B.

$\frac{1}{2}\mathrm{th}$ of initial height

Since it loses 50% of its kinetic energy on striking ground, its PE is also reduced by half.   

R     Ratio of heights

                $\frac{{\mathrm{h}}_1}{{\mathrm{h}}_2} = \frac{100}{100 - 50}$

                 ${\mathrm{h}}_{2 }= \frac{{\mathrm{h}}_1}{2}$

                  ${\mathrm{h}}_2 = \frac{1}{2}$ of the initial height

B.

α = 0

τ   the torque on a given axis is product of moment of inertia and angular acceleration. 

              $\mathrm{\tau } = \mathrm{I} \mathrm{\alpha }$

    I = moment of inertia

    α = angular acceleration

     If     $\mathrm{\tau } = 0 \Rightarrow \mathrm{I\alpha } = 0$

     But       I ≠ 0

                 α = 0

When the torque acting upon a system is zero, the angular momentum is constant and hence conserved. This is analogous to the linear counterpart i.e when the force on the system is zero, the momentum is conserved.

B.

5 m/s

Distance covered towards east 

          = speed × time

          =  2 × 15 = 30 m

Distance covered towards north

         = speed × time

         = 8  × 5 = 40 m

Total displacement

          = $\sqrt{{40}^2 + {30}^2}$

           = 50 m

Average velocity = $\frac{\mathrm{Total} \mathrm{displacement}}{\mathrm{Total} \mathrm{time}}$

                          = $\frac{50}{2 + 8}$

                          = 5 m/s     

B.

A - 4, B - 3

Dimensions of Torque = [M² L² T^-2]

Dimension of angular momentum = [ M¹ L² T^-1]

∴ Relation A-4 ,B -3 is right. 

C.

100%

 Kinetic energy           

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

                 E = $\frac{1}{2}\frac{{\displaystyle {\mathrm{m}}^2{\mathrm{v}}^2}}{\mathrm{m}}$

                 E = $\frac{1}{2}\frac{{\mathrm{p}}^2}{\mathrm{m}}$

                  p² = 2ME

∴                 p ∝ $\sqrt{\mathrm{E}}$

∴                 $\frac{{\mathrm{p}}_1}{{\mathrm{p}}_2} = \sqrt{\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}}$

∴                        = $\sqrt{\frac{\mathrm{E}}{\mathrm{E} + 3\mathrm{E}}}$

                    $\frac{{\mathrm{p}}_1}{{\mathrm{p}}_2} = \frac{1}{2}$

⇒                  p₂ = 2 p₁

                 % increase = $\frac{{\mathrm{p}}_2 - {\mathrm{p}}_1}{{\mathrm{p}}_1}$ × 100

⇒                 $\frac{2{\mathrm{p}}_1 - {\mathrm{p}}_1}{{\mathrm{p}}_1} \times 100$ = 100%