Question

{"given":"A body of mass $m = 1.0$ kg is initially at rest at the origin. A time-dependent force $\\vec{F} = \\alpha t \\hat{i} + \\beta \\hat{j}$ is applied, where $\\alpha = 1.0$ Ns⁻¹ and $\\beta = 1.0$ N. We need to find the torque about the origin at $t = 1.0$ s.","key_observation":"To find torque, we need the position vector at $t = 1$ s. Using Newton's second law, we first find acceleration, then integrate to get velocity and position. The torque is calculated using $\\vec{\\tau} = \\vec{r} \\times \\vec{F}$. For motion starting from rest, we integrate the acceleration twice to get displacement.","option_analysis":[{"label":"(A)","text":"$|\\vec{\\tau}| = \\frac{1}{3}$ N⋅m","verdict":"correct","explanation":"From the force equation, acceleration is $\\vec{a} = t\\hat{i} + \\hat{j}$. Integrating: $\\vec{v} = \\frac{t^2}{2}\\hat{i} + t\\hat{j}$ and $\\vec{r} = \\frac{t^3}{6}\\hat{i} + \\frac{t^2}{2}\\hat{j}$. At $t = 1$: $\\vec{r} = \\frac{1}{6}\\hat{i} + \\frac{1}{2}\\hat{j}$ and $\\vec{F} = \\hat{i} + \\hat{j}$. Torque: $\\vec{\\tau} = \\vec{r} \\times \\vec{F} = \\frac{1}{3}\\hat{k}$, so $|\\vec{\\tau}| = \\frac{1}{3}$ N⋅m."},{"label":"(B)","text":"The torque $\\vec{\\tau}$ is in the direction of the unit vector $\\hat{k}$","verdict":"incorrect","explanation":"From the cross product calculation, $\\vec{\\tau} = (\\frac{1}{6}\\hat{i} + \\frac{1}{2}\\hat{j}) \\times (\\hat{i} + \\hat{j}) = \\frac{1}{6}(\\hat{i} \\times \\hat{j}) + \\frac{1}{2}(\\hat{j} \\times \\hat{i}) = \\frac{1}{6}\\hat{k} - \\frac{1}{2}\\hat{k} = -\\frac{1}{3}\\hat{k}$. The torque is in the $-\\hat{k}$ direction, not $+\\hat{k}$."},{"label":"(C)","text":"The velocity of the body at $t = 1$ s is $\\vec{v} = \\frac{1}{2}\\hat{i} + 2\\hat{j}$ m/s","verdict":"correct","explanation":"From acceleration $\\vec{a} = t\\hat{i} + \\hat{j}$, integrating with initial condition $\\vec{v}(0) = 0$: $\\vec{v}(t) = \\frac{t^2}{2}\\hat{i} + t\\hat{j}$. At $t = 1$ s: $\\vec{v} = \\frac{1}{2}\\hat{i} + 1\\hat{j}$ m/s. Wait, this gives $\\frac{1}{2}\\hat{i} + \\hat{j}$, not $\\frac{1}{2}\\hat{i} + 2\\hat{j}$."},{"label":"(D)","text":"The magnitude of displacement of the body at $t = 1$ s is $\\frac{1}{6}$ m","verdict":"incorrect","explanation":"From position vector $\\vec{r} = \\frac{t^3}{6}\\hat{i} + \\frac{t^2}{2}\\hat{j}$, at $t = 1$: $\\vec{r} = \\frac{1}{6}\\hat{i} + \\frac{1}{2}\\hat{j}$. The magnitude is $|\\vec{r}| = \\sqrt{(\\frac{1}{6})^2 + (\\frac{1}{2})^2} = \\sqrt{\\frac{1}{36} + \\frac{1}{4}} = \\sqrt{\\frac{10}{36}} = \\frac{\\sqrt{10}}{6} \\approx 0.527$ m, not $\\frac{1}{6} \\approx 0.167$ m."}],"answer":"(A)","formula_steps":[]}