The current in a circuit is defined as $I=\frac{dq}{dt}$. The charge (q) flowing through a circuit, as a function of time (t), is given by $q=5t^2-20t+3$. The minimum charge flows through the circuit at:

1. t=4 sec

2. t=2 sec

3. t= 6 sec

4. t=3 sec

A body is moving according to the equation $\mathrm{x}=\mathrm{at}+{\mathrm{bt}}^2-{\mathrm{ct}}^3$ where x represents displacement and a, b and c are constants. The acceleration of the body is: (\(Given: a=\frac{d^2x}{dt^2}\))

1.  $\mathrm{a}+2\mathrm{bt}$

2.  $2\mathrm{b}+6\mathrm{ct}$

3.  $2\mathrm{b}-6\mathrm{ct}$

4.  $3\mathrm{b}-6{\mathrm{ct}}^2$

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

A particle moves along the X-axis so that its X coordinate varies with time t according to the equation $\mathrm{x}=\left(2-5\mathrm{t}+6{\mathrm{t}}^2\right)\mathrm{m}$. The initial velocity of the particle is: (\(Given; v=\frac{dx}{dt}\))

1.  -5 m/s

2.  6 m/s

3.  3 m/s

4.  4 m/s

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$ , if $\mathrm{y} = {\mathrm{t}}^3+\hspace{0.17em}1$ and  $\mathrm{x} = {\mathrm{t}}^2 + 3$.

$1. \frac{{\mathrm{t}}^2}{3}$

$2. \frac{\mathrm{t}}{2}$

$3. \frac{3\mathrm{t}}{2}$

$4. {\mathrm{t}}^2$

The position x of the particle varies with time t as $\mathrm{x}={\mathrm{at}}^2-{\mathrm{bt}}^3$.  The acceleration of the particle will be zero at a time equal to: (\(Given: a=\frac{d^2x}{dt^2}\))

1.  $\frac{\mathrm{a}}{\mathrm{b}}$

2.  $\frac{2\mathrm{a}}{\mathrm{b}}$

3.  $\frac{\mathrm{a}}{3\mathrm{b}}$

4.  Zero

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 :

1. 30.1 m²/s

2. 30.2 m²/s

3. 30.3 m²/s

4. 30.4 m²/s

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 :

1. 0.7

2. 0.6

3. 0.5

4. 0

The maximum or the minimum value of the function y = 25$x^2$+ 5 – 10x is:

1. y_min = 4

2. y_max = 8

3. y_min = 8

4. y_max = 4

The displacement of a particle is given by $\mathrm{y}=\mathrm{a}+\mathrm{bt}+{\mathrm{ct}}^2-{\mathrm{dt}}^4$. The initial velocity and initial acceleration, respectively, are: (\(Given: v=\frac{dx}{dt}~and~a=\frac{d^2x}{dt^2}\))

1.  b, -4d

2.  -d, 2c

3.  b, 2c

4.  2c, -4d