The work done by man is negative of magnitude of decreases in potential energy of chain

$$\eqalign{ & \Delta U = mg\frac{L}{2} - \frac{m}{2}g\frac{L}{4} = 3mg\frac{L}{8} \cr & \therefore W = - \frac{{3mgL}}{8} \cr} $$

Conserve the momentum of third mass with the resultant momentum of 1st and 2nd masses. After getting velocity of third mass, calculate total kinetic energy.

According to question, the third part of mass $$2m$$  will move as shown in the figure, because the total momentum of the system after explosion must remain zero. Let the velocity of third part be $$v'.$$
From the conservation of momentum $$\sqrt 2 \left( {mv} \right) = \left( {2m} \right) \times v' \Rightarrow v' = \frac{v}{{\sqrt 2 }}$$
So, total kinetic energy generated by the explosion
$$\eqalign{ & = \frac{1}{2}m{v^2} + \frac{1}{2}m{v^2} + \frac{1}{2}\left( {2m} \right){{v'}^2} \cr & = m{v^2} + m \times {\left( {\frac{v}{{\sqrt 2 }}} \right)^2} = m{v^2} + \frac{{m{v^2}}}{2} = \frac{3}{2}m{v^2} \cr} $$

$$\eqalign{ & {P_{{\text{generated}}}} = {P_{{\text{input}}}} \times \frac{{90}}{{100}} \cr & = \frac{{mgh}}{t} \times \frac{{90}}{{100}} = \frac{{15 \times 10 \times 60}}{1} \times \frac{{90}}{{100}} \cr & = 8.1\,kW \cr} $$

Consider the blocks shown in the figure to be moving together due to friction between them.

The free body diagrams of both the blocks are shown below.

Work done by static friction on $$A$$ is positive and on $$B$$ is negative.

As the floor exerts a force on the ball along the normal, & no force parallel to the surface, therefore the velocity component along the parallel to the floor remains constant.
Hence $$V\sin \theta = {V^1}\sin {\theta ^1}$$

Power of a body is defined as the rate at which the body can do the work, i.e.
$${\text{Power}} = \frac{{{\text{work}}}}{{{\text{time}}}} = \frac{W}{t}$$
Given, power, $$P = 2\,kW = 2000\,W$$
$$\eqalign{ & W = Mgh = M \times 10 \times 10\,\,\left[ {\because g = 10\,m/{s^2}} \right] \cr & = 100\,M \cr} $$
Time, $$t = 60\,s$$
$$\eqalign{ & \therefore 2000 = \frac{{100M}}{{60}} \cr & \therefore M = 1200\,kg \cr & {\text{and}}\,V = 1200\,L \cr} $$

Net work done in sliding a body up to a height $$h$$ on inclined plane
= Work done against gravitational force + Work done against frictional force
$$\eqalign{ & \Rightarrow W = {W_g} + {W_f}\,......\left( {\text{i}} \right) \cr & {\text{but}}\,W = 300\,J \cr & {W_g} = mgh = 2 \times 10 \times 10 = 200\,J \cr} $$
Putting in Eq. (i), we get
$$\eqalign{ & 300 = 200 + {W_f} \cr & \Rightarrow {W_f} = 300 - 200 = 100\,J \cr} $$

Let us consider a point on the circle

The equation of circle is $${x^2} + {y^2} = {a^2}$$
The force is
$$\eqalign{ & \vec F = K\left[ {\frac{{x\hat i}}{{{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}} + \frac{{y\hat j}}{{{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}}} \right] \cr & \vec F = K\left[ {\frac{{x\hat i}}{{{{\left( {{a^2}} \right)}^{\frac{3}{2}}}}} + \frac{{y\hat j}}{{{{\left( {{a^2}} \right)}^{\frac{3}{2}}}}}} \right] \cr & \vec F = \frac{K}{{{a^3}}}\left[ {x\hat i + y\hat j} \right] \cr} $$
The force acts radially outwards as shown in the figure and the displacement is tangential to the circular path. Therefore the angle between the force and displacement is $${90^ \circ }$$  and $$W=0$$
option (D) is correct.

Let $${V_f}$$  is the final speed of the body.
From questions,
$$\eqalign{ & \frac{1}{2}mV_f^2 = \frac{1}{8}mV_0^2\,\,\,\,\, \Rightarrow {V_f} = \frac{{{V_0}}}{2} = 5\,m/s \cr & F = m\left( {\frac{{dV}}{{dt}}} \right) = - k{V^2}\,\,\,\,\therefore \left( {{{10}^{ - 2}}} \right)\frac{{dV}}{{dt}} = - k{V^2} \cr & \int\limits_{10}^5 {\frac{{dV}}{{{V^2}}}} = - 100K\int\limits_0^{10} {dt} \cr & \frac{1}{5} - \frac{1}{{10}} = 100\,K\left( {10} \right) \cr & {\text{or}},\,\,K = {10^{ - 4}}\,kg\,{m^{ - 1}} \cr} $$

$$F = - \frac{{\partial u}}{{\partial r}}\hat r = \frac{K}{{{r^3}}}\hat r$$
Since particle is moving in circular path
$$\eqalign{ & F = \frac{{m{v^2}}}{r} = \frac{K}{{{r^3}}}\,\,\,\,\, \Rightarrow m{v^2} = \frac{K}{{{r^2}}} \cr & \therefore K.E. = \frac{1}{2}m{v^2} = \frac{K}{{2{r^2}}} \cr} $$
Total energy $$=PE.+K.E.$$
$$ = - \frac{K}{{2{r^2}}} + \frac{K}{{2{r^2}}} = {\text{Zero}}\left[ {\because P.E. = - \frac{K}{{2{r^2}}}{\text{ given}}} \right]$$