A cylindrical tub of radius 12 cm contains water to a depth of 20 cm.

Question:

A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical from ball of radius 9 cm is dropped into the tub and thus the level of water is raised by h cm. What is the value of h?

Solution:

The radius of the cylindrical tub is 12cm. Upon dropping a spherical ball of radius 9cm into the tub, the height of the raised water is cm. Therefore, the volume of the raised water is

$V=\pi \times(12)^{2} \times h$ cubic $\mathrm{cm}$

The volume of the spherical ball is

$V_{1}=\frac{4}{3} \pi \times(9)^{3}$ cubic $\mathrm{cm}$

Since, the volume of the raised water is same as the volume of the spherical ball, we have

$V_{1}=V$

$\Rightarrow \frac{4}{3} \pi \times(9)^{3}=\pi \times(12)^{2} \times h$

$\Rightarrow \quad h=\frac{4 \times(9)^{3}}{3 \times(12)^{2}}$

$\Rightarrow \quad=\frac{27}{4}$

$\Rightarrow \quad=6.75$

Therefore, the height of the raised water is $h=6.75 \mathrm{~cm}$

 

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now