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BITSAT 2009

Exam: BITSAT
Year: 2009
Subject: Physics
Chapter: Current Electricity

BITSAT 2009 · Physics · Current Electricity

The length of a potentiometer wire is . A cell of emf E is balanced...

The length of a potentiometer wire is

\ell

. A cell of emf

E

is balanced at a length

\ell / 3

from the positive end of the wire. If the length of the wire is increased by

\ell / 2.

At what distance will be the same cell give a balance point.

Options

A $2 \ell / 3$
B $\ell / 2$
C $\quad \ell / 6$
D $4 \ell / 3$

Correct answer

B. $\ell / 2$

Step-by-step solution

Potential gradient in the first case

=\frac{\mathrm{E}_{0}}{\ell} \therefore \quad \mathrm{E}=\left(\frac{\ell}{3}\right) \cdot\left(\frac{\mathrm{E}_{0}}{\ell}\right)=\frac{\mathrm{E}_{0}}{3}

Potential gradient in second case =

\frac{E_{0}}{3 \ell / 2} \text{ (x) } \frac{2 E_{0}}{3 \ell}

From equations (i) and (ii),

\frac{\mathrm{E}_{0}}{3}-\left(\frac{2 \mathrm{E}_{0}}{3 \ell}\right) \mathrm{x} \Rightarrow \mathrm{x}=\frac{\ell}{2}

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