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

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