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  • The frequency of an oscillating liquid drop may depend upon radius <img alt="\mathrm{(r)}" sr

The frequency \mathrm{(v)} of an oscillating liquid drop may depend upon radius \mathrm{(r)} of the drop, density \mathrm{(\rho)} of liquid and the surface tension (s) of the liquid as :v=r^{\mathrm{a}} \rho^{\mathrm{b}} \mathrm{s}. The values of \mathrm{a}, \mathrm{b}$ and $\mathrm{c} respectively are

 

Option: 1

\left(-\frac{3}{2}, \frac{1}{2}, \frac{1}{2}\right)
 


Option: 2

\left(\frac{3}{2},-\frac{1}{2}, \frac{1}{2}\right)
 


Option: 3

\left(-\frac{3}{2},-\frac{1}{2}, \frac{1}{2}\right)


Option: 4

\left(\frac{3}{2}, \frac{1}{2},-\frac{1}{2}\right)


Answers (1)

best_answer

\begin{aligned} & \nu \ \ \alpha \ \ \mathrm{r}^a \rho^b \mathrm{~s}^c \\ & T^{-1}=[\mathrm{L}]^{\mathrm{a}}\left[\mathrm{M} \mathrm{L}^{-3}\right]^b\left[\mathrm{MT}^{-2}\right]^c \\ \end{aligned}

\begin{aligned} & T^{-1}=\mathrm{L}^{\mathrm{a}-3 b}, \mathrm{M}^{\mathrm{b}+\mathrm{c}} \mathrm{T}^{-2 c} \\ & -2 \mathrm{c}=-1 \ldots \ldots . . \text { (1) } \\ & \mathrm{c}=1 / 2 \\ \end{aligned}

\begin{aligned} & \mathrm{~b}+\mathrm{c}=0 \ldots \ldots \ldots .(2) \\ \end{aligned}

\begin{aligned} & \mathrm{b}=-1 / 2 \\ & \mathrm{a}-3 \mathrm{~b}=0 \ldots \ldots \ldots .(3) \\ \end{aligned}

\begin{aligned} & \mathrm{a}=3 \mathrm{~b}=\frac{-3}{2} \end{aligned}

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