If $\left|\overrightarrow{{\mathrm{v}}_1}+\overrightarrow{{\mathrm{v}}_2}\right|=\left|\overrightarrow{{\mathrm{v}}_1}-\overrightarrow{{\mathrm{v}}_2}\right|$ and $\overrightarrow{{\mathrm{v}}_1}$ and $\overrightarrow{{\mathrm{v}}_2}$ are non-zero vectors, then :

1. $\overrightarrow{{\mathrm{v}}_1}$ is parallel to $\overrightarrow{{\mathrm{v}}_2}$

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

3. $\overrightarrow{{\mathrm{v}}_1}$ and $\overrightarrow{{\mathrm{v}}_2}$ are mutually perpendicular 

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

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 ° 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

1. 25 N

2. 75 N

3. 87 N

4. 100 N

If vector $\overrightarrow{\mathrm{A}} = \cos \omega t\hat{i} + \sin \omega t\hat{j}$ and $\overrightarrow{\mathrm{B}} = \cos \frac{\omega t}{2}\hat{i} + \sin \frac{\omega t}{2}\hat{j}$ are functions of time, then the value of t at which they are orthogonal to each other will be:

1. $\mathrm{t} = \frac{\mathrm{\pi }}{2\mathrm{\omega }}$

2. $\mathrm{t} = \frac{\mathrm{\pi }}{\mathrm{\omega }}$

3. $\mathrm{t} = 0$

4. $\mathrm{t} = \frac{\mathrm{\pi }}{4\mathrm{\omega }}$

If $\overrightarrow{\mathrm{A}}=\overrightarrow{B}+\overrightarrow{C}$, and the magnitudes of $\overrightarrow{\mathrm{A}},\overrightarrow{B},\overrightarrow{C}$ are 5, 4 and 3 units, respectively. Then the angle between $\overrightarrow{\mathrm{A}} and \overrightarrow{C}$ is:

1. ${\cos }^{-1}\left(\frac{3}{5}\right)$

2. ${\cos }^{-1}\left(\frac{4}{5}\right)$

3. ${\sin }^{-1}\left(\frac{3}{4}\right)$

4. $\frac{\mathrm{\pi }}{2}$

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 $\left(\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\right)$ ?

1. ${\cos }^{-1}\left(-\frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{2({\mathrm{x}}^2-{\mathrm{y}}^2)}\right)$

2. ${\cos }^{-1}\left(-\frac{2({\mathrm{x}}^2-{\mathrm{y}}^2)}{{\mathrm{x}}^2+{\mathrm{y}}^2}\right)$

3. ${\cos }^{-1}\left(-\frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{{\mathrm{x}}^2-{\mathrm{y}}^2}\right)$

4. ${\cos }^{-1}\left(-\frac{{\mathrm{x}}^2-{\mathrm{y}}^2}{{\mathrm{x}}^2+{\mathrm{y}}^2}\right)$

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

1. 6 m/s, 4 m

2. 4 m/s, 6 m

3. 3 m/s, 2 m

4. 2 m/s, 3 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, \(v=\frac{dx}{dt}\))

1. vary with time.

2. be independent of time.

3. be inversely proportional to time.

4. be inversely proportional to acceleration.

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: (\(Here, a=\frac{dv}{dt}\))

1. $\alpha t^2+\beta t$

2. $\alpha t^2+\frac{\beta t}{2}$

3. $\frac{\alpha t^2}{2}+\beta t$

4. $2\alpha t^2+\beta t$