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(B) According to the Doppler effect,when an observer moves towards a stationary source,the observed frequency $n'$ is given by:$n' = n \left( \frac{v

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$2$

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(B) According to the Doppler effect,when an observer moves towards a stationary source,the observed frequency $n'$ is given by:$n' = n \left( \frac{v + v_0}{v} \right)$where $n$ is the source frequency,$v$ is the speed of sound,and $v_0$ is the speed of the observer.For train $A$:$5.5 = 5 \left( \frac{v + v_A}{v} \right) \implies 1.1 = 1 + \frac{v_A}{v} \implies \frac{v_A}{v} = 0.1$ ... $(i)$For train $B$:$6.0 = 5 \left( \frac{v + v_B}{v} \right) \implies 1.2 = 1 + \frac{v_B}{v} \implies \frac{v_B}{v} = 0.2$ ... $(ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{v_B / v}{v_A / v} = \frac{0.2}{0.1} = 2$Therefore,the ratio of the velocity of train $B$ to that of train $A$ is $2$.

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