Home
Class 12
PHYSICS
The bob of a simple pendulum has mass 2g...

The bob of a simple pendulum has mass `2g` and a charge of `5.0muC`. It is at rest in a uniform horizontal electric field of intensity `200V//m`. At equilibrium, the angle that the pendulum makes with the vertical is:
(take `g=10m//s^(2)`)

A

`tan^(-1)(0.2)`

B

`tan^(-1)(0.5)`

C

`tan^(-1)(5.0)`

D

`tan^(-1)(2.0)`

Text Solution

Verified by Experts

The correct Answer is:
B


`Tcos theta=mg , tan theta=(Eq)/(mg)=(200xx5xx10^(-6))/(2xx10^(-3)xx10)=1/2," "T sin theta=Eqimplies" "theta=tan^-10.5`
Promotional Banner

Topper's Solved these Questions

  • JEE MAIN REVISION TEST - 10| JEE -2020

    VMC MODULES ENGLISH|Exercise PHYSICS (SECTION 2)|5 Videos
  • JEE MAIN REVISION TEST - 1 | JEE - 2020

    VMC MODULES ENGLISH|Exercise PHYSICS ( SECTION 2)|5 Videos
  • JEE MAIN REVISION TEST - 12

    VMC MODULES ENGLISH|Exercise PHYSICS|25 Videos

Similar Questions

Explore conceptually related problems

A pendulum bob of mass 80mg and carrying a charge of 2xx 10^(-8)C is at rest in a uniform, horizontal electric field of 20k Vm^-1. Find the tension in the thread.

A pendulum bob of mass m = 80 mg, carrying a charge of q=2xx10^(-8)C , is at rest in a horizontal, uniform electric field of E = 20,000 V/m. The tension T in the thread of the pendulum and the angle alpha it makes with vertical, is (take g=9.8m//s^(2) )

The bob of a pendulum of mass 8 mu g carries an electric charge of 39.2xx10^(-10) coulomb in an electric field of 20xx10^(3) volt/meter and it is at rest. The angle made by the pendulum with the vertical will be

A pendulum bob of mass m charge q is at rest with its string making an angle theta with the vertical in a uniform horizontal electric field E. The tension in the string in

The bob of a simple pendulum of length 1 m has mass 100 g and a speed of 1.4 m/s at the lowest point in its path. Find the tension in the stirng at this instant.

A simple pendulum has a length l & mass of bob m. The bob is given a charge q coulomb. The pendulum is suspended in a uniform horizontal electric field of strength E as shown in figure, then calculate the time period of oscillation when bob is slightly displaced from its mean position.

A simple pendulum is suspended from the ceiling of a car taking a turn of radius 10 m at a speed of 36 km/h. Find the angle made by the string of the pendulum with the vertical if this angle does not change during the turn. Take g=10 m/s^2 .

The bob of a simple pendulum has a mass of 40 g and a positive charge of 4.0 xx 10 ^(-5) C . It makes 20 oscillations in 45 s. A vertical electric field pointing upward and of magnitude 2.5 xx 10^ 4 NC^(-1) is switched on. How much time will it now take to complete 20 oscillations?

A bob of a simple pendulum of mass 40 mg with a positive charge 4 xx 10^(-6) C is oscilliating with time period "T_(1). An electric field of intensity 3.6 xx 10^(4) N/c is applied vertically upwards now time period is T_(2) . The value of (T_(2))/(T_(1)) is (g = 10 m/s^(2))

A simple pendulum is released from rest with the string in horizontal position. The vertical component of the velocity of the bob becomes maximum, when the string makes an angle theta with the vertical. The angle theta is equal to