In an experiment, brass and steel wires of length $1\,m$ each with areas of cross section $1\,mm^2$ are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress requires to produced a new elongation of $0.2\,mm$ is [Given, the Young’s Modulus for steel and brass are respectively $120\times 10^9\,N/m^2$ and $60\times 10^9\,N/m^2$ ]

Q: In an experiment, brass and steel wires of length $1\,m$ each with areas of cross section $1\,mm^2$ are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress requires to produced a new elongation of $0.2\,mm$ is [Given, the Young’s Modulus for steel and brass are respectively $120\times 10^9\,N/m^2$ and $60\times 10^9\,N/m^2$ ]

d $\ell = 1\,M$ $A = {10^{ - 6}}{M^2}$ $stress = \frac{F}{A}$ ${Y_s} = 120 \times {10^9}$ $Stress = \frac{{Stress}}{Y}$ $\Delta \ell = \frac{{\ell \times F}}{{AY}}$ $\Delta {\ell _1} + \Delta {\ell _2} = \frac{{{\ell _1}F}}{{A{Y_1}}} + \frac{{{\ell _2}F}}{{A{Y_2}}} = 0.2 \times {10^{ - 3}}$ $\frac{F}{A} = \frac{{0.2 \times {{10}^{ - 3}}}}{{\frac{\ell }{{{Y_1}}} + \frac{\ell }{{{Y_2}}}}}$ $ = \frac{{0.2 \times {{10}^{ - 3}}}}{{\frac{1}{{120 \times {{10}^9}}} + \frac{1}{{60 \times {{10}^9}}}}} = \frac{{0.2 \times {{10}^{ - 3}} \times {{10}^9} \times 120}}{{1 + 2}}$ $ = \frac{{0.2 \times {{10}^6} \times 120}}{3} = 8 \times {10^6}$

Question

A$1.8\times 10^6\,N/m^2$
B$0.2\times 10^6\,N/m^2$
C$1.2\times 10^6\,N/m^2$
D
None of these

JEE MAIN 2019, Diffcult

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