Question

The time period ( $T$ ) of small oscillations of the surface of a liquid drop depends on its surface tension ( $s$ ), the density ( $\rho$ ) of the liquid and it's mean radius ( $r$ ) as $T=c{s}^x{\rho }^y{r}^z$ . If in the measurement of the mean radius of the drop, the error is $2\%,$ the error in the measurement of surface tension and density both are of $1\%$ . Find the percentage error in measurement of the time period.

Answer 1

Answer: 4

$T=c{s}^x{\rho }^y{r}^z$
$[T]=[s{]}^x[\rho {]}^y[r{]}^z=[M{T}^{-2}{]}^x[M{L}^{-3}{]}^y[L{]}^z\Rightarrow [T]=M^{x+y}{L}^{-3y+z}{T}^{-2x}x+y=0$
$-2x=1\Rightarrow x=-\frac{1}{2},y=\frac{1}{2}-3y+z=0$ , $z=\frac{3}{2}$
$T=c\sqrt{\frac{\rho r^3}{s}}\Rightarrow \frac{\text{ΔT}}{T}=\frac{1}{2}\frac{\text{Δρ}}{\rho }+\frac{3}{2}\frac{\text{Δr}}{r}+\frac{1}{2}\frac{\text{Δs}}{s}=\frac{1}{2}+3+\frac{1}{2}=4$