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