1. Wave speed in a stretched string

For a light string of linear mass density $\mu$, the speed of transverse waves is

$v=\sqrt{\frac{T}{\mu}}$

where $T$ is the tension in the string.

2. When the car is at rest

The heavy ball of mass $M$ hangs vertically, so tension is

$T_1=Mg$

Given wave speed:

$v_1=60\ \text{m s}^{-1}$

So,

$v_1^2=\frac{T_1}{\mu}=\frac{Mg}{\mu}$

Thus,

$\frac{Mg}{\mu}=60^2=3600$

3. When the car accelerates with acceleration $a$

In the accelerating car, the bob is in equilibrium in the non-inertial frame under:

• weight $Mg$ downward,
• pseudo force $Ma$ horizontally backward.

Hence the tension becomes the resultant:

$T_2=M\sqrt{g^2+a^2}$

Given new wave speed:

$v_2=60.5\ \text{m s}^{-1}$

Therefore,

$v_2^2=\frac{T_2}{\mu}=\frac{M\sqrt{g^2+a^2}}{\mu}$

Using $\dfrac{Mg}{\mu}=3600$,

$\frac{T_2}{T_1}=\frac{v_2^2}{v_1^2}$

So,

$\frac{\sqrt{g^2+a^2}}{g}=\left(\frac{60.5}{60}\right)^2$

4. Simplify the ratio

$\frac{60.5}{60}=1.008333\ldots$

Hence,

$\left(\frac{60.5}{60}\right)^2\approx 1.016736$

Thus,

$\sqrt{1+\left(\frac{a}{g}\right)^2}=1.016736$

Squaring both sides,

$1+\left(\frac{a}{g}\right)^2=(1.016736)^2$

$1+\left(\frac{a}{g}\right)^2\approx 1.03375$

Therefore,

$\left(\frac{a}{g}\right)^2\approx 0.03375$

$\frac{a}{g}\approx \sqrt{0.03375}\approx 0.184$

So,

$a\approx 0.184g$

5. Compare with options

• $\dfrac{g}{30}\approx 0.033g$
• $\dfrac{g}{5}=0.2g$
• $\dfrac{g}{10}=0.1g$
• $\dfrac{g}{20}=0.05g$

The closest value is

$\boxed{\frac{g}{5}}$

So the correct option is B.