If 1000 droplets of water of surface tension | vedclass

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(B) Let $r$ be the radius of each small droplet and $R$ be the radius of the large drop.Given: $r = 1\,mm = 10^{-3}\,m$,$n = 1000$,and surface tension

Vedclass · Accepted Answer

$7.92 	imes 10^{-4}\,J$

Vedclass · Answer

(B) Let $r$ be the radius of each small droplet and $R$ be the radius of the large drop.Given: $r = 1\,mm = 10^{-3}\,m$,$n = 1000$,and surface tension $T = 0.07\,N/m$.Since the volume remains constant: $n 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$.$1000 	imes r^3 = R^3 \implies R = 10r = 10 	imes 10^{-3}\,m = 10^{-2}\,m$.The released surface energy $\Delta E$ is the difference between the initial surface area and final surface area multiplied by $T$.$\Delta E = T 	imes (n 	imes 4 \pi r^2 - 4 \pi R^2) = 4 \pi T (n r^2 - R^2)$.Substituting the values: $\Delta E = 4 	imes \frac{22}{7} 	imes 0.07 	imes (1000 	imes (10^{-3})^2 - (10^{-2})^2)$.$\Delta E = 4 	imes 22 	imes 0.01 	imes (1000 	imes 10^{-6} - 100 	imes 10^{-6})$.$\Delta E = 0.88 	imes (10^{-3} - 10^{-4}) = 0.88 	imes (900 	imes 10^{-6}) = 0.88 	imes 9 	imes 10^{-4} = 7.92 	imes 10^{-4}\,J$.

If $1000$ droplets of water of surface tension $0.07\,N/m$,each having the same radius $1\,mm$,combine to form a single drop,the released surface energy in the process is: (Take $\pi = \frac{22}{7}$)

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