To find the surface energy released when eight mercury drops coalesce, we need to compare the initial total surface area of the small drops with the final surface area of the large drop.
Calculating Initial Surface Area
The surface area of a single spherical drop with radius r is given by the formula $A_{small} = 4 \pi r^2$.
There are eight such drops. So, the total initial surface area is:
$ A_{initial} = 8 \times A_{small} = 8 \times (4 \pi r^2) = 32 \pi r^2 $
Calculating Final Surface Area
Let the radius of the larger, combined drop be R. The volume is conserved during coalescence.
The volume of one small drop is $V_{small} = \frac{4}{3} \pi r^3$.
The total volume of the eight small drops is:
$ V_{total\_small} = 8 \times V_{small} = 8 \times \frac{4}{3} \pi r^3 $
The volume of the single large drop is $V_{large} = \frac{4}{3} \pi R^3$.
Equating the volumes:
$ \frac{4}{3} \pi R^3 = 8 \times \frac{4}{3} \pi r^3 $
Simplifying this equation gives:
$ R^3 = 8 r^3 $
Taking the cube root of both sides:
$ R = 2r $
Now, calculate the surface area of this larger drop:
$ A_{final} = 4 \pi R^2 = 4 \pi (2r)^2 = 4 \pi (4r^2) = 16 \pi r^2 $
Calculating Surface Energy Released
Surface energy is calculated as Surface Area multiplied by the surface tension (S).
Initial Surface Energy: $E_{initial} = A_{initial} \times S = 32 \pi r^2 S$
Final Surface Energy: $E_{final} = A_{final} \times S = 16 \pi r^2 S$
The surface energy released is the difference between the initial and final surface energies:
$ \text{Energy Released} = E_{initial} - E_{final} $
$ \text{Energy Released} = 32 \pi r^2 S - 16 \pi r^2 S $
$ \text{Energy Released} = 16 \pi r^2 S $
Conclusion
The surface energy released in this process is $16 \pi r^2 S$. This corresponds to Option B.
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Important Questions from Physical and Thermal Properties of Bulk Matter
The strain-stress plot for materials $A, B, C$ and $D$ is shown in the figure. Which material has the largest Young's modulus ?
- An ideal gas at pressure $P$ and temperature $T$ is expanding such that $P T^{3} = \text{constant}$. The coefficient of volume expansion of the gas is ___________.
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Given $T_3 + T_1 = 2T_2$ and $T_2 - T_1 = \Delta T$, the value of $\Delta L_2$ is _______. - A lift of mass 1600 kg is supported by thick iron wire. If the maximum stress which the wire can withstand is $4 \times 10^8 \text{ N/m}^2$ and its radius is 4 mm, then maximum acceleration the lift can take is _________ $\text{m/s}^2$.
(take $g = 10 \text{ m/s}^2$ and $\pi = 3.14$)
