JEE Main 201912 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 12 April
Shift: Evening Shift
Subject: Physics
Chapter: Mechanical Properties of Fluids

JEE Main 2019 · 12 April · Evening Shift · Physics · Mechanical Properties of Fluids

A solid sphere, of radius R acquires a terminal velocity v _ 1 when...

A solid sphere, of radius R acquires a terminal velocity

{v}_{1}

when falling (due to gravity) through a viscous fluid having a coefficient of viscosity

η .

The sphere is broken into 27 identical solid spheres. If each of these spheres acquires a terminal velocity,

{v}_{2},

when falling through the same fluid, the ratio

\left(\dfrac{{v}_{1}}{{v}_{2}}\right)

equals:

Options

A $\dfrac{1}{9}$
B $27$
C $\dfrac{1}{27}$
D $9$

Correct answer

D. $9$

Step-by-step solution

If $R$ is the radius of initial drop and $r$ is the radius of new drop, then
$\dfrac{4}{3}\pi{R}^{3} = 27 \times \dfrac{4}{3}\pi{r}^{3} ⇒ r = \dfrac{R}{3}$
Terminal velocity,

${v}_{T} ∝ {r}^{2}$
$⇒ \dfrac{{v}_{1}}{{v}_{2}} = \dfrac{{R}^{2}}{{\left(\dfrac{R}{3}\right)}^{2}} = 9$

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