1

IIT-JEE 2010

MCQ (Single Correct Answer)
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
A
B
C
D
2

IIT-JEE 2009

MCQ (Single Correct Answer)
A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is $$\sqrt 3 $$. The inclination θ of this inclined plane from the horizontal plane is gradually increased from $$0^\circ $$. Then
A
at θ = $$60^\circ $$, the block will start sliding down the plane
B
the block will remain at rest on the plane up to certain θ and then it will topple
C
at θ = $$60^\circ $$, the block will start sliding down the plane and continue to do so at higher angles
D
at θ = $$60^\circ $$, the block will start sliding down the plane and on further increasing θ, it will topple at certain θ
3

IIT-JEE 2007

MCQ (Single Correct Answer)
A particle moves in the X - Y plane under the influence of a force such that its linear momentum is $$\overrightarrow p \left( t \right) = A\left[ {\widehat i\cos (kt) - \widehat j\sin (kt)} \right]$$, where A and k are constants. The angle between the force and the momentum is
A
$$0^\circ $$
B
$$30^\circ $$
C
$$45^\circ $$
D
$$90^\circ $$
4

IIT-JEE 2001 Screening

MCQ (Single Correct Answer)
The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle $$\theta $$ should be
A
$$0^\circ $$
B
$$30^\circ $$
C
$$45^\circ $$
D
$$60^\circ $$

Joint Entrance Examination

JEE Advanced JEE Main

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE ME GATE PI GATE EE GATE CE GATE IN

Medical

NEET

CBSE

Class 12