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A one metre steel wire of negligible mass and area of cross-section 0.01 cm² is kept on a smooth horizontal table with one end fixed. A ball of mass 1 kg is attached to the other end. The ball and the wire are rotating with an angular velocity of $ω$ . If the elongation of the wire is 2mm, then $ω$ (in rad s^-1) is $_____$ (Young’s modulus of steel = 2 × 10¹¹Nm^–2)

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answer is 20.

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Detailed Solution

Given,

mass of the ball m= 1kg

length of the wire l=1m

elongation of the wire, $Δ l = 2 mm = 2 \times 10^{- 3} m$

Area $A = 0.01 {cm}^{2} = 0.01 \times 10^{- 4} m$

Young's modulus = $Y = 2 \times 10^{11} {Nm}^{- 2}$

We know that the tension of the wire is,

$Tension T = {mω}^{2} l$

$stress = \frac{tension}{area} = \frac{{mω}^{2} l}{A}$

$Strain = \frac{∆ l}{l} = \frac{stress}{Young' s modulus} \Rightarrow ∆ l = \frac{{mω}^{2} l^{2}}{YA}$

$\Rightarrow ω = \sqrt{\frac{YA ∆ l}{{ml}^{2}}} = \sqrt{\frac{2 \times 10^{11} \times 0.01 \times 10^{- 4} \times 2 \times 10^{- 3}}{1 \times 1^{2}}} = 20 rad / \sec^{- 1}$

Hence the correct answer is 20.

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