AIEEE 2007 Solved Paper

We know that
Rt=R0(1+αt)
where Rt is the resistance of the wire at t∘C,
R0 is the resistance of the wire at 0∘C
and α is the temperature coefficient of resistance
⇒R50=R0(1+50α) ......(i)
R100=R0(1+100α) ......(ii)
From (i),R50−R0=50αR0......(iii)
From (ii),R100−R0=100αR0......(iv)
Dividing (iii) by (iv), we get
‌

R50−R0

R100−R0

=‌

1

2

Here, R50=5Ω and R100=6Ω
∴‌

5−R0

6−R0

=‌

1

2

or, 6−R0=10−2R0
or, R0=4Ω.