Question

In a metre bridge experiment, two unknown resistances X and Y are connected to left and right gaps respectively and the balancing point is obtained at 20 cm from right. What will be the new position of the null point from left if one decides to balance a resistance of 4X against Y? (Assume that length of wire of metre bridge is 120 cm.)

• A. 114 cm
• B. 80 cm
• C. 53.3 cm
• D. 70 m

Answer 1

The correct option is A

114 cm

Initially, balancing point is obtained 20 cm from the right. So, distance of balancing point from left = 120 - 20 = 100 cm.
$\frac{X}{Y}=\frac{l}{120-l}=\frac{100}{120-100}=5\Rightarrow X=5Y$
Now, when balancing 4X against Y, let balancing point be at $l_1$ cm from left.
$\frac{4X}{Y}=\frac{l_1}{120-l_1}\frac{4\times 5Y}{Y}=\frac{l_1}{120-l_1}\therefore l_1=114cm$