Physics Circle home Student Login

Home›Chapters›Surface Tension PYQs›Splitting & Merging of Drops›Question 51

MHT CET 2019 (2 May, Shift 1) · Splitting & Merging of Drops

Splitting & Merging of Drops question from MHT CET 2019 (2 May, Shift 1)

Q51MHT CET 2019 · 2 MAY · SHIFT 1Medium

Splitting & Merging of Drops

Two small drops of mercury each of radius ' RR ' coalesce to form a large single drop. The ratio of the total surface energies before and after the change is

Show answer and solution

Answer: (c)

  1. R′=21/3RR' = 2^{1/3} R (volume conserved).
  2. EbeforeEafter=2⋅4πR2T4π22/3R2T=21/3\frac{E_{before}}{E_{after}} = \frac{2 \cdot 4\pi R^2 T}{4\pi 2^{2/3} R^2 T} = 2^{1/3}
  3. So the answer is (c) 21/3:12^{1/3} : 1.
Report a mistake

More questions on Splitting & Merging of Drops