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JEE Mains 29-Jan-2023 Shift 2 Solved Paper
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© examsnet.com
Question : 21
Total: 90
When two resistance
R
1
and
R
2
connected in series and introduced into the left gap of a meter bridge and a resistance of
10
Ω
is introduced into the right gap, a null point is found at
60
cm
from left side. When
R
1
and
R
2
are connected in parallel and introduced into the left gap, a resistance of
3
Ω
is introduced into the right-gap to get null point at 40
cm
from left end. The product of
R
1
R
2
is _______
Ω
2
[29-Jan-2023 Shift 2]
Your Answer:
Validate
Solution:
R
1
+
R
2
10
=
60
40
=
3
2
⇒
R
1
+
R
2
=
15
Now
R
1
R
2
(
R
1
+
R
2
)
×
3
=
40
60
=
2
3
⇒
R
1
R
2
=
30
© examsnet.com
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