Fill in the blanks:
$(a)$ In electricity consumption,$1$ unit is equal to .......... Joules of work.
$(b)$ $A$ body falling from a height of $10 \ m$ onto hard ground loses $20 \%$ of its energy. It can reach a height of .............
$(c)$ $A$ particle moves in a circular path of radius $a$ under the influence of an attractive force with potential energy $U = -\frac{k}{2r^2}$. Its total energy is = .......
$(d)$ Converting a mass of $1 \ \mu g$ into energy yields ........ energy.

(N/A) $1 \text{ unit} = 1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}$.
$(b)$ Since the body loses $20 \%$ of its energy,it retains $80 \%$ of its initial potential energy.
$mgh_2 = 0.8 \times mgh_1 \Rightarrow h_2 = 0.8 \times 10 \ m = 8 \ m$.
$(c)$ For a circular orbit,the centripetal force $F = \frac{mv^2}{r} = -\frac{dU}{dr}$.
Given $U = -\frac{k}{2r^2}$,then $F = -\frac{d}{dr}(-\frac{k}{2r^2}) = -\frac{k}{r^3}$.
The magnitude of centripetal force is $\frac{mv^2}{r} = \frac{k}{r^3} \Rightarrow \frac{1}{2}mv^2 = \frac{k}{2r^2}$.
Thus,Kinetic Energy $K = \frac{k}{2r^2}$.
Total Energy $E = K + U = \frac{k}{2r^2} - \frac{k}{2r^2} = 0$.
$(d)$ Using $E = mc^2$,where $m = 1 \ \mu g = 10^{-9} \ kg$ and $c = 3 \times 10^8 \ m/s$.
$E = 10^{-9} \times (3 \times 10^8)^2 = 10^{-9} \times 9 \times 10^{16} = 9 \times 10^7 \ J$.