Question 118 of 192

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

A$2.26\times {10}^{–5}J$B$28\times {10}^{–5}J$C$17.5\times {10}^{–5}J$D$5\times {10}^{–5}J$

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Answer: (A) $2.26\times {10}^{–5}J$

As the volume of the mercury will remain same, therefore
$\dfrac{4}{3}\pi {R}^{3}=125\left(\dfrac{4}{3}\pi {r}^{3}\right) \Rightarrow R=5r$
Now, increase in surface energy
$∆U=125\times \left(S\times 4\pi {r}^{2}\right)-\left(S\times 4\pi {R}^{2}\right)$
$=0.45\times 4\pi \times {\left({10}^{-3}\right)}^{2}\left[\dfrac{125}{25}-1\right]$
$=4\times 0.45\times 4\pi \times {10}^{-6}$
$=2.26\times {10}^{-5}J$

JEE Main Physics practice question on Fluid Mechanics (Surface tension), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.