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

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

10 N, 30 N

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

20 N 30 N

answer is B.

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Detailed Solution

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

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