When the resistance of a resistor is increased by 10% in the left gap of a metre bridge then how many times is the increase in its original value

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Solution

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

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