Six particles situated at the corners of a regular hexagon of side \(a\) move at constant speed \(v\). Each particle maintains a direction towards the particle at the next. The time which the particles will take to meet each other is:
1. \(\frac{2 a}{v}~\text{sec}\)
2. \(\frac{a}{v}~\text{sec}\)
3. \(\frac{2 a}{3v}~\text{sec}\)
4. \(\frac{3 a}{v}~\text{sec}\)

A body is projected with velocity $20\sqrt{3}$ m/s with an angle of projection 60$°$ with horizontal. Calculate velocity on that point where body makes an angle 30$°$ with the horizontal.

1. 20 m/s

2. $\frac{20}{\sqrt{3}}<mspace\; width="1em"></mspace>m/s$

3. $10\sqrt{3}<mspace\; width="1em"></mspace>m/s$

4. 10 m/s

In a uniform circular motion, which of the following quantity is not constant

1. Angular momentum

2. Speed

3. Kinetic energy

4. Momentum

A particle is moving with veocity $\overrightarrow{\mathrm{v}}=\mathrm{k}(\mathrm{y}\hat{\mathrm{i}}+\mathrm{x}\hat{\mathrm{j}})$; where k is constant. The general equation for the path is:

1. $\mathrm{y}={\mathrm{x}}^2+\mathrm{constant}$

2. ${\mathrm{y}}^2={\mathrm{x}}^2+\mathrm{constant}$

3. $\mathrm{y}=\mathrm{x}+\mathrm{constant}$

4. xy=constant

A particle is projected with a velocity u making an angle $\mathrm{\theta }$ with the horizontal. At any instant, its velocity v is at right angles to its initial velocity u; then v is:

1.  ucos$\mathrm{\theta }$

2.  utan$\mathrm{\theta }$

3.  ucot$\mathrm{\theta }$

4.  usec$\mathrm{\theta }$

A projectile is given an initial velocity of \(\hat i+2\hat j\). The cartesian equation of its path is: (Take \(g = 10 \ \text{ms}^{ - 2}\)) 

1. \(y=2x-5x^2\)

2. \(y=x-5x^2\) $\mathrm{}$

3. \(4y=2x-5x^2\)$\mathrm{}$

4. \(y=2x-25x^2\)

Ship A is moving westward with a speed of \(10\ \text{km h}^{- 1}\) and Ship B, south of A, is moving northwards with a speed of \(10\ \text{km h}^{- 1}\)${\mathrm{}}^{}$. The time after which the distance between them becomes the shortest is:

1. \(5\) hr

2. \(5 \sqrt2\) hr

3. \(10 \sqrt2\)$\mathrm{}$ hr

4. \(0\) hr

The time taken by the projectile to reach from \(A\) to \(B\) is \(t\). Then the distance \(AB\) is equal to :

1. \(\dfrac{ut}{\sqrt3}\)

2. \(\dfrac{\sqrt3ut}{2}\)

3. \(\sqrt3ut\)

4. \(2ut\)

A particle projected with kinetic energy ${\mathrm{k}}_0$ with an angle of projection $\mathrm{\theta }$. Then the variation of kinetic K with vertical displacement y is

1.  linear

2.  parabolic

3.  hyperbolic

4.  periodic

A river is flowing with a speed of \(1\ \text{km/hr}\). A swimmer wants to go to point \(C\) starting from \(A\). He swims with a speed of \(5\ \text{km/hr}\), at an angle \(\theta\)  with respect to the river. If \(\text {AB = BC} = 400\ \text{m}\), then:

1. The time taken by the man is \(12\) min

2. The time taken by the man is \(8\) min

3. The value of \(\theta\) is \(45^\circ\)

4. The value of \(\theta\) is \(53^\circ\)