Properties of Matter 3 Question 7
13. A drop of liquid of radius $R=10^{-2} m$ having surface tension $S=\frac{0.1}{4 \pi} Nm^{-1}$ divides itself into $K$ identical drops. In this process the total change in the surface energy $\Delta U=10^{-3} J$. If $K=10^{\alpha}$, then the value of $\alpha$ is
(2017 Adv.)
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Answer:
Correct Answer: 13. (a) $\frac{g}{50}$
(b) $\sqrt{\frac{g m _0}{2 A \rho}}$
Solution:
- (a) Mass of water $=($ Volume $)($ density $)$
$$ \begin{array}{ll} \therefore & m _0=(A H) \rho \\ \therefore & H=\frac{m _0}{A \rho} \end{array} $$
Velocity of efflux,
$$ \begin{aligned} v & =\sqrt{2 g H}=\sqrt{2 g \frac{m _0}{A \rho}} \\ & =\sqrt{\frac{2 m _0 g}{A \rho}} \end{aligned} $$
Thrust force on the container due to draining out of liquid from the bottom is given by,
$F=$ (density of liquid) (area of hole)(velocity of efflux) ${ }^{2}$
$$ \begin{aligned} F & =\rho a v^{2} \\ F & =\rho(A / 100) v^{2} \\ & =\rho(A / 100)\left(\frac{2 m _0 g}{A \rho}\right) \\ F & =\frac{m _0 g}{50} \end{aligned} $$
$\therefore$ Acceleration of the container, $a=F / m _0=g / 50$
(b) Velocity of efflux when $75 %$ liquid has been drained out i.e. height of liquid, $h=\frac{H}{4}=\frac{m _0}{4 A \rho}$
