(d)

$(a)$ Mass $m$ is at height $H$ from point $Q$, where potential energy is taken zero.

From geometry of above figure, if at some angle $\theta$, height of mass $m$ above lowest point $Q$ is $h$, then from $\triangle A B C$,

$\cos \theta=\frac{R-h}{R} \Rightarrow h=R(1-\cos \theta)$

Hence, potential energy of $m$ as a function of $\theta$ is

$PE =U(\theta)=m g h$

$\Rightarrow U(\theta)=m g R(1-\cos \theta)$

$(b)$ Kinetic energy at position $\theta=$ Loss of potential energy that occurred in reaching this position

$\Rightarrow$ Kinetic energy is $K(\theta)=m g H-U(\theta)$

$\Rightarrow K(\theta) =m g H-m g R(1-\cos \theta)$

$=m g(H-R(1-\cos \theta))$

$(c)$ For $H << R$

$T=\frac{1}{2 \pi} \sqrt{\frac{R}{g}}$

Time taken to travel from $P$ to $Q$ is one-fourth of this time period. i.e.

$t=\frac{1}{8 \pi} \sqrt{\frac{R}{g}}$

$(d)$ From energy conservation at lowest point, if $m$ has velocity $v$, then

$\frac{1}{2} m v^2 =m g H$

$\Rightarrow m v^2 =2 m g H \text { or } \frac{m v^2}{R}=\frac{2 m g H}{R}$

But this centripetal force is resultant of force of normal reaction $N$ and weight of body.

$\Rightarrow N-m g=\frac{m v^2}{R}$

$\Rightarrow N=m g+\frac{2 m g H}{R}$

$\Rightarrow N=m g(1+\frac{2 H}{R})$

This is the force by block on the concave surface.