A tank has a small hole on its side at a height `y_(1)`. It is filled with a liquid (density `rho`) to a height `y_(2)`. If the absolute pressure at the top of the fluid is `P_(t)`, find the velocity with which it leaves the tank. Assume that the cross-sectional area of the tank is larger as compared to that of the hole.

Options

A. `sqrt((2(P_(t)-P_(0)))/(rho)+2g(y_(2)-y_(1)))`
B. `sqrt((P_(t)-P_(0))/(rho)+2g(y_(2)-y_(1)))`
C. `sqrt((P_(t)-P_(0))/(rho)-2g(y_(2)-y_(1)))`
D. `sqrt((2P_(t))/(rho) +2g(y_(2)-y_(1)))`

Answer

A

Explanation

From equation of continuity, `v_(2) = (A_1)/(A_2) v_(1)`
Since `A_(1) lt lt A_(2), v_(2)` must be very small compared to velocity of efflux at the hole, therefore we can take `v_(2)=0`
Fluid emerging form the holes is open to atmosphereic pressure `P_(0)`. we take two points `A` and `B` at the top of the fluid and at the hole respectively. From Bernoulli's principle,
`P_(0)+1/2 rhov_(1)^(2)+ rho gy_(1) = P_(1) +rho gy_(2)`
Solving for `v_(1)`, we obtain `v_(1) = sqrt((2(p_(t)-p_(0)))/(rho))+2gh`.