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Sanjay01 Kumar
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Sanjay01 Kumar
Asked: March 28, 2025In: Class 12 Physics

Two resistors of resistance, 100 Ω and 200 Ω are connected in parallel in an electric circuit. The ratio of the thermal energy developed in 100 Ω to that in 200 Ω in a given time is

In a parallel circuit, power P = V²/R. Since both resistors have the same voltage, the thermal energy ratio is inversely proportional to resistance. Thus, E100Ω : E 200Ω = 200 : 100 = 2:1. Answer: (b) 2:1.

chapter 3class 12thmcqphysics
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Sanjay01 Kumar
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Sanjay01 Kumar
Asked: March 28, 2025In: Class 12 Physics

Two batteries with emf 12 V and 13 V are connected in parallel across a load resistor of 10 Ω. The internal resistances of the two batteries are 1 Ω and 2 Ω respectively. The voltage across the load lies between

Using the formula for parallel battery circuits: V = 12/1 + 13/2 /1/1 + 1/2 = 12 + 6.5/1.5 = 11.67V Thus, the answer is (a) 11.6 V to 11.7 V.

chapter 3class 12thmcqphysics
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Sanjay01 Kumar
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Sanjay01 Kumar
Asked: March 27, 2025In: Class 12 Physics

A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of the emf is :

Let the emfs be E₁ and E₂. From the potentiometer principle: E₁ + E₂/E₁ – E₂ = 50/10 = 5 Solving, E₁/E₂ = 5/1. Answer: (d) 5:1.

chapter 3class 12thmcqphysics
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Sanjay01 Kumar
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Sanjay01 Kumar
Asked: March 27, 2025In: Class 12 Physics

A resistance of 4 Ω is connected across a cell. Then it is replaced by another resistance of 1 Ω. It is found that power dissipated in resistance in both the cases is 16 watt. Then

Using P = V²R/(R+r)² , equating power for R = 4Ω and R = 1Ω, solving gives r = 2Ω and E = 12V. Maximum power = 18W. Answer: (c).

chapter 3class 12thmcqphysics
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Sanjay01 Kumar
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Sanjay01 Kumar
Asked: March 27, 2025In: Class 12 Physics

The charge flowing in a conductor varies with time as Q = at – bt², The current:

The current is given by I = dQ/dt = a – 2bt. It starts at, reaches a maximum at t = 0 and decreases linearly with time. Thus, the correct answer is (a) decreases linearly with time.

chapter 3class 12thmcqphysics
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