150 spherical marbles, each of diameter $ 1.4 \mathrm{~cm} $ are dropped in a cylindrical vessel of diameter $ 7 \mathrm{~cm} $ containing some water, which are completely immersed in water. Find the rise in the level of water in the vessel.

Given:

150 spherical marbles, each of diameter \( 1.4 \mathrm{~cm} \) are dropped in a cylindrical vessel of diameter \( 7 \mathrm{~cm} \) containing some water, which are completely immersed in water.

To do:

We have to find the rise in the level of water in the vessel.

Solution:

Diameter of each spherical marble $=1.4 \mathrm{~cm}$

This implies,

Radius of each spherical marble $r=\frac{1.4}{2}$

$=0.7 \mathrm{~cm}$

$=\frac{7}{10} \mathrm{~cm}$

Volume of each spherical marble $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \pi \times (\frac{7}{10})^{3}$

$=\frac{1372 \pi}{3000} \mathrm{~cm}^{3}$

Volume of 150 spherical marbles $=\frac{1372 \pi}{3000} \times 150$

$=\frac{1372 \pi}{20}$

$=\frac{343 \pi}{5} \mathrm{~cm}^{3}$

Diameter of the cylindrical vessel $=7 \mathrm{~cm}$

This implies,

Radius of the cylindrical vessel $R=\frac{7}{2} \mathrm{~cm}$

Let the height of the water raised be $h$.
Therefore,

Volume of the water raised $=\pi r_{2}^{2} h$

$=\pi \frac{7}{2} \times \frac{7}{2} h$

$=\frac{49 \pi}{4} h \mathrm{~cm}$

Volume of water raised $=$ Volume of 150 spherical marbles

$\frac{49 \pi}{4} h=\frac{343 \pi}{5}$

$h=\frac{\text { Volume of 150 marbles }}{\pi r_{2}^{2}}$

$=\frac{343 \pi \times 4}{5 \times \pi \times 7^{2}}$

$=\frac{28}{5}$

$=5.6 \mathrm{~cm}$

The rise in the level of water in the vessel is $5.6\ cm$.

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Updated on: 10-Oct-2022

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