If $x$ $=$ $3$ $-4t^2$ $+$ $t^3$, then work done in the first 4s will be:
(Mass of the particle is 3 gram)
1.  384 mJ
2.  168 mJ
3.  192 mJ
4.  None of the above

Two identical balls A and B are moving with velocity $+0.5$ $m{s}^{-1}$ and $-0.3$ $m{s}^{-1}$ respectively. If they collide head on elastically, then their velocities after collision will be:

1. $-0.3$ $m{s}^{-1}$ $and$ $0.5$ $m{s}^{-1}$

2. $+0.5$ $m{s}^{-1}$ $and$ $+$ $0.3$ $m{s}^{-1}$

3. $-0.4$ $m{s}^{-1}$ $and$ $0.3$ $m{s}^{-1}$

4. $-0.3$ $m{s}^{-1}$ $and$ $-0.4$ $m{s}^{-1}$

A bomb of mass 30kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18kg mass is 6ms^–1. The kinetic energy of the other mass is:

1. 524 J

2. 256 J

3. 486 J

4. 324 J

The bob of a simple pendulum having length \(l,\) is displaced from the mean position to an angular position \(\theta\) with respect to vertical. If it is released, then the velocity of the bob at the lowest position will be:

1. \(\sqrt{2 g l \left(\right. 1 - \cos \theta \left.\right)}\)
2. \(\sqrt{2 g l \left(\right. 1 + \cos\theta)}\)
3. \(\sqrt{2 g l\cos\theta}\)
4. \(\sqrt{2 g l}\)

If \(\vec{F} = (60\hat{i} + 15\hat{j}-3\hat{k})\) N and \(\vec{v} = (2\hat{i} - 4\hat{j}+5\hat{k})\) m/s, then instantaneous power is:
1. \(195\) watt
2. \(45\) watt
3. \(75\) watt
4. \(100\) watt

A ball is dropped from a height of \(5~\text {m}.\) If it rebounds up to a height of \(1.8~\text {m},\) then the ratio of velocities of the ball after and before the rebound will be:
1. \(\dfrac{3}{5}\)

2. \(\dfrac{2}{5}\)

3. \(\dfrac{1}{5}\)

4. \(\dfrac{4}{5}\)

A force \(F\) acting on an object varies with distance \(x\) as shown below:

The force is in Newton and \(x\) is in meters. The work done by the force in moving the object from \(x = 0\) to \(x = 6\) m is:

A ball is thrown vertically upward. It has a speed of 10m/sec when it has reached one-half of its maximum height. How high does the ball rise? Take g = 10 m/s²:

1. 5m

2. 15m

3. 10 m

4. 20 m

The kinetic energy of a person is just half of the kinetic energy of a boy whose mass is just half of that person. If the person increases his speed by \(1~\text{m/s},\) then his ​​​​​​​kinetic energy equals to that of the boy, then the initial speed of the person was:
1. \(\left( \sqrt{2}+1 \right)~\text{m/s}\)
2. \(\left( 2+\sqrt{2} \right)~\text{m/s}\)
3. \(2\left( 2+\sqrt{2} \right)~\text{m/s}\)
4. none of the above

A particle of mass m₁ is moving with a velocity v₁ and another particle of mass m₂ is moving with a velocity v₂. Both of them have the same momentum, but their kinetic energies are E₁ and E₂ respectively. If m₁ > m₂ then: 

1. $\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}$

2. ${\mathrm{E}}_1>{\mathrm{E}}_2$

3. ${\mathrm{E}}_1={\mathrm{E}}_2$

4. ${\mathrm{E}}_1<{\mathrm{E}}_2$