$\mathrm{h}=\frac{2\mathrm{Tcos\theta }}{\mathrm{\rho gR}};{\mathrm{h}}_1=\frac{2\times 0.075\times \cos 0\mathrm{°}}{1000\times 10\times 0.2\times {10}^{-3}}$

⇒ h₁ = 75 mm (in T₁) [If we assume entire tube of T₁]

⇒ ${\mathrm{h}}_2=\frac{2\times 0.075\times \cos 60\mathrm{°}}{1000\times 10\times 0.2\times {10}^{-3}}$  = 37.5 mm (in T₂) [If we assume entire tube of T₂]

Option (A) : Since contact angles are different so correction in the height of water column raised in the tube will be different in both the cases.

Option (B) : If joint is 5 cm is above water surface, then lets say water crosses the joint by height h, then:

 $\Rightarrow {\mathrm{P}}_0-\frac{2\mathrm{T}}{\mathrm{r}}+\mathrm{\rho gh}+\mathrm{\rho g}\times 5\times {10}^{-2}$                      

    = P₀

$\Rightarrow \mathrm{cos\theta }=\frac{\mathrm{R}}{\mathrm{r}}\mathrm{·}\mathrm{r}=\frac{\mathrm{R}}{\cos \mathrm{\theta }}$

$\Rightarrow \mathrm{\rho g}(\mathrm{h}+5\times {10}^{-2})=\frac{2\mathrm{Tcos\theta }}{\mathrm{R}}$

$\Rightarrow \mathrm{h}=\frac{2\times 0.075\times \cos 60\mathrm{°}}{0.2\times {10}^{-3}\times 1000\times 10}=5\times {10}^{-2}$

⇒ h = –ve, not possible, so liquid will not cross the interface, but angle of contact at the interface will change, to balance the pressure,

So option (B) is wrong.

Option (C) : If interface is 8 cm above water then water will not even reach the interface, and water will rise till 7.5 cm only in T₁, so option (C) is right.

Option (D) : If interface is 5 cm above the water in vessel, then water in capillary will not even reach the interface. Water will reach only till 3.75 cm, so option (D) is right.