A) \[2.33\times {{10}^{-5}}/{}^\circ C\]
B) \[6.0\times {{10}^{-5}}/{}^\circ C\]
C) \[4.0\times {{10}^{-5}}/{}^\circ C\]
D) \[2.0\times {{10}^{-5}}/{}^\circ C\]
Correct Answer: D
Solution :
[d] \[\ell =5m {{t}_{1}}=0{}^\circ C\] \[{{\ell }_{2}}= 5.01m {{t}_{2}}=100{}^\circ C\] \[\alpha =\frac{{{\ell }_{2}}-{{\ell }_{1}}}{{{\ell }_{1}}\left( {{t}_{2}}-{{t}_{1}} \right)}=\frac{5.01-5}{5\times 100}=2\times {{10}^{-5}}{{/}^{\operatorname{o}}}C\].
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