Q52 / 54 · JEE Advanced 2019 (Paper 2) · Physics

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, 3 and 4 with mass per unit length μ, 2μ, 3μ and 4μ respectively. The instrument is played by vibrating the strings by varying the free length in between the range L0 and 2L0. It is found that in string- 1μ at free length L0 and tension T0 the fundamental mode frequency is f0.
List-I gives the above four strings while list-II lists the magnitude of some quantity.
 
  List I   List II
(a) String -1μ (p) 1
(b) String -22μ (q) 12
(c) String -33μ (r) 12
(d) String -44μ (s) 13
    (t) 316
    (u) 116

The length of the strings 1, 2, 3 and 4 are kept fixed at L0, 3L02, 5L04 and 7L04, respectively. Strings 1, 2, 3 and 4 are vibrated at their 1st, 3rd, 5th and 14th 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 will be.

  • Correct answer
    A
    a-p, b-q, c-t, d-u
  • B
    a-t, b-q, c-r, d-u
  • C
    a-p, b-q, c-r, d-t
  • D
    a-p, b-r, c-t, d-u

Solution

Case 1. L=L0, T=T0, f=f0
f1=12L0 T0μ=f0 (fundamental frequency of vibration in string fixed at both ends)
Case 2. L=3L02, T=T2, f2=f0
f2=32×3L02T22μ=f0⇒f0=12L0T2μ⇒T2=T02
Case 3. L=5L04, T=T3, f3=f0
f3=52×5L04T33μ=f0⇒f0=23L0T3μ⇒T3=3T016
Case 4. L=7L04⇒f4=142×7L04T44μ=f0⇒f0=2L0T4μ⇒T4=T016

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

More Waves and Sound Questions