Question

A small hole is made at the bottom of a symmetrical jar as shown in figure. A liquid is filled into the jar upto a certain height. The rate of descension of liquid is independent of the level of the liquid in the jar. Then the surface of the jar is a surface of revolution of the curve –

• A. y = kx⁴
• B. y = kx²
• C. y = kx³
• D. y = kx⁵

Answer 1

Answer: (A)

y = kx⁴

Answer 2

Let y be the height of liquid at some instant.

Then $–\frac{dy}{dt}$ = constant (given)

From equation of continuity,

(px²) $\left( –\frac{dy}{dt} \right)$ = $a\sqrt{2gy}$ (a = area of hole)

Here, $\pi\left( \frac{–dy}{dt} \right)$, a and g are constants

Hence, squaring the equation, we get y = kx⁴.