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Physics Question 34 – JEE-MAIN 2026

A spherical liquid drop of radius R acquires the terminal velocity v1 when falls through a gas of viscosity η. Now the drop is broken into 64 identical droplets and each droplet acquires terminal velocity v2 falling through the same gas. The ratio of terminal velocities v1/v2 is _______.

When a large drop breaks into smaller droplets, the total volume of the liquid remains constant.

Step 1: Relate the radii of the large drop and small droplets✦ Active

When the large drop breaks into 64 identical smaller droplets, the total volume of the liquid remains constant. Let R be the radius of the large drop and r be the radius of each small droplet.

43πR3=64×43πr3R3=64r3R=(64)1/3rR=4r

Thus, the radius of a small droplet is r=R4.

Step 2: Apply Stokes' Law for terminal velocity○ Expand

The terminal velocity vt of a spherical drop of radius r falling through a viscous medium is given by Stokes' Law:

vt=2r2(ρpρf)g9η

Here, ρp is the density of the liquid, ρf is the density of the gas, g is acceleration due to gravity, and η is the viscosity of the gas. Since the liquid, gas, and gravity are the same for both cases, the term 2(ρpρf)g9η is a constant. Therefore, the terminal velocity is directly proportional to the square of the radius of the drop:

vtr2

So, for the large drop, v1=CR2, and for a small droplet, v2=Cr2, where C is the constant of proportionality.

Step 3: Calculate the ratio of terminal velocities○ Expand

Now, we can find the ratio v1v2:

v1v2=CR2Cr2=R2r2=(Rr)2

Substitute the relationship R=4r from Step 1:

v1v2=(4rr)2=(4)2=16
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