JEE Main 2016 · Physics · Current Electricity · hard

In a metre bridge a 10 Ω resistor is in the left gap and an unknown resistor X in the right gap; the null point is found at 40 cm from the left end.…

Single correct option

In a metre bridge a

10\ \Omega

resistor is in the left gap and an unknown resistor

X

in the right gap; the null point is found at

40\ \mathrm{cm}

from the left end. A

30\ \Omega

resistor is then connected in parallel with

X

. The new null point is at a distance from the left end of

Show answer and solution

Answer: (B) $50\ \mathrm{cm}$

At balance

\dfrac{10}{X}=\dfrac{l}{100-l}

. With

l=40

\dfrac{10}{X}=\dfrac{40}{60}

, so

X=15\ \Omega

. Putting

30\ \Omega

in parallel gives

X'=\dfrac{15\times30}{15+30}=10\ \Omega

. The new condition is

\dfrac{10}{10}=\dfrac{l'}{100-l'}

, so

l'=50\ \mathrm{cm}

. If the

30\ \Omega

were mistakenly put in series,

X'=45\ \Omega

and the null point would shift to about

18\ \mathrm{cm}

, i.e. towards the left, not the right.

Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake