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5.Work, Energy, Power and Collision
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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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