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

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70 cm

80 cm

53.3 cm

114 cm

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{x}{y} = \frac{20}{120 - 20} = \frac{1}{5} \\ x = 5 y; \frac{4 x}{y} = \frac{l}{120 - l}, \\ \frac{4 \times 5 y}{y} = \frac{l}{120 - l}, ∴ l^{'} = 114 cm \end{array}$

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