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Young's Modulus of Elasticity

When the tens...

Question

When the tension in a metal wire is $T_{1}$ its length is $l_{1},$ when the tension is $T_{2}$ its length is $l_{2} .$ The natural length of wire is :

T_{1} l_{1} + T_{2} l_{2}

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

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\frac{l_{1} T_{2} - l_{2} T_{1}}{T_{2} + T_{1}}

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\frac{T_{2}}{T_{1}} (l_{1} + l_{2})

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Solution

The correct option is A $\frac{l_{1} T_{2} - l_{2} T_{1}}{T_{2} - T_{1}}$
$y = \frac{F l}{A △ l}$
y is constant
$∴ △ l \propto F$
$l_{t} - l \propto T_{1}$ and $l_{2} - l \propto T_{2}$
$∴ \frac{l_{1} - l}{l_{2} - l} = \frac{T_{1}}{T_{2}}$

$l_{1} T_{2} - l T_{2} = l_{2} T_{1} - l T_{1}$

$l (T_{1} - T_{2}) = l_{2} T_{1} - l_{1} T_{2}$

$l = \frac{l_{2} T_{1} - L_{1} T_{2}}{T_{1} - T_{2}} \Rightarrow= \frac{l_{1} T_{2} - l_{2} T_{1}}{T_{2} - t_{1}}$

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