# 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:**

Hemisphere is exactly half of a sphere, so the curved surface area is half of the surface area of the sphere.

CSA of the hemisphere with radius r, CSA = 2πr^{2}.

Since the bowl is to be tin-plated from inside, the area to be tin-plated will be equal to the CSA of the hemisphere.

Inner diameter, d = 10.5cm

Inner radius, r = 10.5/2cm = 5.25cm

CSA of hemispherical bowl = 2πr^{2}

= 2 × 22/7 × 5.25cm × 5.25cm

= 173.25 cm^{2}

The cost of tin-plating 100 cm^{2} of the bowl = ₹16

The cost of tin-plating 1 cm^{2} of the bowl = ₹16/100

The cost of tin-plating 173.25 cm^{2 }area of the bowl = (₹16/100) ×173.25 = 27.72

Thus, the cost of tin-plating is ₹ 27.72.

**☛ Check: **NCERT Solutions for Class 9 Maths Chapter 13

**Video Solution:**

## 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.

Class 9 Maths NCERT Solutions Chapter 13 Exercise 13.4 Question 5

**Summary:**

It is given that a hemispherical bowl made of brass has an inner diameter of 10.5 cm. We have found that the cost of tin-plating it on the inside at the rate of ₹ 16 per 100 cm^{2} is ₹27.72.

**☛ Related Questions:**

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- A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Findi) Surface area of the sphereii) Curved surface area of the cylinderiii) Ratio of the areas obtained in (i) and (ii).

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