The position vector of a particle of mass \( 0.5\,\text{kg} \) is given by \( \mathbf{r}(t) = (2t^2\hat{i} - 3t\hat{j} + 4\hat{k})\,\text{m} \). The torque acting on the particle about the origin at time \( t = 2\,\text{s} \) is:

Velocity \( \mathbf{v} = \frac{d\mathbf{r}}{dt} = 4t\hat{i} - 3\hat{j} \); acceleration \( \mathbf{a} = 4\hat{i}\,\text{m/s}^2 \). Force \( \mathbf{F} = m\mathbf{a} = 0.5(4\hat{i}) = 2\hat{i}\,\text{N} \). At \( t = 2\,\text{s} \), \( \mathbf{r} = 8\hat{i} - 6\hat{j} + 4\hat{k} \). Torque \( \boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} = (8\hat{i} - 6\hat{j} + 4\hat{k}) \times 2\hat{i} = -12(-\hat{k}) + 8\hat{j} = -8\hat{j} + 12\hat{k} \). Wait, \( -6\hat{j} \times 2\hat{i} = +12\hat{k} \) and \( 4\hat{k} \times 2\hat{i} = +8\hat{j} \)? Cross product: \( (8\hat{i}-6\hat{j}+4\hat{k})\times 2\hat{i} = -12(-\hat{k}) + 8\hat{j} = 8\hat{j} + 12\hat{k} \). Wait, \( \hat{k} \times \hat{i} = \hat{j} \implies 4\hat{k} \times 2\hat{i} = 8\hat{j} \), and \( -\hat{j} \times \hat{i} = \hat{k} \implies -6\hat{j} \times 2\hat{i} = +12\hat{k} \). Wait, \( \hat{i} \times \hat{j} = \hat{k} \implies \hat{j} \times \hat{i} = -\hat{k} \), so \( (-6)(2)(-\hat{k}) = +12\hat{k} \). No: \( \mathbf{r} \times \mathbf{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 8 & -6 & 4 \ 2 & 0 & 0 \end{vmatrix} = \hat{i}(0) - \hat{j}(0 - 8) + \hat{k}(0 - (-12)) = 8\hat{j} + 12\hat{k} \). Thus the answer is Option 4.