MHT CET · Physics · Mechanical Properties of Fluids
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
- A \(\frac{2}{3}\)
- B \(\sqrt{\frac{2}{3}}\)
- C \(\frac{3}{2}\)
- D \(\sqrt{\frac{3}{2}}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{2}{3}}\)
Step-by-step Solution
Detailed explanation
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
\(\frac{r_A}{r_B}=\sqrt{\frac{8.5 \times 10^3-2.5 \times 10^3}{11.5 \times 10^3-2.5 \times 10^3}}=\sqrt{\frac{6}{9}}=\sqrt{\frac{2}{3}}\)
\(v=\frac{2 r^2(\rho-\sigma) g}{9 \eta}\)
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
\(\frac{r_A}{r_B}=\sqrt{\frac{8.5 \times 10^3-2.5 \times 10^3}{11.5 \times 10^3-2.5 \times 10^3}}=\sqrt{\frac{6}{9}}=\sqrt{\frac{2}{3}}\)
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