JEE Main 2023 — Laws Of Motion Question with Solution

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

FBD of $4\mathrm{kg}$

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

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

FBD of $1\mathrm{kg}$

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

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

From (1) and (2), we get

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

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

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

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

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

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

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

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