If ‘O’ is at equilibrium then the values of the tension T1 and T2 are, (20 N is acting vertically downwards at O).

T11sin30°=T2cos30°;T1=3T2…(1)T1cos30°=20+T2…(2); solving (1) and (2)

If ‘O’ is at equilibrium then the values of the tension T1 and T2 are, (20 N is acting vertically downwards at O).

T11sin30°=T2cos30°;T1=3T2…(1)T1cos30°=20+T2…(2); solving (1) and (2)

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If ‘O’ is at equilibrium then the values of the tension T₁ and T₂ are, (20 N is acting vertically downwards at O).

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10N, 30N

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

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

20N, 30 N

answer is B.

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$T_{11} \sin 30 ° = T_{2} \cos 30 °; T_{1} = \sqrt{3} T_{2} \dots (1)$

$T_{1} \cos 30 ° = 20 + T_{2} \dots (2); solving (1) and (2)$

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If ‘O’ is at equilibrium then the values of the tension T1 and T2 are, (20 N is acting vertically downwards at O).

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If ‘O’ is at equilibrium then the values of the tension T₁ and T₂ are, (20 N is acting vertically downwards at O).