1. Given data

• Mass: $M = 0.2\,\text{kg}$
• Initial position: $\vec r_0 = -l\,\hat i - h\,\hat j = -10\,\hat i -1\,\hat j$
• Initial velocity: $\vec v_0 = 0$
• Constant acceleration: $\vec a = 10\,\hat i\,\text{m/s}^2$

Since acceleration is only along $x$-direction, the particle always remains at $y=-h=-1$

and its motion is one-dimensional along $x$.

2. Position as a function of time

Using $x(t)=x_0+v_{0x}t+\frac12 at^2$ we get $x(t)=-10+\frac12(10)t^2=-10+5t^2$ So, $\vec r(t)=(-10+5t^2)\hat i-1\hat j$

Velocity: $\vec v(t)=at\,\hat i = 10t\,\hat i$

Momentum: $\vec p(t)=M\vec v = 0.2(10t)\hat i=2t\,\hat i$

3. Check option A

At the point $(x=l,y=-h)=(10,-1)$, $-10+5t^2=10$ $5t^2=20$ $t^2=4$ $t=2\,\text{s}$ (positive time is physically valid)

So option A is correct.

4. Torque about the origin

Torque is $\vec \tau = \vec r \times \vec F$ where $\vec F = M\vec a = 0.2\times 10\,\hat i = 2\hat i$

Now, $\vec r = x\hat i - h\hat j$ Thus, $\vec \tau = (x\hat i-h\hat j)\times 2\hat i$ $=2\left[x(\hat i\times \hat i)-h(\hat j\times \hat i)\right]$ Since $\hat i\times \hat i=0$ and $\hat j\times \hat i=-\hat k$, $\vec \tau=2[-h(-\hat k)] = 2h\hat k$ Given $h=1$, $\boxed{\vec \tau = 2\hat k}$ This is constant at all times.

So:

• At $(x=l,y=-h)$, $\vec\tau=2\hat k$ → B is correct.
• At $(x=0,y=-h)$, $\vec\tau=2\hat k$, not $\hat k$ → D is incorrect.

5. Angular momentum about the origin

Angular momentum is $\vec L = \vec r \times \vec p$ At time $t=2\,\text{s}$ (when particle is at $(10,-1)$):

Momentum: $\vec p = 2t\hat i = 4\hat i$

Position: $\vec r = 10\hat i -1\hat j$

Hence, $\vec L=(10\hat i-\hat j)\times 4\hat i$ $=4\left[10(\hat i\times \hat i)- (\hat j\times \hat i)\right]$ $=4\left[0-(-\hat k)\right]=4\hat k$ So option C is correct.

6. Final evaluation of options

• A: Correct
• B: Correct
• C: Correct
• D: Incorrect

Therefore, the correct options are $\boxed{A,\ B,\ C}$