Stone of mass 1 kg tied to the end of a string of length 1m , is whirled in horizontal circle with a uniform angular velocity `2 rad s^(-1)`. The tension of the string is (in newton)

Options

A. 2
B. `1/3`
C. `4`
D. `1/4`

Answer

To find the tension in the string when a stone of mass 1 kg is whirled in a horizontal circle with a uniform angular velocity of 2 rad/s, we can follow these steps:

### Step 1: Identify the given values
- Mass of the stone (m) = 1 kg
- Length of the string (radius of the circle, r) = 1 m
- Angular velocity (ω) = 2 rad/s

### Step 2: Understand the concept of centripetal force
When an object is moving in a circular path, it requires a centripetal force to keep it moving in that path. In this case, the tension in the string provides the necessary centripetal force.

### Step 3: Use the formula for centripetal force
The centripetal force (F_c) can be expressed in terms of mass (m), angular velocity (ω), and radius (r) as:
\[ F_c = m \cdot \omega^2 \cdot r \]

### Step 4: Substitute the known values into the formula
Now, substituting the known values into the formula:
\[ F_c = 1 \, \text{kg} \cdot (2 \, \text{rad/s})^2 \cdot 1 \, \text{m} \]

### Step 5: Calculate the centripetal force
Calculating the values:
\[ F_c = 1 \cdot 4 \cdot 1 = 4 \, \text{N} \]

### Step 6: Conclusion
Since the tension in the string provides the centripetal force, we can conclude that the tension (T) in the string is:
\[ T = 4 \, \text{N} \]

Thus, the tension in the string is **4 Newtons**.

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