NCERT Solutions Class 9th Maths Chapter – 11 Surface Areas and Volumes Exercise 11.2

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

Solution – Formula: Surface area of sphere (SA) = 4πr2

(i) Radius of sphere, r = 10.5 cm
SA = 4×(22/7)×10.52 = 1386
Surface area of sphere is 1386 cm2

(ii) Radius of sphere, r = 5.6cm
Using formula, SA = 4×(22/ 7)×5.62 = 394.24
Surface area of sphere is 394.24 cm2

(iii) Radius of sphere, r = 14cm
SA = 4πr2
= 4×(22/7)×(14)2
= 2464
Surface area of sphere is 2464 cm2

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

Solution
(i) Radius of sphere, r = diameter/2 = 14/2 cm = 7 cm
Formula for Surface area of sphere = 4πr2
= 4×(22/7)×72 = 616
Surface area of a sphere is 616 cm2

(ii) Radius (r) of sphere = 21/2 = 10.5 cm
Surface area of sphere = 4πr2
= 4×(22/7)×10.52 = 1386
Surface area of a sphere is 1386 cm2
Therefore, the surface area of a sphere having diameter 21cm is 1386 cm2

(iii) Radius(r) of sphere = 3.5/2 = 1.75 cm
Surface area of sphere = 4πr2
= 4×(22/7)×1.752 = 38.5
Surface area of a sphere is 38.5 cm2

Solution – Radius of hemisphere, r = 10cm
Formula: Total surface area of hemisphere = 3πr2
= 3×3.14×102 = 942
The total surface area of given hemisphere is 942 cm2.

Solution – Let r1 and r2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So
r1 = 7cm
r2 = 14 cm
Now, Required ratio = (initial surface area)/(Surface area after pumping air into balloon)
= 4πr12/4πr22
= (r1/r2)2
= (7/14)2 = (1/2)2 = ¼
Therefore, the ratio between the surface areas is 1:4.

Solution – Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm
Formula for Surface area of hemispherical bowl = 2πr2
= 2×(22/7)×(5.25)2 = 173.25
Surface area of hemispherical bowl is 173.25 cm2
Cost of tin-plating 100 cm2 area = Rs 16
Cost of tin-plating 1 cm2 area = Rs 16 /100
Cost of tin-plating 173.25 cm2 area = Rs. (16×173.25)/100 = Rs 27.72
Therefore, the cost of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm2 is Rs 27.72.

Solution – Let the radius of the sphere be r.
Surface area of sphere = 154 (given)
Now,
4πr2 = 154
r2 = (154×7)/(4×22) = (49/4)
r = (7/2) = 3.5
The radius of the sphere is 3.5 cm.

Solution – If diameter of earth is said d, then the diameter of moon will be d/4 (as per given statement)
Radius of earth = d/2
Radius of moon = ½×d/4 = d/8
Surface area of moon = 4π(d/8)2
Surface area of earth = 4π(d/2)2

The ratio between their surface areas is 1:16.

Solution – Given:
Inner radius of hemispherical bowl = 5cm
Thickness of the bowl = 0.25 cm
Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm
Formula for outer CSA of hemispherical bowl = 2πr2, where r is radius of hemisphere
= 2×(22/7)×(5.25)2 = 173.25 cm2
Therefore, the outer curved surface area of the bowl is 173.25 cm2.

(i) 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 sphere = 4πr2, where r is the radius of sphere

(ii) Height of cylinder, h = r+r =2r
Radius of cylinder = r
CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h)
= 4πr2

(iii) Ratio between areas = (Surface area of sphere)/(CSA of Cylinder)
= 4πr2/4πr2 = 1/1
Ratio of the areas obtained in (i) and (ii) is 1:1.