Question

The angular momentum of a particle about origin is varying as L = 4t +8 (Sl units) when it moves along a straight line y = x - 4 (x,y in meters).The magnitude of force acting on the particle wouid be:

• A. 1 N
• B. 2 N
• C. $\sqrt{2}N$
• D. $\sqrt{3}N$

Answer 1

The correct option is C $\sqrt{2}N$

The body move in a line

$y=x-4$

It have a slope $\frac{dy}{dx}=1$ and intersection at $(0,-4),(4,0)$

The body moves along line $ABC$ so, force also acting along this line.

$O$ is the perpendicular distance of line $ABC$ then $O$ is

$={\alpha }^{\prime }F$

$=2\sqrt{2}F$

Where

$\frac{O{O}^{\prime }}{OB}=\sin {45}^o$

$\Rightarrow O{O}^{\prime }=4\frac{1}{\sqrt{2}}=2\sqrt{2}$

Again torque about $O$ is

$\frac{dL}{dt}=\frac{d}{dt}(4t+8)$

$\Rightarrow \frac{dL}{dt}=4$

Hence

$4=2\sqrt{2}F$

$\Rightarrow F=\sqrt{2}N$