two rods of length l and mass m are kept perpendicular to each other find distance of centre of mass from the point of their contact

Dear student

Centre of mass of a rod of mass m and length L is at its midpoint i.e. at L/2. It means the whole mass of the rod can be considered at its midpoint.
a) If the two rods are connected perpendicularly by their ends and let this contact point to be origin then centre of mass of this system
 for rod 1: $m_1=m, x_1=\frac{L}{2} and y_1=0$
 for rod 2: $m_2=m, x_2=0 and y_2=\frac{L}{2}$
the X coordinate of centre of mass
 
$$\begin{gathered} X=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2} \\ \Rightarrow X=\frac{m{\displaystyle \frac{L}{2}}+m·0}{m+m} \\ \Rightarrow X=\frac{L}{4} \end{gathered}$$

the Y coordinate of centre of mass

$$\begin{gathered} Y=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2} \\ \Rightarrow Y=\frac{m·0+m{\displaystyle \frac{L}{2}}}{m+m} \\ \Rightarrow Y=\frac{L}{4} \end{gathered}$$

So the Centre of mass of this system is $\left(\frac{L}{4},\frac{L}{4}\right)$
b) If the two rods are connected by their midpoint then centre of mass will be at their contact point.