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Standard VII

Physics

Law of Conservation of Energy

A wire of len...

Question

A wire of length $L$ , area of cross-section $A$ is hanging from a fixed support. The length of the wire becomes $1.1 L$ when a mass $M$ is suspended from its free end. The Young's modulus of wire is given by

\frac{M g}{A}

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\frac{10 M g}{A}

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\frac{40 M g}{A}

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\frac{4 M g}{A}

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Solution

The correct option is B $\frac{10 M g}{A}$
$Y= \frac{Stress}{Strain} = \frac{\frac{M g}{A}}{\frac{Δ L}{L}}$
$Y = \frac{M g L}{A (0.1 L)} = \frac{10 M g}{A}$

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A wire of length L, area of cross-section A is hanging from a fixed support. The length of the wire becomes 1.1L when a mass M is suspended from its free end. The Young's modulus of wire is given by

The correct option is B 10MgAY=StressStrain=MgAΔLL Y=MgLA(0.1L)=10MgA

A wire of length $L$ , area of cross-section $A$ is hanging from a fixed support. The length of the wire becomes $1.1 L$ when a mass $M$ is suspended from its free end. The Young's modulus of wire is given by

The correct option is B $\frac{10 M g}{A}$
$Y= \frac{Stress}{Strain} = \frac{\frac{M g}{A}}{\frac{Δ L}{L}}$
$Y = \frac{M g L}{A (0.1 L)} = \frac{10 M g}{A}$