A tuning fork of frequency 220 ,Hz produces s | vedclass

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(D) At $N.T.P$, the temperature $T_0 = 273 \,K$. The velocity of sound $v_0$ is given by $v_0 = f \lambda_0 = 220 \,Hz 	imes 1.5 \,m = 330 \,m/s$.Whe

Vedclass · Accepted Answer

$0.07$

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(D) At $N.T.P$, the temperature $T_0 = 273 \,K$. The velocity of sound $v_0$ is given by $v_0 = f \lambda_0 = 220 \,Hz 	imes 1.5 \,m = 330 \,m/s$.When the temperature increases to $27^{\circ} C$, the new temperature $T = 273 + 27 = 300 \,K$.The velocity of sound in air is proportional to the square root of the absolute temperature: $\frac{v}{v_0} = \sqrt{\frac{T}{T_0}}$.Substituting the values: $v = 330 	imes \sqrt{\frac{300}{273}} = 330 	imes 1.05 = 346.5 \,m/s$.The new wavelength $\lambda$ is given by $\lambda = \frac{v}{f} = \frac{346.5}{220} = 1.575 \,m$.The increase in wavelength $\Delta \lambda = \lambda - \lambda_0 = 1.575 \,m - 1.5 \,m = 0.075 \,m$.Rounding to the nearest value provided, we get $0.07 \,m$.

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

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