JEE MAIN - Physics (2021 - 26th February Evening Shift - No. 25)
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.
Answer
243
Explanation
$$(27)\left( {{4 \over 3}\pi {r^3}} \right) = {4 \over 3}\pi {R^3}$$
R = 3r
Potential energy of smaller drop :
$${U_1} = {3 \over 5}{{k{q^2}} \over r}$$
Potential energy of bigger drop :
$$U = {3 \over 5}{{k{Q^2}} \over R}$$
$$U = {3 \over 5}{{k{{(27q)}^2}} \over R}$$
$$U = {3 \over 5}k{{(27)(27){q^2}} \over {3r}}$$
$$U = {{(27)(27)} \over 3}\left( {{3 \over 5}{{k{q^2}} \over r}} \right)$$
$$U = 243\,{U_1}$$
R = 3r
Potential energy of smaller drop :
$${U_1} = {3 \over 5}{{k{q^2}} \over r}$$
Potential energy of bigger drop :
$$U = {3 \over 5}{{k{Q^2}} \over R}$$
$$U = {3 \over 5}{{k{{(27q)}^2}} \over R}$$
$$U = {3 \over 5}k{{(27)(27){q^2}} \over {3r}}$$
$$U = {{(27)(27)} \over 3}\left( {{3 \over 5}{{k{q^2}} \over r}} \right)$$
$$U = 243\,{U_1}$$
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