JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A simple pendulum of length \(L\) is placed between the plates of a parallel plate capacitor having electric field \(E,\) as shown in figure. Its bob has mass \(m\) and charge \(q.\) the time period of the pendulum is given by
- A \(2\pi \sqrt {\frac{L}{{\sqrt {{g^2}\, - \,\frac{{{q^2}{E^2}}}{{{m^2}}}} }}} \)
- B \(2\pi \sqrt {\frac{L}{{\left( {g\, + \,\frac{{qE}}{m}} \right)}}} \)
- C \(2\pi \sqrt {\frac{L}{{\left( {g\, - \,\frac{{qE}}{m}} \right)}}} \)
- D \(2\pi \sqrt {\frac{L}{{\sqrt {{g^2}\, + \,{{\left( {\frac{{qE}}{m}} \right)}^2}} }}} \)
Answer & Solution
Correct Answer
(D) \(2\pi \sqrt {\frac{L}{{\sqrt {{g^2}\, + \,{{\left( {\frac{{qE}}{m}} \right)}^2}} }}} \)
Step-by-step Solution
Detailed explanation
\(g_{e f f}=\sqrt{g^{2}+\left(\frac{q E}{m}\right)^{2}}\) \(T=2 \pi \sqrt{\frac{\ell}{g_{e f f}}}\) \(2 \pi \sqrt{\frac{\ell}{\sqrt{g^{2}+\left(\frac{q E}{m}\right)^{2}}}}\)
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