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Question

(A) The frequency of vibration $v$ of a liquid drop is governed by surface tension $\sigma$,density $\rho$,and radius $R$. The given ratio is $\frac{v

Vedclass · Accepted Answer

$k \rho g R^2 / \sigma$

Vedclass · Answer

(A) The frequency of vibration $v$ of a liquid drop is governed by surface tension $\sigma$,density $\rho$,and radius $R$. The given ratio is $\frac{v}{\sqrt{\sigma / \rho R^3}}$.Dimensions of frequency $v = [T^{-1}]$.Dimensions of surface tension $\sigma = [MT^{-2}]$.Dimensions of density $\rho = [ML^{-3}]$.Dimensions of radius $R = [L]$.Let the ratio be $X = \frac{v}{\sqrt{\sigma / \rho R^3}}$.Dimensions of $\sqrt{\frac{\sigma}{\rho R^3}} = \sqrt{\frac{MT^{-2}}{ML^{-3} \cdot L^3}} = \sqrt{\frac{MT^{-2}}{M}} = [T^{-1}]$.Thus,the ratio $\frac{v}{\sqrt{\sigma / \rho R^3}}$ is dimensionless.We check the dimensions of option $(A)$:$\frac{k \rho g R^2}{\sigma} = \frac{[ML^{-3}] \cdot [LT^{-2}] \cdot [L^2]}{[MT^{-2}]} = \frac{[ML^0T^{-2}]}{[MT^{-2}]} = [M^0L^0T^0]$.Since this expression is dimensionless,it is consistent with the dimensional analysis of the given ratio.

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