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In a resonance tube experiment, two succ...

In a resonance tube experiment, two successive resonances are heard at 15 cm 48 cm. End correction will be

A

1.5cm

B

3cm

C

2.5cm

D

1cm

Text Solution

Verified by Experts

The correct Answer is:
A
`rho=(l_2-3l_1)/(2) = (48-3xx15)/)(2)`
`rho = (48-45)/(2) = 3/2 = 1/5cm`
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