As per given figure, a weightless pulley P is attached on a double...

The magnitude of acceleration of both the blocks will be same and is given by

$a = \dfrac{4gsin{60}^{∘} - gsin{30}^{∘}}{4 + 1}$

$= \dfrac{20\sqrt{3} - 5}{5}$

$a =$\left(4\sqrt{3} - 1\right)$m{s}^{- 1}$

Applying Newton's second law for second block (along the direction parallel to the incline on which it is placed)

$T - 5 = 1 \times a$

$⇒ T = 5 + 4\sqrt{3} - 1$

$= 4\left(\sqrt{3} + 1\right)N$