Interpret the setup

The pulley $P$ is fixed at the top where two smooth inclines meet. A single light string passes over the pulley, with masses attached on the two sides:

• left side: mass $m_1 = 4\,\text{kg}$ on incline $60^\circ$
• right side: mass $m_2 = 1\,\text{kg}$ on incline $30^\circ$

Since surfaces are frictionless and the string/pulley are ideal, the tension is same throughout the string.

Find components of weight along the inclines

Along the left incline: $m_1 g \sin 60^\circ = 4 \cdot 10 \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3}\,\text{N}$

Along the right incline: $m_2 g \sin 30^\circ = 1 \cdot 10 \cdot \frac{1}{2} = 5\,\text{N}$

Since $20\sqrt{3} > 5$, the $4\,$kg block moves down its incline and the $1\,$kg block moves up its incline.

Apply Newton's second law to each block

Let the common acceleration be $a$.

For the $4\,$kg block (moving down the plane): $4g\sin 60^\circ - T = 4a$ $20\sqrt{3} - T = 4a \quad ...(1)$

For the $1\,$kg block (moving up the plane): $T - 1g\sin 30^\circ = 1a$ $T - 5 = a \quad ...(2)$

Solve for $T$

From (2): $a = T - 5$

Substitute into (1): $20\sqrt{3} - T = 4(T - 5)$ $20\sqrt{3} - T = 4T - 20$ $20\sqrt{3} + 20 = 5T$ $T = 4\sqrt{3} + 4$

Hence, $T = 4(\sqrt{3}+1)\,\text{N}$

Match with the options

This corresponds to: $\boxed{4(\sqrt{3}+1)\,\text{N}}$

So the correct option is D.