a
as $x=\frac{R(100-\ell)}{\ell}$
$\text { for (1) } \quad x=\frac{1000 \times(100-60)}{40} \approx 667$
$\text { for (2) } \quad x=\frac{100 \times(100-13)}{13} \approx 669$
$\text { for }(\mathrm{3}) \quad x=\frac{10 \times(100-1.5)}{98.5} \approx 656$
$\text { for (4) } \quad x=\frac{1 \times(100-1)}{1} \approx 95$
So reading in serial no. $(4)$ is completely different hence correct answer $(A).$