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# The speed of a wave on a string is 150 m/s when the tension is 120 N. The percentage increase in the tension in order to raise the wave speed by 20 % is?(A) 44 %(B) 40 %(C) 20 %(D) 10 %

Last updated date: 21st Apr 2024
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Hint We should know that tension is described as a pulling force which is transferred between the bodies in the form of axial ways. By axial ways we mean by string, a cable or a chain or it can be any one-dimensional object.

Complete step by step answer
We know that:
$v = \sqrt {\dfrac{T}{m}}$
We can say that:
$v\propto \sqrt T$
$\Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{{{T_1}}}{{{T_2}}}}$
$\Rightarrow \dfrac{{{T_1}}}{{{T_2}}} = {\left( {\dfrac{{{v_1}}}{{{v_2}}}} \right)^2}$
Now we have evaluated to get:
$\dfrac{{{T_2} - {T_1}}}{{{T_1}}} = \dfrac{{v_1^2 - v_1^2}}{{v_1^2}}$
$\Rightarrow {v_2} = {v_1} + \dfrac{{20}}{{100}}{v_1}$
$\Rightarrow {v_2} = \dfrac{{120}}{{100}}{v_1} = \dfrac{6}{5}{v_1}$
Now we have to put the value of ${v_1}$ to get: $\dfrac{6}{5} \times 100$
Now we can write:
$\dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}}=\dfrac{{{(180)}^{2}}-{{(150)}^{2}}}{{{(150)}^{2}}}$
Now we have to evaluate to get:
$\dfrac{{30 \times 330}}{{150 \times 180}} = 0.47$
$\Rightarrow \dfrac{{{T_2} - {T_1}}}{{{T_1}}} \times 100 = 0.44 \times 100 = 44\%$
Hence, we can say that the percentage increase in the tension in order to raise the wave speed by 20 % is 44 %.

Hence the correct answer is option A.

Note We should know that the tension of a body is defined as being equal to the mass of the body multiplied by the gravitational force that is applied on the body plus or minus the mass multiplied by the acceleration. The value of g is taken as 9.8 m/$s^2$, in every case if it is not mentioned to us in the question.