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Home/class 12th/Page 3

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Vatsal Mishra
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Vatsal Mishra
Asked: April 4, 2025In: Class 12 Physics

If the magnetizing field on a ferromagnetic material is increased, its permeability

In ferromagnetic materials, as the magnetizing field increases, permeability initially increases, reaches a maximum and then decreases due to magnetic saturation. So, permeability does not keep increasing. Correct answer: (A) decreases.

class 12thmagnetism and mattermcqphysics
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Vatsal Mishra
Poll
Vatsal Mishra
Asked: April 4, 2025In: Class 12 Physics

A resistance of 980 Ω is connected in series with a voltmeter, after which the scale division becomes 50 times larger. The resistance of voltmeter is

Let voltmeter resistance be R. Since scale becomes 50 times larger, current reduces by 50 times. So, total resistance becomes 50R. Given: 50R = R + 980 ⇒ 49R = 980 ⇒ R = 20 Ω. Answer: (B) 20 Ω.

chapter 4 moving charges and magnetismclass 12thmcqphysics
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Vatsal Mishra
Poll
Vatsal Mishra
Asked: April 4, 2025In: Class 12 Physics

A galvanometer of resistance 3663 Ω gives full scale deflection for a certain current Ig. The value of the resistance of the shunt which when joined to the galvanometer coil will result in 1/34 of the total current passing through the galvanometer is

(A) is correct. If only 1/34 of total current passes through the galvanometer, then shunt resistance Rs = Rg/33 = 3663/33 ≈ 111Ω, which is closest to 100 Ω. So, option (A) is correct.

chapter 4 moving charges and magnetismclass 12thmcqphysics
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Vatsal Mishra
Poll
Vatsal Mishra
Asked: April 4, 2025In: Class 12 Physics

A galvanometer with resistance 1000 Ω gives full scale deflection at 0.1 mA. What value of resistance should be added to 1000 Ω to increase its current range 10 A.

(B) is correct. To convert a galvanometer into an ammeter with a 10 A range, a small shunt resistance is added in parallel. Using Rs = Ig⋅G/I−Ig’ we get Rs ≈ 0.01Ω.

chapter 4 moving charges and magnetismclass 12thmcqphysics
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Answer
Vatsal Mishra
Poll
Vatsal Mishra
Asked: April 4, 2025In: Class 12 Physics

A galvanometer having 20 division scale and 50 Ω resistance is connected in series to a cell of e.m.f. 1.5 V through a resistance of 100 Ω, shows full scale deflection. The figure of merit of the galvanometer in microampere/division is

Total resistance = 100 Ω + 50 Ω = 150 Ω. Current = 1.5 V / 150 Ω = 0.01 A = 10,000 µA. For 20 divisions, current per division = 10,000 µA / 20 = 500 µA/div. Answer: (C) ...

chapter 4 moving charges and magnetismclass 12thmcq
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