$\mathrm{A}=4\mathrm{i}+4\mathrm{j}-4\mathrm{k}$ and $\mathrm{B}=3\mathrm{i}+\mathrm{j}+4\mathrm{k}$, then angle between vectors A and B is:

(1)  $180°$

(2)  $90°$

(3)  $45°$

(4)  $0°$

If vectors \(\overrightarrow{{A}}=\cos \omega t \hat{{i}}+\sin \omega t \hat{j}\) and \(\overrightarrow{{B}}=\cos \left(\frac{\omega t}{2}\right)\hat{{i}}+\sin \left(\frac{\omega t}{2}\right) \hat{j}\) are functions of time. Then, at what value of \(t\) are they orthogonal to one another?
1. \(t = \frac{\pi}{4\omega}\)
2. \(t = \frac{\pi}{2\omega}\)
3. \(t = \frac{\pi}{\omega}\)
4. \(t = 0\)

Six vectors $\overrightarrow{\mathrm{a}}$ through $\overrightarrow{\mathrm{f}}$ have the magnitudes and directions indicated in the figure. Which of the following statements is true? 

1. $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{f}}$

2. $\overrightarrow{\mathrm{d}}+\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{f}}$

3. $\overrightarrow{\mathrm{d}}+\overrightarrow{\mathrm{e}}=\overrightarrow{\mathrm{f}}$

4. $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{e}}=\overrightarrow{\mathrm{f}}$

$\stackrel{}{\mathrm{}}$\(\overrightarrow{A}\) and \(\overrightarrow B\) are two vectors and \(\theta\) is the angle between them. If \(\left|\overrightarrow A\times \overrightarrow B\right|= \sqrt{3}\left(\overrightarrow A\cdot \overrightarrow B\right),\) then the value of \(\theta\) will be:

If a curve is governed by the equation y = sinx, then the area enclosed by the curve and x-axis between x = 0 and x =$\mathrm{ \pi }$ is (shaded region):

              
1. \(1\) unit
2. \(2\) units
3. \(3\) units
4. \(4\) units

The acceleration of a particle starting from rest varies with time according to relation, $a=\alpha t+\beta$. The velocity of the particle at time instant \(t\) is: \(\left(\text{Here,}~ a=\frac{dv}{dt}\right)\)

1. $\alpha t^2+\beta t$

2. $\alpha t^2+\frac{\beta t}{2}$

3. $\frac{\alpha t^2}{2}+\beta t$

4. $2\alpha t^2+\beta t$

The displacement of the particle is zero at \(t=0\) and at \(t=t\) it is \(x\). It starts moving in the \(x\)-direction with a velocity that varies as \(v = k \sqrt{x}\), where \(k\) is constant. The velocity will: (Here, \(v=\frac{dx}{dt}\))

The acceleration of a particle is given as $a=3x^2$.  At t = 0, v = 0 and x = 0. It can then be concluded that the velocity at t = 2 sec will be: (Here, \(a=v\frac{dv}{dx}\))

1.  0.05 m/s

2. 0.5 m/s

3. 5 m/s

4. 50 m/s 

The acceleration of a particle is given by \(a=3t\) at \(t=0\), \(v=0\), \(x=0\). The velocity and displacement at \(t = 2~\text{sec}\) will be:
\(\left(\text{Here,} ~a=\frac{dv}{dt}~ \text{and}~v=\frac{dx}{dt}\right)\)
1. \(6~\text{m/s}, 4~\text{m}\)
2. \(4~\text{m/s}, 6~\text{m}\)
3. \(3~\text{m/s}, 2~\text{m}\)
4. \(2~\text{m/s}, 3~\text{m}\)

The 9 kg block is moving to the right with a velocity of 0.6 m/s on a horizontal surface when a force F, whose time variation is shown in the graph, is applied to it at time t = 0. Calculate the velocity v of the block when t= 0.4s. The coefficient of kinetic fricton is ${\mu }_k=0.3$. [This question includes concepts from Work, Energy & Power chapter]

1. 0.6 m/s

2. 1.2 m/s

3. 1.8 m/s

4. 2.4 m/s