We know that,

$$h = {{2\sigma \cos \theta } \over {r\,\rho g}}$$..... (1)

where r is inner radius of capillary tube; $$\sigma$$ is surface tension; $$\theta$$ is angle between water and wall of capillary tube; h is height of water in capillary tube and g is acceleration due to gravity. From Eq. (1),

$$h \propto {1 \over r}$$

Therefore, for a given material of the capillary tube, h decreases with increase in r.

Hence, option (A) is correct.

h depends on $$\sigma$$, therefore option (B) is incorrect.

In a lift going up with a constant acceleration a, the force due to surface tension remains same but weight of the water increases i.e., W_lift = $$\pi$$r²h$$\rho$$(g + a). Thus, the height of the water rise in a capillary becomes

$$h = {{2\sigma \cos \theta } \over {r\,\rho (g + a)}} = {{2\sigma \cos \theta } \over {r\, \rho{g_{eff}}}}$$,

where g_eff is the effective g in the lift. Hence, h decrease if the experiment is performed in a lift going up with a constant acceleration.

From (1), h $$\propto$$ cos$$\theta$$ and not $$\theta$$, therefore, option (D) is incorrect.