A body is made to revolve along a circle of radius r in horizontal plane at a time period T with the help of light horizontal string such that the tension in the string is F. If the radius of circle is increased to 2r and time period of revolution is decreased to `T//2` then the tension in the string is

Options

A. `F//2`
B. 2F
C. 4F
D. 8F

Answer

To solve the problem step by step, we will analyze the forces acting on the body revolving in a circular path and use the relationships between the radius, time period, and tension in the string.

### Step 1: Understand the initial conditions
- The body is revolving in a circle of radius \( r \) with a time period \( T \).
- The tension in the string is given as \( F \).

### Step 2: Relate tension to centripetal force
The tension \( F \) in the string provides the necessary centripetal force for circular motion. The formula for centripetal force \( F_c \) is given by:
\[ F_c = m \cdot \omega^2 \cdot r \]
where \( m \) is the mass of the body and \( \omega \) is the angular velocity.

### Step 3: Express angular velocity in terms of time period
The angular velocity \( \omega \) can be expressed in terms of the time period \( T \):
\[ \omega = \frac{2\pi}{T} \]
Substituting this into the centripetal force equation gives:
\[ F = m \cdot \left(\frac{2\pi}{T}\right)^2 \cdot r \]

### Step 4: Analyze the new conditions
Now, the radius is increased to \( 2r \) and the time period is decreased to \( \frac{T}{2} \). We need to find the new tension \( F' \).

### Step 5: Calculate the new angular velocity
The new angular velocity \( \omega' \) is:
\[ \omega' = \frac{2\pi}{\frac{T}{2}} = \frac{4\pi}{T} \]

### Step 6: Write the new centripetal force equation
The new centripetal force \( F' \) is given by:
\[ F' = m \cdot \omega'^2 \cdot (2r) \]
Substituting for \( \omega' \):
\[ F' = m \cdot \left(\frac{4\pi}{T}\right)^2 \cdot (2r) \]
This simplifies to:
\[ F' = m \cdot \frac{16\pi^2}{T^2} \cdot (2r) = 32 \cdot m \cdot \frac{\pi^2}{T^2} \cdot r \]

### Step 7: Relate the new tension to the original tension
From the original tension \( F = m \cdot \frac{4\pi^2}{T^2} \cdot r \), we can express \( F' \) in terms of \( F \):
\[ F' = 8 \cdot F \]

### Conclusion
The tension in the string when the radius is increased to \( 2r \) and the time period is decreased to \( \frac{T}{2} \) is:
\[ F' = 8F \]

### Final Answer
The tension in the string is \( 8F \).

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