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Vivek Kohli

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Vivek Kohli
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Vivek Kohli
Asked: February 22, 2025In: Class 12 Physics

The electric power consumed by a 220 V-100 W bulb when operated at 110 V is:

The resistance of the bulb is R = 220/100 = 484Ω. When operated at 110V, power consumed is P ′ = 110²/484 = 25W. Correct answer: (a) 25 W. Class 12th Science Physics NCERT MCQ Questions NCERT Books MCQ Questions ...

class 12thmcqscience
  • 1 1 Answer
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Vivek Kohli
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Vivek Kohli
Asked: February 22, 2025In: Class 2

Two students A and B calculate the charge flowing through a circuit. A concludes that 300 C of charge flows in 1 minute. B concludes that 3.125 x 10¹⁹ electrons flow in 1 second. If the current measured in the circuit is 5 A, then the correct calculation is done by

Current I is given by I = Q/t. Student A: Q = 300C. t = 60s, so I = 300/60 = 5A(correct). Student B: Q = (3.125 × 10¹⁹) × (1.6 × 10⁻¹⁹) = 5C, t = Is, so I ...

class 12thmcqscience
  • 2 2 Answers
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Vivek Kohli
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Vivek Kohli
Asked: February 22, 2025In: Class 12 Physics

When a potential difference V is applied across a conductor at temperature T, the drift velocity of the electrons is proportional to :

The drift velocity vd of electrons in a conductor is given by the equation: vd = eEr/m Since E = V/L, we get: vd = eVr/mL Here, vd is directly proportional to the applied potential difference V. Thus, the correct ...

class 12thscience
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Vivek Kohli
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Vivek Kohli
Asked: February 21, 2025In: Class 12 Physics

In a potentiometer experiment, the balancing length with a cell is 120 cm. When the cell is shunted by a 1Ω resistance, the balancing length becomes 40 cm. The internal resistance of the cell is :

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 ...

class 12thmcqscience
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Vivek Kohli
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Vivek Kohli
Asked: February 21, 2025In: Class 12 Physics

A battery of 15 V and negligible internal resistance is connected across a 50 Ω resistor. The amount of energy dissipated as heat in the resistor in one minute is :

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 = ...

class 12thmcqscience
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