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A small hole is made at the bottom of a symmetric jar (a surface of revolution). Liquid is filled to some height. It is found that the rate at which the liquid level falls is the same regardless of the liquid level. The jar's profile is the surface of revolution of which curve?
- y = k x⁴
- y = k x²
- y = k x³
- y = k x⁵
Correct answer: y = k x⁴
Solution
By continuity, A_surface*(dy/dt) = a*sqrt(2 g y), so dy/dt = a*sqrt(2 g y)/A_surface. For dy/dt to be independent of y, A_surface must scale as sqrt(y). Since A_surface ~ x² (radius x), we need x² ~ y^(1/2), i.e. y ~ x⁴.
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