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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 $T_{1}$ and is $l_{2}$ when the tension is $T_{2}$ . The natural length of the wire is

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Solution

The correct option is C

Let the Area of cross-section of the wire be 'S' and it doesn't' vary as the length changes within the elastic limit.
In the first case, the elongation in the wire= $l_{2} - l$
$\Rightarrow Y_{1} = \frac{(\frac{T_{1}}{s}) l}{(l_{1} - l)}$

Now,the second case, the elongation in the wire $= l_{2} - l$
$\Rightarrow Y_{2} = \frac{(\frac{T_{1}}{s}) l}{(l_{1} - l)}$
Now, $Y_{1} = Y_{2}$ as the material is same
$\Rightarrow \frac{T_{1} l}{s (l_{1} - l)} = \frac{T_{2} l}{s (l_{2} - l)} \Rightarrow \frac{T_{1}}{l_{1} - l} = \frac{T_{2}}{l_{2} - l} T_{1} l_{2} - T_{2} l = T_{2} l_{2} - T_{2} l l (T_{2} - T_{1}) = T_{2} I_{2} - T_{1} I_{2} l = \frac{T_{2} l_{1} - T_{1} l_{2}}{T_{2} - T_{1}}$

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