MHT CET · Physics · Mechanical Properties of Fluids

The velocity of a small ball of mass ' \(M\) ' and density ' \(\mathrm{d}_1\) ' when dropped in a container filled with glycerin becomes constant after some time. If the density of glycerin is ' \(\mathrm{d}_2\) ', the viscous force acting on the ball is ( \(\mathrm{g}\) = acceleration due to gravity)

A \(\operatorname{Mg} \frac{\mathrm{d}_1}{\mathrm{~d}_2}\)
B \(\operatorname{Mgd}_1 \mathrm{~d}_2\)
C \(\operatorname{Mg}\left(\mathrm{d}_1-\mathrm{d}_2\right)\)
D \(\operatorname{Mg}\left(1-\frac{\mathrm{d}_2}{\mathrm{~d}_1}\right)\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\operatorname{Mg}\left(1-\frac{\mathrm{d}_2}{\mathrm{~d}_1}\right)\)

Step-by-step Solution

Detailed explanation

Since the velocity is constant, the net force acting on the ball is zero.
\(
\begin{aligned}
& \mathrm{F}_{\mathrm{V}}+\mathrm{F}_{\mathrm{B}}=\mathrm{Mg} \\
& \mathrm{F}_{\mathrm{V}}+\mathrm{Vd}_2 \mathrm{~g}=\mathrm{Mg}
\end{aligned}
\)
\(
\mathrm{F}_{\mathrm{v}}+\frac{\mathrm{M}}{\mathrm{d}_1} \mathrm{~d}_2 \mathrm{~g}=\mathrm{Mg}
\)
\(
\mathrm{F}_{\mathrm{V}}=\mathrm{Mg}\left(1-\frac{\mathrm{d}_2}{\mathrm{~d}_1}\right)
\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From MHT CET

Explore more questions on app