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