(D) The triple point of water on the Kelvin scale is $T_{K} = 273.15\; K$.
Given that the triple point of water on scale $A$ is $200\; A$,we have $200\; A = 273.15\; K$. Thus,$1\; A = \frac{273.15}{200}\; K$.
Given that the triple point of water on scale $B$ is $350\; B$,we have $350\; B = 273.15\; K$. Thus,$1\; B = \frac{273.15}{350}\; K$.
To find the relation between any temperature $T_{A}$ and $T_{B}$ representing the same physical state,we equate them to the Kelvin scale:
$T_{A} \times \left( \frac{273.15}{200} \right) = T_{B} \times \left( \frac{273.15}{350} \right)$.
Dividing both sides by $273.15$,we get:
$\frac{T_{A}}{200} = \frac{T_{B}}{350}$.
Simplifying the ratio:
$T_{A} = \frac{200}{350} T_{B} = \frac{4}{7} T_{B}$.
Therefore,the relation is $T_{A} = \frac{4}{7} T_{B}$ or $T_{A} : T_{B} = 4 : 7$.