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Two solid sphere of radii $2 \mathrm{~mm}$ and $4 \mathrm{~mm}$ are tied to the two ends of a light string and released in a liquid of specific gravity 1.3 and coefficient of viscosity $1 \mathrm{~Pa}$-s. The string is just taut, when the two spheres are completely in the liquid. If the density of the materials of the two sphere is $2800 \mathrm{kgm}^{-3}$, then the terminal velocity of the system of the sphere is (take, $g=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
$4 \mathrm{cms}^{-1}$
Free body diagram of system is
When system achieves terminal velocity,
$$
\begin{gathered}
F_{\text {net }}=0 \\
\Rightarrow \frac{4}{3} \pi \rho_f g\left(r_{\mathrm{A}}^3+r_{\mathrm{B}}^3\right)+6 \pi \eta v\left(r_A+r_B\right)=\left(m_A+m_B\right) g
\end{gathered}
$$
With values, we get $v=4 \mathrm{cms}^{-1}$
When system achieves terminal velocity,
$$
\begin{gathered}
F_{\text {net }}=0 \\
\Rightarrow \frac{4}{3} \pi \rho_f g\left(r_{\mathrm{A}}^3+r_{\mathrm{B}}^3\right)+6 \pi \eta v\left(r_A+r_B\right)=\left(m_A+m_B\right) g
\end{gathered}
$$
With values, we get $v=4 \mathrm{cms}^{-1}$
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