Mensuration — Exercise 17.3

Find the surface area and volume of a sphere of radius 14 cm.

Answer

Given, r = 14 cm.

Surface area of sphere = 4πr².

Putting values in equation we get,

Surface area of sphere =

$4 \times \dfrac{22}{7} \times (14)^2 = \dfrac{4 \times 22 \times 196}{7} = \dfrac{17248}{7} = 2464$ cm².

Volume of sphere = $\dfrac{4}{3}πr^3$

Putting values in equation we get,

Volume of sphere =

$\dfrac{4}{3} \times \dfrac{22}{7} \times (14)^3 = \dfrac{4 \times 22 \times 2744}{21} = \dfrac{241472}{21} = 11498\dfrac{2}{3}$ cm³.

Hence, the surface area of sphere = 2464 cm² and volume of sphere = $11498\dfrac{2}{3}$ cm³.

Find the surface area and volume of a sphere of diameter 21 cm.

Answer

Given, diameter = 21 cm.

Radius = $\dfrac{\text{Diameter}}{2} = \dfrac{21}{2} = 10.5$ cm.

Surface area of sphere = 4πr².

Putting values in equation we get,

$\text{Surface area of sphere } = 4 \times \dfrac{22}{7} \times (10.5)^2 \\[1em] = \dfrac{4 \times 22 \times 110.25}{7} \\[1em] = \dfrac{9702}{7} = 1386 \text{ cm}^2.$

Volume of sphere = $\dfrac{4}{3}πr^3$

Putting values in equation we get,

$\text{Volume of sphere } = \dfrac{4}{3} \times \dfrac{22}{7} \times (10.5)^3 \\[1em] = \dfrac{4 \times 22 \times 1157.625}{21} \\[1em] = \dfrac{101871}{21} \\[1em] = 4851 \text{ cm}^3.$

Hence, the surface area of sphere = 1386 cm² and volume of sphere = 4851 cm³.

A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm³, find the mass of the shot-put.

Answer

Volume of sphere = $\dfrac{4}{3}πr^3$

Putting values in equation we get,

$\text{Volume of sphere } = \dfrac{4}{3} \times \dfrac{22}{7} \times (4.9)^3 \\[1em] = \dfrac{4 \times 22 \times 117.649}{21} \\[1em] = \dfrac{10353.112}{21} \\[1em] = 493 \text{ cm}^3.$

Since, the density of the metal is 7.8 g per cm³.

Mass = Volume × Density.

So, the mass of the metallic sphere 493 × 7.8 = 3845441.6 g

= $\dfrac{3845441.6}{1000}$ = 3.845 kg.

Hence, the mass of the metallic sphere is approx 3.85 kg.

Find the diameter of a sphere whose surface area is 154 cm².

Answer

Surface area of sphere = 4πr².

Given, surface area = 154 cm².

∴ 4πr² = 154

$4 \times \dfrac{22}{7} \times r^2 = 154 \\[1em] \Rightarrow r^2 = \dfrac{154 \times 7}{88} \\[1em] \Rightarrow r^2 = \dfrac{1078}{88} \\[1em] \Rightarrow r^2 = 12.25 \\[1em] \Rightarrow r = \sqrt{12.25} \\[1em] \Rightarrow r = 3.5 \text{ cm}.$

Diameter = 2 × radius = 2 × 3.5 = 7 cm.

Hence, the diameter of sphere = 7 cm.

Find

(i) the curved surface area.

(ii) the total surface area of a hemisphere of radius 21 cm.

Answer

(i) Given, radius = 21 cm.

Curved surface area of hemisphere = 2πr².

Putting values in equation we get,

$\text{Curved surface area of hemisphere } = 2 \times \dfrac{22}{7} \times (21)^2 \\[1em] = \dfrac{2 \times 22 \times 441}{7} \\[1em] = \dfrac{19404}{7} = 2772 \text{ cm}^2.$

Hence, the curved surface area of hemisphere = 2772 cm².

(ii) Total surface area of hemisphere = 3πr².

Putting values in equation we get,

Total surface area of hemisphere = 3 x $\dfrac{22}{7}$ x (21)²

$= \dfrac{3 \times 22 \times 441}{7} \\[1em] = \dfrac{29106}{7} = 4158 \text{ cm}^2.$

Hence, the total surface area of hemisphere = 4158 cm².

A hemispherical brass bowl has inner-diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹ 16 per 100 cm².

Answer

Given, internal diameter = 10.5 cm.

Internal radius = $\dfrac{\text{Internal diameter}}{2}$

= $\dfrac{10.5}{2} = 5.25$ cm.

Internal curved surface area of hemispherical shell = 2πr².

Putting values in equation we get,

Internal curved surface area of hemisphere = 2 x $\dfrac{22}{7}$ x (5.25)²

$= \dfrac{2 \times 22 \times 27.5625}{7} \\[1em] = \dfrac{1212.75}{7} = 173.25 \text{ cm}^2.$

The cost of tin-plating it on the inside at the rate of ₹ 16/100 cm² or ₹ 0.16/cm².

∴ Cost of tin-plating 173.25 cm² = 173.25 × 0.16 = ₹ 27.72.

Hence, the cost of tin-plating = ₹ 27.72.

The radius of a spherical balloon increases from 7 cm to 14 cm as air is pumped into it. Find the ratio of the surface areas of the balloon in two cases.

Answer

Surface area of sphere = 4πr².

Given, radius in 1st case = 7 cm and in 2nd case = 14 cm.

$\dfrac{\text{Surface area in 1st case}}{\text{Surface area in 2nd case}} = \dfrac{4 \times π \times (7)^2}{4 \times π \times (14)^2} \\[1em] = \dfrac{7 \times 7 }{14 \times 14} \\[1em] = \dfrac{49}{196} \\[1em] = \dfrac{1}{4}$

Hence, the ratio of the surface areas of the balloon in two cases is 1 : 4.

A sphere and a cube have the same surface. Show that the ratio of the volume of the sphere to that of the cube is $\sqrt{6} : \sqrt{π}$.

Answer

Let the side of the cube be a cm and let radius of sphere be r cm.

Surface area of sphere = 4πr².

Surface area of cube = 6a².

Given,
surface area of sphere = surface area of cube.

∴ 4πr² = 6a²

⇒ $\dfrac{r^2}{a^2} = \dfrac{6}{4π}$

⇒ $\dfrac{r}{a} = \sqrt{\dfrac{6}{4π}}$.

Volume of sphere = $\dfrac{4}{3}πr^3$.

Volume of cube = a³.

Ratio of volume of sphere to volume of cube is

$\Rightarrow \dfrac{\text{Volume of sphere}}{\text{Volume of cube}} = \dfrac{\dfrac{4}{3}πr^3}{a^3} \\[1em] = \dfrac{4πr^3}{3a^3} \\[1em] = \dfrac{4π}{3} \times \dfrac{r^3}{a^3} \\[1em] = \dfrac{4π}{3} \times \Big(\dfrac{r}{a}\Big)^3 \\[1em] = \dfrac{4π}{3} \times \Big(\sqrt{\dfrac{6}{4π}}\Big)^3 \\[1em] = \dfrac{4π}{3} \times \dfrac{6}{4π} \times \sqrt{\dfrac{6}{4π}} \\[1em] = \dfrac{4π}{3} \times \dfrac{6}{4π} \times \dfrac{1}{2}\sqrt{\dfrac{6}{π}} \\[1em] = \dfrac{24π}{24π}\sqrt{\dfrac{6}{π}} \\[1em] = \sqrt{\dfrac{6}{π}}.$

Hence proved that the ratio is $\sqrt{6} : \sqrt{π}$.

If the ratio of the radii of two spheres is 3 : 7, find :

(i) the ratio of their volumes.

(ii) the ratio of their surface areas.

Answer

Let the radii of two spheres be 3a and 7a.

(i) Volume of sphere = $\dfrac{4}{3}πr^3$.

$\dfrac{\text{Vol. of Sphere 1}}{\text{Vol. of Sphere 2}} = \dfrac{\dfrac{4}{3}π(3a)^3}{\dfrac{4}{3}π(7a)^3} \\[1em] = \dfrac{\dfrac{4}{3}π \times 27a^3}{\dfrac{4}{3}π \times 343a^3} \\[1em] = \dfrac{27}{343}.$

Hence, the ratio of the volumes of two spheres is 27 : 343.

(ii) Surface area of sphere = 4πr².

$\dfrac{\text{Surface area of Sphere 1}}{\text{Surface area of Sphere 2}} = \dfrac{4π(3a)^2}{4π(7a)^2} \\[1em] = \dfrac{4π \times 9a^2}{4π \times 49a^2} \\[1em] = \dfrac{9}{49}.$

Hence, the ratio of the surface areas of two spheres is 9 : 49.

If the ratio of the volumes of the two spheres is 125 : 64, find the ratio of their surface areas.

Answer

Given,
ratio of the volumes of the two spheres is 125 : 64.

$\therefore \dfrac{\text{Vol. of Sphere 1}}{\text{Vol. of Sphere 2}} = \dfrac{125}{64} \\[1em] \Rightarrow \dfrac{\dfrac{4}{3}π(r_1)^3}{\dfrac{4}{3}π(r_2)^3} = \dfrac{125}{64} \\[1em] \Rightarrow \dfrac{(r_1)^3}{(r_2)^3} = \dfrac{5^3}{4^3} \\[1em] \Rightarrow \dfrac{r_1}{r_2} = \dfrac{5}{4}.$

Surface area of sphere = 4πr².

$\therefore \dfrac{\text{Surface area of Sphere 1}}{\text{Surface area of Sphere 2}} = \dfrac{4π(r_1)^2}{4π(r_2)^2} \\[1em] = \Big(\dfrac{r_1}{r_2}\Big)^2 \\[1em] = \Big(\dfrac{5}{4}\Big)^2 \\[1em] = \dfrac{25}{16}.$

Hence, the ratio of the surface areas of two spheres is 25 : 16.

Find the volume of a sphere whose surface area is 154 cm².

Answer

We know that Surface area of sphere = 4πr².

Given,
Surface area of sphere = 154 cm².

$\therefore 4πr^2 = 154 \\[1em] \Rightarrow 4 \times \dfrac{22}{7} \times r^2 = 154 \\[1em] \Rightarrow r^2 = \dfrac{154 \times 7}{22 \times 4} \\[1em] \Rightarrow r^2 = \dfrac{1078}{88} \\[1em] \Rightarrow r^2 = 12.25 \\[1em] \Rightarrow r = \sqrt{12.25} \\[1em] \Rightarrow r = 3.5 \text{ cm}.$

Volume of sphere = $\dfrac{4}{3}πr^3$

Putting values in equation we get,

Volume of sphere =

$\dfrac{4}{3} \times \dfrac{22}{7} \times (3.5)^3 \\[1em] = \dfrac{4 \times 22 \times 42.875}{21} \\[1em] = \dfrac{3773}{21} \\[1em] = \dfrac{539 \times 7}{3 \times 7} \\[1em] = \dfrac{539}{3} \\[1em] = 179\dfrac{2}{3} cm^3.$

Hence, the volume of sphere = $179\dfrac{2}{3}$ cm³.

If the volume of a sphere is $179\dfrac{2}{3}$ cm³, find its radius and the surface area.

Answer

Volume of sphere = $\dfrac{4}{3}πr^3$.

Given,
Volume of sphere = $179\dfrac{2}{3}$

$\therefore \dfrac{4}{3}πr^3 = 179\dfrac{2}{3} \\[1em] \Rightarrow \dfrac{4}{3} \times \dfrac{22}{7} \times r^3 = \dfrac{539}{3} \\[1em] \Rightarrow \dfrac{88}{21} \times r^3 = \dfrac{539}{3} \\[1em] \Rightarrow r^3 = \dfrac{539 \times 21}{3 \times 88} \\[1em] \Rightarrow r^3 = \dfrac{539 \times 7}{88} \\[1em] \Rightarrow r^3 = \dfrac{3773}{88} \\[1em] \Rightarrow r^3 = 42.875 \\[1em] \Rightarrow r = (42.875)^{\dfrac{1}{3}} \\[1em] \Rightarrow r = 3.5 \text{ cm}.$

Surface area of sphere = 4πr².

Putting values in equation we get,

Surface area of sphere = $4 \times \dfrac{22}{7} \times (3.5)^2$

$= \dfrac{4 \times 22 \times 12.25}{7} \\[1em] = \dfrac{1078}{7} \\[1em] = 154 \text{ cm}^2$

Hence, the radius of the sphere = 3.5 cm and surface area of sphere = 154 cm².

A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain ?

Answer

Volume of hemisphere = $\dfrac{2}{3}πr^3$.

Putting values we get,

$\text{Volume of hemisphere} = \dfrac{2}{3}πr^3 \\[1em] = \dfrac{2}{3}π(3.5)^3 \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times 42.875 \\[1em] = \dfrac{2 \times 22 \times 42.875}{3 \times 7} \\[1em] = \dfrac{1886.5}{21} \\[1em] = \dfrac{18865}{210} \\[1em] = \dfrac{539}{6} \\[1em] = 89\dfrac{5}{6} \text{ cm}^3$

Hence, the volume of water in the hemispherical bowl = $89\dfrac{5}{6} \text{ cm}^3$.

The surface area of a solid sphere is 1256 cm². It is cut into two hemispheres. Find the total surface area and the volume of a hemisphere. Take π = 3.14

Answer

Given,
surface area of the sphere = 1256 cm².

We know that, surface area of sphere = 4πr².

∴ 4πr² = 1256

$\Rightarrow 4 \times 3.14 \times r^2 = 1256 \\[1em] \Rightarrow r^2 = \dfrac{1256}{3.14 \times 4} \\[1em] \Rightarrow r^2 = \dfrac{1256}{12.56} \\[1em] \Rightarrow r^2 = 100 \\[1em] \Rightarrow r = \sqrt{100} \\[1em] \Rightarrow r = 10 \text{ cm}.$

Total surface area of hemisphere = 3πr².

Putting values we get,

Total surface area of hemisphere = $3 \times 3.14 \times (10)^2$

$= 3 \times 3.14 \times 100 \\[1em] = 942 \text{ cm}^2.$

Volume of hemisphere = $\dfrac{2}{3}πr^3$.

Volume of hemisphere = $\dfrac{2}{3} \times 3.14 \times 10^3$

$= \dfrac{2}{3} \times 3.14 \times 1000 \\[1em] = \dfrac{6280}{3} \\[1em] = 2093\dfrac{1}{3} \text{cm}^3.$

Hence, the surface area of hemisphere = 942 cm² and volume of hemisphere = $2093\dfrac{1}{3}\text{ cm}^3.$