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. \(18\cos 75^{\circ}~\text{km/hr}\)

Consider east as positive x-axis, north as positive y-axis and vertically upward direction as z-axis. A helicopter first rises up to an altitude of 100 m than flies straight in north 500 m and then suddenly takes a turn towards east and travels 1000 m east. What is position vector of helicopter. (Take starting point as origin)

(1)  $1000 \hat{\mathrm{i}}-500 \hat{\mathrm{j}}+100 \hat{\mathrm{k}}$

(2)  $1000 \hat{\mathrm{i}}+500 \hat{\mathrm{j}}-100 \hat{\mathrm{k}}$

(3)  $1000 \hat{\mathrm{i}}+500 \hat{\mathrm{j}}+100 \hat{\mathrm{k}}$

(4)  $-1000 \hat{\mathrm{i}}+500 \hat{\mathrm{j}}+100 \hat{\mathrm{k}}$

If rain falls vertically with a velocity ${\mathrm{V}}_{\mathrm{r}}$ wrt wind and wind blows with a velocity ${\mathrm{V}}_{\mathrm{w}}$ from east to west, then a person standing on the roadside should hold the umbrella in the direction

1. $\tan \theta = \frac{V_w}{V_r}$

2. $\tan \theta \mathit{ }\mathit{=}\mathit{ }\frac{{\mathit{V}}_{\mathit{r}}}{{\mathit{V}}_{\mathit{w}}}$

3. $\tan \theta \mathit{ }\mathit{=}\mathit{ }\frac{{\mathit{V}}_{\mathit{r}\mathit{w}}}{\sqrt{\mathit{V}{\mathit{r}}^{\mathit{2}}\mathit{ }\mathit{+}\mathit{ }\mathit{V}{\mathit{w}}^{\mathit{2}}}}$

4. $\tan \theta \mathit{ }\mathit{=}\mathit{ }\frac{{\mathit{V}}_{\mathit{r}}}{\sqrt{\mathit{V}{\mathit{r}}^{\mathit{2}}\mathit{ }\mathit{+}\mathit{ }\mathit{V}{\mathit{w}}^{\mathit{2}}}}$

The component of vector $\overrightarrow{A}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}$ along the direction of $\hat{i}-\hat{j}$ is

(1) $a_x-a_y+a_z$

(2) $a_x-a_y$

(3) $(a_x-a_y)/\sqrt{2}$

(4) $(a_x+a_y+a_z)$

A force is \(60^{\circ}\) inclined to the horizontal. If its rectangular component in the horizontal direction is \(50\) N, then the magnitude of the force in the vertical direction is:

If a unit vector \(\hat j\) is rotated through an angle of \(45^{\circ}\) anticlockwise, then the new vector will be:
1. \(\sqrt{2}\hat i + \sqrt{2}\hat j\)
2. \(\hat i + \hat j\)
3. \(\frac{1}{\sqrt{2}}\hat i + \frac{1}{\sqrt{2}}\hat j\)
4. \(-\frac{1}{\sqrt{2}}\hat i + \frac{1}{\sqrt{2}}\hat j\)

If $\hat{\mathrm{p}}$ is the unit vector in the direction $\overrightarrow{\mathrm{B}}$, then:

1.  $\hat{\mathrm{p}} = \frac{\left|\overrightarrow{\mathrm{B}}\right|}{\overrightarrow{\mathrm{B}}}$

2.  $\hat{\mathrm{p}} = \left|\overrightarrow{\mathrm{B}}\right| \times \overrightarrow{\mathrm{B}}$

3.  $\hat{\mathrm{p}} = \frac{\overrightarrow{\mathrm{B}}}{\left|\overrightarrow{\mathrm{B}}\right|}$

4.  $\hat{\mathrm{p}} = \frac{\overrightarrow{\mathrm{B}}}{\overrightarrow{\mathrm{B}}}$

A force of \(20\) N acts on a particle along a direction, making an angle of \(60^\circ\) with the vertical. The component of the force along the vertical direction will be:

A force of 5N acts on a particle along a direction making an angle of $60°$ with vertical. Its vertical component will be:

1. 10 N

2. 3 N

3. 4 N

4. 2.5 N