Observe the following equilibrium in a 1 L flask.
$$A(g)\rightleftharpoons B(g)$$
At T(K), the equilibrium concentrations of A and B are 0.5 Mand 0.375 M respectively. 0.1 moles of A is added into the flask and heated to T(K) to establish the equilibrium again. The new equilibrium concentrations (in M) of A and B are respectively

The equilibrium reaction is: $$A(g) \rightleftharpoons B(g)$$

Given initial equilibrium concentrations at temperature T(K):
$$[A] = 0.5 \, \text{M}$$ and $$[B] = 0.375 \, \text{M}$$

The equilibrium constant $$K_c$$ is calculated as:
$$K_c = \frac{[B]}{[A]} = \frac{0.375}{0.5} = 0.75$$

Now, 0.1 moles of A are added to the 1 L flask. Since the volume is 1 L, this increases the concentration of A by 0.1 M. The initial amounts before the new equilibrium are:
Moles of A = initial moles + added moles = 0.5 + 0.1 = 0.6 moles
Moles of B = 0.375 moles (unchanged)

Thus, initial concentrations after addition (before new equilibrium) are:
$$[A]_0 = 0.6 \, \text{M}$$
$$[B]_0 = 0.375 \, \text{M}$$

Let $$x$$ be the concentration (in M) of A that reacts to form B. At the new equilibrium:
$$[A] = 0.6 - x$$
$$[B] = 0.375 + x$$

Since $$K_c$$ remains constant at the same temperature:
$$K_c = \frac{[B]}{[A]} = 0.75$$
So, $$\frac{0.375 + x}{0.6 - x} = 0.75$$

Solving for $$x$$:
$$0.375 + x = 0.75 \times (0.6 - x)$$
$$0.375 + x = 0.45 - 0.75x$$
$$x + 0.75x = 0.45 - 0.375$$
$$1.75x = 0.075$$
$$x = \frac{0.075}{1.75} = \frac{75}{1750} = \frac{3}{70} \approx 0.042857 \, \text{M}$$

New equilibrium concentrations:
$$[A] = 0.6 - \frac{3}{70} = \frac{42}{70} - \frac{3}{70} = \frac{39}{70} \approx 0.55714 \, \text{M} \approx 0.557 \, \text{M}$$
$$[B] = 0.375 + \frac{3}{70} = \frac{3}{8} + \frac{3}{70} = \frac{105}{280} + \frac{12}{280} = \frac{117}{280} \approx 0.41786 \, \text{M} \approx 0.418 \, \text{M}$$

Comparing with the options, the new equilibrium concentrations are approximately 0.557 M and 0.418 M, which corresponds to option A.