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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

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

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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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Solution

The correct option is C $\frac{l_{1} T_{2} - l_{2} T_{1}}{T_{2} - T_{1}}$
$Y = \frac{F l}{A △ l}$

Y, l and A are constants
Therefore $\frac{F}{△ l} = c o s t a n t o r △ l \propto F$

Now, $l_{1} - l \propto T_{1} a n d l_{2} - l \propto T_{2}$
Dividing, $\frac{l_{1} - l}{l_{2} - l} = \frac{T_{1}}{T_{2}}$
or $l_{1} T_{2} - l T_{2} = l_{2} T_{1} - l T_{1}$
or $l (T_{1} - T_{2}) = l_{2} T_{1} - l_{1} T_{2}$

or $l = \frac{l_{2} T_{1} - l_{1} T_{2}}{T_{1} - T_{2}}$

or $l = \frac{l_{1} T_{2} - l_{2} T_{1}}{T_{2} - T_{1}}$

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