A) 4 : 9
B) 2 : 3
C) 3 : 2
D) 9 : 4
Correct Answer: B
Solution :
Terminal velocity of drop of radius \[r\] and density \[\rho \] falling in air of density\[\sigma \]is given by \[{{v}_{t}}=\frac{2}{9}\frac{{{r}^{2}}(\rho -\sigma )g}{\eta }\] where, \[\eta =\] viscosity of air Here, \[\frac{{{({{v}_{t}})}_{1}}}{{{({{v}_{t}})}_{2}}}=\frac{4}{9}\] and \[\frac{{{r}_{1}}}{{{r}_{2}}}=?\] \[\therefore \] \[\frac{{{({{v}_{t}})}_{1}}}{{{({{v}_{t}})}_{2}}}=\frac{\frac{2}{9}r_{1}^{2}(\rho -\sigma )g\text{/}\eta }{\frac{a}{2}r_{2}^{2}(\rho -\sigma )/\eta }=\frac{r_{1}^{2}}{r_{2}^{2}}\] \[\Rightarrow \] \[\frac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\frac{{{({{v}_{t}})}_{1}}}{{{({{v}_{t}})}_{2}}}}=\sqrt{\frac{4}{9}}\] \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{2}{3}\] \[{{r}_{1}}:{{r}_{2}}=2:3\]
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