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Chapter 10: Problem 1463

Length of a steel wire is \(11 \mathrm{~m}\) and its mass is \(2.2 \mathrm{~kg}\). What should be the tension in the wire so that the speed of a transverse wave in it is equal to the speed of sound in dry air at \(20^{\circ} \mathrm{C}\) temperature? (A) \(2.31 \times 10^{4} \mathrm{~N}\) (B) \(2.25 \times 10^{4} \mathrm{~N}\) (C) \(2.06 \times 10^{4} \mathrm{~N}\) (D) \(2.56 \times 10^{4} \mathrm{~N}\)

Short Answer

Expert verified
The correct answer is (A) \(2.31 \times 10^{4} \mathrm{~N}\).

Step by step solution

01

Calculate the Linear Density of the Wire

Linear density is defined as mass per unit length. We are given the length (\(L\)) of the steel wire as \(11 \mathrm{~m}\) and its mass (\(M\)) as \(2.2 \mathrm{~kg}\). The linear density (\(\mu\)) can be found using the following equation: \[\mu = \frac{M}{L}\] Plugging in the values, we get: \[\mu = \frac{2.2}{11} = 0.2 \mathrm{~kg/m}\]
02

Find the Speed of Sound in Dry Air at \(20^{\circ} \mathrm{C}\)

The speed of sound in dry air can be calculated using the following equation: \[v = 331.4 + 0.6T\] Where \(v\) is the speed of sound, and \(T\) is the temperature in Celsius. Plugging in the given temperature, we get: \[v = 331.4 + 0.6 \times 20 = 343 \mathrm{~m/s}\]
03

Calculate the Tension in the Wire Using the Wave Speed Formula

The wave speed (\(v\)) in a string can be calculated using the following equation: \[v = \sqrt{\frac{T}{\mu}}\] Where \(T\) is the tension in the string, and \(\mu\) is the linear density. We aim to find the tension that results in a wave speed equal to the speed of sound in dry air (\(343 \mathrm{~m/s}\)). Rearranging the equation to find tension, we get: \[T = v^{2} \times \mu\] Plugging in the values, we get: \[T = 343^{2} \times 0.2 = 23589.4 \mathrm{~N}\]
04

Find the Closest Option

Now that we have calculated the tension in the wire, we can compare our result with the given options. The closest option is: (A) \(2.31 \times 10^{4} \mathrm{~N}\) So, the correct answer is (A).

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Most popular questions from this chapter

A particle having mass \(1 \mathrm{~kg}\) is executing S.H.M. with an amplitude of \(0.01 \mathrm{~m}\) and a frequency of \(60 \mathrm{hz}\). The maximum force acting on this particle is \(\ldots \ldots . . \mathrm{N}\) (A) \(144 \pi^{2}\) (B) \(288 \pi^{2}\) (C) \(188 \pi^{2}\) (D) None of these. (A) \(x=a \sin 2 p \sqrt{(\ell / g) t}\) (B) \(x=a \cos 2 p \sqrt{(g / \ell) t}\) (C) \(\mathrm{x}=\mathrm{a} \sin \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\) (D) \(\mathrm{x}=\mathrm{a} \cos \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\)

If two almost identical waves having frequencies \(\mathrm{n}_{1}\) and \(\mathrm{n}_{2}\), produced one after the other superposes then the time interval to obtain a beat of maximum intensity is \(\ldots \ldots \ldots .\) (A) \(\left\\{1 /\left(\mathrm{n}_{1}-\mathrm{n}_{2}\right)\right\\}\) (B) \(\left(1 / \mathrm{n}_{1}\right)-\left(1 / \mathrm{n}_{2}\right)\) (C) \(\left(1 / \mathrm{n}_{1}\right)+\left(1 / \mathrm{n}_{2}\right)\) (D) \(\left\\{1 /\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)\right\\}\)

When an elastic spring is given a displacement of \(10 \mathrm{~mm}\), it gains an potential energy equal to \(\mathrm{U}\). If this spring is given an additional displacement of \(10 \mathrm{~mm}\), then its potential energy will be.............. (A) \(\mathrm{U}\) (B) \(2 \mathrm{U}\) (C) \(4 \mathrm{U}\) (D) \(\mathrm{U} / 4\).

If the equation of a wave in a string having linear mass density $0.04 \mathrm{~kg} \mathrm{~m}^{-1}\( is given by \)\mathrm{y}=0.02\( \)\sin [2 \pi\\{1 /(0.04)\\}-\\{\mathrm{x} /(0.50)\\}]$, then the tension in the string is \(\ldots \ldots \ldots \ldots\) N. (All values are in \(\mathrm{mks}\) ) (A) \(6.25\) (B) \(4.0\) (C) \(12.5\) (D) \(0.5\)

The periodic time of a S.H.O. oscillating about a fixed point is $2 \mathrm{~s}$. After what time will the kinetic energy of the oscillator become \(25 \%\) of its total energy? (A) \(1 / 12 \mathrm{~s}\) (B) \(1 / 6 \mathrm{~s}\) (C) \(1 / 4 \mathrm{~s}\) (D) \(1 / 3 \mathrm{~s}\).
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