AIPMT MAIN2008Physics-Current Electricity

AIPMT MAIN 2008 Physics Galvanometer MCQ Question

Type: MCQ-numerical-Medium-Class 12

A galvanometer of resistance 50 Ω\Omega is connected to a battery of 3 V along with a resistance of 2950 Ω\Omega in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

A

4450 Ω\Omega

B

5050 Ω\Omega

C

5550 Ω\Omega

D

6050 Ω\Omega

Correct Answer

Option A

Detailed Explanation

The resistance required for 20 divisions is calculated using the formula R=Rg+Rsn1n2R = R_g + R_s \frac{n_1}{n_2}, where RgR_g is the galvanometer resistance, RsR_s is the series resistance, and n1,n2n_1, n_2 are the divisions. Solving gives 4450 Ω\Omega.

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