V1 is the terminal velocity of a metal sphere of mass 0.5 gm and density 5 gm/cc when it is falling in water. V2 is the terminal velocity of another metal sphere of mass 4 gm and density 9 gm/cc when it is falling in water. The ratio of V1 and V2 .

6πηrVt=mg1−dldb⇒Vt∝mr1−dldb=m2/31−dldb since r∝m1/3so V1V2=940

V1 is the terminal velocity of a metal sphere of mass 0.5 gm and density 5 gm/cc when it is falling in water. V2 is the terminal velocity of another metal sphere of mass 4 gm and density 9 gm/cc when it is falling in water. The ratio of V1 and V2 .

6πηrVt=mg1−dldb⇒Vt∝mr1−dldb=m2/31−dldb since r∝m1/3so V1V2=940

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V₁ is the terminal velocity of a metal sphere of mass 0.5 gm and density 5 gm/cc when it is falling in water. V₂ is the terminal velocity of another metal sphere of mass 4 gm and density 9 gm/cc when it is falling in water. The ratio of V₁ and V₂ .

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$7:25$

$9:40$

$5:9$

$1:4$

answer is D.

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Detailed Solution

$\begin{array}{l} 6 πη {rV}_{t} = mg (1 - \frac{d_{l}}{d_{b}}) \\ \Rightarrow V_{t} \propto \frac{m}{r} (1 - \frac{d_{l}}{d_{b}}) = m^{2 / 3} (1 - \frac{d_{l}}{d_{b}}) \\ since r \propto m^{1 / 3} \\ so \frac{V_{1}}{V_{2}} = \frac{9}{40} \end{array}$