A sphere of radius 6 cm is dropped into a cylindrical vessel partly filled with water. The radius of the vessel is 8 cm. If the sphere is submerged completely, then the surface of the water rises by
(a) 4.5 cm
(b) 3
(c) 4 cm
(d) 2 cm
Radius of the sphere = 6 cm.
Volume of the sphere
$=\frac{4}{3} \pi r^{3}$'
$=\frac{4}{3} \pi \times 6 \times 6 \times 6$
and
Radius of the cylinder = 8 cm
Volume of the cylinder
$=\pi r^{2} h$
$=\pi \times 8 \times 8 \times h$
Therefore,
Volume of the sphere = volume of the cylinder
$\frac{4}{3} \pi(6)^{3}=\pi(8)^{2} h$
or
$h=\frac{4 \times 72}{64}=4.5 \mathrm{~cm}$
Hence, the correct answer is choice (a).
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.