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 $\mu, 2 \mu, 3 \mu$ and $4 \mu$ 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$-1 \ (\mu)$ 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.
($1$) If the tension in each string is $T _0$, the correct match for the highest fundamental frequency in $f _0$ units will be,
$(1)$ $I \rightarrow P , II \rightarrow R , III \rightarrow S , IV \rightarrow Q$
$(2)$ $I \rightarrow P , II \rightarrow Q , III \rightarrow T , IV \rightarrow S$
$(3)$ $I \rightarrow Q , II \rightarrow S , III \rightarrow R , IV \rightarrow P$
$(4)$ I $\rightarrow Q , II \rightarrow P , III \rightarrow R$, IV $\rightarrow T$
($2$) The length of the string $1,2,3$ and 4 are kept fixed at $L _0, \frac{3 L _0}{2}, \frac{5 L _0}{4}$ and $\frac{7 L _0}{4}$, respectively. Strings $1,2,3$ and 4 are vibrated at their $1^{\text {tt }}, 3^{\text {rd }}, 5^{\text {m }}$ and $14^{\star}$ 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.
$(1)$ $I \rightarrow P , II \rightarrow Q , III \rightarrow T , IV \rightarrow U$
$(2)$ $I \rightarrow T , II \rightarrow Q , III \rightarrow R$, IV $\rightarrow U$
$(3)$ $I \rightarrow P , II \rightarrow Q , III \rightarrow R , IV \rightarrow T$
$(4)$ I $\rightarrow P , II \rightarrow R , III \rightarrow T , IV \rightarrow U$
List $-I$ gives the above four strings while list $-II$ lists the magnitude of some quantity.
| List $-I$ | List $-II$ |
| $(I)$ String $-1( \mu$ ) | $(P) 1$ |
| $(II)$ String $-2 (2 \mu)$ | $(Q)$ $1 / 2$ |
| $(III)$ String $-3 (3 \mu)$ | $(R)$ $1 / \sqrt{2}$ |
| $(IV)$ String $-4 (4 \mu)$ | $(S)$ $1 / \sqrt{3}$ |
| $(T)$ $3 / 16$ | |
| $(U)$ $1 / 16$ |
$(1)$ $I \rightarrow P , II \rightarrow R , III \rightarrow S , IV \rightarrow Q$
$(2)$ $I \rightarrow P , II \rightarrow Q , III \rightarrow T , IV \rightarrow S$
$(3)$ $I \rightarrow Q , II \rightarrow S , III \rightarrow R , IV \rightarrow P$
$(4)$ I $\rightarrow Q , II \rightarrow P , III \rightarrow R$, IV $\rightarrow T$
($2$) The length of the string $1,2,3$ and 4 are kept fixed at $L _0, \frac{3 L _0}{2}, \frac{5 L _0}{4}$ and $\frac{7 L _0}{4}$, respectively. Strings $1,2,3$ and 4 are vibrated at their $1^{\text {tt }}, 3^{\text {rd }}, 5^{\text {m }}$ and $14^{\star}$ 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.
$(1)$ $I \rightarrow P , II \rightarrow Q , III \rightarrow T , IV \rightarrow U$
$(2)$ $I \rightarrow T , II \rightarrow Q , III \rightarrow R$, IV $\rightarrow U$
$(3)$ $I \rightarrow P , II \rightarrow Q , III \rightarrow R , IV \rightarrow T$
$(4)$ I $\rightarrow P , II \rightarrow R , III \rightarrow T , IV \rightarrow U$
IIT 2019, Advanced
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For fundamental mode
$($image$)$
$f =\frac{ V }{\lambda}=\frac{1}{2 L } \sqrt{\frac{ T }{\mu}}$
For string $(1)$
$f_0=\frac{1}{2 L} \sqrt{\frac{T}{\mu}} \Rightarrow(P)$
For string $(2)$
$f =\frac{1}{2 L } \sqrt{\frac{ T }{2 \mu}}=\frac{ f _0}{\sqrt{2}} \Rightarrow( R )$
For string $(3)$
$f =\frac{1}{2 L } \sqrt{\frac{ T }{3 \mu}}=\frac{ f _0}{\sqrt{3}} \Rightarrow( S )$
For string $(4)$
$f=\frac{1}{2 L} \sqrt{\frac{T}{4 \mu}}=\frac{f_0}{2} \Rightarrow(Q)$
($2$) For string $(1)$
Length of string $= L _0$
It is vibrating in $I ^{ ts }$ harmonic i.e. fundamental mode.
$($image$)$
$f _0=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}} \Rightarrow( P )$
For string $(2)$
Length of string $=\frac{3 L _0}{2}$
It is vibrating in $III ^{\text {rd }}$ harmonic but frequency is still $f _0$.
$f_0=\frac{3 v}{2 L}$
$($image$)$
$f _0=\frac{3}{2\left(\frac{3 L _0}{2}\right)} \sqrt{\frac{ T _2}{2 \mu}}$
$\Rightarrow f _0=\frac{1}{ L _0} \sqrt{\frac{ T _2}{2 \mu}}=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}}$
$\Rightarrow T _2=\frac{ T _0}{2}$
$\Rightarrow \text { (Q) }$
For string $(3)$
Length of string $=\frac{5 L _0}{4}$
It is vibrating in $5^{\text {th }}$ harmonic but frequency is still $f _0$.
$f _0=\frac{5 V }{2 L }$
$($image$)$
$\Rightarrow f _0=\frac{5}{2\left(\frac{5 L _0}{4}\right)} \sqrt{\frac{ T _3}{3 \mu}}=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}}$
$\Rightarrow \frac{2}{ L _0} \sqrt{\frac{ T _3}{3 \mu}}=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}}$
$T _3=\frac{3 T _0}{16} \Rightarrow( T )$
For string $(4)$
Length of string $=\frac{7 L _0}{4}$
It is vibrating in $14^{\text {th }}$ harmonic but frequency is still $f _0$.
$($image$)$
$f _0=\frac{14 v }{2 L }$
$\Rightarrow f _0=\frac{14}{2\left(\frac{7 L _0}{4}\right)} \sqrt{\frac{ T _4}{4 \mu}}=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}}$
$\Rightarrow \frac{4}{ L _0} \sqrt{\frac{ T _4}{4 \mu}}=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}}$
$ \Rightarrow T _4=\frac{ T _0}{16} $
$\Rightarrow( U )$
$($image$)$
$f =\frac{ V }{\lambda}=\frac{1}{2 L } \sqrt{\frac{ T }{\mu}}$
For string $(1)$
$f_0=\frac{1}{2 L} \sqrt{\frac{T}{\mu}} \Rightarrow(P)$
For string $(2)$
$f =\frac{1}{2 L } \sqrt{\frac{ T }{2 \mu}}=\frac{ f _0}{\sqrt{2}} \Rightarrow( R )$
For string $(3)$
$f =\frac{1}{2 L } \sqrt{\frac{ T }{3 \mu}}=\frac{ f _0}{\sqrt{3}} \Rightarrow( S )$
For string $(4)$
$f=\frac{1}{2 L} \sqrt{\frac{T}{4 \mu}}=\frac{f_0}{2} \Rightarrow(Q)$
($2$) For string $(1)$
Length of string $= L _0$
It is vibrating in $I ^{ ts }$ harmonic i.e. fundamental mode.
$($image$)$
$f _0=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}} \Rightarrow( P )$
For string $(2)$
Length of string $=\frac{3 L _0}{2}$
It is vibrating in $III ^{\text {rd }}$ harmonic but frequency is still $f _0$.
$f_0=\frac{3 v}{2 L}$
$($image$)$
$f _0=\frac{3}{2\left(\frac{3 L _0}{2}\right)} \sqrt{\frac{ T _2}{2 \mu}}$
$\Rightarrow f _0=\frac{1}{ L _0} \sqrt{\frac{ T _2}{2 \mu}}=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}}$
$\Rightarrow T _2=\frac{ T _0}{2}$
$\Rightarrow \text { (Q) }$
For string $(3)$
Length of string $=\frac{5 L _0}{4}$
It is vibrating in $5^{\text {th }}$ harmonic but frequency is still $f _0$.
$f _0=\frac{5 V }{2 L }$
$($image$)$
$\Rightarrow f _0=\frac{5}{2\left(\frac{5 L _0}{4}\right)} \sqrt{\frac{ T _3}{3 \mu}}=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}}$
$\Rightarrow \frac{2}{ L _0} \sqrt{\frac{ T _3}{3 \mu}}=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}}$
$T _3=\frac{3 T _0}{16} \Rightarrow( T )$
For string $(4)$
Length of string $=\frac{7 L _0}{4}$
It is vibrating in $14^{\text {th }}$ harmonic but frequency is still $f _0$.
$($image$)$
$f _0=\frac{14 v }{2 L }$
$\Rightarrow f _0=\frac{14}{2\left(\frac{7 L _0}{4}\right)} \sqrt{\frac{ T _4}{4 \mu}}=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}}$
$\Rightarrow \frac{4}{ L _0} \sqrt{\frac{ T _4}{4 \mu}}=\frac{1}{2 L _0} \sqrt{\frac{ T _0}{\mu}}$
$ \Rightarrow T _4=\frac{ T _0}{16} $
$\Rightarrow( U )$
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