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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Some Mechanical Properties of Matter

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Some Mechanical Properties of Matter

A metal wire has a length of l_1 when subjected to a tension T_1 ,...

A metal wire has a length of l1l_1 when subjected to a tension T1T_1, and a length of l2l_2 when the tension is T2T_2. The natural length of the wire is

Options

  1. Al1l2\sqrt{l_1 l_2}
  2. Bl1T2−l2T1T2−T1\dfrac{l_1 T_2 - l_2 T_1}{T_2 - T_1}
  3. Cl1+l22\dfrac{l_1 + l_2}{2}
  4. Dl1T2+l2T1T2+T1\dfrac{l_1 T_2 + l_2 T_1}{T_2 + T_1}

Correct answer

B. l1T2−l2T1T2−T1\dfrac{l_1 T_2 - l_2 T_1}{T_2 - T_1}

Step-by-step solution

Let the natural length of the wire be LL, cross-sectional area be AA, and Young's modulus be YY.

The extension in the wire under tension T1T_1 is Δl1=l1−L\Delta l_1 = l_1 - L.

From Hooke's law, Y=T1LAΔl1⇒T1=YAL(l1−L)Y = \dfrac{T_1 L}{A \Delta l_1} \Rightarrow T_1 = \dfrac{YA}{L}(l_1 - L).

Similarly, the extension under tension T2T_2 is Δl2=l2−L\Delta l_2 = l_2 - L.

T2=YAL(l2−L)T_2 = \dfrac{YA}{L}(l_2 - L).

Dividing the two equations, we get:

T1T2=l1−Ll2−L\dfrac{T_1}{T_2} = \dfrac{l_1 - L}{l_2 - L}

T1l2−T1L=T2l1−T2LT_1 l_2 - T_1 L = T_2 l_1 - T_2 L

L(T2−T1)=l1T2−l2T1L(T_2 - T_1) = l_1 T_2 - l_2 T_1

L=l1T2−l2T1T2−T1L = \dfrac{l_1 T_2 - l_2 T_1}{T_2 - T_1}
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