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

1. \(\frac{\hat{i}   +   2 \hat{j}   +   3 \hat{k}}{6}\)
2. \(\frac{6\hat{j} -4 \hat{k}}{\sqrt{52}}\)
3. \(\frac{6\hat{j} +4 \hat{k}}{\sqrt{52}}\)
4. \(\frac{2\hat{i} - \hat{j}}{\sqrt{5}}\)

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:
1. 12 m/s
2. 12$\sqrt{2}$ m/s
3. 6 m/s
4. \(6\sqrt2\) m/s

If the acceleration \(a(t)= 4t+6\), the velocity of a particle starting from rest is: \(\left(\text{here} ,  a = \frac{d v}{d t}\right)\)
1. \(2t+6\)
2. \(4\)
3. \(0\)
4. \(2t^2+6t\)

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 \(i = (2+3t)~\text{A}\). The charge that crosses through the wire in \(10\) seconds is: \(\left(\text{Instantaneous current,}~i= \frac{dq}{dt} \right)\)
1. \(150~\text{C}\)
2. \(160~\text{C}\)
3. \(170~\text{C}\)
4. None of there

The area of a blot of ink, \(A\), is growing such that after \(t\) seconds, \(A=\left(3t^2+\frac{t}{5}+7\right)\text{m}^2\). Then the rate of increase in the area at \(t = 5~\text{s}\) will be:
1. \(30.1~\text{m}^2/\text{s}\)
2. \(30.2~\text{m}^2/\text{s}\)
3. \(30.3~\text{m}^2/\text{s}\)
4. \(30.4~\text{m}^2/\text{s}\)

A particle starts rotating from rest and its angular displacement is given by \(\theta = \frac{t^2}{40}+\frac{t}{5}\). Then, the angular velocity \(\omega = \frac{d\theta}{dt}\) at the end of \(10~\text{s}\) will be:
1. \(0.7\)
2. \(0.6\)
3. \(0.5\)
4. \(0\)

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

1. $\frac{GMm}{R}$

2. $\frac{2GMm}{R}$

3. $\frac{-GMm}{R}$

4. $\frac{-2GMm}{R}$

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

1. $\frac{Q^2}{C}$

2. $\frac{-Q^2}{2C}$

3. $\frac{-Q^2}{C}$

4. $\frac{Q^2}{2C}$

If the force on an object as a function of displacement is \(F \left(x\right) = 3 x^{2} + x\)$$, what is work as a function of displacement \(w(x)\): \(\left(w= \int f\cdot dx\right)\) Assume \(w(0)= 0\) and force is in the direction of the object's motion.
1. \(\frac{3 x^{3}}{2} + x^{2}\)
2. \(x^{3} + \frac{x^{2}}{2}\)
3. \(6x+1\)
4. \(3 x^{2} + x\)
$$