A particle starting from the point \((1,2)\) moves in a straight line in the XY-plane. Its coordinates at a later time are \((2,3).\) The path of the particle makes with x-axis an angle of:

1.	\(30^\circ\)	2.	\(45^\circ\)
3.	\(60^\circ\)	4.	data is insufficient

A particle is projected from a horizontal plane (x-z plane) such that its velocity vector at time t is given by \(\overrightarrow{\mathrm{v}}=a \hat{i}+(b-c t )\hat{j}.\) Its range on this horizontal plane is given by: 

A river is 1 km wide. The banks are straight and parallel. The current is 5 km/h and is parallel to the banks. A boat has a maximum speed of 3 km/h in still water. In what direction should the boat head so as to arrive at point B directly opposite to its starting point A?

1. directly across the river

2. head $53°$ upstream from the line AB

3. head $37°$ $\mathrm{upstream}$ $\mathrm{from}$ $\mathrm{the}$ $\mathrm{line}$ $\mathrm{AB}$

4. the trip from A to B is not possible with this speed

Two particles A and B, move with constant velocities \(\vec{v_1}\)   and  \(\vec{v_2}\) . At the initial moment their position vector are  \(\vec{r_1}\) and  \(\vec{r_2}\)  respectively. The condition for particles A and B for their collision to happen will be:

1.  $\overrightarrow{{\mathrm{r}}_{1 }}.\overrightarrow{{\mathrm{v}}_1} =\overrightarrow{{\mathrm{r}}_{2 }}.\overrightarrow{{\mathrm{v}}_2}$

2.  $\overrightarrow{{\mathrm{r}}_1} \mathrm{x} \overrightarrow{{\mathrm{v}}_1} =\overrightarrow{{\mathrm{r}}_2} \mathrm{x} \overrightarrow{{\mathrm{v}}_2}$

3. $\overrightarrow{{\mathrm{r}}_1}- \stackrel{\rightharpoonup }{{\mathrm{r}}_2} =\overrightarrow{{\mathrm{v}}_1}- \overrightarrow{{\mathrm{v}}_2}$

4. $\frac{\overrightarrow{{\mathrm{r}}_1}- \overrightarrow{{\mathrm{r}}_2}}{\left|\overrightarrow{{\mathrm{r}}_1}- \overrightarrow{{\mathrm{r}}_2}\right|}= \frac{\overrightarrow{{\mathrm{v}}_2}- \overrightarrow{{\mathrm{v}}_1}}{\left|\overrightarrow{{\mathrm{v}}_2}- \overrightarrow{{\mathrm{v}}_1}\right|}$

A particle moves along a parabolic path y = 9x² in such a way that the x component of the velocity remains constant and has a value of $\frac{1}{3}$ ${\mathrm{ms}}^{-1}$. It can be deduced that the acceleration of the particle will be:

1. $\frac{1}{3}$ $\hat{\mathrm{j}}$ ${\mathrm{ms}}^{-2}$

2. $3$ $\hat{\mathrm{j}}$ ${\mathrm{ms}}^{-2}$

3. $\frac{2}{3}$ $\hat{\mathrm{j}}$ ${\mathrm{ms}}^{-2}$

4. $2$ $\hat{\mathrm{j}}$ ${\mathrm{ms}}^{-2}$

Three balls are thrown from the top of a building with equal speeds at different angles. When the balls strike the ground, their speeds are ${\mathrm{v}}_1,{\mathrm{v}}_2$ $\mathrm{and}$ ${\mathrm{v}}_3$ respectively, then:
           

1. ${\mathrm{v}}_1>{\mathrm{v}}_2>{\mathrm{v}}_3$

2. ${\mathrm{v}}_3>{\mathrm{v}}_2={\mathrm{v}}_1$

3. ${\mathrm{v}}_1={\mathrm{v}}_2={\mathrm{v}}_3$

4. ${\mathrm{v}}_1<{\mathrm{v}}_2<{\mathrm{v}}_3$

Figure below shows a body of mass M moving with a uniform speed v on a circular path of radius, R. What is the change in acceleration in going from P₁ to P₂? 
              

1. zero

2. ${\mathrm{v}}^2/2\mathrm{R}$

3. $2{\mathrm{v}}^2/\mathrm{R}$

4. $\frac{{\mathrm{v}}^2}{\mathrm{R}}\times \sqrt{2}$

A particle is projected with a speed u at an angle $\mathrm{\theta }$ to the horizontal. Radius of curvature at highest point of its trajectory is?

1. $\frac{{\mathrm{u}}^2{\cos }^2\mathrm{\theta }}{2\mathrm{g}}$

2. $\frac{\sqrt{3}{\mathrm{u}}^2{\cos }^2\mathrm{\theta }}{2\mathrm{g}}$

3. $\frac{{\mathrm{u}}^2{\cos }^2\mathrm{\theta }}{\mathrm{g}}$

4. $\frac{\sqrt{3}{\mathrm{u}}^2{\cos }^2\mathrm{\theta }}{\mathrm{g}}$

Two particles A and B are moving in a uniform circular motion in concentric circles of radii $r_A$ $and$ $r_B$ with speeds $v_A$ $and$ $v_B$ respectively. Their time periods of rotation are the same. The ratio of angular speed of A  to that of B will be: