Class 12

Poll

Gopal001 Das

Asked: March 6, 2025In: Class 12 Physics

Two wires A and B of the same material having length in the ratio 1 : 2 and diameter in the ratio 2 : 3 are connected in series with a battery. The ratio of potential differences (Va/Vb) across the two wires respectively is

Resistance is given by R = p L/A . Given La/Lb = 1/2 and da/db = 2/3, so Aa/Ab = 2/3 2 = 4/g. thus, Ra/Rb = (1/2)/(4g) = 9/8. Since V ∝ R in series, Va/Vb = Ra/Rb = ...

Poll

Gopal001 Das

Asked: March 6, 2025In: Class 12 Physics

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

For balance, p/Q = R/Seq . Given S = 3Ω, let shunt resistance be x. Using Seq = 3x/3+x = 2Ω. solving gives x = 2Ω. Answer: (b) 2Ω. Class 12th Science Physics NCERT MCQ Questions NCERT Books MCQ Questions ...

Poll

Gopal001 Das

Asked: March 6, 2025In: Class 12 Physics

In a potentiometer of wire length /, a cell of emf V is balanced at a length U3 from the positive end of the wire. For another cell of emf 1-5 V, the balancing length becomes

In a potentiometer, emf ∝ balancing length. Given v/1.5V = U3/Lx , solving for Lx gives Lx = L/3. Answer: (c) l/3. Class 12th Science Physics NCERT MCQ Questions NCERT Books MCQ Questions Session 2024-2025.

Poll

Janvi Shinha

Asked: February 25, 2025In: Class 12 Physics

In a meter bridge experiment, resistance box (with R = 2 n) 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

In a meter bridge, the balancing condition is R / S = L / (100 – L). Given R = 2 Ω and L = 40 cm, we get 2 / S = 40 / 60. Solving for S, S ...

Poll

Janvi Shinha

Asked: February 25, 2025In: Class 12 Physics

A cell supplies a current of 0-9 A through a 2Ω resistor and a current of 0.3 A through 7Ω resistor. The internal resistance of the cell is :

Let the EMF of the cell be E and internal resistance be r. Using Ohm’s law:For 2Ω resistor: E = 0.9 (2 + r) For 7 resistor E = 0.3 (7 + r) Equating both: 0.9(2 + r) = 0.3 ...

Category: Class 12

Two wires A and B of the same material having length in the ratio 1 : 2 and diameter in the ratio 2 : 3 are connected in series with a battery. The ratio of potential differences (Va/Vb) across the two wires respectively is

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

In a potentiometer of wire length /, a cell of emf V is balanced at a length U3 from the positive end of the wire. For another cell of emf 1-5 V, the balancing length becomes

In a meter bridge experiment, resistance box (with R = 2 n) 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

A cell supplies a current of 0-9 A through a 2Ω resistor and a current of 0.3 A through 7Ω resistor. The internal resistance of the cell is :

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