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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 $l_{t}$ . When the tension is $T_{2}$ , its length is $l_{2}$ . The natural length of wire is

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

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T_{1} l_{1} + i_{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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Solution

The correct option is C $\frac{l_{1} T_{2} - l_{2} T_{1}}{T_{2} - T_{1}}$
When tension $T_{1}$ then length $= l_{1}$
When tension $T_{2}$ then length $= l_{2}$
$△ l = \frac{F l}{A Υ} = \frac{△ l_{1}}{△ l_{2}} = \frac{F_{1}}{F_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{l_{1} - l}{l_{2} - l} = \frac{T_{1}}{T_{2}} \Rightarrow l_{1} T_{2} - l . T_{2} = l_{2} T_{1} - l T_{1}$
Natural length of wire l is $l = \frac{l_{2} T_{1} - l_{1} T_{2}}{T_{2} - T_{1}}$

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