TS EAMCET · Physics · Oscillations
The bob of simple pendulum is hanging vertically down from a fixed identical bob by means of a sting of length \(l\). If both bobs are charged with a charge \(q\) each, time period of the pendulum is : (ignore the radii of the bobs)
- A \(2 \pi \sqrt{\frac{l}{g+\left(\frac{q^2}{l^2 m}\right)}}\)
- B \(2 \pi \sqrt{\frac{l}{g-\left(\frac{q^2}{l^2 m}\right)}}\)
- C \(2 \pi \sqrt{\frac{l}{g}}\)
- D \(2 \pi \sqrt{\frac{l}{g-\left(\frac{q^2}{l}\right)}}\)
Answer & Solution
Correct Answer
(C) \(2 \pi \sqrt{\frac{l}{g}}\)
Step-by-step Solution
Detailed explanation
There will be a electrostatic repulsion between two charged bobs, but it does not affect the motion of pendulum. Thus, time period of pendulum remains same i.e., \(T=2 \pi \sqrt{\frac{l}{g}}\).
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