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MHTCET 26 Apr 2025 Shift 1 Paper

Question : 10 of 150

Marks: +1, -0

Solution:

Concept:

Terminal velocity of a small spherical drop in a viscous medium follows Stokes law, so

v \propto r^2

Explanation:

For a sphere falling with constant velocity, terminal velocity is given by:

v = \frac{2}{9} \frac{(\rho - \sigma) g r^2}{\eta}

Therefore,

v \propto r^2

When 125 identical small drops of radius

r

coalesce into one big drop of radius

R

, volume remains conserved.

\frac{4}{3} \pi R^3 = 125 \times \frac{4}{3} \pi r^3

R^3 = 125 r^3 \Rightarrow R = 5r

Now, using the scaling law:

\frac{v_{\text{big}}}{v_{\text{small}}} = \left(\frac{R}{r}\right)^2 = (5)^2 = 25

Given

v_{\text{small}} = 4\ \text{cm/s} = 0.04\ \text{m/s}

v_{\text{big}} = 25 \times 0.04 = 1.0\ \text{m/s}

Answer:

Option B:

1\ \text{m/s}

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