A) \[1.3\times {{10}^{-3}}{{/}^{\text{o}}}C~\]
B) \[5.2\times {{10}^{-3}}{{/}^{\text{o}}}C\]
C) \[2.6\times {{10}^{-3}}{{/}^{\text{o}}}C\]
D) \[0.26\times {{10}^{-3}}{{/}^{\text{o}}}C\]
Correct Answer: C
Solution :
The volume of the metal at \[{{30}^{o}}C\]is \[{{V}_{30}}=\frac{loss\,of\,weight}{specific\,gravity\,\times g}\] \[=\frac{(45-25)g}{1.5\times g}=13.33c{{m}^{3}}\] Similarly, volume of metal at \[{{40}^{o}}C\]is \[{{V}_{40}}=\frac{(45-27)g}{1.25\times g}=14.40c{{m}^{3}}\] Now, \[{{V}_{40}}={{V}_{30}}[1+\gamma ({{t}_{2}}-{{t}_{1}})]\] or \[\gamma =\frac{{{V}_{40}}-{{V}_{30}}}{{{V}_{30}}({{t}_{2}}-{{t}_{1}})}\] \[=\frac{14.40-13.33}{13.33(40-30)}\] \[=8.03\times {{10}^{-3}}{{/}^{o}}C\] \[\therefore \] Coefficient of linear expansion of the metal is \[\alpha =\frac{\gamma }{3}=\frac{8.03\times {{10}^{-3}}}{3}\] \[\approx 2.6\times {{10}^{-3}}{{/}^{o}}C\]You need to login to perform this action.
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