Question

The figure shows a meter bridge wire $AC$ having uniform cross-section. The length of wire $AC$ is $100cm\text{.}X$ is a standard resistor of $4\Omega$ and $Y$ is a coil. When $Y$ is immersed in melting ice, the null point is at $40cm$ from point $A$. When the coil $Y$ is heated to $100^{\circ }C$ , a $12\Omega$ resistor 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 the coil is $x\times$ $10^{-2}$ SI units. Find the value of $X$ .

Answer 1

Answer: 1

This is linear variation of resistance,
$\frac{i\times 4}{i\times R_0}=\frac{AD}{DC}=\frac{40}{60}=\frac{2}{3}$
$\frac{4}{R_0}=\frac{2}{3}{R}_0=6\Omega R_0=6\frac{R\times 12}{R+12}=6\Omega \Rightarrow R=12\Omega$
The resistance is changed when we change the temperature. Then, the resistance after increasing temperature in terms of coefficient of temperature,
$R=R_0(1+\alpha \Delta T)$
$12=6(1+\alpha \times 100)$
$\alpha =1\times 10^{-2}{}^{\circ }{C}^{-1}$
$X=1$ .