JEE Mains · Physics · STD 11- Mechanical properties of solids
- JEE Mains 2026
- Medium
- MCQ
The two wires \(A\) and \(B\) of equal cross-section but of different materials are joined together. The ratio of Young's modulus of wire \(A\) and wire \(B\) is \(20/11\). When the joined wire is kept under certain tension the elongations in the wires \(A\) and \(B\) are equal. If the length of wire \(A\) is \(2.2\) m, then the length of wire \(B\) is _______ m.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(1.21\)
Step-by-step Solution
Detailed explanation
Given that the wires have equal cross-sectional area \(A\) and are under the same tension \(F\), the elongation \(\Delta L\) is given by:
\(\Delta L = \dfrac{F L}{A Y}\)
Since the elongations in both wires are equal (\(\Delta L_A = \Delta L_B\)):
\(\dfrac{F L_A}{A Y_A} = \dfrac{F L_B}{A Y_B}\)
\(\dfrac{L_A}{Y_A} = \dfrac{L_B}{Y_B}\)
\(L_B = L_A \times \dfrac{Y_B}{Y_A}\)
Substituting the given values \(L_A = 2.2\) m and \(\dfrac{Y_A}{Y_B} = \dfrac{20}{11}\):
\(L_B = 2.2 \times \dfrac{11}{20}\)
\(L_B = \dfrac{24.2}{20} = 1.21\) m
Answer: \(1.21\)
\(\Delta L = \dfrac{F L}{A Y}\)
Since the elongations in both wires are equal (\(\Delta L_A = \Delta L_B\)):
\(\dfrac{F L_A}{A Y_A} = \dfrac{F L_B}{A Y_B}\)
\(\dfrac{L_A}{Y_A} = \dfrac{L_B}{Y_B}\)
\(L_B = L_A \times \dfrac{Y_B}{Y_A}\)
Substituting the given values \(L_A = 2.2\) m and \(\dfrac{Y_A}{Y_B} = \dfrac{20}{11}\):
\(L_B = 2.2 \times \dfrac{11}{20}\)
\(L_B = \dfrac{24.2}{20} = 1.21\) m
Answer: \(1.21\)
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