A) \[-3.75{}^\circ C\]
B) \[-2.75{}^\circ C\]
C) \[2.75{}^\circ C\]
D) \[3.75{}^\circ C\]
Correct Answer: A
Solution :
Given: Coefficient of linear expansion for brass a \[\alpha =19\times {{10}^{-6}}/{{\,}^{o}}C\] Brass sheet at temperature, \[{{t}_{1}}=10{{\,}^{o}}C\] Coefficient of linear expansion for steel sheet \[{{a}_{s}}=11\times {{10}^{-6}}/{{\,}^{o}}C\] \[{{t}_{2}}=?\] Initial area of brass is given by \[{{A}_{b}}={{A}_{0}}(1+\beta \,\Delta t)={{A}_{0}}(1+2\alpha \,\Delta t)\] \[={{A}_{0}}(1+2\times 19\times {{10}^{-6}}\Delta t)\] \[{{A}_{b}}={{A}_{0}}[(1+38\times {{10}^{-6}})({{T}_{2}}-10)]\] ?(i) Initial area of steel is given by \[{{A}_{s}}={{A}_{0}}(1+\beta \,\Delta t)={{A}_{0}}(1+2\alpha \,\Delta t)\] \[={{A}_{0}}(1+2\times 11\times {{10}^{-6}}\Delta t)\] \[{{A}_{s}}={{A}_{0}}[(1+22\times {{10}^{-6}})({{t}_{2}}-20)]\] ?(ii) According to the condition in question \[{{A}_{b}}={{A}_{s}}\] Hence, from Eqs. (i) and (ii), \[{{A}_{0}}[1+38\times {{10}^{-6}}({{t}_{2}}-10)]\] \[={{A}_{0}}[1+22\times {{10}^{-6}}({{t}_{2}}-20)]\] \[\Rightarrow \] \[\frac{{{t}_{2}}-20}{{{t}_{2}}-10}=\frac{38\times {{10}^{-6}}}{22\times {{10}^{-6}}}=\frac{19}{11}\] \[\Rightarrow \] \[11\,{{t}_{2}}-220=19{{t}_{2}}=190\] \[8{{t}_{2}}=-30\Rightarrow {{t}_{2}}=-\frac{30}{8}=-3.75{{\,}^{o}}C\]
You need to login to perform this action.
You will be redirected in
3 sec
