JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE

JEE: JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE

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Q.1. A block of mass m slides along a floor while a force of magnitude θ is applied to it at an angle F as shown in figure. The coefficient of kinetic friction is μκ. Then, the block's acceleration 'a' is given by : (g is acceleration due to gravity)  [2021 6 March (Shift 1)]
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(a) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(b) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(c) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(d) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE

Correct Answer is Option (b)
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
N = mg − f sin θ
F cos θ − μkN = ma
F cos θ − μk(mg − F sin θ) = ma
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.2. A block of 200g mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius 20cm. If the block takes 40s to complete one round, the normal force by the side walls of the groove is : [2021 16 March (Shift 1)]
(a) 0.0314 N
(b) 9.859 × 10−2 N
(c) 6.28 × 10−3 N
(d) 9.859 × 10−4 N

Correct Answer is Option (d)
N = mω2R
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
Given m = 0.2 kg, T = 40 S, R = 0.2 m
Put values in equation (1)
N = 9.859 × 10−4 N


Q.3. Consider a frame that is made up of two thin massless rods A and B as shown in the figure. A vertical force JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE of magnitude 100N is applied at point A of the frame.
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
Suppose the force is JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE resolved parallel to the arms AB and AC of the frame. The magnitude of the resolved component along the arm AC is xN. The value of x, to the nearest integer, is ____ .

[Given: sin(35º) = 0.573, cos(35º) = 0.819
sin(110º) = 0.939, cos(110º) = -0.342]     
[2021 16 March (Shift 1)]

JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
Component along AC
= 100 cos 35N
= 100 × 0.819 N
= 81.9 N
≈ 82 N


Q.4. Statement I: A cyclist is moving on an unbanked road with a speed of 7kmh-1 and takes a sharp circular turn along a path of radius of 2 m without reducing the speed. The static friction coefficient is 0.2 . The cyclist will not slip and pass the curve (g = 9.8 m/s2)
Statement II: If the road is banked at an angle of 45°, cyclist can cross the curve of 2 m radius with the speed of 18.5kmh-1 without slipping.
In the light of the above statements, choose the correct answer from the options given below. [2021 16 March (Shift 2)]
(a) Statement I is incorrect and statement II is correct
(b) Statement I is correct and statement II is incorrect
(c) Both statement I and statement II are false
(d) Both statement I and statement II are true

Correct Answer is Option (d)
Statement I:
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
Vmax = 1-97 m/s
7 km/h = 1.944 m/s
Speed is lower than vmax, hence it can take safe turn.
Statement II:
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
18.5 km/h = 5.14 m/s
Speed is lower than vmax, hence it can take safe turn.


Q.5. A body of mass 2 kg moves under a force of JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE. It starts from rest and was at the origin initially. After 4 s, its new coordinates are (8, b, 20). The value of b is_____. (Round off to the Nearest Integer)  [2021 16 March (Shift 2)]

JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.6. A modem grand-prix racing car of mass m is travelling on a flat track in a circular arc of radius R with a speed v. If the coefficient of static friction between the tyres and the track is μs, then the magnitude of negative lift FL acting downwards on the car is : (Assume forces on the four tyres are identical and g = acceleration due to gravity) [2021 17 March (Shift 1)]
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE

(a) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(b) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(c) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(d) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE

Correct Answer is Option (b)
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.7. Two blocks (m = 0.5 kg and M = 4.5 kg) are arranged on a horizontal frictionless table as shown in figure. The coefficient of static friction between the two blocks is 3/7. Then the maximum horizontal force that can be applied on the larger block so that the blocks move together is_____N. (Round off to the Nearestlnteger)
[Take g as 9.8 ms-2 ]
   [2021 17 March (Shift 1)]

JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.8. A body of mass 1 kg rests on a horizontal floor with which it has a coefficient of static friction 1/√3. It is desired to make the body move by applying the minimum possible force FN. The value of F will be ______(Round off to the Nearest Integer)
[Take g = 10 ms-2]  
[2021 17 March (Shift 2)]

JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.9. A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction between the wood and the floor is 0.5, the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is N.(Round off to the Nearest Integer) [Take g = 10 ms-2]  [2021 17 March (Shift 2)]
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE

JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
N + T = 90
T = μN = 0.5(90 − T)
1.5 T = 45
T = 30


Q.10. A bullet of mass 0.1 kg is fired on a wooden block to pierce through it, but it stops after moving a distance of 50 cm into it. If the velocity of bullet before hitting the wood is 10 m/s and it slows down with uniform deceleration, then the magnitude of effective retarding force on the bullet is 'x' N. The value of x to the nearest integer is ___. [2021 18 March (Shift 1)]

v2 = u2 + 2as
0 = (10)+ 2(−a) (1/2)
a = 100 m/s2
F = ma = (0.1)(100) = 10 N


Q.11. A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to (1 HP = 746 W, g = 10 m/s2 )    [2020]
(a) 1.7 m/s
(b) 1.9 m/s
(c) 1.5 m/s
(d) 2.0 m/s

Correct Answer is Option (b)

Given that M = 2000 kg, g = 10 m/s2, f = 4000 N, P = 60 HP = 60 × 746 J
Total force acting on elevator is:
F = Mg + f
= (2000 x 10) + 4000 = 24000 N
Power of electric motor is given by
P = Fv
⇒ v = P/F
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE 


Q.12. A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of 45° with the vertical. Then F equals (Take g = 10 m/s2 and the rope to be massless)    [2020]
(a) 100 N
(b) 90 N
(c) 70 N
(d) 75 N

Correct Answer is Option (a)

Let tension in the string be T
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
At equilibrium, we have
T sin 450 = F     ..(1)
and    T cos450 = 10 x g    ..(2)
Dividing Eq. (1) by Eq. (2), we get
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE     (∵ g = 10 m/s)
⇒ I = F/100
⇒ F = 100 N


Q.13. Consider a uniform cubical box of side 'a' on a rough floor that is to be moved by applying minimum possible force F at a point 'b' above its centre of mass (see figure). If the coefficient of friction is μ = 0.4, the maximum possible value of 100.b/a for box not to topple before moving is ______.      [2020]

JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
Given that μ = 0.4
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE

Condition for no topple is
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
Friction force is given by
F = μmg
So,
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
So, maximum value of JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
Therefore,
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
So, maximum possible value of 100 × b/a = 75.


Q.14. A spring-mass system (mass m, spring constant k and natural length l) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring-mass system, rotates about its axis with an angular velocity ω, ( k >> mω2) the relative change in the length of the spring is best given by the option     [2020]
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE

Correct Answer is Option (3)

Let the elongation in length of the spring be Δl.
Forces act on the system are:
(i) Centripetal force, which act outward which help in expansion in spring:
Fc = mω2 (l + Δl)
(ii) Spring restoring force, which act inward which opposed the further expansion in spring.
Fs = kΔl
At balanced point, we have:
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
The relative change in the length of the spring is:
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
Since k>> mωthen
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.15. A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force P such that the block does not move downward? (Take g = 10 m/s2)    [2019]
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE

(a) 32 N
(b) 18 N
(c) 23 N
(d) 25 N

Correct Answer is Option (a)

The coefficient of static friction, μ = 0.6
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
The resultant force is
P + μmg cos θ = 3 + mg sin θ
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
Given m = 10 kg, μ = 0.6 and g = 10 m/s2
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
⇒ P + 0.6 × 50√2 = 3 + 50√2
⇒ P + 42.42 = 73.71
⇒ P = 73.71 - 42.42 = 31.29
Therefore, P ≈ 32 N


Q.16. A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 m/s2).     [2019]
(a) 200 N
(b) 140 N

(c) 70 N
(d) 100 N

Correct Answer is Option (d)
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.17. A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here, k is a constant. The value of T is    [2019]
(a) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(b) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(c) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(d) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE 

Correct Answer is Option (b)
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.18. A block kept on a rough inclined plane, as shown in the figure, remains at rest up to a maximum force 2 N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10 N. The coefficient of static friction between the block and the plane is (Take g = 10 m/s2)    [2019]
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(a) √3/2
(b) √3/4
(c) 1/2
(d) 2/3

Correct Answer is Option (a)

Block is at rest then, 2 N force is applied.
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
Now, friction is
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
When 2 N force is applied in downward direction, thus
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
Block is rest (10 N force is applied)
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
When 10 N force is applied in upward direction, then
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
Dividing Eq. (1) by Eq. (2), we get
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.19. Two masses m1 = 5 kg and m2 = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is:    [2018]
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(a) 18.3 g 

(b) 23.3 kg
(c) 43.3 kg
(d) 10.3 kg

Correct Answer is Option (b)
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.20. A given object takes n times more time to slide down a 45° rough inclined plane as it takes to slide down a perfectly smooth 45° incline. The coefficient of kinetic friction between the object and the incline Is: [2018]
(a) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(b) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(c) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(d) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE

Correct Answer is Option (a)

JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
So
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
Sin θ = n2 sinθ − µn2 cosθ
⇒ 1 = n2 − µn2
⇒ 1 − n2 = −µn2
⇒ n2 − 1 = −µn2
⇒ µ = 1 − 1/n2


Q.21. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains the same, the stress in the leg will change by a factor of    [2017]
(a) 81
(b) 1/81
(c) 9
(d) 1/9

Correct Answer is Option (c)
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
∵ Density remains same
So, mass ∝ Volume
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.22. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?   [2017]
(a)
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(b)
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(c)
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(d)
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE

Correct Answer is Option (a)

Acceleration is constant and negative
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.23. A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is:    [2017]
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(a) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(b) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(c) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(d) JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE

Correct Answer is Option (d)
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.24. A conical pendulum of length 1 m makes an angle q = 45° w.r.t. Z-axis and moves in a circle in the XY plane. The radius of the circle is 0.4 m and its center is vertically below O. The speed of the pendulum, in its circular path, will be (Take g = 10 ms-2)    [2017]
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(a) 0.2 m/s
(b) 0.4 m/s
(c) 2 m/s
(d) 4 m/s

Correct Answer is Option (c)
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.25. The machine as shown has 2 rods of length 1 m connected by a pivot at the top. The end of one rod is connected to the floor by a stationary pivot and the end of the other rod has a roller that rolls along the floor in a slot. As the roller goes back and forth, a 2 kg weight moves up and down. If the roller is moving towards right at a constant speed, the weight moves up with a    [2017]
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(a) Speed which is 3/4th of that of the roller when the weight is 0.4 m above the ground
(b) Constant speed
(c) Decreasing speed
(d) Increasing speed

Correct Answer is Option (c)

We know that the velocity is the rate of change in displacement, s with respect to time, t. Here from the diagram we can see that the weight moves up by covering the distance y in time t towards the point C.
Let Vc be the speed (velocity) of the weight moving up towards the point C.
Then, rate of change in y with respect to time is the speed of weight moving up
⇒ dy / dt =  Vc → 1
Vc is the velocity (speed) of the moving up toward point C
Let VA be the velocity (speed) with which the roller moves right, then
⇒ dx / dt =  VA → 2
VA is the velocity (speed) of the moving right
We can understand clearly if we see the diagram
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
From the diagram,
Let us take the triangle AOC, then
⇒ sinθ = opposite side / hypotenuse
⇒ sinθ = y / 1
⇒ sinθ = y
Differentiating above equation,
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
From equation 2 we get
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
Then,
We know that,
⇒ cosθ = adjacent side / hypotenuse
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
⇒ cosθ = x / 2
Differentiating
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
From equation 1 we get
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
From the equation 3 and 4 we get
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
It is given that the roller is moving right, if the roller moves rightwards then the angle θ increases, if θ increases then VC will decrease (from the above relation)
Vc is the speed (velocity) of the weight moving up towards the point C.
So, the speed decreases when the weight moves upward.
Hence the correct answer is option (D) decreasing speed
Note : We are saying that if θ increases then VC will decrease from the relation we got
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
We can see that VC is directly proportional to VA and inversely proportional to θ, so if the θ increases then VC will decrease.


Q.26. A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals m. The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR.
The values of the coefficient of friction m and the distance x (= QR), are, respectively close to [2016]
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
(a) 0.2 and 6.5 m
(b) 0.2 and 3.5 m
(c) 0.29 and 3.5 m
(d) 0.29 and 6.5 m

Correct Answer is Option (c)

 From work energy theorem and given condition:
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.27. A rocket is fired vertically from the earth with an acceleration of 2g, where g is the gravitational acceleration. On an inclined plane inside the rocket, making an angle q. With the horizontal, a point object of mass m is kept. The minimum coefficient of friction μmin between the mass and the inclined surface such that the mass does not move is:    [2016]
(a) tanθ
(b) tan2θ
(c) 3 tanθ
(d) 2 tanθ

Correct Answer is Option (a)

Free body diagram is shown in the figure.
Psuedo force on the object is 2mg in the downward direction as seen from the rocket frame of reference.
Normal force in the direction perpendicular to the inclined plane.
N = 3mg cos(θ)
Maximum frictional force in the direction along the plane,
Fr = μmN cos(θ)
∴ Fr = 3μmmg cos(θ)
This should balance the force on the block along the downward direction of plane.
3mg sin(θ) = 3μmmg cos(θ)
μm = tan(θ)
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.28. Which of the following option correctly describes the variation of the speed v and acceleration 'a' of a point mass falling vertically in a viscous medium that applies a force F = –kv, where 'k' is constant, on the body ?( Graphs are schematic and not drawn to scale)    [2016]
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE 
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE

Correct Answer is Option (4)
a = g – αv
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE


Q.29. Concrete mixture is made by mixing cement, stone and sand in a rotating cylindrical drum. If the drum rotates too fast, the ingredients remain stuck to the wall of the drum and proper mixing of ingredients does not take place. The maximum rotational speed of the drum in revolutions per minute (rpm) to ensure proper mixing is close to (Take the radius of the drum to be 1.25m and its axle to be horizontal):    [2016]
(a) 1.3
(b) 0.4
(c) 27.0
(d) 8.0

Correct Answer is Option (c)
For just complete rotation:
JEE Main Previous year questions (2016-21): Laws of Motion Notes | Study Physics 35 Years JEE Main & Advanced Past year Papers - JEE

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