A cylindrical container with diameter of base $56\ cm$ contains sufficient water to submerge a rectangular solid of iron with dimensions $32\ cm \times 22\ cm \times 14\ cm$. Find the rise in the level of the water when the solid is completely submerged.

 Given:

A cylindrical container with diameter of base $56\ cm$ contains sufficient water to submerge a rectangular solid of iron with dimensions $32\ cm \times 22\ cm \times 14\ cm$.

To do:

We have to find the rise in the level of the water when the solid is completely submerged.

Solution:

Diameter of the cylindrical container $= 56\ cm$

This implies,

Radius $(r) =\frac{56}{2}$

$=28\ cm$

Dimensions of the rectangular solid are $32\ cm \times 22\ cm \times 14\ cm$

Therefore,

Volume of the solid $= lbh$

$= 32 \times 22 \times 14$

$= 9856\ cm^3$

Volume of water in the container $= 9856\ cm^3$

Let $h$ be the level of water.

This implies,

$\pi r^2h = 9856$

$\frac{22}{7} \times 28 \times 28 \times h=9856$

$h=\frac{9856 \times 7}{22 \times 28 \times 28}$

$h=4 \mathrm{~cm}$

The height of the water level is $4 \mathrm{~cm}$.

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Updated on: 10-Oct-2022

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