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The frequency $(v)$ of an oscillating liquid drop may depend upon radius $(r)$ of the drop, density $(\rho)$ of liquid and the surface tension $(s)$ of the liquid as :
$v=r^a \rho^b s^c$. The values of $a, b$ and $c$ respectively are
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$v=r^a \rho^b s^c$. The values of $a, b$ and $c$ respectively are
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The correct answer is:
$\left(-\frac{3}{2},-\frac{1}{2} ; \frac{1}{2}\right)$
$\begin{aligned} & v=\mathrm{r}^{\mathrm{a}} \rho^{\mathrm{b}} \mathrm{s}^{\mathrm{c}} \Rightarrow[v]=[\mathrm{r}]^{\mathrm{a}}[\rho]^{\mathrm{b}}[\mathrm{s}]^{\mathrm{c}} \\ & \Rightarrow \quad\left[\mathrm{T}^{-1}\right]=\left[\mathrm{L}^1\right]^{\mathrm{a}}\left[\mathrm{M}^1 \mathrm{~L}^{-3}\right]^{\mathrm{b}}\left[\frac{\mathrm{MLT}}{\mathrm{L}}\right]^{-2} \\ & \Rightarrow \mathrm{T}^{-1}=\mathrm{M}^{\mathrm{b}+\mathrm{c}} \cdot \mathrm{L}^{\mathrm{a}-3 \mathrm{~b}} \cdot \mathrm{T}^{-2 c} \\ & \Rightarrow \mathrm{b}+\mathrm{c}=0, \mathrm{a}-3 \mathrm{~b}=0,-2 \mathrm{c}=-1 \\ & \Rightarrow \quad \mathrm{c}=\frac{1}{2}, \mathrm{~b}=-\frac{1}{2}, \mathrm{a}-3 \mathrm{~b}=0 ; \\ & \mathrm{a}+\frac{3}{2}=0 \Rightarrow \mathrm{a}=-\frac{3}{2}\end{aligned}$
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