JEE Advanced 2019 · Paper 2 · Physics Q15

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PhysicsList match+3 / -1Question 15 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.

If the tension in each string is $T_0$ , the correct match for the highest fundamental frequency in $f_0$ units 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 → Q, III → T, IV → S

I → P, II → R, III → S, IV → Q

I → Q, II → S, III → R, IV → P

I → Q, II → P, III → R, IV → T

Answer and worked solution

Official answer: B

The highest fundamental occurs at the shortest length $L_0$ :
$f=\frac1{2L_0}\sqrt{\frac{T_0}{\mu_i}}=\frac{f_0}{\sqrt{\mu_i/\mu}}.$

Each string.

String 1 ( $\mu$ ): 1, (P).
String 2 ( $2\mu$ ): $\frac1{\sqrt2}$ , (R).
String 3 ( $3\mu$ ): $\frac1{\sqrt3}$ , (S).
String 4 ( $4\mu$ ): $\frac12$ , (Q).

(B)

Topics:Standing waves in strings Speed of waves

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JEE Advanced 2019 · Paper 2, Physics question 15