Water rises upto a height h in a capillary on the surface of earth in stationary condition. Value of h increases if this tube is taken

The correct option is In a lift going downward with acceleration.

Step-by-Step Explanation:

The height $h$ to which a liquid rises in a capillary tube of radius $r$ is given by Jurin's Law:

$h=\frac{2Tcos\theta }{r\rho g_{\text{eff}}}$

where:
• $T$ is the surface tension of the liquid,
• $\theta$ is the angle of contact,
• $r$ is the radius of the capillary tube,
• $\rho$ is the density of the liquid, and
• $g_{\text{eff}}$ is the effective acceleration due to gravity acting on the system.

From this formula, we can see that under constant liquid properties ($T$, $\rho$, $\theta$) and constant tube radius ($r$), the capillary height $h$ is inversely proportional to the effective acceleration due to gravity $g_{\text{eff}}$:

$h\propto \frac{1}{g_{\text{eff}}}$

Therefore, for the capillary height $h$ to increase, the effective acceleration due to gravity $g_{\text{eff}}$ must decrease.

Let's evaluate the effective gravity in the given scenarios:
1. On the sun: The acceleration due to gravity on the sun's surface is significantly greater than on Earth, which increases $g_{\text{eff}}$ and decreases the height $h$.
2. On poles: Due to Earth's shape and rotation, gravity $g$ is slightly higher at the poles than at the equator. This increase in gravity decreases the height $h$.
3. In a lift going upward with acceleration $a$: The pseudo-force acts downwards, increasing the effective gravity to:

$g_{\text{eff}}=g+a$

Since $g_{\text{eff}}$ increases, the height $h$ decreases.
4. In a lift going downward with acceleration $a$: The pseudo-force acts upwards, reducing the effective gravity to:

$g_{\text{eff}}=g-a$

Since $g_{\text{eff}}$ decreases, the value of the capillary height $h$ increases.