NEET UG Electrostatics & Current Electricity – Topic Test

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Difficulty: 15 medium · 15 hard

Electrostatics & Current Electricity MCQ questions with answers

The first 3 of 30 questions. Unlock the series for the rest, each with a solution.

1. Q1. Consider the following statements concerning an ideal electric dipole of moment p placed in a uniform external electric field E: 1. The net translational force on the dipole is zero. 2. The torque on the dipole is zero when p is antiparallel to E. 3. The potential energy of the dipole is maximum when p is antiparallel to E. Which of these statements is/are correct?

   एकसमान बाह्य विद्युत क्षेत्र E में रखे गए आघूर्ण p के एक आदर्श विद्युत द्विध्रुव के संबंध में निम्नलिखित कथनों पर विचार कीजिए: 1. द्विध्रुव पर नेट स्थानांतरीय बल शून्य होता है। 2. जब p, E के प्रतिसमानांतर होता है तब द्विध्रुव पर बल आघूर्ण शून्य होता है। 3. द्विध्रुव की स्थितिज ऊर्जा तब अधिकतम होती है जब p, E के प्रतिसमानांतर होता है। इनमें से कौन-सा/से कथन सही है/हैं?

   • A. 1 and 2 only / केवल 1 और 2
   • B. 1 and 3 only / केवल 1 और 3
   • C. 2 and 3 only / केवल 2 और 3
   • D. 1, 2 and 3 / 1, 2 और 3
   Solution

   In a uniform field the forces on the two charges are equal and opposite, so the net force is zero (1 correct). Torque = pE sinθ, which vanishes for θ = 180° (2 correct). Potential energy U = −pE cosθ is maximum (+pE) at θ = 180° (3 correct), so all three hold.

   एकसमान क्षेत्र में दोनों आवेशों पर बल बराबर और विपरीत होते हैं, अतः नेट बल शून्य होता है (1 सही)। बल आघूर्ण = pE sinθ, जो θ = 180° के लिए शून्य हो जाता है (2 सही)। स्थितिज ऊर्जा U = −pE cosθ, θ = 180° पर अधिकतम (+pE) होती है (3 सही), अतः तीनों सही हैं।

2. Q2. The electric potential at a distance of 0.3 m from an isolated point charge of 2 nC in vacuum (take 1/4πε₀ = 9 × 10⁹ N·m²/C²) is:

   निर्वात में 2 nC के एक विलगित बिंदु आवेश से 0.3 m दूरी पर विद्युत विभव (1/4πε₀ = 9 × 10⁹ N·m²/C² लें) है:

   • A. 18 V
   • B. 6 V
   • C. 60 V
   • D. 600 V
   Solution

   V = (1/4πε₀)(q/r) = (9 × 10⁹)(2 × 10⁻⁹)/0.3 = 18/0.3 = 60 V.

   V = (1/4πε₀)(q/r) = (9 × 10⁹)(2 × 10⁻⁹)/0.3 = 18/0.3 = 60 V।

3. Q3. The capacitance of an isolated conducting sphere of radius 9 cm placed in vacuum is approximately (take 1/4πε₀ = 9 × 10⁹ N·m²/C²):

   निर्वात में रखे गए 9 cm त्रिज्या के एक पृथक्कृत चालक गोले की धारिता लगभग है (1/4πε₀ = 9 × 10⁹ N·m²/C² लीजिए):

   • A. 90 pF
   • B. 10 pF
   • C. 100 pF
   • D. 1 pF
   Solution

   For an isolated sphere C = 4πε₀R = R/(9×10⁹) = 0.09/(9×10⁹) = 1.0 × 10⁻¹¹ F = 10 pF.

   एक पृथक्कृत गोले के लिए C = 4πε₀R = R/(9×10⁹) = 0.09/(9×10⁹) = 1.0 × 10⁻¹¹ F = 10 pF।

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