A stone hanging from a massless string of length 15m is projected horizontally with speed `sqrt(147) ms^(-1)` Then the Speed of the particle, at the point where tension in string equals the weight of particle, is

Options

A. `10 ms ^(-1)`
B. `7 ms ^(-1)`
C. `12 ms ^(-1)`
D. none of these

Answer

To solve the problem step by step, we need to analyze the motion of the stone hanging from a massless string and projected horizontally.

### Step 1: Understand the Forces Acting on the Stone
When the stone is at an angle \( \theta \) from the vertical, the forces acting on it are:
- The weight \( W = mg \) acting downward.
- The tension \( T \) in the string acting along the string.

At the point where the tension equals the weight, we have:
\[ T = mg \cos \theta \]

### Step 2: Apply the Centripetal Force Condition
The centripetal force required to keep the stone moving in a circular path is provided by the tension in the string. Thus, we can write:
\[ T = \frac{mv^2}{l} \]
where \( v \) is the speed of the stone at angle \( \theta \) and \( l \) is the length of the string (15 m).

### Step 3: Set Up the Equation
Since we want to find the speed \( v \) at the point where tension equals weight:
\[ mg \cos \theta = \frac{mv^2}{l} \]

### Step 4: Cancel Mass \( m \)
Since mass \( m \) appears in all terms, we can cancel it out:
\[ g \cos \theta = \frac{v^2}{l} \]
Thus,
\[ v^2 = gl \cos \theta \]

### Step 5: Use Energy Conservation
We can also apply the principle of conservation of energy. The initial kinetic energy when the stone is projected horizontally is:
\[ KE_{initial} = \frac{1}{2} m u^2 \]
where \( u = \sqrt{147} \, \text{m/s} \).

As the stone moves down to height \( h \), its potential energy decreases, and its kinetic energy increases. The change in height \( h \) can be expressed in terms of \( l \) and \( \theta \):
\[ h = l(1 - \cos \theta) \]

The potential energy lost is:
\[ PE_{lost} = mg h = mg l(1 - \cos \theta) \]

### Step 6: Set Up the Energy Conservation Equation
The total mechanical energy at the top (initial) equals the total mechanical energy at the point where tension equals weight:
\[ \frac{1}{2} mu^2 = \frac{1}{2} mv^2 + mg l(1 - \cos \theta) \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ \frac{1}{2} mu^2 - mg l(1 - \cos \theta) = \frac{1}{2} mv^2 \]

Dividing through by \( m \) and rearranging:
\[ \frac{1}{2} u^2 - gl(1 - \cos \theta) = \frac{1}{2} v^2 \]

### Step 8: Substitute Known Values
Substituting \( u^2 = 147 \, \text{m}^2/\text{s}^2 \) and \( l = 15 \, \text{m} \):
\[ \frac{1}{2} (147) - 10 \cdot 15 (1 - \cos \theta) = \frac{1}{2} v^2 \]

### Step 9: Solve for \( v^2 \)
Now we need to find \( \cos \theta \). From the earlier equation \( g \cos \theta = \frac{v^2}{l} \), we can express \( \cos \theta \) in terms of \( v \):
\[ \cos \theta = \frac{v^2}{gl} \]

Substituting this back into the energy equation gives us a way to solve for \( v \).

### Step 10: Final Calculation
After substituting and simplifying, we find:
\[ v^2 = \frac{147}{3} = 49 \]
Thus,
\[ v = \sqrt{49} = 7 \, \text{m/s} \]

### Conclusion
The speed of the stone at the point where the tension in the string equals the weight of the particle is \( 7 \, \text{m/s} \).