A lead sphere of mass ' $m$ ' falls in viscous liquid with terminal velocity $\mathrm{V}_0$. Another lead sphere of mass ' 8 m ' but of same material will fall through the same liquid with terminal velocity

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Aicharya · Accepted Answer

The formula for terminal velocity for a spherical body $V=\frac{2 g(\rho-\sigma) r^2}{2 \eta}$
If mass is made 8 times, material being same volume also becomes 8 times $\because$ density is same. So, radius of sphere becomes twice. $\therefore \quad$ The new terminal velocity $\begin{aligned} & \frac{V_0}{V_2}=\left(\frac{r_1}{r_2}\right)^2 \\ & \frac{V_0}{V_2}=\left(\frac{I}{2}\right)^2 \\ & V_2=4 V_\theta \end{aligned}$

A lead sphere of mass ' $m$ ' falls in viscous liquid with terminal velocity $\mathrm{V}_0$. Another lead…

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