velocity varies with time as shown.
Net acceleration of the sphere is ($m g-$ Buoyant force$-$Viscous Force) downwards
$a=d \frac{4}{3} \pi R^{3} g-\rho \frac{4}{3} \pi R^{3} g-6 \pi \eta R v$
Hence a varies linearly with velocity. Hence option $\mathrm{C}$ is the correct answer.

$1.$ If the piston is pushed at a speed of $5 \ mms ^{-1}$, the air comes out of the nozzle with a speed of
$(A)$ $0.1 \ ms ^{-1}$ $(B)$ $1 \ ms ^{-1}$ $(C)$ $2 \ ms ^{-1}$ $(D)$ $8 \ ms ^{-1}$
$2.$ If the density of air is $\rho_{ a }$ and that of the liquid $\rho_{\ell}$, then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to
$(A)$ $\sqrt{\frac{\rho_{ a }}{\rho_{\ell}}}$ $(B)$ $\sqrt{\rho_a \rho_{\ell}}$ $(C)$ $\sqrt{\frac{\rho_{\ell}}{\rho_{ a }}}$ $(D)$ $\rho_{\ell}$
Give the answer question $1$ and $2.$
| Column - $\mathrm{I}$ | Column - $\mathrm{II}$ |
| $(a)$Rain drops moves downwards with constant velocity. | $(i)$ Viscous liquids |
| $(b)$ Floating clouds at a height in air. | $(ii)$ Viscosity |
| $(iiii)$ Less density |
(given atmospheric pressure $P_{A}=1.01 \times 10^{5}\,Pa$, density of water $\rho_{ w }=1000\,kg / m ^{3}$ and gravitational acceleration $g=10\,m / s ^{2}$ )