To find the velocity of slider B, we use the geometric relationship and principles of relative motion.

Given:
Slider A's downward velocity, \( v_a = 0.05 \, \text{m/s} \).
The length of the rigid link \( L \) remains constant.

We note that sliders A and B move along perpendicular directions. For the link AB:
At any instance, the squared length of link \( ab \) is given by:

\( (x)^2 + (y)^2 = L^2 \)

Where:
\( x \) and \( y \) are horizontal and vertical displacements of sliders B and A, respectively.

Taking the derivative with respect to time \( t \):

\( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \)

Given \( \frac{dy}{dt} = -0.05 \, \text{m/s} \) (negative since it's downward), we find:

\( x \frac{dx}{dt} = -y \frac{dy}{dt} \)

We solve for \( \frac{dx}{dt} \) to find B's velocity:

\( \frac{dx}{dt} = -\frac{y}{x} \frac{dy}{dt} \)

At the instant when \( y = 0.5 \, \text{m} \) and \( x = 2.0 \, \text{m} \), substituting:

\( \frac{dx}{dt} = -\left(\frac{0.5}{2.0}\right)(-0.05) = 0.0125 \, \text{m/s} \)

The velocity of slider B is 0.0125 m/s, confirming the calculated value falls within the given range of 0.01 to 0.02 m/s.