Home
Class 12
PHYSICS
A simple pendulum consists of a small sp...

A simple pendulum consists of a small sphere of mass m suspended by a thread of length l. The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength E directed vertically upwards. If the electrostatic force acting on the sphere is less than gravitational force then the period of pendulum is

A

`T=2pi[(l)/(g-(qE)/(m))]^((1)/(2))`

B

`T=2pi[(l)/(g)]^((1)/(2))`

C

`T=2pi[(l)/(g+(qE)/(m))]^((1)/(2))`

D

`T=2pi[(ml)/(qE)]^((1)/(2))`

Text Solution

Verified by Experts

The correct Answer is:
A

Net acceleration `g.=g-(qE)/(m)`
So, `T=2pisqrt((l)/(g.)), T=2pisqrt((l)/((g-gE//m)))`
Doubtnut Promotions Banner Mobile Dark
|

Similar Questions

Explore conceptually related problems

A bar magnet of length 2l and pole strength m is placed in a uniform magnrtic field B,inclined a angle theta with the field.What is the force acting on each pole.

A system consist of a uniformly charged sphere and surrounding medium is filled by charge with volume charge density rho=(alpha)/(r ) , where alpha is a positive constant and r is distance from centre of the sphere. It is found that electric field outside sphere is constant and given by E=(3alpha)/(mepsilon_(0)) then m=?

Knowledge Check

  • A uniform ring of mass m is lying at a distance 1.73 a from the centre of a sphere of mass M just over the sphere where a is the small radius of the ring as well as that of the sphere. Then gravitational force exerted on the sphere is

    A
    `("GMm")/(8a^(2))`
    B
    `("GMm")/((1.73 a)^(2))`
    C
    `sqrt(3)("GMm")/(a^(2))`
    D
    `1.73("GMm")/(8a^(2))`

  • A uniform ring of mass m is lying at a distance 1.73 a from the centre of a sphere of mass M just over the sphere where a is the small radius of the ring as well as that of the sphere. Then gravitational force exerted on the sphere is

    A
    `(GMm)/(8 a^(2))`
    B
    `(GM m)/((1 . 73 a)^(2))`
    C
    `sqrt(3) (GM m)/(a^(2))`
    D
    `1 . 73 (GM m)/(8a^(2))`

  • A simple pendulum oscillates about mean position making an angle theta with the vertical. M is the mass of the bob of pendulum, l is the length of string of pendulum. Tension in the string having a velocity v is given by

    A
    `T- mg cos theta= (mv^(2))/(l)`
    B
    `T- mg sin theta= (mv^(2))/(l)`
    C
    `T= mg cos theta`
    D
    `T= (mv^(2))/(l)`

    • Similar Questions

    Explore conceptually related problems

    A simple pendulum consisting of a mass M attached to a string of length L is released from rest at an angle alpha . A pin is located at a distance I below the pivot point. When the pendulum swings, the string hits the pin as shown in figure. The maximum angle theta , which string makes with the vertical after hitting the pin is

    Statement I : A small unchanged solid conducting sphere is supended from a fixed point O by a light insulating string (figure 1). The system is in equilibrium. Now a horizontal unifrom electric field E is switched on As a result the conducting sphere is deflected towards left as shown (figurte 2) Statement II : When a solid conducting sphere having zero net charge is placed in uniform electric field, charges are inducned on the surface of sphere.

    A simple pendulum oscillates about mean position making an angle theta with the vertical. m is the mass of to bob of pendulum. l is the length of string of pendulum. Tension in the string having a velocity v is given by

    Two point masses m each carrying charge -q and +q are attached to the ends of a massless rigid non-conducting rod of length l. The arrangement is placed in a uniform electric field E such that the rod makes a small angle 5^(@) with the field direction. The minimum time needed by the rod to align itself along the field is ..................

    A small sphere of mass m=0.6kg carrying a positive charge q=80muC is connected with a light, flexible and inextensible string of length r=30cm and whirled in a vertical circle. If a horizontal rightward electric field of strength E=10^(5)NC^(-1) exists in the space, then the minimum velocity of the sphere required at the highest point so that it may just complete the circle will be (g=10ms^(-2))

    Home
    Profile