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Question

$(A)$ Radius of the large drop, $R = 1 \,cm = 10^{-2} \,m$.Number of smaller droplets, $n = 10^6$.Surface tension of mercury, $T = 435 	imes 10^{-3}

Vedclass · Accepted Answer

$54.1 	imes 10^{-3} \,J$

Vedclass · Answer

$(A)$ Radius of the large drop, $R = 1 \,cm = 10^{-2} \,m$.Number of smaller droplets, $n = 10^6$.Surface tension of mercury, $T = 435 	imes 10^{-3} \,N/m$.The work done in splitting a large drop of radius $R$ into $n$ smaller droplets is given by the change in surface energy: $W = T 	imes \Delta A$, where $\Delta A$ is the increase in surface area.The surface area of the large drop is $A_1 = 4 \pi R^2$.If $r$ is the radius of each small droplet, then by conservation of volume: $\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3$, which gives $r = R / n^{1/3}$.The total surface area of $n$ small droplets is $A_2 = n 	imes 4 \pi r^2 = n 	imes 4 \pi (R / n^{1/3})^2 = 4 \pi R^2 n^{1/3}$.The increase in area is $\Delta A = A_2 - A_1 = 4 \pi R^2 (n^{1/3} - 1)$.Substituting the values:$W = 4 	imes \pi 	imes (10^{-2})^2 	imes 435 	imes 10^{-3} 	imes ((10^6)^{1/3} - 1)$$W = 4 	imes 3.14159 	imes 10^{-4} 	imes 435 	imes 10^{-3} 	imes (100 - 1)$$W = 4 	imes 3.14159 	imes 435 	imes 10^{-7} 	imes 99$$W \approx 54.1 	imes 10^{-3} \,J$.

$A$ mercury drop of radius $1 \,cm$ is sprayed into $10^6$ droplets of equal size. Calculate the energy expended if surface tension of mercury is $435 \times 10^{-3} \,N/m$.

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