If the angle between vector a and b is an acute angle, then the difference a$-$b is -

(1)  the main diagonal of the parallelogram

(2)  the minor diagonal of the parallelogram

(3)  any of the above

(4)  none of the above

What displacement must be added to the displacement $25 \hat{\mathrm{i}}-6 \hat{\mathrm{j}} \mathrm{m}$  to give a displacement of $7.0 \mathrm{m}$ pointing in the x-direction?

(1)  $18 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}$

(2)  $32 \hat{\mathrm{i}}-13 \hat{\mathrm{j}}$

(3)  $-18 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}$

(4)  $-25 \hat{\mathrm{i}}+13 \hat{\mathrm{j}}$

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. \(100~\text{N}, ~0~\text{N}\)

A man rows a boat at a speed of $\mathrm{}$\(18~\text{km/hr}\) in the northwest direction. The shoreline makes an angle of $\mathrm{}$\(15^{\circ}\) south of west. The component of the velocity of the boat along the shoreline is:
1. \(9~\text{km/hr}\)
2. \(18\frac{\sqrt{3}}{2}~\text{km/hr}\)
3. \(18\cos 15^{\circ}~\text{km/hr}\)
4. \(18\cos 75^{\circ}~\text{km/hr}\)

A bird moves from point (1, -2) to (4, 2). If the speed of the bird is 10 m/sec, then the velocity vector of the bird is:

1. $5\left(\hat{\mathrm{i}}-2\hat{\mathrm{j}}\right)$

2. $5\left(4\hat{\mathrm{i}}+2\hat{\mathrm{j}}\right)$

3. $0.6\hat{\mathrm{i}}+0.8\hat{\mathrm{j}}$

4. $6\hat{\mathrm{i}}+8\hat{\mathrm{j}}$

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. $\sqrt{41} \mathrm{units}$