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 \(t = 2~\text{sec}\) 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. \(2~\text{m/s}, 3~\text{m}\)

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

$\stackrel{}{\mathrm{}}$\(\overrightarrow{A}\) 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 \(\overrightarrow{a}\) is a vector and \(x\) is a non-zero scalar, then which of the following is correct?

A body is moving according to the equation \(x = at +bt^2-ct^3\) where \(x\) represents displacement and \(a, b~\text{and}~c\) are constants. The acceleration of the body is: (\(\text{Given:}~ a=\frac{d^2x}{dt^2}\))
1. \(a+ 2bt\)
2. \(2b+ 6ct\)
3. \(2b- 6ct\)
4. \(3b- 6ct^2\)

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 \(200~\text{N}\) at an angle of $\mathrm{}$\(60^{\circ}\) above the horizontal. Then the horizontal and vertical components of the force will be:
              

1. \(100~\text{N}, ~175~\text{N}\)
2. \(86.6~\text{N}, ~100~\text{N}\)
3. \(100~\text{N}, ~86.6~\text{N}\)
4. \(100~\text{N}, ~0~\text{N}\)

In the given figure

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