1. Capillary rise formula

For a capillary of radius $r$, the rise is

$h=\frac{2T\cos\theta}{\rho g r}$

Given:

• $T=0.075\,\text{N/m}$
• $r=0.2\,\text{mm}=2\times 10^{-4}\,\text{m}$
• $\rho=1000\,\text{kg/m}^3$
• $g=10\,\text{m/s}^2$

So,

$h=\frac{2(0.075)\cos\theta}{1000\cdot 10\cdot 2\times 10^{-4}}$

$=\frac{0.15\cos\theta}{2}=0.075\cos\theta\,\text{m}$

Thus:

• For $\theta=0^\circ$: $h_1=0.075\,\text{m}=7.5\,\text{cm}$
• For $\theta=60^\circ$: $h_2=0.075\times \frac12=0.0375\,\text{m}=3.75\,\text{cm}$

So the two materials individually support equilibrium rises of:

• Tube $T_1$: $7.5\,\text{cm}$
• Tube $T_2$: $3.75\,\text{cm}$

2. Physical idea for joined capillaries

The radius is same throughout, only the material changes. Hence pressure jump at the meniscus depends on the material at the top contact line, i.e. on the part where the meniscus is located.

So if the water meniscus lies in:

• material $T_1$, supported height is $7.5\,\text{cm}$,
• material $T_2$, supported height is $3.75\,\text{cm}$.

The joint matters only because the meniscus may or may not be able to cross into the upper part.

3. Case I

In case I, the lower part is $T_1$ ($\theta=0^\circ$) and the upper part is $T_2$ ($\theta=60^\circ$).

So:

• below the joint, capillary action can support up to $7.5\,\text{cm}$,
• if meniscus enters upper part, then upper material can support only $3.75\,\text{cm}$, which is smaller.

Therefore the liquid cannot remain with meniscus in $T_2$ at a height greater than $3.75\,\text{cm}$. Hence if the joint is above $3.75\,\text{cm}$, the meniscus will stay below the joint, inside $T_1$.

So in case I, actual rise is $7.5\,\text{cm}$ as long as this level is below the joint or at least the meniscus does not need to enter $T_2$.

Option A

Joint is at $8\,\text{cm}$ above water surface.

Since $7.5\,\text{cm}<8\,\text{cm}$, the meniscus stays in $T_1$. Thus rise is

$h=7.5\,\text{cm}$

So A is correct.

Option B

Joint is at $5\,\text{cm}$ above water surface.

If water were to rise above $5\,\text{cm}$, the meniscus would enter $T_2$. But in $T_2$, equilibrium rise possible is only $3.75\,\text{cm}$, so such a higher rise cannot be sustained.

Hence the water level cannot be more than $5\,\text{cm}$ in equilibrium; in fact it will settle with the meniscus at the joint/lower section limit, not above $8.75\,\text{cm}$.

Therefore the statement “more than $8.75\,\text{cm}$” is impossible.

So B is incorrect.

4. Case II

In case II, the lower part is $T_2$ ($\theta=60^\circ$) and the upper part is $T_1$ ($\theta=0^\circ$).

The liquid first rises in $T_2$, which alone can support only $3.75\,\text{cm}$.

If the joint is above $3.75\,\text{cm}$, the meniscus never reaches the upper tube $T_1$.

Option D

Joint is at $5\,\text{cm}$ above water surface. Since $3.75<5$, the meniscus remains in $T_2$. Hence rise is

$h=3.75\,\text{cm}$

So D is correct.

5. Meniscus-weight correction

Including weight of liquid in the meniscus, the corrected capillary rise is

$h=\frac{2T\cos\theta}{\rho g r}-\frac{V_m}{\pi r^2}$

where $V_m$ is the meniscus volume term. This correction depends on the meniscus shape, hence on contact angle $\theta$.

Since the meniscus in case I and case II is formed in different materials (hence with different contact angles, in the relevant equilibria), the correction due to weight of water in the meniscus will be different in the two cases.

So C is correct.

6. Final evaluation of options

• A: Correct
• B: Incorrect
• C: Correct
• D: Correct

Therefore the correct options are:

$\boxed{A,\ C,\ D}$