Physics

# The time of oscillation $T$ of a small drop of liquid depends on radius $r$, density $\rho$ and surface tension $S$. The relation between them is given by

$T\propto \sqrt {\dfrac {\rho r^3}{S}}$

##### SOLUTION
Let $T=S^ar^b{ \rho }^{ c }$
So, $[T]=[M^0L^0T^1]$
$=[M^{1}L^{-1}T^{-2}]^a[L^{1}]^b[ML^{-3}]^c$
$=[M^{a+c}L^{b-3c}T^{-2a}]$

Applying principle of homogeneity,
$a+c=0; b-3c=0; -2a=1$
On solving, we get $a=-1/2. b=3/2, c=1/2$
$T\propto \sqrt {\dfrac {\rho r^3}{S}}$

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Single Correct Medium Published on 18th 08, 2020
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