A) \[1.7\times {{10}^{-4}}/{{\,}^{o}}C\]
B) \[1.7\times {{10}^{-5}}/{{\,}^{o}}C\]
C) \[3.4\times {{10}^{-4}}/{{\,}^{o}}C\]
D) \[3.4\times {{10}^{-5}}/{{\,}^{o}}C\]
Correct Answer: B
Solution :
\[\frac{{{T}_{20}}-{{T}_{0}}}{{{T}_{0}}}=\frac{1}{2}\times \alpha \times 20=10\alpha \] Here, \[{{T}_{20}}\]is the correct time period. The time period at in smaller so that the clock runs fast. The time gained in 24 hours \[=24\,(hrs)\times \alpha \times 10\] \[\therefore \] \[15\,\sec =24\,\text{hrs}\times \alpha \times 10\] or \[\alpha =\frac{15}{86400\times 10}=\frac{1.7\times {{10}^{-5}}}{{{\,}^{o}}C}\] Hence, the correction option is [b].
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