A tank open at the top with a water level of 1 m, as shown in the figure, has a hole at a height of 0.5 m. A free jet leaves horizontally from the smooth hole. The distance X (in m) where the jet strikes the floor is

The correct answer is 1.0.

Step-by-step Explanation:

1. Identify the given parameters from the figure:
- Total height of the water level in the open tank,
$H=1\text{m}$
- Height of the hole from the bottom of the tank,
$h=0.5\text{m}$
- Depth of water above the hole,
$y=H-h=1-0.5=0.5\text{m}$

2. Determine the velocity of efflux (v):
According to Torricelli's Theorem, the horizontal velocity of the jet leaving the smooth hole is given by:

$v=\sqrt{2g(H-h)}$

where $g$ is the acceleration due to gravity.

3. Determine the time of flight (t):
Since the jet leaves horizontally, its initial vertical velocity is zero. The vertical distance it falls to reach the floor is $h=0.5\text{m}$. Using the equation of motion for vertical fall under gravity:

$h=\frac{1}{2}g{t}^2$

Solving for $t$ gives:

$t=\sqrt{\frac{2h}{g}}$

4. Calculate the horizontal distance (X):
The horizontal distance $X$ traveled by the jet is given by the product of the horizontal velocity and the time of flight:

$X=v\times t$

Substituting the expressions for $v$ and $t$:

$X=\sqrt{2g(H-h)}\times \sqrt{\frac{2h}{g}}$

Simplifying the expression:

$X=2\sqrt{h(H-h)}$

5. Substitute the values:

$X=2\sqrt{0.5\times (1-0.5)}$

$X=2\sqrt{0.5\times 0.5}$

$X=2\times 0.5=1.0\text{m}$

Thus, the jet strikes the floor at a horizontal distance of 1.0 m.