The difference in tensions in the string at lowest and highest points in the path of the particle of mass 'm' performing vertical circular motion is :

Options

A. 2mg
B. 4 mg
C. 6mg
D. 8 mg

Answer

To find the difference in tensions in the string at the lowest and highest points of a particle of mass 'm' performing vertical circular motion, we can follow these steps:

### Step 1: Analyze Forces at the Lowest Point
At the lowest point of the circular motion, the forces acting on the mass 'm' are:
- The tension in the string (T_low)
- The weight of the mass (mg)

At this point, the net force provides the centripetal force required for circular motion. Therefore, we can write the following equation:

\[ T_{low} - mg = \frac{mv^2}{r} \]

Where:
- \( T_{low} \) = tension at the lowest point
- \( mg \) = weight of the mass
- \( v \) = speed of the mass at the lowest point
- \( r \) = radius of the circular path

Rearranging gives us:

\[ T_{low} = mg + \frac{mv^2}{r} \]

### Step 2: Analyze Forces at the Highest Point
At the highest point of the circular motion, the forces acting on the mass 'm' are:
- The tension in the string (T_high)
- The weight of the mass (mg)

At this point, the equation for the net force is:

\[ T_{high} + mg = \frac{mv^2}{r} \]

Rearranging gives us:

\[ T_{high} = \frac{mv^2}{r} - mg \]

### Step 3: Calculate the Difference in Tension
Now, we need to find the difference in tension between the lowest and highest points:

\[ \Delta T = T_{low} - T_{high} \]

Substituting the expressions we derived:

\[ \Delta T = \left( mg + \frac{mv^2}{r} \right) - \left( \frac{mv^2}{r} - mg \right) \]

Simplifying this expression:

\[ \Delta T = mg + \frac{mv^2}{r} - \frac{mv^2}{r} + mg \]

\[ \Delta T = 2mg \]

### Final Result
Thus, the difference in tensions in the string at the lowest and highest points in the path of the particle is:

\[ \Delta T = 2mg \]

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