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

The angle that the vector $\mathrm{A}=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}$ makes with the y-axis is:

(1)  ${\tan }^{-1} 3/2$

(2)  ${\tan }^{-1} 2/3$

(3)  ${\sin }^{-1} 2/3$

(4)  ${\cos }^{-1} 3/2$

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 $\mathrm{y}=\mathrm{a}+\mathrm{bt}+{\mathrm{ct}}^2-{\mathrm{dt}}^4$. The initial velocity and initial acceleration, respectively, are: (\(Given: v=\frac{dx}{dt}~and~a=\frac{d^2x}{dt^2}\))

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

The maximum value of the function $7+6\mathrm{x}-9{\mathrm{x}}^2$ is:

1.  8

2.  -8

3.  4

4.  -4

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

1.  a minimum at x=1.

2.  a maximum at x=1.

3.  no extreme point.

4.  no minimum.