Two smooth identical spheres each of radius 125 mm and weight 100 N rest in a horizontal channel having vertical walls. The distance between vertical walls of the channel is 400 mm.

The reaction at the point of contact between two spheres is _________ N (round off to one decimal place).

The correct answer is 125.

Step 1: Understand the Geometry of the System
We are given two identical smooth spheres inside a channel with vertical walls:
- Radius of each sphere,
$R=125\text{mm}$
- Weight of each sphere,
$W=100\text{N}$
- Width of the channel (distance between vertical walls),
$L=400\text{mm}$
- Since the two spheres touch each other, the distance between their centers,
$C_1$ and $C_2$, is equal to the sum of their radii:
$d=2R=2\times 125=250\text{mm}$

Step 2: Determine Center Coordinates and Distances
Let us set the left wall as the vertical axis $x=0$.
- The lower sphere rests on the floor and touches the left wall, so the horizontal position of its center is:
$x_1=R=125\text{mm}$
- The upper sphere touches the right wall at $x=400\text{mm}$, so the horizontal position of its center is:
$x_2=400-R=400-125=275\text{mm}$
- The horizontal distance between the centers is:
$\mathrm{\Delta x}=x_2-x_1=275-125=150\text{mm}$

Step 3: Calculate the Angle of the Line of Centers
Using the Pythagorean theorem, the vertical distance $\mathrm{\Delta y}$ between their centers is:
$\mathrm{\Delta y}=\sqrt{d^2-{\left(\mathrm{\Delta x}\right)}^2}=\sqrt{{250}^2-{150}^2}=\sqrt{62500-22500}=\sqrt{40000}=200\text{mm}$
Let $\theta$ be the angle that the line of centers makes with the horizontal. We can express $\sin \theta$ as:
$\sin \theta =\frac{\mathrm{\Delta y}}{d}=\frac{200}{250}=0.8$

Step 4: Analyze Free-Body Diagram of the Upper Sphere
The upper sphere is in static equilibrium under the action of three forces:
1. Its weight $W=100\text{N}$ acting vertically downwards.
2. The horizontal normal reaction force from the right vertical wall acting to the left.
3. The reaction force $R_C$ at the point of contact with the lower sphere, acting along the line of centers (directed upwards and rightwards at an angle $\theta$ to the horizontal).

Considering the vertical equilibrium of the upper sphere:
$\sum F_y=0$
$R_C\sin \theta -W=0$
$R_C\sin \theta =W$
Substituting the known values:
$R_C\times 0.8=100$
$R_C=\frac{100}{0.8}=125\text{N}$

Thus, the reaction at the point of contact between the two spheres is 125 N.