NEET UG Electrostatics & Current Electricity: Current Electricity – Chapter Test

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Difficulty: 10 medium · 10 hard

Current Electricity MCQ questions with answers

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

1. Q1. A potential difference of 2.0 V is applied across the ends of a conductor of length 1.0 m. If the mobility of the charge carriers in the conductor is 5.0 × 10⁻³ m² V⁻¹ s⁻¹, the drift speed of the carriers is:

   1.0 m लंबाई वाले किसी चालक के सिरों पर 2.0 V का विभवांतर आरोपित किया जाता है। यदि चालक में आवेश वाहकों की गतिशीलता 5.0 × 10⁻³ m² V⁻¹ s⁻¹ है, तो वाहकों की अपवाह चाल है:

   • A. 1.0 × 10⁻² m²/s
   • B. 2.5 × 10⁻³ m/s
   • C. 4.0 × 10⁻³ m/s
   • D. 1.0 × 10⁻² m/s
   Solution

   The field E = V/L = 2.0/1.0 = 2.0 V/m, and drift speed v_d = μE = 5.0 × 10⁻³ × 2.0 = 1.0 × 10⁻² m/s.

   क्षेत्र E = V/L = 2.0/1.0 = 2.0 V/m, तथा अपवाह चाल v_d = μE = 5.0 × 10⁻³ × 2.0 = 1.0 × 10⁻² m/s।

2. Q2. Kirchhoff's loop (voltage) rule for an electrical network is a direct consequence of the conservation of:

   किसी विद्युत नेटवर्क के लिए किरचॉफ का लूप (वोल्टेज) नियम निम्नलिखित में से किसके संरक्षण का प्रत्यक्ष परिणाम है?

   • A. Mass / द्रव्यमान
   • B. Electric charge / विद्युत आवेश
   • C. Linear momentum / रैखिक संवेग
   • D. Energy / ऊर्जा
   Solution

   The loop rule states that the algebraic sum of potential changes around a closed loop is zero, which expresses conservation of energy. (The junction rule expresses conservation of charge.)

   लूप नियम कहता है कि किसी बंद लूप के चारों ओर विभव परिवर्तनों का बीजगणितीय योग शून्य होता है, जो ऊर्जा संरक्षण को व्यक्त करता है। (संधि नियम आवेश संरक्षण को व्यक्त करता है।)

3. Q3. A metallic wire has a resistance of 50 Ω at 0 °C. If the temperature coefficient of resistance of the material is 4.0 × 10⁻³ °C⁻¹, its resistance at 100 °C is:

   एक धात्विक तार का 0 °C पर प्रतिरोध 50 Ω है। यदि पदार्थ के प्रतिरोध का ताप गुणांक 4.0 × 10⁻³ °C⁻¹ है, तो 100 °C पर इसका प्रतिरोध है:

   • A. 70 Ω
   • B. 54 Ω
   • C. 60 Ω
   • D. 90 Ω
   Solution

   Using R = R₀(1 + αΔT) = 50(1 + 4.0×10⁻³ × 100) = 50 × 1.4 = 70 Ω.

   R = R₀(1 + αΔT) = 50(1 + 4.0×10⁻³ × 100) = 50 × 1.4 = 70 Ω का उपयोग करते हुए।

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