The internal resistance r of the cell is given by: r=R l1/l2−1 Given l1 =120 cm, l2 = 40 cm and R=1Ω: r = 1 ×120/40 −1 = 2Ω Answer: (d) 2Ω. Class 12th Science Physics NCERT MCQ Questions NCERT Books ...
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The energy dissipated as heat in a resistor is given by: Q = P × t where power P is: P = V²/R Given: V =15V,R=50Ω, t = 1 minute = 60 seconds. P = 15²/50 = 225/50 = ...
The heat produced by a heater is given by the formula: Q = P × t where: P = 100W,t = 2 minutes = 2 × 60 = 120 seconds Q = 100 × 120 =12,000 J = 12.0 kJ Answer: (c) 12.0 ...
In a potentiometer, the balancing length is proportional to the emf of the cell. Given V balances at l/3, for a cell of emf 1.5V, the new balancing length is (1.5V/V) × (l/3) = 2l/3. Answer: (d) 2l/3. Class 12th ...
For balance in a Wheatstone bridge: P/Q = Rshunted/Seffective Solving for R_shunted, we get 2Ω. Correct answer: (b) 2Ω. Class 12th Science Physics NCERT MCQ Questions NCERT Books MCQ Questions Session 2024-2025.
Using the meter bridge formula R/S = L1/(100 – L1), we substitute R = 2Ω, L1 = 40 cm: 2/S = 40/60 Solving, S = 3Ω. Correct answer: (b) 3Ω. Class 12th Science Physics NCERT MCQ Questions NCERT Books MCQ ...
Using the potentiometer principle: E∝L Initially, the balance length is L₁ = 50 cm. When the cell is shunted by Rₛ = 2 Ω, the new balance length is L₂ = 40 cm. Class 12th Science Physics NCERT MCQ Questions
Let the EMFs of the two cells be E₁ and E₂. In series: E₁ + E₂ ∝ 8 m With one reversed: E₁ – E₂ ∝ 2 m Solving, (E₁ + E₂) / (E₁ – E₂) = 8/2 = 4.
Total resistance is 6Ω, current is 5/3 A, potential drop across the wire is 5V and the potential gradient is 10 mV/cm. Correct answer: (b) 10 mV/cm. Class 12th Science Physics NCERT MCQ Questions NCERT Books MCQ Questions Session 2024-2025.
Using power formulas for series and parallel combinations, the ratio of power consumed by R₁ and R₂ is 1/4. Correct answer: (a) 1/4. Class 12th Science Physics NCERT MCQ Questions NCERT Books MCQ Questions Session 2024-2025.