QJEEM23Ph88aa: JEE Main 2023 Physics - Class 11 JEEM23Ph88aa Solution: As per given figure, a weightless pulley P is attached on a…

Let's consider $T$ tension in string and a acceleration of blocks.

FBD of $4 \mathrm{~kg}$

$4 g \sin \theta-\mathrm{T}=4 \mathrm{a}$

$\frac{4 g \sqrt{3}}{2}-T=4 a$ .........(1)

FBD of $1 \mathrm{~kg}$

$T-g \sin 30^{\circ}=a$

$T-\frac{g}{2}=a \quad\quad...(2)$

From (1) and (2), we get

${{4g\sqrt 3 } \over 2} - {g \over 2} = 5a$

$\Rightarrow$ ${{g\left( {4\sqrt 3 - 1} \right)} \over 2} = 5a$

$\Rightarrow$ $5a = {{10\left( {4\sqrt 3 - 1} \right)} \over 2}$

$\Rightarrow$ $a = \left( {4\sqrt 3 - 1} \right)$ m/s²

By putting value of "a" in equation (2), we get

$T - {{10} \over 2} = \left( {4\sqrt 3 - 1} \right)$

$\Rightarrow$ $T = \left( {4\sqrt 3 - 1} \right) + 5$ = $\left( {4\sqrt 3 + 4} \right)$

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