In a four bar planar mechanism shown in the figure, AB = 5 cm, AD = 4 cm and DC = 2 cm. In the configuration shown, both AB and DC are perpendicular to AD. The bar AB rotates with an angular velocity of 10 rad/s. The magnitude of angular velocity (in rad/s) of bar DC at this instant is

The correct option is 25 (which corresponds to an angular velocity of 25 rad/s).

Let us analyze the four-bar planar mechanism shown in the figure step-by-step.

We are given the lengths of the links as follows:
AB = 5 cm (input link)
AD = 4 cm (fixed link, ground)
DC = 2 cm (output link)

In the given instant/configuration, both AB and DC are perpendicular to AD. This means that link AB is parallel to link DC, and both are perpendicular to the line of centers AD.

The angular velocity of link AB is given as:
$${\omega }_{AB}=10\text{ rad/s}$$
We need to find the magnitude of the angular velocity of the bar DC, denoted as ${\omega }_{CD}$, at this instant.

Let the velocity of point B on link AB be $v_B$. Since B rotates about the fixed point A:
$$v_B={\omega }_{AB}\times AB=10\text{ rad/s}\times 5\text{ cm}=50\text{ cm/s}$$
Since AB is perpendicular to AD, the direction of $v_B$ is horizontal (parallel to AD).

Similarly, let the velocity of point C on link DC be $v_C$. Since C rotates about the fixed point D:
$$v_C={\omega }_{CD}\times CD={\omega }_{CD}\times 2\text{ cm}$$
Since DC is perpendicular to AD, the direction of $v_C$ is also horizontal (parallel to AD).

Now, let us consider the rigid link BC connecting points B and C. The velocity of point C relative to point B must be perpendicular to the line BC because the distance between B and C is constant (rigid body constraint).
In this configuration, since AB and DC are perpendicular to AD, the link BC is inclined. Let us analyze the components of velocities along the line BC. The projection of the velocity of point B along BC must equal the projection of the velocity of point C along BC:

$$v_B\cos \left(\theta \right)=v_C\cos \left(\theta \right)$$
Since the angles that $v_B$ and $v_C$ make with the line BC are identical (both velocities are parallel to AD), we have:

$$v_C=v_B$$
Therefore, the magnitude of the velocity of point C is:

$$v_C=50\text{ cm/s}$$

We can now calculate the angular velocity of link DC using the relation:

$${\omega }_{CD}=\frac{v_C}{CD}$$