A) \[9.6\times {{10}^{-4}}%\]
B) \[9.6\times {{10}^{-2}}%\]
C) \[6.9\times {{10}^{-4}}%\]
D) \[6.9\times {{10}^{-2}}%\]
Correct Answer: B
Solution :
Time period of torsional pendulum \[T=2\pi \sqrt{\frac{1}{C}}\] |
\[\therefore \frac{\Delta T}{T}=-\frac{1}{2}\frac{\Delta I}{I}\] |
As \[I=k{{L}^{2}}\] |
\[\therefore \frac{\Delta I}{I}=\frac{2\Delta L}{L}\] |
Due to small change in temperature \[\Delta \]T, |
\[\Delta L=L\alpha \,\Delta T\] |
\[\therefore \frac{\Delta I}{I}=2\alpha \Delta T\] |
And \[\therefore \frac{\Delta T}{T}=\frac{1}{2}\times 2\alpha \Delta T\] |
=\[\alpha \Delta T=2.4\times {{10}^{-5}}\times (45-5)\] = \[9.6\times {{10}^{-4}}=9.6\times {{10}^{-2}}%\] |
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