1
JEE Advanced 2014 Paper 1 Offline
MCQ (More than One Correct Answer)
+4
-1
In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle $$\theta$$ with the horizontal floor. The coefficient of friction between the wall and the ladder is $$\mu$$1 and that between the floor and the ladder is $$\mu$$2. The normal reaction of the wall on the ladder is N1 and that of the floor is N2. If the ladder is about to slip, then A
$$\mu$$1 = 0, $$\mu$$2 $$\ne$$ 0 and $${N_2}\tan \theta = {{mg} \over 2}$$
B
$$\mu$$1 $$\ne$$ 0, $$\mu$$2 = 0 and $${N_1}\tan \theta = {{mg} \over 2}$$
C
$$\mu$$1 $$\ne$$ 0, $$\mu$$2 $$\ne$$ 0 and $${N_2} = {{mg} \over {1 + {\mu _1}{\mu _2}}}$$
D
$$\mu$$1 = 0, $$\mu$$2 $$\ne$$ 0 and $${N_1}\tan \theta = {{mg} \over 2}$$
2
IIT-JEE 2012 Paper 1 Offline
MCQ (More than One Correct Answer)
+4
-2

A small block of mass 0.1 kg lies on a fixed inclined plane PQ which makes an angle $$\theta$$ with the horizontal. A horizontal force of 1 N acts on the block through its centre of mass as shown in the figure. The block remains stationary if (take g = 10 m/s2) A
$$\theta$$ = 45$$^\circ$$
B
$$\theta$$ > 45$$^\circ$$ and a frictional force ats on the block towards P.
C
$$\theta$$ > 45$$^\circ$$ and a frictional force ats on the block towards Q.
D
$$\theta$$ < 45$$^\circ$$ and a frictional force ats on the block towards Q.
3
IIT-JEE 1986
MCQ (More than One Correct Answer)
+2
-0.5
A reference frame attached to the earth
A
is an inertial frame by definition
B
cannot be an inertial frame because the earth is revolving around the sun
C
is an inertial frame because Newton's laws are applicable in this frame
D
cannot be an inertial frame because the earth rotating about its own axis
4
IIT-JEE 1986
MCQ (More than One Correct Answer)
+2
-0.5
A simple pendulum of length L and mass (bob) M is oscillating in a plane about a vertical line between angular limit $$- \phi$$ and $$+ \phi$$. For an angular displacement $$\theta$$ $$\left( {\left| \theta \right| < \phi } \right)$$, the tension in the string and the velocity of the bob are T and V respectively. The following relations hold good under the above conditions:
A
$$T\cos \theta = Mg$$
B
$$T - Mg\cos \theta = {{M{V^2}} \over L}$$
C
The magnitude of the tangential acceleration of the bob $$\left| {{a_T}} \right| = g\sin \theta$$
D
$$T = Mg\cos \theta$$
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