A) \[4.3\,\times \,{{10}^{-5}}{{/}^{o}}C\]
B) \[2.3\,\times \,{{10}^{-5}}{{/}^{o}}C\]
C) \[5.3\,\times \,{{10}^{-5}}{{/}^{o}}C\]
D) \[3.4\,\times \,{{10}^{-5}}{{/}^{o}}C\]
Correct Answer: B
Solution :
Weight in air\[=46\text{ }g\] \[30g=46g-{{V}_{D{{\rho }_{27}}g}}\] ?(i) \[30.5g=46g-{{V}_{D}}'{{\rho }_{42}}g\] ...(ii) From Eqs. (i) and (ii). \[\frac{V{{'}_{D}}}{{{V}_{D}}}.\frac{{{\rho }_{42}}}{{{\rho }_{27}}}=\frac{15.5}{16}\] \[\frac{{{V}_{D}}(1+\gamma \Delta T)}{{{V}_{D}}}.\frac{1.20}{1.24}=\frac{15.5}{16}(\Delta T=15{}^\circ C)\] \[1+\gamma (15)=\frac{15.5\times 1.24}{1.20\times 16}\] \[\gamma =6.9\times {{10}^{-5}}/{}^\circ C\] Coefficient of linear expansion \[\alpha =\frac{\gamma }{3}=\frac{6.9\times {{10}^{-5}}}{3}=2.3\times {{10}^{-5}}/{}^\circ C\]You need to login to perform this action.
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