The work done to get n smaller equal size spherical drops from a bigger size spherical drop of water is proportional to

(\frac{1}{n^{2 / 3}}) - 1

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(\frac{1}{n^{1 / 3}}) - 1

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n^{1 / 3} - 1

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n^{4 / 3} - 1

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Solution

The correct option is C $n^{1 / 3} - 1$
work= change in potential energy
Initial potential energy=n x T(4) 3.14 $r^{2}$
initial volume= final volume
n x (4/3) 3.14 $r^{3}$ =(4/3)3.14 $R^{3}$
we get, $R^{3}$ = n x $r^{3}$
final potential energy= T(4)3.14 $R^{2}$
work=final potential energy-initial potential energy
on solving we get,
(C) option as the correct answer.

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Similar questions

lim n \to \infty {(\frac{1^{1} \cdot 2^{2} \cdot 3^{3} . . . . . (n - 1)^{n - 1} \cdot n^{n}}{n^{1 + 2 + 3 + . . . . + n}})}^{\frac{1}{n^{2}}}

64

1.5 {m s}^{- 1}

0.1 {ms}^{- 1}

n \in R

n^{\frac{1}{3}} + \frac{1}{n^{\frac{1}{3}}} = 3

(n + \frac{1}{n}, n^{3} + \frac{1}{n^{3}}) = ?

n_{1}

n_{2}

△ E = R_{H} \times c [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]

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