Question

The figure shows a meter bridge wire $AC$ having uniform cross-section. The length of wire $AC$ is $100\,cm\text{.}X$ 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

This is linear variation of resistance,
$\frac{i \times 4}{i \times R_{0}}=\frac{A D}{D C}=\frac{40}{60}=\frac{2}{3}$
$\frac{4}{R_{0}}=\frac{2}{3} R_{0}=6 \,\Omega R_{0}=6 \frac{R \times 12}{R+12}=6 \,\Omega \Rightarrow R=12 \,\Omega$
The resistance is changed when we change the temperature. Then, the resistance after increasing temperature in terms of coefficient of temperature,
$R=R_{0}\left(\right.1+\alpha \Delta T\left.\right)$
$12=6\left(\right.1+\alpha \times 100\left.\right)$
$\alpha=1 \times 10^{-2} \quad{ }^{\circ} C ^{-1}$
$X=1$ .