What is the torque of a force $\overrightarrow{\mathrm{F}}=\left(2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+4\hat{\mathrm{k}} \right)$newton acting at a point $\overrightarrow{\mathrm{r}}=\left(3\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}} \right)$ metre about the origin? (Given: $\overrightarrow{\mathrm{\tau }} =\overrightarrow{\mathrm{r} }\times \overrightarrow{\mathrm{F}}$)

1. $6\hat{\mathrm{i}}-6\hat{\mathrm{j}}+12\hat{\mathrm{k}}$

2. $17\hat{\mathrm{i}}-6\hat{\mathrm{j}}-13\hat{\mathrm{k}}$

3. $-6\hat{\mathrm{i}}+6\hat{\mathrm{j}}-12\hat{\mathrm{k}}$

4. $-17\hat{\mathrm{i}}+6\hat{\mathrm{j}}-13\hat{\mathrm{k}}$

Three non zero vectors $\overrightarrow{\mathrm{A}}, \overrightarrow{\mathrm{B}} \& \overrightarrow{\mathrm{C}}$ satisfy the relation $\overrightarrow{\mathrm{A}}·\overrightarrow{\mathrm{B}}=0 \& \overrightarrow{\mathrm{A}}·\overrightarrow{\mathrm{C}}=0$. Then $\overrightarrow{\mathrm{A}}$ can be parallel to:

(1)  $\overrightarrow{\mathrm{B}}$

(2)  $\overrightarrow{\mathrm{C}}$

(3)  $\overrightarrow{\mathrm{B}}·\overrightarrow{\mathrm{C}}$

(4)  $\overrightarrow{\mathrm{B}}\times \overrightarrow{\mathrm{C}}$

The scalar product of two vectors is 8 and the magnitude of vector product is $8\sqrt{3}$. The angle between them is:

(1)  $30°$

(2)  $60°$

(3)  $120°$

(4)  $150°$

Given: $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0$. Out of the three vectors $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ two are equal in magnitude. The magnitude of the third vector is $\sqrt{2}$ times that of either of the two having equal magnitude. The angles between the vectors are:

(A)  90$°$, 135$°$, 135$°$

(B)  30$°$, 60$°$, 90$°$

(C)  45$°$, 45$°$, 90$°$

(D)  45$°$, 60$°$, 90$°$

Vector $\overrightarrow{\mathrm{A}}$ is of length 2 cm and is 60$°$ above the x-axis in the first quadrant. Vector $\overrightarrow{\mathrm{B}}$ is of length 2 cm and 60$°$ below the x-axis in the fourth quadrant. The sum $\overrightarrow{\mathrm{A}}$+$\overrightarrow{\mathrm{B}}$ is a vector of magnitude -

(A)  2 along + y-axis

(B)  2 along + x-axis

(C)  1 along - x-axis

(D)  2 along - x-axis

Six forces, 9.81 N each, acting at a point are coplanar. If the angle between neighboring forces are equal, then the resultant is

(1)  0 N

(2)  9.81 N

(3)  2$\times$9.81 N

(4)  3$\times$9.81 N

Temperature of a body varies with time as $\mathrm{T}=\left({\mathrm{T}}_0+{\mathrm{at}}^2+\mathrm{b} \mathrm{sint}\right)\mathrm{K}$, where ${\mathrm{T}}_0$ is the temperature in Kelvin at $\mathrm{t}=0 \sec \& \mathrm{a}=2/\mathrm{\pi } \mathrm{K}/{\mathrm{s}}^2 \& \mathrm{b}=-\mathrm{4 }\mathrm{K}$, then the rate of change of temperature $\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)$ at $\mathrm{t}=\mathrm{\pi } \sec$ is:
1. \(8~\text{K}\)
2. \(80~\text{K}\)
3. \(8~\text{K/sec}\)
4. \(80~\text{K/sec}\)

If the distance 's' travelled by a body in time 't' is given by $\mathrm{s}=\frac{\mathrm{a}}{\mathrm{t}}+{\mathrm{bt}}^2$ then the acceleration equals

(1)  $\frac{2\mathrm{a}}{{\mathrm{t}}^3}+2\mathrm{b}$

(2)  $\frac{2\mathrm{s}}{{\mathrm{t}}^3}$

(3)  $2\mathrm{b}-\frac{2\mathrm{a}}{{\mathrm{t}}^3}$

(4)  $\frac{\mathrm{s}}{{\mathrm{t}}^2}$

The velocity of a particle moving on the x-axis is given by $\mathrm{v}={\mathrm{x}}^2+\mathrm{x}$ where v is in m/s and x is in m. Find its acceleration in $\mathrm{m}/{\mathrm{s}}^2$ when passing through the point x=2m.

1.  0

2.  5

3.  11

4.  30

A particle moves in the XY plane and at time t is at the point whose coordinates are $\left({\mathrm{t}}^2, {\mathrm{t}}^3-2\mathrm{t}\right)$. Then at what instant of time, will its velocity and acceleration vectors be perpendicular to each other?

(1)  1/3 sec

(2)  2/3 sec

(3)  3/2 sec

(4)  never