$A$ wire under tension vibrates with a fundamental frequency of $600 \,Hz$. If the length of the wire is doubled, the radius is halved and the wire is made to vibrate under one-ninth the tension. Then the fundamental frequency will become (in $\,Hz$)

$A$ string in a musical instrument is $50 \,cm$ long and its fundamental frequency is $800 \,Hz$. Keeping the tension applied to the string same, the change in the length to produce a sound note of fundamental frequency $1000 \,Hz$ will be: (in $\,cm$)

If the length of a stretched string is reduced by $40 \%$ and the tension is increased by $44 \%$,then the ratio of the final to the initial frequencies of the stretched string is:

$A$ steel wire of length $1 \ m$ and mass $0.1 \ kg$ and having a uniform cross-sectional area of $10^{-6} \ m^2$ is rigidly fixed at both ends. The temperature of the wire is lowered by $20^{\circ} C$. If the wire is vibrating in its fundamental mode,find the frequency (in $Hz$).
$(Y_{\text{steel}} = 2 \times 10^{11} \ N/m^2, \alpha_{\text{steel}} = 1.21 \times 10^{-5} /^{\circ} C)$

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 $2L_0$. 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.

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

$(1)$ If the tension in each string is $T_0$,the correct match for the 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 lengths of the strings $1, 2, 3$ and $4$ are kept fixed at $L_0, 3L_0/2, 5L_0/4$ and $7L_0/4$,respectively. Strings $1, 2, 3$ and $4$ are vibrated at their $1^{st}, 3^{rd}, 5^{th}$ and $14^{th}$ harmonics,respectively,such that all the strings have the same frequency. The correct match for the tension in the four strings in the units of $T_0$ will be.
$(1)$ $I \rightarrow P, II \rightarrow Q, III \rightarrow T, IV \rightarrow U$
$(2)$ $I \rightarrow T, II \rightarrow Q, III \rightarrow R, IV \rightarrow U$
$(3)$ $I \rightarrow P, II \rightarrow Q, III \rightarrow R, IV \rightarrow T$
$(4)$ $I \rightarrow P, II \rightarrow R, III \rightarrow T, IV \rightarrow U$

Two identical strings of length $\ell$ and $2\ell$ vibrate with fundamental frequencies $N$ Hz and $1.5N$ Hz,respectively. The ratio of tensions for the smaller length to the larger length is