\text{If}{\mathrm{ x}}^2-kx-49=\mathrm{0 }\text{and }{x}^2-5kx+11=0\text{(}k>0\text{) have a common root, then the value of}\mathrm{ k}\text{ is:}

The correct option is 15/8.

Let $\alpha$ be the common root of the two given quadratic equations:
1) $x^2-kx-49=0$
2) $x^2-5kx+11=0$

Since $\alpha$ is a common root, it must satisfy both equations. Substituting $x=\alpha$ into both equations, we get:
3) ${\alpha }^2-k\alpha -49=0$
4) ${\alpha }^2-5k\alpha +11=0$

Subtracting equation (4) from equation (3) to eliminate the quadratic term ${\alpha }^2$:
$({\alpha }^2-k\alpha -49)-({\alpha }^2-5k\alpha +11)=0$
$-k\alpha +5k\alpha -49-11=0$
$4k\alpha -60=0$
$4k\alpha =60$
$k\alpha =15$
Since $k>0$, we can express the common root as:
$\alpha =\frac{15}{k}$

Now, we substitute this value of $\alpha$ back into equation (3):
${\left(\frac{15}{k}\right)}^2-k\left(\frac{15}{k}\right)-49=0$
$\frac{225}{k^2}$ - 15 - 49 = 0
$\frac{225}{k^2}$ - 64 = 0
$\frac{225}{k^2}$ = 64
$k^2=\frac{225}{64}$

Taking the square root on both sides, we get:
$k=\pm \sqrt{\frac{225}{64}}$
$k=\pm \frac{15}{8}$

Given the constraint that $k>0$, we select the positive value:
$k=\frac{15}{8}$