JEE Main 2018 · Physics · Current Electricity · medium

In a meter bridge the left gap holds an unknown resistance X and the right gap holds 15 Ω; the null point is at 40 cm from the left end. A second…

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In a meter bridge the left gap holds an unknown resistance

X

and the right gap holds

15\ \Omega

; the null point is at

40\ \mathrm{cm}

from the left end. A second resistance

Y

is then connected in parallel with

X

, and the null point moves to

30\ \mathrm{cm}

from the left end. The value of

Y

Show answer and solution

Answer: (D) $18\ \Omega$

Balance condition:

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

. First case:

X=15\times\dfrac{40}{60}=10\ \Omega

. After adding

Y

in parallel the effective left resistance is

X'=15\times\dfrac{30}{70}=\dfrac{45}{7}\ \Omega

. Then

\dfrac{1}{X'}=\dfrac1X+\dfrac1Y

gives

\dfrac1Y=\dfrac{7}{45}-\dfrac{1}{10}=\dfrac{14-9}{90}=\dfrac1{18}

, so

Y=18\ \Omega

. A parallel resistor always lowers the left resistance, which is why the null point shifts towards the left end.

Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake