Class 12

Poll

Vivek Kohli

Asked: February 21, 2025In: Class 12 Physics

In a potentiometer of wire length l, a cell of emf V is balanced at a length l/3 from the positive end of the wire. For another cell of emf 1-5 V, the balancing length becomes

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 ...

Poll

Vivek Kohli

Asked: February 21, 2025In: Class 12 Physics

In a wheatston’ bridge, P = 2 Ω, Q = 2 Ω, R = 2 Ω and S = 3 Ω. The resistance with which 5 is to be shunted in order that the bridge may be balanced is

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.

Poll

Vivek Kohli

Asked: February 21, 2025In: Class 12 Physics

In a meter bridge experiment, resistance box (with R = 2 Ω) is connected in the left gap and the unknown resistance S in the right gap. If the balancing length be 40 cm, calculate the value of S

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 ...

Poll

Vivek Kohli

Asked: February 21, 2025In: Class 12 Physics

For a cell of emf 2 V, a balance is obtained for 50 cm of the potentiometer wire. If the cell is shunted by a 2 n resistor and the balance is obtained across 40 cm of the wire, then the internal resistance of the cell is:

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

Poll

Vivek Kohli

Asked: February 21, 2025In: Class 12 Physics

Two cells when connected in series are balanced on 8 m on a potentiometer. If the cells are connected with polarities of one of the cell is reversed, they balance on 2 m. The ratio of the cmf’s of the two

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.

Category: Class 12

In a potentiometer of wire length l, a cell of emf V is balanced at a length l/3 from the positive end of the wire. For another cell of emf 1-5 V, the balancing length becomes

In a wheatston’ bridge, P = 2 Ω, Q = 2 Ω, R = 2 Ω and S = 3 Ω. The resistance with which 5 is to be shunted in order that the bridge may be balanced is

In a meter bridge experiment, resistance box (with R = 2 Ω) is connected in the left gap and the unknown resistance S in the right gap. If the balancing length be 40 cm, calculate the value of S

For a cell of emf 2 V, a balance is obtained for 50 cm of the potentiometer wire. If the cell is shunted by a 2 n resistor and the balance is obtained across 40 cm of the wire, then the internal resistance of the cell is:

Two cells when connected in series are balanced on 8 m on a potentiometer. If the cells are connected with polarities of one of the cell is reversed, they balance on 2 m. The ratio of the cmf’s of the two

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