NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas And Volumes Ex 11.2

Exercise 11.2

Assume π = $\frac{22}{7}$ , unless stated otherwise.

1. Find the surface area of a sphere of radius:

i. $\text{10}\text{.5 cm}$

Ans:

Given radius of the sphere $\text{r = 10}\text{.5 cm}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\!\!\times\!\!\text{  }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{10}\text{.5} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{88  }\!\!\times\!\!\text{  1}\text{.5  }\!\!\times\!\!\text{  10}\text{.5} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 1386 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{1386 c}{{\text{m}}^{\text{2}}}$.

ii. $\text{5}\text{.6 cm}$

Ans:

Given radius of the sphere $\text{r = 5}\text{.6 cm}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\!\!\times\!\!\text{  }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{5}\text{.6} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{88  }\!\!\times\!\!\text{  0}\text{.8  }\!\!\times\!\!\text{  5}\text{.6} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 394}\text{.24 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{394}\text{.24 c}{{\text{m}}^{\text{2}}}$.

iii. $\text{14 cm}$

Ans:

Given radius of the sphere $\text{r = 14 cm}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\!\!\times\!\!\text{  }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{14} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{4  }\!\!\times\!\!\text{  44  }\!\!\times\!\!\text{  14} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 2464 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{2464 c}{{\text{m}}^{\text{2}}}$.

2. Find the surface area of a sphere of diameter:

i. $\text{14 cm}$

Ans:

Given diameter of the sphere $\text{= 14 cm}$

So, the radius of the sphere $\text{r = }\frac{\text{14}}{\text{2}}\text{ = 7 cm}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\!\!\times\!\!\text{  }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{7} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{88  }\!\!\times\!\!\text{  7} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 616 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{616 c}{{\text{m}}^{\text{2}}}$.

ii. $\text{21 cm}$

Ans:

Given diameter of the sphere $\text{= 21 cm}$

So, the radius of the sphere $\text{r = }\frac{\text{21}}{\text{2}}\text{ = 10}\text{.5 cm}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\!\!\times\!\!\text{  }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{10}\text{.5} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 1386 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{1386 c}{{\text{m}}^{\text{2}}}$.

iii. $\text{3}\text{.5 m}$

Ans:

Given diameter of the sphere $\text{= 3}\text{.5 m}$

So, the radius of the sphere $\text{r = }\frac{\text{3}\text{.5}}{\text{2}}\text{ = 1}\text{.75 m}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\!\!\times\!\!\text{  }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{1}\text{.75} \right)}^{\text{2}}} \right]\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 38}\text{.5 }{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{38}\text{.5 }{{\text{m}}^{\text{2}}}$.

3. Find the total surface area of a hemisphere of radius $\text{10 cm}$. $\left[ \text{Use  }\!\!\pi\!\!\text{  = 3}\text{.14} \right]$

Ans:

Given the radius of hemisphere $\text{r = 10 cm}$

The total surface area of the hemisphere is the sum of its curved surface area and the circular base.

Total surface area of hemisphere $\text{A = 2 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{ +  }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = 3 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{3  }\!\!\times\!\!\text{  3}\text{.14  }\!\!\times\!\!\text{  }{{\left( \text{10} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 942 c}{{\text{m}}^{\text{2}}}$

Hence, the total surface area of the hemisphere is $\text{942 c}{{\text{m}}^{\text{2}}}$.

4. The radius of a spherical balloon increases from $\text{7 cm}$ to $\text{14 cm}$ as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Ans:

Given the initial radius of the balloon ${{\text{r}}_{1}}\text{ = 7 cm}$

The final radius of the balloon ${{\text{r}}_{2}}\text{ = 14 cm}$

We have to find the ratio of surface areas of the balloon in the two cases.

The required ratio $\text{R = }\frac{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}_{\text{1}}}^{\text{2}}}{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}_{\text{2}}}^{\text{2}}}$

$\Rightarrow \text{R = }{{\left( \frac{{{\text{r}}_{\text{1}}}}{{{\text{r}}_{\text{2}}}} \right)}^{\text{2}}}$

$\Rightarrow \text{R = }{{\left( \frac{\text{7}}{\text{14}} \right)}^{\text{2}}}$

$\Rightarrow \text{R = }\frac{\text{1}}{\text{4}}$

Hence, the ratio of the surface areas of the balloon in both case is $\text{1 : 4}$.

5. A hemispherical bowl made of brass has inner diameter $\text{10}\text{.5 cm}$. Find the cost of tin plating it on the inside at the rate of $\text{Rs}\text{. 16}$ per $\text{100 c}{{\text{m}}^{\text{2}}}$.

Ans:

Given the radius of inner hemispherical bowl $\text{r = }\frac{\text{10}\text{.5}}{\text{2}}\text{ = 5}\text{.25 cm}$

The surface area of the hemispherical bowl $\text{A = 2 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{2  }\!\!\times\!\!\text{  }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{5}\text{.25} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 173}\text{.25 c}{{\text{m}}^{\text{2}}}$

It is given that the cost of tin-plating $\text{100 c}{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 16}$

So, the cost of tin-plating $173.25\text{ c}{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. }\left( \frac{\text{16}}{\text{100}}\text{  }\!\!\times\!\!\text{  173}\text{.25} \right)\text{ = Rs}\text{. 27}\text{.72}$

Hence, the cost of tin-plating the hemispherical bowl is $\text{Rs}\text{. 27}\text{.72}$.

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

Ans:

Let us assume the radius of the sphere be $\text{r}$.

We are given the surface area of the sphere, $\text{A = 154 c}{{\text{m}}^{\text{2}}}$.

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

$\Rightarrow r^{2}=\left(\frac{154 \times 7}{2 \times 22}\right) \mathrm{cm}^{2}$

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

$\Rightarrow \text{r = 3}\text{.5 cm}$

Therefore, the radius of the sphere is $\text{3}\text{.5 cm}$.

7. The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface area.

Ans:

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

So, the diameter of the moon will be $\frac{\text{d}}{\text{4}}$.

The radius of the earth ${{\text{r}}_{\text{1}}}\text{ = }\frac{\text{d}}{\text{2}}$

The radius of the moon ${{\text{r}}_{\text{2}}}\text{ = }\frac{\text{1}}{\text{2}}\text{  }\!\!\times\!\!\text{  }\frac{\text{d}}{\text{2}}\text{ = }\frac{\text{d}}{\text{8}}$

The ratio of surface area of moon and earth $\text{R = }\frac{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}_{\text{2}}}^{\text{2}}}{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}_{\text{1}}}^{\text{2}}}$

$\Rightarrow \text{R = }\frac{\text{4 }\!\!\pi\!\!\text{ }{{\left( \frac{\text{d}}{\text{8}} \right)}^{\text{2}}}}{\text{4 }\!\!\pi\!\!\text{ }{{\left( \frac{\text{d}}{\text{2}} \right)}^{\text{2}}}}$

$\Rightarrow \text{R = }\frac{\text{4}}{\text{64}}$

$\Rightarrow \text{R = }\frac{\text{1}}{\text{16}}$

Therefore, the ratio of surface area of the moon and earth is $\text{1 : 16}$.

8. A hemispherical bowl is made of steel, $\text{0}\text{.25 cm}$ thick. The inner radius of the bowl is $\text{5 cm}$. Find the outer curved surface area of the bowl.

Ans:

Given the inner radius $\text{= 5 cm}$

The thickness of the bowl $\text{= 0}\text{.25 cm}$

So, the outer radius of the hemispherical bowl is $\text{r = }\left( \text{5 + 0}\text{.25} \right)\text{ cm = 5}\text{.25 cm}$

The outer curved surface area of the hemispherical bowl $\text{A = 2 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A =}\left[ \text{ 2  }\!\!\times\!\!\text{  }\frac{\text{2}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{5}\text{.25} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 173}\text{.25 c}{{\text{m}}^{\text{2}}}$

Therefore, the outer curved surface area of the hemispherical bowl is $\text{173}\text{.25 c}{{\text{m}}^{\text{2}}}$.

9. A right circular cylinder just encloses a sphere of radius $\text{r}$ (see figure). Find

i. surface area of the sphere, 

Ans:

The surface area of the sphere is $\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$.

ii. curved surface area of the cylinder, 

Ans:

Given the radius of cylinder $\text{= r}$

The height of cylinder $\text{= r + r = 2r}$

The curved surface area of cylinder $\text{A = 2 }\!\!\pi\!\!\text{ rh}$

$\Rightarrow \text{A = 2 }\!\!\pi\!\!\text{ r }\left( \text{2r} \right)$

$\Rightarrow \text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

Therefore the curved surface area of the cylinder is  $\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$.

iii. ratio of the areas obtained in i. and ii.

The ratio of surface area of the sphere and curved surface area of cylinder  $\text{R = }\frac{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}}{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}}$

$\text{R = }\frac{\text{1}}{\text{1}}$

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

Conclusion

NCERT Solutions for Maths Exercise 11.2 Class 9 Chapter 11 - Surface Areas and Volumes by Vedantu provides clear and simple explanations for calculating the surface areas and volumes of various shapes like cuboids, cubes, cylinders, cones, and spheres. These solutions follow the latest NCERT guidelines and help students understand and apply important geometry concepts. By practicing these problems, students can strengthen their math skills and perform better in exams. Download the solutions in PDF format for easy access and effective learning.

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.

Important Study Materials for Class 9 Maths