If vector \(\overrightarrow{A}   =   \cos \omega t \hat{i}   +   \sin \omega t \hat{j}\) and \(\overrightarrow{B} =\cos \dfrac{\omega t}{2} \hat{i} + \sin \dfrac{\omega t}{2} \hat{j}\) are functions of time, then the value of \(t\) at which they are orthogonal to each other will be the following:
1. \(t = \dfrac{\pi}{2\omega}\)
2. \(t = \dfrac{\pi}{\omega}\)
3. \(t=0\)
4. \(t = \dfrac{\pi}{4\omega}\)

If \(\left|\overrightarrow {v_1}+\overrightarrow {v_2}\right|= \left|\overrightarrow {v_1}-\overrightarrow {v_2}\right|\) and \(\overrightarrow {v_1}\) and \(\overrightarrow {v_2}\) are non-zero vectors, then:
1. \(\overrightarrow {v_1}\) is parallel to \(\overrightarrow {v_2}\)
2. \(\overrightarrow {v_1} = \overrightarrow {v_2}\)
3.  \(\overrightarrow {v_1}\) and \(\overrightarrow {v_2}\) are mutually perpendicular 
4. \(\left|\overrightarrow {v_1}\right|= \left|\overrightarrow {v_2}\right|\)

The component of vector \(\overrightarrow{A} = 3 \hat{i} + \hat{j} + \hat{k}\) along the direction of \(\hat{i} - \hat{j}\) is:
1. \(\sqrt{2}\)
2. \(2\)
3. \(\sqrt{3}\)
4. \(3\)

A force is \(60^{\circ}\) inclined to the horizontal. If its rectangular component in the horizontal direction is \(50\) N, then the magnitude of the force in the vertical direction is:

Component of $3\hat{\mathrm{i}}+4\hat{\mathrm{j}}$ perpendicular to $\hat{\mathrm{i}}+\hat{\mathrm{j}}$ and in the same plane as that of $3\hat{\mathrm{i}}+4\hat{\mathrm{j}}$ is:

1. $\frac{1}{2}\left(\hat{\mathrm{j}}-\hat{\mathrm{i}}\right)$

2. $\frac{3}{2}\left(\hat{\mathrm{j}}-\hat{\mathrm{i}}\right)$

3. $\frac{5}{2}\left(\hat{\mathrm{j}}-\hat{\mathrm{i}}\right)$

4. $\frac{7}{2}\left(\hat{\mathrm{j}}-\hat{\mathrm{i}}\right)$

At what angle must the two forces \((x+y)\) and \((x-y)\) act so that the resultant comes out to be \(\sqrt{x^2+y^2}\)?
1. \(\cos^{-1}\left(-\frac{x^2+y^2}{2(x^2-y^2)}\right )\)
2. \(\cos^{-1}\left(-\frac{2(x^2-y^2)}{(x^2+y^2)}\right )\)
3. \(\cos^{-1}\left(-\frac{x^2+y^2}{x^2-y^2}\right )\)
4. \(\cos^{-1}\left(-\frac{x^2-y^2}{x^2+y^2}\right )\)

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 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 varies with time according to the relation \(a=\alpha t+\beta.\) If the particle starts from rest, what is its velocity at time \(t\)? \(\left(\text{Here,}~ a=\dfrac{dv}{dt}\right)\)
1. \(\alpha t^2+\beta t\)
2. \(\alpha t^2+\beta t/2\)
3. \(\alpha t^2/2+\beta t\)
4. \(2\alpha t^2+\beta t\)

If a curve is governed by the equation \(y=\mathrm{sin}~x,\) then the area enclosed by the curve and \(x-\)axis between \(x = 0\) and \(x =\pi\)$\mathrm{}$ is (shaded region):
           
1. \(1\) unit
2. \(2\) units
3. \(3\) units
4. \(4\) units