Question

The Work done to Break a spherical drop of radius "R" in n drops of equal size is proportional to?
Options:-
$1) \; \dfrac{1}{n^{2/3}} - 1$
$2) \; \dfrac{1}{n^{1/3}} - 1$
$3) \; n^{1/3} - 1$
$4) \; n^{4/3} -1$
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Answer 1

Answer:

A spherical drop of radius R is break into n drops of equal size

let it break into size of radius r

As volume remains same

So, 4/3 π R^3 = n 4/3 π r^3

R^3 = n r^3

R = n^1/3. r

r= R .n^-1/3

Work done = surface tension ×(increase in surface area)

So, work done is proportional to increase in surface area

As surface area proportional to radius^2

So, work proportional to (n r^2 - R^2)

(n R^2 n^-2/3 - R^2)

R^2( n^1/3 -1)

which is proportional to ( n^1/3 -1)

Option c) is correct

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Answer 2

Explanation:

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