question_answer

In an experiment, bras and steel wires of length 1m each with areas of cross section \[1m{{m}^{2}}\]are used. teh 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 required to produce a net elongation of 0.2 mm is : (Given, the Young's Modulus for steel and brass are respectively, \[120\times {{10}^{9}}\text{ }N/{{m}^{2}}\] and \[60\times {{10}^{9}}\text{ }N/{{m}^{2}}\])                                                   [JEE Main 10-4-2019 Afternoon]

A) \[0.2\times {{10}^{6}}\text{ }N/{{m}^{2}}\]

B)       \[4.0\times {{10}^{6}}\text{ }N/{{m}^{2}}\]

C)  \[1.8\times {{10}^{6}}\text{ }N/{{m}^{2}}\] D)       \[1.2\times {{10}^{6}}\text{ }N/{{m}^{2}}\] Correct Answer: B Solution : [b]

\[{{k}_{1}}=\frac{{{y}_{1}}{{A}_{1}}}{{{\ell }_{1}}}=\frac{120\times {{10}^{9}}\times A}{1}\]

\[{{k}_{2}}=\frac{{{y}_{2}}{{A}_{2}}}{{{\ell }_{2}}}=\frac{60\times {{10}^{9}}\times A}{1}\]

\[{{k}_{eq}}=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}\times {{k}_{2}}}=\frac{120\times 60}{180}\times {{10}^{9}}\times A\]

\[{{k}_{eq}}=40\times {{10}^{9}}\times A\]

\[F={{k}_{eq}}(x)\]

\[F=(40\times {{10}^{9}})A.(0.2\times {{10}^{-3}})\]

\[\frac{F}{A}=8\times {{10}^{6}}N/{{m}^{2}}\]

No option is matching. Hence question must be bonus.