The maximum and minimum tension in the string whirling in a circle of radius 2.5 m with constant velocity are in the ratio `5:3` then its velocity is

Options

A. `10(m)/(s)`
B. `7(m)/(s)`
C. `15(m)/(s)`
D. `5(m)/(s)`

Explanation

To solve the problem, we need to analyze the forces acting on the mass at the highest and lowest points of the circular motion and use the given ratio of tensions to find the velocity. Here’s the step-by-step solution:

### Step 1: Understand the Forces Acting on the Mass
When the mass is at the lowest point:
- The tension in the string (T1) acts upwards.
- The weight of the mass (mg) acts downwards.
- The centripetal force required for circular motion is provided by the net force, which is T1 - mg.

At the lowest point, the equation of motion can be written as:
\[ T_1 = mg + m \cdot r \cdot \omega^2 \]
where \( r \) is the radius and \( \omega \) is the angular velocity.

When the mass is at the highest point:
- The tension in the string (T2) acts upwards.
- The weight of the mass (mg) acts downwards.
- The centripetal force is provided by the tension and the weight, so we have:
\[ T_2 + mg = m \cdot r \cdot \omega^2 \]
This can be rearranged to:
\[ T_2 = m \cdot r \cdot \omega^2 - mg \]

### Step 2: Set Up the Ratio of Tensions
We know from the problem that the ratio of maximum tension (T1) to minimum tension (T2) is given as:
\[ \frac{T_1}{T_2} = \frac{5}{3} \]

Substituting the expressions for T1 and T2, we have:
\[ \frac{mg + m \cdot r \cdot \omega^2}{m \cdot r \cdot \omega^2 - mg} = \frac{5}{3} \]

### Step 3: Cross Multiply and Simplify
Cross multiplying gives us:
\[ 3(mg + m \cdot r \cdot \omega^2) = 5(m \cdot r \cdot \omega^2 - mg) \]

Expanding both sides:
\[ 3mg + 3m \cdot r \cdot \omega^2 = 5m \cdot r \cdot \omega^2 - 5mg \]

### Step 4: Rearranging the Equation
Rearranging the terms gives:
\[ 3mg + 5mg = 5m \cdot r \cdot \omega^2 - 3m \cdot r \cdot \omega^2 \]
\[ 8mg = 2m \cdot r \cdot \omega^2 \]

### Step 5: Cancel Out m and Solve for ω²
Dividing both sides by \( 2m \):
\[ \frac{8g}{2} = r \cdot \omega^2 \]
\[ 4g = r \cdot \omega^2 \]
Thus, we find:
\[ \omega^2 = \frac{4g}{r} \]

### Step 6: Relate Velocity to Angular Velocity
We know that:
\[ v = r \cdot \omega \]
Squaring both sides gives:
\[ v^2 = r^2 \cdot \omega^2 \]
Substituting \( \omega^2 \) from above:
\[ v^2 = r^2 \cdot \frac{4g}{r} \]
This simplifies to:
\[ v^2 = 4g \cdot r \]
Taking the square root:
\[ v = \sqrt{4g \cdot r} \]

### Step 7: Substitute Values for g and r
Given:
- \( g = 10 \, \text{m/s}^2 \)
- \( r = 2.5 \, \text{m} \)

Substituting these values:
\[ v = \sqrt{4 \cdot 10 \cdot 2.5} \]
\[ v = \sqrt{100} \]
\[ v = 10 \, \text{m/s} \]

### Final Answer
The velocity of the whirling mass is \( 10 \, \text{m/s} \).

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