MHTCET 25 Apr 2025 Shift 1 Paper

Concept:When identical small drops merge to form one big drop, total volume and total charge are conserved.The potential of the big drop is determined using the new radius and total charge.Explanation:Let the radius of the big drop be $R$ and the charge on each small drop be $q$.The potential of each small drop is given by:$$v = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$$Therefore, the charge on each small drop is:$$q = 4\pi\varepsilon_0 v r$$From conservation of volume:$$n \times \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi R^{3}$$$$\Rightarrow R = r n^{1/3}$$From conservation of charge, the total charge on the big drop is:$$Q = n q$$Now, the potential of the big drop is:$$V_{\text{big}} = \frac{1}{4\pi\varepsilon_0} \frac{n q}{R}$$Substituting $q = 4\pi\varepsilon_0 v r$ and $R = r n^{1/3}$:$$V_{\text{big}} = \frac{1}{4\pi\varepsilon_0} \frac{n (4\pi\varepsilon_0 v r)}{r n^{1/3}}$$$$V_{\text{big}} = v n^{2/3}$$Answer:$$V_{\text{big}} = n^{2/3} v$$Therefore, the correct option is D: $n^{2/3} v$.