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Standard XII

Physics

Young's Modulus of Elasticity

The dimension...

Question

The dimensions of two wires $A$ and $B$ are the same. But their materials are different, Their load- extension graphs are shown. If $Y_{A}$ and $Y_{B}$ are the values of Young's modulus of elasticity of $A$ and $B$ respectively then :

Y_{A} > Y_{B}

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Y_{A} < Y_{B}

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Y_{A} = Y_{B}

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Y_{B} = 2 Y_{A}

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Solution

The correct option is A $Y_{A} < Y_{B}$
$Y = \frac{F L}{Δ l A}$

$Δ l = \frac{F L}{Y A}$

For same load elongation in A is greater than in B therefore

$Δ l_{A} > Δ l_{B}$

$\frac{F L}{Y_{A} A} > \frac{F L}{Y_{B} A}$

$Y_{B} > Y_{A}$

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The dimensions of two wires A and B are the same. But their materials are different, Their load- extension graphs are shown. If YA and YB are the values of Young's modulus of elasticity of A and B respectively then :

The correct option is A YA<YBY=FLΔlAΔl=FLYAFor same load elongation in A is greater than in B therefore ΔlA>ΔlBFLYAA>FLYBAYB>YA

The dimensions of two wires $A$ and $B$ are the same. But their materials are different, Their load- extension graphs are shown. If $Y_{A}$ and $Y_{B}$ are the values of Young's modulus of elasticity of $A$ and $B$ respectively then :

The correct option is A $Y_{A} < Y_{B}$
$Y = \frac{F L}{Δ l A}$

$Δ l = \frac{F L}{Y A}$

For same load elongation in A is greater than in B therefore

$Δ l_{A} > Δ l_{B}$

$\frac{F L}{Y_{A} A} > \frac{F L}{Y_{B} A}$

$Y_{B} > Y_{A}$