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NEET PHYSICSCURRENT ELECTRICITYMedium

Question

In a meter bridge experiment, the null point is at a distance of 30 cm from A. If a resistance of 16 \Omega is connected in parallel with resistance Y, the null point occurs at 50 cm from A. The value of the resistance Y is:

A

112/3 \Omega

B

40/3 \Omega

C

64/3 \Omega

D

48/3 \Omega

Step-by-Step Solution

For a meter bridge, the ratio of resistances in the gaps is equal to the ratio of the balancing lengths: XY=l100l\frac{X}{Y} = \frac{l}{100 - l}.

Case 1: Null point at l1=30l_1 = 30 cm. XY=3010030=3070=37    X=37Y\frac{X}{Y} = \frac{30}{100 - 30} = \frac{30}{70} = \frac{3}{7} \implies X = \frac{3}{7}Y.

Case 2: A resistance of 16 Ω16\ \Omega is connected in parallel with YY. The new equivalent resistance YY' is: Y=16Y16+YY' = \frac{16Y}{16 + Y}. The new null point is at l2=50l_2 = 50 cm. XY=5010050=1    X=Y\frac{X}{Y'} = \frac{50}{100 - 50} = 1 \implies X = Y'.

Solving: substitute XX from Case 1 into the equation from Case 2: 37Y=16Y16+Y\frac{3}{7}Y = \frac{16Y}{16 + Y}. Dividing both sides by YY (since Y0Y \neq 0): 37=1616+Y\frac{3}{7} = \frac{16}{16 + Y} 3(16+Y)=1123(16 + Y) = 112 48+3Y=11248 + 3Y = 112 3Y=64    Y=643 Ω3Y = 64 \implies Y = \frac{64}{3}\ \Omega.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from CURRENT ELECTRICITY. 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.

PHYSICSCURRENT ELECTRICITYbridgeexperimentdistanceresistanceconnected

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