The acceleration of a particle is given by \(a=3t\) at \(t=0\), \(v=0\), \(x=0\). The velocity and displacement at \(t = 2~\text{sec}\) will be:
\(\left(\text{Here,} ~a=\frac{dv}{dt}~ \text{and}~v=\frac{dx}{dt}\right)\)
1. \(6~\text{m/s}, 4~\text{m}\)
2. \(4~\text{m/s}, 6~\text{m}\)
3. \(3~\text{m/s}, 2~\text{m}\)
4. \(2~\text{m/s}, 3~\text{m}\)

At what angle must the two forces \((x+y)\) and \((x-y)\) act so that the resultant comes out to be \(\sqrt{x^2+y^2}\)?
1. \(\cos^{-1}\left(-\frac{x^2+y^2}{2(x^2-y^2)}\right )\)
2. \(\cos^{-1}\left(-\frac{2(x^2-y^2)}{(x^2+y^2)}\right )\)
3. \(\cos^{-1}\left(-\frac{x^2+y^2}{x^2-y^2}\right )\)
4. \(\cos^{-1}\left(-\frac{x^2-y^2}{x^2+y^2}\right )\)

Component of $3\hat{\mathrm{i}}+4\hat{\mathrm{j}}$ perpendicular to $\hat{\mathrm{i}}+\hat{\mathrm{j}}$ and in the same plane as that of $3\hat{\mathrm{i}}+4\hat{\mathrm{j}}$ is:

1. $\frac{1}{2}\left(\hat{\mathrm{j}}-\hat{\mathrm{i}}\right)$

2. $\frac{3}{2}\left(\hat{\mathrm{j}}-\hat{\mathrm{i}}\right)$

3. $\frac{5}{2}\left(\hat{\mathrm{j}}-\hat{\mathrm{i}}\right)$

4. $\frac{7}{2}\left(\hat{\mathrm{j}}-\hat{\mathrm{i}}\right)$

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 starting from rest varies with time according to relation, $a=\alpha t+\beta$. The velocity of the particle at time instant \(t\) is: \(\left(\text{Here,}~ a=\frac{dv}{dt}\right)\)

1. $\alpha t^2+\beta t$

2. $\alpha t^2+\frac{\beta t}{2}$

3. $\frac{\alpha t^2}{2}+\beta t$

4. $2\alpha t^2+\beta t$

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):

              
1. \(1\) unit
2. \(2\) units
3. \(3\) units
4. \(4\) units

A particle is moving along the x-axis. The velocity v of this particle varies with its position x as $\mathrm{v}=\frac{1}{\mathrm{x}}$. Find the velocity of the particle as a function of time t given that at t=0, x=1 and v=$\frac{\mathrm{dx}}{\mathrm{dt}}$.

1. $\mathrm{v}=\sqrt{2\mathrm{t}+1}$

2. $\mathrm{v}=\frac{1}{\sqrt{2\mathrm{t}+1}}$

3. $\mathrm{v}=\frac{1}{\mathrm{t}}$

4. None of these

$\stackrel{}{\mathrm{}}$\(\overrightarrow{A}\) and \(\overrightarrow B\) are two vectors and \(\theta\) is the angle between them. If \(\left|\overrightarrow A\times \overrightarrow B\right|= \sqrt{3}\left(\overrightarrow A\cdot \overrightarrow B\right),\) then the value of \(\theta\) will be:

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:
1. \(8~\text{K}\)
2. \(80~\text{K}\)
3. \(8~\text{K/sec}\)
4. \(80~\text{K/sec}\)