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NCERT solution class 9 chapter 13 Surface Areas and Volumes exercise 13.4 mathematics

EXERCISE 13.4


Page No 225:

Question 1:

Find the surface area of a sphere of radius:

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

Answer:

(i) Radius (r) of sphere = 10.5 cm

Surface area of sphere = 4πr2

Therefore, the surface area of a sphere having radius 10.5cm is 1386 cm2.

(ii) Radius(r) of sphere = 5.6 cm

Surface area of sphere = 4πr2

Therefore, the surface area of a sphere having radius 5.6 cm is 394.24 cm2.

(iii) Radius (r) of sphere = 14 cm

Surface area of sphere = 4πr2

Therefore, the surface area of a sphere having radius 14 cm is 2464 cm2.


Question 2:

Find the surface area of a sphere of diameter:

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

Answer:

(i) Radius (r) of sphere = 

Surface area of sphere = 4πr2

Therefore, the surface area of a sphere having diameter 14 cm is 616 cm2.

(ii) Radius (r) of sphere =

Surface area of sphere = 4πr2

Therefore, the surface area of a sphere having diameter 21 cm is 1386 cm2.

(iii) Radius (r) of sphere = m

Surface area of sphere = 4πr2

Therefore, the surface area of the sphere having diameter 3.5 m is 38.5 m2.


Question 3:

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

Answer:

Radius (r) of hemisphere = 10 cm

Total surface area of hemisphere = CSA of hemisphere + Area of circular end of hemisphere

Therefore, the total surface area of such a hemisphere is 942 cm2.


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.

Answer:

Radius (r1) of spherical balloon = 7 cm

Radius (r2) of spherical balloon, when air is pumped into it = 14 cm

Therefore, the ratio between the surface areas in these two cases is 1:4.


Question 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

Answer:

Inner radius (r) of hemispherical bowl 

Surface area of hemispherical bowl = 2πr2

Cost of tin-plating 100 cm2 area = Rs 16

Cost of tin-plating 173.25 cm2 area  = Rs 27.72

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


Question 6:

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

Answer:

Let the radius of the sphere be r.

Surface area of sphere = 154

∴ 4πr= 154 cm2

Therefore, the radius of the sphere whose surface area is 154 cm2 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 area.

Answer:

Let the diameter of earth be d. Therefore, the diameter of moon will be.

Radius of earth = 

Radius of moon =

Surface area of moon = 

Surface area of earth = 

Required ratio 

Therefore, the ratio between their surface areas will be 1:16


Question 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.

Answer:

Inner radius of hemispherical bowl = 5 cm

Thickness of the bowl = 0.25 cm

∴ Outer radius (r) of hemispherical bowl = (5 + 0.25) cm

= 5.25 cm

Outer CSA of hemispherical bowl = 2πr2

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

Question 9:

A right circular cylinder just encloses a sphere of radius r (see figure). Find

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

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

Answer:

(i) Surface area of sphere = 4πr2

(ii) Height of cylinder = r + r = 2r

Radius of cylinder = r

CSA of cylinder = 2πrh

= 2π(2r)

= 4πr2

(iii) 

Therefore, the ratio between these two surface areas is 1:1


 

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