Question

A cylindrical tub of radius 12 cm contains water up to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. The radius of the ball is

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

Answer 1

Answer: (D)

9 cm

Answer 2

Given radius of the cylinder, r = 12 cm
It is also given that a spherical iron ball is dropped into the cylinder and the water level raised by 6.75 cm
Hence volume of water displaced = volume of the iron ball
Height of the raised water level, h = 6.75 m
Volume of water displaced $=\pi r^{2h}$

$=\pi\times12\times12\times6.75cm^3$

So, Volume of iron ball $=\pi\times12\times12\times6.75cm^3$ ... (1)

But, volume of iron ball $=\frac{4}{3\pi r^3}$ ...(2)
From (1) and (2) we get

$\frac{4}{3\pi r^3}=\pi\times12\times12\times\times6.75$

On solving, we get

$r^3=729$

$r^3=3^6$ Therefore, r' = 9

Radius of the iron ball is 9 cm.

So the correct option is (D)