NCERT Solutions for Class 9 Maths Chapter 11 Surface Area and Volume

• Exercise 11.1: This exercise explains the surface area and volume of a cylinder. This section helps the students to understand how to find the surface area of a right circular cylinder. We need to split the cylinder into two parts: the middle part and the bottom part.

The curved surface area of cylinder= 2πrh

Total surface area of cylinder = πr2(r+h)

• Exercise 11.2: This exercise explains the surface area and volume of a cone.

This section explains about the cone which has a triangle shape and circle shape. By combining these two areas, Class 9 Surfaces Area and Volumes, the formula of the cone is derived as below,

Curved Surface Area of a Cone = ½  × l ×2πr = πrl

Total Surface Area of a Cone = πrl + πr2 = πr (l + r)

• Exercise 11.3: This exercise includes the conversion of one solid from one shape to another. This covers the volume of a cylinder. 

Volume of a Cylinder = πr2h

• Exercise 11.4: This exercise contains word problems that are based on the surface area and volume of 3D shapes. This understands the difference between cylinder and cone. From that, they derived a formula for the volume of a cone is ⅓ πr2h.

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

Exercise (11.1)

1. Diameter of the base of a cone is $\text{10}\text{.5 cm}$ and its slant height is $\text{10 cm}$. Find its curved surface area. $[\text{Assume }\pi ={\displaystyle \frac{\text{22}}{\text{7}}}]$

Ans: We are given the following:

The slant height $\left(\text{l}\right)$ of the cone $\text{= 10 cm}$

The diameter of the base of cone $\text{= 10}\text{.5 cm}$

So, the radius $\left(\text{r}\right)$ of the base of cone $\text{= }{\displaystyle \frac{\text{10}\text{.5}}{\text{2}}}\text{ cm = 5}\text{.25 cm}$

The curved surface area of cone, $\text{A = }\pi \text{ rl}$

$\Rightarrow \text{A = }({\displaystyle \frac{\text{22}}{\text{7}}}\times \text{ 5}\text{.25 }\times \text{ 10})\text{ c}{\text{m}}^{\text{2}}$

$\Rightarrow \text{A = }(\text{22 }\times \text{ 0}\text{.75 }\times \text{ 10})\text{ c}{\text{m}}^{\text{2}}$

$\Rightarrow \text{A = 165 c}{\text{m}}^{\text{2}}$

Therefore, the curved surface area of the cone is $\text{165 c}{\text{m}}^{\text{2}}$.

2. Find the total surface area of a cone, if its slant height is $\text{21 m}$and diameter of its base is $\text{24 m}$. $[\text{Assume }\pi ={\displaystyle \frac{\text{22}}{\text{7}}}]$

Ans: We are given the following:

The slant height $\left(\text{l}\right)$ of the cone $\text{= 21 m}$

The diameter of the base of cone $\text{= 24 m}$

So, the radius $\left(\text{r}\right)$ of the base of cone $\text{= }{\displaystyle \frac{\text{24}}{\text{2}}}\text{ m = 12 m}$

The total surface area of cone, $\text{A = }\pi \text{ r}\left(\text{l + r}\right)$

$\Rightarrow \text{A = }({\displaystyle \frac{\text{22}}{\text{7}}}\times \text{ 12 }\times \left(\text{21 + 12}\right)){\text{m}}^{\text{2}}$

$\Rightarrow \text{A = }({\displaystyle \frac{\text{22}}{\text{7}}}\times \text{ 12 }\times \text{ 33}){\text{m}}^{\text{2}}$

$\Rightarrow \text{A = 1244}\text{.57 }{\text{m}}^{\text{2}}$

Therefore, the total surface area of the cone is $\text{1244}\text{.57 }{\text{m}}^{\text{2}}$.

3. Curved surface area of a cone is $\text{308 c}{\text{m}}^{\text{2}}$ and its slant height is $\text{14 cm}$. Find

(i) Radius of the base and

Ans: It is given that the slant height $\left(\text{l}\right)$ of the cone $\text{= 14 cm}$

The curved surface area of the cone $\text{= 308 c}{\text{m}}^{\text{2}}$

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

We know that curved surface area of the cone $\text{= }\pi \text{ rl}$

$\therefore \pi \text{ rl = 308 c}{\text{m}}^{\text{2}}$

$\Rightarrow ({\displaystyle \frac{\text{22}}{\text{7}}}\times \text{ r }\times \text{ 14})\text{ cm = 308 c}{\text{m}}^{\text{2}}$

$\Rightarrow \text{r = }{\displaystyle \frac{\text{308}}{\text{44}}}\text{ cm}$

$\Rightarrow \text{r = 7 cm}$

Hence, the radius of the base is $\text{7 cm}$.

(ii) Total surface area of the cone. $[\text{Assume }\pi ={\displaystyle \frac{\text{22}}{\text{7}}}]$

Ans: The total surface area of the cone is the sum of its curved surface area and the area of the base.

Total surface area of cone, $\text{A = }\pi \text{ rl + }\pi {\text{r}}^{\text{2}}$

$\Rightarrow \text{A = }[\text{308 + }{\displaystyle \frac{\text{22}}{\text{7}}}\times {\left(\text{7}\right)}^{\text{2}}]\text{ c}{\text{m}}^{\text{2}}$

$\Rightarrow \text{A = }\left[\text{308 + 154}\right]\text{ c}{\text{m}}^{\text{2}}$

$\Rightarrow \text{A = 462 c}{\text{m}}^{\text{2}}$

Hence, the total surface area of the cone is $\text{462 c}{\text{m}}^{\text{2}}$.

4. A conical tent is $\text{10 m}$ high and the radius of its base is $\text{24 m}$. Find

(i) slant height of the tent

Ans:

From the figure we can say that $\text{ABC}$ is a conical tent.

It is given that the height $\left(\text{h}\right)$ of conical tent $\text{= 10 m}$

The radius $\left(\text{r}\right)$ of conical tent $\text{= 24 m}$

Let us assume the slant height as $\text{l}$.

In $\mathrm{\Delta }\text{ ABD}$, we will use Pythagorean Theorem.

$\therefore \text{A}{\text{B}}^{\text{2}}\text{ = AD}{}^{\text{2}}\text{ + B}{\text{D}}^{\text{2}}$

$\Rightarrow {\text{l}}^{\text{2}}\text{ = }{\text{h}}^{\text{2}}\text{ + }{\text{r}}^{\text{2}}$

$$\Rightarrow {\text{l}}^{\text{2}}\text{ = }{\left(\text{10 m}\right)}^{\text{2}}\text{ + }{\left(\text{24 m}\right)}^{\text{2}}$$

$\Rightarrow {\text{l}}^{\text{2}}\text{ = 676 }{\text{m}}^{\text{2}}$

$\Rightarrow \text{l = 26 m}$

The slant height of the tent is $\text{26 m}$.

(ii) cost of canvas required to make the tent, if cost of $\text{1 }{\text{m}}^{\text{2}}$ canvas is $\text{Rs}\text{. 70}$. $[\text{Assume }\pi ={\displaystyle \frac{\text{22}}{\text{7}}}]$

Ans: The curved surface area of the tent, $\text{A = }\pi \text{ rl}$

$\Rightarrow \text{A = }({\displaystyle \frac{\text{22}}{\text{7}}}\times \text{ 24 }\times \text{ 26}){\text{m}}^{\text{2}}$

$\Rightarrow \text{A = }\left({\displaystyle \frac{13728}{7}}\right){\text{m}}^{\text{2}}$

It is given that the cost of $\text{1 }{\text{m}}^{\text{2}}$ of canvas $\text{= Rs}\text{. 70}$

So, the cost of ${\displaystyle \frac{13728}{7}}{\text{m}}^{\text{2}}$ canvas $\text{= Rs}\text{. }({\displaystyle \frac{\text{13728}}{\text{7}}}\times \text{ 70})\text{ = Rs}\text{. 137280}$

Hence, the cost of canvas required to make the tent is $\text{Rs}\text{. 137280}$.

5. What length of tarpaulin $\text{3 m}$ wide will be required to make conical tent of height $\text{8 m}$ and base radius $\text{6 m}$? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately $\text{20 cm}$. $\left[\text{Use }\pi \text{ = 3}\text{.14}\right]$

Ans: We are given the following:

The base radius $\left(\text{r}\right)$ of tent $\text{= 6 m}$

The height $\left(\text{h}\right)$ of tent $\text{= 8 m}$

So the slant height of the tent, $\text{l = }\sqrt{{\text{r}}^{\text{2}}\text{ + }{\text{h}}^{\text{2}}}$ 

$\Rightarrow \text{l = }\left(\sqrt{{\text{6}}^{\text{2}}\text{ + }{\text{8}}^{\text{2}}}\right)\text{ m}$

$\Rightarrow \text{l = }\left(\sqrt{\text{100}}\right)\text{ m}$

$\Rightarrow \text{l = 10 m}$

The curved surface area of the tent, $\text{A = }\pi \text{ rl}$

$\Rightarrow \text{A = }(\text{3}\text{.14 }\times \text{ 6 }\times \text{ 10}){\text{m}}^{\text{2}}$

$\Rightarrow \text{A = 188}\text{.4 }{\text{m}}^{\text{2}}$

It is give the width of tarpaulin $\text{= 3 m}$

Let us assume the length of the tarpaulin sheet required be $\text{x}$.

It is given that there will be a wastage of $\text{20 cm}$.

So, the new length of the sheet $\text{=}\left(\text{x - 0}\text{.2}\right)\text{ m}$

We know that the area of the rectangular sheet required will be the same as curved surface area of the tent.

$\therefore [\left(\text{x - 0}\text{.2}\right)\times \text{ 3}]\text{ m = 188}\text{.4 }{\text{m}}^{\text{2}}$

$\Rightarrow \text{x - 0}\text{.2 m = 62}\text{.8 m}$

$\Rightarrow \text{x = 63 m}$

The length of tarpaulin sheet required is $\text{63 m}$.

6. The slant height and base diameter of a conical tomb are $\text{25 m}$ and $\text{14 m}$ respectively. Find the cost of white-washing its curved surface at the rate of $\text{Rs}\text{. 210}$ per $\text{100 }{\text{m}}^{\text{2}}$. $[\text{Assume }\pi ={\displaystyle \frac{\text{22}}{\text{7}}}]$

Ans: We are given the following:

The base radius $\left(\text{r}\right)$ of tomb $\text{= 7 m}$

The slant height $\left(\text{l}\right)$ of tomb $\text{= 25 m}$

The curved surface area of the conical tomb, $\text{A = }\pi \text{ rl}$

$\Rightarrow \text{A = }({\displaystyle \frac{22}{7}}\times \text{ 7 }\times \text{ 25}){\text{m}}^{\text{2}}$

$\Rightarrow \text{A = 550 }{\text{m}}^{\text{2}}$

It is given that the cost of white-washing $\text{1 }{\text{m}}^{\text{2}}$ area $\text{= Rs}\text{. 210}$

So, the cost of white-washing $550{\text{m}}^{\text{2}}$ area $\text{= Rs}\text{. }({\displaystyle \frac{\text{210}}{\text{100}}}\times \text{ 550})\text{ = Rs}\text{. 1155}$

Hence, the cost of white-washing the curved surface area of a conical tomb is $\text{Rs}\text{. 1155}$.

7. A joker’s cap is in the form of right circular cone of base radius $\text{7 cm}$ and the height $\text{24 cm}$. Find the area of sheet required to make $\text{10}$ such caps. $[\text{Assume }\pi ={\displaystyle \frac{\text{22}}{\text{7}}}]$

Ans: We are given the following:

The base radius $\left(\text{r}\right)$ of conical cap $\text{= 7 cm}$

The height $\left(\text{h}\right)$ of conical cap $\text{= 24 cm}$

So the slant height of the tent, $\text{l = }\sqrt{{\text{r}}^{\text{2}}\text{ + }{\text{h}}^{\text{2}}}$ 

$\Rightarrow \text{l = }\left(\sqrt{{\text{7}}^{\text{2}}\text{ + 2}{\text{4}}^{\text{2}}}\right)\text{ cm}$

$\Rightarrow \text{l = }\left(\sqrt{625}\right)\text{ cm}$

$\Rightarrow \text{l = 25 cm}$

The curved surface area of one conical cap, $\text{A = }\pi \text{ rl}$

$\Rightarrow \text{A = }({\displaystyle \frac{22}{7}}\times \text{ 7 }\times \text{ 25})\text{ c}{\text{m}}^{\text{2}}$

$\Rightarrow \text{A = 550 c}{\text{m}}^{\text{2}}$

So, the curved surface area of $\text{10}$ conical caps $\text{= }(\text{550 }\times \text{ 10})\text{ c}{\text{m}}^{\text{2}}\text{ = 5500 c}{\text{m}}^{\text{2}}$

Therefore, the total area of sheet required is $\text{5500 c}{\text{m}}^{\text{2}}$.

8. A bus stop is barricaded from the remaining part of the road, by using $\text{50}$ hollow cones made of recycled cardboard. Each cone has a base diameter of $\text{40 cm}$ and height $\text{1 m}$. If the outer side of each of the cones is to be painted and the cost of painting is $\text{Rs}\text{. 12}$ per ${\text{m}}^{\text{2}}$, what will be the cost of painting all these cones? 

$\left[\text{Use }\pi \text{ = 3}\text{.14 and take }\sqrt{\text{1}\text{.02}}\text{=1}\text{.02}\right]$

Ans: We are given the following:

The base radius $\left(\text{r}\right)$ of cone $\text{= }{\displaystyle \frac{\text{40}}{\text{2}}}\text{ = 20 cm = 0}\text{.2 m}$

The height $\left(\text{h}\right)$ of cone $\text{= 1 m}$

So the slant height of the cone, $\text{l = }\sqrt{{\text{r}}^{\text{2}}\text{ + }{\text{h}}^{\text{2}}}$ 

$\Rightarrow \text{l = }\left(\sqrt{{\left(\text{0}\text{.2}\right)}^{\text{2}}\text{ + }{\left(\text{1}\right)}^{\text{2}}}\right)\text{ m}$

$\Rightarrow \text{l = }\left(\sqrt{1.04}\right)\text{ m}$

$\Rightarrow \text{l = 1}\text{.02 m}$

The curved surface area of one cone, $\text{A = }\pi \text{ rl}$

$\Rightarrow \text{A = }(\text{3}\text{.14 }\times \text{ 0}\text{.2 }\times \text{ 1}\text{.02}){\text{m}}^{\text{2}}$

$\Rightarrow \text{A = 0}\text{.64056 c}{\text{m}}^{\text{2}}$

So, the curved surface area of $\text{50}$ cones $\text{= }(\text{50 }\times \text{ 0}\text{.64056}){\text{m}}^{\text{2}}\text{ = 32}\text{.028 }{\text{m}}^{\text{2}}$

It is given that the cost of painting $\text{1 }{\text{m}}^{\text{2}}$ area $\text{= Rs}\text{. 12}$

So, the cost of painting $32.028{\text{m}}^{\text{2}}$ area $\text{= Rs}\text{. }(\text{32}\text{.028 }\times \text{ 12})\text{ = Rs}\text{. 384}\text{.336}$

We can also write the cost approximately as $\text{Rs}\text{. 384}\text{.34}$.

Therefore, the cost of painting all the hollow cones is $\text{Rs}\text{. 384}\text{.34}$.

Exercise (11.2)

1. Find the surface area of a sphere of radius: $[\text{Assume }\pi ={\displaystyle \frac{\text{22}}{\text{7}}}]$

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

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

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

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

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

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

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

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

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

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

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

(iii) $\text{14 cm}$ $[\text{Assume }\pi ={\displaystyle \frac{\text{22}}{\text{7}}}]$

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

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

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

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

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

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: $[\text{Assume }\pi ={\displaystyle \frac{\text{22}}{\text{7}}}]$:

(i) $\text{14 cm}$

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

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

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

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

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

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

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 $\text{r = }{\displaystyle \frac{\text{21}}{\text{2}}}\text{ = 10}\text{.5 cm}$

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

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

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

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

(iii) $\text{3}\text{.5 m}$ $[\text{Assume }\pi ={\displaystyle \frac{\text{22}}{\text{7}}}]$

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

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

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

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

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

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 $\text{A = 2 }\pi {\text{r}}^{\text{2}}\text{ + }\pi {\text{r}}^{\text{2}}$

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

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

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

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 ${\text{r}}_1\text{ = 10 cm}$

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 $\text{R = }{\displaystyle \frac{\text{4 }\pi {{\text{r}}_{\text{1}}}^{\text{2}}}{\text{4 }\pi {{\text{r}}_{\text{2}}}^{\text{2}}}}$

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

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

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

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 tinplating it on the inside at the rate of $\text{Rs}\text{. 16}$ per $\text{100 c}{\text{m}}^{\text{2}}$. $[\text{Assume }\pi ={\displaystyle \frac{\text{22}}{\text{7}}}]$

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

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

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

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

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 $\text{= Rs}\text{. }({\displaystyle \frac{\text{16}}{\text{100}}}\times \text{ 173}\text{.25})\text{ = Rs}\text{. 27}\text{.72}$

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 $\text{154 c}{\text{m}}^{\text{2}}$. $[\text{Assume }\pi ={\displaystyle \frac{\text{22}}{\text{7}}}]$

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

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

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

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

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

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

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 ${\displaystyle \frac{\text{d}}{\text{4}}}$.

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

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

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

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

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

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

Therefore, the ratio of surface area of 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. $[\text{Assume }\pi ={\displaystyle \frac{\text{22}}{\text{7}}}]$

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 $\text{A = 2 }\pi {\text{r}}^{\text{2}}$

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

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

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 $\text{4 }\pi {\text{r}}^{\text{2}}$.

(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 $\text{A = 2 }\pi \text{ rh}$

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

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

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

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

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

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

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

Exercise (11.3)

1. Find the volume of the right circular cone with

(i) Radius $\text{6 cm}$, height $\text{7 cm}$ $\left[\text{Assume }\pi \text{ =}{\displaystyle \frac{\text{22}}{\text{7}}}\right]$

Ans: It is given the radius of cone $\text{r = 6 cm}$

The height of the cone $\text{h = 7 cm}$

The volume of the cone $\text{V = }{\displaystyle \frac{\text{1}}{\text{3}}}\pi {\text{r}}^{\text{2}}\text{h}$

$\Rightarrow \text{V = }[{\displaystyle \frac{\text{1}}{\text{3}}}\times {\displaystyle \frac{\text{22}}{\text{7}}}\times {\left(\text{6}\right)}^{\text{2}}\times \text{ 7}]\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = }(\text{12 }\times \text{ 22})\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = 264 c}{\text{m}}^{\text{3}}$

The volume of the right circular cone is $\text{264 c}{\text{m}}^{\text{3}}$.

(ii) Radius $\text{3}\text{.5 cm}$, height $\text{12 cm}$ $\left[\text{Assume }\pi \text{ =}{\displaystyle \frac{\text{22}}{\text{7}}}\right]$

Ans: It is given the radius of cone $\text{r = 3}\text{.5 cm}$

The height of the cone $\text{h = 12 cm}$

The volume of the cone $\text{V = }{\displaystyle \frac{\text{1}}{\text{3}}}\pi {\text{r}}^{\text{2}}\text{h}$

$\Rightarrow \text{V = }[{\displaystyle \frac{\text{1}}{\text{3}}}\times {\displaystyle \frac{\text{22}}{\text{7}}}\times {\left(\text{3}\text{.5}\right)}^{\text{2}}\times \text{ 12}]\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = }(\text{1}\text{.75 }\times \text{ 88})\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = 154 c}{\text{m}}^{\text{3}}$

The volume of the right circular cone is $\text{154 c}{\text{m}}^{\text{3}}$.

2. $\left[\text{Assume }\pi \text{ =}{\displaystyle \frac{\text{22}}{\text{7}}}\right]$ Find the capacity in litres of a conical vessel with

(i) Radius $\text{7 cm}$, slant height $\text{25 cm}$

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

The slant height of the cone $\text{l = 25 cm}$

So, the height of the cone $\text{h = }\sqrt{{\text{l}}^{\text{2}}\text{ - }{\text{r}}^{\text{2}}}$

$\Rightarrow \text{h = }\sqrt{\text{2}{\text{5}}^{\text{2}}\text{ - }{\text{7}}^{\text{2}}}\text{ cm}$

$\Rightarrow \text{h = 24 cm}$

The volume of the cone $\text{V = }{\displaystyle \frac{\text{1}}{\text{3}}}\pi {\text{r}}^{\text{2}}\text{h}$

$\Rightarrow \text{V = }[{\displaystyle \frac{\text{1}}{\text{3}}}\times {\displaystyle \frac{\text{22}}{\text{7}}}\times {\left(\text{7}\right)}^{\text{2}}\times \text{ 24}]\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = }(\text{154 }\times \text{ 8})\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = 1232 c}{\text{m}}^{\text{3}}$

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

So, the capacity of the conical vessel $\text{= }{\displaystyle \frac{\text{1232}}{\text{1000}}}\text{ = 1}\text{.232 litres}$

Therefore, the capacity of the conical vessel is $\text{1}\text{.232 litres}$.

(ii) height $\text{12 cm}$, slant height $\text{13 cm}$ $\left[\text{Assume }\pi \text{ =}{\displaystyle \frac{\text{22}}{\text{7}}}\right]$

Ans: It is given the height of cone $\text{h = 12 cm}$

The slant height of the cone $\text{l = 13 cm}$

So, the radius of the cone $\text{r = }\sqrt{{\text{l}}^{\text{2}}\text{ - }{\text{h}}^{\text{2}}}$

$\Rightarrow \text{r = }\sqrt{\text{1}{\text{3}}^{\text{2}}\text{ - 1}{\text{2}}^{\text{2}}}\text{ cm}$

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

The volume of the cone $\text{V = }{\displaystyle \frac{\text{1}}{\text{3}}}\pi {\text{r}}^{\text{2}}\text{h}$

$\Rightarrow \text{V = }[{\displaystyle \frac{\text{1}}{\text{3}}}\times {\displaystyle \frac{\text{22}}{\text{7}}}\times {\left(\text{5}\right)}^{\text{2}}\times \text{ 12}]\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = }(\text{4 }\times {\displaystyle \frac{\text{22}}{\text{7}}}\times \text{ 25})\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = }{\displaystyle \frac{2200}{7}}\text{ c}{\text{m}}^{\text{3}}$

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

So, the capacity of the conical vessel $\text{= }{\displaystyle \frac{\text{2200}}{\text{7}}}\times {\displaystyle \frac{\text{1}}{\text{1000}}}\text{ = 0}\text{.314 litres}$

Therefore, the capacity of the conical vessel is $\text{0}\text{.314 litres}$.

3. The height of a cone is $\text{15 cm}$. It its volume is $\text{1570 c}{\text{m}}^{\text{3}}$, find the diameter of its base. $\left[\text{Use }\pi \text{ = 3}\text{.14}\right]$

Ans: It is given the height of cone $\text{h = 12 cm}$

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

The volume of the cone is $\text{V = 1570 c}{\text{m}}^{\text{3}}$

We know the formula for the volume of the cone $\text{= }{\displaystyle \frac{\text{1}}{\text{3}}}\pi {\text{r}}^{\text{2}}\text{h}$

$\therefore {\displaystyle \frac{\text{1}}{\text{3}}}\pi {\text{r}}^{\text{2}}\text{h = 1570 c}{\text{m}}^{\text{3}}$

$\Rightarrow [{\displaystyle \frac{\text{1}}{\text{3}}}\times {\displaystyle \frac{\text{22}}{\text{7}}}\times {\left(\text{r}\right)}^{\text{2}}\times \text{ 12}]\text{ cm = 1570 c}{\text{m}}^{\text{3}}$

$\Rightarrow {\text{r}}^{\text{2}}\text{ = 100 c}{\text{m}}^{\text{2}}$

$$\Rightarrow \text{r = 10 cm}$$

Diameter of base $$\text{= 2r = 20 cm}$$

Therefore, the diameter of the cone is  $$\text{20 cm}$$.

4. If the volume of right circular cone of height $\text{9 cm}$ is $\text{48 }\pi \text{ c}{\text{m}}^{\text{3}}$, find the diameter of its base.

Ans: It is given the height of cone $\text{h = 9 cm}$

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

The volume of the cone is $\text{V = 48 }\pi \text{ c}{\text{m}}^{\text{3}}$

We know the formula for the volume of the cone $\text{= }{\displaystyle \frac{\text{1}}{\text{3}}}\pi {\text{r}}^{\text{2}}\text{h}$

$\therefore {\displaystyle \frac{\text{1}}{\text{3}}}\pi {\text{r}}^{\text{2}}\text{h = 48 }\pi \text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow [{\displaystyle \frac{\text{1}}{\text{3}}}\times \pi \times {\left(\text{r}\right)}^{\text{2}}\times \text{ 9}]\text{ cm = 48 }\pi \text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow {\text{r}}^{\text{2}}\text{ = 16 c}{\text{m}}^{\text{2}}$

$$\Rightarrow \text{r = 4 cm}$$

Diameter of base $$\text{= 2r = 8 cm}$$

Therefore, the diameter of the base of the cone is $$\text{8 cm}$$.

5. A conical pit of top diameter $\text{3}\text{.5 m}$ is $\text{12 m}$ deep. What is the capacity in kilolitres? $\left[\text{Assume }\pi \text{ =}{\displaystyle \frac{\text{22}}{\text{7}}}\right]$

Ans: It is given the height of conical pit $\text{h = 12 m}$

The radius of conical pit $\text{r = }{\displaystyle \frac{\text{3}\text{.5}}{\text{2}}}\text{ m = 1}\text{.75 m}$

We know the volume of the conical pit $\text{V = }{\displaystyle \frac{\text{1}}{\text{3}}}\pi {\text{r}}^{\text{2}}\text{h}$

$\Rightarrow \text{V = }[{\displaystyle \frac{\text{1}}{\text{3}}}\times {\displaystyle \frac{\text{22}}{\text{7}}}\times {\left(\text{1}\text{.75}\right)}^{\text{2}}\times \text{ 12}]{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = 38}\text{.5 }{\text{m}}^{\text{3}}$

We know that $\text{1 kilolitre = 1 }{\text{m}}^{\text{3}}$

So, the capacity of the pit $\text{= }(\text{38}\text{.5 }\times \text{ 1})\text{ kilolitres = 38}\text{.5 kilolitres}$

Therefore, the capacity of the conical pit is $\text{38}\text{.5 kilolitres}$.

6. The volume of a right circular cone is $\text{9856 c}{\text{m}}^{\text{3}}$. If the diameter of the base is $\text{28 cm}$, find

(i) Height of the cone

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

So, the radius $\text{r = }{\displaystyle \frac{\text{28}}{\text{2}}}\text{ = 14 cm}$

Let us assume the height of the cone be $\text{h}$.

The volume of the cone is $\text{V = 9856 c}{\text{m}}^{\text{3}}$

We know the formula for the volume of the cone $\text{= }{\displaystyle \frac{\text{1}}{\text{3}}}\pi {\text{r}}^{\text{2}}\text{h}$

$\therefore {\displaystyle \frac{\text{1}}{\text{3}}}\pi {\text{r}}^{\text{2}}\text{h = 9856 c}{\text{m}}^{\text{3}}$

$\Rightarrow [{\displaystyle \frac{\text{1}}{\text{3}}}\times {\displaystyle \frac{22}{7}}\times {\left(\text{14}\right)}^{\text{2}}\times \text{ h}]\text{ c}{\text{m}}^{\text{2}}\text{ = 9856 c}{\text{m}}^{\text{3}}$

$$\Rightarrow \text{h = }\left({\displaystyle \frac{\text{9856 }\times \text{ 21}}{\text{22 }\times \text{ 196}}}\right)\text{ cm}$$

$$\Rightarrow \text{h = 48 cm}$$

Therefore, the height of the cone is $$\text{48 cm}$$.

(ii) Slant height of the cone

Ans: The slant height of the cone $\text{l = }\sqrt{{\text{h}}^{\text{2}}\text{ + }{\text{r}}^{\text{2}}}$

$\Rightarrow \text{l = }\sqrt{\text{4}{\text{8}}^{\text{2}}\text{ + 1}{\text{4}}^{\text{2}}}\text{ cm}$

$\Rightarrow \text{l = }\sqrt{\text{2304 + 196}}\text{ cm}$

$\Rightarrow \text{l = 50 cm}$

Therefore, the slant height of the cone is $\text{50 cm}$.

(iii) Curved surface area of the cone. $\left[\text{Assume }\pi \text{ =}{\displaystyle \frac{\text{22}}{\text{7}}}\right]$

Ans: The curved surface area of the cone $\text{A = }\pi \text{ rl}$

$\Rightarrow \text{A = }({\displaystyle \frac{\text{22}}{\text{7}}}\times \text{ 14 }\times \text{ 50})\text{ c}{\text{m}}^{\text{2}}$

$\Rightarrow \text{A = 2200 c}{\text{m}}^{\text{2}}$

Therefore, the curved surface area of the cone is $\text{2200 c}{\text{m}}^{\text{2}}$.

7. A right triangle $\mathrm{\Delta }\text{ ABC}$ with sides $\text{5 cm}$,$\text{12 cm}$ and $$\text{13 cm}$$ is revolved about the side $\text{12 cm}$. Find the volume of the solid so obtained.

Ans: We will draw the given figure.

If the triangle is revolved about the side $\text{12 cm}$, we will get a cone with:

Radius $\text{r = 5 cm}$

Slant height $\text{l = 13 cm}$

Height $\text{h = 12 cm}$

We know the volume of the cone $\text{V = }{\displaystyle \frac{\text{1}}{\text{3}}}\pi {\text{r}}^{\text{2}}\text{h}$

$\Rightarrow \text{V = }[{\displaystyle \frac{\text{1}}{\text{3}}}\times \pi \times {\left(\text{5}\right)}^{\text{2}}\times \text{ 12}]\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = 100 }\pi \text{ c}{\text{m}}^{\text{3}}$

Therefore, the volume of the cone will be $\text{100 }\pi \text{ c}{\text{m}}^{\text{3}}$.

8. If the triangle $\mathrm{\Delta }\text{ ABC}$ in the Question $\text{7}$ above is revolved about the side $\text{5 cm}$, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions $\text{7}$ and $\text{8}$.

Ans:

If the triangle is revolved about the side $\text{5 cm}$, we will get a cone with:

Radius $\text{r = 12 cm}$

Slant height $\text{l = 13 cm}$

Height $\text{h = 5 cm}$

We know the volume of the cone $\text{V = }{\displaystyle \frac{\text{1}}{\text{3}}}\pi {\text{r}}^{\text{2}}\text{h}$

$\Rightarrow \text{V = }[{\displaystyle \frac{\text{1}}{\text{3}}}\times \pi \times {\left(\text{12}\right)}^{\text{2}}\times \text{ 5}]\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = 240 }\pi \text{ c}{\text{m}}^{\text{3}}$

Therefore, the volume of the cone will be $\text{240 }\pi \text{ c}{\text{m}}^{\text{3}}$.

The ratio of volume of cone from previous question an the one we obtained above $\text{= }{\displaystyle \frac{\text{100 }\pi }{\text{240 }\pi }}\text{ = }{\displaystyle \frac{\text{5}}{\text{12}}}\text{ = 5 : 12}$

Therefore, the required ratio is $\text{5 : 12}$.

9. A heap of wheat is in the form of a cone whose diameter is $\text{10}\text{.5 m}$and height is $\text{3 m}$. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

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

Ans: It is given that diameter of the heap $\text{= 10}\text{.5 m}$

So, the radius of heap $\text{r = }{\displaystyle \frac{\text{10}\text{.5}}{\text{2}}}\text{ = 5}\text{.25 m}$

Height of heap $\text{h = 3 m}$

We know the volume of the cone $\text{V = }{\displaystyle \frac{\text{1}}{\text{3}}}\pi {\text{r}}^{\text{2}}\text{h}$

$\Rightarrow \text{V = }[{\displaystyle \frac{\text{1}}{\text{3}}}\times {\displaystyle \frac{22}{7}}\times {\left(\text{5}\text{.25}\right)}^{\text{2}}\times \text{ 3}]{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = 86}\text{.625 }{\text{m}}^{\text{3}}$

Hence, the volume of heap is $\text{86}\text{.625 }{\text{m}}^{\text{3}}$.

The area of canvas required is same as curved surface area of the cone.

$\therefore \text{A = }\pi \text{ rl}$

$\Rightarrow \text{A = }\pi \text{ r}\sqrt{{\text{h}}^{\text{2}}\text{ + }{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }{\displaystyle \frac{\text{22}}{\text{7}}}\times \text{ 5}\text{.25 }\times \sqrt{{\left(\text{3}\right)}^{\text{2}}\text{ + }{\left(\text{5}\text{.25}\right)}^{\text{2}}}{\text{m}}^{\text{2}}$

$\Rightarrow \text{A = }({\displaystyle \frac{\text{22}}{\text{7}}}\times \text{ 5}\text{.25 }\times \text{ 6}\text{.05}){\text{m}}^{\text{2}}$

$\Rightarrow \text{A = 99}\text{.825 }{\text{m}}^{\text{2}}$

Therefore, to protect the heap from the rain, the amount of canvas required is $\text{99}\text{.825 }{\text{m}}^{\text{2}}$.

Exercise (11.4)

1. $\left[\text{Assume }\pi \text{ =}{\displaystyle \frac{\text{22}}{\text{7}}}\right]$ 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 $\text{V = }{\displaystyle \frac{\text{4}}{\text{3}}}\pi {\text{r}}^{\text{3}}$

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

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

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

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

(ii) $\text{0}\text{.63 m}$ $\left[\text{Assume }\pi \text{ =}{\displaystyle \frac{\text{22}}{\text{7}}}\right]$

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

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

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

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

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

2. $\left[\text{Assume }\pi \text{ =}{\displaystyle \frac{\text{22}}{\text{7}}}\right]$ 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 $\text{r = }{\displaystyle \frac{\text{28}}{\text{2}}}\text{ = 14 cm}$

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

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

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

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

(ii) $\text{0}\text{.21 m}$ $\left[\text{Assume }\pi \text{ =}{\displaystyle \frac{\text{22}}{\text{7}}}\right]$

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

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

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

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

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

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 $\text{8}\text{.9 g per c}{\text{m}}^{\text{3}}$? $\left[\text{Assume }\pi \text{ =}{\displaystyle \frac{\text{22}}{\text{7}}}\right]$

Ans: It is given the diameter of metallic ball $\text{= 4}\text{.2 cm}$

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

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

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

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

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

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

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

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

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 earth be $\text{d}$.

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

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

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

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

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

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

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

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

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

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

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

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

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

5. How many litres of milk can a hemispherical bowl of diameter $\text{10}\text{.5 cm}$ can hold? $\left[\text{Assume }\pi \text{ =}{\displaystyle \frac{\text{22}}{\text{7}}}\right]$

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

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

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

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

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

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

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

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. $\left[\text{Assume }\pi \text{ =}{\displaystyle \frac{\text{22}}{\text{7}}}\right]$

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 $\text{V = }{\displaystyle \frac{\text{2}}{\text{3}}}\pi \left({\text{R}}^{\text{3}}\text{ - }{\text{r}}^{\text{3}}\right)$

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

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

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

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 $\text{154 c}{\text{m}}^{\text{2}}$. $\left[\text{Assume }\pi \text{ =}{\displaystyle \frac{\text{22}}{\text{7}}}\right]$

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

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

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

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

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