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A container of cross-section area 'S' an...

A container of cross-section area `'S'` and height `'h'` is filled with mercury up to the brim. Then the container is sealed airtight and a hole of small cross section area `'S//n'` (where `'n'` is a positive constant) is puched in its bottom. Find out the time interval upto which the mercury will come out from the bottom hole.
[Take the atmospheric pressure to be equal to `h_(0)` height of mercury column: `h gt h_(0)`]

Text Solution

Verified by Experts

The correct Answer is:
`(1)/(4)overset(T)underset(0)(int)dt ; t = nsqrt((2)/(g)(h - h_(0)))`
Let the velocity of effiux of mercury coming out of hole be `v` at an instant mercury level in container is `y`. aT same instant the speed of top surface of fluid `v`. From equilibrium of continuity
`(S)/(n)v = Sv ….(1)`
`:' (S)/(n) lt lt S :. v gt gt v`
applying Bernoulli's theorem between `A` and `B`
`Yeg + (1)/(2)ev^(3) = P_(atm) + (1)/(2) ev_(2)`
`:' v lt lt v` higher powers of `v` can
be neglected and `P_(atm) = h_(0)eg`
`:. v = sqrt(2g(Y - h_(0))) ......(2)`
Hence mercury flows out of both till `y = h_(0)`.
from equation `(1)`
`v = -(dy)/(dt) = (1)/(h) v = (1)/(h)sqrt(2g(Y - h_(0)))`
or `(-dy)/(sqrt(2gY - h_(0))) = (1)/(h) dt`
integrading between limits
at `l = 0 y = n` and `t = T, Y = h_(0) , -overset(h_(0))underset(n)(int)(-dy)/(sqrt(2g(Y - h_(0)))) = (1)/(h) = (1)/(n)overset(T)underset(0)(int)dt , t = nsqrt((2)/(g)(h - h_(0)))`
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