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A string of length 1m is fixed at one end and carries a mass of 100g at the other end. The string make $2/ π$ revolution per second around the vertical axis through the fixed end. If angle of inclination of the string with the vertical is ${cos}^{-1} 5/8$ the linear velocity of the mass is nearly.

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Detailed Solution

Using $u = r ω = r \times 2 π f = 1 \sin θ \times 2 π f$

We get $(∴ r = 1 \sin θ)$

$u = 1 \times 0.78 \times 2 π \times \frac{2}{π} = 3.2 m s^{- 1} ≃ 3 m / s$

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