A.

$\frac{\mathrm{Mg} \sqrt{{\left({\displaystyle \frac{\mathrm{L}}{2}}\right)}^2 + {\left( \mathrm{H} - \mathrm{h} \right)}^2}}{2 \left(\mathrm{H} - \mathrm{h}\right)}$

At equilibrium 

          Mg = 2 T sinθ    

                T = $\frac{\mathrm{M} \mathrm{g}}{2 \mathrm{sin\theta }}$                                  .......(i)

Here,       sinθ = $\frac{\left(\mathrm{H} - \mathrm{h}\right)}{\sqrt{{\left({\displaystyle \raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^2 + {\left(\mathrm{H} - \mathrm{h}\right)}^2}}$

From equaton (i)

               T = $\frac{\mathrm{Mg} \sqrt{\left({\displaystyle \raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right) + {\left(\mathrm{H} - \mathrm{h} \right)}^2}}{2 \left(\mathrm{H} - \mathrm{h}\right)}$

D.

Elevator is accelerating down with g.

Elevator is accelerating down with g.

B.

t

As given the displacement of a body is given to be proportional to the cube of time elapsed. 

              s  ∝ t³

                s = k t³

velocity:-  Velocity of an object is the rate of change of its position with respect to time. 

             v = $\frac{\mathrm{ds}}{\mathrm{dt}}$

                 = $\frac{\mathrm{d} \left( \mathrm{k} {\mathrm{t}}^3 \right)}{\mathrm{dt}}$

           v = 3 kt²

 Acceleration:- Acceleration is the rate of change of velocity with respect to time.

             $\mathrm{\alpha } = \frac{\mathrm{dv}}{\mathrm{dt}}$

                 = $\frac{\mathrm{d}}{\mathrm{dt}}\left( 3 \mathrm{k} {\mathrm{t}}^2 \right)$

              $\mathrm{\alpha }$ = 6 kt

∴              $\mathrm{\alpha } \propto \mathrm{t}$

A.

t^3/2

      P = constant

       $\frac{∆\mathrm{W}}{∆\mathrm{t}}$ = P

     ΔW = P Δt

By using work-energy theorem

    $\frac{1}{2} {\mathrm{mv}}^2 = \mathrm{Pt}$

     v² = $\sqrt{\frac{2\mathrm{P}}{\mathrm{m}}}$ t^1/2 dt

Integrating both sides

    ${\int }_0^{\mathrm{x}}\mathrm{dx} = \sqrt{\frac{2\mathrm{P}}{\mathrm{m}}} {\int }_0^{\mathrm{t}}$t^1/2 dt

         x = $\sqrt{\frac{2\mathrm{P}}{\mathrm{m}}} \left(\frac{2}{3}\right)$ t^3/2

∴      x  ∝  t^3/2

     displacement ∝ (time)^3/2

C.

Solid sphere

The total mechanical energy of a system is conserved if the forces doing work on it. 

Law of conservation of mechanical energy: the total amount of mechanical energy, in a closed system in the absence of dissipative forces ( eg. friction, air resistance), remains constant. This means that potential energy can become kinetic energy or vice versa, but energy cannot disappear. 

According to conservation of mechanical energy 

       mgh = $\frac{1}{2} \mathrm{m} {\mathrm{v}}^2 \left(1 + \frac{{\mathrm{k}}^2}{{\mathrm{R}}^2}\right)$

⇒       v² = $\left(\frac{2 \mathrm{gh}}{1 + {\displaystyle \frac{{\mathrm{k}}^2}{{\mathrm{R}}^2}}}\right)$

It is independent of the mass of the rolling body.

For a ring

          k² = R²

          v_ring = $\sqrt{\frac{2 \mathrm{gh}}{1 + 1}}$

For solid cylinder

        k² = $\sqrt{\frac{2 \mathrm{gh}}{1 + 1}}$

        k² = $\sqrt{\mathrm{g} \mathrm{h}}$

For solid cylinder

            k² = $\frac{{\mathrm{R}}^2}{2}$

     v_cylinder = $\sqrt{\frac{2 \mathrm{gh}}{1 + {\displaystyle \frac{1}{2}}}}$

                 = $\sqrt{\mathrm{g} \mathrm{h}}$

For a solid sphere

      v_sphere = $\sqrt{\frac{2 \mathrm{gh}}{1 + {\displaystyle \frac{2}{5}}}}$

        v_sphere = $\sqrt{\frac{10 \mathrm{gh}}{7}}$

Among the given three bodies the solid sphere has the greatest and the ring has the least velocity at the bottom of the inclined plane.