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
If a young man of mass \(60\) kg stands on the floor of a lift which is accelerating downwards at \(1~\text{m/s}^2\), then the reaction of the floor of the lift on the man will be: \(\left(g = 9.8~\text{m/s}^2 \right)\)
1. | \(528\) N | 2. | \(540\) N |
3. | \(546\) N | 4. | None of these |
If \(\mu\) between block \(A\) and inclined plane is \(0.5\) and that between block \(B\) and the inclined plane is \(0.8\), then the normal reaction between blocks \(A\) and \(B\) will be:
1. \(180\) N
2. \(216\) N
3. \(0\)
4. None of these
A particle is on a smooth horizontal plane. A force \(F\) is applied, whose \((F\text-t)\) graph is given.
Consider the following statements.
(a) | At time \(t_1\), acceleration is constant. |
(b) | Initially the particle must be at rest. |
(c) | At time \(t_2\), acceleration is constant. |
(d) | The initial acceleration is zero. |
Select the correct statement(s):
1. | (a), (c) | 2. | (a), (b), (d) |
3. | (c), (d) | 4. | (b), (c) |
A coin placed on a rotating table just slips if it is placed at a distance \(4r\) from the center. On doubling the angular velocity of the table, the coin will just slip when the distance from the centre is equal to:
1. \(4r\)
2. \(2r\)
3. \(r\)
4. \(\frac{r}{4}\)
A massless string of length \(1\) m fixed at one end carries a mass of \(2\) kg at the other end. The string makes \(\frac{2}{\pi}\) rev/s around the axis through the fixed end as shown in the figure. The tension on the string will be:
1. | \(32\) N | 2. | \(3\) N |
3. | \(16\) N | 4. | \(4\) N |
Three blocks \(A\), \(B\) and \(C\) of mass \(3M\), \(2M\) and \(M\) respectively are suspended vertically with the help of springs \(\mathrm{PQ}\) and \(\mathrm{TU}\) and a string \(\mathrm{RS}\) as shown in fig. The acceleration of blocks \(A\), \(B\) and \(C\) are \(a_{1} , a_{2}~ \text{and}~ a_{3}\) respectively.
The value of acceleration \(a_{1}\) at the moment string \(\mathrm{RS}\) is cut will be:
1. \(g\) downward
2. \(g\) upward
3. more than \(g\) downward
4. zero
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
1. |
\(\overrightarrow N+\overrightarrow T+\overrightarrow W=0\) |
2. | \(T^2=N^2+W^2\) |
3. | \(T = N + W\) | 4. | \(N = W \tan \theta\) |
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