The bob of a pendulum was released from a horizontal position. The length of the pendulum is 10m. 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 ms^(–2)`]

Options

A. `6 sqrt 5 ms^(-1)`
B. `5 sqrt 6 ms^(-1)`
C. `5 sqrt 5 ms^(-1)`
D. `2 sqrt 5 ms^(-1)`

Answer

To solve the problem, we will use the principle of conservation of energy. The pendulum bob is released from a horizontal position, which means it starts with potential energy and no kinetic energy. As it swings down, some of this potential energy is converted into kinetic energy, but we also have to account for the energy dissipated due to air resistance.

### Step-by-step Solution:

1. **Identify Initial Potential Energy:**
The initial potential energy (PE) when the bob is at the horizontal position (height = length of the pendulum) is given by:
\[ PE = mgh \]
where:
- \( m \) = mass of the bob (we will see it cancels out)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 10 \, \text{m} \) (length of the pendulum)

So,
\[ PE = mg \cdot 10 \]

2. **Calculate Energy Dissipated:**
Since 10% of the energy is dissipated due to air resistance, the energy available for conversion to kinetic energy (KE) is:
\[ KE = 90\% \text{ of } PE = 0.9 \cdot PE = 0.9 \cdot mg \cdot 10 \]

3. **Set Up the Energy Conservation Equation:**
At the lowest point of the swing (point B), all the potential energy that is not dissipated will be converted into kinetic energy. The potential energy at the lowest point is zero, so we have:
\[ 0.9 \cdot mg \cdot 10 = \frac{1}{2} mv^2 \]

4. **Cancel the Mass \( m \):**
Since \( m \) appears on both sides of the equation, we can cancel it out:
\[ 0.9 \cdot g \cdot 10 = \frac{1}{2} v^2 \]

5. **Substitute the Value of \( g \):**
Now substituting \( g = 10 \, \text{m/s}^2 \):
\[ 0.9 \cdot 10 \cdot 10 = \frac{1}{2} v^2 \]
This simplifies to:
\[ 90 = \frac{1}{2} v^2 \]

6. **Solve for \( v^2 \):**
Multiply both sides by 2:
\[ 180 = v^2 \]

7. **Take the Square Root:**
Finally, take the square root to find \( v \):
\[ v = \sqrt{180} = \sqrt{36 \cdot 5} = 6\sqrt{5} \, \text{m/s} \]

### Final Answer:
The speed with which the bob arrives at the lowest point is \( 6\sqrt{5} \, \text{m/s} \).