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PhysicsFluidViscosityEasy2 minPYQ_2020

PhysicsEasysingle choice

Two drops, of the same radius, are falling through air with a steady velocity of $5 {cm s}^{- 1}$ . If the two drops coalesce, the terminal velocity would be

Options:

Answer:

Solution:

If $R$ is radius of bigger drop formed, then
$\frac{4}{3} π R^{3} = 2 \times \frac{4}{3} π r^{3} or R = 2^{1 / 3} r$
As, $v_{0} \propto r^{2}$
∴ $\frac{{(v_{0})}_{1}}{v_{0}} = \frac{R^{2}}{r^{2}} = \frac{(2^{1 / 3} r)^{2}}{r^{2}} = 2^{2 / 3}$
or ${(v_{0})}_{1} = v_{0} 2^{2 / 3}$
$5 \times {(4)}^{1 / 3}$
$= (5) \times {(2)}^{2 / 3}$
$= 5 \times {(4)}^{1 / 3} {cm s}^{-1}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:FluidSubtopic:Viscosity

⏱ 2mℹ️ Source: PYQ_2020

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