JEE (MAIN)-2025 (Online)
Physics Memory Based
Answer & Solutions
MORNING SHIFT EN DATE : 22-01-2025
LL
Disclaimer
A
The questions and solutions provided for JEE Main 2025 Session-1 are based on students' memory. While every
effort has been made to ensure accuracy, there may be discrepancies or variations from the actual exam. These
materials are intended for reference and educational purposes only. They do not represent the official question
paper or solutions provided by the exam conducting authority.
We recommend cross-verifying with official sources and using this content as a supplementary resource for your
preparation.
Offline Corporate Address : "SANKALP" CP-6, Indra Vihar Kota (Rajasthan), India 324005
Online Corporate Address : One Biz Square, A-12 (a), Road No. 1, Indraprastha Industrial Area, Kota - 324005 (Raj.)
+91-9513736499 | +91-7849901001 | | www.allen.in
, MEMORY BASED QUESTIONS JEE–MAIN EXAMINATION – JANUARY, 2025
(Held On Wednesday 22nd January, 2025) TIME : 9 : 00 AM to 12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
4. A parallel plate capacitor of capacitance 40µF is
SECTION-A
connected to a 100 V power supply now the
1. Find the dimensional formula of
𝐵
. intermediate space between the plates is filled with a
𝜇0
dielectric material of dielectric constant k = 2. due to the
(1) [AL–1] (2) [AL]
introduction of dielectric the extra charge and the
(3) [MALT1] (4) [MALT–1]
change in electrostatic energy in the capacitor
Ans. (1)
respectively are
Sol. B = 0ni
B N
(1) 2 µC and 0.4 J (2) 2 µC and 0.2 J
= Li
μ0 (3) 4 µC and 0.2 J (4) 8 µC and 0.2 J
B
[ ] = [AL−1 ] Ans. (4)
μ0
2. Two metals Cs and Li have work function Cs = 1eV
and Li = 2.5eV. When light of wavelength = 550 nm
(1) Cs
(3) Both
(2) Li
(4) None
EN
falls on both plates, which plate will show photoelectric
effect.
Sol.
40F
q1 -q1
100V
q1 = 40 × 100μC = 4mC
1
K=2
80F
100V
U1 = × 10 × 10−6 × 104 = 0.2J
Ans. (1) 2
LL
Sol. E =
1242ev−nm
= 2.25ev q2 = 8mC
550 nm
U2 = 0.4J
3. Two batteries are connected in parallel.
q2 − q1 = 4mC
Statement-1 : Net emf is less than the emf of any one of
U2 − U1 = 0.2J
them.
5. Statement-1: VSD is always less than MSD.
Statement-2 : Net internal resistance is less than Statement-2: Vernier constant is product of number of
individual resistance. MSD × No. of VSD.
A
(1) Statement-1 is true and Statement-2 is false (1) Statement-1 is true and Statement-2 is false
(2) Statement-1 is false and Statement-2 is true (2) Statement-1 is false and Statement-2 is true
(3) Both Statements are true
(3) Both Statements are true
(4) Both Statements are false
(4) Both Statements are false
Ans. (4)
Ans. (2) Sol. Theoretical
E E
( 1+ 2)
r1 r 2 6. Entire YDSE setup is immersed in medium.
Sol. Eeq = 1 1 =E
+
r1 r 2 Assertion : Fringe width reduces
1 1
= r +r =
1 2 Reason : Speed of light decreases while frequency
req 1 2 r remains constant
r
req = 2 (1) Assertion is correct and Reason is false
(2) Both Assertion and Reason are correct
(3) Assertion is false and Reason is correct
(4) Both Assertion and Reason are false
1
Physics Memory Based
Answer & Solutions
MORNING SHIFT EN DATE : 22-01-2025
LL
Disclaimer
A
The questions and solutions provided for JEE Main 2025 Session-1 are based on students' memory. While every
effort has been made to ensure accuracy, there may be discrepancies or variations from the actual exam. These
materials are intended for reference and educational purposes only. They do not represent the official question
paper or solutions provided by the exam conducting authority.
We recommend cross-verifying with official sources and using this content as a supplementary resource for your
preparation.
Offline Corporate Address : "SANKALP" CP-6, Indra Vihar Kota (Rajasthan), India 324005
Online Corporate Address : One Biz Square, A-12 (a), Road No. 1, Indraprastha Industrial Area, Kota - 324005 (Raj.)
+91-9513736499 | +91-7849901001 | | www.allen.in
, MEMORY BASED QUESTIONS JEE–MAIN EXAMINATION – JANUARY, 2025
(Held On Wednesday 22nd January, 2025) TIME : 9 : 00 AM to 12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
4. A parallel plate capacitor of capacitance 40µF is
SECTION-A
connected to a 100 V power supply now the
1. Find the dimensional formula of
𝐵
. intermediate space between the plates is filled with a
𝜇0
dielectric material of dielectric constant k = 2. due to the
(1) [AL–1] (2) [AL]
introduction of dielectric the extra charge and the
(3) [MALT1] (4) [MALT–1]
change in electrostatic energy in the capacitor
Ans. (1)
respectively are
Sol. B = 0ni
B N
(1) 2 µC and 0.4 J (2) 2 µC and 0.2 J
= Li
μ0 (3) 4 µC and 0.2 J (4) 8 µC and 0.2 J
B
[ ] = [AL−1 ] Ans. (4)
μ0
2. Two metals Cs and Li have work function Cs = 1eV
and Li = 2.5eV. When light of wavelength = 550 nm
(1) Cs
(3) Both
(2) Li
(4) None
EN
falls on both plates, which plate will show photoelectric
effect.
Sol.
40F
q1 -q1
100V
q1 = 40 × 100μC = 4mC
1
K=2
80F
100V
U1 = × 10 × 10−6 × 104 = 0.2J
Ans. (1) 2
LL
Sol. E =
1242ev−nm
= 2.25ev q2 = 8mC
550 nm
U2 = 0.4J
3. Two batteries are connected in parallel.
q2 − q1 = 4mC
Statement-1 : Net emf is less than the emf of any one of
U2 − U1 = 0.2J
them.
5. Statement-1: VSD is always less than MSD.
Statement-2 : Net internal resistance is less than Statement-2: Vernier constant is product of number of
individual resistance. MSD × No. of VSD.
A
(1) Statement-1 is true and Statement-2 is false (1) Statement-1 is true and Statement-2 is false
(2) Statement-1 is false and Statement-2 is true (2) Statement-1 is false and Statement-2 is true
(3) Both Statements are true
(3) Both Statements are true
(4) Both Statements are false
(4) Both Statements are false
Ans. (4)
Ans. (2) Sol. Theoretical
E E
( 1+ 2)
r1 r 2 6. Entire YDSE setup is immersed in medium.
Sol. Eeq = 1 1 =E
+
r1 r 2 Assertion : Fringe width reduces
1 1
= r +r =
1 2 Reason : Speed of light decreases while frequency
req 1 2 r remains constant
r
req = 2 (1) Assertion is correct and Reason is false
(2) Both Assertion and Reason are correct
(3) Assertion is false and Reason is correct
(4) Both Assertion and Reason are false
1
