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Question

# The length of the wire shown in figure between the pulley and fixed support is 1.5 m and mass is 12.0 g. The frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulley and the fixed support, is: [Assume the pulley acts as a fixed support and consider g=10 m/s2]

A
10 Hz
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B
30 Hz
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C
100 Hz
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D
70 Hz
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Solution

## The correct option is C 100 HzThe string behaves as if fixed at both ends. Standing wave with 2 loops can be considered as ⇒λ=L We know that Frequency, f=Velocityλ=vλ From FBD of block: Tension in the string is T= Tension =18g N Mass per unit length of the string is μ=12×10−31.5 kg/m Velocity of wave in the string is v=√Tμ ⇒v=√18×10×1.512×10−3=150 m/s ∴f=vλ=vL=1501.5=100 Hz

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