Problem 476 - JEE Physics

Question

If the water falls from a dam into a turbine wheel \(19.6 \mathrm{~m}\) below, then the velocity of water at the turbine is \(\ldots \ldots\) \(\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(9.8 \mathrm{~m} / \mathrm{s}\) (B) \(19.6 \mathrm{~m} / \mathrm{s}\) (C) \(39.2 \mathrm{~m} / \mathrm{s}\) (D) \(98.0 \mathrm{~m} / \mathrm{s}\)

Step-by-Step Solution

Answer

The velocity of the water at the turbine is approximately \(19.6\) m/s. The correct answer is (B).

Step 1: Identify the variables

Here, we need to identify the known and unknown variables in the problem. The known variables are the height of the falling water (h) which is 19.6 m, and the acceleration due to gravity (g) which is 9.8 m/s². The unknown variable is the final velocity of water at the turbine (v).

Step 2: Choose the appropriate kinematic equation

Since we are given height and acceleration, and need to find final velocity, we can choose the equation: \(v^2 = u^2 + 2gh\), where u is the initial velocity (0 m/s as the water starts at rest), g is the acceleration due to gravity, and h is the height of the water.

Step 3: Plug in the values

Now, we will substitute the given values into the equation: \(v^2 = 0^2 + 2 (9.8) (19.6)\).

Step 4: Calculate the velocity

By calculating the values, we end up with the equation: \(v^2 = 2 (9.8) (19.6)\), and therefore: \(v^2 = 384.16\). Next, take the square root of both sides to find the final velocity (v): \[v = \sqrt{384.16}\]. Finally, we get the velocity of the water at the turbine: \(v ≈ 19.6\) m/s. Among the given options, (B) is closest to our calculated value: 19.6 m/s.

Key Concepts

Equations of motion

In kinematics, the equations of motion are vital for describing the behavior of objects in motion. They are mathematical formulas that relate an object's velocity, position, acceleration, and time. In this exercise, where water is falling from a dam, we utilize one specific equation of motion:

• \(v^2 = u^2 + 2gh\)

Here:

• \(v\) is the final velocity of the water at the turbine,
• \(u\) is the initial velocity (which is 0 m/s, as the water starts from rest),
• \(g\) is the acceleration due to gravity, and
• \(h\) represents the height from which the water falls.

This equation helps us calculate the final velocity of the water when it reaches the turbine. By setting the initial velocity \(u\) to 0, as the water starts falling from a height at rest, we simplify the equation to \(v^2 = 2gh\), making it straightforward to find \(v\).

Gravitational acceleration

Gravitational acceleration is a key factor in the motion of freely falling objects. On Earth, this constant is approximately \(9.8 \, \text{m/s}^2\), meaning that any object in free fall accelerates towards the ground at this rate, regardless of its mass.
Understanding gravitational acceleration is crucial when analyzing scenarios like water falling from a dam. It helps predict how fast an object will move as it falls.
In our example, knowing the height of \(19.6 \) meters and using gravitational acceleration (\(g = 9.8 \, \text{m/s}^2\)), we can apply it directly into our motion equation to find the velocity of the water when it hits the turbine. This acceleration is the reason why, after falling freely under gravity from a specific height, the water reaches the turbine with a velocity of approximately \(19.6 \, \text{m/s}\).

Energy conservation

The principle of energy conservation states that energy cannot be created or destroyed; it can only be transformed from one form to another. In the context of water falling from a dam, this principle helps us understand how potential energy is converted into kinetic energy.

• At the top, water possesses potential energy due to its height \(h\).
• As water falls under gravity, this potential energy is converted into kinetic energy, which is the energy of motion.

When the water reaches the turbine, all of its initial potential energy is converted into kinetic energy (assuming no losses), which can be expressed as:\[\text{Potential Energy}_{\text{top}} = \text{Kinetic Energy}_{\text{bottom}}\]Given by the equation \(mgh = \frac{1}{2}mv^2\). Here, \(m\) is the mass of the water, which cancels out, simplifying the understanding of the energy conversion to \(gh = \frac{1}{2}v^2\). Solving this shows why the final velocity of the water (\(v \approx 19.6 \, \text{m/s}\)) aligns with the energy transformation principles.