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In Millikan's oil drop experiment, a cha...

In Millikan's oil drop experiment, a charged drop of mass `1.8 xx 10^(-14)` kg is stationary between its plates. The distance between the plates is 0.90 cm. The potential difference is 2.0 kV. The number of electrons on the drop is:

A

500

B

5

C

5

D

zero

Text Solution

Verified by Experts

The correct Answer is:
C
`E = (V)/(d) = (2000)/( 0. 9 xx 10 ^(-2)) = (2)/(9) xx 10 ^(6) V // m`
Since oil drop is stationary, therefore, (ne) E = mg
`n = ( mg)/(cE) =(1.8 xx 10 ^(-14)xx 10 xx 9)/( 1. 6 xx 10 ^(-19) xx 2 xx 10 ^(6))= 5`
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Knowledge Check

  • In Millikan's oil drop experiment, a charged drop of mass 1.8 xx106(-14) kg is stationary between its plates. The distance between the plates is 0.90 cm. The potential difference is 2.0kV. The number of electrons on the drop is:

    A
    500
    B
    50
    C
    5
    D
    zero

  • In 'Millikan's oil drop experiment' an oil drop carrying a charge Q is held stationary by a p.d. 2400V between the plates. To keep a drop of half the radius stationary the potential difference had to be made 600 V. What is the charge on the second drop?

    A
    `(3Q)/(2)`
    B
    `(Q)/(4)`
    C
    Q
    D
    `(Q)/(2)`

  • A cathode ray tube contains pair of parallel metal plates 1.0 cm apart and 3.0 cm long. A narrow horizontal beam of electrons with a velocity of 3 xx 10^7 ms^(-1) is passed down the tube mid way between the two plates. When the potential difference of 550 V is maintained across the plates, it is found that the electron beam is so deflected that it just strikes the end of one of the plates. Then the specific charge of an electron in C/kg is

    A
    `1.8 xx 10^9`
    B
    `1.8 xx 10^(11)`
    C
    `3.6 xx 10^(12)`
    D
    `3.6 xx10^(14)`

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