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Question

An aluminum wire of length 60 cm is joined to a steel wire of length 80 cm and stretched between two fixed supports. The tension produced is 40 N. Cross-sectional area is 1 mm2 (steel) and 3 mm2 (aluminum). Minimum frequency of the tuning fork which can produce standing waves with the joint as a node is nearly equal to

(density of Al=2.6 g cc1 and density of steel =7.8 g cc1)

A
90 Hz
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B
145 Hz
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C
180 Hz
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D
250 Hz
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Solution

The correct option is C 180 Hz
frequency of wire oscillating is f=n2lTμ=n2lTAρ
Since frequency remains same for both the wires,
n2l1TA1ρ1=P2l2TA2ρ2
nP=43

Taking the frequency expression for the aluminium wire and for minimum frequency P=3
f=32(0.6)40106×2.6×103×3
=10.440×1032.6×3
=1000.423.9=180 Hz

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