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JEE Advanced 2018 Paper 2
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© examsnet.com
Question : 10
Total: 54
A moving coil galvanometer has 50 turns and each turn has an area
2
×
10
−
4
m
2
.The magnetic field produced by the magnet inside the galvanometer is 0.02T . The torsional constant of the suspension wire is
10
4
N
m
r
a
d
−
1
.When a current flows through the galvanometer, a full scale deflection occurs if the coil rotates by 0.2rad.The resistance of the coil of the galvanometer is
50
Ω
This galvanometer is to be converted into an ammeter capable of measuring current in the range
0
−
1.0
A
.for this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance, in ohms,is…..
Your Answer:
Validate
Solution:
n
=
50
turns
A
=
2
×
10
−
4
m
2
B
=
0.02
T
K
=
10
−
4
m
2
Q
m
=
0.2
r
a
d
R
g
=
50
Ω
I
A
=
0
−
1.0
A
τ
=
M
B
=
C
θ
,
M
=
n
I
A
B
I
N
A
=
C
θ
;
0.02
×
1
×
50
×
2
×
10
−
4
=
10
−
4
×
0.210
I
g
=
0.1
A
For galvanometer, resistance is to be connected to ammeter in shunt.
I
g
×
R
g
=
(
I
−
I
g
)
S
0.1
×
50
=
(
1
−
0.1
)
S
S
=
50
9
=
5.55
© examsnet.com
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