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-1 (μ) 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.
The length of the string 1,2,3 and 4 are kept fixed at L0,23L0,45L0 and 47L0, 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.
(A)I→P, II→Q, III→T, IV→U✓ Correct
(B)I→T, II→Q, III→R, IV→U
(C)I→P, II→Q, III→R, IV→T
(D)I→P, II→R, III→T, IV→U
Answer: (A)
Solution
For string 1 (length L0, 1st harmonic):
f0=2L01μT1
With T1=T0 → (P)
For string 2 (length 23L0, 3rd harmonic):
f0=2(3L0/2)32μT2=L012μT2
Comparing with string 1: T2=2T0 → (Q)
For string 3 (length 45L0, 5th harmonic):
f0=2(5L0/4)53μT3=L023μT3
Solving: T3=163T0 → (T)
For string 4 (length 47L0, 14th harmonic):
f0=2(7L0/4)144μT4=L044μT4
Solving: T4=16T0 → (U)
Therefore: I→P, II→Q, III→T, IV→U
JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Single correct option