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)  $0$

Coordinates of a moving particle are given by $\mathrm{x}={\mathrm{ct}}^2$ and $\mathrm{y}={\mathrm{bt}}^2$. The speed of the particle is given by

(1)  $2\mathrm{t} \left(\mathrm{c}+\mathrm{b}\right)$

(2)  $2\mathrm{t} \sqrt{{\mathrm{c}}^2-{\mathrm{b}}^2}$

(3)  $\mathrm{t} \sqrt{{\mathrm{c}}^2+{\mathrm{b}}^2}$

(4)  $2\mathrm{t} \sqrt{{\mathrm{c}}^2+{\mathrm{b}}^2}$

The x and y components of vector $\overrightarrow{\mathrm{A}}$ are 4m and 6m respectively. The x, y components of vector $\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$ are 10m and 9m respectively. The length of $\overrightarrow{\mathrm{B}}$ is ______ and angle that $\overrightarrow{\mathrm{B}}$ makes with the x axis is given by _______.

(A)  $\sqrt{0.45}\mathrm{m}, {\tan }^{-1}\left(1/2\right)$

(B)  $\sqrt{4.5}\mathrm{m}, {\tan }^{-1}\left(1/2\right)$

(C)  $\sqrt{45}\mathrm{m}, {\tan }^{-1}\left(1/2\right)$

(D)  $4\sqrt{5}\mathrm{m}, {\tan }^{-1}\left(1/2\right)$

A particle travels with speed 50 m/s from the point (3, 7) in a direction $7\hat{\mathrm{i}}-24\hat{\mathrm{j}}$. Find its position vector after 3 seconds.

1.  $151\hat{\mathrm{i}}+45\hat{\mathrm{j}}$

2.  $45\hat{\mathrm{i}}-137\hat{\mathrm{j}}$

3.  $151\hat{\mathrm{i}}-45\hat{\mathrm{j}}$

4.  $4.5\hat{\mathrm{i}}-151\hat{\mathrm{j}}$

If \(\overrightarrow{A}={2}\hat{i}+\hat{j}\;{\&}\;\overrightarrow{B}=\hat{i}{-}\hat{j},\)  then the components of \(\overrightarrow {A}\) along with \(\overrightarrow {B}\) & perpendicular to \(\overrightarrow {B}\) respectively will be:
1. \(\frac{\hat{i} - \hat{j}}{2} ,~\frac{3}{2}\left(\hat i+\hat j\right)\)
2. \(\frac{\hat{i} - \hat{j}}{2} ,~-\frac{2}{3}\left(\hat i+\hat j\right)\)
3. \(\frac{\hat{i} - \hat{j}}{2} ,~-\frac{3}{2}\left(\hat i-\hat j\right)\)
4. \(\frac{\hat{i} - \hat{j}}{2} ,~\frac{2}{3}\left(\hat i-\hat j\right)\)