According to question, we have
Let the tension at point $$A$$ be $${T_A}.$$  So, from Newton’s second law $${T_A} - mg = \frac{{mv_c^2}}{R}$$
Energy at point $$A = \frac{1}{2}mv_0^2\,......\left( {\text{i}} \right)$$
Energy at point $$C$$ is $$\frac{1}{2}mv_c^2 + mg \times 2R\,......\left( {{\text{ii}}} \right)$$

Applying Newton's 2^nd law at point $$C$$
$${T_c} + mg = \frac{{mv_c^2}}{R}$$
To complete the loop $${T_c} \geqslant 0$$
$$\eqalign{ & {\text{So,}}\,\,mg = \frac{{mv_c^2}}{R} \cr & \Rightarrow {v_c} = \sqrt {gR} \,......\left( {{\text{iii}}} \right) \cr} $$
From Eqs. (i) and (ii) by conservation of energy
$$\eqalign{ & \frac{1}{2}mv_0^2 = \frac{1}{2}mv_c^2 + 2mgR \cr & \Rightarrow \frac{1}{2}mv_0^2 = \frac{1}{2}mgR + 2mgR \times 2\,\left( {\because {v_C} = \sqrt {gR} } \right) \cr & \Rightarrow v_0^2 = gR + 4gR \cr & \Rightarrow {v_0} = \sqrt {5gR} \cr} $$

Given, the potential energy of a particle in a force field, $$U = \frac{A}{{{r^2}}} - \frac{B}{{{r^1}}}$$
For stable equilibrium, $$F = - \frac{{dU}}{{dr}} = 0$$
$$\eqalign{ & = \frac{{dU}}{{dr}} = - 2A{r^{ - 3}} + B{r^{ - 2}} \cr & 0 = - \frac{{2A}}{{{r^3}}} + \frac{B}{{{r^2}}}\,\,\left( {{\text{As}}\,\,\frac{{ - dU}}{{dr}} = 0} \right) \cr & {\text{or}}\,\,\frac{{2A}}{r} = B \cr} $$
The distance of particle from the centre of the field $$r = \frac{{2A}}{B}$$

Initial length of the spring,
$$\eqalign{ & {\ell _i} = \sqrt {{{\left( {2R} \right)}^2} + {{\left( {1.5R} \right)}^2}} = 2.5R \cr & \therefore {x_i} = 2.5R - 2R = 0.5R. \cr & {\text{Now}}\,\,\frac{1}{2}kx_i^2 + mg\left( {1.5R} \right) = \frac{1}{2}m{v^2} \cr & {\text{or}}\,\,\frac{1}{2}k{\left( {0.5R} \right)^2} + mg\left( {1.5R} \right) \cr & = \frac{1}{2}m{v^2}\,\,\left[ {k = \frac{{4mg}}{R}} \right] \cr & \therefore v = 2\sqrt {gR} \cr} $$

$$W = F\,s\,\cos {90^ \circ } = {\text{zero}}$$

Work done against gravity $$ = mg\sin \theta \times d$$
$$ = 2 \times 10 \times 10 = 200\,J\left( {d\sin \theta = 10} \right)$$
Actual work done $$= 300\,J$$
Work done against friction $$ = 300 - 200 = 100\,J$$

Length of hanging part $$ = \frac{L}{n}$$
Mass of hanging part $$ = \frac{M}{n}$$
Weight of hanging part $$ = \frac{Mg}{n}$$
Let ‘C’ be the centre of mass of the hanging part.

The hanging part can be assumed to be a particle of weight $$\frac{Mg}{n}$$  at a distance $$\frac{L}{n}$$  below the table top. The work done in lifting it to the table top is equal to increase in its potential energy.
$$\eqalign{ & \therefore W = \left( {\frac{{Mg}}{n}} \right)\,\,\left( {\frac{L}{n}} \right) \cr & \therefore W = \frac{{MgL}}{{{n^2}}} \cr} $$

As we know that, relation between kinetic energy and momentum is given by
$$KE = \frac{{{p^2}}}{{2m}}$$
If $${p_1} = {p_2}$$  for two bodies
$$\eqalign{ & {\text{so,}}\,\,K{E_1} \propto \frac{1}{{{m_1}}} \cr & {\text{and}}\,\,K{E_2} \propto \frac{1}{{{m_2}}} \cr} $$
Therefore, ratio of two masses is given by
$$\frac{{{m_1}}}{{{m_2}}} = \frac{{K{E_2}}}{{K{E_1}}} = \frac{1}{4}\,\,\,\left[ {\because \frac{{K{E_1}}}{{K{E_2}}} = \frac{4}{1}} \right]$$

As surface is smooth so work done against friction is zero. Also the displacement and force of gravity are perpendicular so work done against gravity is zero.

Work done on the body is gain in the kinetic energy. Acceleration of the body is $$a = \frac{V}{T}.$$
Velocity acquired in time $$t$$ is $$v = at = \frac{V}{T}t$$
$$K.E.$$  acquired $$ \propto {v^2}.$$   That is work done $$ \propto \frac{{{V^2}{t^2}}}{{{T^2}}}$$

Given, force $$F = 3\hat i + \hat j$$
$$\eqalign{ & {r_1} = \left( {2\hat i + \hat k} \right)m\,\,{\text{and}}\,\,{r_2} = \left( {4\hat i + 3\hat j - \hat k} \right)m \cr & \therefore s = {r_2} - {r_1} = \left( {4\hat i + 3\hat j - \hat k} \right) - \left( {2\hat i + \hat k} \right) \cr & = \left( {2\hat i + 3\hat j - 2\hat k} \right)m \cr & \therefore W = F \cdot S = \left( {3\hat i + \hat j} \right) \cdot \left( {2\hat i + 3\hat j - 2\hat k} \right) \cr & = 3 \times 2 + 3 + 0 = 6 + 3 = 9\,J \cr} $$