R.A. Millikan set up as apparatus in which a tiny, charged oil drop placed in an electric field E could be balanced by adjusting E until the electric force on the drop was equal and opposite to its weight. He measured the drop radius by observing the limiting speed that the drop attained when the drop fell in air with the electric field turned off. Millikan determined r by cutting off the electric field and measuring the terminal speed v_t of the drop as it fell. Now, the magnitude of charge on the drop is given by

He charged the oil drops by irradiating them with burst of x-rays. The radius of drop is 1.64 x 10 ^-4 cm, E at balance is equal to 1.92 x 10⁵ N/C and density of the oil is 0.851 gm/cm³

1.         What is the nature of charge if E is in downward direction?

What is the charge on the drop? (A) (B) (C) (D)

If oil in the experiment is replaced with another liquid of density just half that of oil, then number of electrons in the drop would be

What is the expression for the charge in terms of potential difference? (Assume the air drag is negligible)

(A) (B) (C) (D)

A charged oil drop in a Millikan oil drop apparatus is observed to fall through 1.00 mm at a constant speed in 39.3s if VAB = 0. The same drop can be held at rest between the two plates separately by 1.00mm if VAB = 9.16V. Then find the radius of the drop

(A) (B) (C) (D)