JEE Main Electrostatics and Current Electricity: Current Electricity – Chapter Test
JEE Main Electrostatics and Current Electricity: Current Electricity – Chapter Test is an online chapter test in the JEE Main pattern, covering Electrostatics and Current Electricity. Attempt it on SeatBeat with a timer like the real exam, then see a detailed solution for every question and your score and weak-topic analysis.
Topics in this test
| Topic | Questions |
|---|---|
| Electrostatics and Current Electricity | 20 |
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.
Q1. Three resistors of 2 Ω, 3 Ω and 6 Ω are connected in parallel and the combination is joined in series with a 4 Ω resistor. The total resistance of the network is:
2 Ω, 3 Ω और 6 Ω के तीन प्रतिरोधक समांतर क्रम में जुड़े हैं और यह संयोजन एक 4 Ω प्रतिरोधक के साथ श्रेणीक्रम में जोड़ा गया है। नेटवर्क का कुल प्रतिरोध है:
- A. 4 Ω
- B. 5 Ω
- C. 6 Ω
- D. 11 Ω
Solution
Parallel combination: 1/R = 1/2 + 1/3 + 1/6 = 1, so R = 1 Ω. Adding the series 4 Ω gives 5 Ω.
समांतर संयोजन: 1/R = 1/2 + 1/3 + 1/6 = 1, अतः R = 1 Ω। श्रेणी में 4 Ω जोड़ने पर 5 Ω मिलता है।
Q2. The temperature coefficient of resistance of a conductor is defined using the relation R_T = R₀(1 + αΔT). For a material whose resistance increases from 100 Ω at 0 °C to 100.4 Ω at 10 °C, the value of α is:
किसी चालक के प्रतिरोध का ताप गुणांक संबंध R_T = R₀(1 + αΔT) का उपयोग करके परिभाषित किया जाता है। एक ऐसे पदार्थ के लिए जिसका प्रतिरोध 0 °C पर 100 Ω से बढ़कर 10 °C पर 100.4 Ω हो जाता है, α का मान है:
- A. 4 × 10⁻³ °C⁻¹
- B. 4 × 10⁻⁵ °C⁻¹
- C. 4 × 10⁻⁴ °C⁻¹
- D. 4 × 10⁻² °C⁻¹
Solution
α = (R_T − R₀)/(R₀ ΔT) = 0.4/(100×10) = 4×10⁻⁴ per °C.
α = (R_T − R₀)/(R₀ ΔT) = 0.4/(100×10) = 4×10⁻⁴ प्रति °C।
Q3. Assertion (A): In the balanced condition of a potentiometer, the measurement of EMF of a cell is independent of the internal resistance of that cell. Reason (R): At balance no current is drawn from the cell being measured, so there is no potential drop across its internal resistance.
अभिकथन (A): पोटेंशियोमीटर की संतुलित अवस्था में, किसी सेल के विद्युत वाहक बल (EMF) का मापन उस सेल के आंतरिक प्रतिरोध से स्वतंत्र होता है। कारण (R): संतुलन पर मापे जा रहे सेल से कोई धारा नहीं ली जाती, अतः उसके आंतरिक प्रतिरोध पर कोई विभव पात नहीं होता।
- A. Both A and R are true and R is the correct explanation of A / A और R दोनों सही हैं तथा R, A की सही व्याख्या है
- B. A is true but R is false / A सही है परंतु R गलत है
- C. A is false but R is true / A गलत है परंतु R सही है
- D. Both A and R are true but R is not the correct explanation of A / A और R दोनों सही हैं परंतु R, A की सही व्याख्या नहीं है
Solution
At balance the galvanometer shows no deflection, meaning zero current through the test cell, so no Ir drop occurs and the measured value equals the true EMF — both A and R are true and R explains A.
संतुलन पर गैल्वेनोमीटर में कोई विक्षेप नहीं होता, अर्थात परीक्षण सेल से कोई धारा नहीं बहती, अतः कोई Ir पात नहीं होता और मापा गया मान वास्तविक EMF के बराबर होता है — A और R दोनों सही हैं तथा R, A की व्याख्या करता है।
Frequently asked questions
How many questions are in the JEE Main Electrostatics and Current Electricity: Current Electricity – Chapter Test?
20 questions for 20 marks, to be answered in 20 minutes.
Is there negative marking in the JEE Main Electrostatics and Current Electricity: Current Electricity – Chapter Test?
Yes — 0.25 marks are deducted for each wrong answer, as in the real exam pattern.
Which topics does the JEE Main Electrostatics and Current Electricity: Current Electricity – Chapter Test cover?
Electrostatics and Current Electricity (20 questions).
Is the JEE Main Electrostatics and Current Electricity: Current Electricity – Chapter Test free?
It is part of the JEE Main 2027 Mock Test Series (104 tests, ₹299). You can first try Basic Concepts & Atomic Structure – Topic Test for free.
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