A tank is filled upto a height h with a liquid and is placed on a platform of height h from the ground. To get maximum range xₘ a small hole is punched at a distance of y from the free surface of the liquid. Then

Correct Answer: xₘ = 2h

To find the maximum horizontal range of the liquid jet, we can analyze the motion of the liquid exiting the hole using Torricelli's law and project kinematics.

1. Determine the height of the hole:
The tank is filled with liquid up to a height h and is placed on a platform of height h. Therefore, the free surface of the liquid is at a height of h + h = 2h from the ground.
A small hole is punched at a distance y below the free surface. Thus, the height of the hole above the ground level (H) is given by:
$H=2h-y$

2. Velocity of efflux:
According to Torricelli's theorem, the horizontal velocity v of the liquid flowing out of the hole at a depth y from the free surface is:
$v=\sqrt{2gy}$
where g is the acceleration due to gravity.

3. Time of flight:
Once the liquid leaves the hole horizontally, it undergoes projectile motion. The time t taken by the liquid to reach the ground from height H is:
$t=\sqrt{\frac{2H/g}{}}=\sqrt{\frac{2(2h-y)}{g}}$

4. Horizontal range:
The horizontal range x is the product of the horizontal velocity and the time of flight:
$x=v\cdot t$
Substituting the values of v and t:
$x=\sqrt{2gy}\cdot \sqrt{\frac{2(2h-y)}{g}}$
Simplifying the expression:
$x=2\sqrt{y(2h-y)}$

5. Maximizing the range:
To find the condition for maximum range xₘ, we differentiate the quantity under the square root with respect to y and equate it to zero:
$\frac{d}{dy}\left[y\right(2h-y\left)\right]=0$
$\frac{d}{dy}[2hy-y^2]=0$
$2h-2y=0\Rightarrow y=h$
Thus, the maximum range is achieved when the hole is punched at a depth y = h from the free surface.

6. Calculating maximum range (xₘ):
Substituting y = h back into the range equation:
$x_m=2\sqrt{h(2h-h)}$
$x_m=2\sqrt{h^2}=2h$
Therefore, the maximum range xₘ is equal to 2h.