NEET UG Electrostatics & Current Electricity: Electrostatic Potential and Capacitance – Chapter Test

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Difficulty: 9 medium · 11 hard

Electrostatic Potential and Capacitance MCQ questions with answers

The first 3 of 20 questions. Unlock the series for the rest, each with a solution.

1. Q1. Match List I (combination of identical capacitors, each of capacitance C) with List II (equivalent capacitance). List I: A. Two in series; B. Two in parallel; C. Three in series; D. Three in parallel. List II: 1. C/2; 2. 2C; 3. C/3; 4. 3C.

   सूची I (समान संधारित्रों का संयोजन, प्रत्येक की धारिता C) का सूची II (तुल्य धारिता) से मिलान करें। सूची I: A. दो श्रेणीक्रम में; B. दो समांतर क्रम में; C. तीन श्रेणीक्रम में; D. तीन समांतर क्रम में। सूची II: 1. C/2; 2. 2C; 3. C/3; 4. 3C।

   • A. A-3, B-4, C-1, D-2
   • B. A-1, B-2, C-4, D-3
   • C. A-2, B-1, C-4, D-3
   • D. A-1, B-2, C-3, D-4
   Solution

   Two in series give C/2, two in parallel give 2C, three in series give C/3, and three in parallel give 3C.

   दो श्रेणीक्रम में C/2 देते हैं, दो समांतर क्रम में 2C देते हैं, तीन श्रेणीक्रम में C/3 देते हैं, और तीन समांतर क्रम में 3C देते हैं।

2. Q2. Among the following electrical quantities, which one is the odd one out being a scalar quantity?

   निम्नलिखित विद्युत राशियों में से कौन-सी एक अदिश राशि होने के कारण भिन्न है?

   • A. Electric dipole moment / विद्युत द्विध्रुव आघूर्ण
   • B. Electric force / विद्युत बल
   • C. Electric potential / विद्युत विभव
   • D. Electric field intensity / विद्युत क्षेत्र तीव्रता
   Solution

   Electric potential is a scalar quantity, whereas electric field intensity, electric dipole moment and electric force are all vectors.

   विद्युत विभव एक अदिश राशि है, जबकि विद्युत क्षेत्र तीव्रता, विद्युत द्विध्रुव आघूर्ण तथा विद्युत बल सभी सदिश हैं।

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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