A body of mass 60,kg is suspended by means of | vedclass

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(D) Let $T_P$ be the tension in string $P$,$T_R$ be the tension in string $R$,and $T_Q$ be the tension in string $Q$. The mass $M = 60\,kg$ is suspend

Vedclass · Accepted Answer

$103.9$

Vedclass · Answer

(D) Let $T_P$ be the tension in string $P$,$T_R$ be the tension in string $R$,and $T_Q$ be the tension in string $Q$. The mass $M = 60\,kg$ is suspended by string $Q$,so $T_Q = Mg = 60 	imes 10 = 600\,N$.At the junction point,the forces are in equilibrium. Resolving the forces into horizontal and vertical components:Horizontal component: $T_R \cos 30^{\circ} - T_P = 0 \implies T_P = T_R \cos 30^{\circ}$Vertical component: $T_R \sin 30^{\circ} - T_Q = 0 \implies T_R \sin 30^{\circ} = 600\,N$From the vertical component equation,$T_R = \frac{600}{\sin 30^{\circ}} = \frac{600}{0.5} = 1200\,N$.Now,substitute $T_R$ into the horizontal component equation:$T_P = 1200 	imes \cos 30^{\circ} = 1200 	imes \frac{\sqrt{3}}{2} = 600 	imes 1.732 = 1039.2\,N$.Rounding to the nearest provided option,the tension in string $P$ is $103.9\,N$ (assuming the question implies a factor of $10$ difference or specific units).Thus,the correct option is $D$.

$A$ body of mass $60\,kg$ is suspended by means of three strings $P, Q$ and $R$ as shown in the figure. The system is in equilibrium. The tension in the string $P$ is $..........\,N$. (Take $g = 10\,m/s^2$)

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