Question

The cost of painting outer curved surface of a hollow hemispherical iron bowl at the rate of ₹ 2 per cm square is ₹ 1256. If the thickness of the iron is 1 cm, find the cost of painting the inner curved surface of the bowl. (Take $\pi$ = 3.14)

• A. ₹ 1076.32
• B. ₹ 2034.72
• C. ₹ 1017.36
• D. ₹ 508.68

Answer 1

The correct option is C

₹ 1017.36

The cost of painting outer curved surface of a hollow hemispherical iron bowl at the rate of ₹ 2 per cm square is ₹ 1256.

So, outer curved surface area
$=\frac{1256}{2}$
$=628{cm}^2$

We know, curved surface area of hemisphere of radius 'r' is $2\pi r^2.$
$\Rightarrow 2\pi r^2=628c{m}^2$
$\Rightarrow$$r^2=100{cm}^2$
$\Rightarrow r=10cm$

Given, thickness of iron = 1 cm.
Therefore, inner radius = 9 cm.

Then, inner curved surface area
$=2\pi (9{)}^2$
$=508.68{cm}^2$

$\therefore$ The cost of painting = 508.6 $\times$ 2
= ₹ 1017.36