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(A) The condition for resonance in a closed pipe is given by $L_n = \frac{(2n-1)v}{4f}$,where $n$ is the resonance order,$v$ is the speed of sound,and

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$x > 54$

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(A) The condition for resonance in a closed pipe is given by $L_n = \frac{(2n-1)v}{4f}$,where $n$ is the resonance order,$v$ is the speed of sound,and $f$ is the frequency of the tuning fork.For the first resonance in winter $(n=1)$: $L_1 = \frac{v_w}{4f} = 18\,cm$.For the second resonance in summer $(n=2)$: $L_2 = \frac{3v_s}{4f} = x\,cm$.Since the temperature in summer is higher than in winter,the speed of sound in summer $(v_s)$ is greater than in winter $(v_w)$,i.e.,$v_s > v_w$.From the first resonance,we have $v_w = 4f 	imes 18 = 72f$.Substituting $v_s > v_w$ into the second resonance equation: $x = \frac{3v_s}{4f} > \frac{3v_w}{4f} = 3 	imes 18 = 54\,cm$.Therefore,$x > 54$.

While measuring the speed of sound by performing a resonance column experiment,a student gets the first resonance condition at a column length of $18\,cm$ during winter. Repeating the same experiment during summer,she measures the column length to be $x\,cm$ for the second resonance. Then

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