Q 11-04-121JEE MainJEE Main 2023 (12 Apr, Shift 1)Easy
Three forces $F_1=10\ \text{N}$, $F_2=8\ \text{N}$, $F_3=6\ \text{N}$ are acting on a particle of mass $5\ \text{kg}$. The forces $F_2$ and $F_3$ are applied perpendicularly so that particle remains at rest. If the force $F_1$ is removed, then the acceleration of the particle is
Answer: (D) $2\ \text{m s}^{-2}$
$F_2$ and $F_3$ combine to $\sqrt{64+36}=10\ \text{N}$, which balances $F_1$. Removing $F_1$ leaves a net $10\ \text{N}$: $a=\dfrac{10}{5}=2\ \text{m s}^{-2}$.
Solution by Sreeraj P, M.Sc Physics