Q139 / 150 · MHT CET 2022 (07 Aug Shift 2) · Physics

A uniform metal wire has length ' $L$ ', mass ' $M$ ' and cross-sectional area ' $A$ '. It is under tension ' $T$ ' and ' $V$ ' is the speed of transverse wave along the wire. The density of the wire

  • A
    $\frac{A T}{V^2}$
  • B
    $\frac{T}{A^2 V}$
  • Correct answer
    C
    $\frac{T}{V^2 A}$
  • D
    $\frac{V^2}{A^2 T}$

Solution

For transverse waves on a wire $V=\sqrt{\frac{T}{m}}$ The mass per unit length can be written as $m=\frac{M}{L}=\frac{A L \rho}{L}=A \rho$ $\begin{aligned} & \therefore V=\sqrt{\frac{T}{m}}=\sqrt{\frac{T}{A \rho}} \\ & \therefore V^2=\frac{T}{A \rho} \\ & \therefore A=\frac{T}{V^2 \rho}\end{aligned}$

Hence, the correct answer is option: 'C'.

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