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← JEE Main 2022 · 27 Jul · Shift 2
PhysicsSingle correct+4 / -1Question 37 of 90

JEE Main 2022 · 27 Jul · Shift 2, Physics question 37

A steel wire of length 3.2 m (Ys=2.0×1011Y_s=2.0\times10^{11} Nm−2^{-2}) and a copper wire of length 4.4 m (Yc=1.1×1011Y_c=1.1\times10^{11} Nm−2^{-2}), both of radius 1.4 mm are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm. The load applied, in Newton, will be : (Given π=227\pi=\dfrac{22}7)

Choose one option
Answer and worked solution

Official answer: D

Series wires carry the same F. 1.4×10−3=Fπ(1.4×10−3)2(3.22×1011+4.41.1×1011)1.4\times10^{-3}=\frac F{\pi(1.4\times10^{-3})^2}\left(\frac{3.2}{2\times10^{11}}+\frac{4.4}{1.1\times10^{11}}\right).

Solve. F×9.09×10−6=1.4×10−3F\times9.09\times10^{-6}=1.4\times10^{-3}, so F=154F=154 N.

(D)