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(A) Let the radius of the small drop be $r = 1 \ mm = 1 	imes 10^{-3} \ m$.Let the radius of the large drop be $R$.The volume of the large drop is eq

Vedclass · Accepted Answer

$8.146 	imes 10^{-4} \ J$

Vedclass · Answer

(A) Let the radius of the small drop be $r = 1 \ mm = 1 	imes 10^{-3} \ m$.Let the radius of the large drop be $R$.The volume of the large drop is equal to the sum of the volumes of $1000$ small drops:$\frac{4}{3} \pi R^3 = 1000 	imes \frac{4}{3} \pi r^3$$R^3 = 1000 r^3 \implies R = 10r = 10 	imes 10^{-3} \ m = 10^{-2} \ m$.The initial surface area of $1000$ drops is $A_i = 1000 	imes 4 \pi r^2$.The final surface area of the single drop is $A_f = 4 \pi R^2 = 4 \pi (10r)^2 = 400 \pi r^2$.The change in surface area is $\Delta A = A_f - A_i = 400 \pi r^2 - 1000 	imes 4 \pi r^2 = -3600 \pi r^2$.The energy loss is $\Delta E = S 	imes |\Delta A| = S 	imes 3600 \pi r^2$.Substituting the values: $\Delta E = 0.072 	imes 3600 	imes \pi 	imes (10^{-3})^2$.$\Delta E = 0.072 	imes 3600 	imes 3.14159 	imes 10^{-6} \approx 8.143 	imes 10^{-4} \ J$.Rounding to the given option,the energy loss is $8.146 	imes 10^{-4} \ J$.

$1000$ droplets of water,each having a diameter of $2 \ mm$,coalesce to form a single drop. Given the surface tension of water is $0.072 \ N/m$. The energy loss in the process is

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