If the time period t of a drop of liquid of density d, radius r, vibrating under surface tension s is given by the formula $t=\sqrt{d^a r^b s^c}$  and if a = 1, c = - 1, then b =

Given $t=d^{a/2} r^{b/2} s^{c/2}$. Substituting dimensions, we· have . 

$$\begin{gathered} \left(T\right)={\left(M{L}^{-3}\right)}^{a/2} {\left(L\right)}^{b/2} {\left(M{T}^{-2}\right)}^{c/2} \\ =M^{\left(a+c\right)/2} L^{\left(-3a/2+b/2\right)} T^{-c} \end{gathered}$$

 Equating powers of L, we have, $-\frac{3}{2}a +\frac{b}{2}=0$. Given a = 1. 

$\therefore -\frac{3}{2}+\frac{b}{2}=0$ or b = 3