A) 5s/day
B) 10.3s/day
C) 20.6 s/day
D) 20 min/day
Correct Answer: B
Solution :
As, \[{{T}_{2}}={{T}_{1}}\sqrt{{{l}_{2}}/{{l}_{1}}}\] \[={{T}_{1}}\times \sqrt{\frac{{{l}_{1}}[\{1+12\times {{10}^{-6}}\times (40-20)\}]}{{{l}_{1}}}}\]\[={{T}_{1}}\sqrt{1+240\times {{10}^{-6}}}\] \[\therefore \] \[\frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}}={{(1+240\times {{10}^{-6}})}^{1/2}}-1\] \[=120\times {{10}^{-6}}\] or \[{{T}_{2}}-{{T}_{1}}=120\times {{10}^{-6}}\times (24\times 60\times 60)s\] \[=10.36\,s/day\] It will be a loss as \[{{T}_{2}}>{{T}_{1}}.\]
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