\times\text{  }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{d}}^{\text{3}}}$

The ratio of volume of moon and that of earth $\text{= }\dfrac{\dfrac{\text{1}}{\text{512}}\text{  }\times\text{  }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{d}}^{\text{3}}}}{\dfrac{\text{1}}{\text{8}}\text{  }\times\text{  }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{d}}^{\text{3}}}}\text{ = }\dfrac{\text{1}}{\text{64}}$

So, $\dfrac{\text{Volume of moon}}{\text{Volume of earth}}\text{=}\dfrac{\text{1}}{\text{64}}$

$\Rightarrow \text{Volume of moon = }\left( \dfrac{\text{1}}{\text{64}} \right)\text{ Volume of earth}$

Therefore, the volume of moon is $\dfrac{\text{1}}{\text{64}}$ times the volume of earth.

5. How many litres of milk can a hemispherical bowl of diameter $\text{10}\text{.5 cm}$ can hold?

Ans: It is given the diameter of the hemispherical bowl $\text{= 10}\text{.5 cm}$.

So, the radius of the bowl $\text{r = }\dfrac{\text{10}\text{.5}}{\text{2}}\text{ = 5}\text{.25 cm}$.

The volume of the hemispherical bowl $\text{V = }\dfrac{\text{2}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{V =}\left[ \text{ }\dfrac{\text{2}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{5}\text{.25} \right)}^{\text{3}}} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 303}\text{.1875 c}{{\text{m}}^{\text{3}}}$

We know that $\text{1000 c}{{\text{m}}^{\text{3}}}\text{ = 1 litre}$

So, the capacity of the bowl $\text{= }\dfrac{\text{303}\text{.1875}}{\text{1000}}\text{ = 0}\text{.303 litre}$

Therefore, the volume of the hemispherical bowl is $\text{0}\text{.303 litre}$.

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

Ans: The inner radius of hemispherical tank $\text{r = 1 m}$

The thickness of iron sheet $\text{= 1 cm = 0}\text{.01 m}$.

So, the outer radius of the hemispherical tank $\text{R = }\left( \text{1 + 0}\text{.01} \right)\text{ = 1}\text{.01 m}$

The volume of iron sheet required to make the tank $\text{V = }\dfrac{\text{2}}{\text{3}}\text{ }\pi\text{ }\left( {{\text{R}}^{\text{3}}}\text{ - }{{\text{r}}^{\text{3}}} \right)$

$\Rightarrow \text{V = }\dfrac{\text{2}}{\text{3}}\times \dfrac{22}{7}\times \left( {{\left( 1.01 \right)}^{\text{3}}}\text{ - }{{\left( 1 \right)}^{\text{3}}} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\dfrac{\text{44}}{\text{21}}\text{  }\times\text{  }\left( \text{1}\text{.030301 - 1} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 0}\text{.06348 }{{\text{m}}^{\text{3}}}$

Therefore, the volume of iron sheet required to make the hemispherical tank is $\text{0}\text{.06348 }{{\text{m}}^{\text{3}}}$.

7. Find the volume of a sphere whose surface area is $\text{154 c}{{\text{m}}^{\text{2}}}$. 

Ans: Let us assume the radius of the sphere is $\text{r}$.

It is given the surface area of the sphere $\text{= 154 c}{{\text{m}}^{\text{2}}}$.

$\therefore \text{4 }\pi\text{ }{{\text{r}}^{\text{2}}}\text{ = 154 c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{r}}^{\text{2}}}\text{ = }\left( \dfrac{\text{154  }\times\text{  7}}{\text{4  }\times\text{  22}} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{r}}^{\text{2}}}\text{ = }\left( \dfrac{\text{49}}{\text{4}} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{r = }\left( \dfrac{\text{7}}{\text{2}} \right)\text{ cm}$

The volume of the sphere $\text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{V = }\left[ \dfrac{\text{4}}{\text{3}}\text{  }\text{ }\times\text{ }\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\text{ }\times\text{ }\text{  }{{\left( \dfrac{\text{7}}{\text{2}} \right)}^{\text{3}}} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\left[ \dfrac{\text{49  }\text{ }\times\text{ }\text{  11}}{\text{3}} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 179}\text{.67 c}{{\text{m}}^{\text{3}}}$

Therefore, the volume of the sphere is $\text{179}\text{.67 c}{{\text{m}}^{\text{3}}}$.

8. A dome of a building is in the form of a hemisphere. From inside, it was whitewashed at the cost of $\text{Rs}\text{. 498}\text{.96}$. If the cost of white-washing is $\text{Rs}\text{. 2}\text{.00}$per square meter, find the

(i) Inside surface area of the dome,

Ans: It is given that it costs $\text{Rs}\text{. 2}\text{.00}$ to whitewash an area $\text{= 1 }{{\text{m}}^{\text{2}}}$

So, it costs $\text{Rs}\text{. 498}\text{.96}$ to whitewash an area $\text{= }\dfrac{\text{498}\text{.96}}{\text{2}}\text{ }{{\text{m}}^{\text{2}}}\text{ = 249}\text{.48 }{{\text{m}}^{\text{2}}}$.

Therefore, the inner surface area of the dome is $\text{249}\text{.48 }{{\text{m}}^{\text{2}}}$.

(ii) Volume of the air inside the dome. $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

Ans: Let us assume the radius of the hemispherical dome is $\text{r}$.

We obtained the curved surface area of the inner dome $\text{= 249}\text{.48 }{{\text{m}}^{\text{2}}}$

$\therefore \text{2}{{\text{ }\pi\text{ }}^{\text{2}}}\text{ = 249}\text{.48 }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{2  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\text{r}}^{\text{2}}}\text{ = 249}\text{.48 }{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{r}}^{\text{2}}}\text{ = }\left( \dfrac{\text{249}\text{.48  }\times\text{  7}}{\text{2  }\times\text{  22}} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{r}}^{\text{2}}}\text{ = 39}\text{.69 }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{r = 6}\text{.3 m}$

Volume of hemispherical dome $\text{V = }\dfrac{\text{2}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{V = }\left[ \dfrac{\text{2}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{6}\text{.3} \right)}^{\text{3}}} \right]\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 523}\text{.908 }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 523}\text{.9  }{{\text{m}}^{\text{3}}}\left( \text{approximately} \right)$

Therefore, the volume of air inside the hemispherical dome is $\text{523}\text{.9 }{{\text{m}}^{\text{3}}}$.

9. Twenty-seven solid iron spheres, each of radius $\text{r}$ and surface area $\text{S}$ are melted to form a sphere with surface area $\text{{S}'}$. Find the 

(i) radius $\text{{r}'}$ of the new sphere, 

Ans: It is given the radius of one iron sphere $\text{= r}$.

The volume of one iron sphere $\text{= }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

So, the volume of $\text{27}$ iron spheres $\text{= 27  }\times\text{  }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

These spheres are melted to form one big sphere.

Let us assume the radius of this new sphere is $\text{{r}'}$.

The volume of new iron sphere $\text{= }\dfrac{\text{4}}{\text{3}}\text{  }\pi\text{ }{{\text{{r}'}}^{\text{3}}}$

We can now equate the volumes.

$\Rightarrow \dfrac{\text{4}}{\text{3}}\text{  }\pi\text{ }{{\text{{r}'}}^{\text{3}}}\text{ = 27  }\times\text{  }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow {{\text{{r}'}}^{\text{3}}}\text{ = 27}{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{{r}' = 3r}$

Therefore, the radius of the new sphere is $\text{3r}$.

(ii) ratio of $\text{S}$ and $\text{{S}'}$. 

Ans: The surface area of an iron sphere of $\text{r}$is $\text{S = 4 }\pi\text{ }{{\text{r}}^{\text{2}}}$.

The surface area of an iron sphere of $\text{{r}'}$is $\text{{S}' = 4 }\pi\text{ }{{\text{{r}'}}^{\text{2}}}$.

$\Rightarrow \text{{S}' = 4 }\pi\text{ }{{\left( \text{3r} \right)}^{\text{2}}}$

$\Rightarrow \text{{S}' = 36 }\pi\text{ }{{\text{r}}^{\text{2}}}$

The ratio of $\dfrac{\text{S}}{{\text{{S}'}}}\text{ = }\dfrac{\text{4 }\pi\text{ }{{\text{r}}^{\text{2}}}}{\text{36 }\pi\text{ }{{\text{r}}^{\text{2}}}}\text{ = }\dfrac{\text{1}}{\text{9}}\text{ = 1 : 9}$

Therefore, the required ratio is $\text{1: 9}$.

10. A capsule of medicine is in the shape of a sphere of diameter $\text{3}\text{.5 mm}$. How much medicine $\left( \text{in m}{{\text{m}}^{\text{3}}} \right)$ is needed to fill this capsule? 

Ans: It is given that the diameter of the capsule $\text{= 3}\text{.5 mm}$.

So, the radius will be $\text{r = }\left( \dfrac{\text{3}\text{.5}}{\text{2}} \right)\text{ = 1}\text{.75 mm}$.

Volume of spherical capsule $\text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{V = }\left[ \dfrac{\text{4}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{1}\text{.75} \right)}^{\text{3}}} \right]\text{ m}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 22}\text{.458 m}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 22}\text{.46 m}{{\text{m}}^{\text{3}}}\left( \text{approx} \right)$

Hence, the amount of medicine required to fill the capsule is $\text{22}\text{.46 m}{{\text{m}}^{\text{3}}}$.

Conclusion

NCERT Class 9 Maths Chapter 11 Exercise 11.4 - Surface Areas and Volumes helps you understand the key concepts of calculating surface areas and volumes of cylinders and cones. It is important to focus on the step-by-step solutions provided to understand the methods used. Practicing these problems will enhance your understanding and improve your problem-solving skills. By mastering these exercises, you can perform well in your exams. Vedantu provides these solutions to support your learning journey.

Class 9 Maths Chapter 11: Exercises Breakdown

CBSE Class 9 Maths Chapter 11 Other Study Materials

Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.