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Test: Newton’s Second Law of Motion - NEET MCQ


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10 Questions MCQ Test Physics Class 11 - Test: Newton’s Second Law of Motion

Test: Newton’s Second Law of Motion for NEET 2024 is part of Physics Class 11 preparation. The Test: Newton’s Second Law of Motion questions and answers have been prepared according to the NEET exam syllabus.The Test: Newton’s Second Law of Motion MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Newton’s Second Law of Motion below.
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Test: Newton’s Second Law of Motion - Question 1
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A constant force acting on a body of mass 3 kg changes its speed from 2 m/s to 3.5 m/s in 10 second. If the direction of motion of the body remains unchanged, what is the magnitude and direction of the force?

Detailed Solution for Test: Newton’s Second Law of Motion - Question 1

Given : Mass, m = 3 kg;

Initial Velocity, u = 2m/s;

Final velocity, v = 3.5 m/s

Time taken, t = 10 seconds

v = u+at

3.5 = 2 +a x 10

F = ma = 3 x 0.15 = 0.45 N

Since the applied force increase the speed of the body, it acts along the direction of motion.

Test: Newton’s Second Law of Motion - Question 2
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A simple pendulum is oscillating in a vertical plane. If resultant acceleration of bob of mass m at a point A is in horizontal direction, find the tangential force at this point in terms of tension T and mg.

Detailed Solution for Test: Newton’s Second Law of Motion - Question 2


When the acceleration of bob is horizontal, net vertical force on the bob will be zero.
T cos θ – mg = 0
The tangential force at that instant is 

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Test: Newton’s Second Law of Motion - Question 3
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A painter is applying force himself to raise him and the box with an  cceleration of 5 m/s2 by a massless rope and pulley arrangement as shown in figure. Mass of painter is 100 kg and that of box is 50 kg. If g = 10 m/s2, then : 

Detailed Solution for Test: Newton’s Second Law of Motion - Question 3

For the whole system, 2T – 1500 = 150 x 5 ⇒ T = 1125 N
For the person, T – 1000 + N = 100 x 5 N = 1500 – 1125 = 375 N 

Test: Newton’s Second Law of Motion - Question 4
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Two blocks ‘A’ and ‘B’ each of mass ‘m’ are placed on a smooth horizontal surface. Two horizontal  force F and 2F are applied on the two blocks ‘A’ and ‘B’ respectively as shown in figure. The block A does not slide on block B. Then the normal reaction acting between the two blocks is : 
Detailed Solution for Test: Newton’s Second Law of Motion - Question 4

Test: Newton’s Second Law of Motion - Question 5
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A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is . The inclination q of this inclined plane from the horizontal plane is gradually increased from 0º. Then                                                                                                 [jee 2009]
Detailed Solution for Test: Newton’s Second Law of Motion - Question 5

Test: Newton’s Second Law of Motion - Question 6
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A block of mass m is on an inclined plane of angle q. The coefficient of friction betwen the block and the plane is m and tan q > m. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to the positive. As P is varied from P = mg (sin q _ m cos q ) to Pz = mg (sin q + m cos q), the frictional force f versus P graph will look like              [jee 2010]

                                 
Detailed Solution for Test: Newton’s Second Law of Motion - Question 6


Test: Newton’s Second Law of Motion - Question 7
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A block is moving on an inclined plane making an angle 45º with horizontal and the coefficient of friciton is μ. the force required to just push it up the inclined plane is 3 times the force requried to just prevent it from sliding down. If we define N = 10μ, then N is

[jee 2011]

Detailed Solution for Test: Newton’s Second Law of Motion - Question 7


Test: Newton’s Second Law of Motion - Question 8
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In cgs system the gravitational unit of force is gram weight (g wt).Then 1g wt (or g f) =_______ dyne.
Detailed Solution for Test: Newton’s Second Law of Motion - Question 8

For SI to CGS conversion,

1 Kg = 103 g

1 m = 102 cm

=> In CGS System,

1 N (Kg.m/s2) = 105 dynes (g.cm/s2)

=> 1 g.wt = 9.8 m/s2 x 102 cm / 1 s2 x 10-3 g x 10-2 cm / 1 s x 105 dynes

= 980 dynes (g.cm/s2)

Test: Newton’s Second Law of Motion - Question 9
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The dimensional formula of Plancks’s constant and angular momentum are
Detailed Solution for Test: Newton’s Second Law of Motion - Question 9

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 Second Law of Motion - Question 10
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Three forces 2570_image008 act on an object of mass m = 2 kg. The acceleration of the object in m/s2 is:
Detailed Solution for Test: Newton’s Second Law of Motion - Question 10

Force vector follows the principle of superposition which says all the force vectors can be vectorially added if applied on one point to get the net force vector. Hence we get
F = F1 + F2 + F3 
= (-2 + 3) i + (1 - 2) j
 F = i - j = ma
Thus we get a = (i - j) /m
= (i - j) / 2

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