A) \[1.3\times {{10}^{-3}}/{}^\circ C\]
B) \[5.2\times {{10}^{-3}}/{}^\circ C\]
C) \[2.6\times {{10}^{-3}}/{}^\circ C\]
D) \[0.26\times {{10}^{-3}}/{}^\circ C\]
Correct Answer: C
Solution :
[c] \[{{V}_{1}}=\frac{4\left( 5-25 \right)g}{{{d}_{1}}}\]and \[{{V}_{2}}=\frac{\left( 45-27 \right)g}{{{d}_{2}}}\] \[\therefore \,\,\,\,\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{20}{18}\times \frac{{{d}_{2}}}{{{d}_{1}}}\] or \[\frac{{{V}_{1}}}{{{V}_{1}}\left( 1+{{r}_{metal}}\times 10 \right)}=\frac{20}{18}\times \left[ \frac{{{d}_{1}}}{\left( 1+{{r}_{\ell }}\times 10 \right){{d}_{1}}} \right]\] After simplifying, we get \[\alpha =\frac{{{\gamma }_{metal}}}{{}}=2.6\times {{10}^{-3}}{{/}^{o}}C\]
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