JEE MAIN - Physics (2022 - 28th July Morning Shift - No. 22)
The diameter of an air bubble which was initially $$2 \mathrm{~mm}$$, rises steadily through a solution of density $$1750 \mathrm{~kg} \mathrm{~m}^{-3}$$ at the rate of $$0.35 \,\mathrm{cms}^{-1}$$. The coefficient of viscosity of the solution is _________ poise (in nearest integer). (the density of air is negligible).
Answer
11
Explanation
$$F = 6\pi \eta rv$$
$${4 \over 3}\pi {r^3}{\rho _l}g = 6\pi \eta rv$$
$$\eta = {{2{r^2}{\rho _l}g} \over v}$$
$$ = {{2 \times {{(2 \times {{10}^{ - 3}})}^2} \times 1750 \times 10} \over {9 \times 3.5 \times {{10}^{ - 3}} \times 4}}$$
$$ = 11$$ poise
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