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A charged oil drop is suspended in a uniform field of 3 × 10^4 ~V / m so that it neither falls nor rises. The charge on the drop will be (take the…

JEE Main 2004 — Physics Electromagnetism

2004mcqeasy

A charged oil drop is suspended in a uniform field of 3×104 V/m3 \times 10^4 \mathrm{~V} / \mathrm{m}3×104 V/m so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge =9.9×10−15 kg=9.9 \times 10^{-15} \mathrm{~kg}=9.9×10−15 kg and g=\mathrm{g}=g= 10 m/s210 \mathrm{~m} / \mathrm{s}^210 m/s2 )

Official previous-year question

Held on 30 Apr 2004 · Verified 6 Jul 2026.

Options

  1. A

    3.3×10−18C3.3 \times 10^{-18} \mathrm{C}3.3×10−18C

  2. B

    3.2×10−18C3.2 \times 10^{-18} \mathrm{C}3.2×10−18C

  3. C

    1.6×10−18C1.6 \times 10^{-18} \mathrm{C}1.6×10−18C

  4. D

    4.8×10−18C4.8 \times 10^{-18} \mathrm{C}4.8×10−18C

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