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A charged oil drop is suspended in a uniform field at 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 and g=10 m/s²)
Options
(a) 3.3×10⁻¹⁸ C
(b) 3.2×10⁻¹⁸ C
(c) 1.6×10⁻¹⁸ C
(d) 4.8×10⁻¹⁸ C
Correct Answer:
3.3×10⁻¹⁸ C
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- When three identical bulbs of 60 W-200 V rating are connected in series to a 200 V
- The time by a photoelectron to come out after the photon strikes is approximately
- The current through an inductor changes from 3 A to 2 A is 1 m/s. If induced e.m.f.
- The period of oscillation of a mass M suspended from a spring of negligible mass is T
- In a given reaction, ᴢXᴬ → ᴢ+1Yᴬ → ᴢ-1Kᴬ⁻⁴ → ᴢ-1Kᴬ⁻⁴
Topics: Dual Nature of Matter and Radiation (150)
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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