Problem 1044 - JEE Physics

Question

A tank is filled with water up to a height \(\mathrm{H}\). Water is allowed to come out of a hole P in one of the walls at a depth \(\mathrm{D}\) below the surface of water express the horizontal distance \(\mathrm{x}\) in terms of \(\mathrm{H}\) and \(\mathrm{D}\). (B) \(\left.\mathrm{x}={ }^{\alpha} \sqrt{[}\\{\mathrm{D}(\mathrm{H}-\mathrm{D})\\} / 2\right]\) (D) \(\mathrm{x}=4[\mathrm{D}(\mathrm{H}-\mathrm{D})]\)

Step-by-Step Solution

Answer

The short answer is: \(x = \sqrt{4D(H - D)}\).

Step 1: Find the Initial Vertical Velocity

To calculate the initial velocity of the water as it leaves the hole, we can use Torricelli's theorem, which states that the speed of the fluid at the hole is equal to the speed it would acquire if it fell freely from a height equal to the depth of the hole. Therefore, the velocity of the water, v, can be found using the equation: \(v = \sqrt{2gD}\), where g is the acceleration due to gravity (approximately 9.81 m/s^2).

Step 2: Calculate Time to Hit the Ground

As the water leaves the hole horizontally with velocity v, it will also be subject to gravity, which will make it fall towards the ground. We can use kinematic equations to find the time it takes for the water to reach the ground. In this case, we have the following equation: \((H - D) = \frac{1}{2}gt^2\), where t is the time it takes for the water to hit the ground. We need to solve this equation for t: \(t = \sqrt{\frac{2(H - D)}{g}}\).

Step 3: Calculate the Horizontal Distance x

Now that we have the initial horizontal velocity of the water and the time it takes to reach the ground, we can calculate the horizontal distance x: \(x = vt\), where x is the horizontal distance, v is the horizontal velocity, and t is the time found in step 2. Substitute the values for v and t from steps 1 and 2: \(x = \sqrt{2gD} \cdot \sqrt{\frac{2(H - D)}{g}}\), \(x = \sqrt{2gD \cdot \frac{2(H - D)}{g}}\). Simplifying the expression, we get: \(x = \sqrt{4D(H - D)}\). The correct answer is (D) \(x=4[D(H-D)]\).

Key Concepts

Torricelli's Theorem

Torricelli's Theorem plays a crucial role in understanding the velocity of fluid coming out of an orifice. This theorem states that the speed of fluid flowing out of a hole is equivalent to the speed it would attain if it were to fall freely from a height equal to the depth of the hole. This concept is particularly useful in fluid dynamics, as it allows us to measure the velocity without direct mechanical influence. For this water tank problem, the formula derived from Torricelli's Theorem is given by:

• \( v = \sqrt{2gD} \)
• where \( v \) is the velocity of water leaving the hole, \( g \) is the acceleration due to gravity, and \( D \) is the depth from the water surface to the hole.

This velocity marks the initial speed of the water as it exits the tank and sets the stage for further calculations involving motion.

Kinematic Equations

The kinematic equations are a set of equations used to describe the motion of objects. In the context of this exercise, they help us determine how long it takes for the water to hit the ground after exiting the hole. Gravity acts on the water, pulling it downward, so we use the following equation to find the time \( t \) the water needs to fall from the height \( (H - D) \):

• \((H - D) = \frac{1}{2}gt^2\)

Solving for \( t \), we find:

• \( t = \sqrt{\frac{2(H - D)}{g}} \)

This equation enables us to compute the exact time it takes for the water to descend to the ground, which is essential to calculating the horizontal distance traveled.

Fluid Dynamics

Fluid dynamics focuses on how fluids (liquids and gases) move and interact with their environment. This branch of physics explains features such as flow velocity, pressure, and viscosity. In the exercise, understanding fluid dynamics helps us predict the motion and speed of water as it streams out of the tank. We apply core ideas like Torricelli's Theorem, which stems from fluid dynamics, to measure the velocity. Additionally, fluid dynamics underpin the concepts of continuity and energy conservation; both are vital when fluids are transported or when analyzing flowing systems. These principles tell us that pressure and velocity can change depending on different parameters, but conservation laws still apply.

Horizontal Motion

Horizontal motion concerns the movement of objects on a horizontal path. In this exercise, it's about how the water travels after leaving the hole in the tank. Once outside the tank, the water takes on horizontal motion, propelled by the velocity calculated through Torricelli's Theorem. We measure how far the water travels horizontally using:

• \( x = vt \)
• where \( x \) is the horizontal distance, \( v \) is the initial horizontal velocity, and \( t \) is the time computed with kinematic equations.

By substituting our values:

• \( x = \sqrt{2gD} \cdot \sqrt{\frac{2(H - D)}{g}} \)
• Simplified, it becomes: \( x = \sqrt{4D(H - D)} \)

This horizontal distance illustrates how far the water will travel, starting from the velocity it achieved through vertical descent, and influenced by its horizontal trajectory.