Two metal spheres are falling through a liquid of density $2.5 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$ with…

Two metal spheres are falling through a liquid of density $2.5 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$ with the same uniform speed. The density of material of first sphere and second sphere is $11.5 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$ and $8.5 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$ respectively. The ratio of the radius of first sphere to that of second sphere is
  1. $\frac{2}{3}$
  2. $\sqrt{\frac{2}{3}}$
  3. $\frac{3}{2}$
  4. $\sqrt{\frac{3}{2}}$

Solution

Terminal velocity is given by $v=\frac{2 r^2(\rho-\sigma) g}{9 \eta}$
As the velocity is the same, $\begin{array}{ll} \therefore & r_A^2\left(\rho_A-\sigma\right)=r_B^2\left(\rho_B-\sigma\right) \\ \therefore & \frac{r_A}{r_B}=\sqrt{\frac{\rho_B-\sigma}{\rho_A-\sigma}} \end{array}$
Substituting the given values, we get $\frac{r_A}{r_B}=\sqrt{\frac{8.5 \times 10^3-2.5 \times 10^3}{11.5 \times 10^3-2.5 \times 10^3}}=\sqrt{\frac{6}{9}}=\sqrt{\frac{2}{3}}$

Asked in: MHT CET 2024 (10 May Shift 2)

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