A) 50000
B) 5000
C) 10000
D) 1000
Correct Answer: B
Solution :
\[{{T}_{0}}=2\pi \sqrt{\frac{m}{k}}\] \[=\frac{2\pi }{\sqrt{10}}\] \[A={{A}_{0}}{{e}^{-t/\gamma }}\] \[\therefore \]for\[A=\frac{{{A}_{0}}}{e},t=\gamma \] \[t=\gamma =\frac{2m}{b}=\frac{2m}{\frac{{{B}^{2}}{{\ell }^{2}}}{R}}={{10}^{4}}s\] \[\therefore \] No of oscillation\[\frac{t}{{{T}_{0}}}=\frac{{{10}^{4}}}{2\pi /\sqrt{10}}\approx 5000\].You need to login to perform this action.
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