Toppling will take place about right most point so,

F ×(√3a/2) = mg × a/2

⇒ F= mg/√3

As the object slides with constant speed friction force is balancing mgsinθ. Torque of Normal reaction and  friction force about centre of object will cancel each other.

τ=mg.sinθ × a/2 = (mg a sinθ)/2

Torque of gravitational force mg about O is mgcosθ ×L/2

Torque = I × α

mgcosθ ×L/2 = (mL²/3) × α

α = (3gcosθ)/2

The angular momentum

$L$

of a particle about the origin is given by:

$$L = m v r \sin\theta$$

where:

• 
  $m = 5$ 

  g (mass of the particle),

• 
  $v = 3\sqrt{2}$ 

  cm/s (speed of the particle),

• 
  $r = 2\sqrt{5}$ 

  cm (perpendicular distance from the origin),

• 
  $\theta = 90^\circ$ 

  (since the velocity is along a straight line parallel to the x-axis, the perpendicular distance is directly used).

Since

$\sin 90^\circ = 1$

, the equation simplifies to:

$$L = m v r$$

Substituting the given values:

$$L = (5) \times (3\sqrt{2}) \times (2\sqrt{5})$$

$$L = 5 \times 3 \times 2 \times \sqrt{10}$$

$$L = 30 \sqrt{10} \text{ g-cm²/s}$$

Thus, the magnitude of the angular momentum is:

$$\mathbf{30\sqrt{10} \text{ g-cm²/s}}$$

Angular momentum $L = mvr_{\perp}$. The line equation is $X - Y + 4 = 0$.
The perpendicular distance $r_{\perp}$ from the origin $(0,0)$ to the line is $\frac{|0 - 0 + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
Thus, $L = 5 \times (3\sqrt{2}) \times (2\sqrt{2}) = 60 \text{ units}$.

For a bound elliptical orbit, total energy $E$ is always negative. $T$ and $V$ vary with distance, and $L$ is conserved in both magnitude and direction as Torque is Zero. Gravitational force is passing through Center of Rotation so Toque is zero.