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 at which level of liquid goes down 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 -

Engineering

Physics

Equation of Continuity

Question

y = kx³

y = kx⁵

y = kx⁴

y = kx²

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Let y be the height of liquid at some instant.
Then – $\frac{dy}{dt}$ = constant (given)
From equation of continuity
(πx²) $(- \frac{dy}{dt})$ = a $\sqrt{2 gy}$ (a = area of hole)
Here, π $(\frac{- dy}{dt})$ , a and g are constants
Hence, squaring the equation, we get y = kx⁴.

Please subscribe our Youtube channel to unlock this solution.