A tuning fork of frequency $220 \mathrm{~Hz}$ produces sound waves of wavelength $1.5 \mathrm{~m}$ in air at…

A tuning fork of frequency $220 \mathrm{~Hz}$ produces sound waves of wavelength $1.5 \mathrm{~m}$ in air at N.T.P. The increase in wavelength when the temperature of air is $27^{\circ} \mathrm{C}$ is nearly $\left(\sqrt{\frac{300}{273}}=1.05\right)$
  1. $0.06 \mathrm{~m}$
  2. $0.10 \mathrm{~m}$
  3. $0.09 \mathrm{~m}$
  4. $0.07 \mathrm{~m}$

Solution

$\begin{aligned} & \mathrm{v}_0=\mathrm{f} \lambda_0=220 \times 1.5 \\ & \mathrm{v}_0=330 \mathrm{~m} / \mathrm{s} \end{aligned}$ We know, $\begin{aligned} \frac{\mathrm{v}}{\mathrm{v}_0} & =\sqrt{\frac{\mathrm{T}}{\mathrm{T}_0}} \\ \Rightarrow \mathrm{v} & =330 \sqrt{\frac{300}{273}}=330 \times 1.05 \\ \mathrm{v} & =346.1 \mathrm{~m} / \mathrm{s} \\ \lambda \quad \lambda & =\frac{\mathrm{v}}{\mathrm{f}}=\frac{346.1}{220} \\ \lambda & =1.57 \mathrm{~m} \end{aligned}$ $\therefore \quad$ The increase, in wavelength is: $\begin{aligned} & \Delta \lambda=\lambda-\lambda_0=1.57-1.5 \\ & \Delta \lambda=0.07 \mathrm{~m} \end{aligned}$

Asked in: MHT CET 2023 (12 May Shift 2)

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