Home
Class 12
PHYSICS
A small metallic bob of mass m of a simp...

A small metallic bob of mass `m` of a simple pendulum of length `l` is suspended from a silk threas between two parallel charged plates.The electric field intensity `E` act vertically downwards . If bob is given a charge `+ q`,time period of oscillation
(i) `T = 2 pi sqrt((l)/((g - (qE)/(m))))` , if the lower plate is charged positively
(ii) `T = 2 pi sqrt((l)/((g - (qE)/(m))))`,if the lower plate is charged negatively
(iii) `T = 2 pi sqrt((l)/((g + (qE)/(m))))`,if the lower plate is charged positively
(iv) `T = 2 pi sqrt((l)/((g + (qE)/(m))))` , if the lower plate is charged negatively

A

`2pisqrt(L//g)`

B

`2pisqrt((L)/(g+qE//m))`

C

`2pisqrt((L)/(g-qE//m))`

D

`2pisqrt((L)/(g-qE//m))`

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Topper's Solved these Questions

  • ELECTROSTATICS - I

    MOTION|Exercise EXERCISE - 1 (Section C)|8 Videos

  • ELECTROSTATICS - I

    MOTION|Exercise EXERCISE - 1 (Section D)|8 Videos

  • ELECTROSTATICS - I

    MOTION|Exercise EXERCISE - 1 (Section A)|10 Videos

  • ELECTROSTATICS

    MOTION|Exercise Exercise -3 Section (B) Previous Year Problems | JEE Main|30 Videos

  • ELECTROSTATICS-II

    MOTION|Exercise Exercise - 4 (Level - II)|30 Videos