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NEET PHYSICSMedium

Twenty seven drops of same size are charged at 200 V200 \text{ V} each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

A

660 V660 \text{ V}

B

1320 V1320 \text{ V}

C

1520 V1520 \text{ V}

D

1980 V1980 \text{ V}

Step-by-Step Solution

Electric potential due to a charged sphere = kQR\frac{kQ}{R}. For smaller drop, V=kqr=200 VV = \frac{kq}{r} = 200 \text{ V}. As volume remains the same, (43πr3)×27=43πR3⇒R=273r=3r(\frac{4}{3}\pi r^3) \times 27 = \frac{4}{3}\pi R^3 \Rightarrow R = \sqrt[3]{27} r = 3r. Using charge conservation, Q=27qQ = 27q. Vbig drop=kQR=k(27q)3r=9(kqr)=9×220=1980 VV_{\text{big drop}} = \frac{kQ}{R} = \frac{k(27q)}{3r} = 9(\frac{kq}{r}) = 9 \times 220 = 1980 \text{ V}.

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