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Standard XII

Physics

Friction and Vertical Circular Motion Problems

A stone of ma...

Question

A stone of mass $1 k g$ tied to a light inextensible string of length $L = \frac{10}{3} m$ , whirling in a circular path in a vertical plane. The ratio of maximum tension in the string to the minimum tension in the string is $4$ . If g is taken to be $10 m / s^{2}$ the speed of the stone at the highest point of the circle is

10 m / s

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5 \sqrt{2} m / s

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10 \sqrt{3} m / s

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20 m / s

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Solution

The correct option is A $10 m / s$
$T h e r a t i o t o t h e m a m i m u m T_{2} {t o t h e m i n i m u m t e n s i o n T}_{1} i s - \frac{T_{2}}{T_{1}} = 4 T_{2} = 4 T_{1} N o w t h e d i f f e r e n c e b e t w e e n t h e t w o 6 m g; H e n c e w e h a v e T_{2} - T_{1} = 6 m g ∴ 4 T_{1} - T_{1} = 6 m g 3 T_{1} = 6 m g T_{1} = 2 m g N o w t e n s i o n a t t h e t o p o f t h e c i r c l e i s - T_{1 + m g} = \frac{{m v}_{1}^{2}}{r} = \frac{{m v}_{1}^{2}}{L} 2 m g + m g = \frac{{m v}_{1}^{2}}{\frac{10}{3}} v_{1}^{2} = 3 g \frac{10}{3} = 10 g v_{1} = \sqrt{10 g} = \sqrt{10 \times 10} = 10 m / s$

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A stone of mass 1 kg tied to a light inextensible string of length L=103 m, whirling in a circular path in a vertical plane. The ratio of maximum tension in the string to the minimum tension in the string is 4. If g is taken to be 10 m/s2 the speed of the stone at the highest point of the circle is

The correct option is A 10 m/sTheratiotothemamimumT2totheminimumtensionT1is−T2T1=4T2=4T1Nowthedifferencebetweenthetwo6mg;HencewehaveT2−T1=6mg∴4T1−T1=6mg3T1=6mgT1=2mgNowtensionatthetopofthecircleis−T1+mg=mv21r=mv21L2mg+mg=mv21103v21=3g103=10gv1=√10g=√10×10=10m/s