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:

From the provided images, we can analyze the given setup of the open tank:
1. The total height of the water level in the tank from the floor is labeled as 1 m.
2. The hole (orifice) on the side of the tank is located at a height of 0.5 m from the floor.
3. The depth of the water column above the hole is labeled as $h$ in the second image.

First, we calculate the depth of the hole below the free water surface ($h$):

$h=1\text{m}-0.5\text{m}=0.5\text{m}$

According to Torricelli's Law, the horizontal velocity ($v$) of the water jet emerging from the hole is given by:

$v=\sqrt{2gh}$

where $g$ is the acceleration due to gravity.

Once the jet leaves the hole horizontally, it behaves like a projectile under gravity. The time ($t$) it takes for the water to fall vertically from a height of $y=0.5\text{m}$ to the floor is given by the equations of motion:

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

Solving for $t$:

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

The horizontal distance ($X$) traveled by the jet before striking the floor is the product of its horizontal velocity and the time of flight:

$X=v\times t$

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

$X=\sqrt{2gh}\times \sqrt{\frac{2y}{g}}$

Simplifying the expression by cancelling gravity ($g$):

$X=2\sqrt{h\times y}$

Substituting the values $h=0.5\text{m}$ and $y=0.5\text{m}$ into the equation:

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

Therefore, the horizontal distance $X$ where the jet strikes the floor is 1.0 m.