A certain vector in the xy-plane has an x-component of \(4\) m and a y-component of \(10\) m. It is then rotated in the xy-plane so that its x-component is doubled. Then, its new y-component will be: (approximately)
1. \(20\) m
2. \(7.2\) m
3. \(5.0\) m
4. \(4.5\) m

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

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

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

1. 60^o

2. 45^o

3. 30^o

4. 90^o

If \(\vec{a}\) is a vector and \(x\) is a non-zero scalar, then which of the following is correct?
1. \(x\vec{a}\) is a vector in the direction of \(\vec{a}\)
2. \(x\vec{a}\) is a vector collinear to \(\vec{a}\)
3. \(x\vec{a}\) and \(\vec{a}\) have independent directions
4. \(x\vec{a}\) is a vector perpendicular to \(\vec{a}\)

A body is moving according to the equation $\mathrm{x}=\mathrm{at}+{\mathrm{bt}}^2-{\mathrm{ct}}^3$ where x represents displacement and a, b and c are constants. The acceleration of the body is: (\(Given: a=\frac{d^2x}{dt^2}\))

1.  $\mathrm{a}+2\mathrm{bt}$

2.  $2\mathrm{b}+6\mathrm{ct}$

3.  $2\mathrm{b}-6\mathrm{ct}$

4.  $3\mathrm{b}-6{\mathrm{ct}}^2$

The linear velocity of a rotating body is given by $\overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{\omega }}\times \overrightarrow{\mathrm{r}}$, where $\mathrm{\omega }$ is the angular velocity and r is the radius vector. The angular velocity of a body, $\overrightarrow{\mathrm{\omega }}=\hat{\mathrm{i}}-2\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ and their radius vector is  $\overrightarrow{\mathrm{r}}=4\hat{\mathrm{j}}-3\hat{\mathrm{k}}, \mathrm{then\; value\; of} \left|\overrightarrow{\mathrm{v}}\right|$ will be:

1.  $\sqrt{29} \mathrm{units}$

2.  $31 \mathrm{units}$

3.  $\sqrt{37} \mathrm{units}$

4.  $\sqrt{41} \mathrm{units}$

A child pulls a box with a force of $200 \mathrm{N}$ at an angle of $60°$above the horizontal. Then the horizontal and vertical components of the force will be:
              

1. 100 N, 175 N

2.  86.6 N, 100 N  

3.  100 N, 86.6 N

4. 100 N, 0 N

In the given figure

1.  Angle between $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is $110°$

2.  Angle between $\overrightarrow{\mathrm{C}}$ and $\overrightarrow{\mathrm{D}}$ is $60°$

3.  Angle between $\overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{C}}$ is $110°$

4.  Angle between $\overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{C}}$ is $70°$

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