Two mercury drops,each with same radius r,mer | vedclass

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(D) Let the radius of the bigger drop be $R$. Since the volume remains constant during the merger of two drops,we have:$2 	imes \frac{4}{3} \pi r^3 =

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$2^{8/3} \pi r^2 T$

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(D) Let the radius of the bigger drop be $R$. Since the volume remains constant during the merger of two drops,we have:$2 	imes \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$$R^3 = 2r^3 \Rightarrow R = 2^{1/3} r$The surface energy $E$ of the bigger drop is given by the product of surface tension $T$ and its surface area $A = 4 \pi R^2$:$E = T 	imes 4 \pi R^2$Substituting $R = 2^{1/3} r$ into the equation:$E = T 	imes 4 \pi (2^{1/3} r)^2$$E = T 	imes 4 \pi 	imes 2^{2/3} r^2$$E = 4 	imes 2^{2/3} \pi r^2 T$Since $4 = 2^2$,we have $2^2 	imes 2^{2/3} = 2^{2 + 2/3} = 2^{8/3}$.Therefore,$E = 2^{8/3} \pi r^2 T$.

Two mercury drops,each with same radius $r$,merge to form a bigger drop. If $T$ is the surface tension of mercury,then the surface energy of the bigger drop is given by

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