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(C) Given: Length of the tube $L = 0.8 \ m$,Frequency $f = 500 \ Hz$,Speed of sound $v = 340 \ m/s$.First,calculate the wavelength $\lambda$ of the so

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$2$

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(C) Given: Length of the tube $L = 0.8 \ m$,Frequency $f = 500 \ Hz$,Speed of sound $v = 340 \ m/s$.First,calculate the wavelength $\lambda$ of the sound wave:$\lambda = \frac{v}{f} = \frac{340}{500} = 0.68 \ m$.The resonance condition for a tube closed at one end is $L_n = \frac{(2n-1)\lambda}{4}$,where $n = 1, 2, 3, \dots$.We need to find the number of values of $n$ such that $L_n \le 0.8 \ m$.For $n=1$: $L_1 = \frac{\lambda}{4} = \frac{0.68}{4} = 0.17 \ m$.For $n=2$: $L_2 = \frac{3\lambda}{4} = 3 	imes 0.17 = 0.51 \ m$.For $n=3$: $L_3 = \frac{5\lambda}{4} = 5 	imes 0.17 = 0.85 \ m$.Since $L_3 = 0.85 \ m > 0.8 \ m$,the third resonance cannot be formed within the tube length.Therefore,only $2$ resonances will be heard.

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