If time period (T) of vibration of a liquid drop depends on surface tension (S) , radius (r) of drop

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1.Units, Dimensions and Measurement

medium

If the time period $(T)$ of vibration of a liquid drop depends on surface tension $(S)$, radius $(r)$ of the drop and density $(\rho )$ of the liquid, then the expression of $T$ is

A

$T = k\sqrt {\rho {r^3}/S} $

B

$T = k\sqrt {{\rho ^{1/2}}{r^3}/S} $

C

$T = k\sqrt {\rho {r^3}/{S^{1/2}}} $

D

None of these

Solution

(a) Let $T \propto {S^x}{r^y}{\rho ^z}$

by substituting the dimension of $[T] = [T]$

$[S] = [M{T^{ – 2}}],\,[r] = [L],\,[\rho ] = [M{L^{ – 3}}]$

and by comparing the power of both the sides

$x = – 1/2,\,y = 3/2,\,z = 1/2$

so $T \propto \sqrt {\rho {r^3}/S} \Rightarrow T = k\sqrt {\frac{{\rho {r^3}}}{S}} $

Standard 11

Physics

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