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Standard XI

Physics

Young's Modulus of Elasticity

The length of...

Question

The length of a steel wire when stretched by a force $F_{1}$ is $l_{1}$ and when stretched by a force $F_{2}$ is $l_{2}$ .The unstretched length of the wire is

$\frac{l_{1} F_{2} - l_{2} F_{1}}{F_{2} - F_{1}}$

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$\frac{l_{2} F_{2} - l_{1} F_{1}}{F_{2} - F_{1}}$

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$\frac{F_{1} (l_{2} - l_{1})}{F_{2} - F_{1}}$

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$\frac{F_{2} (l_{2} - l_{1})}{F_{2} - F_{1}}$

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Solution

The correct option is A
$\frac{l_{1} F_{2} - l_{2} F_{1}}{F_{2} - F_{1}}$

$Δ l = \frac{F l}{Y A} \Rightarrow Δ l \propto F$

Therefore,

$l_{1} - l_{o} = F_{1}$ and $l_{2} - l_{o} = F_{2}$

$\Rightarrow l_{o} = \frac{l_{1} F_{2} - l_{2} F_{1}}{F_{2} - F_{1}}$

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The length of a steel wire when stretched by a force F1 is l1 and when stretched by a force F2 is l2.The unstretched length of the wire is

The correct option is A l1F2−l2F1F2−F1 Δl=FlYA⇒Δl∝F Therefore, l1−lo=F1 and l2−lo=F2 ⇒lo=l1F2−l2F1F2−F1

The length of a steel wire when stretched by a force $F_{1}$ is $l_{1}$ and when stretched by a force $F_{2}$ is $l_{2}$ .The unstretched length of the wire is

The correct option is A
$\frac{l_{1} F_{2} - l_{2} F_{1}}{F_{2} - F_{1}}$

$Δ l = \frac{F l}{Y A} \Rightarrow Δ l \propto F$

Therefore,

$l_{1} - l_{o} = F_{1}$ and $l_{2} - l_{o} = F_{2}$

$\Rightarrow l_{o} = \frac{l_{1} F_{2} - l_{2} F_{1}}{F_{2} - F_{1}}$