JEE Main Physics Class 12 Current Electricity Part 1 Questions

Resistance of a metal conductor at temperature t°C is given by
Rt=R0(1+αt)
R0 is the resistance of the wire at 0°C
and α is the temperature coefficient of resistance.
Resistance at 50°C,R50=R0(1+50α) .......(i)
Resistance at 100°C,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