$\mathrm{F}=-\left(\frac{G \frac{4}{3} \pi r^{3}(3 \rho) m}{r^{2}}\right)$
$\mathrm{ma}=-4 \pi G \rho m r$
$\mathrm{a}=-4 \pi G \rho r$
so $\omega=\sqrt{4 \pi G \rho}=\frac{2 \pi}{T}$
or $\mathrm{T}=2 \pi \cdot \sqrt{\frac{1}{4 \pi G \rho}}=\sqrt{\frac{\pi}{G \rho}}$
now time for $A$ to $B$ $\frac{1}{2} \sqrt{\frac{\pi}{G \rho}}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
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.
| $List-I$ | $List-II$ |
| $(I)$ 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 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 $\rightarrow P , II \rightarrow R , III \rightarrow T , IV \rightarrow U$
