JEE Advanced 2019 Paper 1

Option 4 :(Ig)G=(I)(RA)
(2×10−6)10=(1×10−3)RA
∴Resistance of ammeter RA=0.02Ω
(4) is correct
Option 3:V=(Ig).RV
=50kΩ
∴ (3) is wrong
Option 2:R1=‌‌

R⋅RV

R+RV

,R2=RA
Reading of ammeter = I
Reading of voltmeter V = IR
Measured value of resistance is Rm=‌‌

V

I

=R1
=‌‌

R.RV

R+RV

=‌‌

103×5×104

103+5×104

=‌‌

5×104

M

=980.4Ω
∴ (2) is correct
Option 1: measured value of R depends on R and RV only, but independent of RA, r values.
∴ Measured value of resistance will again be 980.4Ω only