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 $t = \sqrt{r^{2 b} s^{c} d^{a / 2}}$ It is observed that the time period is directly proportional to $\sqrt{\frac{d}{s}} .$ The value of b should therefore be :

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3 /4

2 /3

$\sqrt{3}$

3 /2

answer is C.

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Detailed Solution

$\begin{array}{l} t = \sqrt{r^{2 b} s^{c} d^{\frac{a}{2}}}; T^{2} = r^{2 b} s^{c} d^{\frac{a}{2}} \\ T^{2} = L^{2 b} {({MT}^{- 2})}^{c} {({ML}^{- 3})}^{\frac{a}{2}} \\ M^{0} L^{0} T^{2} = M^{\frac{a}{2} + c} L^{\frac{- 3 a}{2} + 2 b} T^{- 2 c} \\ - 2 c = 2 \Rightarrow c = - 1 \\ \frac{a}{2} + c = 0 \Rightarrow a = - 2 c = 2 \\ \frac{- 3 a}{2} + 2 b = 0 \Rightarrow 2 b = \frac{3 a}{2} \\ \Rightarrow b = \frac{3 a}{4} = \frac{3}{4} \times 2 = \frac{3}{2} \end{array}$