Question

The figure shows a meter bridge wire AC having
uniform area of cross-section. X is a standard
resistance of 42 and Y is a resistance coil. When
Y is immersed in melting ice the null point is at
40 cm from point A. When the coil Y is heated to
100*C, a resistance of 100 has to be connected
in parallel with Y. in order to keep the bridge
balanced at the same point. The temperature
coefficient of resistance of coil is​

Answer 1

Answer:

ρt=6.3×10−4K−1

Explanation:

Missing figure is attached

Given the length of wire is 100 cm

Taking the first case, when null pointer is at 40 cm

X(100l)=Y(l) ...(i)

Y = 4 x (100-40) / 40

Y = 6

Taking the second case, when coil Y is at 100 degrees and a parallel resistor is connected

1 / Req = 1/ Y’ + 1/100

1 / 6 = 1/ Y’ + 1/100

Solving this we get,

Y’ = 600 / 94

Now we know that;

Y′=Y(1+ρ1×100)

ρt=6.3×10−4K−1