A) \[\frac{1}{\alpha }\]
B) \[2\alpha \]
C) \[\frac{\alpha }{2}\]
D) \[\alpha \]
Correct Answer: D
Solution :
\[{{t}_{0}}=2=2\pi \sqrt{\frac{{{\ell }_{0}}}{g}}\] \[t=2\pi \sqrt{\frac{{{\ell }_{0}}(1+\alpha T)}{g}}=2{{(1+\alpha T)}^{1/2}}=2+\alpha T\] \[t-{{t}_{0}}=\alpha T\Rightarrow \Delta t=\alpha T\]You need to login to perform this action.
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