A) 1 :1
B) 1 :4
C) 1 : 16
D) 1 : 32
Correct Answer: D
Solution :
Terminal velocity is attained when gravitational force is equal and opposite to atmospheric drag. It is given by \[v=\frac{2}{9}\frac{{{r}^{2}}(\rho -\sigma )}{\eta }g\] where\[\eta \]is coefficient of viscosity, r the radius, \[\rho \]and\[\sigma \]the density. \[\therefore \] \[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{r_{1}^{2}}{r_{2}^{2}}={{\left( \frac{1}{2} \right)}^{2}}=\frac{1}{4}\] Ratio of masses of the two hail stones is \[\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{\frac{4}{3}\pi r_{1}^{3}.\rho }{\frac{4}{3}\pi r_{2}^{3}.\rho }=\frac{1}{8}\] The ratio of momenta \[=\frac{{{p}_{1}}}{{{p}_{2}}}\] \[=\frac{{{m}_{1}}{{v}_{1}}}{{{m}_{2}}{{v}_{2}}}=\frac{1}{8}\times \frac{1}{4}=\frac{1}{32}\] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{t}^{2}}}+\frac{k}{m}y=0\]
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