NCERT Solutions For Class 9 Math Chapter- 13 Exercise – 13.4

NCERT Solutions For Class 9 Math Chapter – 13 Exercise – 13.4

 

Q1. Find the surface area of a sphere of radius:

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

Solution:-

 

(i)  r = 10.5 cm

Surface area of the sphere = 4πr2

= 4 x 22/7 x (10.5)2

= 1386 cm2

 

(ii) R = 5.6 cm

Surface area of sphere = 4πr2

= 4 x 22/7 x (5.6)2

= 394.24 cm2.

 

(iii) R = 14 cm

Surface area of the sphere = 4πr2

= 4 x 22/7 x (14)2

= 2464 cm2

 

Q2. Find the surface area of the sphere of diameter:

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

 Solution:-

 

(i) D = 14 cm

R

= 14/2 = 7 cm

Surface area =  4πr2

= 4x 22/7 x 7  x 7

= 616 cm2

 

(ii) Diameter d = 21 cm

R = 21/2  cm

Surface area = 4πr2

4 x 22/7 x 21/2 x 21/2

= 1386 cm2

 

(iii) D = 3.5 cm

R = 3.5 cm/2

= 1.75

Surface area = 4πr2

4 x 22/7 x 1.75 x 1.75

= 38.5 cm2

 

Q3. Find the total surface are of the hemisphere of radius 10 cm.

 

 

 

 

 

Solution:-

 

TSA of hemisphere = 3πr2

TSA = 3πr2

= 3 x 3.14x10x10

= 942 cm2

 

Q4. 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 two caes.

Solution:-

 

R1 = 7 cm

R2 = 14 cm

Surface area before pumping the air = 4πr12

Surface area of the balloon after pumping = 4πr22

Ratio of the surface area of the balloon = CSA1/CSA2

= 4πr12/4πr22

= r12/r22

= (r1/r2)2

= (7/14)2

= (1/2)2

= ¼

 

Q5. A hemispherical bowl made of brass  has inner diameter 10.5 cm Find the cost of tinplating it on the inside at the rate of Rs 16 per 100 cm2.

 Solution:-

 

Inner diameter D = 10.5 cm

Inner radius = 10.5 cm/2

= 5.25 cm

 

CSA of hemispherical bowl = 2πr2

= 2 x 22/7 x 5.25 x 5.25

= 173.25 cm2

 

The cost of tinplating the bowl at Rs 16 per 100 cm2 = 173.25/100 x 16

= Rs 27.72

 

Q6. Find the radius of a sphere whose surface area is 154 cm2.

Solution:-

 

Surface Area = 4πr2 = 154 cm2

R2 = 154 cm2/4π

R2 = 154 cm2/4 x 7/22

R2 = 49/4 cm2

R = √49/4 cm2

R = 7/2 cm

R = 3.5 cm

 

Q7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find ratio of their surfaces area

Solution:-

 

Radius of the earth be R.

Radius of the moon be r.

 

Diameter of the moon = ¼ x diameter of the earth

Radius of the moon = ¼ x radius of the earth

r = ¼ x R

r/R = ¼

 

Surface area of the earth = 4πR2

Surface area of the moon

Ratio of their surface areas = 4πr2/4Πr2

= r2/R2

= (1/4)2

= 1/16

 

Q8. 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.

 

 

 

 

Solution:-

 

Inner radius of the bowl = 5 cm

Thickness of steel = 0.25 cm

Outer radius of the bowl = R = 5 cm + 0.25 cm = 5.25 cm

 

Outer CSA of the hemisphere = 2ΠR2

= 2 X 22/7 X 5.25 X 5.25

= 173.25 cm2.

 

Q9. 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).

 Solution:-

 

Radius of the sphere = Radius of the cylinder = r

Height of the cylinder = Diameter of the sphere = 2r

 

(i) Surface area of the sphere = 4πr2

(ii) Curved surface area of the cylinder = 2πrh

= 2π x 2r

= 4πr2

 

(iii) Ratio of the areas obtained in (i) and (ii) = 4πr2/4πr2

= 1/1