Q109 / 125 · BITSAT 2013 · Physics

Two waves of wavelengths $99 \mathrm{~cm}$ and $100 \mathrm{~cm}$ both travelling with velocity $396 \mathrm{~m} / \mathrm{s}$ are made to interfere. The number of beats produced by them per second is

  • A
    1
  • B
    2
  • Correct answer
    C
    4
  • D
    8

Solution

Velocity of wave $\mathrm{v}=\mathrm{n} \lambda$ where $n=$ frequency of wave $\Rightarrow n=\frac{v}{\lambda}$ $\mathrm{n}_{2}=\frac{\mathrm{v}_{2}}{\lambda_{2}}=\frac{396}{100 \times 10^{-2}}=396 \mathrm{~Hz}$ no. of beats $=\mathrm{n}_{1}-\mathrm{n}_{2}=4$

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

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