A cylindrical vessel of inner radius 4 cm contains water up to some height. A solid sphere of radius 3 cm is lowered into the water until it is completely immersed. The water level in the vessel will rise by:

Correct option is C

Given:
Inner radius of the cylindrical vessel,  r = 4 cm.
Radius of the solid sphere, R = 3 cm

Formula Used:
Volume of the sphere =$$\frac{4}{3} \pi R^3$$​
Volume of water displaced = Volume of the sphere.
Rise in water level h  in the cylinder =$$\frac{\text{Volume of water displaced}}{\text{Base area of the cylinder}}$$​​
Base area of the cylinder =$$\pi r^2$$​
Solution:
$$\text{Volume of the sphere} = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi (27) = 36 \pi \, \text{cm}^3$$​
When the sphere is completely immersed, it displaces an equal volume of water. Therefore, the volume of water displaced is $$36 \pi \, \text{cm}^3$$​
$$\text{Base area of the cylinder} = \pi r^2 = \pi (4)^2 = 16 \pi \, \text{cm}^2$$​
​$$h = \frac{\text{Volume of water displaced}}{\text{Base area of the cylinder}} = \frac{36 \pi}{16 \pi} = \frac{36}{16} = \frac{9}{4}\text{cm}$$​​