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Young's Modulus of Elasticity

The length of...

Question

The length of a metal wire is $l_{1}$ when the tension in it is $F_{1}$ and $l_{2}$ when the tension in it is $F_{2}$ . The natural length of the wire is

\frac{l_{1} F_{1} + l_{2} F_{2}}{F_{1} + F_{2}}

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

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

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

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Solution

The correct option is C $\frac{l_{1} F_{2} - l_{2} F_{1}}{F_{2} - F_{1}}$
We have
$l_{1} - l = \frac{F_{1} l}{A Y}$
and
$l_{2} - l = \frac{F_{2} l}{A Y}$
Thus we get
$\frac{l_{1} - l}{F_{1}} = \frac{l_{2} - l}{F_{2}}$
or
$F_{2} l_{1} - F_{2} l = F_{1} l_{2} - F_{1} l$
or
$l = \frac{F_{2} l_{1} - F_{1} l_{2}}{F_{2} - F_{1}}$

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