09/04/2024
Morning
Corporate Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 | Ph.: 011-47623456
Memory Based
Answers & Solutions
Time : 3 hrs. for M.M. : 300
JEE (Main)-2024 (Online) Phase-2
(Physics, Chemistry and Mathematics)
IMPORTANT INSTRUCTIONS:
(1) The test is of 3 hours duration.
(2) This test paper consists of 90 questions. Each subject (PCM) has 30 questions. The maximum marks
are 300.
(3) This question paper contains Three Parts. Part-A is Physics, Part-B is Chemistry and Part-C is
Mathematics. Each part has only two sections: Section-A and Section-B.
(4) Section - A : Attempt all questions.
(5) Section - B : Attempt any 05 questions out of 10 Questions.
(6) Section - A (01 – 20) contains 20 multiple choice questions which have only one correct answer.
Each question carries +4 marks for correct answer and –1 mark for wrong answer.
(7) Section - B (21 – 30) contains 10 Numerical value based questions. The answer to each question
should be rounded off to the nearest integer. Each question carries +4 marks for correct answer and
–1 mark for wrong answer.
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, JEE (Main)-2024 : Phase-2 (09-04-2024)-Morning
PHYSICS
SECTION - A m 2
(1) .d (2) m 2 .d
Multiple Choice Questions: This section contains 20 2
multiple choice questions. Each question has 4 choices
3m 2 d
(1), (2), (3) and (4), out of which ONLY ONE is correct. (3) (4) 2m 2d
2
Choose the correct answer: Answer (1)
1. The dimension of latent heat is Sol. W = KE
(1) [M0L2T–1] (2) [M0L2T–2] 1
( ) ( )
2 2
W = m d − 0
(3) [M0LT–2] (4) [M–1L2T–2] 2
Answer (2)
m2d
2 –2 W =
Q ML T
Sol. L =
2
=
M M 4. Two persons are pulling a rope towards themselves
= L T
2 –2 with force of 200 N each. If Young’s modulus is
2 × 1011 N/m2 and area of cross is 2 cm2 for the
2. In the pulley-block system shown, the pulley and rope. The elongation in the rope is, if distance of
the block are ideal. If the acceleration of the blocks their holding the rope is 2 m.
g (1) 10 m (2) 20 m
is , find m1 : m2 (Given m2 > m1)
8
(3) 5 m (4) 40 m
Answer (1)
Fl
Sol. = l
YA
200 2
= l
2 1011 2 10 –4
(1) 7 : 9 (2) 5 : 7 100 × 10–7 = l
(3) 3 : 4 (4) 9 : 11 10 × 10–6 m = l
Answer (1) 5. A galvanometer having resistance of 200 shows
(m2 − m1)g g full deflection at 20 A. If the galvanometer has to
Sol. a = =
(m1 + m2 ) 8 measure current up to 200 mA, the shunt resistance
required is
m1 7
= 200 200
m2 9 (1) (2)
99 999
3. Velocity of a particle of mass m as a function of
20
displacement x is given by v = x . (3) (4) 200 × 999
99
Work done to move it from x = 0 to x = d is Answer (2)
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, JEE (Main)-2024 : Phase-2 (09-04-2024)-Morning
Rs Sol. E = mc2
Sol. 20 A = 20mA
Rs + 200 1 10−3 9 1016
= eV
Rs + 200 = 1000 Rs 1.6 10−19
= 5.625 × 1032 eV
200
Rs = = 5.6 × 1026 MeV
999
6. A particle oscillating simple harmonic motion such 9. Find equivalent resistance between terminal is A
that its speed and acceleration at distance 2 m from and B for the given network.
mean position are 4 m/s and 16 m/s2 respectively.
(1) 10 m (2) 6m
(3) 8m (4) 3m
Answer (2)
Sol. 4 = A2 − 4 …(i)
16 = 22 …(ii)
from (i) and (ii)
A= 6m (1) 16 (2) 20
7. Assertion : Object at radius of curvature of
(3) 15 (4) 19
biconvex lens forms image at same distance an
other side of lens. Answer (4)
Reason : Image of a real object formed by concave Sol. Simplifying the circuit, we get
lens is always virtual and erect.
(1) Assertion and reason are correct and reason is
correct explanation of assertion.
(2) Assertion and reason are correct but reason is
not correct explanation of assertion.
(3) Assertion is correct but Reason is incorrect 15
RAB = 14 + = 19
(4) Assertion is incorrect but Reason is correct 3
Answer (2) 10. A person covers first half of the distance with 6 m/s
and rest half of the distance is covered with 9 m/s
1 1 1
Sol. Theoretical − = & | v | = | u |= 2f and 15 m/s in two equal time intervals. Find average
v u f
8. The equivalent energy of 1 gm mass is equal to speed of the journey.
(1) 8.3 × 1026 MeV (2) 5.6 × 1026 MeV (1) 12 m/s (2) 9 m/s
(3) 8.3 × 1012 MeV (4) 5.6 × 1012 MeV (3) 10 m/s (4) 8 m/s
Answer (2) Answer (4)
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Morning
Corporate Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 | Ph.: 011-47623456
Memory Based
Answers & Solutions
Time : 3 hrs. for M.M. : 300
JEE (Main)-2024 (Online) Phase-2
(Physics, Chemistry and Mathematics)
IMPORTANT INSTRUCTIONS:
(1) The test is of 3 hours duration.
(2) This test paper consists of 90 questions. Each subject (PCM) has 30 questions. The maximum marks
are 300.
(3) This question paper contains Three Parts. Part-A is Physics, Part-B is Chemistry and Part-C is
Mathematics. Each part has only two sections: Section-A and Section-B.
(4) Section - A : Attempt all questions.
(5) Section - B : Attempt any 05 questions out of 10 Questions.
(6) Section - A (01 – 20) contains 20 multiple choice questions which have only one correct answer.
Each question carries +4 marks for correct answer and –1 mark for wrong answer.
(7) Section - B (21 – 30) contains 10 Numerical value based questions. The answer to each question
should be rounded off to the nearest integer. Each question carries +4 marks for correct answer and
–1 mark for wrong answer.
-1-
, JEE (Main)-2024 : Phase-2 (09-04-2024)-Morning
PHYSICS
SECTION - A m 2
(1) .d (2) m 2 .d
Multiple Choice Questions: This section contains 20 2
multiple choice questions. Each question has 4 choices
3m 2 d
(1), (2), (3) and (4), out of which ONLY ONE is correct. (3) (4) 2m 2d
2
Choose the correct answer: Answer (1)
1. The dimension of latent heat is Sol. W = KE
(1) [M0L2T–1] (2) [M0L2T–2] 1
( ) ( )
2 2
W = m d − 0
(3) [M0LT–2] (4) [M–1L2T–2] 2
Answer (2)
m2d
2 –2 W =
Q ML T
Sol. L =
2
=
M M 4. Two persons are pulling a rope towards themselves
= L T
2 –2 with force of 200 N each. If Young’s modulus is
2 × 1011 N/m2 and area of cross is 2 cm2 for the
2. In the pulley-block system shown, the pulley and rope. The elongation in the rope is, if distance of
the block are ideal. If the acceleration of the blocks their holding the rope is 2 m.
g (1) 10 m (2) 20 m
is , find m1 : m2 (Given m2 > m1)
8
(3) 5 m (4) 40 m
Answer (1)
Fl
Sol. = l
YA
200 2
= l
2 1011 2 10 –4
(1) 7 : 9 (2) 5 : 7 100 × 10–7 = l
(3) 3 : 4 (4) 9 : 11 10 × 10–6 m = l
Answer (1) 5. A galvanometer having resistance of 200 shows
(m2 − m1)g g full deflection at 20 A. If the galvanometer has to
Sol. a = =
(m1 + m2 ) 8 measure current up to 200 mA, the shunt resistance
required is
m1 7
= 200 200
m2 9 (1) (2)
99 999
3. Velocity of a particle of mass m as a function of
20
displacement x is given by v = x . (3) (4) 200 × 999
99
Work done to move it from x = 0 to x = d is Answer (2)
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, JEE (Main)-2024 : Phase-2 (09-04-2024)-Morning
Rs Sol. E = mc2
Sol. 20 A = 20mA
Rs + 200 1 10−3 9 1016
= eV
Rs + 200 = 1000 Rs 1.6 10−19
= 5.625 × 1032 eV
200
Rs = = 5.6 × 1026 MeV
999
6. A particle oscillating simple harmonic motion such 9. Find equivalent resistance between terminal is A
that its speed and acceleration at distance 2 m from and B for the given network.
mean position are 4 m/s and 16 m/s2 respectively.
(1) 10 m (2) 6m
(3) 8m (4) 3m
Answer (2)
Sol. 4 = A2 − 4 …(i)
16 = 22 …(ii)
from (i) and (ii)
A= 6m (1) 16 (2) 20
7. Assertion : Object at radius of curvature of
(3) 15 (4) 19
biconvex lens forms image at same distance an
other side of lens. Answer (4)
Reason : Image of a real object formed by concave Sol. Simplifying the circuit, we get
lens is always virtual and erect.
(1) Assertion and reason are correct and reason is
correct explanation of assertion.
(2) Assertion and reason are correct but reason is
not correct explanation of assertion.
(3) Assertion is correct but Reason is incorrect 15
RAB = 14 + = 19
(4) Assertion is incorrect but Reason is correct 3
Answer (2) 10. A person covers first half of the distance with 6 m/s
and rest half of the distance is covered with 9 m/s
1 1 1
Sol. Theoretical − = & | v | = | u |= 2f and 15 m/s in two equal time intervals. Find average
v u f
8. The equivalent energy of 1 gm mass is equal to speed of the journey.
(1) 8.3 × 1026 MeV (2) 5.6 × 1026 MeV (1) 12 m/s (2) 9 m/s
(3) 8.3 × 1012 MeV (4) 5.6 × 1012 MeV (3) 10 m/s (4) 8 m/s
Answer (2) Answer (4)
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