MCQ
A source and an observer approach each other with same velocity $50 \mathrm{~m} / \mathrm{s}$. If the apparent frequency is $435 \mathrm{sec}$, then the real frequency is
  • $320 \mathrm{~sec}$
  • B
    $360 \mathrm{~sec}$
  • C
    $390 \mathrm{~sec}$
  • D
    $420 \mathrm{~sec}$

Answer

Correct option: A.
$320 \mathrm{~sec}$
$n^{\prime}=n\left[\frac{v+v_O}{v-v_S}\right] ;$ Here $v=332 \mathrm{~m} / \mathrm{s}$ and $v_0=v_s=50 \mathrm{~m} / \mathrm{s}$
$\Rightarrow 435=n\left[\frac{332+50}{332-50}\right] \Rightarrow n=321.12 \mathrm{sec}^{-1} \approx 320 \mathrm{sec}^1$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free