JEE Main 2023 · Physics · Current Electricity · medium

In a metre bridge, an unknown resistance X is in the left gap and a 15 Ω resistance in the right gap. The balance point is found at 40 cm from the…

Numerical answer

In a metre bridge, an unknown resistance

X

is in the left gap and a

15\ \Omega

resistance in the right gap. The balance point is found at

40\ \mathrm{cm}

from the left end. A

10\ \Omega

resistor is now connected in parallel with

X

, leaving the right gap unchanged. The new balance length from the left end (in cm) is

Show answer and solution

Answer: 25

Balance condition:

\dfrac{X}{15}=\dfrac{l}{100-l}=\dfrac{40}{60}

, so

X=10\ \Omega

. With

10\ \Omega

in parallel, the left gap becomes

X'=\dfrac{10\times10}{20}=5\ \Omega

. New balance:

\dfrac{l}{100-l}=\dfrac{5}{15}=\dfrac13

, so

3l=100-l

and

l=25\ \mathrm{cm}

. Placing the

10\ \Omega

in series instead would shift the balance point the other way, to

57.1

cm.

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