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

A motor boat of mass m moving along a lake with velocity ${\mathrm{V}}_0$. At t=0, the engine of the boat is shut down. Magnitude of resistance force offered to the boat is equal to rV. (V is instantaneous speed). What is the total distance covered till it stops completely? $\left[\mathrm{Hint}: \mathrm{F}\left(\mathrm{x}\right)=\mathrm{mV}\frac{\mathrm{dV}}{\mathrm{dx}}=-\mathrm{rV}\right]$

(1)  ${\mathrm{mV}}_0/\mathrm{r}$

(2)  $3 {\mathrm{mV}}_0/2\mathrm{r}$

(3)  ${\mathrm{mV}}_0/2\mathrm{r}$

(4)  $2{\mathrm{mV}}_0/\mathrm{r}$

A particle is moving along positive \(x\text-\)axis. Its position varies as \(x = t^3-3t^2+12t+20,\) where \(x\) is in meters and \(t\) is in seconds. The velocity of the particle when its acceleration zero is:
1. \(1~\text{m/s}\)
2. \(3~\text{m/s}\)
3. \(6~\text{m/s}\)
4. \(9~\text{m/s}\)