It is very important to have the idea of any examination to get success in that. The best way to have that idea is to have a look at previous year Questions of that examination. That's why we have collected some previous year Questions with Solutions of JEE Main, JEE Advance, NEET, IIT JEE, AIEEE, AIIMS, WBJEE, EAMCET, Karnataka CET, CPMT, Kerala CET, MP PMT and Other State JE Exams so that you can get clear pattern and level of question that will be asked in forthcoming JEE Main, JEE Advance, IIT JEE, AIEEE, AIIMS, WBJEE, EAMCET, Karnataka CET, CPMT, Kerala CET, MP PMT and Other State JE Exams.
JEE Previous Year's Questions With Solutions
1. A particle of mass 0.3 kg is
subjected to a force F = – kx with k = 15 N/m. What will be its initial acceleration
if it is released from a point 20 cm away from the origin? (AIEEE 2005)
(A) 3 m/s2
(B) 5 m/s2
(C) 10 m/s2
(D) 15 m/s2.
Sol: (C)
Given: mass, m = 0.3 kg; k = 15 N/m; x = 20 cm = 0.2 m.
We have, force F = – kx
Or, F = 15 * 0.2 = 3 N.
And, acceleration = F/m = 3/0.3 = 10 m/s2.
2. A block is kept on a frictionless
inclined surface with angle of inclination α. The incline is given an acceleration
‘a’ to keep the block stationary. Then ‘a’ is equal to: (AIEEE 2005)
(A) g tan α
(B) g/tan α
(C) g cosec α
(D) g.
Sol: (A)
The equation of motion of the block along the inclination (a
= 0):
→ ma cos α – mg sin α = ma
Or, ma cos α – mg sin α = 0
Or, a = g tan α.
3. A player caught a cricket ball of
mass 150 g moving at a rate of 20 m/s. If the catching process is completed in
0.1 s, the force of the blow exerted by the ball on the hand of the player is
equal to: (AIEEE 2006)
(A) 3 N
(B) 150 N
(C) 30 N
(D) 300 N.
Sol: (C)
Given: mass, m = 150 g = 0.15 kg; initial velocity, u = 20
m/s; final velocity, v = 0; time taken, t = 0.1 s.
Acceleration, a = (v – u)/t = – 200 m/s2
So, force exerted by the ball on the hand of the player is =
ma = 0.15 * 200 = 30 N.
4. A ball of mass 0.2 kg is thrown
vertically upwards by applying a force by hand. If the hand moves 0.2 m which
applying the force and the ball goes upto 2 m height further, find the
magnitude of the force. Consider g = 10 m/s2: (AIEEE 2006)
(A) 4 N
(B) 16 N
(C) 22 N
(D) 20 N.
Sol: (D)
Given: S = 0.2 m; mass, m = 0.2 kg; height, h = 2 m.
→Work done by hand = Potential energy of the ball
Or, FS = mgh
Or, F = mgh/S = (0.2 * 10 * 2)/0.2 = 20 N.
5. A block of mass m is connected to another
of mass M by a spring (massless) of spring constant k. The blocks are at rest
and the spring is unstretched. Then a constant force F starts acting on the
block of mass M to pull it. Find the force of the block of mass m: (AIEEE 2007)
(A) mF/M
(B) MF/ (M + m)
(C) Fm/ (M + m)
(D) F (M + m)/m.
Sol: (C)
Taking M and m are together as a system. External force
acting on the system is F.
Therefore, acceleration a = F/ (M + m).
So, force on the block of mass m = ma = Fm/ (M + m).
6. A 1 N pendulum bob is held at an
angle θ from the vertical by a 2 N horizontal force F as shown in the
figure. The tension in the string supporting the pendulum bob (in Newton) is: (EAMCET 2011)
(A) 1
(B) √5
(C) 2/cos θ
(D) Cos θ.
Sol: (B)
Given: F = 2 N; weight, mg = 1 N.
Let the tension in the string be T.
Force balance along vertical direction:
→ T cos θ = mg
Or, T cos θ = 1 ------- (1)
Force balance along horizontal direction:
→ T sin θ = F
Or, T sin θ = 2 ------- (2)
Solving equation (1) and (2)
We get, tan θ = 2
And sin θ = 2/√5, cos θ = 1/√5.
Therefore, T = √5.
7. The maximum tension a rope can
withstand is 60 kg wt. The ratio of maximum acceleration with which two boys of
masses 20 kg and 30 kg can climb up the rope at the same is: (EAMCET 2011)
(A) 3 : 2
(B) 4 : 3
(C) 2 : 1
(D) 1 : 2.
Sol: (C)
Given: maximum tension of rope, T = 60 kg-wt = 60g N.
The equation of motion of boy of mass 20 kg:
→ Ma1 = T – Mg
Or, 20a1 = 60g – 20g
Or, a1 = 2g.
The equation of motion of boy of mass 30 kg:
→ Ma2 = T – Mg
Or, 30a2 = 60g – 30g
Or, a2 = g.
So, the ratio of maximum acceleration is a1 : a2
= 2 : 1.
8. Choose the correct statement
(1) The position of centre of mass of a system is dependent
on the choice of co-ordinate system.
(2) Newton’s second law of motion is applicable to the centre
of mass of the system.
(3) When no external force acts on a body, the acceleration
of centre of mass is zero.
(4) Internal forces can change the state of centre of mass. (EAMCET 2012)
(A) Both (1) and (2) are correct
(B) Both (2) and (3) are wrong
(C) Both (1) and (3) are wrong
(D) Both (1) and (4) are wrong.
Sol: (D)
Statement (1) is wrong.
Statement (2) is correct.
Statement (3) is correct.
Statement (4) is wrong.
Therefore, option (D) is correct.
10. Two wooden blocks of masses M and m
are placed on a smooth horizontal surface as shown in figure. If a force P is
applied to the system as shown in figure such that the mass m remains
stationary with respect to block of mass M, then the magnitude of the force P
is: (EAMCET 2013)
(A) g tan β
(B) mg cos β
(C) (M + m) g cosec β
(D) (M + m) g tan β.
Sol: (C)
Acceleration of the system, a = P/ (M + m).
The equation of motion of the block along the inclination (a’
= 0):
→ ma cos β – mg sin β = ma’
Or, (Pm cos β)/ (M + m) – mg sin β = 0
Or, P = (M + m) g tan α.
11. A mass M kg is suspended by a
weightless string. The horizontal force required to hold the mass at 600
with the vertical is: (EAMCET 2014)
(A) Mg
(B) Mg √3
(C) Mg (√3 + 1)
(D) Mg/ √3.
Sol: (B)
Let the tension in the string be T and horizontal force be F.
Force balance along vertical direction:
→ T cos 600 = Mg
Or, T/2 = Mg ------- (1)
Force balance along horizontal direction:
→ T sin 600 = F
Or, T √3/2 = F ------- (2)
Solving equation (1) and (2)
We get, F = Mg √3.
12. A force (2i + j
– k)
N acts on a body which is initially at rest. At the end of 20 s the velocity of
the body is (4i + 2j – 2k) m/s, then the
mass of the body is: (EAMCET 2015)
(A) 4.5 kg
(B) 5 kg
(C) 8 kg
(D) 10 kg.
Sol: (D)
Magnitude of force, F = √ [22 + 12 + (–
1)2] = √6 N.
Magnitude of final velocity, v = √ [42 + 22
+ (– 2)2] = √24 m/s.
Initial velocity, u = 0; time taken, t = 20 s.
Acceleration, a = (v – u)/t = √24/20.
We know, force = mass * acceleration
Or, √6 = mass * √24/20
Or, mass = 10 kg.
13. A cricket ball of mass 0.25 kg with
speed 10 m/s collides with a bat and returns with same speed within 0.01 s. The
force acted on bat is: (WBJEE 2011)
(A) 500 N
(B) 50 N
(C) 250 N
(D) 25 N.
Sol: (A)
Given: mass, m = 0.25 kg; time, t = 0.01 s; change in
velocity, ΔV = 10 – (–10) = 20 m/s.
Acceleration, a = ΔV/t = 20/0.01 = 2000 m/s2.
Force, F = ma = 0.25 * 2000 = 500 N.
14. Three blocks of mass 4 kg, 2 kg, 1
kg respectively are in contact on a frictionless table as shown in the figure.
If a force of 14 N is applied on the 4 kg block, the contact force between the 4
kg and the 2 kg block will be: (WBJEE
2012)
(A) 2 N
(B) 6 N
(C) 8 N
(D) 14 N.
Sol: (A)
Taking three blocks as a system (mass, M = 4 + 2 + 1 = 7 kg);
force, F = 14 N:
Acceleration, a = F/M = 14/7 = 2 m/s2.
From the F.B.D. of 4 kg block:
→ 14 – F4,2 = ma
Or, F4,2 = 14 – 4 * 2
Or, F4,2 = 6 N.
15. A mass of 1 kg is suspended by means
of a thread. The system is (i) lifted up with an acceleration of 4.9 m/s2,
(ii) lowered with an acceleration of 4.9 m/s2. The ratio of tension
in the first and second case is: (WBJEE
2016)
(A) 3:1
(B) 1:2
(C) 1:3
(D) 2:1
Sol: (A)
(i) The equation of motion of the block when lifted up with
an acceleration of 4.9 m/s2:
→ Ma = T1 – Mg
Or, 1 * 4.9 = T1 – 1 * 9.8
Or, T1 = 14.7 N.
(ii) The equation of motion of the block when lowered with an
acceleration of 4.9 m/s2:
→ Ma = Mg – T2
Or, 1 * 4.9 = 1 * 9.8 – T2
Or, T2 = 4.9 N.
So, the ratio of tension in the first and second case is 3:1.
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