Fluid Mechanics

JEE Main level, single correct, medium

187 of 192

Question 187 of 192

In an experiment to verify Stokes law, a small spherical ball of radius rr and density ρ\rho falls under gravity through a distance hh in air before entering a tank of water. If the terminal velocity of the ball inside water is same as its velocity just before entering the water surface, then the value of hh is proportional to: (ignore viscosity of air)

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Answer: (A) r4{r}^{4}

The velocity of ball just before striking the surface of water is given by using kinematical equation of motion under the gravity
v2=u2−2gy{v}^{2}={u}^{2}-2gy
⇒v2=−2g(−h)\Rightarrow {v}^{2}=-2g(-h)
⇒v=2gh...(1)\Rightarrow v=\sqrt{2gh}...(1)
Now, the terminal velocity of spherical ball inside the water tank is given by,
vT=29r2(ρ−σ)gη...(2){v}_{T}=\dfrac{2}{9}\dfrac{{r}^{2}(\rho -\sigma )g}{\eta }...(2)
After falling through hh the velocity of ball should be equal to its terminal velocity.
From equation (1) and (2) we get,
⇒2gh=29r2(ρ−σ)gη\Rightarrow \sqrt{2gh}=\dfrac{2}{9}\dfrac{{r}^{2}(\rho -\sigma )g}{\eta }
Squaring the above equation, we get
⇒h=281r4g(ρ−σ)2η2\Rightarrow h=\dfrac{2}{81}\dfrac{{r}^{4}g(\rho -\sigma {)}^{2}}{{\eta }^{2}}
⇒h∝r4\Rightarrow h\propto {r}^{4}

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