If the angle between the unit vectors $\hat{\mathrm{a}}$ and $\hat{\mathrm{b}}$ is $60°$, then $|\hat{\mathrm{a}}-\hat{\mathrm{b}}|$ is :

(1)  0

(2)  1

(3)  2

(4)  4

For the figure -

1.  A+B=C

2.  B+C=A

3.  C+A=B

4.  A+B+C=0

Two constant forces ${\overrightarrow{\mathrm{F}}}_1=2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+3\hat{\mathrm{k}} \left(\mathrm{N}\right)$ and ${\overrightarrow{\mathrm{F}}}_2=\hat{\mathrm{i}}+\hat{\mathrm{j}}-2\hat{\mathrm{k}} \left(\mathrm{N}\right)$ act on a body and displace it from the position ${\overrightarrow{\mathrm{r}}}_1=\hat{\mathrm{i}}+2\hat{\mathrm{j}}-2\hat{\mathrm{k}} \left(\mathrm{m}\right)$ to the position ${\overrightarrow{\mathrm{r}}}_2=7\hat{\mathrm{i}}+10\hat{\mathrm{j}}+5\hat{\mathrm{k}} \left(\mathrm{m}\right)$. What is the work done W? $[Given : W =\overrightarrow{F}.∆\overrightarrow{r}]$

(A)  9 Joule

(B)  41 Joule

(C)  -3 Joule

(D)  None of these

A vector A points, vertically upward and, B points towards north. The vector product $\mathrm{A}\times \mathrm{B}$ is -

1.  along west

2.  along east

3.  zero

4.  vertically downward

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. $\sqrt{41} \mathrm{units}$

The displacement of a particle is given by \(y = a+bt+ct^2-dt^4\). The initial velocity and initial acceleration, respectively, are: \(\left(\text{Given:}~ v=\frac{dx}{dt}~\text{and}~a=\frac{d^2x}{dt^2}\right)\)
1. \(b, -4d\)
2. \(-d, 2c\)
3. \(b, 2c\)
4. \(2c, -4d\)

The momentum is given by p=4t+1, the force at t=2s is-$[Given: F=\frac{dP}{dt}]$

(A)  4 N

(B)  8 N

(C)  10 N

(D)  15 N

If the momentum of a particle is given by P=(180-8t) kg m/s, then its force will be-$[Given: F=\frac{dP}{dt}]$

(1)  Zero

(2)  8 N

(3)  -8 N

(4)  4 N

If \(f \left(x\right) = x^{2} - 2 x + 4\), then \(f(x)\) has:

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

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