The resistances in left and right gap of a meter brigdge are 20 `omega` and 30 `omega` respecitively when the resistance in the left gap is reduced to half its value then balance point shifts by
A
15 cm to the right
B
15 cm to the left
C
20 cm to the right
D
20 cm to the left
Text Solution
Verified by Experts
The correct Answer is:
a
Balance condition in meter bridge is given by `(P)/(Q)=(l)/(100-l)` where P is the resistance in left gap and Q is the resistance in right gap and L is length of wire from one end (i.e left end ) where null or balance point is obtained here `20/30 =(l)/(100-l) rarr l =40 cm` Now `P=(20)/(2)=10` cm `therefore 10/230 =(l)/(100-I)rarr I=25 cm` so balance point shifts by (40-25) cm i.e 15 cm to the left
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