A car is moving towards an approaching bike, the bike is moving with a velocity 30ms^-1 . Frequency of horn sounded by the car is 100Hz, while it is observed by bike rider as 120Hz. The actual duration of horn is 6s. Find the time interval for which the bike rider hears it. (Take velocity of sound in air as 330ms^-1)

Velocity of bike ${\mathrm{V}}_{\mathrm{b}}=30{\mathrm{ms}}^{-1}\text{ Let the velocity of car be }{\mathrm{V}}_{\mathrm{e}}\cdot {\mathrm{f}}_{\text{obs }}={\mathrm{f}}_{\text{actual }}\times \frac{{\mathrm{V}}_{\mathrm{so}}+{\mathrm{V}}_{\mathrm{b}}}{{\mathrm{V}}_{\mathrm{so}}-{\mathrm{V}}_{\mathrm{c}}}$
$120\mathrm{Hz}=100\mathrm{Hz}\times \frac{330+30}{330-{\mathrm{V}}_{\mathrm{e}}}\Rightarrow {\mathrm{V}}_{\mathrm{e}}=30{\mathrm{ms}}^{-1}$
Distance between initial and final pulse of horn $={\mathrm{V}}_{\mathrm{so}}\times \mathrm{\Delta t}-{\mathrm{V}}_{\mathrm{e}}\mathrm{\Delta t}\left(\mathrm{\Delta t}\text{ is duration of emission of sound }\right)=(330-30)\times 6=1800\mathrm{m}$
This wave train travels with 330ms^-1 and cross bike rider moving with velocity of 30ms^-1 towards it. 
Time taken $=\frac{\text{ length of wave train }}{{\mathrm{V}}_{\mathrm{so}}+{\mathrm{V}}_{\mathrm{b}}}=\frac{1800}{330+30}=\frac{180}{36}=5\mathrm{s}$