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A particle P of mass m attached to a vertical axis by two strings AP and BP of length 1m each . The separation $A B = ℓ .$ P rotates around the axis with an angular velocity $ω$ . The tension in the two strings are $T_{1} a n d T_{2}$

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$T_{1} = T_{2}$

$T_{1} + T_{2} = m ω^{2} ℓ$

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

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

answer is B.

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Detailed Solution

$Δ A B P$ is equilateral. Let $T_{1}$ and $T_{2}$ be the tension in PA and PB, respectively

$T_{1} \cos 60^{0} = T_{2} \cos 60^{0} + m g$

$o r T_{1} - T_{2} = 2 m g$

$T_{1} \cos 30^{0} + T_{2} \cos 30^{0} = m ω^{2} ℓ \cos 30^{0}$

$o r T_{1} + T_{2} = m ω^{2} ℓ$

$∴ T_{2} = \frac{1}{2} m [ω^{2} ℓ - 2 g]$

for $T_{2} \geq 0, ω \geq \sqrt{2 g / ℓ}$

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