JEE MAIN - Physics (2024 - 31st January Evening Shift - No. 12)
A small spherical ball of radius $$r$$, falling through a viscous medium of negligible density has terminal velocity '$$v$$'. Another ball of the same mass but of radius $$2 r$$, falling through the same viscous medium will have terminal velocity:
$$4 \mathrm{v}$$
$$2 \mathrm{~V}$$
$$\frac{v}{4}$$
$$\frac{\mathrm{v}}{2}$$
Explanation
Since density is negligible hence Buoyancy force will be negligible
At terminal velocity.
$$\mathrm{Mg} =6 \pi \eta \mathrm{rv}$$
$$\mathrm{V} \propto \frac{1}{\mathrm{r}} \quad$$ (as mass is constant)
Now, $$\frac{\mathrm{v}}{\mathrm{v}^{\prime}}=\frac{\mathrm{r}^{\prime}}{\mathrm{r}}$$
$$r^{\prime}=2 \mathrm{r}$$
So, $$v^{\prime}=\frac{v}{2}$$
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