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QJEEM21Ph5e38: JEE Main 2021 Physics - Class 12 JEEM21Ph5e38 Solution: 27 similar drops of mercury are maintained at 10V each. All…

27 similar drops of mercury are maintained at 10V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is __________ times that of a smaller drop.

Step-by-Step Verified Solution

(27)(43πr3)=43πR3(27)\left( {{4 \over 3}\pi {r^3}} \right) = {4 \over 3}\pi {R^3}

R = 3r

Potential energy of smaller drop :

U1=35kq2r{U_1} = {3 \over 5}{{k{q^2}} \over r}

Potential energy of bigger drop :

U=35kQ2RU = {3 \over 5}{{k{Q^2}} \over R}

U=35k(27q)2RU = {3 \over 5}{{k{{(27q)}^2}} \over R}

U=35k(27)(27)q23rU = {3 \over 5}k{{(27)(27){q^2}} \over {3r}}

U=(27)(27)3(35kq2r)U = {{(27)(27)} \over 3}\left( {{3 \over 5}{{k{q^2}} \over r}} \right)

U=243 U1U = 243\,{U_1}

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