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Question

A galvanometer of resistance 50 \Omega is connected to a battery of 3 V along with a resistance of 2950 \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

5050 \Omega

B

5550 \Omega

C

6050 \Omega

D

4450 \Omega

Step-by-Step Solution

  1. Initial State: The total resistance in the circuit is the sum of the galvanometer resistance (GG) and the initial series resistance (R1R_1). Rtotal1=G+R1=50+2950=3000ΩR_{total1} = G + R_1 = 50 + 2950 = 3000 \, \Omega The current (I1I_1) flowing through the galvanometer is given by Ohm's Law: I1=VRtotal1I_1 = \frac{V}{R_{total1}} This current produces a deflection of θ1=30\theta_1 = 30 divisions. Since deflection is proportional to current (IθI \propto \theta), we have I1=k30I_1 = k \cdot 30.

  2. Final State: The deflection is reduced to θ2=20\theta_2 = 20 divisions. The new current (I2I_2) is: I2=k20I_2 = k \cdot 20 Taking the ratio of currents: I2I1=2030=23\frac{I_2}{I_1} = \frac{20}{30} = \frac{2}{3}

  3. Calculation: Let the new series resistance be R2R_2. The new total resistance is Rtotal2=G+R2R_{total2} = G + R_2. Since voltage VV is constant, current is inversely proportional to resistance (I1/RI \propto 1/R): I2I1=Rtotal1Rtotal223=300050+R2\frac{I_2}{I_1} = \frac{R_{total1}}{R_{total2}} \Rightarrow \frac{2}{3} = \frac{3000}{50 + R_2} 2(50+R2)=90002(50 + R_2) = 9000 50+R2=450050 + R_2 = 4500 R2=4450ΩR_2 = 4450 \, \Omega .

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOVING CHARGES AND MAGNETISM. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOVING CHARGES AND MAGNETISMgalvanometerresistanceconnectedbatteryresistance

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