MCQ
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 $......\times 10^{-5}\,J$
- ✓$2.26$
- B$28$
- C$17.5$
- D$5$
✓
Answer
Correct option: A.
$2.26$
a
Initial surface energy $=0.45 \times 4 \pi\left(10^{-3}\right)^2$
Initial surface energy $=0.45 \times 4 \pi\left(10^{-3}\right)^2$
$\frac{4}{3} \pi\left(10^{-3}\right)^3=125 \times \frac{4 \pi}{3} R_{\text {new }}^3$
$10^{-3}=5 R_{\text {new }}$
$R_{\text {new }}=\frac{10^{-3}}{5}\,m$
So, final surface energy $=0.45 \times 125 \times 4 \pi\left(\frac{10^{-3}}{5}\right)^2$
Increase in energy $=0.45 \times 4 \pi \times\left(10^{-3}\right)^2\left[\frac{125}{25}-1\right]$
$=4 \times 0.45 \times 4 \pi \times 10^{-6}$
$=2.26 \times 10^{-5}\,J$
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