Sum of the digits of a two-digit number is 11.

Let the unit's digit be $x$. Then, the ten's digit $=11-x$

$\therefore$ Number $=10(11-x)+x=110-10 x+x=110-9 x$

Number obtained by interchanging the digits $=10 x+(11-x)=9 x+11$

According to the question,

$9 x+11-(110-9 x)=9$

$\Rightarrow$ $9 x+11-110+9 x=9$

$\Rightarrow \quad 18 x=9-11+110$

$\Rightarrow$ $18 x=108$

$\Rightarrow$ $x=\frac{108}{18}$

$\Rightarrow$ $x=6$

Hence, unit's digit $=6$ and ten's digit $=11-6=5$ Therefore, the required number is 56 .