Question

the bob of a simple pendulum has a mass of 40g and a positive charge of 4*10^-6

Answer 1

Your question is incomplete .
A complete question is ------> The bob of a simple pendulum has a mass of 40g and a positive charge of 4 x (10)^(-6) C. It makes 20 oscillations in 45 sec. A vertical electric field pointing upward and of magnitude 2.5 x (10)^(4) N/C is switched on. How much time will now it take to complete 20 oscillations. ?

Solution :-
Let mass of Bob , m = 40g = 0.04 Kg
charge of Bob, q = 4 × 10⁻⁶C
we know, according to Coulombs law,
F = qE
∵ F = ma [ Newton's law ]
ma = qE
a = qE/m , here E is electric field intensity
Now, put value of q , E and m
a = 4 × 10⁻⁶ × 2.5 × 10⁴/0.04
= 2.5 m/s²
Because electrostatic force act on upward direction so,
effective acceleration = g - a
= 10 - 2.5 = 7.5 m/s² [ assume acceleration due to gravity, g = 10m/s²]

Now, time period is given by,
T = $\bold{2\pi\sqrt{\frac{l}{a_{eff}}}}$

Case 1 :-
T = $\bold{2\pi\sqrt{\frac{l}{10}}}$ [ because effective acceleration in case 1 is g and g = 10 m/s² ] ------(1)
Case 2 :- T' = $\bold{2\pi\sqrt{\frac{l}{7.5}}}$ -------(2)
Now, T/T' = √(7.5/10) = √(3/4) = √3/2
T' = 2T/√3
T is given 45 sec
so, T' = 2 × 45/√3 = 30√3 sec = 30 × 1.732 = 3 × 17.32 = 51.96 ≈ 52 sec

Hence, answer is 52 sec

Answer 2

Answer:

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