Ex 13.4 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Question 1. Find the surface area of a sphere of radius: (i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
 Solution: (i) Here r = 10.5 cm
∴ Surface area of the sphere = 4πr²

= 4 x (22/7) x (10.5)² cm²

= 4 x (22/7) x (105/10) x (105/10) cm²

= 22 x 3 x 21 cm² = 1386 cm²

(ii) Here r = 5.6 cm

(iii) Here, r = 14 cm

∴ Surface area = 4πr² = 4 x (22/7) (14)² cm²

= 4 x (22/7) x 14 x 14 cm²

= 4 x 22 x 2 x 14 cm² = 2464 cm²

Question 2. Find the surface area of a sphere of diameter: (i) 14 cm (ii) 21 cm (iii) 3.5 m
 Solution: (i) Here, Diameter = 14 cm
⇒ Radius = (14/2)= 7 cm
∴ Surface area = 4πr² = 4 x (22/7) x (7)² cm²
= 4 x (22/7) x 7 x 7 cm² = 88 x 7 cm² = 616 cm²

(ii) Here, Diameter = 21 cm ⇒ r = (21/2) cm

∴ Surface area = 4πr² = 4 x (22/7) (21/2)² cm²

= 4 x (22/7) x (21/2) x (21/2) cm²

= 22 x 3 x 21 cm² = 1386 cm²

(iii) Here, Diameter = 3.5 m

Question 3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
 Solution: Here, radius (r) = 10 cm

∴

Plane surface area of the hemisphere  = πr² = 3.14 x (10)² cm²

= (314/100) x 10 x 10 cm²

= 314 cm²

∴ Total surface area = 628 cm² + 314 cm² = 942 cm²

REMEMBER

(i) Curved surface area of a hemisphere = 2πr²
(ii) Total surface area of a hemisphere = 3πr² 

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.
 Solution: Case I: When radius (r₁) = 7 cm

= 4 x 22 x 14 x 2 cm² = 2464 cm²
The required ratio   

⇒ The required ratio = 1 : 4

Alternate method

Question 5. 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 ₹ 16 per 100 cm².
 Solution: Inner diameter of the hemisphere = 10.5 m

∴    Radius (r) = 10.5/2 cm = 105/20 cm

∵ Curved surface area of a hemisphere = 2πr²
∴ Inner curved surface area of hemispherical bowl

Cost of tinplating:

Question 6. Find the radius of a sphere whose surface area is 154 cm².
 Solution: Let the radius of the sphere be ‘r’ cm.

∴   Surface area = 4πr²

⇒               4πr² = 154

⇒              4 x 22/7 x r² = 154

Thus, the required radius of the sphere is 3.5 cm

 Question 7. The diameter of the moon is apπroximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.
 Solution: Let the radius of the earth = r
∴ Radius of the moon = (r/4)
∵ Surface area of a sphere = 4πr² 
Since, the earth as well as the moon are considered to be spheres.
∴ Surface area of the earth = 4πr²

or [Surface area of the moon] : [Surface area of the earth] = 1 : 16
Thus, the required ratio = 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.
 Solution: Inner radius (r) = 5cm
Thickness = 0.25 cm
∴ Outer radius (R) = [5.00 + 0.25] cm = 5.25 cm
∴ Outer surface area of the bowl = 2πr²

Question 9. A right circular cylinder just encloses a sphere of radius r (see the 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).

Solution: (i) For the sphere: Radius = r
∴ Surface area of the sphere = 4 πr²

(ii) For the right circular cylinder:
∵ Radius of the cylinder = Radius of the sphere
∴ Radius of the cylinder = r
Height of the cylinder = Diameter of the sphere
⇒ Height of the cylinder (h) = 2r
Since, curved surface area of a cylinder = 2πrh = 2πr(2r) = 4πr²

VOLUME OF A CUBOID

All solid bodies occupy space. The amount of space occupied by a solid is called its volume. The units of volume are cubic centimetres (written as cm³) or cubic metres (written as m³).
Note: If an object is hollow then its interior is empty, and can be filled with fluid (air or some liquid) which will take the shape of the container. The volume of the substance that can fill the interior is called the capacity of the container.

REMEMBER

I . Volume of cuboid = [Base area] x [height]
II. Volume of the cube = Edge x Edge x Edge = (side)³