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JEE Main Physics Class 12 Current Electricity Part 2 Questions
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© examsnet.com
Question : 50
Total: 100
In the given potentiometer circuit arrangement, the balancing length
A
C
is measured to be
250
c
m
. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes
400
c
m
. The ratio of the emf of two cells,
ε
1
ε
2
is :
[25 Jul 2021 Shift 2]
5
3
8
5
4
3
3
2
Validate
Solution:
E
1
=
k
l
1
.........(i)
E
1
+
E
2
=
k
l
2
........(ii)
E
1
E
1
+
E
2
=
l
1
l
2
=
250
400
=
5
8
8
E
1
=
5
E
1
+
5
E
2
3
E
1
=
5
E
2
E
1
E
2
=
5
3
© examsnet.com
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