Question

A point mass M is released from rest and slides down a spherical bowl of radius R from a height H as shown in the figure below. The surface of the bowl is smooth (no friction). The velocity of mass at the bottom of the bowl is

• A. $\sqrt{gH}$
• B. $\sqrt{2gR}$
• C. $\sqrt{2gH}$
• D. 0

Answer 1

The correct option is C $\sqrt{2gH}$


Loss of P.E = Gain in K.E.
$Mgh=\frac{1}{2}M{v}^2$
$v=\sqrt{2gH}$