Wave pulse can travel along a tense string like a violin spring. Aseries of experiments showed that the wave velocity V of a pulse depends on the following quantities, the tension T of the string, the cross-section area

Correct answer: A. V=k T A Let V=kT^ a A^ b ^ c , k= dimensional constant Writing dimension on both we side aligned [LT^ -1 ]&=[MLT^ -2 ]^ a [~L^ 2 ]^ b [ML^ -3 ]^ c &=[M^ a+c L^ a+2 ~b-3 c T^ -2 a ] aligned Comparing power on both sides we have array l a+c=0, a+2 b-3 c=1, -2 a=-1 a=(1)/(2),c=-(1)/(2) =-(1)/(2) =k (T)/(A ) array

BITSAT 2013

Exam: BITSAT
Year: 2013
Subject: Physics
Chapter: Units and Dimensions

BITSAT 2013 · Physics · Units and Dimensions

Wave pulse can travel along a tense string like a violin spring....

Wave pulse can travel along a tense string like a violin spring. Aseries of experiments showed that the wave velocity

\mathrm{V}

of a pulse depends on the following quantities, the tension T of the string, the cross-section area A of the string and then as per unit volume

\rho

of the string. Obtain an expression for

V

in terms of the

\mathrm{T}, \mathrm{A}

and

\rho

using dimensional analysis.

Options

A $\mathrm{V}=\mathrm{k} \sqrt{\frac{\mathrm{T}}{\mathrm{A} \rho}}$
B $\mathrm{V}=\mathrm{k} \sqrt{\frac{\mathrm{T}}{\mathrm{A}}}$
C $V=k \sqrt{\frac{A \rho}{T}}$
DNone of these

Correct answer

A. $\mathrm{V}=\mathrm{k} \sqrt{\frac{\mathrm{T}}{\mathrm{A} \rho}}$

Step-by-step solution

Let

\mathrm{V}=\mathrm{kT}^{\mathrm{a}} \mathrm{A}^{\mathrm{b}} \rho^{\mathrm{c}}

\mathrm{k}=

dimensional constant Writing dimension on both we side

\begin{aligned}\left[\mathrm{LT}^{-1}\right]&=\left[\mathrm{MLT}^{-2}\right]^{a}\left[\mathrm{~L}^{2}\right]^{\mathrm{b}}\left[\mathrm{ML}^{-3}\right]^{\mathrm{c}}\\ &=\left[\mathrm{M}^{\mathrm{a}+\mathrm{c}} \mathrm{L}^{\mathrm{a}+2 \mathrm{~b}-3 \mathrm{c}} \mathrm{T}^{-2 \mathrm{a}}\right] \end{aligned}

Comparing power on both sides we have

\begin{array}{l}a+c=0, a+2 b-3 c=1, \quad-2 a=-1 \\ \therefore \quad a=\frac{1}{2},c=-\frac{1}{2} \Rightarrowb=-\frac{1}{2} \thereforeV=k \sqrt{\frac{T}{A \rho}} \end{array}

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