NEET Laws of Motion MCQs & PYQs

Q: A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is

Maximum Friction force action on an object is μN. Here μ=0.2 and Normal reaction force will be equal to applied force i.e 10 N., So, friction force is 0.2*10 = 2N

Q: An insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of friction between the insect and the surface is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle a with the vertical, the maximum possible value of α is given by :

For Equilibrium, ⇒ μmg cos θ = mg sin θ ⇒ μ = tan θ = 1/3 ⇒ cot θ = 3

Practice NEET Laws of Motion Questions

Question 1:

difficult

Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A 15 kg weight is attached to the rope at the mid-point which now no longer remains horizontal. The minimum tension required to completely straighten the rope is:

For Equilibrium condition,

2T cos θ = mg ⇒ T = mg/(2 cos θ)

As, we try to make string horizontal θ approaches 90°. so Tension will become T= mg/2 cos (90°)

i.e T becomes infinite

💬 Discuss this Question

Question 2:

difficult

A block is kept on the frictionless inclined surface with angle of inclination . The incline is given an acceleration a to keep the block stationary w.r.t incline plane. Then a is equal to

Taking incline plane as frame of reference, Pseudo force ma act towards right.

component of Pseudo force up the incline is m.a.cos α which is balanced by m.g.sin α. so,

m.a.cos α = m.g.sin α

⇒a= g tan α

💬 Discuss this Question

Question 3:

difficult

A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is

Maximum Friction force action on an object is μN.

Here μ=0.2 and Normal reaction force will be equal to applied force i.e 10 N.,

So, friction force is 0.2*10 = 2N

💬 Discuss this Question

Question 4:

difficult

A painter is raising himself and the crate on which he stands with an acceleration of 5m/s² by a massless rope–and–pulley arrangement. Mass of painter is 100 kg and that of the crate is 50 kg. If g = 10 m/s², then :

Let us assume tension force in the wire is T. Then net force acting on man and crate in upward direction is 2T. Gravitational force acting in downward direction is (50+100) g = 1500 N. Writing equation of motion for the system.

2T- 1500 = 150 × a

⇒ 2T- 1500 = 150 × 5

⇒ 2T = 750 +1500 = 2250

⇒ T = 1125 N

💬 Discuss this Question

Question 5:

difficult

An insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of friction between the insect and the surface is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle a with the vertical, the maximum possible value of α is given by :

For Equilibrium,

⇒ μmg cos θ = mg sin θ

⇒ μ = tan θ = 1/3

⇒ cot θ = 3

💬 Discuss this Question

Question 6:

difficult

A rod AB is shown in figure. End A of the rod is fixed on the ground. Block is moving with velocity √3 m/s towards right. The velocity of end B of rod when rod makes an angle of 60⁰ with the ground is:

💬 Discuss this Question

Question 7:

difficult

A \( 0.5 \text{ kg} \) ball moving with a speed of \( 12 \text{ m/s} \) strikes a hard wall at an angle of \( 30^{\circ} \) with the wall. It is reflected with the same speed at the same angle. If the ball is in contact with the wall for \( 0.25 \) seconds, the average force acting on the wall is (2006)

Change in momentum \( \Delta P = 2mv \sin\theta \) (perpendicular to wall). \( \Delta P = 2 \times 0.5 \text{ kg} \times 12 \text{ m/s} \times \sin 30^{\circ} = 6 \text{ Ns} \). Average force \( F = \Delta P / \Delta t = 6 \text{ Ns} / 0.25 \text{ s} = 24 \text{ N} \).

💬 Discuss this Question