A uniform chain of length \(L\) and mass \(M\) is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If \(g\) is acceleration due to gravity, the work required to pull the hanging part on the table is:
1. \(MgL\)
2. \(MgL/3\)
3. \(MgL/9\)
4. \(MgL/18\)
A particle of mass 'm' is moving in a horizontal circle of radius 'r' under a centripetal force equal to –K/r2, where K is a constant. The total energy of the particle will be:
1.
2.
3.
4.
A force \(F = -k(y\hat i +x\hat j)\) (where \(k\) is a positive constant) acts on a particle moving in the \(xy\text-\)plane. Starting from the origin, the particle is taken along the positive \(x\text-\)axis to the point \((a,0)\) and then parallel to the \(y\text-\)axis to the point \((a,a)\). The total work done by the force on the particle is:
1. \(-2ka^2\)
2. \(2ka^2\)
3. \(-ka^2\)
4. \(ka^2\)
A lorry and a car moving with the same K.E. are brought to rest by applying the same retarding force, then:
1. Lorry will come to rest in a shorter distance
2. Car will come to rest in a shorter distance
3. Both will come to rest in a same distance
4. None of the above
The relationship between force and position is shown in the given figure (in a one-dimensional case). The work done by the force in displacing a body from \(x = 1\) cm to \(x = 5\) cm is:
1. \(20\) ergs
2. \(60\) ergs
3. \(70\) ergs
4. \(700\) ergs
The graph between the resistive force \(F\) acting on a body and the distance covered by the body is shown in the figure. The mass of the body is \(25\) kg and the initial velocity is \(2\) m/s. When the distance covered by the body is \(4\) m, its kinetic energy would be:
1. \(50\) J
2. \(40\) J
3. \(20\) J
4. \(10\) J
The relationship between the force F and the position x of a body is as shown in the figure. The work done in displacing the body from x = 1 m to x = 5 m will be:
1. | 30 J | 2. | 15 J |
3. | 25 J | 4. | 20 J |
A block of mass \(10\) kg, moving in the \(x\text-\)direction with a constant speed of \(10\) ms-1, is subjected to a retarding force \(F=0.1x\) J/m during its travel from \(x =20\) m to \(30\) m. Its final kinetic energy will be:
1. | \(475\) J | 2. | \(450\) J |
3. | \(275\) J | 4. | \(250\) J |
A ball is thrown vertically downwards from a height of 20 m with an initial velocity vo. It collides with the ground, loses 50% of its energy in a collision and rebounds to the same height. The initial velocity vo is: (Take g = 10 ms-2)
1. 14 ms-1
2. 20 ms-1
3. 28 ms-1
4. 10 ms-1