Question

If the time period ($T$) of vibrating a liquid drop depends on surface tension ($S$), radius ($r$) of the drop and density $\left(\rho \right)$ of the liquid, then the expression of $T$ is :

• A. $T=K\sqrt{\frac{\rho r^3}{S}}$
• B. $T=K\sqrt{\frac{\rho {{}^{1/2}r}^3}{S}}$
• C. $T=K\sqrt{\frac{\rho r^3}{S^{1/2}}z}$
• D. none

Answer 1

Answer: (A)

$T=K\sqrt{\frac{\rho r^3}{S}}$

$T\propto S^a{r}^b{\rho }^c$
$M^0{L}^0{T}^1=\left(M^1{T}^{-2}{)}^a\right(L^1{)}^b(M{L}^{-3}{)}^c$
$a+c=0,b-3c=0,-2a=1$
$c=\frac{1}{2},$ $b=\frac{3}{2}$
$a=-\frac{1}{2}\Rightarrow T\alpha$ $S^{-\frac{1}{2}}$$r^{\frac{3}{2}}$ $p^{\frac{1}{2}}$
$T=K\sqrt{\frac{p{r}^3}{s}}$