JEE MAIN 2025 Practice Material PDF
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2025
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This is practice material for the JEE Main 2025 physics exam. It contains multiple-choice questions and numerical problems covering topics in physics, such as thermodynamics and thermal radiation.
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SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Sri Chaitanya IIT Academy.,India. A.P T.S KARNATAKA TAMILNADU MAHARASTRA DELHI RANCHI...
SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Sri Chaitanya IIT Academy.,India. A.P T.S KARNATAKA TAMILNADU MAHARASTRA DELHI RANCHI A right Choice for the Real Aspirant ICON Central Office - Madhapur - Hyderabad Sr.Super60_ Nucleus & Sterling PHYSICS RPTM-01(N&S) (SINGLE CORRECT ANSWER TYPE) This section contains 20 multiple choice questions. Each question has 4 options (1), (2), 3) and 4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and –1 in all other cases. 1. Statement I: The collision frequency of hydrogen molecules in a closed vessel is doubled when its temperature is doubled. Statement II: All solids expand on heating. 1) Statement I is true and Statement II is false 2) Statement I is false and Statement II is true 3) Both the statements are true. 4) Both the statements are false. Key :-4 Sol:- collision frequency is proportional to square-root of absolute temperature and some solids contract on heating. 2. The linear coefficient of expansion of metallic pendulum which looses 10s at 300 C and gains 5s per day at 150 C is approximately ( 0 C 1 ) 1) 1.12 105 2) 2.3 105 3) 4.6 105 4) 5 105 Key :-2 dT1 1 dl 1 1 5 Sol:- t (t 15) T 2 l 2 2 86400 dT2 10 1 15 1 1 (t 30) 15 T 86400 2 86400 2 43200 3. A piece of metal weight 46gm in air, when it is immersed in the liquid of specific gravity 1.24 at 270 C it weighs 30 gm. When the temperature of liquid is raised to 420 C the metal piece weight 30.5 gm, specific gravity of the liquid at 420 C is 1.20, then the linear expansion of the metal will be 1) 3.316 105 / 0C 2) 2.316 105 / 0C 3) 4.316 105 / 0C 4) 5.316 105 / 0C Key :-2 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 2 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Sol:- Loss of weight at 27ºC is = 46 – 30 = 16 = V1 × 1.24 l × g …(i) Loss of weight at 42ºC is = 46 – 30.5 = 15.5 = V2 × 1.2 l × g …(ii) 16 V1 1. 24 Now dividing (i) by (ii), we get = 15. 5 V2 1.2 But V2 = 1 + 3 (t2 – t1) = 15. 5 1.24 = 1.001042 V1 16 1.2 3 (42º – 27º) = 0.001042 = 2.316 × 10–5/ºC 4. In an exhaust pump using a barrel of certain volume, the pressure falls to half of its initial value in just two strokes. The percent fraction of volume of gas removed with each stroke is approximately 1) 20 2) 30 3) 40 4) 50 Key :-3 P0 P P0 V Sol:- Pn 0 0.414 41.4% V n 2 V V (1 ) (1 )2 V V 5. A typical dark nebula is about 20 light years in diameter (shaped spherically ) and contains about 50 hydrogen atoms per cubic centimetre (monatomic hydrogen and not H 2 ) at a temperature of about 20K (A light year = 9.46 1015 m Boltzman’s constant 1.38 1023 JK 1). The pressure inside a dark nebula is: 1) 1.4 1014 Pa 2) 2.8 1014 Pa 3) 1.4 1016 Pa 4) 2.8 1016 Pa Key :-1 Sol:- p nkT 50 106 1.38 10 23 20 1.38 1014 6. Assertion (A):- When a trough of water is placed in a very large evacuated room, part of it evaporates and the remaining freezes to ice. Reason(R):- A liquid boils at its boiling point temperature but can evaporate at any temperature. 1) Both (A) and (R) are true and (R) is the correct explanation of (A). 2) Both (A) and (R) are true and (R) is NOT the correct explanation of (A). 3) (A) is true and (R) is false. 4) (A) is false and (R) is true. Key :-1 n1 n 2 n1 n2 Sol:- mix 1 1 1 2 1 (NUMERICAL VALUE TYPE) Section-II contains 10 Numerical Value Type questions. Attempt any 5 questions only. First 5 attempted questions will be considered if more than 5 questions attempted. The Answer should be within 0 to 9999. If the Answer is in Decimal then round off to the nearest Integer value (Example i,e. If answer is above 10 and less than 10.5 round off is 10 and If answer is from 10.5 and less than 11 round off is 11). Marking scheme: +4 for correct answer, 0 if not attempt and -1 in all other cases. JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 3 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 7. A cube of side (a) and of coefficient of linear expansion S is floating in a liquid of coefficent of real expansion (γ L ) at 00 C. Even if the temperature of the system (cube and liquid) is raised depth of the cube submerged in the liquid remains same. L / S is equal to _____ Key :-2 Sol:- In both cases mass of liquid displaced is same. 0 a 2h 0 a 2 (1 2S t )h t a 2 (1 2S t )h L 2S 1 Lt 8. A thin rod of negligible mass and area of cross-section 4 107 m2 suspended vertically from one end, has a length of L m at 1000 C. The rod is cooled to 00 C , but prevented from contracting by attaching a mass at the lower end. Find the mass attached in kg. (Given : Young’s modulus of the rod = 1011 N / m2 , coefficient of linear expansion = 105 K 1 , g 10 ms2 ) Key :-4 YAt 1011 4 10 7 10 5 100 Sol:- F mg YAt m 4kg g 10 9. In a thermodynamic process helium gas obeys the law TP 2/5 constant. The heat given to the gas when the temperature of 2 moles of the gas is raised from T to 4T is xRT. (R is the universal gas constant). Find the value of x. Key :-0 1 x 1 x 2 Sol:- Since in polytropic process T P x cons tan t x 5 5 x of monoatomic gas given process is adiabatic Q=0 3 26. The rate of collisions of the molecules against the vessels wall N 12 n v T PV 1 cons tan t x 1 V 3R R C C 2R 2 1 x 10. A gas consisting of monatomic molecules (degrees of freedom = 3) was expanded in a polytropic process so that the rate of collisions of the molecules against the vessel’s wall JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 4 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 did not change. The molar heat capacity of the gas in the process is kR. The value of ' k ' is. Key :-2 Sol:- 11. The figure given below shows the cooling curve of pure wax material after heating. It cools from A to B and solidifies along BD. If L and C are respective values of latent heat and the specific heat of the liquid wax, the ratio L/C is. Key :-20 L Sol:- BD slope at B 4 10 / 2 20 C 12. The net work done(in J) in a thermodynamic cycle that expands isobarically at pressure of two atmospheres (2 105 Pa) followed by isothermal expansion to a pressure of 1atmosphere and volume 2litres, isobaric compression to 1 litre and isothermal compression to the original state, is. ln 2 0.7 Key :-70 P2 Sol:- P2 (V4 V3 ) PV 2 4 ln P1 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 5 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 P2 P1 (V2 V1 ) PV 1 1 ln P1 But PV 1 1 and PV 2 3 PV 2 4 PV 1 2 P2 P2 PV 1 2 ln PV 1 1 ln P1 P1 P P1 ln 2 (V2 V1 ) (105 ) ln 2(1103 ) 70 J P1 13. What amount of heat( in J ) is to be transferred to nitrogen in the isobaric heating process for that gas to perform work of 2J? Key :-7 du f Sol:- for icobaric process dw 2 14. 1gm of ice at 00 C is mixed with 1gm of water at 1000 C the resulting temperature( in 0 C) will be. Li 80cal / g and s 1cal / g 0C ) Key :-10 Li 80 w 100 CW Sol:- mix 1 10 C 2 2 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 6 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 RPTM-02(N&S) (SINGLE CORRECT ANSWER TYPE) This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and –1 in all other cases. 1. Assertion : Like light radiations, thermal radiations are also electromagnetic radiation. Reason : The thermal radiations require no medium for propagation. 1) If both assertion and reason are true and the reason is the correct explanation of the assertion. 2) If both assertion and reason are true b ut reason is not the correct explanation of the assertion. 3) If assertion is true but reason is false. 4) If the assertion and reason both are false. Key:- 2 Sol:- Light radiations and thermal radiations both belongs to electromagnetic spectrum. Light radiations belongs to visible region while thermal radiation belongs to infrared region of EM spectrum. Also EM radiations requires no medium for propagation. (NUMERICAL VALUE TYPE) Section-II contains 10 Numerical Value Type questions. Attempt any 5 questions only. First 5 attempted questions will be considered if more than 5 questions attempted. The Answer should be within 0 to 9999. If the Answer is in Decimal then round off to the nearest Integer value (Example i,e. If answer is above 10 and less than 10.5 round off is 10 and If answer is from 10.5 and less than 11 round off is 11). Marking scheme: +4 for correct answer, 0 if not attempt and -1 in all other cases. 2. There is certain mass of hydrogen gas in a container closed with a frictionless piston as shown in figure. The gas is heated for 25 s by an electric heater of resistors 50 built in the container using a power supply of 220 volts. While the gas expands at constant pressure its temperature increases by 250o C. If efficiency of electric heater is 75%, then find the mass of gas approximately (in gram ) in container 220V Key:-5 V2 Sol:- 0.75 t nC pT n 2.496 R JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 7 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 3. In a three dimensional coordinate system (OXYZ) , a concave mirror of radius of curvature 40 cm is placed at x 80cm. An object placed at origin is given a velocity v 9iˆ 6 ˆj 3kˆ cm / s.If the magnitude of velocity of image is N cm/sec then find 0 N Key:-6 V2 V V Sol:- VIX VoxVIY Voy , Viz Voz U2 U U 1 4. Figure shows a parabolic graph between T and (T= temperature and V=Volume) for a v gas undergoing an adiabatic process. If the ratio of rms velocity of molecules and speed of sound in the gas at the same temperature is n , then the find the value of n. T T1 1 / V1 1/V Key:-2 3RT 1 M 3 2 Sol:- T 2 TV 1/2 const v3/ 2 Ratio V RT v v M 5. Between two isotherm we have a cycle as shown. The work done by the gas during the cycle is 15N (in J). Find the value of N. (Take T1 127o C , T2 16o C , n 1mole ) JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 8 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Key:-5 P V Sol:- 2 2 P2V1 PV 1 2 P1 V1 W P2 V2 V1 P1 V2 V1 nRT2 nRT1 P2V1 PV 1 2 nRT2 nRT1 P2 . P3 V2 V1 6. Find the total number of distinct images formed it the angle between two plain mirrors is 70 degrees and object is placed between the reflecting faces of mirror equidistant from both the mirrors. Key:-6 Sol:- 7. Intensity at A due to source is I, without concave mirror. If the intensity at A after placing concave mirror as shown is nI then find the value of n? Key:-2 P Sol:- I 4 10 2 1 1 1 V . f v u Intensity at A due to reflection = I Total I I 2 I JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 9 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 8. A block plane surface at a constant high temperature T1 is parallel to another black plane surface at a constant lower temperature T2. Between the plate is vacuum. In order to reduce the heat flow due to radiation, a heat shield consisting of two thin black plates, thermally isolated from each other is placed between the warm and the cold surfaces and parallel to these. After sometime stationary conditions are obtained. If the new stationary heat flow becomes 1/n the rate of heat flow in the absence of heat shield then find n? Neglect the effects due to finite size of the surfaces. Key:-3 Sol:- Under steady state conditions the net rate of heat flow per unit area is the same everywhere H T14 T34 H T34 T44 H T44 T24 Adding these three equations, we get 3H T14 T24 H 0 Here H 0 T14 T24 is the rate of heat flow per unit area in the absence of the shields. JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 10 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 RPTM-03(N&S) (SINGLE CORRECT ANSWER TYPE) This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and –1 in all other cases. 1. Assertion : Like light radiations, thermal radiations are also electromagnetic radiation. Reason : The thermal radiations require no medium for propagation. 1) If both assertion and reason are true and the reason is the correct explanation of the assertion. 2) If both assertion and reason are true b ut reason is not the correct explanation of the assertion. 3) If assertion is true but reason is false. 4) If the assertion and reason both are false. Key:- 2 Sol:- Light radiations and thermal radiations both belongs to electromagnetic spectrum. Light radiations belongs to visible region while thermal radiation belongs to infrared region of EM spectrum. Also EM radiations requires no medium for propagation. (NUMERICAL VALUE TYPE) Section-II contains 10 Numerical Value Type questions. Attempt any 5 questions only. First 5 attempted questions will be considered if more than 5 questions attempted. The Answer should be within 0 to 9999. If the Answer is in Decimal then round off to the nearest Integer value (Example i,e. If answer is above 10 and less than 10.5 round off is 10 and If answer is from 10.5 and less than 11 round off is 11). Marking scheme: +4 for correct answer, 0 if not attempt and -1 in all other cases. 2. There is certain mass of hydrogen gas in a container closed with a frictionless piston as shown in figure. The gas is heated for 25 s by an electric heater of resistors 50 built in the container using a power supply of 220 volts. While the gas expands at constant pressure its temperature increases by 250o C. If efficiency of electric heater is 75%, then find the mass of gas approximately (in gram ) in container 220V Key:-5 V2 Sol:- 0.75 t nC pT n 2.496 R 3. In a three dimensional coordinate system (OXYZ) , a concave mirror of radius of curvature 40 cm is placed at x 80cm. An object placed at origin is given a velocity JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 11 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 v0 9iˆ 6 ˆj 3kˆ cm / s.If the magnitude of velocity of image is N cm/sec then find N Key:-6 V2 V V Sol:- VIX Vox VIY Voy , Viz Voz U2 U U 1 4. Figure shows a parabolic graph between T and (T= temperature and V=Volume) for a v gas undergoing an adiabatic process. If the ratio of rms velocity of molecules and speed of sound in the gas at the same temperature is n , then the find the value of n. T T1 1 / V1 1/V Key:-2 3RT 1 M 3 2 Sol:- T 2 TV 1/2 const v3/ 2 Ratio V RT v v M 5. Between two isotherm we have a cycle as shown. The work done by the gas during the cycle is 15N (in J). Find the value of N. (Take T1 127o C , T2 16o C , n 1mole ) Key:-5 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 12 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 P2 V Sol:- 2 P2V1 PV 1 2 P1 V1 W P2 V2 V1 P1 V2 V1 nRT2 nRT1 P2V1 PV 1 2 nRT2 nRT1 P2 . P3 V2 V1 6. Find the total number of distinct images formed it the angle between two plain mirrors is 70 degrees and object is placed between the reflecting faces of mirror equidistant from both the mirrors. Key:-6 Sol:- 7. Intensity at A due to source is I, without concave mirror. If the intensity at A after placing concave mirror as shown is nI then find the value of n? Key:-2 P Sol:- I 4 10 2 1 1 1 V . f v u Intensity at A due to reflection = I Total I I 2 I JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 13 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 8. A block plane surface at a constant high temperature T1 is parallel to another black plane surface at a constant lower temperature T2. Between the plate is vacuum. In order to reduce the heat flow due to radiation, a heat shield consisting of two thin black plates, thermally isolated from each other is placed between the warm and the cold surfaces and parallel to these. After sometime stationary conditions are obtained. If the new stationary heat flow becomes 1/n the rate of heat flow in the absence of heat shield then find n? Neglect the effects due to finite size of the surfaces. Key:-3 Sol:- Under steady state conditions the net rate of heat flow per unit area is the same everywhere H T14 T34 H T34 T44 H T44 T24 Adding these three equations, we get 3H T14 T24 H 0 Here H 0 T14 T24 is the rate of heat flow per unit area in the absence of the shields. JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 14 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 GTM-01(N) (SINGLE CORRECT ANSWER TYPE) This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and –1 in all other cases. 1. Statement-I: Due to rapid pumping of tyres, air inside the tyres is hotter than atmospheric air Statement-II: Adiabatic process occurs at very high rate 1) Statement-I is true; Statement-II is false 2) Statement-I is false; Statement-II is true; 3) Statement-I & Statement-II are true 4) Statement-I & Statement-II are false Key: 3 Sol: Conceptual 2. An inclined plane is placed on a horizontal smooth surface. The plane is struck by an elastic ball whose velocity is horizontal just before the impact. The ball bounces off the inclined plane and then lands on it again at the point of first impact. Find the ratio of the masses of the ball and the inclined plane. (Angle 30 ) 1) 2 2) 3 3) 4 4) 5 Key: 1 Sol: Conservation of linear momentum in horizontal direction m1V1 m1 m2 V2 Coefficient of restitution = 1 V y cos V1 sin Component of velocity of ball along the inclined plane remain same V1 cos V2 cos V y sin Solving m1 cot 2 1 m2 3. Assertion: The resistivity of a semiconductor increases in temperature. Reason: In a conducting solid, the rate of collisions between free electrons and ions increases with increase of temperature. 1) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion 2) If both Assertion and Reason are true but Reason is not correct explanation of the Assertion JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 15 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 3) If Assertion is true but Reason is false 4) If Assertion is false but the Reason is true Key: 4 Sol: Resistivity of semiconductor decreases with temperature. The atoms of a semiconductor vibrate with large amplitudes at higher temperatures there by increasing conductivity not resistivity. 4. A tank of cross sectional area A0 is filled with a liquid. A small orifice of area A A A0 is present at the bottom of tank. At a moment the height of liquid in the tank is ‘H’, what is acceleration of top layer of liquid at this moment? H 2 A 1) g 2) A g 3) g 4) depends on H A0 A0 Key: 3 Sol: Velocity of efflux v 2 gh AV A Velocity of top layer 2 gh Av A0 d A acceleration top layer 2 gh dx Av A 1 dh 2g Av 2 h dt A 1 A 2g 2 gh A0 2 h A0 2 A g A 0 Correct option is (3) 5. Statement 1: A metallic surface is irradiated by a monochromatic light of frequency v v0 (the threshold frequency). The maximum kinetic energy and the JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 16 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 stopping potential are Kmax and V0 respectively. If the frequency incident on the surface is doubled, both the Kmax and V0 are also doubled. Statement 2: The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light 1) Statement 1 is True, Statement 2 is False. 2)Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1. 3)Statement 1 is False, Statement 2 is True. 4)Statement 1 is True, Statement 2 is True; Statement 2 is not the correct explanation for Statement 1. Key: 3 Sol: Conceptual 6. A certain quaternary star system consists of three stars, each of mass m, moving in same circular orbit about a stationary central star of mass M. The three identical stars orbit in same sense and are symmetrically located with respect to each other (The centre of all stars lie in one plane). Considering gravitational force of all remaining bodies on every star, the time period of each of three stars is r3 r3 1) 2 2) 2 m m GM GM 3 3 r3 r3 3) 2 4) 2 G M 3m G M 3m Key: 2 Sol: The distance between the orbiting stars is d 2r cos30 3r. The net inward force on orbiting stars is Gm 2 GMm Gm 2 mv 2 cos30 cos30 d2 r2 d2 r m 4 2r 3 r3 G M or T 2 p 3 T2 G M m 3 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 17 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 7. A rod is made of uniform material and has non-uniform cross section. It is fixed at both the ends as shown and heated at mid-section. Which of the following statement is correct? 1) Force of compression in the rod will be maximum at mid-section 2) Compressive stress in the rod will be maximum at left end 3) Since rod is fixed at both the ends, its length will remain unchanged. Hence, no strain will be induced in it 4) None of these Key: 2 Sol: The compressive force in the rod will be same, therefore, compressive stress will be maximum at this end. Hence, option (b) along is correct 8. A linearly polarized electromagnetic wave given as E E i cos kz t is incident normally on a perfectly reflecting infinite wall at z a. Assuming that the material of the wall is optically inactive, the reflected wave will be given as 1) E r E i cos kz t 2) E r E i cos kz t 3) E r E i cos kz t 4) E r E i sin kz t Key: 2 Sol: Since the EM wave is incident normally on the reflecting wall, it will reflect back in the same direction with i i and z z creating the additional phase of c. We have given the equation of the incident wave, E E i cos kz t . Therefore, we can write the reflected wave equation as, E E i cos k z t E E i cos kz t E E i cos kz t We know the identity, cos cos Using this identity in the above equation, we get, E E i cos kz t 9. The following equations represent transverse waves z1 A cos 3Kx t …(i) z2 A cos 4 Ky t …(ii) JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 18 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 A wave is formed by the superposition of waves (i) and (ii). This wave is 3 1) Traveling in the direction making an angle tan1 with +ve x–axis 4 4 2) Traveling in the direction making an angle tan 1 with +ve x–axis 3 3 3) Traveling in the direction making an angle tan 1 with +ve y–axis 4 4 4) Traveling in the direction making an angle tan 1 with +ve y–axis 3 Key: 2 Sol: Equation of resultant wave is 3Kx 4Ky 3Kx 4Ky Z z1 z 2 2A cos cos t 2 2 3Kx 4Ky Equation of wavefront is t C (C is constant) 2 y C’ 4K x C’ 3K So at any given instant of time the equation of wave front will be 3Kx + 4 Ky = C , Wave travels perpendicular to the wave front also with increase in C' 4 4 time both x and y should increase, as tan 3K , tan 1. C' 3 3 4K (NUMERICAL VALUE TYPE) Section-II contains 10 Numerical Value Type questions. Attempt any 5 questions only. First 5 attempted questions will be considered if more than 5 questions attempted. The Answer should be within 0 to 9999. If the Answer is in Decimal then round off to the nearest Integer value (Example i,e. If answer is above 10 and less than 10.5 round off is 10 and If answer is from 10.5 and less than 11 round off is 11). Marking scheme: +4 for correct answer, 0 if not attempt and -1 in all other cases. 10. The spool shown in the figure is placed on a rough horizontal surface has inner radius r and outer radius R. The angle between the applied force F (tangential to the circle of radius r) and the horizontal can be varied. The critical angle for which the spool does not roll and remains stationary is , then n is___(if R 2 r ) n JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 19 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Key: 3 Sol: If spool is not to translate F cos f 1 ,If spool is not to rotate Fr = fR.. (2) fR From eq. (1) and (2) we get static friction cos f r r r or cos or cos 1 R R F sin F R r F cos f 11. An particle with K.E. T= 5.3 Mev initiates a nuclear reaction Be 9 2 4 C 12 0 n1 with energy yield Q=+5.7 Mev. Then the K.E. of neutron outgoing at the right angle to the motion direction of the particle will be [initially 68 Be9 is at rest] (in MeV), then n is ____ n Key: 8 Sol: Be9 , n c12 Taking components along X and Y-axis of momentum before and after collision P Pc cos 0 Pn Pc sin (or) P2 Pn2 Pc2 4T Tn 12Tc T Tn P2 Tc T K.E. 3 12 2m and Q Tc Tn T 12. Seven homogeneous bricks, each of length L, are arranged as shown in figure. Each brick is displaced with respect to the one in contact by L/10. The x-coordinate of the centre of 22 mass relative to the origin O shown is L. Find the value of n. 5n JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 20 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 O X Key: 7 L 6 7 4 22 L Sol: 1 7 5 5 5 35 13. Two metal bars are fixed vertically and are connected on the top by a capacitor C. A sliding conductor of length and mass m slides with its ends in contact with the bars. The arrangement is placed in a uniform horizontal magnetic field directed normal to the plane of the figure. The conductor is released from rest. Find the displacement of the conductor (in metre) after 2s. Assume that CB22 4m (the mass of the conductor) and g 10 m / s 2. C Key: 4 Sol: Due to the motion of the conductor in magnetic field, an e.m.f. is induced in it. As a result, a current flows through the conductor. According to Lenz’s law, a force Bi (due to induced current) opposes the motion of the conductor. Let at some instant t, velocity of the conductor be v. The net accelerating force on conductor is F mg Bi …(1) Here, induced e.m.f. Bv Charge on the capacitor, q Ce C Bv Since, v is increasing, the charge and hence the current through the capacitor is also dq dv increasing. The current through capacitor is given by ic CBf …(2) dt dt dv dv From equations (1) and (2), we get m mg B CB dt dt JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 21 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 dv dv dv m mg B 22C m B 22C mg dt dt dt dv mg 1 2 mgt 2 a x t at 4 dt m B 2 2C 2 2 m B 2 2C 14. A highly accurate determination of g has a quoted error of 6 parts in 109. If the dependence of g on geographical latitude at sea level is given by g g 0 1 sin 2 where is the latitude and is a dimensionless constant with a value of 0.0055.The northward displacement near latitude 450 which gives a change in g equal to the quoted error (radius of earth R 6.4 106 m ) is (0.1)n m, where n is_____(approximately) Key: 7 dg Sol: 2 g0 sin cos So, d dg dg d 1.09 106 radian 2 g 0 sin cos g0 sin 2 15. A stationary He ion emitted a photon corresponding to the first line of the Lyman series. The photon liberates electron from a stationary hydrogen atom in the ground state. n The velocity of the liberated electron is 3.1 10 m/s. Find n (You can make necessary approximations) Key: 6 1 1 Sol: given by E 13.6 . eV... 1 n2 n2 1 2 This transition energy is shared between recoiling helium ion and photon From conservation of energy. E jf K.E..... 2 Where K.E. is kinetic energy of recoiling atom andhf is energy of photon From conservation of momentum, P Photon P He... 3 2 E ... 4 2E 1 1 2 mc 1 1 E 13.6 eV 40.8eV... 5 2 2 1 2 1 2 6 mv 40.8 13.6 eV 37.2eV v 3.1 10 m / s x6 2 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 22 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 16. An ideal string is wrapped several times on a solid cylinder of mass 4 kg and radius 1 m. The pulleys are ideal and the surface between block and ground is smooth. If the torque 80 acting on the cylinder is N-m, then find the value of n. n Solid cylinder 4 kg Block R 8 kg F = 40 N Smooth Key: 9 Sol: FBD of cylinder and block are as shown by Newton’s laws 40 2T f s 8a (1) T f s 4a / 2 (2) Subtracting equation (2) from equation (1) 40 3T 6a T 40 6a / 3 Also, by I , we get 3a T R fs R I 2R T f s 3Ia / 2 R 2 40 6a / 3 40 12a / 3 3Ia / 2R 2 a 80 / 18 9I / 2R 2 As I I 3a / 2R I I 3a / 2 R 3I 80 3 80 80 N m 2R 9I 18 9 9 18 2 2R JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 23 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 RPTM-04(S) (SINGLE CORRECT ANSWER TYPE) This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and –1 in all other cases. 1. If molar heat capacity of the given process (as shown in figure) is C, then 1) C < CV 2) C = 0 3) C > CV 4) C = CV Key: 3 Sol: Conceptual 2. A light wave is incident normally on a glass slab of refractive index 1.5. If 4% of light gets reflected and the amplitude of the electric field of the incident light is 30V/m, then the amplitude of the electric field for the wave propagating in the glass medium will be: 1) 10 V/m 2) 24 V/m 3) 30 V/m 4) 6 V/m Key: 2 Sol: 96 Prefracted P1 100 96 K 2 At2 K1 Ai2 100 96 r2 At2 r1 Ai2 100 96 1 30 2 At2 100 3 2 64 At (30)2 24 100 3. Two identical coherent sources produce a zero order bright fringe on a screen. If is the band width, the minimum distance between two points on either side of the bright fringe where the intensity is half that of maximum intensity is 1) / 2 2) / 4 3) / 3 4) / 6 Key: 2 Sol: JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 24 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 4. Intensity observed in an interference pattern I I0 sin 2 . At 30 intensity 2 I 5 0.002. The percentage error in angle if I 0 20 w / m is 4 4 3 1) 4 3 10 2 % 2) 102 % 3) 102 % 4) 3 10 2 % Key: 3 Sol: Conceptual 5. In the adjacent diagram, CP represents a wave front and AO and BP, the corresponding two rays. Find the condition on for constructive interference at P between the ray BP and reflected rays OP 3 1) cos 2) cos 2d 4d 4 3) sec cos 4) sec cos d d Key: 2 Sol: PR d PO d sec , and CO PO cot 2 d sec cos 2 Optical path difference between the two rays is x CO PO d sec d sec cos 2 2 Path difference between the two rays is 2n , n 0,1, 2,3... Condition for constructive interference should be ……… x n , n 0,1, 2,3... d d sec 1 cos 2 2 cos 2cos 2 2 cos 4d JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 25 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 (NUMERICAL VALUE TYPE) Section-II contains 10 Numerical Value Type questions. Attempt any 5 questions only. First 5 attempted questions will be considered if more than 5 questions attempted. The Answer should be within 0 to 9999. If the Answer is in Decimal then round off to the nearest Integer value (Example i,e. If answer is above 10 and less than 10.5 round off is 10 and If answer is from 10.5 and less than 11 round off is 11). Marking scheme: +4 for correct answer, 0 if not attempt and -1 in all other cases. 6. Figure shows two coherent sources S1 S 2 vibrating in same phase. AB is a straight wire lying at a far distance from the sources S1 and S 2. Let 103 BOA 0.12. d How many bright spots will be seen on the wire, including points A and B. Key: 3 Sol: Say ‘n’ fringes are present in the region shown by ‘y n. D 0.06 n 103 y n d y tan 0.06 D 180 d n 180 0.06 1 3 Hence; only one maxima above and below point O. So total 3 bright spots will be present (including point ‘O’ i.e. the central maxima). 7. Parallel monochromatic beam is falling normally on two slits S1 and S2 separated by d as shown in figure. By some mechanism, the separation between the slits S 3 and S4 can be changed. The intensity is measured at the point P which is at the common perpendicular D bisector of S1S2 and S3S4. When z the intensity measured at P is I and when 2d 4D z. Intensity is xI Find x. d JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 26 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Key: 2 D Sol: Where Z OS4 as shown 2d 2 4 If intensity at 'p' is I then intensity of light at S3 and S4 is I/4 & I/4 Path difference S4P-S3P=0. So, intensity of slits S1 and S2 I I1 I 2 intensity of s1 and s2 8 D yd z d 4 D 1 d IF Z 4. 4 , p 2 d D 2D d 2 D 2 I I 2I 1 .2d 4 , I3 I 4 COS 4 8 8 8 2 1 1 I I at I P 2. cos 0 I p 2 I 2 2 2 2 8. Find the angular width (in degrees) of the central bright maximum in the Fraunhofer diffraction pattern of a slit of width 12 105 cm when the slit is illuminated by monochromatic light of wavelength 6000Å. Key: 60 Sol: Here sin a Where is the angular width of the central maximum. a 12 10 5 cm 12 107 m 6000A 6 107 m 6 107 sin 0.5 30 12 107 Angular width of the central maximum. 2 60 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 27 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 9. Figure shows a steel rod joined to a brass rod. Each of the rods has length of 31 cm and area of cross-section 0.20 cm². The junction is maintained at a constant temperature 50C and the two ends are maintained at 100C. The amount of heat taken out from the 2 cold junction in 10 minutes after the steady state is reached is n 10 J. Find 'n'. The thermal conductivities are K steel 46 W / m C and K brass =109 W / m C. Key: 3 Sol: Let T 100 C T0 50 A Heat T (T T0 ) ( K S K B ) 10 60 0.2 104 (100 50) (46 109) 10 60 300 J 31 102 10. In Fraunhofer diffraction due to a narrow slit a screen is placed 2 m away from the lens to obtain the pattern. If the slit width is 0.2 mm and the first minima lie 5 mm on either side of the central maximum, find the wavelength of light. (in angstroms) Key: 5000 Sol: In the case of Fraunhofer diffraction at a narrow rectangular aperature, x a sin n , n 1, a sin sin D ax ax , D D Here a 0.2 mm 00.2 cm, x 5 mm 0.5 cm 0.02 0.5 5 D 2m 200 cm , 5 10 cm, L 5000 A 200 11. The focal length of objective and eye lens of a microscope are 4 cm and 8 cm respectively. If the least distance of distinct vision is 24 cm and object distance is 4.5 cm from the objective lens. The final image is formed at near point, then the magnitude of magnifying power of the microscope will be JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 28 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Key: 32 1 1 1 1 1 1 Sol: find ue ; find f e fe ve ue f e ve ue v0 ve m m0 me u0 u e 12. Two vessels A and B, thermally insulated, contain an ideal monoatomic gas. A small tube fitted with a valve connects these vessels. Initially the vessel A has 2 litres of gas at 300 5 2 K and 2 10 N m pressure while vessel B has 4 litres of gas at 350 K and 4 105 N m 2 pressure. The valve is now opened and the system reaches equilibrium in pressure and temperature. The new pressure will be 310 93 10n N / m2 Find n. Key: 5 Sol: Conceptual 13. A hemispherical portion of the surface of a solid glass sphere (µ = 1.5) of radius 10 cm (surrounding is air) is silvered to make the inner side reflecting. An object is placed on the axis of the hemisphere at a distance 30cm from the centre of the sphere. The light from the object is refracted at the unsilvered part, then reflected from the silvered part and again refracted at the unsilvered part. What is distance (in cm) of final image from pole of reflecting surface. Key: 7 Sol: Conceptual JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 29 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 RPTM-04/05(N&S) (SINGLE CORRECT ANSWER TYPE) This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and –1 in all other cases. 1. Assertion (A): Air pressure in a car tyre increases during driving. Reason (R): Absolute zero degree temperature is not zero energy temperature 1) Both A and R are true and R is the correct explanation of A 2) Both A and R true but R is NOT the correct explanation of A 3) A is true but R is false 4) A is false and R is also false Key: 2 Sol: When a car is in motion, the temperature of the tyre rises at absolute zero, the translation motion of molecules ceases but other form of molecular energy do not become zero. Therefore absolute zero temperature is not temperature of zero energy. (NUMERICAL VALUE TYPE) Section-II contains 10 Numerical Value Type questions. Attempt any 5 questions only. First 5 attempted questions will be considered if more than 5 questions attempted. The Answer should be within 0 to 9999. If the Answer is in Decimal then round off to the nearest Integer value (Example i,e. If answer is above 10 and less than 10.5 round off is 10 and If answer is from 10.5 and less than 11 round off is 11). Marking scheme: +4 for correct answer, 0 if not attempt and -1 in all other cases. 2. The value of ‘g’ at a particular point near the pole of Earth is 10m s 2. Suppose the earth suddenly shrinks uniformly to half its present radius without losing any mass. The value of ‘g’ at the same point(in m s 2 ) (assuming that the distance of the point from the centre of earth does not change) will now be Key: 10 Sol: As distance from center and total mass remain same, there is no change in acceleration due to gravity 3. Assume that the acceleration due to gravity on the surface of the moon is 0.2 times the acceleration due to gravity on the surface of the earth. If Re is the maximum range of a projectile on the earth’s surface, the maximum range on the surface of the moon for the same velocity of projection is nRe. The value of n is (for small velocities) Key: 5 u 2 sin 2 Sol: Range of projectile R g 1 Rm ge R 1 R If u and are constant then R m Rm e 5Re g Re g m Re 0.2 0.2 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 30 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 4. Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each 1 GM particle is (1 2 n ). The value of n is 2 R Key: 2 Mv 2 GM 2 GM 2 Sol: 2 F cos 45 F ' (From figure) Where F and F ' 2 R 2R 4R2 2 GM 2 GM 2 Mv 2 GM 2 1 1 Mv 2 2 2 2 R 2 4R R R 4 2 GM 2 4 1 GM v (1 2 2) R 4 2 2 R 5. 2 kg of ice at – 20°C is mixed with 5 kg of water at 20°C in an insulating vessel having a negligible heat capacity. Calculate the final mass of water (in kg) remaining in the container. It is given that the specific heats of water and ice are 1 Kcal/kg / °C and 0.5K cal / kg / C while the latent heat of fusion of ice is 80 Kcal / kg Key: 6 Sol: Initially ice will absorb heat to raise it's temperature to 0 oC then it's melting takes place If mi = Initial mass of ice, mi ' Mass of ice that melts and mW Initial mass of water By Law of mixture Heat gained by ice = Heat lost by water mi c (20) mi ' L mwcw 2 0.5(20) mi ' 80 5 1 20 mi ' 1kg So final mass of water = Initial mass of water + Mass of ice that melts 5 1 6 kg. 6. A slab of glass of thickness 6cm and refractive index 1.5 is placed in front of a concave mirror, the faces of the slab being perpendicular to the principal axis of the mirror. If the radius of curvature of the mirror is 40cm and the reflected image coincides with the object, then the distance of the object from the mirror (in cm) is Key: 42 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 31 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 1 1 Sol: x 1 t 1 6 2cm 1.5 Distance of object from mirror 42cm 7. A point object O is placed on the principal axis of a convex lens of focal length 20cm at a distance of 40cm to the left of it. The diameter of the lens is 10cm. If the eye is placed 60cm to the right of the lens at a distance h below the principal axis, then the maximum value of h to see the image will be ‘x’ cm. The value of 10x is Key: 25 Sol: In the following ray diagram 's , ABC and CDE are symmetric AB DE 5 h So, h 2.5cm BC CD 40 20 8. In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the sits. If the screen is moved by 5 102 m towards the slits, the change in fringe width is 3 105 m If separation between the slits is 103 m , the wavelength of light used is (in A ). Key: 6000 D 1 D1 2 D1 D2 2 2 Sol: D 1 d 2 D2 2 D2 D D2 d2 3 10 5 2 10 3 6 10 7 m 6000 A 5 102 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 32 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 RPTM-05/06(N&S) (NUMERICAL VALUE TYPE) Section-II contains 10 Numerical Value Type questions. Attempt any 5 questions only. First 5 attempted questions will be considered if more than 5 questions attempted. The Answer should be within 0 to 9999. If the Answer is in Decimal then round off to the nearest Integer value (Example i,e. If answer is above 10 and less than 10.5 round off is 10 and If answer is from 10.5 and less than 11 round off is 11). Marking scheme: +4 for correct answer, 0 if not attempt and -1 in all other cases. 1. In the figure shown S is a large non-conducting sheet of uniform charge density .A rod R of length l and uniformly distributed total mass ‘m’. It is parallel to the sheet and hinged at its mid-point. The linear charge densities on the upper and lower half are shown x in the figure. The angular acceleration of the rod just after it is released is. Find ym 0 ( x y) S R R Key: 5 Sol: The upper and lower half of the rod are placed in uniform electric field of large sheet. Hence the magnitude of force and its effective point of application on upper and lower l l ml 2 half of rod are as shown in figure. The torque on rod is 0 2 2 2 12 l 2 2 0 S /2 2 2 0 3 Angular acceleration is 2m 0 2. Two small dipoles of moment p are placed as shown in figure. The force between the two nkp 2 dipoles is.Find n 4 x JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 33 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Key: 3 Sol: Conceptual 3. An infinitely long thin wire, having a uniform charge density per unit length of 5 nC/m, is passing through a spherical shell of radius 1m, as shown in the figure. A 10 nC charge is distributed uniformly over the spherical shell. If the configuration of the charges remains static, the magnitude of the potential difference between points P and R, in Volt, is ___. 1 [Given :In SI units 9 109 Nm 2C 2 ] loge 2 0.7 4 0 P R 0.2 5 4 Key: 320 Sol: Due to wire VR 4 2k dV E. dx dV dx 2 9 109 5 109 4 0.7 252V x Vp 0.5 kQ kQ 3 3 Due to sphere VR VP kQ 9 109 10 109 = - 67.5V 4 1 4 4 V V1 V2 VP VR 252 67.5 319.5 , Answer rounded off to 320 4. The distance between charges +q and –q is 2l and between +4q and -4q is 4l.The ql electrostatic potential at point P at a distance r from centre O is 109V. Where r2 1 the value of is ______. (Use 9 109 Nm 2C 2 , l r ) 4 0 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 34 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 P r 60 +4q -q O +q -4 q 2 4 Key: 63 K Sol: V V1 V2 ( q (2) cos 60 4 q (4)cos120) r2 5. Two persons A and B wear glasses of optical powers (in air) P1 2 D and P2 1D respectively. The glasses have refractive index 1.5.Now they jump into a swimming pool and look at each other. B appears to be present at distance 2 m (from A) to A. ‘A’ appears to be present at distance 1 m (from B) to B. The refractive index of water in the x swimming pool is.Find x 5 Key: 6 1 1 1 1 1 1 Sol: ( s 1 1) ( 1 1) f' R1 R2 f R1 R2 Find f ' in water & equate object distance 6. Twelve infinite long wire of uniform linear charge density () are passing along the twelve edges of a cube. The electric flux through any face of cube is .Find y y 0 Key: 2 Sol: cube =Flux due to single wire from whole cube cube Similarly four wires out of twelve will have same contribution and eight will 8 0 have zero face1 4 8 0 2 0 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 35 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 RPTM-06/07(N&S) (SINGLE CORRECT ANSWER TYPE) This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and –1 in all other cases. 1. Statement-I: Monoatomic, diatomic and polyatomic gases are adiabatically compressed P such that compression ratio is 2 . then monoatomic gas will have maximum final P1 volume out of these three gases. (Assume equal volume and pressure initially) Statement-II: Monoatomic gas has least degree of freedom. 1) statement 1 is true and statement 2 is true 2) statement 1 is true and statement 2 is false 3) statement 1 is false and statement 2 is true 4) statement 1 is false and statement 2 is false Key: 1 Sol: Monoatomic gas has least degree of freedom 2. Statement (A): The difference between new torchlight cell and an old one is due to increase in internal resistance. Statement (B): At 0 kelvin specific resistance of a perfect insulator is infinity. 1) Both A and B are true 2) A is true B is false 3) A is false but B is true 4) Both A and B are false Key: 1 Sol: in a torchlight cell the internal resistance keeps increasing with usage due to decreasing concentration of the electrolyte 3. In a standard YDSE the region between screen and slits is immersed in a liquid whose 5 T refractive index varies with time T as l until it reaches a steady state value of 2 4 5. A glass plate of thickness 36 m and refractive index 3/2 is introduced in front of 4 one of the slits. Find the speed of central maxima when it is at ‘O’? (take d = 2mm and D = 1m) 1) 2 103 ms 1 2) 3 103 ms 1 3) 4 103 ms 1 4) 5 103 ms 1 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 36 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Key: 2 Sol: path difference x S 2 P liq g l t S1 P liquid x S 2 P S1P liq g l t yd 1 S 2 P S1 P air g l t x l g l t D For central maxima x 0. 3 5 T D t yd 2 2 4 0 l g l t y D 5 T d 2 4 The time when y become zero is D 4 T t 0 D 4 T t 0 4 T 0 T 4sec d 10 T dy Dt d 4 T tD d 4 T 10 T 1 Speed of central maxima V V dx d dT 10 T d dT (NUMERICAL VALUE TYPE) Section-II contains 10 Numerical Value Type questions. Attempt any 5 questions only. First 5 attempted questions will be considered if more than 5 questions attempted. The Answer should be within 0 to 9999. If the Answer is in Decimal then round off to the nearest Integer value (Example i,e. If answer is above 10 and less than 10.5 round off is 10 and If answer is from 10.5 and less than 11 round off is 11). Marking scheme: +4 for correct answer, 0 if not attempt and -1 in all other cases. 4. Two cells A and B each of 2V are connected in series to an external resistance R = 1ohm. The internal resistance of A is rA = 1.9ohm and B is rB = 0.9ohm. Find the potential difference between the terminals of A. Key: 0 Sol: VA E A irA 5. When a gas filled in a closed vessel is heated by raising the temperature by 10 C , its pressure increases by 0.4%. The initial temperature of the gas is_____ K Key: 250 P T 1100 Sol: PT ; 0.4 ; T 250 K P T T 6. A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ)____ Key: 6 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 37 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 1 C1C2 V1 V2 2 Sol: loss of energy 2 C1 C2 7. A uniform heating wire of resistance 36 is connected across a potential difference of 240 V. the wire is then cut into half and a potential difference of 240 V is applied across each half separately. The ratio of power dissipation in the first case to the total power in the second case would be 1:x where x is Key: 4 V 2 240 240 2 2 P1 1 Sol: P1 When wire is cut into half R ' 18 P2 R1 36 9 P2 4 8. 4.0g of a gas occupies 22.4 liters at NTP. The specific heat capacity of the gas at constant volume is 5.0JK 1mol 1. If the speed of sound in this gas at NTP is 952ms 1 , then the heat capacity at constant pressure is (Take gas constant R 8.3 JK 1mol 1 )____ JK 1mol 1 Key: 8 Cv Mv 2 Sol: Cp 8 J / K.mol RT 9. The volume V of a given mass of ideal monoatomic gas changes with temperature T according to the relation V KT 2/3. The work done when temperature changes by 90 K will be xR. The value of x is_____[R= universal gas constant] Key: 60 RT 2 Sol: W pdV W dV W nR 90 60 R KT 2/3 3 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 38 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 GTM-02(N) (SINGLE CORRECT ANSWER TYPE) This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and –1 in all other cases. 1. A radioactive sample of decay constant starts decaying at time t = 0. The instant of time at which probability of survival of a nucleus is twice the probability of it having decayed is 3 2 ln ln ln 3 ln 2 2) 2 4) 3 1) 3) Key: 2 3 ln 2 2 Sol: e t 2 1 e rt ,3e t 2 e t t 3 2. The output of the following logic gate combination is 1 with the given input as shown, with one unknown input X, then 1) X is 0 only 2) X is 1 only 3) Output is independent of X 4) Output 1 is not possible for any value of X Key: 3 Sol: Output is independent of X 3. In a vertical plane other than magnetic meridian the angle of dip is 300. If the plane is rotated by 900 keeping it vertical then angle of dip becomes 450. The true dip at that place is 1 1 1) tan 1 2 2) cot 1 3 3) cot 1 4) tan 1 2 2 Key: 4 Sol: cot 2 cot 2 1 cot 2 2 ,cot 2 3 1 1 cot 1 2 tan 1 2 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 39 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 4. Statement – 1 : A nucleus having energy E1 decays by emission to a daughter nucleus having energy E2 but the rays are emitted with a continuous energy spectrum having end point energy E1 E2 Statement – 2 : To conserve energy and momentum in decay, at least three particles must take part in transformation. 1) Statement – 1 is true, Statement – 2 is true 2) Statement – 1 is false, Statement – 2 is false 3) Statement – 1 is false, Statement – 2 is true 4) Statement – 1 is true, Statement – 2 is false Key: 1 Sol: Statement – 1 is true, Statement – 2 is true 5. Assertion (A): Non-polar materials do not have any permanent dipole moment Reason (R): When a non-polar material is placed in a electric field, the centre of the positive charge distribution of it’s individual atom or molecule coinsides with the centre of the negative charge distribution. In the light of above statements, choose the most appropriate answer from the options given below: 1) Both A and R are true and R is correct explanation of A 2) Both A and R are true and R is not correct explanation of A 3) A is true and R is false 4) A is false and R is true Key: 3 Sol: S1 : In nonpolar molecules, centre of +ve charge coincides with centre of –ve charge, hence netdipole moment is comes to zero. S2 : Wen non polar material is placed in external field, centre of charges does not coincide, hence give non zero moment in field (NUMERICAL VALUE TYPE) Section-II contains 10 Numerical Value Type questions. Attempt any 5 questions only. First 5 attempted questions will be considered if more than 5 questions attempted. The Answer should be within 0 to 9999. If the Answer is in Decimal then round off to the nearest Integer value (Example i,e. If answer is above 10 and less than 10.5 round off is 10 and If answer is from 10.5 and less than 11 round off is 11). Marking scheme: +4 for correct answer, 0 if not attempt and -1 in all other cases. 6. A hollow sphere of inner radius R and outer radius 2R is made from a material of thermal conductivity K. It is surrounded by another hollow sphere of inner radius 2R and outer radius 3R made by material of thermal conductivity 2K. The inside of smaller sphere and outside of bigger sphere are maintained at 0ºC and 70ºC respectively. Temperature of interface in steady state, will be_______. Key: 60 or 333 Sol: Let temperature of interface is 0 70 1 1 1 1 1 1 4K R 2R 4 2K 2R 3R 70 1 1 R 6R JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 40 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 6 70 7 6 70 = 60ºC. 7. Figure shows a soap film formed between two square figures made of a uniform wire. The bigger square is held while keeping it in horizontal plane and the smaller square is slowly allowed to drop vertically. It reaches an equillibrium state after dropping a height h. Let surface tension of soap=T. Mass per unit length of the wire = λ. Acceleration due n ga to gravity = g. Given h 2 4T 2 2 g 2 Find the integer value of ‘n’? Key: 3 3a Sol: tan 2 4aT cos 4a g 2h g 3 ga cos h 2T 2 4T 2 2g 2 8. As shown in figure, a block of mass M and base area A slides on a horizontal table on oil film of thickness h and coefficient of viscosity . The mass m is released from rest at time t = 0. Find constant speed of the system of blocks in m/s if m 2kg , A 100 cm 2 , 1 poise , g=10m/s2 h 0.1mm Key: 2 Sol: constant speed of the system of blocks in m/s is 2 JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 41 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 9. A plane electromagnetic wave travelling in a non-magnetic medium is in given by E 9 108 NC 1 sin 9 108 rad S 1 t 6m 1 x where x is in meter and ‘t’ is in second. The dielectric constant of the medium is Key: 4 9 108 Sol: 1.5 108 ms 1 k 6 c 3 108 Refractive index n 2 1.5 108 Also n r r , For a non-magnetic medium r 1 Therefore n r r n 2 2 2 4 Dielectric constant k r 4 10. A soap bubble of radius r is blown up to form a bubble of radius 2r under isothermal conditions. If is the surface tension of soap solution, the energy spent in doing so is 4 x r 2. Find ‘x’. Key: 6 Sol: Surface area of bubble of radius r 4 r 2. Surface are of bubble of radius 2 r 4 2 r 16 r 2. Therefore, increase in surface area 16 r 2 4 r 2 12 r 2. since a 2 bubble has two surface, the total increase in surface are 24 r 2 Energy spent = work done = 24 r 2 11. A mass of 20kg is hanging with support of two strings of same linear mass density. Now pulses are generated in both strings at same time near the joint at mass. If ratio of time, a taken by a pulse travel through string 1 to that taken by pulse on string 2 is then fill b c the value of a b c g 10 m / s 2 Key: 14 Sol: JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 42 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 4T1 5T2 3 T1 T2 5 5 4 T v 3 4 v 1 t1 10 8m v2 4 5 3 t1 l1 / v1 8 4 3 t 2 10 6m 5 t 2 l2 / v 2 6 2 3 3 3 12. A rod made of glass, refractive index and of square across-section, is bent into the 2 shape shown in figure. A parallel beam of light falls normally on the plane flat surface A. Referring to the diagram d is the width of a side and R is the radius of inner semicircle. Find the maximum value of ratio d/R so that all light entering the glass through surface A emerges from the glass through surface B. Fill 10(d/R). Key: 5 Sol: Figure shows three rays 1, 2, 3 incident on plane face A. We can see that angle of incidence at curved surface is least for ray 3. If ray 3 reflects at the curved surface, then all the rays will reflect as their angle of incidence is greater than angle 3 Hence the required condition is 3 critical so sin 3 sin critical R 2 d 1 d 1 or dR 3 R 2 R max imum 2 13. A plano convex lens has thickness 4 cm. When it is placed on a horizontal table with curved surface in contact, the apparent depth of the lens is found to be 3cm and if lens is inverted its apparent depth is found to be 25/8 cm. Find the magnitude radius of curvature of the lens (in cm). JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Page 43 SRI CHAITANYA IIT ACADEMY, INDIA JEE MAIN 2025 BRAHMANDA PRACTICE MATERIAL-1 Key: 25 Sol: 4cm 4 App depth 3cm = 3 We use refraction formula for apparent depth as 4 24 1 ; u 4cm; R R 2 1; v cm 3 8 4 1 2 1 2 1 8 4 3 1 1 8 1 v u R 25 3 4 R 3R 3 25 75 R 25cm 14. A satellite is revolving around earth in a circular orbit of radius 4R. Due to ejection of a rocket the speed of satellite is reduced to half as a result of which satellite starts falling toward earth. Find the angle (in degree) at which it will hit the earth surface Key: 37 or 53 Sol: With respect to earth, we conserve energy and angular momentum of satellite as by v angular momentum m 4R m v1 cos R 2 GMe 2 V1 cos .........1 4R 1 v GMem 1 2
