The figure shows a meter bridge circuit with AB=100cm,X=12Ω and R=18Ω, and the jockey J is at the balance point. If R is now made 8Ω, through what distance (in cm) will J have to be moved to obtain a balanced bridge ?
A
20.0
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B
20
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C
20.00
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Solution
In case 1:
X=12Ω & R=18Ω
Let the balanced point be at lcm from point A.
AJ=l;BJ=(100−l)cm
Applying balance condition,
XR=l100−l
1218=l100−l
3l=200−2l
∴l=2005=40cm
Now in case 2:
X=12Ω and R′=8Ω
XR′=l′100−l′
128=l′100−l′
2l′=300−3l′
∴l′=3005=60cm
So, Δl=l′−l=60−40=20cm
Hence, J has to be moved by 20cm in order to achieve the balanced point.