A stone of mass $1 k g$ tied to a light inextensible string of length $l = \frac{10}{3} m$ is rotating in a circular path of radius $l$ in a vertical plane. If the ratio of maximum tension and minimum tension in the string is $4$ , the speed of the stone at the highest point of the circle would be $(g = 10 m s^{- 2})$

20 m s^{- 1}

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10 \sqrt{3} m s^{- 1}

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5 \sqrt{2} m s^{- 1}

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10 m s^{- 1}

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Solution

The correct option is D $10 m s^{- 1}$
$\begin{matrix} G i v e n, s u p p o s e : max . t e n s i o n (T_{1}) a n d m i m . t e n s i o n (T_{2}) . \frac{T_{1}}{T_{2}} = 4 T_{1} = 4 T_{2} N o w a s d i f f b e t w e e n max . t e n s i o n a n d m i m . t e n s i o n f o r a b o d y i n v e r t i c a l c i r c l e i s : T_{1} - T_{2} = 6 m g 4 T_{2} - T_{2} = 6 m g 3 T_{2} = 6 m g T_{2} = 2 m g N o w, H i g e s t p o i n t min . t e n s i o n c a n b e : T_{2} = \frac{m v^{2}}{r} - m g 2 m g = \frac{m v^{2}}{r} - m g 3 m g = \frac{m v^{2}}{r} ∴ v = \sqrt{3 g r} H e r e, r = L = \frac{10}{3} v = \sqrt{3 \times 10 \times \frac{10}{3}} v = \sqrt{100} m / s ∴ 10 m s^{- 1} s o t h a t t h e c o r r e c t o p t i o n i s D . \end{matrix}$

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Q. A stone of mass

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Q. 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

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