Q133 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Physics

When the area of cross-section of a stretched wire is halved and tension is doubled, the speed of propagation of transverse waves along it becomes $k$ times the initial speed. Then, $k$

  • A
    1
  • B
    4
  • Correct answer
    C
    2
  • D
    8

Solution

Speed of transverse wave in stretched wire is given as $$ v=\sqrt{\frac{T}{m}} $$ where, $T=$ tension in the wire and $\quad m=$ mass per unit length of wire. $$ \begin{aligned} & \therefore \quad v=\sqrt{\frac{T}{M / l}} \quad\left[\because m=\frac{M}{l}\right] \\ & =\sqrt{\frac{T l}{M}}=\sqrt{\frac{T l}{V \cdot \rho}} \\ & {[\because M=V \cdot \rho]} \\ & =\sqrt{\frac{T l}{A \cdot l \cdot \rho}} \\ & {[\because V=A \cdot l]} \\ & v=\sqrt{\frac{T}{A \rho}} \Rightarrow v \propto \sqrt{\frac{T}{A}} \\ & \Rightarrow \quad \frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1} \times \frac{A_1}{A_2}} \\ & \text { Given, } \quad A_2=\frac{A_1}{2}, T_2=2 T_1 \\ & \end{aligned} $$ $\therefore$ From Eq. (i), we get $$ \begin{aligned} & \frac{v_2}{v_1}=\sqrt{\frac{2 T_1}{T_1} \times \frac{A_1}{\frac{A_1}{2}}}=\sqrt{4} \\ & v_2=2 v_1 \Rightarrow K=2 \end{aligned} $$

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

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