A stone of mass 900g is tied to a string and moved in a vertical circle of radius 1m making 10 rpm. The tension in the string, when the stone is at the lowest point is (if `pi^2 = 9.8` and `g = 9.8 m//s^2`)

Options

A. 97 N
B. `9.8` N
C. `8.82`N
D. `17.8` N

Answer

To find the tension in the string when the stone is at the lowest point of its vertical circular motion, we can follow these steps:

### Step 1: Convert mass to kilograms
The mass of the stone is given as 900 grams. To convert this to kilograms, we divide by 1000:
\[ m = \frac{900 \text{ g}}{1000} = 0.9 \text{ kg} \]

### Step 2: Calculate the weight of the stone
The weight (W) of the stone can be calculated using the formula:
\[ W = mg \]
Given that \( g = 9.8 \, \text{m/s}^2 \):
\[ W = 0.9 \text{ kg} \times 9.8 \text{ m/s}^2 = 8.82 \text{ N} \]

### Step 3: Calculate the angular velocity (ω)
The stone is moving at 10 revolutions per minute (rpm). To convert this to radians per second, we use the formula:
\[ \omega = 2\pi f \]
where \( f \) is the frequency in revolutions per second. Since 10 rpm means:
\[ f = \frac{10 \text{ rev}}{60 \text{ s}} = \frac{1}{6} \text{ rev/s} \]
Now substituting \( f \) into the formula for \( \omega \):
\[ \omega = 2\pi \left(\frac{1}{6}\right) = \frac{\pi}{3} \text{ rad/s} \]

### Step 4: Calculate the centripetal acceleration
The centripetal acceleration (a_c) can be calculated using:
\[ a_c = \omega^2 r \]
where \( r = 1 \text{ m} \):
\[ a_c = \left(\frac{\pi}{3}\right)^2 \times 1 = \frac{\pi^2}{9} \text{ m/s}^2 \]

### Step 5: Write the equation for tension (T)
At the lowest point of the circular motion, the tension in the string (T) must counteract both the weight of the stone and provide the necessary centripetal force. The equation can be written as:
\[ T = W + m a_c \]
Substituting the values we have:
\[ T = mg + m \left(\frac{\pi^2}{9}\right) \]
\[ T = m \left(g + \frac{\pi^2}{9}\right) \]

### Step 6: Substitute the known values
Now substituting \( m = 0.9 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( \pi^2 = 9.8 \):
\[ T = 0.9 \left(9.8 + \frac{9.8}{9}\right) \]
Calculating \( \frac{9.8}{9} \):
\[ \frac{9.8}{9} = 1.0889 \text{ (approximately)} \]
Thus:
\[ T = 0.9 \left(9.8 + 1.0889\right) = 0.9 \times 10.8889 \approx 9.8 \text{ N} \]

### Final Answer
The tension in the string when the stone is at the lowest point is approximately:
\[ T \approx 9.8 \text{ N} \]