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. `g/30`
B. `g/20`
C. `g/5`
D. `g/10`

Explanation

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the wave speed in the string at rest
The speed of transverse waves in the string when the car is at rest is given as \( V_1 = 60 \, \text{m/s} \). The wave speed in a string is given by the formula:
\[ V = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string and \( \mu \) is the linear mass density of the string.

At rest, the tension \( T \) is simply the weight of the heavy ball:
\[ T = Mg \]
Thus, we can write:
\[ 60 = \sqrt{\frac{Mg}{\mu}} \tag{1} \]

### Step 2: Understand the wave speed in the string when the car is accelerating
When the car accelerates with acceleration \( a \), the effective weight of the ball changes due to the acceleration. The effective gravitational acceleration becomes:
\[ g' = \sqrt{g^2 + a^2} \]
Thus, the tension in the string when the car is accelerating is:
\[ T' = Mg' \]
Substituting this into the wave speed formula gives:
\[ V_2 = \sqrt{\frac{Mg'}{\mu}} = \sqrt{\frac{M\sqrt{g^2 + a^2}}{\mu}} \]
It is given that \( V_2 = 60.5 \, \text{m/s} \). Therefore, we can write:
\[ 60.5 = \sqrt{\frac{M\sqrt{g^2 + a^2}}{\mu}} \tag{2} \]

### Step 3: Set up the ratio of the two equations
Now, we will divide equation (2) by equation (1):
\[ \frac{60.5}{60} = \frac{\sqrt{M\sqrt{g^2 + a^2}/\mu}}{\sqrt{Mg/\mu}} \]
This simplifies to:
\[ \frac{60.5}{60} = \sqrt{\frac{\sqrt{g^2 + a^2}}{g}} \]

### Step 4: Square both sides
Squaring both sides gives:
\[ \left(\frac{60.5}{60}\right)^2 = \frac{\sqrt{g^2 + a^2}}{g} \]
Let’s denote \( x = \frac{60.5}{60} \):
\[ x^2 = \frac{\sqrt{g^2 + a^2}}{g} \]
Rearranging gives:
\[ \sqrt{g^2 + a^2} = g x^2 \]

### Step 5: Square again
Squaring both sides again:
\[ g^2 + a^2 = g^2 x^4 \]
Rearranging gives:
\[ a^2 = g^2 x^4 - g^2 = g^2 (x^4 - 1) \]
Thus:
\[ a = g \sqrt{x^4 - 1} \]

### Step 6: Calculate \( x \)
Calculating \( x \):
\[ x = \frac{60.5}{60} \approx 1.00833 \]
Calculating \( x^4 \):
\[ x^4 \approx (1.00833)^4 \approx 1.0335 \]
Thus:
\[ x^4 - 1 \approx 0.0335 \]

### Step 7: Calculate \( a \)
Now substituting back:
\[ a \approx g \sqrt{0.0335} \approx g \cdot 0.1833 \approx 0.1833g \]

### Conclusion
The value of \( a \) in terms of \( g \) is approximately \( 0.1833g \).