A solid wooden toy is in the form of a hemisphere surmounted by a cone of same radius. The radius of hemisphere is $ 3.5 \mathrm{~cm} $ and the total wood used in the making of toy is $ 166 \frac{5}{6} \mathrm{~cm}^{3} $. Find the height of the toy. Also, find the cost of painting the hemispherical part of the toy at the rate of $ ₹ 10 $ per $ \mathrm{cm}^{2} . $ (Take $ \left.\pi=22 / 7\right) $.

Given:

A solid wooden toy is in the form of a hemisphere surmounted by a cone of same radius.

The radius of hemisphere is \( 3.5 \mathrm{~cm} \) and the total wood used in the making of toy is \( 166 \frac{5}{6} \mathrm{~cm}^{3} \). 

To do:

We have to find the height of the toy and the cost of painting the hemispherical part of the toy at the rate of \( ₹ 10 \) per \( \mathrm{cm}^{2} . \) 

Solution:

Volume of the wood in making the toy $=166 \frac{5}{6} \mathrm{~cm}^{3}$

$=\frac{1001}{6} \mathrm{~cm}^{3}$

Radius of the hemisphere $=3.5 \mathrm{~cm}$

Let $h$ be the height of the conical part.

This implies,

Volume of the wood used $=\frac{1}{3} \pi r^{2} h+\frac{2}{3} \pi r^{3}$

$=\frac{1}{3} \times \frac{22}{7} \times(3.5)^{2} \times h+\frac{2}{3} \times \frac{22}{7} \times(3.5)^{3}$

$=\frac{22}{21}(3.5)^{2}[h+2 \times 3.5]$

$=\frac{22}{21} \times 12.25(h+7)$

$=\frac{38.5}{3}(h+7)$

Therefore,

$\frac{1001}{6}=\frac{38.5}{3}(h+7)$

$\frac{1001(3)}{6(38.5)}=(h+7)$

$h=13-7$

$h=6 \mathrm{~cm}$

Total height of the toy $=6+3.5$

$=9.5 \mathrm{~cm}$

Surface area of the hemispherical part $=2 \pi r^{2}$

$=2 \times \frac{22}{7} \times (3.5)^2$

$=\frac{2 \times 22}{7} \times (\frac{7}{2})^2 $

$=77 \mathrm{~cm}^{2}$

Cost of painting the toy per $\ cm^2 =Rs.\ 10$

Total cost of painting the hemispherical part of the toy $=Rs.\ 77 \times 10$

$=Rs.\ 770$ 

The cost of painting the hemispherical part of the toy is $Rs.\ 770$.

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Updated on: 10-Oct-2022

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