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JEE Advanced 2019 Paper 2

Section: Physics

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Question : 15 of 54

Marks: +1, -0

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

\mu, 2\mu, 3\mu

4\mu

respectively. The instrument is played by vibrating the strings by varying the free length in between the range

L_0

2L_0

It is found that in string-1 (µ) at free length

L_0

T_0

the fundamental mode frequency is

f_0

List-I gives the above four strings while list-II lists the magnitude of some quantity.

List I	List II
I) String-1(µ)	P) 1
II) String 2(2µ)	Q) 1/2
III) String-3(3µ)	R) $\frac{1}{\sqrt{2}}$
IV) String-4(4µ)	S) $\frac{1}{\sqrt{3}}$
	T) 3/16
	U) 1/16

If the tension in each string is

T_0

the correct match for the highest fundamental frequency in

f_0

	I	II	III	IV
1)	P	Q	T	S
2)	Q	P	R	T
3)	P	R	S	Q
4)	Q	S	R	P

[JEE Adv 2019 P2]

1
2
3
4

Solution:

f_{o} = \sqrt{\;\frac{T_{o}}{\mu}} \times \;\frac{1}{2L_{o}}

f_{1} = f_{o}

f_{2} = \;\frac{f_{o}}{\sqrt{2}}

f_{3} = \;\frac{f_{o}}{\sqrt{3}}

f_{4} = \;\frac{f_{o}}{2}

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