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5.Work, Energy, Power and Collision
hard
$A$ particle of mass $m$ is released from $a$ height $H$ on $a$ smooth curved surface which ends into a vertical loop of radius $R$, as shown Choose the correct alternative $(s)$ if $H = 2R$
A
The particles reaches the top of the loop with zero velocity
B
The particle cannot reach the top of the loop
C
The particle break off at $a$ height $R < H < 2R$
D
Both $(B)$ and $(C)$
Solution
Velocity on reaching the base of circle is $u=\sqrt{2 g H}=\sqrt{2 g 2 R}=\sqrt{4 g R}$
This is less than $\sqrt{5 g R} .$ Hence, the particle cannot complete the circle. But this velocity is more than $\sqrt{2 g R}$ so the particle will break off the loop at a height above the centre before reaching the top.
Standard 11
Physics
