JEE Main Electrostatics and Current Electricity – Topic Test

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Difficulty: 15 medium · 15 hard

Electrostatics and Current Electricity MCQ questions with answers

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

1. Q1. A dielectric slab of dielectric constant K is placed in a uniform external electric field of magnitude E0 (the field that would exist due to the free charges alone, in the absence of the dielectric). The magnitude of the induced (bound) surface charge density that appears on the faces of the slab perpendicular to the field is:

   K परावैद्युतांक वाले एक परावैद्युत स्लैब को E0 परिमाण (वह क्षेत्र जो परावैद्युत की अनुपस्थिति में केवल मुक्त आवेशों के कारण उपस्थित होता) के एक समान बाह्य विद्युत क्षेत्र में रखा जाता है। क्षेत्र के लंबवत स्लैब के फलकों पर प्रकट होने वाले प्रेरित (बद्ध) पृष्ठ आवेश घनत्व का परिमाण है:

   • A. ε0E0(K - 1)
   • B. ε0E0(1 - 1/K)
   • C. ε0E0K
   • D. ε0E0/K
   Solution

   Inside the dielectric the net field is reduced to E0/K because the bound charges produce an opposing field; writing E = E0 - σb/ε0 = E0/K and solving gives σb = ε0E0(1 - 1/K).

   परावैद्युत के अंदर, बद्ध आवेशों द्वारा उत्पन्न विपरीत क्षेत्र के कारण निकाय क्षेत्र घटकर E0/K हो जाता है; E = E0 - σb/ε0 = E0/K लिखकर हल करने पर σb = ε0E0(1 - 1/K) प्राप्त होता है।

2. Q2. A uniform wire of length 2 m and cross-sectional area 0.5 mm² has a resistance of 20 Ω. The resistivity of the material of the wire is:

   2 m लंबाई और 0.5 mm² अनुप्रस्थ-काट क्षेत्रफल के एक समान तार का प्रतिरोध 20 Ω है। तार के पदार्थ की प्रतिरोधकता है:

   • A. 1 × 10⁻⁵ Ω·m
   • B. 2 × 10⁻⁶ Ω·m
   • C. 2.5 × 10⁻⁶ Ω·m
   • D. 5 × 10⁻⁶ Ω·m
   Solution

   Resistivity ρ = RA/L = 20 × (0.5 × 10⁻⁶)/2 = 5 × 10⁻⁶ Ω·m.

   प्रतिरोधकता ρ = RA/L = 20 × (0.5 × 10⁻⁶)/2 = 5 × 10⁻⁶ Ω·m।

3. Q3. A potentiometer is used to compare the EMFs of two cells. The balancing lengths for the two cells are 250 cm and 400 cm respectively. The ratio of their EMFs E₁ : E₂ is:

   दो सेलों के विद्युत वाहक बलों (EMF) की तुलना करने के लिए एक पोटेंशियोमीटर का उपयोग किया जाता है। दोनों सेलों के लिए संतुलन लंबाइयाँ क्रमशः 250 cm और 400 cm हैं। उनके विद्युत वाहक बलों का अनुपात E₁ : E₂ है:

   • A. 8 : 5
   • B. 5 : 8
   • C. 25 : 16
   • D. 2 : 3
   Solution

   At balance EMF is proportional to balancing length, so E₁ : E₂ = 250 : 400 = 5 : 8.

   संतुलन पर EMF, संतुलन लंबाई के समानुपाती होता है, अतः E₁ : E₂ = 250 : 400 = 5 : 8।

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