The length of metallic wire is $l_{1}$ when tension in it is $T_{1}$ . It is $l_{2}$ when the tension is T₂. The original length of the wire will be:

Option 1 -

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

Option 2 -

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

Option 3 -

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

Option 4 -

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

2 Views | Posted 2 weeks ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 weeks ago

Correct Option - 3

Detailed Solution:

Stress = Y × strain

-> $\frac{T_{1}}{A} = Y \times - \frac{(l_{1} - l)}{l} (i)$

$\frac{T_{2}}{A} = Y \times \frac{(l_{2} - l)}{l} (i i)$

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

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

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The length of metallic wire is $l_{1}$ when tension in it is $T_{1}$ . It is $l_{2}$ when the tension is T₂. The original length of the wire will be:

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The length of metallic wire is l 1 when tension in it is T 1 . It is l 2 when the tension is T2. The original length of the wire will be:

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The length of metallic wire is $l_{1}$ when tension in it is $T_{1}$ . It is $l_{2}$ when the tension is T₂. The original length of the wire will be: