The diameter of one drop of water is 0.2 cm. The work done in breaking one drop into 1000 droplets will be (Given, `S_("water") = 7 xx 10^(-2) Nm^(-1)`)A. `7.9 xx 10^(-6) J`B. `5.92 xx 10^(-6) J`C. `2.92 xx 10^(-6) J`D. `1.92 xx 10^(-6)J`

The diameter of one drop of water is 0.2 cm. The work done in breaking

1.	The diameter of one drop of water is 0.2 cm. The work done in breaking one drop into 1000 droplets will be (Given, `S_("water") = 7 xx 10^(-2) Nm^(-1)`)A. `7.9 xx 10^(-6) J`B. `5.92 xx 10^(-6) J`C. `2.92 xx 10^(-6) J`D. `1.92 xx 10^(-6)J`
Answer» Correct Answer - A Let R and r be the radii of big drop and each smaller drop, respectively `:.` Volume of big drop = Volume of smaller drops `(4)/(3)pi R^(3) = 1000 xx (4)/(3) pi xx r^(3)` `rArr r = (R)/(10) = (0.1 xx 10^(-2))/(10) = 0.1 xx 10^(-4) m` (`because R = 0.1 xx 10^(-2) m`) The work done in breaking the big drop into 100 small droplets is `W = S(1000 xx 4pi r^(2) - 4pi R^(2))` `= S xx 4pi (1000 xx 10^(-8) - 10^(-6)) = 7.9 xx 10^(-6) J`

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The diameter of one drop of water is 0.2 cm. The work done in breaking one drop into 1000 droplets will be (Given, `S_("water") = 7 xx 10^(-2) Nm^(-1)`)A. `7.9 xx 10^(-6) J`B. `5.92 xx 10^(-6) J`C. `2.92 xx 10^(-6) J`D. `1.92 xx 10^(-6)J`

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