Answer:
Option C
Explanation:
For balanced meter bridge
$\frac{X}{R}=\frac{l}{(100-l)}$ ( where , R=90 Ω)
$\therefore $ $\frac{X}{90}=\frac{40}{(100-40)}$
$\therefore $ $X=60$
$X=R\frac{l}{(100-l)}$
$\frac{\triangle X}{X}=\frac{\triangle l}{l}+\frac{\triangle l}{100-l}$
$=\frac{0.1}{40}+\frac{0.1}{60}$
$\triangle X=0.25$
So, X= $60\pm 0.25$ Ω