A simple pendulum with a bob of mass $‘m’$ oscillates from $A$ to $ C$ and back to $A$ such that $PB$ is $H.$ If the acceleration due to gravity is $‘g’$, then the velocity of the bob as it passes through $B$ is
A$mgH$
B$\sqrt {2gH} $
C$\sqrt {gH} $
D
Zero
AIPMT 1995, Medium
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B$\sqrt {2gH} $
b (b) At $B$, the velocity is maximum using conservation of mechanical energy
$\Delta PE = \Delta KE$
==> $mgH = \frac{1}{2}m{v^2}$
==> $v = \sqrt {2gH} $
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