Question

Figure shows a meter bridge wire $AC$ with a uniform cross-section. The length of wire $AC$ is $100\text{cm}$. $X$ is a standard resistor of $4\Omega$ and $Y$ is a coil. When $Y$ is immersed in melting ice the null point is at $40\text{cm}$ 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 $Y$ is $x\times {10}^{-2}$. The value of $x$ is

Answer 1

According to given condition for null point at $40cm$,
$\frac{i\times 4}{i\times R_0}=\frac{AD}{DC}=\frac{40}{60}=\frac{2}{3}$
$\frac{4}{R_0}=\frac{2}{3}\Rightarrow R_0=6\Omega$
Here, $R_0$ is resistance of coil $Y$ at $0^{\circ }C$
In second case,
When $12\Omega$ is connected in parallel with coil $Y$, null point is at same position. So,
$\left[\frac{R\times 12}{R+12}\right]=6\Rightarrow R=12$
Here $R$ is resistance of coil $Y$ at ${100}^{\circ }C$
Now due to thermal expansion
$R=R_0[1+\alpha \Delta T]\Rightarrow 12=6[1+\alpha (100\left)\right]$
$\Rightarrow \alpha =1\times {10}^{-2}$