AIPMT 2010 Physics and Chemistry Paper

When the two way key is switched off, then
The current flowing in the resistors R and X is I=1A...(i)
When the key between the terminals 1 and 2 is plugged in, then
Potential difference across R=IR=kℓ1...(ii)
where k is the potential gradient across the potentiometer wire
When the key between the terminals 1 and 3 is plugged in, then
Potential difference across (R+X)=I(R+X)=kℓ2 .....(iii)
From equation (ii), we get
R=‌

kℓ1

I

=‌

kℓ1

1

=kℓ1Ω
From equation (iii), we get
R+X=‌

kℓ2

I

=‌

kℓ2

1

=kℓ1    Using (i)
X=kℓ2−R
=kℓ2−kℓ1
=k(ℓ2−ℓ1)Ω