In \(1.0~\text{s}\), a particle goes from point \(A\) to point \(B\), moving in a semicircle of radius \(1.0~\text{m}\) (see figure). The magnitude of the average velocity is:
1. | \(3.14~\text{m/s}\) | 2. | \(2.0~\text{m/s}\) |
3. | \(1.0~\text{m/s}\) | 4. | zero |
The angle turned by a body undergoing circular motion depends on the time as given by the equation, \(\theta = \theta_{0} + \theta_{1} t + \theta_{2} t^{2}\). It can be deduced that the angular acceleration of the body is?
1. \(\theta_1\)
2. \(\theta_2\)
3. \(2\theta_1\)
4. \(2\theta_2\)
A particle is moving eastwards with velocity of \(5\) m/s. In \(10\) seconds the velocity changes to \(5\) m/s northwards. The average acceleration in this time is?
1. | zero |
2. | \(\frac{1}{\sqrt{2}}~ \text{m/s}^2\) toward north-west |
3. | \(\frac{1}{\sqrt{2}}~\text{m/s}^2\) toward north-east |
4. | \(\frac{1}{2}~\text{m/s}^2 \) toward north-west |
A vector is turned without a change in its length through a small angle The value of and are, respectively:
1. | \(0, ad\theta\) | 2. | \(a d\theta, 0\) |
3. | \(0,0\) | 4. | None of these |
A particle is moving such that its position coordinates \((x,y)\) are \((2\) m, \(3\) m) at time \(t=0,\) \((6\) m, \(7\) m) at time \(t=2\) s and \((13\) m, \(14\) m) at time \(t=5\) s. Average velocity vector \((v_{avg})\) from \(t=0\) to \(t=5\) s is:
1. | \(\frac{1}{5}\left ( 13\hat{i}+14\hat{j} \right )\) | 2. | \(\frac{7}{3}\left ( \hat{i}+\hat{j} \right )\) |
3. | \(2\left ( \hat{i}+\hat{j} \right )\) | 4. | \(\frac{11}{5}\left ( \hat{i}+\hat{j} \right )\) |
The coordinates of a moving particle at a time \(t\), are given by, \(x= 5\sin 10t, y = 5\cos 10t\). It can be deduced that the speed of the particle will be:
1. | \(25\) units | 2. | \(50\) units |
3. | \(10\) units | 4. | \(30\) units |
What determines the nature of the path followed by a particle:
1. Speed
2. Velocity
3. Acceleration
4. Both (2) and (3)
Certain neutron stars are believed to be rotating at about \(1\) rev/s. If such a star has a radius of \(20\) km, the acceleration of an object on the equator of the star will be:
1. | \(20 \times 10^8 ~\text{m/s}^2\) | 2. | \(8 \times 10^5 ~\text{m/s}^2\) |
3. | \(120 \times 10^5 ~\text{m/s}^2\) | 4. | \(4 \times 10^8 ~\text{m/s}^2\) |
A particle starts from the origin at t=0 and moves in the x-y plane with a constant acceleration 'a' in the y direction. Its equation of motion is . The x component of its velocity (at t=0) will be:
1. variable
2.
3.
4.
Three particles are moving with constant velocities \(v_1 ,v_2\) and \(v\) respectively as given in the figure. After some time, if all the three particles are in the same line, then the relation among \(v_1 ,v_2\) and \(v\) is:
1. \(v =v_1+v_2\)
2. \(v= \sqrt{v_{1} v_{2}}\)
3. \(v = \frac{v_{1} v_{2}}{v_{1} + v_{2}}\)
4. \(v=\frac{\sqrt{2} v_{1} v_{2}}{v_{1} + v_{2}}\)