A) \[2\pi \sqrt{\frac{l}{g}}\]
B) \[2\pi \sqrt{\frac{l}{\sqrt{g+\frac{qE}{m}}}}\]
C) \[2\pi \sqrt{\frac{l}{\sqrt{g-\frac{qE}{m}}}}\]
D) \[2\pi \sqrt{\frac{l}{\sqrt{{{g}^{2}}-{{\left( \frac{qE}{m} \right)}^{2}}}}}\]
Correct Answer: D
Solution :
Time period of simple pendulum in air \[T=2\pi \frac{\sqrt{l}}{g}\] When it is suspended between vertical plates of a charged parallel plate capacitor, then accelertion due to electric field, \[a=\frac{qE}{m}\] This acceleration is acting horizontally and acceleration due to gravity is acting vertically So, effective acceleration \[g=\sqrt{{{g}^{2}}+{{a}^{2}}}=\sqrt{{{g}^{2}}+{{\left( \frac{qE}{m} \right)}^{2}}}\] Hence, \[T=2\pi \sqrt{\frac{l}{\sqrt{{{g}^{2}}+{{\left( \frac{qE}{m} \right)}^{2}}}}}\]You need to login to perform this action.
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