1. Given data

• Mass of disc: $M = 4\,\text{kg}$
• Radius of disc: $R = 10\,\text{cm} = 0.1\,\text{m}$
• Mass of hanging block: $m = 2\,\text{kg}$
• Acceleration due to gravity: $g = 10\,\text{m s}^{-2}$

We need to find the tension $T$ in the cord.

2. Moment of inertia of the disc

For a uniform disc about its central axle,

$I = \frac{1}{2}MR^2$

So,

$I = \frac{1}{2}(4)(0.1)^2 = 2 \times 0.01 = 0.02\,\text{kg m}^2$

3. Relation between linear and angular acceleration

Since the cord does not slip,

$a = \alpha R$

where $a$ is the downward acceleration of the block and $\alpha$ is the angular acceleration of the disc.

4. Equation for the hanging block

Taking downward direction as positive:

$mg - T = ma$

Substitute values:

$2 \cdot 10 - T = 2a$

$20 - T = 2a \qquad ...(1)$

5. Rotational equation for the disc

The tension provides torque on the disc:

$TR = I\alpha$

Using $\alpha = \frac{a}{R}$,

$TR = I\frac{a}{R}$

$T = \frac{Ia}{R^2}$

Now,

$\frac{I}{R^2} = \frac{0.02}{(0.1)^2} = \frac{0.02}{0.01} = 2$

Hence,

$T = 2a \qquad ...(2)$

6. Solve equations (1) and (2)

From (2), substitute into (1):

$20 - 2a = 2a$

$20 = 4a$

$a = 5\,\text{m s}^{-2}$

Then,

$T = 2a = 2 \times 5 = 10\,\text{N}$

7. Final answer

The tension in the cord is

$\boxed{10\,\text{N}}$