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

                          

1. $\overrightarrow{\mathrm{B}}$ = $\overrightarrow{\mathrm{A}}$ + $\overrightarrow{\mathrm{C}}$

2. $\overrightarrow{\mathrm{A}}$ = $\overrightarrow{\mathrm{B}}$ + $\overrightarrow{\mathrm{C}}$

3. $\overrightarrow{\mathrm{C}}$ = $\overrightarrow{\mathrm{A}}$ + $\overrightarrow{\mathrm{B}}$

4. All of these

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.  $\frac{\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 3\hat{\mathrm{k}}}{6}$

2.  $\frac{6\hat{\mathrm{j}} - 4\hat{\mathrm{k}}}{\sqrt{52}}$

3.  $\frac{6\hat{\mathrm{j}} + 4\hat{\mathrm{k}}}{\sqrt{52}}$

4.  $\frac{2\hat{\mathrm{i}} - \hat{\mathrm{j}}}{\sqrt{5}}$

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: $\left(here, a=\frac{dv}{dt}\right)$

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:  $\left(\mathrm{Given}, \mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}\right)$

1. 30

2. 31

3. 32

4. 33

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

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