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(B) Given,frequency of source,$f_s = 5 	ext{ kHz}$.For the passenger in train $A$,the observed frequency is $f_A = 5.5 	ext{ kHz}$.Using the Doppler

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$2$

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(B) Given,frequency of source,$f_s = 5 	ext{ kHz}$.For the passenger in train $A$,the observed frequency is $f_A = 5.5 	ext{ kHz}$.Using the Doppler effect formula for an observer moving towards a stationary source: $f_A = f_s \left( \frac{v + v_A}{v} ight)$,where $v$ is the speed of sound.$5.5 = 5 \left( 1 + \frac{v_A}{v} ight) \implies 1.1 = 1 + \frac{v_A}{v} \implies \frac{v_A}{v} = 0.1$  --- $(i)$For the passenger in train $B$,the observed frequency is $f_B = 6 	ext{ kHz}$.Similarly,$f_B = f_s \left( \frac{v + v_B}{v} ight)$.$6 = 5 \left( 1 + \frac{v_B}{v} ight) \implies 1.2 = 1 + \frac{v_B}{v} \implies \frac{v_B}{v} = 0.2$  --- $(ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{v_B / v}{v_A / v} = \frac{0.2}{0.1} = 2$.Thus,the ratio of the speed of train $B$ to that of train $A$ is $2$.

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