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Q.
The magnitude of the normal reaction that acts on the block
at the point $Q$ is
JEE AdvancedJEE Advanced 2013Work, Energy and Power
Solution:
At point $Q$, component of weight along $P Q$ (radially outwards) is
$m g \cos 60^{\circ}$ or $\frac{m g}{2}$.
Normal reaction is radially inwards
$N-\frac{m g}{2} =\frac{m v^{2}}{R} $
or $ N=\frac{m g}{2}+\frac{m v^{2}}{R} $
$=\frac{1 \times 10}{2}+\frac{1 \times(10)^{2}}{40}=7.5 \,N$