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(C) Let $T_1$ and $T_2$ be the tensions in string $1$ and string $2$,respectively. For the mass to be in equilibrium:$T_1 \sin(37^\circ) + T_2 \sin(53

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$\frac{4\sqrt{4}}{3\sqrt{3}}$

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(C) Let $T_1$ and $T_2$ be the tensions in string $1$ and string $2$,respectively. For the mass to be in equilibrium:$T_1 \sin(37^\circ) + T_2 \sin(53^\circ) = mg = 20 	imes 10 = 200\ N$$T_1 \cos(37^\circ) = T_2 \cos(53^\circ) \Rightarrow T_1 	imes (4/5) = T_2 	imes (3/5) \Rightarrow T_1 = (3/4) T_2$Substituting $T_1$ in the vertical equilibrium equation:$(3/4) T_2 	imes (3/5) + T_2 	imes (4/5) = 200 \Rightarrow (9/20 + 16/20) T_2 = 200 \Rightarrow (25/20) T_2 = 200 \Rightarrow T_2 = 160\ N$$T_1 = (3/4) 	imes 160 = 120\ N$The lengths of the strings are $L_1 = 10 \sin(53^\circ) = 10 	imes (4/5) = 8\ m$ and $L_2 = 10 \sin(37^\circ) = 10 	imes (3/5) = 6\ m$.The wave speed is $v = \sqrt{T/\mu}$. Thus,$v_1 = \sqrt{T_1/\mu}$ and $v_2 = \sqrt{T_2/\mu}$.The time taken is $t = L/v$. Therefore,$t_1/t_2 = (L_1/v_1) / (L_2/v_2) = (L_1/L_2) 	imes \sqrt{T_2/T_1}$.$t_1/t_2 = (8/6) 	imes \sqrt{160/120} = (4/3) 	imes \sqrt{4/3} = \frac{4\sqrt{4}}{3\sqrt{3}}$.

$A$ mass of $20\ kg$ is hanging with the support of two strings of the same linear mass density. Now,pulses are generated in both strings at the same time near the joint at the mass. The ratio of the time taken by a pulse to travel through string $1$ to that taken by a pulse on string $2$ is:

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