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

A force $\overrightarrow{\mathrm{F}}=6\hat{\mathrm{i}}-8\hat{\mathrm{j}}+10\hat{\mathrm{k}}$ Newton produces acceleration $1 \mathrm{m}/{\mathrm{s}}^2$ in a body. The mass of the body is (in kg)

(1)  $6\hat{\mathrm{i}}-8\hat{\mathrm{j}}+10\hat{\mathrm{k}}$

(2)  $100$

(3)  $10\sqrt{2}$

(4)  $10$

Given the vectors

$\overrightarrow{\mathrm{A}}=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}-\hat{\mathrm{k}}$
$\overrightarrow{\mathrm{B}}=3\hat{\mathrm{i}}-2\hat{\mathrm{j}}-2\hat{\mathrm{k}}$
$\& \overrightarrow{\mathrm{C}}=\mathrm{p}\hat{\mathrm{i}}+\mathrm{p}\hat{\mathrm{j}}+2\mathrm{p}\hat{\mathrm{k}}$

Find the angle between $\left(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}\right) \& \overrightarrow{\mathrm{C}}$

(A)  $\mathrm{\theta }={\cos }^{-1}\left(\frac{2}{\sqrt{3}}\right)$

(B)  $\mathrm{\theta }={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

(C)  $\mathrm{\theta }={\cos }^{-1}\left(\frac{\sqrt{2}}{3}\right)$

(D)  none of these

The vector having a magnitude of 10 and perpendicular to the vector $3\hat{\mathrm{i}}-4\hat{\mathrm{j}}$ is- 

1. $4\hat{\mathrm{i}}-3\hat{\mathrm{j}}$

2. $5\sqrt{2} \hat{\mathrm{i}}-5\sqrt{2} \hat{\mathrm{j}}$

3. $8\hat{\mathrm{i}}+6\hat{\mathrm{j}}$

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

A force $\overrightarrow{\mathrm{F}}=3\hat{\mathrm{i}}+\mathrm{c}\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ acting on a particle causes a displacement $\overrightarrow{\mathrm{d}}=-4\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}$. If the work done is 6J then the value of 'c' is- $[Given: W=\overrightarrow{F.}\overrightarrow{d}]$

1. 12 

2. 0

3. 6

4. 1

A particle moving along a straight line according to the law $\mathrm{x}=\mathrm{At}+{\mathrm{Bt}}^2+{\mathrm{Ct}}^3$, where x is its position measured from a fixed point on the line and t is the time elapsed till it reaches position x after starting from the fixed point. Here A, B and C are positive constants.

(1)  Its velocity at t=0 is A

(2)  Its acceleration at t=0 is B

(3)  Its velocity at t=0 is B

(4)  Its acceleration at t=0 is C

If the velocity of a particle moving on x-axis is given by $\mathrm{v}=3{\mathrm{t}}^2-12\mathrm{t}+6$. At which time is the acceleration of particle zero?

1.  $2$ sec

2.  $2+\sqrt{2}$ sec

3.  $2-\sqrt{2}$ sec

4.  zero

Momentum of a body moving in a straight line is $\mathrm{p}=\left({\mathrm{t}}^2+2\mathrm{t}+1\right) \mathrm{kg} \mathrm{m}/\mathrm{s}$. Find the force acting on a body at t=2 sec

(1)  6 N

(2)  8 N

(3)  4 N

(4)  2 N

A particle moves along straight line such that at time t its position from a fixed point O on the line is $\mathrm{x}=3{\mathrm{t}}^2-2$. The velocity of the particle when t=2 is:

(A)  $8 {\mathrm{ms}}^{-1}$

(B)  $4 {\mathrm{ms}}^{-1}$

(C)  $12 {\mathrm{ms}}^{-1}$

(D)  $0$