Question

Please solve the q 195?

Answer 1

Below

Explanation:

First of all, only (1) & (3) can qualify as answers on the basis of dimensional analysis.

Apply Bernoulli's Theorem to the holes, where #2 is the lower hole:

`(P + rho g h + 1/2 rho v^2)_1 = (P + rho g h + 1/2 rho v^2)_2`

`P_1 = P_2 = P_(atm)`, and `(h_1, h_2) =(h, 0)`

`implies rho g h + 1/2 rho v_1^2 = 1/2 rho v_2^2 color(red)(implies v_2^2 - v_1^2 = 2 g h )`

If we now think what happens in time `delta t` at each hole of cross-sectional area `a`. A small volume equal to `delta V = a * v \ delta t` is emmitted from the tank.

So for that small volume:

• `delta m = rho * a \ v \ delta t`

and it adds momentum, `delta p`, to the existing flow, where:

• `delta p = delta m * v = rho a v \ delta t * v = rho a v^2 \ delta t`

`implies ((delta p)/( delta t))_(t to 0) equiv dot p = rho a v^2 \ `

This is significant because Newton's 2nd Law says that:

• `sum mathbf F = mathbf dot p`

The force imbalance is:

`F_2 - F_1 = ((dp)/(dt))_2 - ((dp)/(dt))_1 = rho a (v_2^2 - v_1^2)`

`= 2 rho a g h`