C

Step 1: Identify the System
We have two blocks; A with mass $m_A = 10$ kg and B with mass $m_B = 40$ kg connected by a string passing over a frictionless pulley.

Step 2: Analyze Forces
For block A, which is on a frictionless slope of 45 degrees, we can break down the forces acting on it:
- The gravitational force acting on block A ($F_{gA} = m_A g = 10 imes 10 = 100$ N) can be resolved into components parallel and perpendicular to the incline. The component parallel to the incline is $F_{||A} = F_{gA} imes rac{1}{ oot{2}} = 100 imes rac{1}{ oot{2}}$ N.

For block B, which is hanging vertically, the force is simply the weight of the block: $F_{gB} = m_B g = 40 imes 10 = 400$ N.

Step 3: Set Up Equations
According to Newton's second law, we can write:
1. For block A: $T - F_{||A} = m_A a $
2. For block B: $F_{gB} - T = m_B a $

Substituting the known values:
For A:
$T - (100/ oot{2}) = 10a$
For B:
$(400 - T) = 40a$

Step 4: Solve the Equations
From the equations:
1. $T = 10a + (100/ oot{2})$
2. $T = 400 - 40a$
Equating these two expressions for tension $T$ gives:
$10a + (100/ oot{2}) = 400 - 40a$
Combine like terms:
$50a = 400 - (100/ oot{2})$
Now solving for $a$:
$a = (400 - (100/ oot{2}))/50$.
Calculating the value of $a$, we arrive at $8$ m/s².

Step 5: Calculate the Tension
Using the value of $a$, plug it back into one of the tension equations:
Using $T = 400 - 40a$ and substituting $a = 8$:
$T = 400 - 40(8) = 400 - 320 = 80$ N.

Conclusion
Therefore, the tension in the string is N.