A mercury drop of radius 10^{-3} m is broken | vedclass

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(A) Initial radius $R = 10^{-3} \ m$. Surface tension $T = 0.45 \ Nm^{-1}$.Initial surface area $A_i = 4 \pi R^2$.Initial surface energy $E_i = T 	im

Vedclass · Accepted Answer

$2.26$

Vedclass · Answer

(A) Initial radius $R = 10^{-3} \ m$. Surface tension $T = 0.45 \ Nm^{-1}$.Initial surface area $A_i = 4 \pi R^2$.Initial surface energy $E_i = T 	imes 4 \pi R^2$.Let the radius of each small droplet be $r$. Since volume is conserved,$\frac{4}{3} \pi R^3 = 125 	imes \frac{4}{3} \pi r^3$.$R^3 = 125 r^3 \implies R = 5r \implies r = \frac{R}{5} = \frac{10^{-3}}{5} \ m$.Final surface area $A_f = 125 	imes 4 \pi r^2 = 125 	imes 4 \pi \left(\frac{R}{5}ight)^2 = 125 	imes 4 \pi \frac{R^2}{25} = 5 	imes 4 \pi R^2$.Gain in surface energy $\Delta E = E_f - E_i = T(A_f - A_i) = T(5 	imes 4 \pi R^2 - 4 \pi R^2) = T(4 	imes 4 \pi R^2) = 16 \pi T R^2$.Substituting values: $\Delta E = 16 	imes 3.14159 	imes 0.45 	imes (10^{-3})^2$.$\Delta E = 16 	imes 3.14159 	imes 0.45 	imes 10^{-6} \approx 22.619 	imes 10^{-6} \ J = 2.26 	imes 10^{-5} \ J$.

$A$ mercury drop of radius $10^{-3} \ m$ is broken into $125$ equal size droplets. Surface tension of mercury is $0.45 \ Nm^{-1}$. The gain in surface energy is $...... \times 10^{-5} \ J$.

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