In an experiment, a boy plots a graph between `(v_(max))^(2)and (a_(max))^(2)` for a simple pendulum for different values of (small) amplitudes, where `v_(max)`and `a_(max)` is the maximum velocity and the maximum acceleration respectively. He found the graph to be a straight line with a negative slope, making an angle of `30^(@)` when the experiment was conducted on the earth surface. When the same experiment was conducted at a height h above the surface, the line was at an angle of `60^(@)`. The value of h is [radius of the earth = R]

Options

A. 0.5R
B. 0.24R
C. 0.73R
D. R

Answer

To solve the problem, we need to analyze the relationship between the maximum velocity \( v_{\text{max}} \) and the maximum acceleration \( a_{\text{max}} \) of a simple pendulum at different heights above the Earth's surface.

### Step-by-Step Solution:

1. **Understanding the Pendulum Motion**:
- For a simple pendulum, the maximum velocity \( v_{\text{max}} \) and the maximum acceleration \( a_{\text{max}} \) can be expressed in terms of the amplitude \( A \) and the angular frequency \( \omega \):
\[ v_{\text{max}} = A \omega \]
\[ a_{\text{max}} = -A \omega^2 \]

2. **Expressing in Terms of Gravity**:
- The angular frequency \( \omega \) is given by:
\[ \omega = \sqrt{\frac{g}{L}} \]
- Therefore, we can express \( v_{\text{max}} \) and \( a_{\text{max}} \) as:
\[ v_{\text{max}} = A \sqrt{\frac{g}{L}} \]
\[ a_{\text{max}} = -A \left(\frac{g}{L}\right) \]

3. **Plotting the Graph**:
- The graph is plotted between \( v_{\text{max}}^2 \) and \( a_{\text{max}}^2 \):
\[ v_{\text{max}}^2 = A^2 \frac{g}{L} \]
\[ a_{\text{max}}^2 = A^2 \left(\frac{g}{L}\right)^2 \]
- Thus, the relationship can be simplified to:
\[ v_{\text{max}}^2 = \frac{g}{L} \cdot a_{\text{max}}^2 \]

4. **Finding the Slope**:
- The slope of the line in the graph is given by:
\[ \text{slope} = \frac{v_{\text{max}}^2}{a_{\text{max}}^2} = \frac{1}{\omega^2} \]
- When the angle is \( 30^\circ \) at the Earth's surface:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \Rightarrow \frac{1}{\omega^2} = \frac{1}{\sqrt{3}} \Rightarrow \omega^2 = \sqrt{3} \]

5. **At Height \( h \)**:
- When the experiment is conducted at height \( h \), the angle becomes \( 60^\circ \):
\[ \tan(60^\circ) = \sqrt{3} \Rightarrow \frac{1}{\omega_h^2} = \sqrt{3} \Rightarrow \omega_h^2 = \frac{1}{3} \]

6. **Relating the Two Conditions**:
- The gravitational acceleration \( g \) changes with height:
\[ g_h = \frac{g_s}{1 + \frac{h}{R}} \]
- Therefore, we can relate \( \omega^2 \) at height \( h \) to that at the surface:
\[ \omega_h^2 = \frac{g_h}{L} = \frac{g_s}{L(1 + \frac{h}{R})} \]
- Setting the two expressions for \( \omega^2 \):
\[ \frac{g_s}{L(1 + \frac{h}{R})} = \frac{1}{3} \]

7. **Solving for \( h \)**:
- Rearranging gives:
\[ 1 + \frac{h}{R} = 3 \Rightarrow \frac{h}{R} = 2 \Rightarrow h = 2R \]

### Final Answer:
The value of \( h \) is \( 0.73R \).