Calculate the electric field on the axis of a very long uniformly charged, thin rod at a distance r from one end. The charge per unit length of the rod is $\lambda$. [This question includes concepts from 12th syllabus]

If rain falls vertically with a velocity ${\mathrm{V}}_{\mathrm{r}}$ wrt wind and wind blows with a velocity ${\mathrm{V}}_{\mathrm{w}}$ from east to west, then a person standing on the roadside should hold the umbrella in the direction

The component of vector $\overrightarrow{A}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}$ along the direction of $\hat{i}-\hat{j}$ is

A force is 60 ° inclined to the horizontal. If its rectangular component in the horizontal direction is 50 N, then the magnitude of the force in the vertical direction is

For the given figure, which of the following is true?

The value of the unit vector, which is perpendicular to both $\mathrm{A} = \hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 3\hat{\mathrm{k}} \mathrm{and} \mathrm{B} = \hat{\mathrm{i}} - 2\hat{\mathrm{j}} - 3\hat{\mathrm{k}} \mathrm{is\; equal\; to}$:

The instantaneous velocity (defined as $v=\frac{ds}{dt}$) at time $t=\frac{\mathrm{\pi }}{2}$ of a particle, whose position equation is given as  s(t)=12 tan$\left(\frac{t}{2}+\mathrm{\pi }\right)$m, is

If the acceleration a(t) = 4t+6, the velocity of a particle starting from rest is: $\left(here, a=\frac{dv}{dt}\right)$

A force of $F\left(x\right)=2x^2+3$ N is applied to an object. How much work is done, in Joules, moving the object from x=1 to x=4 meters? $\left(work={\int }_a^{b}F\left(x\right) dx\right)$

A car has a certain displacement between 0 seconds and 2 seconds. If we defined its velocity as v(t)=6t-5, then the displacement in meters is: $\left(\mathrm{Here}, \mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}\right)$

The relation between time t and distance x is $t=a{x}^2+bx$, where a and b are constants, the acceleration is $\left(here, v=\frac{dx}{dt} and a=\frac{d^{2}x}{d{t}^2}\right)$

Given velocity v(t)=$\frac{5}{2}\mathrm{t}+3$. Assume s(t) is measured in meters and t is measured in seconds. If s(0)=0, the position s(4) at t=4s is:  $\left(\mathrm{Given}, \mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}\right)$

The current through a wire depends on time as i =(2+3t) A.The charge that crosses through the wire in 10 seconds is: $\left(\mathrm{Instantaneous} \mathrm{current}, \mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}\right)$

The area of a blot of ink, A, is growing such that after t seconds, \(A=3t^2+\frac{t}{5}+7~m^2\). Then the rate of increase in the area at t= 5s will be :

A particle starts rotating from rest and its angular displacement is given by $\mathrm{\theta }=\frac{{\mathrm{t}}^2}{40}+\frac{\mathrm{t}}{5}$. Then, the angular velocity $\left(\mathrm{\omega }=\frac{\mathrm{d\theta }}{\mathrm{dt}}\right)$ at the end of 10 s will be :

The value of ${\int }_{x=\infty }^{x=R}\frac{GMm}{x^2} dx$ is

${\int }_0^Q\frac{q}{C}dq$, where C is a constant, can be expressed as:

If the force on an object as a function of displacement is $F\left(x\right)=3x^2+x$, what is work as a function of displacement $w\left(x\right) \left(W=\int \left(fdx\right)\right)$. Assume W(0)=0 and force is in the direction of the object's motion.

The velocity of a rocket, in metres per second, t seconds after it was launched is modelled by v (t) = 2t. What is the total distance travelled by the rocket during the first four seconds of its launch?

The work done by gravity exerting an acceleration of -10 m/${\mathrm{s}}^2$ for a 10 kg block down 5 m from its original position with no initial velocity is:

$({\mathrm{F}}_{\mathrm{grav}}=\mathrm{mass} \times \mathrm{acceleration} \mathrm{and} \mathrm{W}={\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{F}(\mathrm{x}\left)\mathrm{dx}\right)$

Water flows into a container of 1000 L at a rate of (180+3t) gal/min for an hour, where t is measured in minutes. Find the amount of water that flows into the pool during the first 20 minutes.

If v(t)=3t-1 and x(2)=1, then the original position function is: 
Hint: $\left(v\left(t\right)=\frac{ds}{dt}\right)$

If charge flown through a wire is given by q=3sin(3t), then-current flown through the wire at $\mathrm{t}=\frac{\mathrm{\pi }}{9}$ seconds is: $\left(\mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}\right)$

A weight hanging from a spring is stretched down 3 cm beyond its rest position and released at time t=0 to bob up and down. Its position at any later time t is s=3cos(t). Then its velocity at time t is $\left(velocity =\frac{ds}{dt}\right)$

The position of a particle is given by $s\left(t\right)=\frac{2t^2+1}{t+1.}$ Then, at t = 2, its velocity is :$\left(v_{inst}=\frac{ds}{dt}\right)$

The instantaneous velocity at t=$\frac{\mathrm{\pi }}{2}$ of a particle whose positional equation is given by $x\left(t\right)=12 {\cos }^2\left(t\right)$ is $\left(v_{inst}=\frac{dx}{dt}\right)$ -

If acceleration of a particle is given as a(t)=sin(t)+2t.Then the velocity of the particle will be:

(acceleration $\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}$)

If x=3tan(t) and y=sec(t), then the value of $\frac{d^{2}y}{d{x}^2} at t=\frac{\mathrm{\pi }}{4}$ is:

A particle's position as a function of time is given by $\mathrm{x}=-{\mathrm{t}}^2+6\mathrm{t}+3$.The maximum value of the position co-ordinate of the particle is:

A gas undergoes a process where pressure P=$3V^4$, at every instant, here V is the volume. Find an expression for the bulk modulus (B) in terms of volume if it is related to P and V as $B=-\frac{VdP}{dV}$.