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Q. No.What is the value of linear velocity if \(\overrightarrow{\omega} = 3\hat{i} - 4\hat{j} + \hat{k}\) and \(\overrightarrow{r} = 5\hat{i} - 6\hat{j} + 6\hat{k}\):
| 1. | \(6 \hat{i}+2 \hat{j}-3 \hat{k} \) |
| 2. | \(-18 \hat{i}-13 \hat{j}+2 \hat{k} \) |
| 3. | \(4 \hat{i}-13 \hat{j}+6 \hat{k}\) |
| 4. | \(6 \hat{i}-2 \hat{j}+8 \hat{k}\) |
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 car moves on a circular path such that its speed is given by \(v= Kt\), where \(K\) = constant and \(t\) is time. Also given: radius of the circular path is \(r\). The net acceleration of the car at time \(t\) will be:
1. \(\sqrt{K^{2} +\left(\frac{K^{2} t^{2}}{r}\right)^{2}}\)
2. \(2K\)
3. \(K\)
4. \(\sqrt{K^{2} + K^{2} t^{2}}\)
A stone tied to the end of a \(1\) m long string is whirled in a horizontal circle at a constant speed. If the stone makes \(22\) revolutions in \(44\) seconds, what is the magnitude and direction of acceleration of the stone?
| 1. | \(\pi^2 ~\text{ms}^{-2} \) and direction along the tangent to the circle. |
| 2. | \(\pi^2 ~\text{ms}^{-2} \) and direction along the radius towards the centre. |
| 3. | \(\frac{\pi^2}{4}~\text{ms}^{-2} \) and direction along the radius towards the centre. |
| 4. | \(\pi^2~\text{ms}^{-2} \) and direction along the radius away from the centre. |
| 1. | parallel to the position vector. |
| 2. | at \(60^{\circ}\) with position vector. |
| 3. | parallel to the acceleration vector. |
| 4. | perpendicular to the position vector. |
A car is moving at a speed of \(40\) m/s on a circular track of radius \(400\) m. This speed is increasing at the rate of \(3\) m/s2. The acceleration of the car is:
1. \(4\) m/s2
2. \(7\) m/s2
3. \(5\) m/s2
4. \(3\) m/s2
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 moves with constant speed \(v\) along a circular path of radius \(r\) and completes the circle in time \(T\). The acceleration of the particle is:
1. \(2\pi v / T\)
2. \(2\pi r / T\)
3. \(2\pi r^2 / T\)
4. \(2\pi v^2 / T\)
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