The coordinates of a particle with respect to origin in a given reference frame is (1, 1, 1) meters. If a force of \vecF = \hati - \hatj + \hatk acts on the particle, then the magnitude of torque (with respect to origin) in z-direction is \underline\hspace2cm .

Rotational Motion PYQ — JEE Main 2025 · 29 Jan · Shift 1 · Q69 with Solution

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Correct answer: 2

Given data

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

Force acting on the particle: $\vec F = \hat i - \hat j + \hat k$
Torque about the origin

Torque is given by: $\vec \tau = \vec r \times \vec F$

So,
$\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix}$$$
Expand the determinant
$- \hat j\begin{vmatrix}1 & 1 \\ 1 & 1\end{vmatrix} + \hat k\begin{vmatrix}1 & 1 \\ 1 & -1\end{vmatrix}$$ Now calculate each minor: $$\begin{vmatrix}1 & 1 \\ -1 & 1\end{vmatrix} = 1\cdot 1 - 1\cdot(-1) = 2$$ $$\begin{vmatrix}1 & 1 \\ 1 & 1\end{vmatrix} = 1\cdot 1 - 1\cdot 1 = 0$$ $$\begin{vmatrix}1 & 1 \\ 1 & -1\end{vmatrix} = 1\cdot(-1) - 1\cdot 1 = -2$$ Therefore, $$\vec \tau = 2\hat i - 0\hat j - 2\hat k = 2\hat i - 2\hat k$$$
Torque in the z-direction

The z-component of torque is: $\tau_z = -2$

Its magnitude is: $|\tau_z| = 2$
Final answer

$\boxed{2}$
Comparison with stored answer

Stored correct answer = $2$

Our derived answer = $2$

Hence, the answer matches the stored correct answer.