JEE Advanced 2019 · Paper 2 · Physics Q16

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PhysicsList match+3 / -1Question 16 of 54

Passage

Answer Q.15 and Q.16 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 $\mu$ , $2\mu$ , $3\mu$ and $4\mu$ respectively. The instrument is played by vibrating the strings by varying the free length in between the range $L_0$ and $2L_0$ . It is found that in string-1 ( $\mu$ ) 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.

The length of the strings 1, 2, 3 and 4 are kept fixed at $L_0$ , $\dfrac{3L_0}2$ , $\dfrac{5L_0}4$ , and $\dfrac{7L_0}4$ , 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 $T_0$ will be,

List-I

I
String-1 ( $\mu$ )
II
String-2 ( $2\mu$ )
III
String-3 ( $3\mu$ )
IV
String-4 ( $4\mu$ )

List-II

P
1
Q
$1/2$
R
$1/\sqrt2$
S
$1/\sqrt3$
T
$3/16$
U
$1/16$

I → P, II → R, III → T, IV → U

I → P, II → Q, III → T, IV → U

I → P, II → Q, III → R, IV → T

I → T, II → Q, III → R, IV → U

Answer and worked solution

Official answer: B

Each string vibrates at $f_n=\dfrac n{2L}\sqrt{\dfrac T\mu}$ , and all must equal string 1's fundamental $f_0=\dfrac1{2L_0}\sqrt{\dfrac{T_0}\mu}$ .

Each string.

String 1 ( $n=1$ , $L_0$ , $\mu$ ): $T=T_0$ , (P).
String 2 ( $n=3$ , $\frac32L_0$ , $2\mu$ ): $\frac1{L_0}\sqrt{\frac T{2\mu}}=f_0$ , so $T=\frac{T_0}2$ , (Q).
String 3 ( $n=5$ , $\frac54L_0$ , $3\mu$ ): $\frac2{L_0}\sqrt{\frac T{3\mu}}=f_0$ , so $T=\frac{3T_0}{16}$ , (T).
String 4 ( $n=14$ , $\frac74L_0$ , $4\mu$ ): $\frac4{L_0}\sqrt{\frac T{4\mu}}=f_0$ , so $T=\frac{T_0}{16}$ , (U).

(B)

Topics:Standing waves in strings Speed of waves

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JEE Advanced 2019 · Paper 2, Physics question 16