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Standard XII

Physics

Units and Dimensions

A weightless ...

Question

A weightless thread can support tension upto $30 N$ . A particle of mass $0.5 k g$ is tied to it and is revolved in a circle of radius $2 m$ in a vertical plane. If $g = 10 m / s^{2}$ , then the maximum angular velocity of the stone will be

5 r a d / s

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\sqrt{30} r a d / s

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\sqrt{60} r a d / s

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10 r a d / s

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Solution

The correct option is A $5 r a d / s$
$T - m g = m ω^{2} r T_{m a x} = m g + m ω^{2} r ω^{2} = (\frac{T - m g}{m r}) = (\frac{30 - 5}{0.5 (2)}) = 5 r a d / s e c$

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A weightless thread can support tension upto 30 N. A stone of mass 0.5 kg is tied to it and is revolved in a circular path of radius 2 m in a vertical plane. If g= 10m/ $s^{2}$ , then the maximum angular velocity of the stone will be

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A weightless thread can support tension upto 30N. A particle of mass 0.5kg is tied to it and is revolved in a circle of radius 2m in a vertical plane. If g=10m/s2, then the maximum angular velocity of the stone will be

The correct option is A 5rad/sT−mg=mω2rTmax=mg+mω2rω2=(T−mgmr)=(30−50.5(2))=5rad/sec

A weightless thread can support tension upto $30 N$ . A particle of mass $0.5 k g$ is tied to it and is revolved in a circle of radius $2 m$ in a vertical plane. If $g = 10 m / s^{2}$ , then the maximum angular velocity of the stone will be

The correct option is A $5 r a d / s$
$T - m g = m ω^{2} r T_{m a x} = m g + m ω^{2} r ω^{2} = (\frac{T - m g}{m r}) = (\frac{30 - 5}{0.5 (2)}) = 5 r a d / s e c$