Newton's Law Of Motion MCQ Level - 1 - IIT JAM MCQ

Free Body diagram in the frame of the car

It is given that a bob is hanging over a pulley inside a car through a string the second end of the string is in the hand of a person standing in the car the car is moving with constant acceleration ‘a’ directed horizontally. Other end of the string is pulled with constant acceleration ‘a’ (relative to car) vertically. We need to find the tension in the string if the θ is constant.

The force on the bob will be in the opposite direction of the motion of the car and also a tension in the string is present.

The forces on the bob of mass m is given by,

It can be seen in the figure that in the y’-y’ direction,

The tension in the string is equal to T = m√ a² + g² + ma. The correct answer for this problem is option D.

The man will first press the platform to get extra normal force required for the jump. Hence initially the reading will be greater than 60 kg wt and then decrease to zero.

The correct answer is: Becomes > 60 kg-wt initially and then decreases to zero

The forces acting on the block A are: The normal reaction (N) by the block B in the upward direction, the weight of block (mg) A in the downward direction and the resultant force (ma) in the downward direction. 

The simplified diagram of the block with forces is given below:

The normal reaction can be calculated from equation for the resultant force given below:


N = 4N
The correct answer is: 4N

The equation of height = h = ut + 1 / 2 at²

Since the boxes are at rest initially, then the equation becomes,

⇒  a should be same in both case, because h and t are same in both cases as given

In figure, all the forces acting on the block are shown. The weight (mg) acts downwards and the force F₁ is applied in the upward direction and the acceleration is also produced in the upward direction, so the equation for resultant force will give the value of F₁.

In figure, all the forces acting on the block are shown. The weight (mg) acts downwards and the force F₂ is applied in the upward direction and the acceleration is also produced in the upward direction, so the equation for resultant force will give the value of F₂.

The correct answer is: (i)

Free body diagram of pulley is shown in figure. Pulley is equilibrium under four forces. Three forces are shown in figure and the fourth, which is equal and opposite to the resultant of these three forces, in the force applied by the clamp on the pulley (say F).

From the F.B.D we obtain

Therefore, the force F is equal and opposite of R as shown in figure.

The correct answer is: 

The length of the rope on the left side is L₁ and on the right side is L₂. The total length of the rope is L.
L = L₁ + L₂
Total mass of the rope is M.
The mass per unit length of the rope is M/L
So, the mass of the rope on the left side is m₁=L₁M/L
Mass of the rope on the right side is m₂=L₂ML

Weight of the rope on the left side is, m₁g=L₁Mg/L
Weight of the rope on the right side is, m₂g=L₂Mg/L
Now, taking the component of the forces on the ropes,
On the left side of the rope,
m₁gsinα − T = m₁a
L₁Msinα/L−T = L₁Ma/L
And, on the right side of the rope,
T − m₂gsinβ = m₂a
T− L₂Mgsinβ/L = L₂Ma/L
Adding the above two equations,
L₁Mgsinα/L − L₂Mgsinβ/L = L₁Ma/L+L₂Ma/L
(L₁sinα − L₂sinβ)g = (L₁+L₂)a=La
a =  (L₁sinα − L₂sinβ)g/a
Now, since both ends of the rope are in the same horizontal line or in the same vertical height, the components,
L₁sinα = L₂sinβ
So, we can write,
a=0

Let us assume that the block A of mass m is tied to the spring. However at t = 0s, the spring is not compressed or extended.

Let T be the tension in the string going over pulley.   Just after releasing the force on block A, both blocks move with the same acceleration. The tension is same in the rope on both sides.  The equations of motion for both blocks are: 

 T − mg = ma

2 mg − T= 2ma

So  mg = 3ma

=>a = g/3

So at t = 0, block A moves up initially with an acceleration g/3 and block B moves down with acceleration g/3.

As shown in the FBD resolving the forces along the vertical and horizontal axis in the final position we have 

Putting in the values and solving we get 
Putting in the value of cotθ as we have the option C as the correct one.

Reason: Newton's third law of motion.  If a body exerts a force on another, there is an equal and opposite force called a reaction on the first body by the second. In Downward moving elevator (with a uniform acceleration), the man experiences an upward force. This reduces his weight.
In case of uniform velocity, Force= mass x acceleration is zero since acceleration is zero.

The correct answer is: