Q129 / 180 · NEET 2026 (Cancelled) · Physics
The magnitude and direction of the acceleration produced in a body of mass $5$ kg when two mutually perpendicular forces $8$ N and $6$ N act on it, are respectively :
- A$20$ m s$^{-2}$; $\tan^{-1}(4/3)$ with $8$ N force
- B$2$ m s$^{-2}$; $\tan^{-1}(3/4)$ with $6$ N force
- C$2$ m s$^{-2}$; $\tan^{-1}(4/3)$ with $8$ N force
- Correct answerD$2$ m s$^{-2}$; $\tan^{-1}(3/4)$ with $8$ N force
Solution
Given forces are $F_1 = 8$ N and $F_2 = 6$ N acting perpendicular to each other.
The magnitude of the net force is $F = \sqrt{F_1^2 + F_2^2} = \sqrt{8^2 + 6^2} = 10$ N.
The magnitude of acceleration is $a = \dfrac{F}{m} = \dfrac{10}{5} = 2$ m s$^{-2}$.
Let $\theta$ be the angle made by the net force with the $8$ N force.
$\tan \theta = \dfrac{F_2}{F_1} = \dfrac{6}{8} = \dfrac{3}{4}$
$\theta = \tan^{-1}\left(\dfrac{3}{4}\right)$ with the $8$ N force.
Answer: $2$ m s$^{-2}$; $\tan^{-1}(3/4)$ with $8$ N force
The magnitude of the net force is $F = \sqrt{F_1^2 + F_2^2} = \sqrt{8^2 + 6^2} = 10$ N.
The magnitude of acceleration is $a = \dfrac{F}{m} = \dfrac{10}{5} = 2$ m s$^{-2}$.
Let $\theta$ be the angle made by the net force with the $8$ N force.
$\tan \theta = \dfrac{F_2}{F_1} = \dfrac{6}{8} = \dfrac{3}{4}$
$\theta = \tan^{-1}\left(\dfrac{3}{4}\right)$ with the $8$ N force.
Answer: $2$ m s$^{-2}$; $\tan^{-1}(3/4)$ with $8$ N force
Hence, the correct answer is option: 'D'.