Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

System of Particle and Rotational Motion

Question:

The moment of the force, F = 4 i + 5 j - 6 k at (2, 0, –3), about the point (2, –2, –2), is given by :

Options:

- 8 i - 4 j - 7 k

- 4 i - j - 8 k

- 7 i - 4 j - 8 k

- 7 i - 8 j - 4 k

Correct Answer:

- 7 i - 4 j - 8 k

Explanation:

$\vec{\tau} = (\vec{r} - \vec{r_0}) x \vec{F}$

(r - ro) = (2 i - 3 k) - (2 i - 2 j - 2 k) = 2 j - k

$\vec{\tau}$ = (2 j - k) $\times$ (4 i + 5 j - 6 k) = - 7 i - 4 j - 8 k