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

Question

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 $μ, 2 μ, 3 μ$ and $4 μ$ respectively. The instrument is played by vibrating the strings by varying the free length in between the range $L_{0}$ and $2 L_{0} .$ It is found that in string- $1 (μ)$ at free length $L_{0}$ and tension $T_{0}$ 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
$(a)$	String $- 1 (μ)$	$(p)$	$1$
$(b)$	String $- 2 (2 μ)$	$(q)$	$\frac{1}{2}$
$(c)$	String $- 3 (3 μ)$	$(r)$	$\frac{1}{\sqrt{2}}$
$(d)$	String $- 4 (4 μ)$	$(s)$	$\frac{1}{\sqrt{3}}$
		$(t)$	$\frac{3}{16}$
		$(u)$	$\frac{1}{16}$

If the tension in each string is $T_{0},$ the correct match for the highest fundamental frequency in $f_{0}$ units will be,

TestHub · Accepted Answer

Fundamental frequency is maximum when length is minimum i.e. $L_{0},$
Case $1 . L = L_{0}, T = T_{0}, f = f_{0}; f_{1} = \frac{1}{2 L_{0}} \sqrt{\frac{T_{0}}{μ}} = f_{0}$ (fundamental frequency of vibration in string fixed at both ends)
Case $2 . f_{2} = \frac{1}{L_{0}} \sqrt{\frac{T_{2}}{2 μ}} = \frac{f_{0}}{\sqrt{2}}$
Case $3 . f_{3} = \frac{1}{L_{0}} \sqrt{\frac{T_{2}}{3 μ}} = \frac{f_{0}}{\sqrt{3}}$
Case $4 . f_{4} = \frac{1}{L_{0}} \sqrt{\frac{T_{2}}{4 μ}} = \frac{f_{0}}{2}$

Physics - Wave on String Question with Solution | TestHub

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JEE Advanced (2028)