A mass of 100 gm is tied to one end of a string 2 m long. The body is revolving in a horizontal circle making a maximum of 200 revolutions per min. The other end of the string is fixed at the centre of the circle of revolution. The maximum tension that the string can bear is (approximately)

Options

A. `8.76` N
B. `8.94` N
C. `89.42` N
D. `87.64` N

Explanation

To solve the problem, we need to calculate the maximum tension in the string when the mass is revolving in a horizontal circle. The tension in the string provides the necessary centripetal force for the circular motion. Here are the steps to find the maximum tension:

### Step 1: Convert the mass from grams to kilograms
The mass of the object is given as 100 grams. To convert this to kilograms:
\[ \text{mass} (m) = \frac{100 \text{ gm}}{1000} = 0.1 \text{ kg} \]

**Hint:** Remember that 1 kg = 1000 grams.

### Step 2: Convert the revolutions per minute to revolutions per second
The maximum frequency (n) is given as 200 revolutions per minute. To convert this to revolutions per second:
\[ n = \frac{200 \text{ rev/min}}{60 \text{ sec/min}} \approx 3.33 \text{ rev/sec} \]

**Hint:** To convert from minutes to seconds, divide by 60.

### Step 3: Calculate the angular velocity (ω)
The angular velocity (ω) in radians per second can be calculated using the formula:
\[ \omega = 2\pi n \]
Substituting the value of n:
\[ \omega = 2\pi \times 3.33 \approx 20.94 \text{ rad/sec} \]

**Hint:** Remember that there are \(2\pi\) radians in one revolution.

### Step 4: Use the formula for maximum tension
The maximum tension (T_max) in the string can be calculated using the formula for centripetal force:
\[ T_{\text{max}} = m \omega^2 r \]
Where:
- \(m = 0.1 \text{ kg}\)
- \(r = 2 \text{ m}\)
- \(\omega \approx 20.94 \text{ rad/sec}\)

Substituting the values:
\[ T_{\text{max}} = 0.1 \times (20.94)^2 \times 2 \]

### Step 5: Calculate T_max
Calculating \( (20.94)^2 \):
\[ (20.94)^2 \approx 438.56 \]
Now substituting back:
\[ T_{\text{max}} = 0.1 \times 438.56 \times 2 \approx 87.71 \text{ N} \]

### Final Answer
The maximum tension that the string can bear is approximately \(87.71 \text{ N}\).

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