Question

A particle of mass $m$ is projected in vertical plane at an angle of $\theta$ with the horizontal at a speed $V_0$. Match the physical quantities in column I with its variation in Column II
$$\begin{array}{cccc}& \text{Column I}& & \text{Column - II}\\ \text{(A)}& \text{Magnitude of momentum}& \text{(p)}& \text{Continuously increasing linearly with time}\\ \text{(B)}& \text{Magnitude of angular momentum about}& \text{(q)}& \text{First decreases and then increases}\\ & \text{point of projection}& & \\ \text{(C)}& \text{Distance of the particle from point of projection}& \text{(r)}& \text{Continuously increasing but not linearly with time}\\ \text{(D)}& \text{Magnitude of Torque about point of projection}& \text{(s)}& \text{increasing linearly with horizontal displacement}\end{array}$$

• A. $A\to q;B\to r;C\to r;D\to p,s$
• B. $A\to p;B\to r;C\to r;D\to q$
• C. $A\to p,s;B\to q;C\to r;D\to r$
• D. $A\to s;B\to p,s;C\to q;D\to r$

Answer 1

The correct option is A $A\to q;B\to r;C\to r;D\to p,s$

The speed of the particle first decreases and then increases and so does the magnitude of linear momentum.

$\frac{d}{dt}\overrightarrow{L}=\overrightarrow{\tau }$
$\tau =mgx=mg{V}_0\cos \theta .t$
$\Delta L=\int mgxdt=\int mg{V}_0\cos \theta .tdt\Delta L=mg{V}_0\cos \theta \frac{t^2}{2}$
With time and horizontal displacement, magnitude of torque linearly increases.

Angular momentum increases as a funtion of $t^2$ and as function of $x^2$.

The distance of the particle from point of projection continuously increases but non linearly.