## 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 = 4πr^{2}

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, r_{1} = 7 cm

Final radius of the balloon, r_{2} = 14 cm

Initial surface area of the balloon = S_{1}

Final surfacg area of the balloon = S_{2
}

S_{1} : S_{2} = 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 cm^{2}.

Solution:

d = 10.5 cm

r =5.25 cm

Rate of tinplating = \(\frac{16}{100}\) = ₹ 0.16 per cm^{2}

Inner curved surface area of bow l = 2 πr^{2}

ICSA = 2 x \(\frac{22}{7}\) x 5.25 x 5.25 = 173.25 cm^{2}

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 cm^{2}.

Solution:

Surface area of sphere = 4 πr2

154 = 4 x \(\frac{22}{7}\) x r^{2}

\(\frac{154×7}{4×22}\) = r^{2}

⇒ r^{2} = 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:

d_{1} = \(\frac{1}{4}\)d_{2}

Let .S_{1}, and .S_{2}, be the surface area of moon and earth respectively.

S_{1} = 4 πr_{1}^{2}

S_{2} = 4 πr_{2}^{2}

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:

r_{1} = 5 cm

t (thickness) = 0.25 cm

r_{2} = r_{1} + 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 cm^{2}

Question 9.

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

- surface area of the sphere.
- curved surface area of the cylinder.
- 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 S_{1} = 4πr^{2}

2. CSA of a cylinder,= 2πrh

= 2πr x 2r = 4πr^{2}

3.