JEE Main 2024 8 April Shift 2 Physics question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 8 April Shift 2 exam from 3 PM to 6 PM. The Physics question paper for JEE Main 2024 8 April Shift 2 includes 30 questions divided into 2 sections, Section 1 with 20 MCQs and Section 2 with 10 numerical questions. Candidates must attempt any 5 numerical questions out of 10. The memory-based JEE Main 2024 question paper pdf for the 8 April Shift 2 exam is available for download using the link below.
Related Links:
- JEE Main 2025 Question Paper pdf with solutions
- JEE Main Previous Years Question Paper with Solution PDF
JEE Main 2024 8 April Shift 2 Physics Question Paper PDF Download
| JEE Main 2024 Physics Question Paper | JEE Main 2024 Physics Answer Key | JEE Main 2024 Physics Solutions |
|---|---|---|
| Download PDF | Download PDF | Download PDF |
JEE Main 8 Apr Shift 2 2024 Physics Questions with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 31. A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is: (1) √6m (2) 2m (3) 1m (4) √5m |
(2) 2m | Using the work-energy theorem, the total work done by all forces is equal to the change in kinetic energy (ΔKE) of the system. The equation is |
| 32. In a hypothetical fission reaction: The identity of emitted particles (R) is: (1) Proton (2) Electron (3) Neutron (4) γ-radiations |
(3) Neutron | To identify the emitted particles, we verify the conservation of atomic number (Z) and mass number (A). The missing mass corresponds to three neutrons (R = neutrons). |
| 33. If ε₀ is the permittivity of free space and E is the electric field, then ε₀E² has the dimensions: (1) [M₀L⁻²TA] (2) [ML⁻¹T⁻²] (3) [M⁻¹L⁻³T⁴A²] (4) [ML²T⁻²] |
(2) [ML⁻¹T⁻²] | The electric field formula, along with the permittivity, leads to the dimensional analysis of ε₀E². Simplifying the electric field expressions gives the dimensions [ML⁻¹T⁻²]. |
| 34. The position of the image formed by the combination of lenses is: (1) 30 cm (right of third lens) (2) 15 cm (left of second lens) (3) 30 cm (left of third lens) (4) 15 cm (right of second lens) |
(1) 30 cm (right of third lens) | The image positions are calculated lens by lens using the lens formula. The final image is formed 30 cm to the right of the third lens. |
| 35. A plane progressive wave is given by: y = 2 cos 2π(330t − x) m. The frequency of the wave is: (1) 165 Hz (2) 330 Hz (3) 660 Hz (4) 340 Hz |
(2) 330 Hz | The angular frequency ω is related to the frequency f by ω = 2πf. Given the wave equation, ω = 2π × 330, hence the frequency is 330 Hz. |
| 36. A thin circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its center and perpendicular to its plane with angular velocity ω. If another disc of same dimensions but of mass M/2 is placed gently on the first disc co-axially, then the new angular velocity of the system is: (1) 4/5ω (2) 5/4ω (3) 2/3ω (4) 3/2ω |
(3) 2/3ω | Using the law of conservation of angular momentum, we calculate the combined moment of inertia of both discs and apply it to find the new angular velocity, which comes out to be 2/3ω. |
| 37. A cube of ice floats partly in water and partly in kerosene oil. The ratio of volume of ice immersed in water to that in kerosene oil (specific gravity of kerosene oil = 0.8, specific gravity of ice = 0.9) is: (1) 8 : 9 (2) 5 : 4 (3) 9 : 10 (4) 1 : 1 |
(4) 1 : 1 | Using the principle of floatation, the weight of the ice cube is equal to the total upthrust from both water and kerosene oil. The ratio of the immersed volumes in both fluids comes out to be 1:1. |
| 38. Given below are two statements: Statement (I): The mean free path of gas molecules is inversely proportional to the square of molecular diameter. Statement (II): Average kinetic energy of gas molecules is directly proportional to absolute temperature of gas. In the light of the above statements, choose the correct answer from the options given below: (1) Statement I is false but Statement II is true (2) Statement I is true but Statement II is false (3) Both Statement I and Statement II are false (4) Both Statement I and Statement II are true |
(4) Both Statement I and Statement II are true | Statement I is true as the mean free path (λ) is inversely proportional to the square of molecular diameter (d). Statement II is also true as the average kinetic energy (KE) of gas molecules is directly proportional to the absolute temperature (T), according to the kinetic theory of gases. |
| 39. Two satellites A and B orbit a planet in circular orbits having radii 4R and R, respectively. If the speed of A is 3v, the speed of B will be: (1) 4/3v (2) 3v (3) 6v (4) 12v |
(3) 6v | Using the orbital speed formula v = √(GM/R), the speed of satellite B can be found by comparing the ratios of the radii of A and B. Since the speed of A is 3v, the speed of B comes out to be 6v. |
| 40. A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of the magnetic field at a = 2a and 2a from the axis of the wire is: (1) 1 : 4 (2) 4 : 1 (3) 1 : 1 (4) 3 : 4 |
(3) 1 : 1 | Using Ampère's law for a current-carrying wire, the magnetic field inside and outside the wire can be calculated. The ratio of the magnetic fields at a distance of a and 2a from the center of the wire comes out to be 1:1. |
| 41. The angle of projection for a projectile to have the same horizontal range and maximum height is: (1) tan⁻¹(1) (2) tan⁻¹(4) (3) tan⁻¹(1/4) (4) tan⁻¹(1/2) |
(2) tan⁻¹(4) | The horizontal range is u² sin 2θ / g, and the maximum height is u² sin² θ / 2g. By equating the two expressions, we find that the angle of projection θ should be tan⁻¹(4) for the range and height to be the same. |
| 42. Water boils in an electric kettle in 20 minutes after being switched on. Using the same main supply, the length of the heating element should be ... to ... times of its initial length if the water is to be boiled in 15 minutes: (1) Increased, 3/4 (2) Increased, 4/3 (3) Decreased, 3/4 (4) Decreased, 4/3 |
(3) Decreased, 3/4 | The resistance of the heating element is inversely proportional to its length. Using the time-power relation, we find that the length should be decreased to 3/4 of its original length to heat the water in 15 minutes. |
| 43. A capacitor has air as dielectric medium and two conducting plates of area 12 cm², and they are 0.6 cm apart. When a slab of dielectric having area 12 cm² and 0.6 cm thickness is inserted between the plates, one of the conducting plates has to be moved by 0.2 cm to keep the capacitance the same as in the previous case. The dielectric constant of the slab is: (1) 1.50 (2) 1.33 (3) 0.66 (4) 1 |
(1) 1.50 | The capacitance before and after inserting the dielectric is calculated using the formula for capacitance. The change in capacitance with dielectric insertion gives the dielectric constant as 1.50. |
| 44. A given object takes n times the time to slide down a 45° rough inclined plane as it takes to slide down an identical perfectly smooth 45° inclined plane. The coefficient of kinetic friction between the object and the surface of the inclined plane is: (1) 1 − 1/n² (2) 1 − n² (3) √(1 − 1/n²) (4) √(1 − n²) |
(1) 1 − 1/n² | Using the relation between acceleration and time of descent for both smooth and rough planes, the coefficient of kinetic friction is found to be 1 − 1/n². |
| 45. A coil of negligible resistance is connected in series with a 90Ω resistor across a 120V, 60Hz supply. A voltmeter reads 36V across the resistor. The inductance of the coil is: (1) 0.76H (2) 2.86H (3) 0.286H (4) 0.91H |
(1) 0.76H | The total impedance of the circuit is calculated using the voltmeter readings. The inductive reactance is then determined and used to find the inductance of the coil, which comes out to be 0.76H. |
| 46. There are 100 divisions on the circular scale of a screw gauge of pitch 1mm. With no measuring quantity in between the jaws, the zero of the circular scale lies 5 divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found that 4 linear scale divisions are clearly visible while 60 divisions on the circular scale coincide with the reference line. The diameter of the wire is: (1) 4.65mm (2) 4.55mm (3) 4.60mm (4) 3.35mm |
(2) 4.55mm | The least count of the screw gauge is 0.01mm. The zero error is +0.05mm. The reading is calculated as the sum of the linear scale reading, the circular scale reading multiplied by the least count, minus the zero error. The diameter comes out to be 4.55mm. |
| 47. A proton and an electron have the same de Broglie wavelength. If Kp and Ke are the kinetic energies of the proton and electron respectively, then choose the correct relation: (1) Kp > Ke (2) Kp = Ke (3) Kp = Ke² (4) Kp < Ke |
(4) Kp < Ke | The de Broglie wavelength is inversely proportional to the momentum of the particle. Since the proton has a greater mass than the electron, for the same de Broglie wavelength, the kinetic energy of the proton will be less than that of the electron. |
| 48. The least count of a vernier caliper is 1/20N cm. The value of one division on the main scale is 1mm. Then the number of divisions of the main scale that coincide with N divisions of the vernier scale is: (1) 2N − 1 / 20N (2) 2N − 1 / 2 (3) (2N − 1) (4) 2N − 1 / 2N |
(2) 2N − 1 / 2 | The least count is calculated as the difference between one main scale division and one vernier scale division. Using the relation between least count, main scale divisions, and vernier scale divisions, the result is 2N − 1 / 2. |
| 49. If M₀ is the mass of isotope ¹²₅B, Mp and Mn are the masses of proton and neutron, then the nuclear binding energy of the isotope is: (1) (5Mp + 7Mn − M₀)c² (2) (M₀ − 5Mp)c² (3) (M₀ − 12Mn)c² (4) (M₀ − 5Mp − 7Mn)c² |
(1) (5Mp + 7Mn − M₀)c² | The binding energy is given by the mass defect (∆m) multiplied by c². The mass defect is the difference between the sum of the masses of the protons and neutrons and the actual mass of the nucleus. The nuclear binding energy is (5Mp + 7Mn − M₀)c². |
| 50. A diatomic gas (γ = 1.4) does 100 J of work in an isobaric expansion. The heat given to the gas is: (1) 350 J (2) 490 J (3) 150 J (4) 250 J |
(1) 350 J | For an isobaric process, the heat supplied is the sum of the work done and the change in internal energy. Using the formula Q = ∆U + W, we calculate the heat as 350 J. |
| 51. The coercivity of a magnet is 5×10³ A/m. The amount of current required to be passed in a solenoid of length 30 cm and number of turns 150, so that the magnet gets demagnetised when inside the solenoid, is ... A. | (1) 10 A | Using the relation for coercivity, Hc = ni, where n is the number of turns per unit length and i is the current, the required current to demagnetize the magnet is 10 A. |
| 52. Small water droplets of radius 0.01 mm are formed in the upper atmosphere and fall with a terminal velocity of 10 cm/s. Due to condensation, if 8 such droplets are coalesced and form a larger drop, the new terminal velocity will be ... cm/s. | (1) 40 cm/s | The terminal velocity (Vt) of a droplet is proportional to R². When 8 droplets coalesce, the radius increases by a factor of 2. Hence, the terminal velocity increases by a factor of 4, giving a new terminal velocity of 40 cm/s. |
| 53. If the net electric field at point P along the Y-axis is zero, then the ratio of |q₂/q₃| is 8/5√x, where x = ... | (5) 5 | The electric field contributions from charges q₂ and q₃ are balanced. The geometry of the charge configuration and the formula for the electric field give x = 5. |
| 54. A heater is designed to operate with a power of 1000W in a 100V line. It is connected in combination with a resistance of 10Ω and a resistance R, to a 100V mains as shown in the figure. For the heater to operate at 62.5W, the value of R should be ... Ω. | (5) 5 Ω | Using the power and voltage relations for the circuit, the value of R is calculated as 5 Ω for the heater to operate at 62.5W. |
| 55. An alternating emf E = 110√2 sin(100t)V is applied to a capacitor of 2µF. The RMS value of current in the circuit is ... mA. | (22) 22 mA | The RMS current is calculated using the formula I_rms = E_rms / X_C, where X_C is the capacitive reactance. Using the given values, the RMS value of current is 22 mA. |
| 56. Two slits are 1mm apart, and the screen is located 1m away from the slits. A light wavelength 500 nm is used. The width of each slit to obtain 10 maxima of the double-slit pattern within the central maximum of the single-slit pattern is ... ×10⁻⁴ m. | (2) 2 × 10⁻⁴ m | Using the relation between the slit width and the number of maxima within the central maximum, we find the width of each slit to be 2 × 10⁻⁴ m. |
| 57. A body of mass 0.2 kg executes simple harmonic motion along the x-axis with a frequency of (25/π) Hz. At the position x = 0.04m, the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation is ... cm. | (2) 6 cm | Using the total energy of the harmonic oscillator (T.E. = K.E. + P.E.), we calculate the amplitude A of oscillation as 6 cm. |
| 58. A potential divider circuit is connected with a DC source of 20V, a light-emitting diode (LED) with a glow voltage of 1.8V, and a zener diode with a breakdown voltage of 3.2V. The length (PR) of the resistive wire is 20 cm. The minimum length of PQ to just glow the LED is ... cm. | (1) 5 cm | Using the voltage divider principle, the voltage across PQ is calculated to be 16.8V, which corresponds to a length of 5 cm for PQ to just glow the LED. |
| 59. A body of mass M thrown horizontally with velocity v from the top of a tower of height H touches the ground at a distance of 100m from the foot of the tower. A body of mass 2M thrown at a velocity v/2 from the top of a tower of height 4H will touch the ground at a distance of ... m. | (2) 100 m | The horizontal distance for both bodies is determined by their horizontal velocity and time of flight. Since the new body is thrown with half the velocity and from four times the height, the horizontal distance remains the same at 100m. |
| 60. A circular table is rotating with an angular velocity of ω rad/s about its axis (see figure). There is a smooth groove along a radial direction on the table. A steel ball is gently placed at a distance of 1m on the groove. All surfaces are smooth. If the radius of the table is 3m, the radial velocity of the ball with respect to the table at the time the ball leaves the table is x√2ω m/s, where the value of x is ... | (2) 2 | The centripetal acceleration and velocity of the ball are calculated using the rotational motion equations. The ball's radial velocity when it leaves the table is given by 2√2ω m/s, with x = 2. |
JEE Main 2024 Question Paper Session 1 (January)
Those appearing for JEE Main 2024 can use the links below to practice and keep track of their exam preparation level by attempting the shift-wise JEE Main 2024 question paper provided below.
| Exam Date and Shift | Physics Question Paper PDF |
|---|---|
| JEE Main 2024 Physics Question Paper Jan 27 Shift 1 | Check Here |
| JEE Main 2024 Physics Question Paper Jan 27 Shift 2 | Check Here |
| JEE Main 2024 Physics Question Paper Jan 29 Shift 1 | Check Here |
| JEE Main 2024 Physics Question Paper Jan 29 Shift 2 | Check Here |
| JEE Main 2024 Physics Question Paper Jan 30 Shift 1 | Check Here |
| JEE Main 2024 Physics Question Paper Jan 30 Shift 2 | Check Here |
| JEE Main 2024 Physics Question Paper Jan 31 Shift 1 | Check Here |
| JEE Main 2024 Physics Question Paper Jan 31 Shift 2 | Check Here |
| JEE Main 2024 Physics Question Paper Feb 1 Shift 1 | Check Here |
| JEE Main 2024 Physics Question Paper Feb 1 Shift 2 | Check Here |
Also Check:
| JEE Main 2024 Paper Analysis | JEE Main 2024 Answer Key |
| JEE Main 2024 Cutoff | JEE Main 2024 Marks vs Rank |
JEE Main 2024 8 April Shift 2 Physics Question Paper by Coaching Institute
| Coaching Institutes | Question Paper with Solutions PDF |
|---|---|
| Aakash BYJUs | Download PDF |
| Vedantu | Download PDF |
| Reliable Institute | To be updated |
| Resonance | To be updated |
| Sri Chaitanya | To be updated |
| FIIT JEE | To be updated |
JEE Main 2024 8 April Shift 2 Physics Paper Analysis
JEE Main 2024 8 April Shift 2 Physics paper analysis is updated here with details on the difficulty level of the exam, topics with the highest weightage in the exam, section-wise difficulty level, etc.
JEE Main 2024 Physics Question Paper Pattern
| Feature | Question Paper Pattern |
|---|---|
| Examination Mode | Computer-based Test |
| Exam Language | 13 languages (English, Hindi, Assamese, Bengali, Gujarati, Kannada, Malayalam, Marathi, Odia, Punjabi, Tamil, Telugu, and Urdu) |
| Sectional Time Duration | None |
| Total Marks | 100 marks |
| Total Number of Questions Asked | 30 Questions |
| Total Number of Questions to be Answered | 25 questions |
| Type of Questions | MCQs and Numerical Answer Type Questions |
| Section-wise Number of Questions | 20 MCQs and 10 numerical type, |
| Marking Scheme | +4 for each correct answer |
| Negative Marking | -1 for each incorrect answer |
Read More:
- JEE Main 2024 question paper pattern and marking scheme
- Most important chapters in JEE Mains 2024, Check chapter wise weightage here
Comments