In a resonance tube, we get first and second resonances at 27 cm and 83 cm. Then gap between open end and node of pressure at open end is

1 cm

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2 cm

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3 cm

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4 cm

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Solution

The correct option is A 1 cm
$e + l_{1} = \frac{λ}{4}$ , $l_{2} + e = \frac{3 λ}{4}$
where e = end correction = gap between end and pressure node.
$l_{2} + e - 3 l_{1} - 3 e = 0$
$(l_{2} - 3 l_{1}) = 2 e$
$e = \frac{l_{2} - 3 l_{1}}{2} = \frac{83 - 81}{2} = 1 cm$

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In a resonance tube, we get first and second resonances at 27 cm and 83 cm. Then gap between open end and node of pressure at open end is

The correct option is A 1 cme+l1=λ4, l2+e=3λ4 where e = end correction = gap between end and pressure node. l2+e−3l1−3e=0 (l2−3l1)=2e e=l2−3l12=83−812=1 cm

The correct option is A 1 cm
$e + l_{1} = \frac{λ}{4}$ , $l_{2} + e = \frac{3 λ}{4}$
where e = end correction = gap between end and pressure node.
$l_{2} + e - 3 l_{1} - 3 e = 0$
$(l_{2} - 3 l_{1}) = 2 e$
$e = \frac{l_{2} - 3 l_{1}}{2} = \frac{83 - 81}{2} = 1 cm$