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Water is filled to height h in a fixed v...

Water is filled to height `h` in a fixed vertical cylinder placed on horizontal surface. At time `t = 0` a small hole a drilled at a height `h//4` from bottom of cylinder as shown. The cross section area of hole is a and the cross-section area of cylinder is `A` such that `A gtgt a`.

The duration of time for which water flows out of hole is.

A

`(A)/(a) sqrt((3h)/(2g))`

B

`(a)/(A) sqrt((3h)/(2g))`

C

`(A)/(a) sqrt((2h)/(3g))`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
The initial velocity of water coming out of hole is horizontal and hole is at a height `(h)/(4)` from ground. Hence time taken by water to reach ground is `t = sqrt((2(h//4))/(g))` which remains constant
`:. x = vt` where v is velocity of effux…
Since v decreases with time x will decrease

Let y be the height of water surface above hole
`:. -(dy)/(dt)=(av)/(A)=(asqrt(2gy))/(A)`
`:. underset(3h//4)overset(0)(int) (dy)/(sqrt(2gy))= underset(0) overset(t)(int) -(a)/(A) dt`
`:. t = (A)/(a) sqrt((3h)/(2g))`
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