Home
Class 11
PHYSICS
An oil drop of radius r and density r is...

An oil drop of radius r and density r is held stationary in a uniform vertically upwards electric field 'E'. If `p_(0)=(ltp)` is the density of air and e is quanta of charge, then the drop has–

A

`(4pir^(3)(p-p_(0)))/(3eE)` excess electrons

B

`(4pir^(2)(p-p_(0)))/(eE)`excess electrons

C

deficiency of `(4pir^(3)(p-p_(0))g)/(3eE)` electrons

D

deficiency of `(4pir^(2)(p-p_(0))g)/(eE)`electrons

Text Solution

Verified by Experts

Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • DUAL NATURE OF MATTER AND RADIATION

    DISHA|Exercise PHYSICS|45 Videos

  • ELECTROMAGNETIC INDUCTION

    DISHA|Exercise PHYSICS|45 Videos

Similar Questions

Explore conceptually related problems

An oil drop of radius 2 mm with a density 3 g cm^(-3) is held stationary under a constant electric field 3.55 xx 10^5 V m^(-1) in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess ? Consider g=9.81 m//s^2

An oil drop of radiuss 2mm with density 3g/cm^-3 is held stationary under a constant vec(E) = 3.35 x 10^5 V/m in the milikan's drop experiment. What is number of excess electron that oil drop will possess (g=9.81)

Knowledge Check

  • An oil drop of radius r carrying a charge q remain stationary in the presence of electric field of intensity E. If the density of oil rho , then

    A
    `E=4/3pi r^(3)rhogq`
    B
    `E=4/3pir^(3)rho g`
    C
    `E=4/3pir^(3)rho g//q`
    D
    `E=4/3pir^(3)rho//g^(3)`

  • An oil drop of mass m and charge +q is balanced in vacuum by a uniform electric field of intensity E. the direction of this field should be

    A
    vertically up
    B
    vertically down
    C
    horizontal
    D
    inclined at `45^@` to the horizontal.

  • An oil drop of mass m and charge +q is balanced in vaccum by a uniform electric field of intensity E . the direction of this field should be

    A
    Vertically upwards
    B
    verticallu down
    C
    horizontal perpendicular to the direction of motion of the elctron beam
    D
    inclined at `45^(0)` to the horizontal

    • Similar Questions

    Explore conceptually related problems

    An oil drop of radiuss 2mm with density 3(g/(cm^3)) is held stationary under a constant vec(E) = 3.35 xx 10^5 V/m in the milikan's drop experiment. What is number of excess electron that oil drop will possess (g=9.81)m/s^2

    A uniform non conducting ring has mass m and radius R. tow point charges q and q' are fixed on its circumference at a separation of sqrt(2)R the ring remains in equilibrium in air with its plane vertical in a region where exists a uniform vertically upward electric field E. Given E=(4mg)/(7q) (a) Find angle theta in equilibrium position (see figure). (b) The ring is given a small rotation in the plane of the figure and released. Will it perform oscillations?

    An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 xx 10 ^(4) NC^(-1) in Millikan's oil drop experiment. The density of the oils is 1.26 g cm ^(-3) Estimate the radius of the drop (g= 9.81 m s ^(-2) and e = 1.60 xx 10 ^(-19) C )

    An oil drop of mass m fall through air with a terminal velocity in the presence of upward electric field of intensity E . the drop carries a charge +q . R is the viscous drag force and f is the buoyancy force . Them for the motion of the drop.

    An oil drop of 10 excess electron is held stationary under a consatnt electric field of 3.6xx10^(4)NC^(-1) in Millikan's oil drop experiment. The density of oil is 1.26gcm^(-3) . Radius of the oil drop is (Take, g=9.8ms^(-2), e=1.6xx10^(-19)C )

    Home
    Profile