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

1.	\(x\overrightarrow{a}\) is a vector in the direction of \(\overrightarrow{a}\).
2.	\(x\overrightarrow{a}\) is a vector collinear to \(\overrightarrow{a}\).
3.	\(x\overrightarrow{a}\) and \(\overrightarrow{a}\) have independent directions.
4.	\(x\overrightarrow{a}\) is a vector perpendicular to \(\overrightarrow{a}\).

The vector \(\overrightarrow b\) which is collinear with the vector \(\overrightarrow a = \left(2, 1, -1\right)\) and satisfies the condition \(\overrightarrow a. \overrightarrow b=3\) is:
1. \(\left(1, \frac{1}{2}, \frac{-1}{2}\right)\)
2. \(\left(\frac{2}{3}, \frac{1}{3}, \frac{-1}{3}\right)\)
3. \(\left(\frac{1}{2}, \frac{1}{4}, \frac{-1}{4}\right)\)
4. \(\left(1, 1, 0\right)\)

If a, b and c are three non-zero vectors such that $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0$, then the value of $\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}.\overrightarrow{\mathrm{a}}$ will be:

If $\stackrel{}{\mathrm{}}$\(\overrightarrow P +\overrightarrow Q\) = \(\overrightarrow P-\overrightarrow Q\) and \(\theta\) is the angle between \(\overrightarrow {P}\) and \(\overrightarrow {Q}\), then:
1. \(\theta = 0^{\circ}\)
2. \(\theta = 90^{\circ}\)
3. \(P=0\)
4. \(Q=0\)

What is the torque of a force $\overrightarrow{\mathrm{F}}=\left(2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+4\hat{\mathrm{k}} \right)$newton acting at a point $\overrightarrow{\mathrm{r}}=\left(3\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}} \right)$ metre about the origin? (Given: $\overrightarrow{\mathrm{\tau }} =\overrightarrow{\mathrm{r} }\times \overrightarrow{\mathrm{F}}$)

1. $6\hat{\mathrm{i}}-6\hat{\mathrm{j}}+12\hat{\mathrm{k}}$

2. $17\hat{\mathrm{i}}-6\hat{\mathrm{j}}-13\hat{\mathrm{k}}$

3. $-6\hat{\mathrm{i}}+6\hat{\mathrm{j}}-12\hat{\mathrm{k}}$

4. $-17\hat{\mathrm{i}}+6\hat{\mathrm{j}}-13\hat{\mathrm{k}}$

The momentum of a body moving in a straight line is $\mathrm{p}=\left({\mathrm{t}}^2+2\mathrm{t}+1\right) \mathrm{kg} \mathrm{m}/\mathrm{s}$. Force acting on the body at t=2 sec will be: \(\left(\text{Given:}~ F=\frac{dp}{dt}\right)\)
1. 6 N
2. 8 N
3. 4 N
4. 2 N

Temperature of a body varies with time as $\mathrm{T}=\left({\mathrm{T}}_0+{\mathrm{at}}^2+\mathrm{b} \mathrm{sint}\right)\mathrm{K}$, where ${\mathrm{T}}_0$ is the temperature in Kelvin at $\mathrm{t}=0 \sec \& \mathrm{a}=2/\mathrm{\pi } \mathrm{K}/{\mathrm{s}}^2 \& \mathrm{b}=-\mathrm{4 }\mathrm{K}$, then the rate of change of temperature $\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)$ at $\mathrm{t}=\mathrm{\pi } \sec$ is:
1. \(8~\text{K}\)
2. \(80~\text{K}\)
3. \(8~\text{K/sec}\)
4. \(80~\text{K/sec}\)

$\stackrel{}{\mathrm{}}$\(\overrightarrow{A}\) 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:

A particle is moving along the x-axis. The velocity v of this particle varies with its position x as $\mathrm{v}=\frac{1}{\mathrm{x}}$. Find the velocity of the particle as a function of time t given that at t=0, x=1 and v=$\frac{\mathrm{dx}}{\mathrm{dt}}$.

1. $\mathrm{v}=\sqrt{2\mathrm{t}+1}$

2. $\mathrm{v}=\frac{1}{\sqrt{2\mathrm{t}+1}}$

3. $\mathrm{v}=\frac{1}{\mathrm{t}}$

4. None of these