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

The maximum value of the function \(7 + 6 x - 9 x^{2}\) is:
1. \(8\)
2. \(-8\)
3. \(4\)
4. \(-4\)

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

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)  $2 \overrightarrow{\mathrm{BA}}$