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  • If the time period t of the oscillation of a drop of liquid of density d, radius r, vibrating under surface tension s is given by the formula <img alt="\large t = \sqrt{r^{2b}\: \: s^{c}\: \: d^{\f

If the time period t of the oscillation of a drop of liquid of density d, radius r, vibrating under surface tension s is given by the formula \large t = \sqrt{r^{2b}\: \: s^{c}\: \: d^{\frac{a}{2}}}. It is observed that the time period is directly proportional to \large \sqrt{\frac{d}{s}}, The value of b should therefore be :

Option: 1

\frac{3}{4}
 


Option: 2

\sqrt{3}


Option: 3

\frac{3}{2}

 


Option: 4

\frac{2}{3}


Answers (1)

As given

 [t]=[r^b][\sqrt{\frac{d}{s}}]

\begin{aligned} &[T]=\left[L^{b}\right]\left[\sqrt{\frac{M / L^{3}}{M T^{-2}}}\right] \\ \Rightarrow & L^{b} / L^{3 / 2}=1 \\ b=& 3 / 2 \end{aligned}

Posted by

Sumit Saini

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