Question

A wire \(P\) of length \(L\) made of a material of density \(D\) and Young's modulus \(Y\) elongates by \(2.0\) mm under its own weight. If another wire \(Q\) made of a material of density \[ \frac{2D}{3} \] and Young's modulus \[ \frac{3Y}{4} \] elongates by \(1.0\) mm under its own weight, then the length of the wire \(Q\) is

• A. \(0.75L\)
• B. \(0.50L\)
• C. \(1.25L\)
• D. \(1.75L\)

Hint:

For a uniform wire suspended vertically, \[ \boxed{ \Delta l=\frac{\rho gL^2}{2Y}. } \] Notice that the elongation is proportional to \[ \boxed{\frac{\rho L^2}{Y}.} \]

Answer 1

The Correct Option is A

Step 1: Use the formula for elongation due to self-weight. The elongation of a wire under its own weight is \[ \Delta l = \frac{\rho gL^2}{2Y}, \] where \[ \rho \] is the density and \[ Y \] is Young's modulus.

Step 2: Write the elongation for each wire. For wire \(P\), \[ 2 = \frac{DgL^2}{2Y}. \] For wire \(Q\), \[ 1 = \frac{\left(\frac{2D}{3}\right)gL_Q^2} {2\left(\frac{3Y}{4}\right)}. \] Simplifying, \[ 1 = \frac{4DgL_Q^2}{9Y}. \]

Step 3: Find the length of wire \(Q\). Taking the ratio, \[ \frac{1}{2} = \frac{\frac{4DgL_Q^2}{9Y}} {\frac{DgL^2}{2Y}} = \frac{8L_Q^2}{9L^2}. \] Hence, \[ L_Q^2 = \frac{9}{16}L^2. \] Therefore, \[ L_Q = \frac34L = 0.75L. \] Hence, \[ \boxed{0.75L}. \] Thus, \[ \boxed{(A)} \] is the correct answer.