The linear acceleration of centre of mass will be $$a = \frac{F}{m},$$   wherever the force is applied. Hence, the acceleration will be same whatever the value of $$h$$ may be.

The cubical block is in equilibrium.
For translational equilibrium
$$\eqalign{ & (a)\,\,\,\,\sum {{F_x}} = 0 \Rightarrow F = N \cr & (b)\,\,\,\,\sum {{F_y}} = 0 \Rightarrow f = mg \cr} $$

For Rotational Equilibrium
$$\sum {{\tau _c} = 0} $$
Where $$ {{\tau _c} = } $$   torque about c.m.
Torque created by frictional force ($$f$$ ) about $$C=f × a$$   in clockwise direction.
There should be another torque which should counter this torque. The normal reaction $$N$$  on the block acts as shown. This will create a torque $$N \times b$$   in the anticlockwise direction.
Such that $$f × a=N × b$$
Note : The normal force does not act through the centre of the body always. The point of application of normal force depends on all the forces acting on the body.

$$\eqalign{ & {y_{cm}} = \frac{{\int {ydm} }}{{\int {dm} }} \cr & = \frac{{\int\limits_0^h {\pi {r^2}dy\rho \times y} }}{{\frac{1}{3}\pi {R^2}h\rho }} \cr & = \frac{{3h}}{4} \cr} $$

To have linear motion, the force $${\vec F}$$  has to be applied at centre of mass.
i.e. the point $$‘P ’$$ has to be at the centre of mass
$$y = \frac{{{m_1}{y_1} + {m_2}{y_2}}}{{{m_1} + {m_2}}} = \frac{{m \times 2\ell + 2m \times \ell }}{{3m}} = \frac{{4\ell }}{3}$$

Torque working on the bob of mass m is, $$\tau = mg \times \ell \sin \theta .$$    (Direction parallel to plane of rotation of particle)

As $$\tau $$ is perpendicular to $$\vec L,$$ direction of $$L$$ changes but magnitude remains same.

Angular momentum
$$\eqalign{ & L = I\omega \cr & {\text{where}}\,I = m{r^2}{\text{and}}\,\omega = \frac{v}{r} \cr & L = m{r^2} \times \frac{v}{r} \cr & L = mvr\,....\left( {\text{i}} \right) \cr} $$
Centripetal force
$$\eqalign{ & F = \frac{{m{v^2}}}{r} \cr & F = \frac{m}{r}.{\left( {\frac{L}{{mr}}} \right)^2}\left[ {{\text{From}}\,{\text{Eq}}.\left( {\text{i}} \right)} \right] \cr & F = \frac{{m{L^2}}}{{r \cdot {m^2} \cdot {r^2}}} \cr & F = \frac{{{L^2}}}{{m{r^3}}} \cr} $$

Apply conservation of angular momentum
The angular momentum of a disc of moment of inertia $${{I_1}}$$ and rotating about its axis with angular velocity $$\omega $$ is $${L_1} = {I_1}\omega $$
When a round disc of moment of inertia $${I_2}$$ is placed on first disc, then angular momentum of the combination is $${L_2} = \left( {{I_1} + {I_2}} \right)\omega '$$
In the absence of any external torque, angular momentum remains conserved i.e., $${L_1} = {L_2}$$
$$\eqalign{ & {I_1}\omega = \left( {{I_1} + {I_2}} \right)\omega ' \cr & \Rightarrow \omega ' = \frac{{{I_1}\omega }}{{{I_1} + {I_2}}} \cr} $$

When solid cylinder rolls down on an inclined plane, then it has both rotational and translational kinetic energy

Total kinetic energy $$K = {K_{{\text{rot}}}} + {K_{{\text{trans}}}}$$
or $$K = \frac{1}{2}I{\omega ^2} + \frac{1}{2}m{v^2}$$
where, $$I =$$  moment of inertia of solid cylinder about its axis
$$ = \frac{1}{2}m{r^2}$$
$$\eqalign{ & \therefore K = \frac{1}{2}\left( {\frac{1}{2}m{r^2}} \right){\omega ^2} + \frac{1}{2}m{v^2} \cr & = \frac{1}{4}m{v^2} + \frac{1}{2}m{v^2}\,\,\left( {{\text{as}}\,v = r\omega } \right) \cr & = \frac{3}{4}m{v^2} \cr} $$
Now, gain in $$KE =$$  loss in $$PE$$
$$\eqalign{ & \therefore \frac{3}{4}m{v^2} = mgh \cr & \Rightarrow v = \sqrt {\left( {\frac{4}{3}gh} \right)} \cr} $$

Considering the normal reaction of the floor and wall to be $$N$$  and with reference to the figure.

By vertical equilibrium
$$N + N\sin {30^ \circ } = 1.6\,g\,\,\, \Rightarrow N = \frac{{3.2g}}{3}\,\,.....(i)$$
By horizontal equilibrium
$$f = N\cos {30^ \circ } = \frac{{\sqrt 3 }}{2}N = \frac{{16\sqrt 3 }}{3}\,{\text{From }}(i)$$
Taking torque about $$A$$  we get
$$\eqalign{ & 1.6g \times AB = N \times x \cr & 1.6g \times \frac{\ell }{2}\cos {60^ \circ } = \frac{{3.2g}}{3} \times x \cr & \therefore \frac{{3\ell }}{8} = x\,\,.....(ii) \cr & {\text{But }}\cos {30^ \circ } = \frac{h}{x} \cr & \therefore x = \frac{h}{{\cos {{30}^ \circ }}}\,\,\,.....(iii) \cr & {\text{From }}(ii)\,\,{\text{and }}(iii) \cr & \frac{h}{{\cos {{30}^ \circ }}} = \frac{{3\ell }}{8} \cr & \therefore \frac{h}{\ell } = \frac{{3\sqrt 3 }}{{16}} \cr} $$

$$\eqalign{ & {x_i} = \frac{{2 \times 2 + 2 \times 0}}{{2 + 2}} = 1\,m. \cr & {\text{and}}\,{x_f} = \frac{{1 \times 2 + 3 \times 0}}{{1 + 3}} = 0.5\,m. \cr & \therefore \Delta x = 0.5\;m \cr} $$