Q185 / 200 · Manipal (MET) 2013 · Physics

The speed of a wave on a string is $150 \mathrm{~m} / \mathrm{s}$ when the tension is 120 N . The percentage increase in the tension in order to raise the wave speed by $20 \%$ is

  • Correct answer
    A
    44 %
  • B
    40 %
  • C
    20 %
  • D
    10 %

Solution

We know that $v=\sqrt{\frac{T}{m}}$ $v \propto \sqrt{T}$ $\begin{aligned} & \frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}} \\ & \frac{T_1}{T_2}=\left(\frac{v_1}{v_2}\right)^2 \\ & \frac{T_2-T_1}{T_1}=\frac{v_1^2-v_1^2}{v_1^2} \\ & v_2=v_1+\frac{20}{100} v_1=\frac{120}{100} v_1=\frac{6}{5} v_1 \\ &= \frac{6}{5} \times 150\end{aligned}$ $\begin{aligned} \frac{T_2-T_1}{T_1} & =\frac{(180)^2-(150)^2}{(150)^2} \\ & =\frac{30 \times 330}{150 \times 180}=0.47 \\ & =\frac{T_2-T_1}{T} \times 100=0.44 \times 100=44 \%\end{aligned}$

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

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