# The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle θ should be:        1. 0° 2. 30° 3. 45° 4. 60°

Subtopic:  Tension & Normal Reaction |
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A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. The force on the pulley by the clamp is given by:

1. $\sqrt{2}Mg$

2. $\sqrt{2}mg$

3. $\sqrt{{\left(M+m\right)}^{2}+{m}^{2}}g$

4.$\sqrt{{\left(M+m\right)}^{2}+{M}^{2}}g$

Subtopic:  Tension & Normal Reaction |
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A block B is placed on top of block A. The mass of block B is less than the mass of block A. Friction exists between the blocks, whereas the ground on which block A is placed is assumed to be smooth. A horizontal force F, increasing linearly with time begins to act on B. The acceleration aA and aB of blocks A and B respectively are plotted against t. The correctly plotted graph is:

1.
2.
3.
4.

Subtopic:  Friction |
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The variation of momentum with the time of one of the bodies in a two-body collision is shown in fig. The instantaneous force is the maximum corresponding to the point-

1. P

2. Q

3. R

4. S

Subtopic:  Newton's Laws |
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A uniform chain of length L hangs partly from a table which is kept in equilibrium by friction. If the maximum length that can be supported without slipping is l, then the coefficient of friction between the table and the chain is:

1. $\frac{l}{L}$

2. $\frac{l}{L+l}$

3. $\frac{l}{L-l}$

4. $\frac{L}{L+l}$

Subtopic:  Friction |
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Block A has a mass of 10 kg. Between block A and the table, the coefficient of static friction is 0.2, and the coefficient of kinetic friction is also 0.2. The required mass of B to start the motion will be:

1. 2 kg

2. 2.2 kg

3. 4.8 kg

4. 200 gm

Subtopic:  Friction |
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A block of mass m lying on a rough horizontal plane is acted upon by a horizontal force P and another force Q inclined at an angle θ to the vertical. The block will remain in equilibrium if the coefficient of friction between it and the surface is:

1. $\frac{\left(P+Q\mathrm{sin}\theta \right)}{\left(mg+Q\mathrm{cos}\theta \right)}$

2. $\frac{\left(P\mathrm{cos}\theta +Q\right)}{\left(mg-Q\mathrm{sin}\theta \right)}$

3. $\frac{\left(P+Q\mathrm{cos}\theta \right)}{\left(mg+Q\mathrm{sin}\theta \right)}$

4. $\frac{\left(P\mathrm{sin}\theta -Q\right)}{\left(mg-Q\mathrm{cos}\theta \right)}$

Subtopic:  Friction |
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The elevator is ascending. Together, the lift and the passenger weigh 1500 kg. The variation in the speed of the lift is as given in the graph. The tension in the rope pulling the lift at t = 11th sec will be:

1. 17400 N

2. 14700 N

3. 12000 N

4. Zero

Subtopic:  Application of Laws |
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A particle moves in the XY-plane under the action of a force F such that the components of its linear momentum p at any time t are ${p}_{x}=2\mathrm{cos}t$, ${p}_{y}=2\mathrm{sin}t$. The angle between F and p at time t will be:

1. 90°

2. 0°

3. 180°

4. 30°

Subtopic:  Newton's Laws |
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Two stones of masses m and 2m are whirled in horizontal circles, the heavier one in a radius $\frac{r}{2}$ and the lighter one in the radius r. The tangential speed of lighter stone is n times that of heavier stone when they experience the same centripetal forces. The value of n is:

1. 2

2. 3

3. 4

4. 1

Subtopic:  Uniform Circular Motion |
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