If ‘O’ is at equilibrium then the values of the tension $$T1$$ and $$T2$$ are x, y. If $$20$$ N is vertically down. Then x, y are

$$T1Sin300$$ $$=$$ $$T2Cos300$$ $$T2T1=$$$$tan$$ $$300=13$$ $$20+T2Sin300=T1Cos300$$ $$20$$ $$=$$ $$T132-T22=T2$$ $$=$$ y $$T1=203N$$ $$=$$ x

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If ‘O’ is at equilibrium then the values of the tension T₁ and T₂ are x, y. If 20 N is vertically down. Then x, y are

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20 N 30 N

20 \sqrt{3} N

, 20 N

20 \sqrt{3} N, 20 \sqrt{3} N

10 N, 30 N

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detailed solution

Correct option is B

$T_{1} Sin 30^{0} = T_{2} Cos 30^{0} \frac{T_{2}}{T_{1}} = \tan 30^{0} = \frac{1}{\sqrt{3}} 20 + T_{2} Sin 30^{0} = T_{1} Cos 30^{0} 20 = T_{1} \frac{\sqrt{3}}{2} - \frac{T_{2}}{2} = T_{2} = y T_{1} = 20 \sqrt{3} N = x$

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If ‘O’ is at equilibrium then the values of the tension T1 and T2 are x, y. If 20 N is vertically down. Then x, y are

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If ‘O’ is at equilibrium then the values of the tension T₁ and T₂ are x, y. If 20 N is vertically down. Then x, y are