A block of mass $0.1\,\text{kg}$ is held against a wall applying a horizontal force of $5\,\text{N}$ on the block. If the coefficient of friction between the block and the wall is $0.5$, the magnitude of the frictional force acting on the block is :

Step 1: Identify the given parameters.

The given values for the block and the applied force are:

$$m = 0.1\ \text{kg}$$ $$F = 5\ \text{N}$$ $$\mu = 0.5$$

Step 2: Determine the normal force.

For horizontal equilibrium, the normal force $N$ exerted by the wall must balance the applied horizontal force $F$.

$$N = F$$ $$N = 5\ \text{N}$$

Step 3: Calculate the maximum possible static friction.

The maximum static frictional force $f_{\max}$ is determined by the normal force and the coefficient of static friction.

$$f_{\max} = \mu N$$ $$f_{\max} = 0.5 \times 5$$ $$f_{\max} = 2.5\ \text{N}$$

Step 4: Calculate the downward gravitational force.

The weight of the block acting vertically downwards is calculated using $g = 9.8\ \text{m/s}^{2}$.

$$W = mg$$ $$W = 0.1 \times 9.8$$ $$W = 0.98\ \text{N}$$

Step 5: Determine the actual frictional force.

Since the downward gravitational force is less than the maximum static friction, the block remains at rest, and the actual static frictional force $f$ must exactly balance the weight.

$$W < f_{\max}$$ $$f = W$$ $$f = 0.98\ \text{N}$$

Answer: The magnitude of the frictional force acting on the block is $0.98\ \text{N}$.