The momentum of a body moving in a straight line is . Force acting on the body at t=2 sec will be: \(\left(\text{Given:}~ F=\frac{dp}{dt}\right)\)
1. 6 N
2. 8 N
3. 4 N
4. 2 N
Temperature of a body varies with time as , where is the temperature in Kelvin at , then the rate of change of temperature at is:
1. \(8~\text{K}\)
2. \(80~\text{K}\)
3. \(8~\text{K/sec}\)
4. \(80~\text{K/sec}\)
\(\overrightarrow{A}\) 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:
1. | \(60^{\circ}\) | 2. | \(45^{\circ}\) |
3. | \(30^{\circ}\) | 4. | \(90^{\circ}\) |
A particle is moving along the x-axis. The velocity v of this particle varies with its position x as . Find the velocity of the particle as a function of time t given that at t=0, x=1 and v=.
1.
2.
3.
4. None of these
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 = is (shaded region):
1. \(1\) unit
2. \(2\) units
3. \(3\) units
4. \(4\) units
The acceleration of a particle starting from rest varies with time according to relation, . The velocity of the particle at time instant \(t\) is: \(\left(\text{Here,}~ a=\frac{dv}{dt}\right)\)
1.
2.
3.
4.
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, \(v=\frac{dx}{dt}\))
1. | vary with time. |
2. | be independent of time. |
3. | be inversely proportional to time. |
4. | be inversely proportional to acceleration. |
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
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. \(\cos^{-1}\left(-\frac{x^2-y^2}{x^2+y^2}\right )\)