1
JEE Advanced 2014 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

A block of mass m1 = 1 kg another mass m2 = 2 kg, are placed together (see figure) on an inclined plane with angle of inclination $$\theta$$. Various values of $$\theta$$ are given in List I. The coefficient of friction between the block m1 and the plane is always zero. The coefficient of static and dynamic friction between the block m2 and the plane are equal to $$\mu$$ = 0.3. In List II expressions for the friction on the block m2 are given. Match the correct expression of the friction in List II with the angles given in List I, and choose the correct option. The acceleration due to gravity is denoted by g.

[Useful information : tan (5.5$$^\circ$$) $$\approx$$ 0.1; tan (11.5$$^\circ$$) $$\approx$$ 0.2; tan (16.5$$^\circ$$) $$\approx$$ 0.3]

JEE Advanced 2014 Paper 2 Offline Physics - Laws of Motion Question 8 English

List I List II
P. $$\theta = 5^\circ $$
1. $${m_2}g\sin \theta $$
Q. $$\theta = 10^\circ $$
2. $$({m_1} + {m_2})g\sin \theta $$
R. $$\theta = 15^\circ $$
3. $$\mu {m_2}g\cos \theta $$
S. $$\theta = 20^\circ $$
4. $$\mu ({m_1} + {m_2})g\cos \theta $$

A
P-1, Q-1, R-1, S-3
B
P-2, Q-2, R-2, S-3
C
P-2, Q-2, R-2, S-4
D
P-2, Q-2, R-3, S-3
2
IIT-JEE 2011 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-0.75

A ball of mass (m) 0.5 kg is attached to the end of a string having length (L) 0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in radian/s) is

IIT-JEE 2011 Paper 1 Offline Physics - Laws of Motion Question 23 English

A
9
B
18
C
27
D
36
3
IIT-JEE 2010 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-0.75

A block of mass m is on an inclined plane of angle θ. The coefficient of friction between the block and the plane is μ and tan θ > μ. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from P1 = mg(sinθ − μ cosθ) to P2 = mg(sinθ + μ cosθ), the frictional force f versus P graph will look like

IIT-JEE 2010 Paper 1 Offline Physics - Laws of Motion Question 24 English
A
IIT-JEE 2010 Paper 1 Offline Physics - Laws of Motion Question 24 English Option 1
B
IIT-JEE 2010 Paper 1 Offline Physics - Laws of Motion Question 24 English Option 2
C
IIT-JEE 2010 Paper 1 Offline Physics - Laws of Motion Question 24 English Option 3
D
IIT-JEE 2010 Paper 1 Offline Physics - Laws of Motion Question 24 English Option 4
4
IIT-JEE 2009 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1
A piece of wire is bent in the shape of a parabola y = kx2 (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration $$a$$. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the y-axis is
A
$${a \over {gk}}$$
B
$${a \over {2gk}}$$
C
$${{2a} \over {gk}}$$
D
$${a \over {4gk}}$$
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