A particle is moving along the path y = ${\mathrm{x}}^2$ from x = 0 m to x = 2 m. Then the distance traveled by the particle is:

1.  4 m

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

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

4.  $<\sqrt{20}<mspace\; width="1em"></mspace>\mathrm{m}$

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. \(\dfrac{2 a}{v}\ \text{seconds}\)

2. \(\dfrac{a}{v}~\text{seconds}\)

3. \(\dfrac{2 a}{3v}~\text{seconds}\)

4. \(\dfrac{3 a}{v}~\text{seconds}\)

A body is projected with velocity \(20\sqrt3 \text{  m/s}\) with an angle of projection \(60^°\) with the horizontal. The velocity at the point where the body makes an angle \(30^°  \)with the horizontal is:

1. \(20 \text{  m/s}\)

2. \(\dfrac{20}{\sqrt3} \text{  m/s}\)

3. \(10\sqrt3 \text{  m/s}\)

4. \(10 \text{  m/s}\)

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 velocity \(\vec v=k(y\hat i+x \hat j)\), where \(k\) is constant. The general equation for the path is:

1. \(y=x^2+\text{constant}\)

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

3. \(y=x+\text{constant}\)$\mathrm{}$

4. \(xy=\text{constant}\)

A particle is projected with a velocity \(u\) making an angle \(θ\) with the horizontal. At any instant, its velocity \(v\) is at right angles to its initial velocity \(u\); then \(v\) is:

1.  \(u \cos \theta \)

2.  \(u \tan θ\)

3.  \( u \cot θ\)

4.  \(u \sec θ\)

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, \(100 \ \text{km}\)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