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A particle P of mass m attached to a vertical axis by two strings AP and BP of length L each. The separation AB=L, P rotates around the axis with an angular velocity $ω$ The tensions in the strings $AP & BP are T_{1} and T_{2}$ respectively, then :

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$T_{1} = T_{2}$

$T_{1} + T_{2} = m ω^{2} L$

$T_{1} - T_{2} = 2 mg$

BP will remain taut only if $ω \geq \sqrt{2 g / L}$

answer is B, C, D.

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Detailed Solution

$(b) T_{1} \cos 30 + T_{2} \cos 30 = m ω^{2} r$

$(T_{1} + T_{2}) \frac{\sqrt{3}}{2} = m ω^{2} . Lcos 30^{0}$

$T_{1} + T_{2} = m ω^{2} L$

$(c) T_{1} \sin 30 - T_{2} \sin 30 = m g; T_{1} - T_{2} = 2 mg$

$(d) T_{1} = T_{2} + 2 mg; T_{2} = m ω^{2} L - T_{1}$

$T_{2} = m ω^{2} L - T_{2} - 2 mg$

$2 T_{2} = m ω^{2} L - 2 mg; T_{2} = \frac{m ω^{2} L}{2} - mg$

For string BP to be Taut, $T_{2} \geq 0; \frac{m ω^{2} L}{2} - mg \geq 0$

$m \frac{ω^{2} L}{2} \geq m g; ω \geq \sqrt{\frac{2 g}{L}}$

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