Surface Areas and Volumes ⇒⇒ Exercise 11.2

Question 1

Find the surface area of a sphere of radius:
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm

Solution :

The surface area of a sphere is calculated using the formula :$ 4{ \pi}× r^2$

(i) Radius = 10.5 cm

Surface area of a sphere = $ 4{ \pi}× r^2$.

$$ 4 × { 22 \over 7} × (10.5)^2 $$

$$ 4 × { 22 \over 7} × 10.5 × 10.5 $$

$$ 4 × 22 × 1.5 × 10.5 $$

$$⇒ 1386 cm^2 $$

The surface area of the sphere with a radius of 10.5 cm = $ {1386 } cm^2 $

(ii) Radius = 5.6 cm

Surface area of a sphere = $ 4{ \pi}× r^2$.

$$ 4 × { 22 \over 7} × (5.6)^2 $$

$$ 4 × { 22 \over 7} × 5.6 × 5.6 $$

$$ 4 × 22 × 0.8 × 5.6 $$

$$⇒ 394.24 cm^2 $$

The surface area of the sphere with a radius of 5.6 cm = $ {394.24 } cm^2 $

(iii) Radius = 14 cm

Surface area of a sphere = $ 4{ \pi}× r^2$.

$$ 4 × { 22 \over 7} × (14)^2 $$

$$ 4 × { 22 \over 7} × 14 × 14 $$

$$ 4 × 22 × 2 × 14 $$

$$⇒ 2464 cm^2 $$

The surface area of the sphere with a radius of 14 cm = $ {2464 } cm^2 $

Question 2

Find the surface area of a sphere of diameter:
(i) 14 cm (ii) 21 cm (iii) 3.5 m

Solution :

The surface area of a sphere is calculated using the formula :$ 4{ \pi}× r^2$

(i) Diameter of sphere = 14 cm

Given, Diameter of the sphere = 14 cm

Therefore, The radius is half of the diameter.

$$ { { 1 \over 2} × 14 } $$

$$⇒ r = {7 }cm $$

Surface area of a sphere = $ 4{ \pi}× r^2$.

$$ 4 × { 22 \over 7} × (7)^2 $$

$$ 4 × { 22 \over 7} × 7 × 7 $$

$$ 4 × 22 × 7 $$

$$⇒ 616 cm^2 $$

The Surface area of a sphere = $ {616 } cm^2 $

(i) Diameter of sphere = 21 cm

Given, Diameter of the sphere = 21 cm

Therefore, The radius is half of the diameter.

$$ { { 1 \over 2} × 21 } $$

$$⇒ r = { 10.5} cm $$

Surface area of a sphere = $ 4{ \pi}× r^2$.

$$ 4 × { 22 \over 7} × (10.5)^2 $$

$$ 4 × { 22 \over 7} × 10.5 × 10.5 $$

$$ 4 × 22 × 1.5 × 10.5 $$

$$⇒ 1386 cm^2 $$

The Surface area of a sphere = $ {1386 } cm^2 $

(i) Diameter of sphere = 3.5 cm

Given, Diameter of the sphere = 3.5 cm

Therefore, The radius is half of the diameter.

$$ { { 1 \over 2} × 3.5 } $$

$$⇒ r = { 1.75} cm $$

Surface area of a sphere = $ 4{ \pi}× r^2$.

$$ 4 × { 22 \over 7} × (1.75)^2 $$

$$ 4 × { 22 \over 7} × 1.75 × 1.75 $$

$$ 4 × 22 × 0.25 × 1.75 $$

$$⇒ 38.5 cm^2 $$

The Surface area of a sphere = $ {38.5 } cm^2 $

Question 3

Find the total surface area of a hemisphere of radius 10 cm.(Use π = 3.14)

Solution :

Radius of hemisphere (r) = 10 cm

Total surface area of hemisphere = $ 3{ \pi}× r^2$.

$$ 3 × 3.14 × (10)^2 $$

$$⇒ 3 × 3.14 × 10 × 10 $$

$$ ⇒ 942 cm^2 $$

Total surface area of hemisphere = = $ {942 } cm^2 $

Question 4

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

Solution :

The surface area of a sphere is calculated using the formula :$ 4{ \pi}× r^2$

Case I : when Radius = 7 cm

Surface area of a sphere = $ 4{ \pi}× r^2$.

$$ 4{ \pi} × (7)^2 $$

Case II : when Radius = 14 cm

Surface area of a sphere = $ 4{ \pi}× r^2$.

$$ 4{ \pi} × (14)^2 $$

Now, Required ratio = SA in first case / SA in second case

$$⇒ {{4{ \pi} × (7)^2} \over {4{ \pi} × (14)^2} } $$

$$⇒ { {7 × 7} \over { 14 × 14} } $$

$$⇒ { 1 \over 4} $$

The ratio of the surface areas of the balloons = 1: 4

Question 5

A hemispherical bowl made of brass has an inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 $cm^2$. (Assume π = 22/7)

Solution :

Given, Diameter of the sphere = 10.5 cm

Therefore, The radius is half of the diameter.

$$ { { 1 \over 2} × 10.5 } $$

$$⇒ r = {5.25 }cm $$

Radius of hemisphere (r) = 5.25 cm

Surface area of hemispherical bowl = $ 2{ \pi}× r^2$.

$$ 2 × { 22 \over 7} × (5.25)^2 $$

$$⇒ 2 × { 22 \over 7} × 5.25 × 5.25 $$

$$⇒ 2 × 22 × 0.75 × 5.25 $$

$$ ⇒ 173.25 cm^2 $$

Surface area of hemispherical bowl = = $ {173.25 } cm^2 $

Find the Cost of tin-plating

Cost of tin-plating per 100 $cm^2$ = Rs. 16

Cost of tin-plating for 173.25 $cm^2$ area=

$$⇒ {{ 16 \over 100} × 173.25 } $$

$$⇒ {27.72 } $$

Therefore, the cost of tin-plating the inner side of the hemispherical bowl = Rs. 27.72

Question 6

Find the radius of a sphere whose surface area is 154 $cm^2$. (Assume π = 22/7)

Solution :

Surface area of sphere = 154 $cm^2$

Let the radius of the sphere be r.

Surface area = = $ 4{ \pi}× r^2$

$$ 154 = 4 × { 22 \over 7} × (r)^2 $$

$$⇒ (r)^2 = {{154 × 7} \over {22 × 4}} $$

$$ ⇒ (r)^2 ={{7 × 7} \over {4}} $$

$$ ⇒ (r)^2 =({{7 } \over {2}})^2 $$

$$ ⇒ r ={7 \over 2} = 3.5 cm. $$

The radius of the sphere is 3.5 cm

Question 7

The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Solution :

Given that the diameter of the moon is approximately one-fourth of the diameter of the earth:

If diameter of earth is said d, then radius of Earth (r) = d/2

Since the diameter is twice the radius (d = 2r), this also means the radius of the moon is one-fourth of the radius of the earth .

Radius of moon = ${ 1 \over 4} × { d \over 2}$

Radius of moon = ${ d \over 8} $

Surface area of a sphere = $ 4{ \pi}× r^2$.

Since, the earth as well as the moon are considered to be spheres.

The surface area of the moon :

$$⇒ {{4{ \pi} × ({ d \over 8})^2} } $$

The surface area of the earth :

$$⇒ {{4{ \pi} × ({ d \over 2})^2} } $$

Ratio of their Surface areas =

$$⇒ {{4{ \pi} × ({ d \over 8})^2} } \over {{4{ \pi} × ({ d \over 2})^2} }$$

$$⇒ { 4 \over {64}}$$

$$⇒ { 1 \over {16}}$$

The ratio of the surface areas of the moon and the earth is 1:16.

Question 8

A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. (Assume π =22/7)

Solution :

Find the Outer Radius of the hemispherical bowl

The outer radius of the bowl is the sum of its inner radius and the thickness of the steel.

$$⇒ { r = 5 cm + 0.25 cm = 5.25 }cm $$

Find the Outer Curved Surface Area

Surface area of hemispherical bowl = $ 2{ \pi}× r^2$.

$$ 2 × { 22 \over 7} × (5.25)^2 $$

$$⇒ 2 × { 22 \over 7} × 5.25 × 5.25 $$

$$⇒ 2 × 22 × 0.75 × 5.25 $$

$$ ⇒ 173.25 cm^2 $$

Outer Curved Surface Area of hemispherical bowl = $ {173.25 } cm^2 $

Question 9

A right circular cylinder just encloses a sphere of radius r (see fig.).
(i) Find surface area of the sphere,
{ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).

Solution :

(i) Surface area of the sphere = $ 4{ \pi}× r^2$, where r is the radius of sphere

(ii) For cylinder :

∵ Radius of the cylinder = Radius of the sphere

∴ Radius of the cylinder base = r

Height of the cylinder = Diameter of the sphere

[As sphere touches both upper and lower surface of cylinder]

Height of the cylinder (h) = r + r = 2r

CSA of the cylinder formula = 2πrh

$ = 2πr× (2r) $

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

(iii) Ratio between areas = $ { {(Surface Area Of Sphere)} \over {(CSA Of Cylinder)} }$

$$⇒ { { 4{\pi} r^2 } \over {{ 4{\pi} r^2 }} }$$

$$⇒ { 1 \over 1 }$$