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A mass m kg is whirled in a vertical plane by tying it at the end of a flexible wire of length l and area of cross section 'A'. When the mass is at its lowest position the strain produced in the wire is, if Young’s modulus of the wire is 'Y' (if velocity at lowest point is V= $\sqrt{5 g ℓ}$ )

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\frac{A Y}{6 m g}

\frac{6 m g}{A Y}

\frac{5 m g}{A Y}

\frac{A Y}{5 m g}

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detailed solution

Correct option is B

$A t l o w e s t p o i n t, T - m g = \frac{m v^{2}}{r}$

$T = m g + \frac{m (5 g r)}{r} = 6 m g$

$e = \frac{6 m g}{A Y}$

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A mass m kg is whirled in a vertical plane by tying it at the end of a flexible wire of length l and area of cross section 'A'. When the mass is at its lowest position the strain produced in the wire is, if Young’s modulus of the wire is 'Y' (if velocity at lowest point is V=5gℓ )

Unlock the full solution & master the concept.

Ready to Test Your Skills?

free study material

A mass m kg is whirled in a vertical plane by tying it at the end of a flexible wire of length l and area of cross section 'A'. When the mass is at its lowest position the strain produced in the wire is, if Young’s modulus of the wire is 'Y' (if velocity at lowest point is V= $\sqrt{5 g ℓ}$ )