JEE Advanced 2014 Paper 1 Offline | Laws of Motion Question 11 | Physics

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

JEE Advanced 2014 Paper 1 Offline Physics - Laws of Motion Question 10 English

JEE Advanced 2014 Paper 1 Offline Physics - Laws of Motion Question 10 English

$$\mu$$₁ = 0, $$\mu$$₂ $$\ne$$ 0 and $${N_2}\tan \theta = {{mg} \over 2}$$

$$\mu$$₁ $$\ne$$ 0, $$\mu$$₂ = 0 and $${N_1}\tan \theta = {{mg} \over 2}$$

$$\mu$$₁ $$\ne$$ 0, $$\mu$$₂ $$\ne$$ 0 and $${N_2} = {{mg} \over {1 + {\mu _1}{\mu _2}}}$$

$$\mu$$₁ = 0, $$\mu$$₂ $$\ne$$ 0 and $${N_1}\tan \theta = {{mg} \over 2}$$

IIT-JEE 2012 Paper 1 Offline

MCQ (More than One Correct Answer)

-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)

IIT-JEE 2012 Paper 1 Offline Physics - Laws of Motion Question 7 English

$$\theta$$ = 45$$^\circ$$

$$\theta$$ > 45$$^\circ$$ and a frictional force ats on the block towards P.

$$\theta$$ > 45$$^\circ$$ and a frictional force ats on the block towards Q.

$$\theta$$ < 45$$^\circ$$ and a frictional force ats on the block towards Q.

IIT-JEE 1986

MCQ (More than One Correct Answer)

-0.5

A reference frame attached to the earth

is an inertial frame by definition

cannot be an inertial frame because the earth is revolving around the sun

is an inertial frame because Newton's laws are applicable in this frame

cannot be an inertial frame because the earth rotating about its own axis

IIT-JEE 1986

MCQ (More than One Correct Answer)

-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:

$$T\cos \theta = Mg$$

$$T - Mg\cos \theta = {{M{V^2}} \over L}$$

The magnitude of the tangential acceleration of the bob $$\left| {{a_T}} \right| = g\sin \theta $$

$$T = Mg\cos \theta $$

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Questions Asked from MCQ (Multiple Correct Answer)

JEE Advanced 2019 Paper 2 Offline (1) JEE Advanced 2014 Paper 1 Offline (1) IIT-JEE 2012 Paper 1 Offline (1) IIT-JEE 1986 (2) IIT-JEE 1982 (1)

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