If $\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}}=\overrightarrow{\mathrm{P}}-\overrightarrow{\mathrm{Q}}$ and $\mathrm{\theta }$ is the angle between $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$, then

1.  $\mathrm{\theta }=0°$

2.  $\mathrm{\theta }=90°$

3.  $\mathrm{P}=0$

4.  $\mathrm{Q}=0$

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

The momentum of a body moving in a straight line is $\mathrm{p}=\left({\mathrm{t}}^2+2\mathrm{t}+1\right) \mathrm{kg} \mathrm{m}/\mathrm{s}$. Force acting on the body at t=2 sec will be: (\(Given: F=\frac{dp}{dt}\))

1.  6 N

2.  8 N

3.  4 N

4.  2 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.  $8 \mathrm{K}$

2.  ${80}^{} \mathrm{K}$

3.  $8 \mathrm{K}/\sec$

4.  ${80}^{} \mathrm{K}/\sec$

$\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ are two vectors and θ is the angle between them. If $\left|\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}\right|=\sqrt{3}\left(\overrightarrow{A}.\overrightarrow{B}\right)$, then the value of θ will be:

1. 60^o

2. 45^o

3. 30^o

4. 90^o

A particle is moving along the x-axis. The velocity v of this particle varies with its position x as $\mathrm{v}=\frac{1}{\mathrm{x}}$. Find the velocity of the particle as a function of time t given that at t=0, x=1 and v=$\frac{\mathrm{dx}}{\mathrm{dt}}$.

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

If a curve is governed by the equation y=sinx, then the area enclosed by the curve and x-axis between x =0 and x =$\mathrm{\pi }$ is (shaded region) :

1. 1 unit

2. 2 units

3. 3 units

4. 4 units

The acceleration of a particle starting from rest varies with time according to relation, $a=\alpha t+\beta$. The velocity of the particle at time instant t is: (\(Here, a=\frac{dv}{dt}\))

1. $\alpha t^2+\beta t$

2. $\alpha t^2+\frac{\beta t}{2}$

3. $\frac{\alpha t^2}{2}+\beta t$

4. $2\alpha t^2+\beta t$

The displacement of the particle is zero at t=0 and at t=t it is x. It starts moving in the x-direction with a velocity that varies as $v=k\sqrt{x}$, where k is constant. The velocity will : (Here, \(v=\frac{dx}{dt}\))

1. vary with time.

2. be independent of time.

3. be inversely proportional to time.

4. be inversely proportional to acceleration.