A Zener of breakdown voltage V _ Z = 8V and maximum Zener current,...

Minimum value of ${R}_{L}$ for which the diode is shorted is

$\dfrac{{R}_{L}}{{R}_{L} + 100} \times 10 = 8 ⇒ {R}_{L} = 400Ω$

For maximum value of ${R}_{L}$, current through diode is $10mA$

So, ${i}_{R} = {i}_{{R}_{L}} + {I}_{ZM}$

$⇒ \dfrac{2}{100} = \dfrac{8}{{R}_{L}} + 10 \times {10}^{- 3} ⇒ 10 \times {10}^{- 3} = \dfrac{8}{{R}_{L}} ⇒ {R}_{L} = 800Ω$

So, $\dfrac{{R}_{Lmax}}{{R}_{Lmin}} = 2$