Physics Circle home Student Login

Home›Chapters›Surface Tension PYQs›Splitting & Merging of Drops›Question 21

MHT CET 2025 (19 Apr, Shift 1) · Splitting & Merging of Drops

Splitting & Merging of Drops question from MHT CET 2025 (19 Apr, Shift 1)

Q21MHT CET 2025 · 19 APR · SHIFT 1Medium

Splitting & Merging of Drops

When a big drop of water is formed from ' nn ' small drops of water, the energy loss is ' 3 E ' where ' EE ' is the energy of the bigger drop. The radius of the bigger drop is ' R ' and that of smaller drop is ' rr ' then the value of ' nn ' is

Show answer and solution

Answer: (b)

  1. E=4πR2TE = 4\pi R^2 T (big drop). Energy lost =4πT(nr2−R2)= 4\pi T(n r^2 - R^2).
  2. 4πT(nr2−R2)=3⋅4πR2T⇒nr2=4R24\pi T(n r^2 - R^2) = 3 \cdot 4\pi R^2 T \Rightarrow n r^2 = 4R^2
  3. So the answer is (b) n=4R2r2n = \frac{4R^2}{r^2}.
Report a mistake

More questions on Splitting & Merging of Drops