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.

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

A vector $\overrightarrow{\mathrm{A}}$ is directed along $30°$ west of north direction and another vector $\overrightarrow{\mathrm{B}}$ along $15°$ south of east. Their resultant cannot be in ____________ direction.

(1)  North

(2)  East

(3)  North-East

(4)  South

ABCD is a quadrilateral. Forces $\overrightarrow{\mathrm{BA}}, \overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{CD}} \& \overrightarrow{\mathrm{DA}}$ act at a point. Their resultant is 

(A)  $2 \overrightarrow{\mathrm{AB}}$

(B)  $2 \overrightarrow{\mathrm{DA}}$

(C)  zero vector

(D)  $2 \overrightarrow{\mathrm{BA}}$

The maximum and minimum magnitude of the resultant of two vectors are 17 units and 7 units respectively. Then the magnitude of resultant of the vectors when they act perpendicular to each other is:

(1)  14

(2)  16

(3)  18

(4)  13

A vector $\overrightarrow{\mathrm{A}}$ makes an angle of $20°$ and $\overrightarrow{\mathrm{B}}$ makes an angle of $110°$ with the X-axis. The magnitudes of these vectors are 3 m and 4 m respectively. Find the magnitude of the resultant.

(A)  $1$

(B)  $5\sqrt{2}$

(C)  $5$

(D)  $7$

A particle is moving westward with a velocity ${\overrightarrow{\mathrm{v}}}_i=5 \mathrm{m}/\mathrm{s}.$ Its velocity changed to ${\overrightarrow{\mathrm{v}}}_f=5\mathrm{m}/\mathrm{s}$ northward. The change in velocity vector $\left(∆\overrightarrow{\mathrm{V}}={\overrightarrow{\mathrm{v}}}_f-{\overrightarrow{\mathrm{v}}}_i\right)$ is:

(1)  $5\sqrt{2} \mathrm{m}/\mathrm{s}$ towards north east

(2)  $5 \mathrm{m}/\mathrm{s}$ towards north west

(3)  zero

(4)  $5\sqrt{2} \mathrm{m}/\mathrm{s}$towards north west

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