A pendulum of length $2\,m$ lift at $P$. When it reaches $Q$, it losses $10\%$ of its total energy due to air resistance. The velocity at $Q$ is .... $m/sec$
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(a) If $v$ is velocity of pendulum at $Q$
and $10\%$ energy is lost while moving from $P$ to $Q$
Hence, by applying conservation of between $P$ and $Q$
$\frac{1}{2}m{v^2} = 0.9\,(mgh)$
and $10\%$ energy is lost while moving from $P$ to $Q$
Hence, by applying conservation of between $P$ and $Q$
$\frac{1}{2}m{v^2} = 0.9\,(mgh)$
==>${v^2} = 2 \times 0.9 \times 10 \times 2$
==> $v = 6\,m/sec$
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