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NEET PHYSICSMedium

An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55×1042.55 \times 10^4 N/C. The density of the oil is 1.26 g/cm31.26 \text{ g/cm}^3. The radius of the drop is:

A

9.82×10−49.82 \times 10^{-4} mm

B

9.82×10−79.82 \times 10^{-7} mm

C

8.92×10−48.92 \times 10^{-4} mm

D

8.92×10−78.92 \times 10^{-7} mm

Step-by-Step Solution

For the oil drop to remain stationary, the upward electric force must balance the downward gravitational force: qE=mgqE = mg.

  1. Calculate charge q=ne=12×1.6×10−19 Cq = ne = 12 \times 1.6 \times 10^{-19} \text{ C}.
  2. Express mass in terms of density (ρ\rho) and radius (rr): m=ρ×43πr3m = \rho \times \frac{4}{3}\pi r^3. (Note: Convert density to SI units: 1.26 g/cm3=1.26×103 kg/m31.26 \text{ g/cm}^3 = 1.26 \times 10^3 \text{ kg/m}^3).
  3. Equate forces: qE=ρ43πr3gqE = \rho \frac{4}{3}\pi r^3 g.
  4. Solve for rr: r=(3qE4πρg)1/3r = \left( \frac{3qE}{4\pi \rho g} \right)^{1/3}. Substituting values: r=(3×12×1.6×10−19×2.55×1044×3.14×1260×9.81)1/3≈9.82×10−7 mr = \left( \frac{3 \times 12 \times 1.6 \times 10^{-19} \times 2.55 \times 10^4}{4 \times 3.14 \times 1260 \times 9.81} \right)^{1/3} \approx 9.82 \times 10^{-7} \text{ m}.
  5. Convert to mm: 9.82×10−7 m=9.82×10−4 mm9.82 \times 10^{-7} \text{ m} = 9.82 \times 10^{-4} \text{ mm}. (See NCERT Physics Class 12, Exercise 1.25).
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