MHTCET 15 May 2024 Shift 1 Paper

Concept:When a large liquid drop splits into smaller droplets, its total surface area increases, which increases the surface energy.The change in surface energy is given by $\Delta E = T \times (\text{final area} - \text{initial area})$.Explanation:The initial surface area of the mercury drop of radius $R$ is $4\pi R^{2}$.So, the initial surface energy is $E_i = 4\pi R^{2} T$.Each small droplet has radius $r$ and surface area $4\pi r^{2}$.For 1000 droplets, the total final surface area is $1000 \times 4\pi r^{2} = 4000\pi r^{2}$.Thus, the final surface energy is $E_f = 4000\pi r^{2} T$.The change in surface energy is $\Delta E = E_f - E_i = 4000\pi r^{2} T - 4\pi R^{2} T$.Since volume is conserved: $\frac{4}{3}\pi R^{3} = 1000 \times \frac{4}{3}\pi r^{3}$.This gives $R^{3} = 1000 r^{3}$, hence $r = \frac{R}{10}$.Substitute $r = \frac{R}{10}$: $\Delta E = 4000\pi \left(\frac{R}{10}\right)^{2} T - 4\pi R^{2} T$.This simplifies to $\Delta E = 40\pi R^{2} T - 4\pi R^{2} T = 36\pi R^{2} T$.Answer:The correct option is D: $36 \pi R^{2} T$.