QJEEAdv18Ph6db0: JEE Advanced 2018 Physics - Class 11 JEEAdv18Ph6db0 Solution: A uniform capillary tube of inner radius is dipped vertically…

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.