Question

Two equal resistance are connected in the gaps of a metre bridge if the resistance in the left is increased by 10% the balancing point shifts

1.10% to right

2.10% to left

3.9.6%to right

4.4.8% to right

Answer 1

Since a Metre-Bridge has wire of 100 cm, so if the 2 resistances are equal, the balance point is at 50 cm.

Now if the left resistance is increased by 10%, it becomes 1.1 times its original value.

So balance point L :

1.1 R/ R = L / (100 - L)

110 - 1.1 L = L

110 = 2.1 L

L = 52.4 cm approx.

So the balance point L has shifted right by 2.4 cm

So in terms of %, it has shifted right by (2.4 / 50 x 100) %

= 4.8 % to the right