MHT-CET Physics and Chemistry Full Test-7 PHY-Que-2 Solution

Q.2. Eight small drops of mercury each of radius ‘r’, coalesce to form a large single drop. The ratio of total surface energy before and after the change is

A. 2 : 1

B. 1 : 1

C. 1 : 4

D. 1 : 8

Answer :- A. 2 : 1

Explanation :-

When mercury drops coalesce to form a larger drop, the volume is conserved. The total initial volume of the eight small drops is equal to the final volume of the large single drop.

For a single small drop with radius r, its volume is:

V_small= \frac{4}{3}\pi R^{3}

For eight such drops, the total initial volume is:

V_initial=8× \frac{4}{3}\pi R^{3}= \frac{32}{3}\pi R^{3}

Let the radius of the large drop be R. Since the volume is conserved:

V_large= \frac{4}{3}\pi R^{3}=\frac{32}{3}\pi R^{3}

Equating the volumes:

\frac{4}{3}\pi R^{3}=\frac{32}{3}\pi R^{3}

Solving for R gives:

R³=8r³⟹R=2r

The total surface energy of the drops depends on the surface area and the surface tension (σ). The formula for surface energy is:

Surface Energy =σ × Surface Area

The surface area of a single small drop is:

A_small=4πr²

The total initial surface area of the eight small drops is:

A_initial=8 × 4πr²=32πr²

The surface area of the large drop is:

A_large=4πR²=4π(2r)²=16πr²

Thus, the initial total surface energy of the small drops is:

E_initial=σ × 32πr²

The surface energy of the large drop is:

E_large=σ×16πr²

The ratio of the total surface energy before and after the amalgamation is:

\frac{E_{initial}}{E_{large}}=\frac{\sigma \times 32\pi r^{2}}{\sigma \times 16\pi r^{2}}=\frac{32}{16}=2

Therefore, the ratio is:

Option A: 2 : 1