RS AGGARWAL CLASS 9 CHAPTER 15 VOLUMES AND SURFACE AREA OF SOLIDS EXERCISE 15D

  EXERCISE 15D

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Question 1:

Find the volume and surface area of a sphere whose radius is:
(i) 3.5 cm
(ii) 4.2 cm
(iii) 5 m

Answer 1:

(i) Radius of the sphere = 3.5 cm
Now, volume = $\frac{4}{3}\mathrm{\pi }{r}^3$
                     $$\begin{gathered} =\frac{4}{3}\times \frac{22}{7}\times 3.5\times 3.5\times 3.5 \\ =179.67 {\mathrm{cm}}^3 \end{gathered}$$

∴ Surface area = $4\mathrm{\pi }{r}^2$
                       $$\begin{gathered} =4\times \frac{22}{7}\times 3.5\times 3.5 \\ =154 {\mathrm{cm}}^2 \end{gathered}$$

(ii) Radius of the sphere=4.2 cm
Now, volume = $\frac{4}{3}\mathrm{\pi }{r}^3$
                      $$\begin{gathered} =\frac{4}{3}\times \frac{22}{7}\times 4.2\times 4.2\times 4.2 \\ =310.46 {\mathrm{cm}}^3 \end{gathered}$$

∴ Surface area = $4\mathrm{\pi }{r}^2$
                         $$\begin{gathered} =4\times \frac{22}{7}\times 4.2\times 4.2 \\ =221.76 {\mathrm{cm}}^2 \end{gathered}$$

(iii) Radius of sphere=5 m
Now, volume = $\frac{4}{3}\mathrm{\pi }{r}^3$
                      =$$\begin{gathered} \frac{4}{3}\times \frac{22}{7}\times 5^3 \\ =523.81 {\mathrm{cm}}^3 \end{gathered}$$

∴ Surface area = $4\mathrm{\pi }{r}^2$
                       $$\begin{gathered} =4\times \frac{22}{7}\times 5^2 \\ = 314.29 {\mathrm{cm}}^2 \end{gathered}$$

Question 2:

The volume of a sphere is 38808 cm³. Find its radius and hence its surface area.

Answer 2:

Volume of the sphere = 38808 cm³
Suppose that r cm is the radius of the given sphere.

$$\begin{gathered} \therefore \frac{4}{3}\mathrm{\pi }{r}^{\mathit{ }\mathit{3}}=38808 \\ \Rightarrow r^{\mathit{3}}=\frac{38808\times 3\times 7}{4\times 22}=9261 \\ \Rightarrow r\mathit{ }=\sqrt[3]{9261} =21 \mathrm{cm} \end{gathered}$$

∴ Surface area of the sphere =$4\mathrm{\pi }{r}^2$
                                            $$\begin{gathered} =4\times \frac{22}{7}\times 21\times 21 \\ =5544 {\mathrm{cm}}^2 \end{gathered}$$

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Question 3:

Find the surface area of a sphere whose volume is 606.375 m³.

Answer 3:

Volume of the sphere = 606.375 m³

$$\begin{gathered} \mathrm{Then} \frac{4}{3}\mathrm{\pi }{r}^3=606.375 \\ \Rightarrow r^3=\frac{606.375\times 3\times 7}{4\times 22}=144.703 \\ \Rightarrow r\mathit{ }=5.25 \mathrm{m} \end{gathered}$$

∴ Surface area = $4{\mathrm{\pi r}}^2$
                      $$\begin{gathered} =4\times \frac{22}{7}\times 5.25\times 5.25 \\ =346.5 {\mathrm{m}}^2 \end{gathered}$$

Question 4:

Note Take $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
Find the volume of a sphere whose surface area is 154 cm².

Answer 4:

Let the radius of the sphere be r cm.

Surface area of the sphere = 154 cm²

$$\begin{gathered} \therefore 4\mathrm{\pi }{r}^2=154 \\ \Rightarrow 4\times \frac{22}{7}\times r^2=154 \\ \Rightarrow r=\sqrt{\frac{154\times 7}{4\times 22}}=\sqrt{12.25} \\ \Rightarrow r=3.5 \mathrm{cm} \end{gathered}$$
∴ Volume of the sphere = $\frac{4}{3}\mathrm{\pi }{r}^3=\frac{4}{3}\times \frac{22}{7}\times {\left(3.5\right)}^3\approx 179.67 {\mathrm{m}}^3$

Thus, the volume of the sphere is approximately 179.67 m³.

Question 5:

The surface area of a sphere is (576π) cm². Find its volume.

Answer 5:

Surface area of the sphere = (576π) cm<sup>2
Suppose that r cm is the radius of the sphere.
$$\begin{gathered} \mathrm{Then} 4\mathrm{\pi }{r}^2=576\mathrm{\pi } \\ \Rightarrow r^2=\frac{576}{4}=144 \\ \Rightarrow r=12 \mathrm{cm} \end{gathered}$$

 $$\begin{gathered} \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{sphere}=\frac{4}{3}\times \mathrm{\pi }\times 12\times 12\times 12 {\mathrm{cm}}^3 \\ =2304\mathrm{\pi } {\mathrm{cm}}^3 \end{gathered}$$</sup>

Question 6:

How many lead shots, each 3 mm in diameter, can be made from a cuboid with dimensions (12 cm × 11 cm × 9 cm)?

Answer 6:

Here, l = 12 cm, b = 11 cm and h = 9 cm

$$\begin{gathered} \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cuboid} =l\times b\times h \\ =12\times 11\times 9 \\ = 1188 {\mathrm{cm}}^3 \end{gathered}$$

Radius of one lead shot = 3 mm= $\frac{0.3}{2}\mathrm{cm}$

 $$\begin{gathered} \mathrm{Volume} \mathrm{of} \mathrm{one} \mathrm{lead} \mathrm{shot} =\frac{4}{3}\times \frac{22}{7}\times {\left(\frac{0.3}{2}\right)}^3 \\ =\frac{11\times 9}{7000} \\ =0.014 {\mathrm{cm}}^3 \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Number} \mathrm{of} \mathrm{lead} \mathrm{shots} =\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cuboid}}{\mathrm{volume} \mathrm{of} \mathrm{one} \mathrm{lead} \mathrm{shot}} \\ =\frac{1188}{0.014} \\ =84857.14\approx 84857 \end{gathered}$$

Question 7:

How many lead balls, each of radius 1 cm, can be made from a sphere of radius 8 cm?

Answer 7:

Radius of the sphere = 8 cm

Volume of the sphere = $\frac{4}{3}\mathrm{\pi }{r}^3 =\frac{4}{3}\times \frac{22}{7}\times 8^3=2145.52 {\mathrm{cm}}^3$
Radius of one lead ball = 1 cm
Volume of one lead ball = $\frac{4}{3}\times \frac{22}{7}\times 1^3=4.19 {\mathrm{cm}}^3$

∴ Number of lead balls = $\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{sphere}}{\mathrm{volume} \mathrm{of} \mathrm{one} \mathrm{lead} \mathrm{ball}}=\frac{2145.52}{4.19}=512.05\approx 512$

Question 8:

A solid sphere of radius 3 cm is melted and then cast into smaller spherical balls, each of diameter 0.6 cm. Find the number of small balls thus obtained.

Answer 8:

Radius of the solid sphere = 3 cm

Volume of the solid sphere = $\frac{4}{3}\mathrm{\pi }{r}^3$
                                         = $\frac{4}{3}\times \frac{22}{7}\times 3^3 {\mathrm{cm}}^3$

Diameter of the spherical ball = 0.6 cm
Radius of the spherical ball = 0.3 cm

Volume of the spherical ball = $\frac{4}{3}\times \frac{22}{7}\times {\left(0.3\right)}^3 {\mathrm{cm}}^3$

Now, number of small spherical balls = $\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{sphere}}{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{spherical} \mathrm{ball}}$
                                                        $$\begin{gathered} =\frac{{\displaystyle \frac{4}{3}}\mathrm{\pi }\times 27}{{\displaystyle \frac{4}{3}}\mathrm{\pi }\times \left(0.3^3\right)} \\ =1000 \end{gathered}$$

∴ The number of small balls thus obtained is 1000

Question 9:

A metallic sphere of radius 10.5 cm is melted and then recast into smaller cones, each of radius 3.5 cm and height 3 cm. How many cones are obtained?

Answer 9:

Radius of the metallic sphere, r = 10.5 cm
Volume of the sphere = $\frac{4}{3}\mathrm{\pi }{r}^3\mathit{=}\frac{4}{3}\mathrm{\pi }(10.5{)}^3 {\mathrm{cm}}^3$
Radius of each smaller cone, r₂ = 3.05 cm
Height of each smaller cone = 3 cm
$\mathrm{Volume} \mathrm{of} \mathrm{each} \mathrm{smaller} \mathrm{cone} =\frac{1}{3}\mathrm{\pi }{r_2}^{2}h=\frac{1}{3}\mathrm{\pi }(3.05{)}^2\times 3 {\mathrm{cm}}^3$
Number of cones obtained = $\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{sphere}}{\mathrm{volume} \mathrm{of} \mathrm{each} \mathrm{smaller} \mathrm{cone}}$
                                        $$\begin{gathered} =\frac{{\displaystyle \frac{4}{3}}\mathrm{\pi }{r}^{\mathit{3}}}{{\displaystyle \frac{1}{3}}\mathrm{\pi }{r_2}^{\mathit{2}}h} \\ =\frac{4\times 10.5\times 10.5\times 10.5}{3.5\times 3.5\times 3} \\ =126.006\approx 126 \end{gathered}$$

∴ 126 cones are obtained.

Question 10:

How many spheres 12 cm in diameter can be made from a metallic cylinder of diameter 8 cm and height 90 cm?

Answer 10:

Diameter of each sphere, d=12 cm
Radius of each sphere, r=6 cm
$\mathrm{Volume} \mathrm{of} \mathrm{each} \mathrm{sphere}=\frac{4}{3}\mathrm{\pi }{r}^3\mathit{=}\frac{\mathit{4}}{\mathit{3}}\mathrm{\pi }(6{)}^3 {\mathrm{cm}}^3$
Diameter of base of  the cylinder, D=8 cm
Radius of base of cylinder, R=4 cm
Height of the cylinder, h = 90 cm
$\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cylinder}=\mathrm{\pi }{R}^{\mathit{2}}h\mathit{=}\mathrm{\pi }(4{)}^2\times 90 {\mathrm{cm}}^3$

Number of spheres= $\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cylinder}}{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{sphere}}$
                              $$\begin{gathered} =\frac{\mathrm{\pi }{R}^{2}h}{{\displaystyle \frac{4}{3}}\pi r^3} \\ =\frac{4^2\times 90\times 3}{4\times 6^3} \\ =\frac{12\times 90}{216} \\ =5 \end{gathered}$$

∴ Five spheres can be made.

Question 11:

The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. Find the length of the wire.

Answer 11:

Let the length of the wire be h cm.
Radius of the wire, r = 1 mm = 0.1 cm
Radius of the sphere, R = 3 cm
Now, volume of the sphere = volume of the cylindrical wire
$$\begin{gathered} \Rightarrow \frac{4}{3}\mathrm{\pi }{R}^3=\mathrm{\pi }{r}^{\mathit{2}}h \\ \Rightarrow 4\times 3^2={\left(0.1\right)}^2\times h \\ \Rightarrow h=\frac{4\times 9}{0.1\times 0.1}=3600 \mathrm{cm}= 36 \mathrm{m} \end{gathered}$$

∴ Length of the wire = 36 m

Question 12:

The diameter of a copper sphere is 18 cm. It is melted and drawn into a long wire of uniform cross section. If the length of the wire is 108 m, find its diameter.

Answer 12:

Radius of the copper sphere, R = 9 cm
Length of the wire, h = 108 m = 10800 cm
Volume of the sphere = volume of the wire
Suppose that r cm is the radius of the wire.

$$\begin{gathered} \mathrm{Then} \frac{4}{3}\mathrm{\pi }{R}^3=\mathrm{\pi }{r}^{2}h \\ \Rightarrow \frac{4}{3}\times 9^3=r^2\times 10800 \\ \Rightarrow r^2=\frac{4\times 729}{3\times 10800}=\frac{4\times 81}{3\times 1200}=\frac{9}{100} \\ \Rightarrow r=\frac{3}{10}=0.3 \mathrm{cm} \end{gathered}$$

∴ Diameter of the wire = 0.6 cm

Question 13:

A sphere of diameter 15.6 cm is melted and cast into a right circular cone of height 31.2 cm. Find the diameter of the base of the cone.

Answer 13:

Radius of the sphere, r = 7.8 cm
Height of the cone, h = 31.2 cm
Suppose that R cm is the radius of the cone.
Now, volume of the sphere = volume of the cone
$$\begin{gathered} \Rightarrow \frac{4}{3}\mathrm{\pi }{r}^3=\frac{1}{3}\mathrm{\pi }{R}^{\mathit{2}}h \\ \Rightarrow 4\times {\left(7.8\right)}^3=R^2\times 31.2 \\ \Rightarrow R^{\mathit{2}}=\frac{4\times 7.8\times 7.8\times 7.8}{31.2}=60.84 \\ \Rightarrow R = 7.8 \mathrm{cm} \end{gathered}$$

∴ The diameter of the cone is 15.6 cm.

Question 14:

A spherical cannonball 28 cm in diameter is melted and cast into a right circular cone would, whose base is 35 cm in diameter. Find the height of the cone.

Answer 14:

Radius of the spherical cannonball, R = 14 cm
Radius of the base of the cone, r = 17.5 cm
Let h cm be the height of the cone.
Now, volume of the sphere = volume of the cone
$$\begin{gathered} \Rightarrow \frac{4}{3}\mathrm{\pi }{R}^3=\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h \\ \Rightarrow 4\times {14}^3={\left(17.5\right)}^2\times h \\ \Rightarrow h=\frac{4\times 14\times 14\times 14}{17.5\times 17.5}=35.84 \mathrm{cm} \end{gathered}$$

∴ The height of the cone is 35.84 cm.

Question 15:

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. Find the radius of the third ball.

Answer 15:

Radius of the original spherical ball = 3 cm
Suppose that the radius of third ball is r cm.
Then volume of the original spherical ball = volume of the three spherical balls
$$\begin{gathered} \Rightarrow \frac{4}{3}\mathrm{\pi }\times 3^3=\frac{4}{3}\mathrm{\pi }\times 1.5^3+\frac{4}{3}\mathrm{\pi }\times 2^3+\frac{4}{3}\mathrm{\pi }\times r^3 \\ \Rightarrow 27 =3.375+8+r^3 \\ \Rightarrow r^3=27-11.375 = 15.625 \\ \Rightarrow r = 2.5 \mathrm{cm} \end{gathered}$$

∴ The radius of the third ball is 2.5 cm.

Question 16:

The radii of two spheres are in the ratio 1 : 2. Find the ratio of their surface areas.

Answer 16:

Suppose that the radii are r and 2r.
Now, ratio of the surface areas  = $\frac{4\mathrm{\pi }{r}^2}{4\mathrm{\pi }{\left(2r\right)}^2}=\frac{r^2}{4r^2}=\frac{1}{4}$
                                                     = 1:4

∴ The ratio of their surface areas is 1 : 4.

Question 17:

The surface areas of two spheres are in the ratio 1 : 4. Find the ratio of their volumes.

Answer 17:

Suppose that the radii of the spheres are r and R.
We have:
$$\begin{gathered} \frac{4\mathrm{\pi }{r}^2}{4{\mathrm{\pi R}}^2}=\frac{1}{4} \\ \Rightarrow \frac{r}{R}=\sqrt{\frac{1}{4}}=\frac{1}{2} \end{gathered}$$

Now, ratio of the volumes = $\frac{{\displaystyle \frac{4}{3}}\mathrm{\pi }{r}^3}{{\displaystyle \frac{4}{3}}\mathrm{\pi }{R}^3}={\left(\frac{r}{R}\right)}^3={\left(\frac{1}{2}\right)}^3 =\frac{1}{8}$

∴ The ratio of the volumes of the spheres is 1 : 8.

Question 18:

A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?

Answer 18:

Suppose that the radius of the ball is r cm.
Radius of the cylindrical tub =12 cm
Depth of the tub= 20 cm
Now, volume of the ball = volume of water raised in the cylinder
$$\begin{gathered} \Rightarrow \frac{4}{3}\mathrm{\pi }{r}^3=\mathrm{\pi }\times {12}^2\times 6.75 \\ \Rightarrow r^3=\frac{144\times 6.75\times 3}{4}=36\times 6.5\times 3=729 \\ \Rightarrow r = 9 \mathrm{cm} \end{gathered}$$

∴ The radius of the ball is 9 cm.

Question 19:

A cylindrical bucket with base radius 15 cm is filled with water up to a height of 20 cm. A heavy iron spherical ball of radius 9 cm is dropped into the bucket to submerge completely in the water. Find the increase in the level of water.

Answer 19:

Let h cm be the increase in the level of water.
Radius of the cylindrical bucket = 15 cm
Height up to which water is being filled = 20 cm
Radius of the spherical ball = 9 cm
Now, volume of the sphere = increased in volume of the cylinder
$$\begin{gathered} \Rightarrow \frac{4}{3}\mathrm{\pi }\times 9^3=\mathrm{\pi }\times {15}^2\times h \\ \Rightarrow h=\frac{4\times 729}{3\times 15\times 15}=\frac{4\times 27}{25}=4.32 \mathrm{cm} \end{gathered}$$

∴ The increase in the level of water is 4.32 cm.

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Question 20:

The outer diameter of a spherical shell is 12 cm and its inner diameter is 8 cm. Find the volume of metal contained in the shell. Also, find its outer surface area.

Answer 20:

Outer radius of the spherical shell = 6 cm
Inner radius of the spherical shell = 4 cm

$$\begin{gathered} \mathrm{Volume} \mathrm{of} \mathrm{metal} \mathrm{contained} \mathrm{in} \mathrm{the} \mathrm{shell} =\frac{4}{3}\times \frac{22}{7}\left(6^3-4^3\right) \\ =\frac{88}{21}\times \left(216-64\right) \\ =\frac{88}{21}\times 152 \\ =636.95 {\mathrm{cm}}^3 \end{gathered}$$


$$\begin{gathered} \therefore \mathrm{Outer} \mathrm{surface} \mathrm{area} =4\times \frac{22}{7}\times 6\times 6 \\ =452.57 {\mathrm{cm}}^2 \end{gathered}$$

Question 21:

A hollow spherical shell is made of a metal of density 4.5 g per cm³. If its internal and external radii are 8 cm and 9 cm respectively, find the weight of the shell.

Answer 21:

Internal radius of the hollow spherical shell, r= 8 cm
External radius of the hollow spherical shell, R= 9 cm

Volume of the shell =  $\frac{4}{3}\mathrm{\pi }\left(R^3-r^3\right)$
                            $$\begin{gathered} =\frac{4}{3}\mathrm{\pi }\left(9^3-8^3\right) \\ =\frac{4}{3}\times \frac{22}{7}\times \left(729-512\right) \\ =\frac{4\times 22\times 217}{21} \\ =\frac{88\times 31}{3} \\ =\frac{2728}{3}{\mathrm{cm}}^3 \end{gathered}$$

Weight of the shell = volume of the shell $\times$ density per cubic cm
                             = $\frac{2728}{3}\times 4.5\approx 4092 \mathrm{g} = 4.092 \mathrm{kg}$

∴ Weight of the shell = 4.092 kg

Question 22:

A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm. Find the radius of the base of the cone.

Answer 22:

Radius of the hemisphere = 9 cm
Height of the right circular cone = 72 cm
Suppose that the radius of the base of the cone is r cm.
Volume of the hemisphere = volume of the cone

$$\begin{gathered} \Rightarrow \frac{2}{3}\mathrm{\pi }\times 9^3=\frac{1}{3}\mathrm{\pi }\times r^2\times 72 \\ \Rightarrow r^2=\frac{2\times 9\times 9\times 9}{72}=\frac{81}{4} \\ \Rightarrow r\mathit{ }=\frac{9}{2}=4.5 \mathrm{cm} \end{gathered}$$

∴ The radius of the base of the cone is 4.5 cm.

Question 23:

A hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3 cm and height 4 cm. How many bottles are required to empty the bowl?

Answer 23:

Internal radius of the hemispherical bowl = 9 cm
Radius of a cylindrical shaped bottle = 1.5 cm
Height of a bottle = 4 cm

Number of bottles required to empty the bowl  = $\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{hemispherical} \mathrm{bowl}}{\mathrm{volume} \mathrm{of} \mathrm{a} \mathrm{cylindrical} \mathrm{shaped} \mathrm{bottle}}$
                                                                        $$\begin{gathered} =\frac{{\displaystyle \frac{2}{3}}\mathrm{\pi }\times 9^3}{\mathrm{\pi }\times 1.5^2\times 4} \\ =\frac{2\times 9\times 9\times 9}{3\times 1.5\times 1.5\times 4} \\ =54 \end{gathered}$$

∴ 54 bottles are required to empty the bowl.

Question 24:

A hemispherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. Find the volume of steel used in making the bowl.

Answer 24:

Internal radius of the hemispherical bowl = 4 cm
Thickness of a the bowl = 0.5 cm
Now, external radius of the bowl = (4 + 0.5 ) cm = 4.5 cm

Now, volume of steel used in making the bowl =  volume of the shell
                                                                       $$\begin{gathered} =\frac{2}{3}\mathrm{\pi }\left(4.5^3-4^3\right) \\ =\frac{2}{3}\times \frac{22}{7}\times \left(91.125-64\right) \\ =\frac{2}{3}\times \frac{22}{7}\times 27.125 \\ = 56.83 {\mathrm{cm}}^3 \end{gathered}$$

∴ 56.83 cm^₃ of steel is used in making the bowl .

Question 25:

Note Take $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Answer 25:

Inner radius of the bowl, r = 5 cm

Let the outer radius of the bowl be R cm.

Thickness of the bowl = 0.25 cm         (Given)

∴ R − r = 0.25 cm

⇒ R = 0.25 + r = 0.25 + 5 = 5.25 cm

∴ Outer curved surface area of the bowl = $2\mathrm{\pi }{r}^2=2\times \frac{22}{7}\times {\left(5.25\right)}^2$ = 173.25 cm²

Thus, the outer curved surface area of the bowl is 173.25 cm².

Question 26:

Note Take $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹ 32 per 100 cm².

Answer 26:

Inner radius of the bowl, r = $\frac{10.5}{2}$ = 5.25 cm

∴ Inner curved surface area of the bowl = $2\mathrm{\pi }{r}^2=2\times \frac{22}{7}\times {\left(5.25\right)}^2$ = 173.25 cm²

Rate of tin-plating = ₹ 32 per 100 cm²

∴ Cost of tin-plating the bowl on the inside

= Inner curved surface area of the bowl × Rate of tin-plating

$=173.25\times \frac{32}{100}$

= ₹ 55.44

Thus, the cost of tin-plating the bowl on the inside is ₹ 55.44.

Question 27:

Note Take $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
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?

Answer 27:

Let the radius of the moon and earth be r units and R units, respectively.

∴ 2r = $\frac{1}{4}$ × 2R        (Given)

⇒ $r=\frac{R}{4}$            .....(1)

$\therefore \frac{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{moon}}{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{earth}}=\frac{{\displaystyle \frac{4}{3}\mathrm{\pi }{r}^3}}{{\displaystyle \frac{4}{3}\mathrm{\pi }{R}^3}}=\frac{{\displaystyle {\left({\displaystyle \frac{R}{4}}\right)}^3}}{R^3}=\frac{{\displaystyle 1}}{64}$             [Using (1)]

Thus, the volume of the moon is $\frac{1}{64}$ of the volume of the earth.

Question 28:

Note Take $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of the hemisphere?

Answer 28:

Let the radius of the solid hemisphere be r units.

Numerical value of surface area of the solid hemisphere = $3\mathrm{\pi }{r}^2$

Numercial value of volume of the solid hemisphere = $\frac{2}{3}\mathrm{\pi }{r}^3$

It is given that the volume and surface area of the solid hemisphere are numerically equal.

$$\begin{gathered} \therefore \frac{2}{3}\mathrm{\pi }{r}^3=3\mathrm{\pi }{r}^2 \\ \Rightarrow 2r=9 \mathrm{units} \end{gathered}$$
Thus, the diameter of the hemisphere is 9 units.