JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Fluid Mechanics

A tiny spherical oil drop carrying a net charge

q

is balanced in still air with a vertical uniform electric field of strength

\frac{81 π}{7} \times 1 0^{5} Vm^{- 1}

. When the field is switched off, the drop is observed to fall with terminal velocity

2 \times 1 0^{- 3} m s^{- 1}

. Given

g = 9.8 m s^{- 2}

, viscosity of the air

= 1.8 \times 1 0^{- 5} Ns m^{- 2}

and the density of oil

= 900 kg m^{- 3}

, the magnitude of

q

Answer: (D)

Solution

When balanced:

q E = m g = \frac{4}{3} π R^{3} ρ g

At terminal velocity:

6 π η R v_{T} = m g = \frac{4}{3} π R^{3} ρ g

From the terminal velocity equation:

R = \frac{9 η v _{T}}{2 ρ g}

Substituting values:

R = \frac{9 \times 1.8 \times 1 0 ^{- 5} \times 2 \times 1 0 ^{- 3}}{2 \times 900 \times 9.8} \approx 1 0^{- 6} m

Now,

q = \frac{m g}{E} = \frac{\frac{4}{3} π R ^{3} ρ g}{E}

Substituting and calculating gives

q = 8.0 \times 1 0^{- 19}

C.

The answer is **(D)

8.0 \times 1 0^{- 19}

C**.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Single correct option