Question

A ball of weight W is dropped vertically into a lubricant oil that offers a viscous resistance proportional to the square of the velocity of the particle. The buoyancy of the oil is negligible, and the resistance is R when the velocity is 1 cm/s. What uniform velocity will the particle finally attain? (in cm/s)

• A. RW
• B. $\frac{W}{R}$
• C. $\sqrt{\frac{W}{R}}$
• D. $\sqrt{\frac{R}{W}}$

Answer 1

The correct option is C $\sqrt{\frac{W}{R}}$

The free body diagram of the ball is shown below.



The forces acting on the ball are weight W vertically down and viscous resistance F upwards. The acceleration of the ball is taken downwards.

It is given that, $F\alpha v^2$

$F=k{v}^2.....\left(1\right)$

$\text{Also given, when}v=1cm/s,F=R$

$i.e.,R=k\times 1$

$ork=RN{s}^2/m^2$

$\therefore \left(1\right)\Rightarrow F=R{v}^2$

The equation of motion is

$\frac{W}{g}\ddot{x}=\text{Net force in the direction as}\ddot{x}$

$=W-F$

When the ball attains uniform velocity,
$\ddot{x}=0$

i.e.,$0=W-F$

or F = W

$i.e.,R{v}^2=W$

$v=\sqrt{\frac{W}{R}}cm/s$