A bob of pendulum of mass 50 g is suspended by string with the roof of an elevator. If the lift is flying with a uniform acceleration of `5m s^(-2)` the tension in the string is `(g = 10m s^(-2))`

Options

A. 0.5 N
B. 0.225 N
C. 0.75 N
D. 0.025 N

Answer

To find the tension in the string of the pendulum bob suspended in an elevator that is accelerating upwards, we can follow these steps:

### Step 1: Identify the forces acting on the bob
The forces acting on the bob are:
- The gravitational force (weight) acting downwards: \( F_g = mg \)
- The tension in the string acting upwards: \( T \)
- A pseudo force acting downwards due to the upward acceleration of the elevator: \( F_{pseudo} = ma \)

### Step 2: Convert mass from grams to kilograms
The mass of the bob is given as 50 g. To convert this to kilograms:
\[ m = \frac{50 \text{ g}}{1000} = 0.05 \text{ kg} \]

### Step 3: Write the equations of motion
Since the elevator is accelerating upwards, the net force acting on the bob can be expressed as:
\[ T - mg - ma = 0 \]
Rearranging gives:
\[ T = mg + ma \]

### Step 4: Substitute the values
We know:
- \( g = 10 \, \text{m/s}^2 \)
- \( a = 5 \, \text{m/s}^2 \)
- \( m = 0.05 \, \text{kg} \)

Now, substituting these values into the equation for tension:
\[ T = m(g + a) = 0.05 \, \text{kg} \times (10 \, \text{m/s}^2 + 5 \, \text{m/s}^2) \]

### Step 5: Calculate the tension
Calculating the expression:
\[ T = 0.05 \, \text{kg} \times 15 \, \text{m/s}^2 = 0.05 \times 15 = 0.75 \, \text{N} \]

### Final Answer
The tension in the string is \( T = 0.75 \, \text{N} \).

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