The circuit shown in figure represents 

when A is connected to C and B is grounded, ${\mathrm{D}}_1$ does not conduct but ${\mathrm{D}}_2$ conducts, so $V_0=5-\frac{5}{R}\times R=0$

Again when A is grounded and B is connected to C, ${\mathrm{D}}_1$ conducts but ${\mathrm{D}}_2$ does not conduct. 

So, ${\mathrm{V}}_0=5-\left(\frac{5}{\mathrm{R}}\right)\times \mathrm{R}=0$. 

when A and B both are grounded, ${\mathrm{D}}_1$ and ${\mathrm{D}}_2$ both conduct and ${\mathrm{V}}_0=5-\left(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$\mathrm{R}$}\right.\right)\mathrm{R}=0$

When A and B both are connected to C, ${\mathrm{D}}_1$ and ${\mathrm{D}}_2$  both do not conduct. In this case, ${\mathrm{V}}_0=5\mathrm{V}$

Hence the circuit represents an AND gate.