JEE Main Joint Entrance Examination - Main — 120 Questions
and Answers Already Graded A+ Premium Exam Tested And
Verified
Subject Area Physics, Chemistry, and Mathematics
Description This exam assesses foundational knowledge and problem-solving skills in
physics, chemistry, and mathematics at a level suitable for engineering aspirants.
It covers topics from the JEE Main syllabus, including mechanics,
electromagnetism, organic chemistry, calculus, and algebra.
Expected Grade A+
Total Questions 50
Duration 3 hours
Learning Outcomes 1. Apply Newton's laws and conservation principles to solve complex mechanics
problems.
2. Analyze electromagnetic phenomena and circuit behavior using Maxwell's
equations and circuit laws.
3. Interpret organic reaction mechanisms and predict products under given
conditions.
4. Solve advanced problems in calculus, including limits, integrals, and
differential equations.
5. Demonstrate proficiency in algebra, including complex numbers and matrices.
Accreditation This exam adheres to the standards of the Joint Entrance Examination (JEE)
Main, recognized by the Indian Institutes of Technology and National Institutes of
Technology.
Page 1
,1. A particle moves along a circular path of radius R with a tangential acceleration
a_t = kt, where k is a constant. At t=0, the particle starts from rest. The magnitude of
the net acceleration at time t is:
A. "(k²t² + (k²t t)/(R²))
B. "(k²t² + (k²t t)/(4R²))
C. "(k²t² + (k²t v)/(9R²))
D. "(k²t² + (k²t v)/(R²))
Answer: B. "(k²t² + (k²t t)/(4R²))
Tangential acceleration a_t = kt, so tangential velocity v = "+ kt dt = (1/2)kt². Centripetal
acceleration a_c = v²/R = (k²t t)/(4R). Net acceleration = "(a_t² + a_c²) = "(k²t² +
k²t t/(4R²)).
2. A uniform rod of mass M and length L is pivoted at one end. It is released from
rest in a horizontal position. The angular acceleration of the rod immediately after
release is:
A. 3g/(2L)
B. 2g/(3L)
C. g/L
D. 3g/L
Answer: A. 3g/(2L)
The torque about the pivot due to gravity is Ä = Mg(L/2). Moment of inertia about the
end is I = (1/3)ML². Angular acceleration ± = Ä/I = (MgL/2) / (ML²/3) = (3g)/(2L).
3. A charged particle with charge q and mass m enters a uniform magnetic field B
with velocity v perpendicular to the field. The particle moves in a circular path of
radius R. If the velocity is doubled, the new radius is:
A. R/2
B. R
C. 2R
D. 4R
Answer: C. 2R
The centripetal force is provided by the magnetic force: qvB = mv²/R, so R = mv/(qB).
If v is doubled, R becomes 2R.
Page 2
,4. In the circuit shown, the switch S is closed at t=0. The capacitor C is initially
uncharged. The time constant of the circuit for charging is:
A. RC
B. 2RC
C. RC/2
D. RC/"2
Answer: B. 2RC
The circuit consists of two resistors R in series with the capacitor, so total resistance =
2R. Time constant Ä = (2R)C = 2RC.
5. The product of the reaction between 2-methylpropene and HBr in the presence of
peroxide is:
A. 1-bromo-2-methylpropane
B. 2-bromo-2-methylpropane
C. 1,2-dibromo-2-methylpropane
D. 2-bromo-1-methylpropane
Answer: A. 1-bromo-2-methylpropane
In the presence of peroxide, HBr adds anti-Markovnikov due to free radical
mechanism. The bromine attaches to the less substituted carbon, giving
1-bromo-2-methylpropane.
6. The number of stereoisomers possible for 2,3-dichlorobutane is:
A. 2
B. 3
C. 4
D. 6
Answer: B. 3
2,3-dichlorobutane has two chiral centers, but due to meso compound (when the two
chiral centers have opposite configurations and the molecule has a plane of symmetry),
there are 3 stereoisomers: (R,R), (S,S), and (R,S) meso.
Page 3
, 7. The equilibrium constant Kp for the reaction N ‚(g) + 3H ‚(g) !Ì 2NH ƒ(g) is 4.0 ×
10 { t at 400°C. If the total pressure at equilibrium is 10 atm and the mole fraction of
N ‚ is 0.2, the mole fraction of NH ƒ is:
A. 0.1
B. 0.2
C. 0.4
D. 0.6
Answer: A. 0.1
Let x be mole fraction of NH ƒ. Then mole fractions: N ‚ = 0.2, H ‚ = 1 - 0.2 - x = 0.8 - x.
Partial pressures: p_NH ƒ = 10x, p_N ‚ = 2, p_H ‚ = 10(0.8-x) = 8-10x. Kp =
(p_NH ƒ)²/(p_N ‚ * (p_H ‚)³) = (100x²)/(2*(8-10x)³) = 4×10 { t. Solving gives x "H 0.1.
8. If the function f(x) = x³ + ax² + bx + c has a local maximum at x = -1 and a local
minimum at x = 3, then the values of a and b are:
A. a = -3, b = -9
B. a = 3, b = -9
C. a = -3, b = 9
D. a = 3, b = 9
Answer: A. a = -3, b = -9
f'(x) = 3x² + 2ax + b. f'(-1)=0 => 3 - 2a + b = 0; f'(3)=0 => 27 + 6a + b = 0. Solving gives
a = -3, b = -9.
9. The integral "+ €^À (x sin x) / (1 + cos² x) dx equals:
A. À²/4
B. À²/2
C. À/2
D. À
Answer: A. À²/4
Use property "+ €^a f(x) dx = "+ €^a f(a-x) dx. Let I = "+ €^À (x sin x)/(1+cos² x) dx. Then I
= "+ €^À ((À-x) sin x)/(1+cos² x) dx. Adding, 2I = À "+ €^À sin x/(1+cos² x) dx = À [
-arctan(cos x) ] €^À = À(À/2) = À²/2, so I = À²/4.
Page 4
and Answers Already Graded A+ Premium Exam Tested And
Verified
Subject Area Physics, Chemistry, and Mathematics
Description This exam assesses foundational knowledge and problem-solving skills in
physics, chemistry, and mathematics at a level suitable for engineering aspirants.
It covers topics from the JEE Main syllabus, including mechanics,
electromagnetism, organic chemistry, calculus, and algebra.
Expected Grade A+
Total Questions 50
Duration 3 hours
Learning Outcomes 1. Apply Newton's laws and conservation principles to solve complex mechanics
problems.
2. Analyze electromagnetic phenomena and circuit behavior using Maxwell's
equations and circuit laws.
3. Interpret organic reaction mechanisms and predict products under given
conditions.
4. Solve advanced problems in calculus, including limits, integrals, and
differential equations.
5. Demonstrate proficiency in algebra, including complex numbers and matrices.
Accreditation This exam adheres to the standards of the Joint Entrance Examination (JEE)
Main, recognized by the Indian Institutes of Technology and National Institutes of
Technology.
Page 1
,1. A particle moves along a circular path of radius R with a tangential acceleration
a_t = kt, where k is a constant. At t=0, the particle starts from rest. The magnitude of
the net acceleration at time t is:
A. "(k²t² + (k²t t)/(R²))
B. "(k²t² + (k²t t)/(4R²))
C. "(k²t² + (k²t v)/(9R²))
D. "(k²t² + (k²t v)/(R²))
Answer: B. "(k²t² + (k²t t)/(4R²))
Tangential acceleration a_t = kt, so tangential velocity v = "+ kt dt = (1/2)kt². Centripetal
acceleration a_c = v²/R = (k²t t)/(4R). Net acceleration = "(a_t² + a_c²) = "(k²t² +
k²t t/(4R²)).
2. A uniform rod of mass M and length L is pivoted at one end. It is released from
rest in a horizontal position. The angular acceleration of the rod immediately after
release is:
A. 3g/(2L)
B. 2g/(3L)
C. g/L
D. 3g/L
Answer: A. 3g/(2L)
The torque about the pivot due to gravity is Ä = Mg(L/2). Moment of inertia about the
end is I = (1/3)ML². Angular acceleration ± = Ä/I = (MgL/2) / (ML²/3) = (3g)/(2L).
3. A charged particle with charge q and mass m enters a uniform magnetic field B
with velocity v perpendicular to the field. The particle moves in a circular path of
radius R. If the velocity is doubled, the new radius is:
A. R/2
B. R
C. 2R
D. 4R
Answer: C. 2R
The centripetal force is provided by the magnetic force: qvB = mv²/R, so R = mv/(qB).
If v is doubled, R becomes 2R.
Page 2
,4. In the circuit shown, the switch S is closed at t=0. The capacitor C is initially
uncharged. The time constant of the circuit for charging is:
A. RC
B. 2RC
C. RC/2
D. RC/"2
Answer: B. 2RC
The circuit consists of two resistors R in series with the capacitor, so total resistance =
2R. Time constant Ä = (2R)C = 2RC.
5. The product of the reaction between 2-methylpropene and HBr in the presence of
peroxide is:
A. 1-bromo-2-methylpropane
B. 2-bromo-2-methylpropane
C. 1,2-dibromo-2-methylpropane
D. 2-bromo-1-methylpropane
Answer: A. 1-bromo-2-methylpropane
In the presence of peroxide, HBr adds anti-Markovnikov due to free radical
mechanism. The bromine attaches to the less substituted carbon, giving
1-bromo-2-methylpropane.
6. The number of stereoisomers possible for 2,3-dichlorobutane is:
A. 2
B. 3
C. 4
D. 6
Answer: B. 3
2,3-dichlorobutane has two chiral centers, but due to meso compound (when the two
chiral centers have opposite configurations and the molecule has a plane of symmetry),
there are 3 stereoisomers: (R,R), (S,S), and (R,S) meso.
Page 3
, 7. The equilibrium constant Kp for the reaction N ‚(g) + 3H ‚(g) !Ì 2NH ƒ(g) is 4.0 ×
10 { t at 400°C. If the total pressure at equilibrium is 10 atm and the mole fraction of
N ‚ is 0.2, the mole fraction of NH ƒ is:
A. 0.1
B. 0.2
C. 0.4
D. 0.6
Answer: A. 0.1
Let x be mole fraction of NH ƒ. Then mole fractions: N ‚ = 0.2, H ‚ = 1 - 0.2 - x = 0.8 - x.
Partial pressures: p_NH ƒ = 10x, p_N ‚ = 2, p_H ‚ = 10(0.8-x) = 8-10x. Kp =
(p_NH ƒ)²/(p_N ‚ * (p_H ‚)³) = (100x²)/(2*(8-10x)³) = 4×10 { t. Solving gives x "H 0.1.
8. If the function f(x) = x³ + ax² + bx + c has a local maximum at x = -1 and a local
minimum at x = 3, then the values of a and b are:
A. a = -3, b = -9
B. a = 3, b = -9
C. a = -3, b = 9
D. a = 3, b = 9
Answer: A. a = -3, b = -9
f'(x) = 3x² + 2ax + b. f'(-1)=0 => 3 - 2a + b = 0; f'(3)=0 => 27 + 6a + b = 0. Solving gives
a = -3, b = -9.
9. The integral "+ €^À (x sin x) / (1 + cos² x) dx equals:
A. À²/4
B. À²/2
C. À/2
D. À
Answer: A. À²/4
Use property "+ €^a f(x) dx = "+ €^a f(a-x) dx. Let I = "+ €^À (x sin x)/(1+cos² x) dx. Then I
= "+ €^À ((À-x) sin x)/(1+cos² x) dx. Adding, 2I = À "+ €^À sin x/(1+cos² x) dx = À [
-arctan(cos x) ] €^À = À(À/2) = À²/2, so I = À²/4.
Page 4
