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AP EAMCET · PHYSICS · Mechanical Properties of Solids

Two wires \(A\) and \(B\) of same cross-section are connected end to end. When same tension is created in both wires, the elognation in \(B\) wire is twice the elongation in \(A\) wire. If \(L_A\) and \(L_B\) are the initial lengths of the wires \(A\) and \(B\) respectively, then
(Young's modulus of material of wire \(A=2 \times 10^{11} \mathrm{Nm}^{-2}\) and Young's modulus of material of wire \(B=1.1 \times 10^{11} \mathrm{Nm}^{-2}\) ).

  1. A \(\frac{L_A}{L_B}=\frac{10}{11}\)
  2. B \(\frac{L_A}{L_B}=\frac{4}{5}\)
  3. C \(\frac{L_A}{L_B}=\frac{9}{11}\)
  4. D \(\frac{L_A}{L_B}=\frac{3}{7}\)
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Correct Answer

(A) \(\frac{L_A}{L_B}=\frac{10}{11}\)

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Detailed explanation

The given situation is shown below. Here, elongation in wire \(B\) is twice of elongation in wire \(A\) on the application of same tension \(T\). i.e \(\Delta L_B=2 \Delta L_A\) Young's modulus of wire \(A\) and \(B\) are given as \(Y_A=\frac{T L_A}{A \cdot \Delta L_A}\) and…
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