A stone of mass 1 kg is tied to the end of a massless string of length 1 m. If the breaking tension of the string is 400 N, then the maximum linear velocity the stone can have without breaking the string, while rotating in a horizontal plane, is

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Solution

Q.
A stone of mass 1 kg is tied to the end of a massless string of length 1 m. If the breaking tension of the string is 400 N, then the maximum linear velocity the stone can have without breaking the string, while rotating in a horizontal plane, is [2023]

1 20 ${ms}^{- 1}$

2 40 ${ms}^{- 1}$

3 400 ${ms}^{- 1}$

4 10 ${ms}^{- 1}$

Ans.
(1)

$T \cos θ = m g and T \sin θ = \frac{m v^{2}}{l^{2} \sin θ}$

$\cos θ = \frac{m g}{T}$ ...(i)

$\sin 2 θ = \frac{m v^{2}}{T l^{2}}$ ...(ii)

$From (i) and (ii): l = {(\frac{m g}{T})}^{2} + \frac{m v^{2}}{T l^{2}}$

$l = {(\frac{10}{400})}^{2} + \frac{v^{2}}{400}$

$\Rightarrow v = 20 m/s$

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