JEE Advanced (Single Correct MCQs): Laws of Motion - JEE MCQ

F = ma

Limiting frictional force,  f_l = µ_sN = 0.5 × 5 = 2.5 N. But force tending to produce relative motion is the weight (W) of the block which is less than f_l. Therefore, the frictional force is equal to the weight, the magnitude of the frictional force f has to balance the weight 0.98 N acting downwards.

Therefore the frictional force = 0.98 N.

Since the body presses the surface with a force N hence according to Newton's third law the surface presses the body with a force N. The other force acting on the body is its weight mg.

For circular motion to take place, a centripetal force is required which is provided by (mg + N).

where r is the radius of curvature at the top.
If the surface is smooth then on applying conservation of mechanical energy, the velocity of the body is always same at the top most point. Hence, N and r have inverse relationship. From the figure it is clear that r is minimum for first figure, therefore N will be maximum.
Note : If we do not assume the surface to be smooth, we cannot reach to a conclusion.

KEY CONCEPT : For  the maximum possible value of α, mg sin α will also be maximum and equal to the frictional force.
In this case f is the limiting friction. The two forces acting on the insect are mg and N. Let us resolve mg into two components.
mg cos α balances N.
mg sin α is balanced by the frictional force.

∴ N = mg cos α
f = mg sin α
But f = µN = µ mg cos α

∴ µ mg cos α = mg sin α ⇒ cot α = 

The tension in both strings will be same due to symmetry.

For equilibrium in vertical direction for body B we have

[∵ T = mg, (at equilibrium]

At equilibrium T = Mg

F.B.D. of pulley
F₁ = (m + M) g
The resultant force on pulley is

The forces acting on the block are shown. Since the block is not moving forward for the maximum force F applied, therefore
F cos 60° = f = µN ... (i) (Horizontal  Direction)
Note : For maximum force F, the frictional force is the limiting friction = µN]
and F sin 60° + mg = N... (ii)
From (i) and (ii)

Let ω be the angular frequency of the system.
The maximum acceleration of the system,

The force of friction provides this acceleration.

In situation 1, the tension T has to hold both the masses 2m and m therefore,
T = 3mg
In situation 2, when the string is cut, the mass m is a freely falling body and its acceleration due to gravity is g.
For mass 2m, just after the string is cut, T remains 3mg because of the extension of string.

∴ 3mg – 2mg = 2m × a    ∴ g/2 = a

The acceleration of mass m is due to the force T cos θ

For the block to slide, the angle of inclination should be equal to the angle of repose, i.e.,

Therefore, option (a) is wrong.
For the block to topple, the condition of the block will be as shown in the figure.

For this, θ < 60°. From this we can conclude that the block will topple at lesser angle of inclination. Thus the block will remain at rest on the plane up to a certain anlgle θ and then it will topple.

As tan θ > μ, the block has a tendency to move down the incline. Therfore a force P is applied upwards along the incline. Here, at equilibrium P + f = mg sin θ   ⇒ f = mg sin θ - P

Now as P increases, f decreases linearly with respect to P.
When P = mg sin θ, f = 0.
When P is increased further, the block has a tendency to move upwards along the incline.
Therefore the frictional force acts downwards along the incline.
Here, at equilibrium P = f + mg sin θ

∴ f = P – mg sinθ
Now as P increases, f increases linearly w.r.t P.
This is represented by graph (a) .

Here, the horizontal component of tension provides the necessary centripetal force.
∴ T sin θ = mrω²

From (i) and (ii)

While turning the maximum allowed speed is: 
V _max ​ = √μrg ​ 
where, 
μ is coefficient of friction, 
r is radius of the turn and 
g acceleration due to gravity. 
For the inner wheel the radius of the turn is less than the outer wheel and therefore its maximum speed is also less than for the outer wheel. Thus the inner wheel leaves the ground first.