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Properties of Matter question

2023 · 6 Apr · Shift 1 · Q48

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Properties of Matter
2023 · 6 Apr · Shift 1 · Q48

Properties of Matter question

2023 · 6 Apr · Shift 1 · Q48

JEE MainPhysicsProperties of MatterMCQ+4 / −1

A small ball of mass

\mathrm{M}

and density

\rho

is dropped in a viscous liquid of density

\rho_{0}

. After some time, the ball falls with a constant velocity. What is the viscous force on the ball ?

A
$\mathrm{F}=\mathrm{Mg}\left(1-\frac{\rho_{\mathrm{O}}}{\rho}\right)$
B
$\mathrm{F}=\mathrm{Mg}\left(1+\frac{\rho}{P_{o}}\right)$
C
$\mathrm{F}=\mathrm{Mg}\left(1+\frac{\rho_{\mathrm{o}}}{\rho}\right)$
D
$F=M g\left(1 \pm \rho \rho_{0}\right)$

View written solutionFree

Correct answer: A

Identify the situation

When the ball has fallen for some time and moves with constant velocity, its acceleration becomes zero. Hence the net force on the ball is zero.

So, downward force = total upward forces.

Forces acting on the ball

Weight of the ball downward:
$W = Mg$
Buoyant force upward:
$B = \rho_0 V g$ where $V$ is the volume of the ball.
Viscous force upward: let it be $F$ .

Thus, at terminal velocity, $Mg = F + \rho_0 V g$

So, $F = Mg - \rho_0 V g$

Express volume in terms of mass and density of the ball

Given density of the ball is $\rho$ , $\rho = \frac{M}{V} \Rightarrow V = \frac{M}{\rho}$

Substitute into the expression for $F$ : $F = Mg - \rho_0 \left(\frac{M}{\rho}\right) g$

$F = Mg\left(1 - \frac{\rho_0}{\rho}\right)$

Match with options

This matches Option A: $F = Mg\left(1-\frac{\rho_0}{\rho}\right)$

Conclusion

The viscous force on the ball is $\boxed{F = Mg\left(1-\frac{\rho_0}{\rho}\right)}$

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