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A stone of mass 2 kg is whirled in a horizontal circle attached at the end of 1.5m long string. If the string makes an angle of 30° with vertical, compute its period. (g = 9.8 m/s^{2})

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#### Solution

Given: m = 2 kg

l = 1.5 m

Θ = 30°

g = 9.8 m/s^{2}

To find: Period (T)

= antilog[0.3592]

∴ T = 2.2875 s

Period of revolution is 2.287 s

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