A body of mass 60,kg is suspended by means of three strings P, Q and R as shown in figure is in equi

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4-1.Newton's Laws of Motion

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A body of mass $60\,kg$ is suspended by means of three strings $P, Q$ and $R$ as shown in the figure is in equilibrium. The tension in the string $P$ is $..........g\,N$

A

$130.9$

B

$60$

C

$50$

D

$103.9$

(AIIMS-2016)

Solution

(d)

Horizontal component $= T _1 \cos 30^{\circ}- T =0$

Vertical component $=T_1 \sin 30^{\circ}- mg =0$

$\therefore \frac{ T _1 \cos 30^{\circ}}{ T _1 \sin 30^{\circ}}=\frac{ T }{ mg }$

$T = mg \cot 30^{\circ}$

$=60 \times 10 \times 1.732$

$=103.9\,gN$

Standard 11

Physics

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