If two identical mercury drops are combined to form a single drop, then its temperature will
A. decrease
B. increase
C. remains the same
D. none of these

Hint: First try to find out the volume of the two mercury drop in both the cases, first a single drop each and then when two of that drop get combined to form a single drop. Then equate both the initial volume and final volume. At last we get the radius of the combined drop in terms of initial radius.

Complete answer:
First start with finding the volume of the two identical mercury drops.
Let initial volume be
$V = 2 \times \dfrac{4}{3}\pi \left( {{R^2}} \right)$ (equation 1)
Where R is initial radius of two identical drop.

Now final volume be
$V = \dfrac{4}{3}\pi {\left( {{R_1}} \right)^2}$ (equation 2)

Equating both the equation 1 and equation 2;
${R_1} ={2^{\frac {1}{3}}}R$

Work done will be,
$W = T\left[{2 \times 4\pi {R^2} - 4\pi {{\left( {2^{\frac {1}{3}}R}\right)}^2}} \right]$
$W = 4\pi {R^2}T\left(2-{2^{\frac{2}{3}}}\right)$

There will just be a very slight temperature rise. The combined drop's surface area will be less than the two drops' initial combined surface area. The surface tension does a little bit of work as it draws in the extra surface.

According to the definition, the system is adiabatic, hence the energy must stay inside the mercury. The drop will initially wiggle. But soon, that motion will slow down due to the mercury's viscosity. That will very slightly increase the temperature.

Therefore, the work done is greater so energy will also increase. Similarly, temperature will also increase.

Hence the correct answer is Option B.

Note:When two or more small mercury drops come into touch with one another, they unite to form a bigger drop because to the strong force of cohesion that acts within the drops and surface tension. Smaller surface area and lower surface tension result in a bigger drop.