MHTCET 19 Apr 2026 Shift 1 Paper

Concept:When identical charged drops merge, their charges add up while the total volume remains conserved, changing the radius and hence the electric potential of the resulting drop.Explanation:Let the radius of each small spherical drop be $r$ and its charge be $q$.The potential of a charged spherical drop is:$$V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$$When $n$ identical drops coalesce, the total charge becomes:$$Q = nq$$Volume is conserved during coalescence, so if the big drop has radius $R$:$$\frac{4}{3}\pi R^3 = n\left(\frac{4}{3}\pi r^3\right)$$Simplifying gives:$$R = n^{1/3} r$$The potential of the big drop is therefore:$$V' = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R}$$Substituting $Q = nq$ and $R = n^{1/3}r$:$$V' = \frac{1}{4\pi\varepsilon_0} \frac{nq}{n^{1/3}r} = n^{2/3}\left(\frac{1}{4\pi\varepsilon_0}\frac{q}{r}\right)$$Recognizing that the bracket equals $V$:$$V' = n^{2/3} V$$Answer:The potential of the big drop is $(n^{2/3})V$, which matches Option D.