c
For balanced meter bridge
$\frac{X}{R}=\frac{\ell}{(100-\ell)}$
$\frac{X}{40}=\frac{90}{60} \Rightarrow X=60 \Omega$
$X=R \frac{\ell}{(100-\ell)}$
$\frac{\Delta X}{X}=\frac{\Delta \ell}{\ell}+\frac{\Delta \ell}{100-\ell}=\frac{0.1}{40}+\frac{0.1}{60}$
$\Delta X=0.25$
$\text { so अत: } X=(60 \pm 0.25) \Omega$