JEE MAIN - Physics (2022 - 30th June Morning Shift - No. 5)
An air bubble of negligible weight having radius r rises steadily through a solution of density $$\sigma$$ at speed v. The coefficient of viscosity of the solution is given by :
$$\eta = {{4r\sigma g} \over {9v}}$$
$$\eta = {{2{r^2}\sigma g} \over {9v}}$$
$$\eta = {{2\pi {r^2}\sigma g} \over {9v}}$$
$$\eta = {{2{r^2}\sigma g} \over {3\pi v}}$$
Explanation
Air bubble moves with constant speed v. So net force = 0.
$$\therefore$$ Buoyant Force = Viscous force
$$ \Rightarrow {F_b} = {F_v}$$
$$ \Rightarrow \sigma \times {4 \over 3}\pi {r^3}g = 6\pi nrv$$
$$ \Rightarrow n = {{2\sigma {r^2}g} \over {9v}}$$
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