A heavy ball is mass M is suspended from the ceiling of a car by a light string of mass `m(m lt lt M)`. When the car is at rest, the speed of transverse waves in the string is `60ms^(-1)`. When the car has acceleration a, the wave-speed increases to `60.5ms^(-1)`. The value of a, in terms of gravitational acceleration g, is closest to:

Options

A. (a)`(g)/(30)`
B. (b)`(g)/(20)`
C. (c)`(g)/(5)`
D. (d)`(g)/(10)`

Explanation

To solve the problem, we need to analyze the wave speed in the string under two different conditions: when the car is at rest and when it is accelerating.

### Step-by-Step Solution:

1. **Understanding Wave Speed in the String**:
The speed of transverse waves in a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( v \) is the wave speed, \( T \) is the tension in the string, and \( \mu \) is the linear mass density of the string.

2. **Case 1: Car at Rest**:
When the car is at rest, the only force acting on the ball is its weight, which creates tension in the string:
\[ T_1 = Mg \]
Therefore, the wave speed when the car is at rest is:
\[ v_1 = 60 \, \text{m/s} = \sqrt{\frac{Mg}{\mu}} \tag{1} \]

3. **Case 2: Car Accelerating**:
When the car is accelerating with acceleration \( a \), the effective gravitational force acting on the ball increases. The tension in the string becomes:
\[ T_2 = M(g + a) \]
The wave speed in this case is:
\[ v_2 = 60.5 \, \text{m/s} = \sqrt{\frac{M(g + a)}{\mu}} \tag{2} \]

4. **Setting Up the Ratio**:
We can set up a ratio of the two wave speeds:
\[ \frac{v_1}{v_2} = \frac{\sqrt{\frac{Mg}{\mu}}}{\sqrt{\frac{M(g + a)}{\mu}}} \]
Simplifying this gives:
\[ \frac{60}{60.5} = \sqrt{\frac{g}{g + a}} \]

5. **Squaring Both Sides**:
Squaring both sides to eliminate the square root:
\[ \left(\frac{60}{60.5}\right)^2 = \frac{g}{g + a} \]

6. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 60^2 (g + a) = 60.5^2 g \]

7. **Expanding and Rearranging**:
Expanding and rearranging the equation:
\[ 3600g + 3600a = 60.5^2 g \]
\[ 3600a = 60.5^2 g - 3600g \]

8. **Calculating \( a \)**:
Now, we can solve for \( a \):
\[ a = \frac{(60.5^2 - 3600)}{3600} g \]

9. **Calculating Numerical Values**:
Calculate \( 60.5^2 \):
\[ 60.5^2 = 3660.25 \]
Therefore:
\[ a = \frac{(3660.25 - 3600)}{3600} g = \frac{60.25}{3600} g \]
\[ a \approx 0.01674 g \]

10. **Final Approximation**:
To find the value of \( a \) in terms of \( g \):
\[ a \approx \frac{g}{60} \text{ (approximately)} \]

### Final Answer:
The value of \( a \) in terms of gravitational acceleration \( g \) is closest to \( \frac{g}{60} \).