Q87 / 90 · JEE Main 2023 (29 Jan Shift 2) · Physics

When two resistance R1 and R2 connected in series and introduced into the left gap of a meter bridge and a resistance of 10 Ω is introduced into the right gap, a null point is found at 60 cm from left side. When R1 and R2 are connected in parallel and introduced into the left gap, a resistance of 3 Ω is introduced into the right-gap to get null point at 40 cm from left end. The product of R1 R2 is ______ Ω2

Correct answer: 30

Solution

A meter bridge consists of a wire of length 1 m and of uniform cross-sectional area stretched taut and clamped between two thick metallic strips bent at right angles with two gaps across which resistors are to be connected. It works on the principle of the Wheatstone bridge. Therefore, we can write

R1+R210=6040⇒R1+R2=15 Ω

Then for the second case, we can write 

R1R2R1+R2×3=4060⇒R1R2=30 Ω2

Hence, the correct answer is 30.

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