Metal spheres, each of radius $ 2 \mathrm{~cm} $, are packed into a rectangular box of internal dimension $ 16 \mathrm{~cm} \times 8 \mathrm{~cm} \times 8 \mathrm{~cm} $ when 16 spheres are packed the box is filled with preservative liquid. Find the volume of this liquid. [Use $ \pi=669 / 213] $

Given:

Metal spheres, each of radius \( 2 \mathrm{~cm} \), are packed into a rectangular box of internal dimension \( 16 \mathrm{~cm} \times 8 \mathrm{~cm} \times 8 \mathrm{~cm} \) when 16 spheres are packed the box is filled with preservative liquid.

To do:

We have to find the volume of the liquid.

Solution:

Dimensions of the rectangular box are $16 \mathrm{~cm} \times 8 \mathrm{~cm} \times 8 \mathrm{~cm}$

This implies,

Volume of the rectangular box $=16 \times 8 \times 8$

$=1024 \mathrm{~cm}^{3}$

Radius of each metal sphere $r=2 \mathrm{~cm}$

Volume of each metal sphere $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \times \frac{669}{213} \times(2)^{3}$

$=\frac{4}{3} \times \frac{669}{213} \times 8$

$=\frac{7136}{213} \mathrm{~cm}^{3}$

This implies,

Volume of 16 metal spheres $= \frac{7136}{213}\times 16$

$=536.0375 \mathrm{~cm}^{3}$

Volume of liquid filled $=$ Volume left after packing 16 spheres

$=1024-536.0375$

$=487.9625 \mathrm{~cm}^{3}$
The volume of the liquid is $487.96 \mathrm{~cm}^{3}$.

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Updated on: 10-Oct-2022

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