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Test: Newton's laws of Motion (June 4) - NEET MCQ


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10 Questions MCQ Test - Test: Newton's laws of Motion (June 4)

Test: Newton's laws of Motion (June 4) for NEET 2024 is part of NEET preparation. The Test: Newton's laws of Motion (June 4) questions and answers have been prepared according to the NEET exam syllabus.The Test: Newton's laws of Motion (June 4) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Newton's laws of Motion (June 4) below.
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Test: Newton's laws of Motion (June 4) - Question 1

We slip on a rainy day due to ______.

Detailed Solution for Test: Newton's laws of Motion (June 4) - Question 1
  • It is usually seen that during rainy days we tend to slip because rainwater acts as a lubricant between our feet and the ground.  

  • This lubrication (rainwater) converts dry friction into fluid friction. 

  • That is, the friction between feet and ground reduces, making us slip.

Therefore, due to less friction, we slip on a rainy day.

Test: Newton's laws of Motion (June 4) - Question 2

Passengers in a bus lean forward as bus suddenly stops. This is due to

Detailed Solution for Test: Newton's laws of Motion (June 4) - Question 2

When the bus moves, the passenger's body comes into a state of motion but when the bus stops the lower part of the body which is in contact with the floor comes into a state of rest whereas the upper part of the body still remains in the state of motion and because of this the upper part of the body falls in the forward direction. This is due to the presence of inertia. 

Hence the correct answer is C.

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Test: Newton's laws of Motion (June 4) - Question 3

A particle of mass 'm' original at rest, is subjected to a force whose direction is constant but whose magnitude varies with according to the relation
 

Where F0​ and T are constant.
Then speed of the particle after a time 2T is:

Detailed Solution for Test: Newton's laws of Motion (June 4) - Question 3

Given: 


Speed of the particle after a time 2T, 



Test: Newton's laws of Motion (June 4) - Question 4

The dimensional formula of Plancks’s constant and angular momentum are

Detailed Solution for Test: Newton's laws of Motion (June 4) - Question 4

Planck's Constant (h)  = 6.626176 x 10-34  m2 kg/s
So, Unit of planck constant=  m2 kg/s

Dimensions =M L2 T −1   ________ (1)

Angular momentum l = mvr

Where, m-mass

v-velocity

r-radius

Dimensions of angular momentum  =  M L T −1 L   = M  L2  T −1  _______________ (2)
From (1) and (2). 

Planck's constant and angular momentum have the same dimensions.

Test: Newton's laws of Motion (June 4) - Question 5

Which of the following quantity has the unit of newton-second?

Detailed Solution for Test: Newton's laws of Motion (June 4) - Question 5

Net external force F = dp/dt, so using dimensional analysis, unit of momentum should be the same as unit of force times unit of time which is Newton second.

Test: Newton's laws of Motion (June 4) - Question 6

 If newton is the unit of force in MKS system then, in CGS system 1N= ______ dyne

Detailed Solution for Test: Newton's laws of Motion (June 4) - Question 6

1N = kg m /sec
1 dyne = gram cm/sec
Thus 1N in CGS system = 1000 gram 100 cm/sec
That is 1N = 100000 dyne

Test: Newton's laws of Motion (June 4) - Question 7

Ten one rupee coins are put on top of each other on a table. Each coin has a mass m. The reaction of the 6th coin (counted from the bottom) on the 7th coin is 

Detailed Solution for Test: Newton's laws of Motion (June 4) - Question 7

The sixth coin is under the weight of four coins baove it. Hence,
Reaction of the 6th coin on the 7th coin = Force on the 6th coin due to 7th coin
= 4mg

Test: Newton's laws of Motion (June 4) - Question 8

A cork of mass 10 g is floating on water The net force acting on this cork is

Detailed Solution for Test: Newton's laws of Motion (June 4) - Question 8

When the cork is floating, its weight is balanced by the upthrust. Therefore, net force on the cork is zero.

Test: Newton's laws of Motion (June 4) - Question 9

 A stone of mass 1 kg is lying on the floor of a train which is accelerating with 1 m s-2. The net force acting on the stone is

Detailed Solution for Test: Newton's laws of Motion (June 4) - Question 9

Here, mass of the stone, m = 1 kg
As the stone is lying on the floor of the train, its acceleration is same as that of train.
∴ Force acting on the stone F = ma = (1kg)(1 m s-2)
= 1 N

Test: Newton's laws of Motion (June 4) - Question 10

A rocket with a lift-off mass 2 x 104 kg is blasted upwards with an initial acceleration of 5 m s-2. The initial thrust of the blast is (Take g = 10 m s-2)

Detailed Solution for Test: Newton's laws of Motion (June 4) - Question 10

Here, m = 2 × 104kg
Initial acceleration, a = 5ms-2
Initial thrust = upthrust required to impart acceleration a + upthrust required to overcome gravitational pull
∴ F  = m(a + g) = (2 × 104kg)(5 + 10)ms−2
= 3 × 105N

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