Question

Two cars are moving on two perpendicular roads towards a crossing with uniform speeds of $72km/hr$ and $36km/hr$. If first car blows horn of frequency $280Hz$, then the frequency of horn heard by the driver of second car when line joining the cars make ${45}^{\circ }$ angle with the roads; will be

• A. $280Hz$
• B. $321Hz$
• C. $298Hz$
• D. $289Hz$

Answer 1

The correct option is C $298Hz$

We know that: $n^{\prime }=n\left(\frac{v+v_B\cos \theta }{v-v_A\cos \theta }\right)$.

Given,
Speed of first car,
$v_A=72km/hr=72\times \frac{5}{18}=20m/\text{sec}$

Speed of second car,
$v_B=36km/hr=36\times \frac{5}{18}=10m/\text{sec}$

Actual frequency, $n=280Hz$
$\theta =45°$

Then the frequency of horn heard by the driver of second car,
$n^{\prime }=280\left(\frac{340+10\cos \cos {45}^{\circ }}{340-20\cos \cos {45}^{\circ }}\right)$

$n^{\prime }=280\left(\frac{340+10/\sqrt{2}}{340-20/\sqrt{2}}\right)$

$n^{\prime }=280\left(\frac{340+7.07}{340-14.14}\right)$

$n^{\prime }=280\left(\frac{347.07}{325.86}\right)$

$n^{\prime }=280\left(1.065\right)$

$n^{\prime }=298Hz$

Final answer: (b)