A container of large uniform cross sectional area `A`, resting on horizontal surface, holds two immiscible non viscous and incompressible liquids of density `d` and `2d`, each of height `H/2` as shown in the figure. The lower density liquid is open to the atmosphere having pressure `P_(0)`. A tiny hole of area `(s(sltltA)` is punched on the vertical side of the container at a height `h(hltH/2)`. Determine

a.the initial speed of efflux of the liquid at the hole
b. the horizontal disance `x` travelled by the liquid initially
c. the height `h_m` at which at the hole should be punched so that the liquid travels the maximum distance `x_(m)` initially. also calculate `x_(m)` (neglect air resistance in calculations).

Answer

(ii) (a) ` sqrt((3H - 4h ) (g)/(2))` (b) `sqrt(h (3H- 4h)) ` (c ) ` h_m = (3)/(8) H ` (d) ` x_m = (3H)/(4)`

Explanation

(ii) (a) `sqrt((3H - 4h)(g)/(2))` (b) `sqrt(h (3H - 4h)) ` (c ) ` h_m = (3)/(8) H ` (d) `x _m= (3H)/(4 ) `
(ii) (a) Applying Bernoulli’s theorem, ` p _ 0 + d g (( H)/(2)) + 2dg (( H)/(2) - h ) = p _ 0 + (1)/(2) (2d ) v ^ 2 `
Here, v is velocity of efflux at 2. Solving this, we get ` v = sqrt((3H - 4h ) (g)/(2))`
(b) Time taken to reach the liquid to the botton will be ` t = sqrt (2h//g ) `
` therefore ` Horizontal distance x travelled by the liquid is ` x = vt = sqrt((3H - 4h ) (g)/(2) ) (sqrt((2h)/(g)) ) `
` x = sqrt (h (3H - 4h )) `
(c) For x to be maximum ` (dx )/(dh ) = 0 ` or ` (1)/(2 sqrt(h (2H - 4h )) ) (3H - 8 h ) = 0 or h = (3H )/(8) `
Therefore, x will be maximum at ` h = (3H)/( 8) `
The maximum value of x will be ` x_m = sqrt(((3H)/(8)) [3H- 4 ((3H)/(8))] ) , x _ m = (3)/(4) H `