1. Let the increasing A.P. with 11 terms be $b_1, b_2, \dots, b_{11}$ with first term $a$ and common difference $d$.

   So the terms are: $a,\ a+d,\ a+2d,\ \dots,\ a+10d$

2. The variance does not change if we add or subtract the same constant from all terms.

   Hence, the variance of $a, a+d, a+2d, \dots, a+10d$ is the same as the variance of $0, d, 2d, \dots, 10d$

   That is, $\text{Var}(0,d,2d,\dots,10d)=d^2\,\text{Var}(0,1,2,\dots,10)$

3. Now find the variance of numbers $0,1,2,\dots,10$.

   Their mean is $\bar{x}=\frac{0+1+2+\cdots+10}{11}=\frac{55}{11}=5$

4. Variance is $\sigma^2=\frac{1}{11}\sum_{k=0}^{10}(k-5)^2$

   Compute: $(-5)^2+(-4)^2+(-3)^2+(-2)^2+(-1)^2+0^2+1^2+2^2+3^2+4^2+5^2$ $=25+16+9+4+1+0+1+4+9+16+25=110$

   Therefore, $\text{Var}(0,1,2,\dots,10)=\frac{110}{11}=10$

5. Hence the given variance is $d^2 \cdot 10 = 90$

   So, $d^2=9$ $d=\pm 3$

6. Since the A.P. is increasing, the common difference must be positive.

   Therefore, $d=3$

Final Answer

The common difference is $\boxed{3}$