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.

1. Understanding the Configuration from the Given Diagram:
From the given schematic of the four-bar planar mechanism:
- Link 1 is the fixed ground link $AD$ of length $4\text{cm}$.
- Link 2 is the input link $AB$ of length $5\text{cm}$ rotating about the fixed pivot $A$ with an angular velocity of ${\omega }_2=10\text{rad/s}$.
- Link 3 is the coupler link $BC$ connecting link 2 and link 4.
- Link 4 is the output link $DC$ of length $2\text{cm}$ rotating about the fixed pivot $D$.
- Right-angle indicators at pivots $A$ and $D$ show that both $AB$ and $DC$ are perpendicular to the horizontal ground link $AD$ at this instant.

2. Locating the Instantaneous Centers (I-centers):
According to the Kennedy-Aronhold Theorem, the three instantaneous centers $I_{12}$, $I_{14}$, and $I_{24}$ must lie on a straight line:
- $I_{12}$ is located at pivot $A$.
- $I_{14}$ is located at pivot $D$.
- Therefore, the instantaneous center $I_{24}$ lies on the line passing through $A$ and $D$ (the horizontal x-axis).

Similarly, the three instantaneous centers $I_{23}$, $I_{34}$, and $I_{24}$ must lie on a straight line:
- $I_{23}$ is located at joint $B$.
- $I_{34}$ is located at joint $C$.
- Therefore, $I_{24}$ lies on the line passing through $B$ and $C$.

Thus, $I_{24}$ is the intersection point of the horizontal line $AD$ and the line $BC$.

3. Calculating the Coordinates and Distances:
Let us set up a Cartesian coordinate system with pivot $A$ at the origin $(0,0)$:
- Pivot $A=(0,0)$
- Pivot $D=(4,0)$
- Joint $B=(0,5)$
- Joint $C=(4,2)$

The equation of the line passing through $B(0,5)$ and $C(4,2)$ is given by:

$y-5=\frac{2-5}{4-0}(x-0)$

$y-5=-\frac{3}{4}x$

To find the intersection of this line with the horizontal line $AD$ (where $y=0$):

$0-5=-\frac{3}{4}x\Rightarrow x=\frac{20}{3}\text{cm}$

Hence, the coordinates of $I_{24}$ are $(\frac{20}{3},0)$.

Now, calculate the distances of $I_{24}$ from $A$ ($I_{12}$) and $D$ ($I_{14}$):

$I_{12}{I}_{24}=\frac{20}{3}-0=\frac{20}{3}\text{cm}$

$I_{14}{I}_{24}=\frac{20}{3}-4=\frac{8}{3}\text{cm}$

4. Applying the Angular Velocity Ratio Theorem:
The relation between the angular velocities using the instantaneous center is:

${\omega }_2\times \left(I_{12}{I}_{24}\right)={\omega }_4\times \left(I_{14}{I}_{24}\right)$

${\omega }_4={\omega }_2\times \frac{I_{12}{I}_{24}}{I_{14}{I}_{24}}$

Substitute the known values into the equation:

${\omega }_4=10\times \frac{20/3}{8/3}$

${\omega }_4=10\times \frac{20}{8}=10\times 2.5=25\text{rad/s}$

Thus, the magnitude of the angular velocity of bar $DC$ at this instant is $25\text{rad/s}$.