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In a resonance tube with tuning fork of frequency 512Hz, first resonance occurs at water level equal to 30.3 cm and second resonance occurs at 63.7 cm. The maximum possible error in the speed of sound is (A) 51.2 cm/s (B) 102.4 cm/s (C) 204.8 cm/s (D) 153.6 cm/s

 In a resonance tube with tuning fork of frequency 512Hz, first resonance occurs at water level equal to 30.3 cm and second resonance occurs at 63.7 cm. The maximum possible error in the speed of sound is
(A) 51.2 cm/s (B) 102.4 cm/s (C) 204.8 cm/s (D) 153.6 cm/s 

Grade:Select Grade

2 Answers

Hitesh Soni
32 Points
8 years ago
we have got two equations here:
1st eq. is-
l1 + e = v/4f.........(1)
2nd eq.is
l2 + e = 3v/4f........(2)
Apply (2) – (1)
l2 - l1 = 3v/4f – v/4f = 2v/4f = v/2f
delta(l2 - l1) / (l2 - l1) = delta(v) / v
delta(v) = 2f delta(l2 - l1) = 2f(deltal1 + deltal2) …......for maximum error
delta(v) = 2 x 512 x (0.1 + 0.1) = 2 x 512 x 0.2
delta(v) = 204.8 cm/s
Hence (c) is correct.
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

we have got two equations here:
1steq. is-
l1 + e = v/4f.........(1)
2ndeq.is
l2 + e = 3v/4f........(2)
Apply (2) – (1)
l2-l1=3v/4f – v/4f = 2v/4f = v/2f
delta(l2-l1) / (l2-l1) = delta(v) / v
delta(v) = 2f delta(l2-l1) = 2f(deltal1+ deltal2) …......for maximum error
delta(v) = 2 x 512 x (0.1 + 0.1) = 2 x 512 x 0.2
delta(v) = 204.8 cm/s

Thanks and Regards

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