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Question

The length of the metallic wire is l1 when the tension in it is T1 It is l2 when the tension is T2. The original length of the wire will be:


A

l1+l22

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B

T1l1-T2l2T2-T1

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C

T2l1+T1l2T2+T1

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D

T2l1-T1l2T2-T1

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Solution

The correct option is D

T2l1-T1l2T2-T1


Step 1: Given that:

The length of the first metallic wire is l1

The tension of the first metallic wireT1

The length of the second metallic wire is l2

The tension of the second metallic wire is T2

Step 3: Draw the required diagram:

Step 2: Formula used:

The general formula of the young modulus in terms of stress and strain=StressStrain

Stress can be calculated as -stress=FA

where Fis the force exerted by the object under tension, A is the cross-sectional area

Strain will be calculated as- Strain=∆ll

∆l is the change in length and l is the actual length.

Thus putting values we get young's modulus as-

y=FA∆ll

Step 3: Calculating the original length:

Therefore for the first wire's tension T1:

Using the formula of young's modulus-

For the first metallic wire-

Substituting the known values-

y=T1Al-l1l……………………(1)

Now for the second wire's tension T2

Substituting the known values-

y=T2Al-l2l………………………(2)

Step 4: Now equating the equations:

Therefore, equating equation (1) and equation (2) we get-

⇒T1Al-l1l=T2Al-l2l⇒T1l-l1=T2l-l2⇒T1l-T1l2=T2l-T2l1⇒l(T1-T2)=T1l2-T2l1⇒l=T2l1-T1l2T2-T1

Therefore the original length of the wire will be T2l1-T1l2T2-T1.

Hence, option D is the correct answer.


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