A block \(B\) of mass \(3\) kg is kept on block \(A\) of mass \(5\) kg in a lift accelerating upward with an acceleration of \(g.\) The reaction by \(A\) on \(B\) is:
1. | \(10g\) | 2. | \(16g\) |
3. | \(4g\) | 4. | \(6g\) |
If the force acting on a system is zero, the quantity which remains constant is:
1. | Force | 2. | Linear momentum |
3. | Speed | 4. | Kinetic energy |
Two blocks, \(A\) and \(B\), of masses \(2m\) and \(4m\) are connected by a string. The block of mass \(4m\) is connected by a spring (massless). The string is suddenly cut. The ratio of the magnitudes of accelerations of masses \(2m\) and \(4m\) at that instant will be:
1. | \(1:2\) | 2. | \(2:1\) |
3. | \(1:4\) | 4. | \(4:1\) |
The reading of spring balance in the depicted figure will be:
1. | \(0\) N | 2. | \(20\) N |
3. | \(10\) N | 4. | \(5\) N |
The angle of banking for a cyclist taking a turn at a curve is given by \(\tan\theta = \frac{v^{n}}{rg}\) where symbols have their usual meaning. The value of \(n\) is:
1. | \(1\)
|
2. | \(2\)
|
3. | \(3\)
|
4. | \(4\) |
1. | \(36\) km/h | 2. | \(54\) km/h |
3. | \(72\) km/h | 4. | \(90\) km/h |
A \(100\) kg gun fires a ball of \(1\) kg horizontally from a cliff at a height of \(500\) m. It falls on the ground at a distance of \(400\) m from the bottom of the cliff. The recoil velocity of the gun is: (Take \(g=10\) m/s2)
1. \(0.2\) m/s
2. \(0.4\) m/s
3. \(0.6\) m/s
4. \(0.8\) m/s
1. |
\(\overrightarrow N+\overrightarrow T+\overrightarrow W=0\) |
2. | \(T^2=N^2+W^2\) |
3. | \(T = N + W\) | 4. | \(N = W \tan \theta\) |
Calculate the reading of the spring balance shown in the figure: (take \(g=10\) m/s2)
1. \(60\) N
2. \(40\) N
3. \(50\) N
4. \(80\) N
In the diagram, a \(100\) kg block is moving up with constant velocity. Find out the tension at the point \(P\).
1. \(1330\) N
2. \(490\) N
3. \(1470\) N
4. \(980\) N