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 \(B=1.1 \times 10^{11} \mathrm{Nm}^{-2}\) ).
- A \(\frac{L_A}{L_B}=\frac{10}{11}\)
- B \(\frac{L_A}{L_B}=\frac{4}{5}\)
- C \(\frac{L_A}{L_B}=\frac{9}{11}\)
- D \(\frac{L_A}{L_B}=\frac{3}{7}\)
Answer & Solution
Correct Answer
(A) \(\frac{L_A}{L_B}=\frac{10}{11}\)
Step-by-step Solution
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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