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NEET
Exam
NEET
Subject
Physics
Chapter
Waves and Sound

NEET · Physics · Waves and Sound

Answer the following by appropriately matching the lists based on...

Answer the following by appropriately matching the lists based on the information given in the paragraph.
A musical instrument is made using four different metal strings, 1,2,31, 2, 3 and 44 with mass per unit length μ,2μ,3μ\mu, 2\mu, 3\mu and 4μ4\mu respectively. The instrument is played by vibrating the strings by varying the free length in between the range L0{L}_{0} and 2L0.2{L}_{0} . It is found that in string- 11\left(\mu\right)atfreelengthat free length{L}_{0}andtensionand tension{T}_{0}thefundamentalmodefrequencyisthe fundamental mode frequency is{f}_{0} .<br/>List−Igivestheabovefourstringswhilelist−IIliststhemagnitudeofsomequantity.<br/> <tableborder="1pxsolid"><thead><tr><th> </th><th><strongstyle="text−align:center;">ListI</strong></th><th> </th><th><strongstyle="text−align:center;">ListII</strong></th></tr></thead><tbody><tr><td><br/>List-I gives the above four strings while list-II lists the magnitude of some quantity.<br/> <table border="1px solid"> <thead> <tr> <th> </th> <th><strong style="text-align: center;">List I</strong></th> <th> </th> <th><strong style="text-align: center;">List II</strong></th> </tr> </thead> <tbody> <tr> <td>( a )</td><td>String</td> <td>String- 1\left(\mu\right)</td><td></td> <td>( p )</td><td></td> <td>1</td></tr><tr><td></td> </tr> <tr> <td>( b )</td><td>String</td> <td>String- 2\left(2\mu\right)</td><td></td> <td>( q )</td><td></td> <td>\dfrac{1}{2}</td></tr><tr><td></td> </tr> <tr> <td>( c )</td><td>String</td> <td>String- 3\left(3\mu\right)</td><td></td> <td>( r )</td><td></td> <td>\dfrac{1}{\sqrt{2}}</td></tr><tr><td></td> </tr> <tr> <td>( d )</td><td>String</td> <td>String- 4\left(4\mu\right)</td><td></td> <td>( s )</td><td></td> <td>\dfrac{1}{\sqrt{3}}</td></tr><tr><td> </td><td> </td><td></td> </tr> <tr> <td> </td> <td> </td> <td>( t )</td><td></td> <td>\dfrac{3}{16}</td></tr><tr><td> </td><td> </td><td></td> </tr> <tr> <td> </td> <td> </td> <td>( u )</td><td></td> <td>\dfrac{1}{16}</td></tr></tbody></table><br/>Thelengthofthestrings</td> </tr> </tbody></table><br/>The length of the strings1, 2, 3andand4arekeptfixedatare kept fixed at{L}_{0}, \dfrac{3{L}_{0}}{2}, \dfrac{5{L}_{0}}{4}andand\dfrac{7{L}_{0}}{4},respectively.Stringsrespectively. Strings1, 2, 3andand4arevibratedattheirare vibrated at their{1}^{st}, {3}^{rd}, {5}^{th} and 14th{14}^{th} harmonics, respectively such that all the strings have same frequency. The correct match for the tension in the four strings in the units of T0{T}_{0} will be.

Options

  1. Aa−p,b−q,c−t,d−ua - p, b - q, c - t, d - u
  2. Ba−t,b−q,c−r,d−ua - t, b - q, c - r, d - u
  3. Ca−p,b−q,c−r,d−ta - p, b - q, c - r, d - t
  4. Da−p,b−r,c−t,d−ua - p, b - r, c - t, d - u

Correct answer

A. a−p,b−q,c−t,d−ua - p, b - q, c - t, d - u

Step-by-step solution

Case 1.L=L0,T=T0,f=f01 . L = {L}_{0}, T = {T}_{0}, f = {f}_{0}
f1=12L0T0μ=f0{f}_{1} = \dfrac{1}{2{L}_{0}}\sqrt{\dfrac{{T}_{0}}{\mu}} = {f}_{0} (fundamental frequency of vibration in string fixed at both ends)
Case 2.L=3L02,T=T2,f2=f02 . L = \dfrac{3{L}_{0}}{2}, T = {T}_{2}, {f}_{2} = {f}_{0}
f2=32×3L02T22μ=f0⇒f0=12L0T2μ⇒T2=T02{f}_{2} = \dfrac{3}{2 \times \dfrac{3{L}_{0}}{2}}\sqrt{\dfrac{{T}_{2}}{2\mu}} = {f}_{0} ⇒ {f}_{0} = \dfrac{1}{\sqrt{2}{L}_{0}}\sqrt{\dfrac{{T}_{2}}{\mu}} ⇒ {T}_{2} = \dfrac{{T}_{0}}{2}
Case 3.L=5L04,T=T3,f3=f03 . L = \dfrac{5{L}_{0}}{4}, T = {T}_{3}, {f}_{3} = {f}_{0}
f3=52×5L04T33μ=f0⇒f0=23L0T3μ⇒T3=3T016{f}_{3} = \dfrac{5}{2 \times \dfrac{5{L}_{0}}{4}}\sqrt{\dfrac{{T}_{3}}{3\mu}} = {f}_{0} ⇒ {f}_{0} = \dfrac{2}{\sqrt{3}{L}_{0}}\sqrt{\dfrac{{T}_{3}}{\mu}} ⇒ {T}_{3} = \dfrac{3{T}_{0}}{16}
Case 4.L=7L04⇒f4=142×7L04T44μ=f0⇒f0=2L0T4μ⇒T4=T0164 . L = \dfrac{7{L}_{0}}{4} ⇒ {f}_{4} = \dfrac{14}{2 \times \dfrac{7{L}_{0}}{4}}\sqrt{\dfrac{{T}_{4}}{4\mu}} = {f}_{0} ⇒ {f}_{0} = \dfrac{2}{{L}_{0}}\sqrt{\dfrac{{T}_{4}}{\mu}} ⇒ {T}_{4} = \dfrac{{T}_{0}}{16}
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