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In a resonance tube experiment, the first resonance is obtained for 10 cm of air column and the second for 32 cm. The end correction for this apparatus is equal to ......... 

(A) 1.9 cm 

(B) 0.5 cm 

(C) 2 cm 

(D) 1.0 cm

1 Answer

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Best answer

The correct option is  (D) 1.0 cm.

Explanation:

end correction = {(ℓ2 – 3ℓ1) / 2} 

Here ℓ1 = 10 cm 

2 = 32 cm 

∴ end correction = [{32 – (3 × 10)} / 2] = 1 cm

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