MHT CET · Physics · Waves and Sound
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 \(\left(\sqrt{\frac{300}{273}}=1.05\right)\)
- A \(0.06 \mathrm{~m}\)
- B \(0.10 \mathrm{~m}\)
- C \(0.09 \mathrm{~m}\)
- D \(0.07 \mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(0.07 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \mathrm{v}_0=\mathrm{f} \lambda_0=220 \times 1.5 \\
& \mathrm{v}_0=330 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
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:
\(
\begin{aligned}
& \Delta \lambda=\lambda-\lambda_0=1.57-1.5 \\
& \Delta \lambda=0.07 \mathrm{~m}
\end{aligned}
\)
\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:
\(
\begin{aligned}
& \Delta \lambda=\lambda-\lambda_0=1.57-1.5 \\
& \Delta \lambda=0.07 \mathrm{~m}
\end{aligned}
\)
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