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JEE Advanced 2018 Paper 2

Section: Physics

Question : 10 of 54

Marks: +1, -0

2 \times 10^{-4} \text{m}^2

10^{4} \text{Nmrad}^{-1}

50 \Omega

0-1.0 \text{A}

Solution:

n=50 \text{turns}\;\;\;\;\; A=2\times10^{-4}\text{m}^2\;\;\;\;\;B=0.02\text{T}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;K=10^{-4}\text{m}^2

Q_m=0.2 \text{rad}\;\;\;R_g=50 \Omega\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;I_A=0-1.0 \text{A}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \tau=MB=C\theta,\ M=nIA

BINA=C\theta;\;\;\;0.02\times1\times50\times2\times10^{-4}=10^{-4}\times0.210\;\;\;\;I_g=0.1 \text{A}

For galvanometer, resistance is to be connected to ammeter in shunt.

I_g \times R_g = (I-I_g)S\;\;\;0.1\times50=(1-0.1)S\;\;\;S=\frac{50}{9}=5.55

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