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Q.

The body of mass 5kg is attached to a wire 0.2m long its breaking strength is 5×10⁷Nm^-2 It can be rotated in a horizontal circle with maximum angular velocity 5rads-1, then the area of cross-section of the wire is 5 × 10^–xm² then x = $_____$

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answer is 7.

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Detailed Solution

Given,

$Mass m = 5 kg, Length l = 0.2 m$

$Angular velocity ω = 5 rad / s, Breaking stress p = 5 \times 10^{7} {Nm}^{2}$

We know that,

breaking force = centripetal force

${mω}^{2} r = p \times A \Rightarrow A = \frac{{mω}^{2} r}{p} = \frac{5 \times 5 \times 5 \times 0.2 }{5 \times 10^{7}} = 5 \times 10^{- 7} m^{2}$

Hence the correct answer is 7.

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The body of mass 5kg is attached to a wire 0.2m long its breaking strength is 5×107Nm-2 It can be rotated in a horizontal circle with maximum angular velocity 5rads-1, then the area of cross-section of the wire is 5 × 10–xm2 then x = _____