A) \[2\times {{10}^{-5}}{{l}^{0}}C\]
B) \[3\times {{10}^{-5}}{{l}^{0}}C\]
C) \[6\times {{10}^{-5}}{{l}^{0}}C\]
D) \[1.8\times {{10}^{-4}}{{l}^{0}}C\]
Correct Answer: A
Solution :
| Increase in volume of ball \[\Delta V=V\gamma \Delta T\] |
| \[\frac{\Delta V}{V}=\gamma \Delta T\] |
| \[\frac{\Delta V}{V}=\frac{0.18}{100}\]and \[\Delta T=30{{\,}^{0}}C\] |
| \[\therefore \]\[\gamma =\frac{\Delta V}{V\Delta t}=\frac{0.18}{100\times 30}=6\times {{10}^{-5}}\]per \[^{0}C\] |
| \[\alpha =\frac{\gamma }{3}=\frac{6\times {{10}^{-5}}}{3}=2\times {{10}^{-5}}\]per \[{{\,}^{0}}C\] |
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