Q. No.
1. \(4~\text N\)
2. \(6~\text N\)
3. \(10~\text N\)
4. zero
When a body of mass \(m\) just begins to slide as shown, match List-I with List-II:
| List-I | List-II | ||
| (a) | Normal reaction | (i) | \(P\) |
| (b) | Frictional force \((f_s)\) | (ii) | \(Q\) |
| (c) | Weight \((mg)\) | (iii) | \(R\) |
| (d) | \(mg \mathrm{sin}\theta ~\) | (iv) | \(S\) |
Choose the correct answer from the options given below:
| (a) | (b) | (c) | (d) | |
| 1. | (ii) | (i) | (iii) | (iv) |
| 2. | (iv) | (ii) | (iii) | (i) |
| 3. | (iv) | (iii) | (ii) | (i) |
| 4. | (ii) | (iii) | (iv) | (i) |
A ball of mass \(0.15~\text{kg}\) is dropped from a height \(10~\text{m}\), strikes the ground, and rebounds to the same height. The magnitude of impulse imparted to the ball is \((g=10 ~\text{m}/\text{s}^2)\) nearly:
| 1. | \(2.1~\text{kg-m/s}\) | 2. | \(1.4~\text{kg-m/s}\) |
| 3. | \(0~\text{kg-m/s}\) | 4. | \(4.2~\text{kg-m/s}\) |
Two bodies of mass, \(4~\text{kg}\) and \(6~\text{kg}\), are tied to the ends of a massless string. The string passes over a pulley, which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (\(g\)) is:
| 1. | \(\dfrac{g}{2}\) | 2. | \(\dfrac{g}{5}\) |
| 3. | \(\dfrac{g}{10}\) | 4. | \(g\) |
A truck is stationary and has a bob suspended by a light string in a frame attached to the truck. The truck suddenly moves to the right with an acceleration of \(a.\) In the frame of the truck, the pendulum will tilt:
| 1. | to the left and the angle of inclination of the pendulum with the vertical is \(\text{sin}^{-1} \left( \dfrac{a}{g} \right )\) |
| 2. | to the left and the angle of inclination of the pendulum with the vertical is \(\text{cos}^{-1} \left ( \dfrac{a}{g} \right )\) |
| 3. | to the left and the angle of inclination of the pendulum with the vertical is \(\text{tan}^{-1} \left ( \dfrac{a}{g} \right )\) |
| 4. | to the left and the angle of inclination of the pendulum with the vertical is \(\text{tan}^{-1} \left ( \dfrac{g}{a} \right )\) |
A block of mass \(m\) is placed on a smooth inclined wedge \(ABC\) of inclination \(\theta\) as shown in the figure. The wedge is given an acceleration '\(a\)' towards the right. The relation between \(a\) and \(\theta\) for the block to remain stationary on the wedge is:
1. \(a = \dfrac{g}{\mathrm{cosec }~ \theta}\)
2. \(a = \dfrac{g}{\sin\theta}\)
3. \(a = g\cos\theta\)
4. \(a = g\tan\theta\)
A balloon with mass \(m\) is descending down with an acceleration \(a\) (where \(a<g\)). How much mass should be removed from it so that it starts moving up with an acceleration \(a\)?
| 1. | \( \frac{2 m a}{g+a} \) | 2. | \( \frac{2 m a}{g-a} \) |
| 3. | \( \frac{m a}{g+a} \) | 4. | \( \frac{m a}{g-a}\) |
Three blocks with masses \(m\), \(2m\), and \(3m\) are connected by strings as shown in the figure. After an upward force \(F\) is applied on block \(m\), the masses move upward at constant speed \(v\). What is the net force on the block of mass \(2m\)? (\(g\) is the acceleration due to gravity)
| 1. | \(2~mg\) | 2. | \(3~mg\) |
| 3. | \(6~mg\) | 4. | zero |
1. \(9680~\text{N}\)
2. \(11000~\text{N}\)
3. \(1200~\text{N}\)
4. \(8600~\text{N}\)
The mass of a lift is \(2000\) kg. When the tension in the supporting cable is \(28000\) N, then its acceleration is:
(Take \(g=10\) m/s2)
| 1. | \(30\) ms-2 downwards | 2. | \(4\) ms-2 upwards |
| 3. | \(4\) ms-2 downwards | 4. | \(14\) ms-2 upwards |
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