The torque due to the force (2 \hati+\hatj+2 \hatk) about the origin, acting on a particle whose position vector is (\hati+\hatj+\hatk) , would be

Rotational Motion PYQ — JEE Main 2025 · 22 Jan · Shift 2 · Q53 with Solution

View written solutionFree

Correct answer: B

Given data

Position vector of the particle: $\vec r = \hat i + \hat j + \hat k$

Force acting on the particle: $\vec F = 2\hat i + \hat j + 2\hat k$
Formula for torque about the origin

Torque is given by the cross product: $\vec \tau = \vec r \times \vec F$
Compute the cross product
$\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 1 \\ 2 & 1 & 2 \end{vmatrix}$$ Expanding: $$\vec \tau = \hat i \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} - \hat j \begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix} + \hat k \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}$$$
Evaluate each determinant
- For $\hat i$ : $1\cdot 2 - 1\cdot 1 = 1$
- For $\hat j$ : $1\cdot 2 - 1\cdot 2 = 0$
- For $\hat k$ : $1\cdot 1 - 1\cdot 2 = -1$
Therefore, $\vec \tau = \hat i - \hat k$
Match with options

$\hat i - \hat k$ corresponds to Option B.
Comparison with stored answer

Stored correct answer is B, which matches our derived answer.