Gujarati
5.Work, Energy, Power and Collision
medium

A particle slides from the top of a smooth hemispherical surface of radius $R$ which is fixed on a horizontal surface. If it separates from the hemisphere at a height $h$ from the horizontal surface, then the speed of the particle is

A

$\sqrt{(2 g(R-h))}$

B

$\sqrt{(2 g(R+h))}$

C

$\sqrt{2 g R}$

D

$\sqrt{2 g h}$

(KVPY-2019)

Solution

(a)

Condition given in question is shown below.

Let $v=$ speed of particle when it separates from hemisphere.

As there is no friction, loss of potential energy appears in form of kinetic energy of particle.

$\therefore m g(R-h) =\frac{1}{2} m v^2$

$\Rightarrow v =\sqrt{ } 2 g(R-h)$

Standard 11
Physics

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