Question

A cylindrical tube of radius $12$cm contains water upto a depth of $20$cm. A spherical iron ball is dropped into the tube and thus the level of water is raised by $6.75$cm. The radius of the ball is __________.

• A. $4.5$cm
• B. $6$cm
• C. $7.25$cm
• D. $9$cm

Answer 1

Answer: (D)

$9$cm

Volume of cylinder $V_c=\pi r^{2}h$

Thus, volume of water in cylinder $V_1=\pi \times 144\times 20=2880\pi c{m}^3$

After dropping the spherical ball in cylinder new volume

$V_2=144\times 20\pi +\frac{4}{3}\pi r^3=\pi \times 144\times 26.75$

$\Rightarrow \frac{4}{3}{r}^3=144\left(6.75\right)$

$r^3=729\therefore r=9cm$