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JEE Main 202312 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
12 April
Shift
Morning Shift
Subject
Physics
Chapter
Mathematics in Physics

JEE Main 2023 · 12 April · Morning Shift · Physics · Mathematics in Physics

Three forces F _ 1 = 10N, F _ 2 = 8N, F _ 3 = 6N are acting on a...

Three forces F1=10N,F2=8N,F3=6N{F}_{1} = 10N, {F}_{2} = 8N, {F}_{3} = 6N are acting on a particle of mass 5kg5kg. The forces F2{F}_{2} and F3{F}_{3} are applied perpendicularly so that particle remains at rest. If the force F1{F}_{1} is removed, then the acceleration of the particle is

Options

  1. A7ms–27m{s}^{– 2}
  2. B0.5ms–20 . 5m{s}^{– 2}
  3. C4.8ms–24 . 8m{s}^{– 2}
  4. D2ms–22m{s}^{– 2}

Correct answer

D. 2ms–22m{s}^{– 2}

Step-by-step solution

When the force F1{F}_{1} is removed, the particle starts accelerating because of the resultant force between F2{F}_{2} and F3{F}_{3}.

The resultant force can be calculated as follows:

{F=F22+F32=10N\begin{cases}F & = & \sqrt{{{F}_{2}}^{2} + {{F}_{3}}^{2}} \\ & = & 10N\end{cases}

Hence, the acceleration of the particle is given by

{a=10N5kg=2ms–2\begin{cases}a & = & \dfrac{10N}{5kg} \\ = & 2m{s}^{– 2}\end{cases}

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