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) |
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 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. \(2mg\)
2. \(3mg\)
3. \(6mg\)
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 |
A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall at an angle of with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 s, the average force acting on the wall is:
1. 48 N
2. 24 N
3. 12 N
4. 96 N
A block of mass \(m\) is placed on a smooth wedge of inclination \(\theta\). The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block (\(g\) is the acceleration due to gravity) will be:
1. \(mg~\mathrm{sin\theta}\)
2. \(mg\)
3. \(\frac{mg}{\mathrm{cos\theta}}\)
4. \(mg~\mathrm{cos\theta}\)
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