Q129 / 150 · MHT CET 2023 (12 May Shift 1) · Physics

41 tuning forks are arranged in increasing order of frequency such that each produces 5 beats/second with next tuning fork. If frequency of last tuning fork is double that of frequency of first fork. Then frequency of first and last fork is

  • A
    $400,200 \mathrm{~Hz}$
  • Correct answer
    B
    $200,400 \mathrm{~Hz}$
  • C
    $100,200 \mathrm{~Hz}$
  • D
    $205,410 \mathrm{~Hz}$

Solution

Let Frequency of $1^{\text {st }}$ tuning fork be $=\mathrm{n}_1$ $\therefore \quad$ frequency of $41^{\text {st }}$ tuning fork $=n_{41}$ Now, $$ \begin{array}{ll} & \mathrm{n}_{41}=\mathrm{n}_1+(41-1) \times 5 \\ & \text { But, } \mathrm{n}_{41}=2 \mathrm{n}_1 \\ \therefore \quad & 2 \mathrm{n}_1=\mathrm{n}_1+200 \\ \therefore \quad & \mathrm{n}_1=200 \mathrm{~Hz} \\ \therefore \quad & \mathrm{n}_{41}=400 \mathrm{~Hz} \end{array} $$

Hence, the correct answer is option: 'B'.

Relevant Questions