Properties Of Solids And Liquids Question 356

Question: 8 mercury drops coalesce to form one mercury drop, the energy changes by a factor of [DCE 2000]

Options:

A) 1

B) 2

C) 4

D) 6

Show Answer

Answer:

Correct Answer: C

Solution:

As volume remains constant therefore $ R={{n}^{1/3}}r $

$ \frac{\text{Energy of big drop}}{\text{Energy of small drop}}=\frac{4\pi R^{2}T}{4\pi r^{2}T}=\frac{R^{2}}{r^{2}} $

$ ={{(8)}^{2/3}}=4 $



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