A mercury drop of 1, cm radius is broken into | vedclass

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Question

(A) The radius of the large drop is $R = 1 \, cm = 10^{-2} \, m$.The number of small drops is $n = 10^6$.The surface tension is $T = 35 	imes 10^{-3}

Vedclass · Accepted Answer

$4.4 	imes {10^{ - 3}} \, J$

Vedclass · Answer

(A) The radius of the large drop is $R = 1 \, cm = 10^{-2} \, m$.The number of small drops is $n = 10^6$.The surface tension is $T = 35 	imes 10^{-3} \, N/cm = 35 	imes 10^{-3} \, N / (10^{-2} \, m) = 3.5 \, N/m$.When a large drop of radius $R$ is broken into $n$ small drops of radius $r$,the volume remains conserved: $\frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3$,which implies $r = R / n^{1/3}$.The energy required is equal to the increase in surface area multiplied by the surface tension: $W = T 	imes (A_{final} - A_{initial})$.$W = T 	imes (n \cdot 4\pi r^2 - 4\pi R^2) = 4\pi T R^2 (n^{1/3} - 1)$.Substituting the values: $W = 4 	imes 3.14 	imes 3.5 	imes (10^{-2})^2 	imes ((10^6)^{1/3} - 1)$.$W = 4 	imes 3.14 	imes 3.5 	imes 10^{-4} 	imes (10^2 - 1) = 43.96 	imes 10^{-4} 	imes 99 \approx 4.35 	imes 10^{-3} \, J$.Rounding to the nearest provided option,the energy used is $4.4 	imes 10^{-3} \, J$.

$A$ mercury drop of $1\, cm$ radius is broken into ${10^6}$ small drops. The energy used will be (surface tension of mercury is $35 \times {10^{ - 3}} \, N/cm$).

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