Question

The bob of a pendulum was released from a horizontal position. The length of the pendulum is 10 m. If it dissipates 10% of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is: [Use, \( g : 10 \, \text{m/s}^2 \)]

• A. \( 6\sqrt{5} \, \text{ms}^{-1} \)
• B. \( 5\sqrt{6} \, \text{ms}^{-1} \)
• C. \( 5\sqrt{5} \, \text{ms}^{-1} \)
• D. \( 2\sqrt{5} \, \text{ms}^{-1} \)

Answer 1

The Correct Option is A

Given:  
- Length of the pendulum, \( \ell = 10 \, \text{m} \)  
- Gravitational acceleration, \( g = 10 \, \text{m/s}^2 \)  
- 10% of the initial energy is dissipated against air resistance.

Step 1. Calculate the initial energy:  
  Since the bob is released from a horizontal position, the initial potential energy (at the top) is:  
  
  \(\text{Initial energy} = mg\ell\)
  

Step 2. Calculate the energy at the lowest point:  
  Since 10% of the initial energy is dissipated, only 90% of the initial energy is available at the lowest point.  
 
 \(\text{Energy at the lowest point} = \frac{9}{10} mg\ell\)
  
Step 3. Relate energy to speed at the lowest point:  
  At the lowest point, this energy is entirely kinetic, so:  

  \(\frac{9}{10}mg\ell = \frac{1}{2}mv^2\)
  
  Simplify by canceling \( m \) from both sides:  

  \(\frac{9}{10} \times 10 \times 10 = \frac{1}{2}v^2\)
  
 \(v^2 = 180\)

  \(v = \sqrt{180} = 6\sqrt{5} \, \text{m/s}\)
 

Thus, the speed of the bob at the lowest point is \( 6\sqrt{5} \, \text{m/s} \).

The Correct Answer is: \( 6\sqrt{5} \, \text{m/s} \)