The vector sum of two vectors P and Q is minimum when the angle $\mathrm{\theta }$ between their positive directions, is

If the vector sum of two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is maximum, then the angle $\mathrm{\theta }$ between two vectors will be:

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

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: $\overrightarrow{\mathrm{\tau }} =\overrightarrow{\mathrm{r} }\times \overrightarrow{\mathrm{F}}$)

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:

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

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:

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 -

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

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:

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

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.

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?

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? $\left[\mathrm{Hint}: \mathrm{F}\left(\mathrm{x}\right)=\mathrm{mV}\frac{\mathrm{dV}}{\mathrm{dx}}=-\mathrm{rV}\right]$

A particle is moving along positive x-axis. Its position varies as $\mathrm{x}={\mathrm{t}}^3-3{\mathrm{t}}^2+12\mathrm{t}+20$, where x is in meters and t is in seconds.

Velocity of the particle when its acceleration zero is

Two forces ${\overrightarrow{\mathrm{F}}}_1=2\hat{\mathrm{i}}+2\hat{\mathrm{j}} \mathrm{N}$ and ${\overrightarrow{\mathrm{F}}}_2=3\hat{\mathrm{j}}+4\hat{\mathrm{k}} \mathrm{N}$ are acting on a particle.

The resultant force acting on particle is:

(A)  $2\hat{\mathrm{i}}+5\hat{\mathrm{j}}+4\hat{\mathrm{k}}$

(B)  $2\hat{\mathrm{i}}-5\hat{\mathrm{j}}-4\hat{\mathrm{k}}$

(C)  $\hat{\mathrm{i}}-3\hat{\mathrm{j}}-2\hat{\mathrm{k}}$

(D)  $\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$

$\mathrm{A}=4\mathrm{i}+4\mathrm{j}-4\mathrm{k}$ and $\mathrm{B}=3\mathrm{i}+\mathrm{j}+4\mathrm{k}$, then angle between vectors A and B is:

If vectors $\overrightarrow{\mathrm{A}}=\mathrm{cos\omega t}\hat{\mathrm{i}}+\mathrm{sin\omega t}\hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{B}}=\cos \frac{\mathrm{\omega t}}{2}\hat{\mathrm{i}}+\sin \frac{\mathrm{\omega t}}{2}\hat{\mathrm{j}}$ are functions of time.Then, at what value of t are they orthogonal to one another?

Six vectors $\overrightarrow{\mathrm{a}}$ through $\overrightarrow{\mathrm{f}}$ have the magnitudes and directions indicated in the figure. Which of the following statements is true? 

$\overrightarrow{\mathrm{A}}$ 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:

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) :

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: (\(Here, a=\frac{dv}{dt}\))

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, \(v=\frac{dx}{dt}\))

The acceleration of a particle is given as $a=3x^2$. At t = 0, v = 0 and x = 0. It can then be concluded that the velocity at t = 2 sec will be: (Here, \(a=v\frac{dv}{dx}\))

The acceleration of a particle is given by a=3t  at t=0, v=0, x=0. The velocity and displacement at t = 2 sec will be: (\(Here, a=\frac{dv}{dt}~ and~v=\frac{dx}{dt}\))

The 9 kg block is moving to the right with a velocity of 0.6 m/s on a horizontal surface when a force F, whose time variation is shown in the graph, is applied to it at time t = 0. Calculate the velocity v of the block when t= 0.4s. The coefficient of kinetic fricton is ${\mu }_k=0.3$. [This question includes concepts from Work, Energy & Power chapter]

The relationship between force and position is shown in the figure given (in one dimensional case). Find the work done by the force in displaying a body from x= 1 cm to x= 5cm is [This question includes concepts from Work, Energy and Power chapter]

The graph between the resistive force F acting on a body and the distance covered by the body is shown in the figure. The mass of the body is 25 kg and initial velocity is 2 m/s. When the distance covered by the body is 4 m, its kinetic energy would be  [This question includes concepts from Work, Energy & Power chapter]

A constant force F is applied on a body. The power (P) generated is related to the time elapsed (t) as [This question includes concepts from Work, Energy and Power chapter]

Electric field is given by E $=\frac{100}{x^2}$ . Find the potential difference between x= 10 and x= 20 m. [This question includes concepts from 12th syllabus]