As per given figure, a weightless pulley $P$ is attached on a double inclined frictionless surface. The tension in the string (massless) will be (if $g = 10 m s^{- 2}$ )

Question

As per given figure, a weightless pulley P is attached on a double inclined frictionless surface. The tension in the string (massless) will be (if g = 10 &#160; m &#160; s - 2 )

Aicharya · Accepted Answer

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

$a = \frac{4 g \sin 60^{\circ} - g \sin 30^{\circ}}{4 + 1}$

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

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

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

$T - 5 = 1 \times a$

$\Rightarrow T = 5 + 4 \sqrt{3} - 1$

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

As per given figure, a weightless pulley P is attached on a double inclined frictionless surface. The…

Solution