NEET-XII-Physics
14: Some Mechanical Properties of Matter
- #24A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury = 0.465 J m-2.Ans : Given:
Initial radius of mercury drop R = 2 mm = 2 × 10-3 m
Surface tension of mercury T = 0.465 J/m2
Let the radius of a small drop of mercury be r.
As one big drop is split into 8 identical droplets:
volume of initial drop = 8 × (volume of a small drop)
`` \left(\frac{4}{3}\right)\pi {R}^{3}=\left(\frac{4}{3}\right)\,\mathrm{\,\pi \,}{r}^{3}\times 8``
`` ``
Taking cube root on both sides of the above equation:
`` r=\frac{R}{2}=10``-3 m
Surface energy = T × surface area
∴ Increase in surface energy = TA' - TA
= (8 × 4 πr2 - 4πR2) T
`` =4\pi T\left[8\times \left(\frac{{R}^{2}}{4}\right)-{R}^{2}\right]``
`` =4\pi T{R}^{2}``
= 4 × (3.14) × (0.465) × (4 × 10-6)
= 23.36 × 10-6
= 23.4 μJ
Hence, the required increase in the surface energy of the mercury droplets is 23.4 μJ.
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