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.
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(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.
(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.
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) ...
The correct answer is (A) E = 0, B = 0. Since the proton moves with constant velocity without changing direction or speed, no net force acts on it. Hence, both electric and magnetic fields must be zero to maintain ...