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Standard XII
Physics
Time Period Independent of Amplitude
Question

The bob of a simple pendulum has a mass of 60 g and a positive charge of 6×10−5C. It makes 30 oscillations is 50 s above earth's surface. A vertical electric field pointing upward and a magnitude 5×104N/C is switched on. How much time will it now take to complete 60 oscillations? (g=10 m/s2).

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Solution
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Q1
The bob of a simple pendulum has a mass of 60 g and a positive charge of 6×10−5C. It makes 30 oscillations is 50 s above earth's surface. A vertical electric field pointing upward and a magnitude 5×104N/C is switched on. How much time will it now take to complete 60 oscillations? (g=10 m/s2).
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Q2
The bob of a simple pendulum has a mass of $$40$$g and a positive charge of $$4.0\times 10^{-6}$$C. It makes $$20$$ oscillations in $$45$$s. A vertical electric field pointing upward and of magnitude $$2.5\times 10^4N C^{-1}$$ is switched on. How much time will it now take to complete $$20$$ oscillations?
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Q3
The bob of a simple pendulum has mass 40 grams and positive charge 4*10 to power -6 C . It makes 20 oscillations in 45 seconds . A vertical electric field pointing upwards is switched on and is of magnitude 2.5 * 10 to power 4 NC-1 . How much time will it take to complete 20 oscillations ?
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Q4
A bob of a simple pendulum of mass 40gm with a positive charge 4×10−6 is oscillating with time period ‘T1’. An electric field of intensity 3.6×104 N/c is applied vertically upwards now time period is T2. The value of ′T′2T1 is (g = 10 m/s2)

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Q5
The bob of a simple pendulum has mass $$2g$$ and a charge of $$5.0\mu C$$. It is at rest in a uniform horizontal electric field of intensity $$2000\ V/m$$. At equilibrium, the angle that the pendulum makes with the vertical is : (take $$g = 10\ m/s^{2})$$.
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