MP Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

MP Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Assume π = \(\frac{22}{7}\), unless stated otherwise.

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

Question 2.
Find the surface area of a sphere of diameter:

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

Solution:
(i) Here, Diameter = 14 cm

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

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:
Initial radius of the balloon, r₁ = 7 cm
Final radius of the balloon, r₂ = 14 cm
Initial surface area of the balloon = S₁
Final surfacg area of the balloon = S₂

S₁ : S₂ = 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 ₹ 16 per 100 cm².
Solution:
d = 10.5 cm
r =5.25 cm
Rate of tinplating = \(\frac{16}{100}\) = ₹ 0.16 per cm²
Inner curved surface area of bow l = 2 πr²
ICSA = 2 x \(\frac{22}{7}\) x 5.25 x 5.25 = 173.25 cm²
Cost of tin-plating = Rate x ICSA
= 0.16 x 173.25 = ₹ 27.72

Question 6.
Find the radius of a sphere whose surface area is 154 cm².
Solution:
Surface area of sphere = 4 πr2
154 = 4 x \(\frac{22}{7}\) x r²
\(\frac{154×7}{4×22}\) = r²
⇒ r² = 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 areas.
Solution:
d₁ = \(\frac{1}{4}\)d₂

Let .S₁, and .S₂, be the surface area of moon and earth respectively.
S₁ = 4 πr₁²
S₂ = 4 πr₂²

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:
r₁ = 5 cm
t (thickness) = 0.25 cm
r₂ = r₁ + t
= 5 + 0.25 = 5.25 cm
OCSA of bowl = 2x \(\frac{22}{7}\) x 5.25 x 5.25
= \(\frac{44}{7}\) x 5.25 x 5.25 = 173.25 cm²

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

1. surface area of the sphere.
2. curved surface area of the cylinder.
3. ratio of the areas obtained in (i) and (ii).

Solution:
As the sphere just encloses a right circular cylinder, the height of the cylinder is equal to the diameter of the sphere.
∴ Height of cylinder h = 2r
where r is the radius of sphere

1. Surface area of sphere S₁ = 4πr²

2. CSA of a cylinder,= 2πrh
= 2πr x 2r = 4πr²

3.

MP Board Class 9th Maths Solutions