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Standard XII

Physics

Torque about a Point

A light wire ...

Question

A light wire of length $L$ , area of cross section $A$ is hanging from fixed support. If a mass $M$ is suspended from its free end, then strain produced in the wire will be (symbols have their usual meaning)

\frac{M g}{2 Y A}

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\frac{M g}{Y A}

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\frac{2 M g}{Y A}

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Zero

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Solution

The correct option is B $\frac{M g}{Y A}$
Young’s modulus of elasticity is
$Y = \frac{Stress}{Strain}$
$Strain (\frac{Δ L}{L}) = \frac{M g}{A Y}$

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A light wire of length L, area of cross section A is hanging from fixed support. If a mass M is suspended from its free end, then strain produced in the wire will be (symbols have their usual meaning)

The correct option is B MgYAYoung’s modulus of elasticity is Y=StressStrain Strain(ΔLL)=MgAY

A light wire of length $L$ , area of cross section $A$ is hanging from fixed support. If a mass $M$ is suspended from its free end, then strain produced in the wire will be (symbols have their usual meaning)

The correct option is B $\frac{M g}{Y A}$
Young’s modulus of elasticity is
$Y = \frac{Stress}{Strain}$
$Strain (\frac{Δ L}{L}) = \frac{M g}{A Y}$