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# Test: Force & Newton's Laws of Motion - 1

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## 30 Questions MCQ Test Physics Class 11 | Test: Force & Newton's Laws of Motion - 1

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Test: Force & Newton's Laws of Motion - 1 - Question 1

### A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. If a force of 2.8 N is applied on the block parallel to floor, the force of friction between the block and floor is (Take g = 10 m/s2)

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 1

Weight of the block is 20N thus maximum friction that can act is 0.4 x 20 = 8N
But as the external force acting, F = 2.8N < 8N,
We get f = F = 2.8N

Test: Force & Newton's Laws of Motion - 1 - Question 2

### A block has been placed on an inclined plane. The slope angle θ of the plane is such that the block slides down the plane at a constant speed. The coefficient of kinetic friction is equal to

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 2 At a constant speed means, acceleration is 0. Therefore, using formula
a = g( sinθ - ucosθ)
0 = g(sinθ - u cosθ)
ucosθ = sinθ
u = sinθ/cosθ
u= tanθ
Hence D is correct.

Test: Force & Newton's Laws of Motion - 1 - Question 3

### A body of mass 60 kg is dragged with just enough force to start moving on a rough surface with coefficients of static and kinetic frictions 0.5 and 0.4 respectively. On applying the same force, what is the acceleration?

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 3 Test: Force & Newton's Laws of Motion - 1 - Question 4

`A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g = 9.8 m/s2 the resulting acceleration of the slab will be Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 4

Block will slip on the slab as F > flim Hence kinetic friction acts between block and slab, Test: Force & Newton's Laws of Motion - 1 - Question 5

If the coefficient of friction of a surface is μ = √3, then the angle of inclination of a plane to make a block on it to just start sliding is

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 5 Test: Force & Newton's Laws of Motion - 1 - Question 6

If A and B moves with acceleration ‘a’ as shown in diagram, calculate acceleration of C with respect to B Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 6  Test: Force & Newton's Laws of Motion - 1 - Question 7

A block takes twice as much time to slide down a 45o rough inclined plane as it takes to slide down a similar smooth plane. The coefficient of friction is

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 7  Test: Force & Newton's Laws of Motion - 1 - Question 8

The acceleration that is produced by a 15N force in a mass of 8 kg will be equal to

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 8

As we know that F = ma
20 = 8a
=> 20/8 = a
a = 1.87 ms-2

Test: Force & Newton's Laws of Motion - 1 - Question 9

A block of mass 6 kg is being pulled by force 24 N as shown. If coefficient of friction between block and the surface is 0.6, frictional force acting on the block is Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 9 Test: Force & Newton's Laws of Motion - 1 - Question 10

In figure, coefficient of friction between m1 and m2 is m and that between m1 and the wall is zero. A force F is pressing the system against the wall. Minimum value of force required to hold the system in equilibrium is Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 10

There is no force which can balance the weight of m1.

Test: Force & Newton's Laws of Motion - 1 - Question 11

A body moves along a circular path of radius 5 m. The coefficient of friction between the surface of the path and the body is 0.5. The angular velocity in rad/s with which the body should move so that it does not leave the path is (g = 10 m/s2)

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 11 Test: Force & Newton's Laws of Motion - 1 - Question 12

In figure, coefficient of friction between m2 and the horizontal surface is μ, minimum value of m so that the system has zero acceleration, is Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 12

For zero acceleration,

Putting force = Frictional force   m1g = μ(m2 + m)g

Test: Force & Newton's Laws of Motion - 1 - Question 13

A uniform rope of length l lies on a table. If the coefficient of friction is μ, then the maximum length l1 of the part of this rope which can overhang from the edge of the table without sliding down is

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 13 Test: Force & Newton's Laws of Motion - 1 - Question 14

A man of mass m stands on a platform of equal mass m and pulls himself by two ropes passing over pulleys as shown in figure. If he pulls each rope with a force equal to half his weight, his upward acceleration would be Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 14 Test: Force & Newton's Laws of Motion - 1 - Question 15

The car A is used to pull a load B with the pulley arrangement shown. If A has a forward velocity v determine an expression for the upward velocity vB of the load in terms of vA and θ.θ is angle between  string and horizontal Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 15

Let θ angle between string and horizontal, T is tension in string  Test: Force & Newton's Laws of Motion - 1 - Question 16

A man slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional force between man and poles is equal to (in terms of man’s weight w)

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 16 Test: Force & Newton's Laws of Motion - 1 - Question 17

A cart of mass M has a block of mass m attached to its as shown in the figure. The coefficient of friction between the block and cart is μ. The minimum acceleration of the cart so that the block m does  not fall is Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 17

Let the acceleration of the cart be a, then normal reaction on cart is N = ma and frictional force between
block and cart is f = μN = μma.

For equilibrium, μma = mg ⇒ a = g/μ

Test: Force & Newton's Laws of Motion - 1 - Question 18

A block is placed on a rough surface and an external horizontal force F is applied on it. The force of friction ‘f’ exerted by the floor on the block is measured for different values of F. The graph between F and f is

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 18

Theoretical

Test: Force & Newton's Laws of Motion - 1 - Question 19

If the coefficient of friction between an insect and bowl surface is μ and the radius of the bowl is r, the maximum height to which the insect can crawl in the bowl is

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 19   Test: Force & Newton's Laws of Motion - 1 - Question 20

A block rests on a rough inclined plane making an angle of 30o with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (Take g = 10 m/s2)

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 20

mg sin θ = 10
m x 10 x (1/2) = 10
m = 2 kg

Test: Force & Newton's Laws of Motion - 1 - Question 21

Pulleys and strings are massless. The horizontal surface is smooth. What is the acceleration of the block Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 21

T = ma,  F = 2T

Test: Force & Newton's Laws of Motion - 1 - Question 22

A ball of mass 0.5 kg is attached to the end of a string having length 0.5 m. The hall is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball in (rad/s)

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 22

T sin θ = mrω2;  r = l sin θ

Test: Force & Newton's Laws of Motion - 1 - Question 23

In the pulley-block arrangement shown in figure, find the relation between acceleration of block A and B Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 23

x1 + x3 = l

Differentiating with respect to time,

we get,  v1 + v3 = 0

Again differentiating with respect to time,  Differentiating with respect to time,  v1 + v2 – 2v3 = 0

Again differentiating with respect to time, a1 + a2 – 2a3 = 0 Test: Force & Newton's Laws of Motion - 1 - Question 24

Frictional force between two bodies

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 24

Friction does not always oppose motion. For eg when a car is accelerating forward , the tyres push against the road in a backward direction. Friction pushes itself in forward and the car also pushes forward. Thus friction and motion are in the same direction.
Similarly when a runner accelerates forward the shoe pushes backward, friction pushes forward and runner pushes forward. Thus both are in same direction
Hence force of friction does not always oppose motion or the tendency of motion.

Test: Force & Newton's Laws of Motion - 1 - Question 25

The limiting friction between two surface does not depend

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 25

Conceptual

Test: Force & Newton's Laws of Motion - 1 - Question 26

A body is sliding down a rough inclined plane of inclination 60o. If the angle of repose is 30o, the acceleration of the body (g = 10 m/s2) in m/s2 is

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 26

Angle of repose is the maximum angle of inclination when the downward force on the block is exactly balance by the frictional force.

i.e. F = μR = μmgcosθ = mgsinθ, where, θ is the angle of repose.

Solving this we will get:

μ = tan300 = 1/(3)1/2.

Now when it will start moving then the acceleration of the box will be:

a = g(sinα−μcosα)

where α=60o

∴  a=10(0.866 − 1/2(3)½]=5.77 m/s2

Test: Force & Newton's Laws of Motion - 1 - Question 27

A cyclist of mass 30 kg exerts a force of 250 N to move his cycle. acceleration is 4 ms−2. force of friction between road and tyres will be

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 27

F = ma

so a = 250/30

a = 8.34

now the result is 4mt/secsqure

so reduction is 4.34

friction force will b 4.34 ×30

F = 130.2 N

Test: Force & Newton's Laws of Motion - 1 - Question 28

A conveyor belt is moving at a constant speed of 2 m/s. A box is gently dropped on it. The coefficient of friction between them is m  = 0.5. The distance that the box will move relative to belt before coming to rest on it, taking g = 10 m/s2, is

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 28  Test: Force & Newton's Laws of Motion - 1 - Question 29

A block of weight 100 N is pushed by a force F on a horizontal rough plane moves with an acceleration 1 m/s2, when force is doubled its acceleration becomes 10 m/s2. The coefficient of friction is  (g = 10 m/s2)

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 29 Test: Force & Newton's Laws of Motion - 1 - Question 30

What is the smallest radius of a circle at which a bicyclist can travel if his speed is 7 m/s and the coefficient of static friction between tyres and road is 0.25

Detailed Solution for Test: Force & Newton's Laws of Motion - 1 - Question 30

Centripetal force = Frictional force
⇒ mv2/r​=μmg
∴r=v2​/μg

∴r=72​/0.25×10
∴r=19.6 m≈20 m

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