Millikan’s famous oil drop experiment which brought him a Nobel Prize in Physics will always shine brightly in the memory of any one who loves Physics. Let us consider the following M.C.Q. related to Millikan’s experiment:

In the case of the first drop, we have

w = (400/d)×Q ……….. (1)

where ‘w’ is the apparent weight of the first drop and ‘d’ is the separation between the plates.

In the case of the second drop, the apparent weight is w/8 since the volume (and hence the mass) of the drop is reduced to one-eighth

[Note that the volume is directly proportional to the cube of the radius]

Therefore we have,

w/8 = (600/d)×q ………... (2)

where ‘q’ is the charge on the second drop. Dividing eq(1) by eq(2), we obtain q = Q/12.

Consider now the following simple question which is meant for high lighting the

Remember that the minimum electric charge in nature is the electronic charge, which is equal to 1.6×10

The number of electrons passing per second through any section of a conductor or a region of space to produce a given current is a constant and is independent of the voltage.

Consider the following M.C.Q. which appeared in the I.I.T. screening test paper of 2002:

As the current is 3.2mA, the charge reaching the target per second is 3.2 millicoulomb. Since the electronic charge is 1.6×10

The accelerating voltage of 15kV is just a distraction in the question.

**In an oil drop experiment (Millikan’s), an oil drop carrying a charge Q is held stationary between the plates by applying a potential difference of 400V. To keep another drop of half the radius stationary, the potential difference had to be increased to 600V. The charge on the second drop is**

(a) Q/24

(b) Q/12

(c) Q/6

(d) 3Q/2

(e) 2Q/3(a) Q/24

(b) Q/12

(c) Q/6

(d) 3Q/2

(e) 2Q/3

In the case of the first drop, we have

w = (400/d)×Q ……….. (1)

where ‘w’ is the apparent weight of the first drop and ‘d’ is the separation between the plates.

In the case of the second drop, the apparent weight is w/8 since the volume (and hence the mass) of the drop is reduced to one-eighth

[Note that the volume is directly proportional to the cube of the radius]

Therefore we have,

w/8 = (600/d)×q ………... (2)

where ‘q’ is the charge on the second drop. Dividing eq(1) by eq(2), we obtain q = Q/12.

Consider now the following simple question which is meant for high lighting the

*quantum**nature*of electric charge:**An experimenter obtained the following values for the charge (in coulomb) on five different drops in Millikan’s oil drop experiment. Which one is the most unlikely value?**

(a) 3.2×10

(b) 4.8×10

(c) 1.6×10

(d) 8×10

(e) 5.6×10(a) 3.2×10

^{-19}(b) 4.8×10

^{-18}(c) 1.6×10

^{-17}(d) 8×10

^{-18}(e) 5.6×10

^{-19}.Remember that the minimum electric charge in nature is the electronic charge, which is equal to 1.6×10

^{-19}coulomb. All charges will be integral multiples of this minimum value. So, the unlikely value is 5.6×10^{-19}coulomb [option (e)].The number of electrons passing per second through any section of a conductor or a region of space to produce a given current is a constant and is independent of the voltage.

Consider the following M.C.Q. which appeared in the I.I.T. screening test paper of 2002:

**The potential difference applied to an X-ray tube is 15kV and the current through it is 3.2mA. Then the number of electrons striking the target per second is**

(a) 2×10

(b) 5×10

(c) 1×10

(d) 4×10(a) 2×10

^{16}(b) 5×10

^{6}(c) 1×10

^{17}(d) 4×10

^{15}As the current is 3.2mA, the charge reaching the target per second is 3.2 millicoulomb. Since the electronic charge is 1.6×10

^{-19}coulomb, the number of electrons striking the target per second is (3.2×10^{-3})/(1.6×10^{-19}) = 2×10^{16}.The accelerating voltage of 15kV is just a distraction in the question.

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