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

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

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

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

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

For the given figure, which of the following is true?

1. $\overrightarrow{\mathrm{B}}$ = $\overrightarrow{\mathrm{A}}$ + $\overrightarrow{\mathrm{C}}$

2. $\overrightarrow{\mathrm{A}}$ = $\overrightarrow{\mathrm{B}}$ + $\overrightarrow{\mathrm{C}}$

3. $\overrightarrow{\mathrm{C}}$ = $\overrightarrow{\mathrm{A}}$ + $\overrightarrow{\mathrm{B}}$

4. All of these

The value of the unit vector, which is perpendicular to both A = \(\hat{i} + 2\hat{j} + 3 \hat{k}\) and B = \(\hat{i} - 2\hat{j} - 3 \hat{k}\) $\mathrm{is\; equal\; to}$:

1.  $\frac{\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 3\hat{\mathrm{k}}}{6}$

2.  $\frac{6\hat{\mathrm{j}} - 4\hat{\mathrm{k}}}{\sqrt{52}}$

3.  $\frac{6\hat{\mathrm{j}} + 4\hat{\mathrm{k}}}{\sqrt{52}}$

4.  $\frac{2\hat{\mathrm{i}} - \hat{\mathrm{j}}}{\sqrt{5}}$

The area of a blot of ink, A, is growing such that after t seconds, \(A=3t^2+\frac{t}{5}+7~m^2\). Then the rate of increase in the area at t= 5s will be :

1. 30.1 m²/s

2. 30.2 m²/s

3. 30.3 m²/s

4. 30.4 m²/s

The current in a circuit is defined as $I=\frac{dq}{dt}$. The charge (q) flowing through a circuit, as a function of time (t), is given by $q=5t^2-20t+3$. The minimum charge flows through the circuit at:

1. t=4 sec

2. t=2 sec

3. t= 6 sec

4. t=3 sec

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