A stone of mass 250 gram , attached at the end of a string of length 1.25 m is whirled in a horizontal circle at a speed of 5 m/s . What is the tension in the string ?

Options

A. `2.5N`
B. 5N
C. 6N
D. 8N

Explanation

To find the tension in the string when a stone is whirled in a horizontal circle, we can use the concept of centripetal force. The tension in the string provides the necessary centripetal force to keep the stone moving in a circular path.

### Step-by-Step Solution:

1. **Convert Mass to Kilograms**:
The mass of the stone is given as 250 grams. We need to convert this to kilograms since the standard unit of mass in physics is kilograms.
\[ \text{Mass (m)} = 250 \text{ grams} = \frac{250}{1000} \text{ kg} = 0.25 \text{ kg} \]

**Hint**: Remember to convert grams to kilograms by dividing by 1000.

2. **Identify the Radius of the Circle**:
The length of the string is given as 1.25 m, which is the radius (r) of the circular path.
\[ r = 1.25 \text{ m} \]

**Hint**: The radius of the circular motion is equal to the length of the string.

3. **Calculate the Centripetal Force**:
The formula for centripetal force (F_c) is given by:
\[ F_c = \frac{m v^2}{r} \]
where:
- \( m \) is the mass of the stone (0.25 kg),
- \( v \) is the speed (5 m/s),
- \( r \) is the radius (1.25 m).

Plugging in the values:
\[ F_c = \frac{0.25 \times (5)^2}{1.25} \]
\[ F_c = \frac{0.25 \times 25}{1.25} = \frac{6.25}{1.25} = 5 \text{ N} \]

**Hint**: Use the centripetal force formula to find the force acting towards the center of the circular path.

4. **Determine the Tension in the String**:
In this case, the tension in the string (T) is equal to the centripetal force required to keep the stone moving in a circle. Thus:
\[ T = F_c = 5 \text{ N} \]

**Hint**: The tension in the string provides the necessary centripetal force for circular motion.

### Final Answer:
The tension in the string is **5 Newtons**.

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