Exercise 11.2

Assume π = 227, unless stated otherwise.

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

(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm

Instructions

We know: Surface area of a sphere (SA) =

(i) r = 10.5 cm: SA = 4 × 227 × 10.52 = cm2

(ii) r = 5.6 cm: SA = 4 × 227 × 5.62 = cm2 (Round off to two decimal places)

(iii) r = 14cm: SA = 4 × 227 × 142 = cm2

We have found all the answers.

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

(i) 14 cm (ii) 21 cm (iii) 3.5 m

Instructions

Formula for the surface area of sphere =

(i) 14 cm: We have r = cm which gives: SA = 4 × 227 × 72 = cm2

(ii)21 cm: We have r = cm which gives: SA = 4 × 227 × 10.52 = cm2

(iii) 3.5 m: We have r = cm which gives: SA = 4 × 227 × 1.752 = cm2

We have found all answers.

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

Instructions

Total surface area of the hemisphere = where r = cm

Total surface area of the hemisphere = × × = cm2

The total surface area of the given hemisphere is 942 cm2.

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.

Instructions

Finding out the ratio

We know: curved surface area of a sphere =
We need to find the ratio of the initial and final surface area.
That is to find the ratio shown above.
Let the initial radius be r1 while the final radius is r2. Substituting these variables, we get the ratio:
Putting the values of r1 and r2 into the ratio, we get:
We have found the answer.

5. 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 Rs. 16 per 100 cm2

Instructions

We have: Inner radius of hemispherical bowl (r) = cm with surface area of hemispherical bowl =

Surface Area = 2×227×5.252 = cm2 (Round off to two decimal places)

Cost of tin-plating 100 cm2 area = Rs which gives us cost of tin-plating 1 cm2 area = Rs

Thus, total cost of tin-plating = Rs. 16×173.25100 = Rs (Round off to two decimal places)

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

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

Instructions

We have surface area = cm2 with the formula being

So, 4πr2 = 154 ⇒ r2 = 154×74×22 = 494

Thus, r = cm

We have found the answer.

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

Instructions

If the diameter of the earth is said d, then the diameter of the moon will be .

So, radius of earth = while radius of moon =

Thus, surface area of moon =

And surface area of earth =

Required Ratio = 4πd824πd22 =

The ratio between their surface areas is 1:16.

8. 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.

Instructions

Finding out the outer surface

We know: curved surface area of a hemisphere =
The outer radius will be cm
Substituting the values into the formula
We find the surface area to be cm2
We have found the answer.

9. A right circular cylinder just encloses a sphere of radius r. Find:

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

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

Instructions

Finding out the surface area and ratio

We know: curved surface area of a sphere = where r - radius of sphere
Since, the sphere is enclosed in a cylinder: radius of cylinder is equal to while height of cylinder is
Curved surface area of cylinder = where r and h are radius and height of cylinder.
Substituting the values of r and h into the formula, we get:
Thus, the ratio of the sphere and cylinder is
We have found the desired answer.

Chapter 11: Surface Areas and Volumes

Glossary

Exercise 11.2

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Chapter 11: Surface Areas and Volumes

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Glossary

Exercise 11.2