SURFACE TENSION_JEE MAIN_2023_S1_01.02.23 - Competishun – Live JEE & NEET Courses and Exam Prep

JEE MAIN_2023

S1_01.02.23

Question

A mercury drop of radius $10^(–3)$ m is broken into 125 equal size droplets. Surface tension of mercury is 0.45 $Nm^(–1)$. The gain in surface energy is:

Select the correct option:

$2.26 \times 10^{-5} \mathrm{~J}$

$28 \times 10^{-5} \mathrm{~J}$

$17.5 \times 10^{-5} \mathrm{~J}$

$5 \times 10^{-5} \mathrm{~J}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Physics

Medium

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