A) \[3.316\times {{10}^{-5}}{{/}^{0}}C\]
B) \[2.316\times {{10}^{-5}}{{/}^{0}}C\]
C) \[4.316\times {{10}^{-5}}{{/}^{0}}C\]
D) None of these
Correct Answer: B
Solution :
[b] Loss of weight at \[27{}^\circ C\]is \[=46-30=16={{V}_{1}}\times 1.24{{\rho }_{l}}\times g...(i)\] Loss of weight at \[42{}^\circ C\]is \[=46-30.5=15.5={{V}_{2}}\times 1.2{{\rho }_{l}}\times g...(ii)\] Now dividing (i) by (ii), we get\[\frac{16}{15.5}=\frac{{{V}_{1}}}{{{V}_{2}}}\times \frac{1.24}{1.2}\] But\[\frac{{{V}_{1}}}{{{V}_{2}}}=1+3\alpha ({{t}_{2}}-{{t}_{1}})=\frac{15.5\times 1.24}{16\times 1.2}=1.001042\] \[\Rightarrow 3\alpha (42{}^\circ -27{}^\circ )=0.001042\Rightarrow \alpha =2.316\times {{10}^{-5}}/{}^\circ C\]
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