Question

Surface of vessel is a surface of revolution of the curve $y = kx^n$ vessel contains a liquid of density $\rho$. A small hole is made at the bottom of the vessel. If rate of fall of liquid surface in the vessel is independent of level of liquid in vessel then, the value of n is ____.

Answer 1

4

Answer 2

Solution:

1. Volume Calculation:
   For a vessel formed by rotating the curve $y = kx^n$ about the y-axis, at a height $y = h$ we have

   $$x = \left(\frac{h}{k}\right)^{1/n}.$$

   Thus the cross-sectional area at height $h$ is

   $$S(h) \propto x^2 \propto h^{2/n}.$$

   Consequently, the volume up to height $h$ is

   $$V \propto \int_0^h S(h')\, dh' \propto h^{1 + 2/n}.$$

2. Rate of Change of Volume:
   Differentiating with respect to $h$, we get

   $$\frac{dV}{dh} \propto h^{2/n}.$$

3. Outflow Rate (Torricelli's Law):
   A small hole at the bottom gives a velocity $v=\sqrt{2gh}$, so the outflow rate is

   $$\frac{dV}{dt} = -A\sqrt{2gh}.$$

   Using the chain rule:

   $$\frac{dV}{dt} = \frac{dV}{dh}\frac{dh}{dt} \propto h^{2/n}\frac{dh}{dt}.$$

   Equating the $h$-dependency from both sides:

   $$h^{2/n}\frac{dh}{dt} \propto h^{1/2}.$$

4. Condition for Constant Rate of Fall:
   For $\frac{dh}{dt}$ to be independent of $h$, we require

   $$\frac{2}{n} = \frac{1}{2}.$$

   Solving,

   $$n = 4.$$

Minimal Explanation:
Using the relation $V \propto h^{1+2/n}$, we find $\frac{dV}{dh} \propto h^{2/n}$. From the discharge $\frac{dV}{dt} \propto \sqrt{h}$, equate exponents: $2/n=\frac{1}{2}$, so $n=4$.

Answer:
$n = 4$