JEE Main 2024 4th April Morning Shift 1 PDF

JEE | NEET | Class 8 - 10 Download eSaral App  −   ( ) =  − −    ()   − −  ...

JEE | NEET | Class 8 - 10 Download eSaral App  −   ( ) =  − −    ()   − −     = =  −     −  −    = + =     = = 7 12 5 = 12 + + 7 14 = + + = = + +    +    ( ) ( ) − + = = = JEE-Main-04-04-2024 Morning Shift www.esaral.com 1  JEE | NEET | Class 8 - 10 Download eSaral App   +    = +   = +  = + +  = + +  = + +  =      +  +              +  =  −              =  −     = = =   + − − + = = =   = − +  +     − + = ( + ) −( + )  ( ) −( ) → + +  JEE-Main-04-04-2024 Morning Shift www.esaral.com 2  JEE | NEET | Class 8 - 10 Download eSaral App + − + → + − +    −  + − +   = + − +    −    −    =    = − − − = = = = =  −   − +  +     −     −   − −  − −   − − +  − −         −    −   Shift JEE-Main-04-04-2024 Morning   www.esaral.com 3  JEE | NEET | Class 8 - 10 Download eSaral App − −  − −       −   − − +  − + − − −  − +        =  +          = −         =  −  − −  − − −  +  −  =  − +  +  +  −  = − +  +  +  −  = − + + = + = + =    −  =     − =    = −     JEE-Main-04-04-2024 Morning Shift www.esaral.com 4  JEE | NEET | Class 8 - 10 Download eSaral App    =  ( )= +  =  = = − + ( )= + +             = =  −    ( ) =   =     −   JEE-Main-04-04-2024 Morning Shift www.esaral.com 5  JEE | NEET | Class 8 - 10 Download eSaral App  β   β =. → →     () + =    + = − − + + = + =     ( + )  + = + + + −   − = =     JEE-Main-04-04-2024 Morning Shift www.esaral.com 6  JEE | NEET | Class 8 - 10 Download eSaral App − =  = − = −  − = −  = − = − ( − ) − − (− ) − =− − − − =− − ( −  − − ) + −   =− − + − − = − − − −  − = + − = + −    = =        − + + − ++ + = − − + − + − + − − + − + −       = + − = − = + + JEE-Main-04-04-2024 Morning Shift www.esaral.com 7  JEE | NEET | Class 8 - 10 Download eSaral App  − =    +  =      +  =   + =   − =  +  −   − = − = − = −  − − = = +  = = JEE-Main-04-04-2024 Morning Shift www.esaral.com 8  JEE | NEET | Class 8 - 10 Download eSaral App + − − − − = = = = − −     − − + + + = = =  =  =  +  +  = − + − + − = − − ( + + + + ) − ( + + ) = ( )=   − + + = + + + + = ( )+( + + + )= + ( + )( + + + + +  =  =  − =−       ( + )− (+ )  + − =  + − =  − − + =  − − + − + =  − − − − − =  − ( − − )= = − − =  = JEE-Main-04-04-2024 Morning Shift www.esaral.com 9  JEE | NEET | Class 8 - 10 Download eSaral App + − = +   =  −  ( + + )−   +   = − + + −       +   +   +   +  = −   +   + −                       −− + + −    + =− + + + + + + − + +  − + + − −   + =− + + + + − − + = − − + + + + − − + = + − +   JEE-Main-04-04-2024 Morning Shift www.esaral.com 10  JEE | NEET | Class 8 - 10 Download eSaral App PHYSICS 1. A metallic wire of uniform mass density having mass M and length l is bent to form a semicircle. A point mass m is kept at the centre of the semicircle. Find the gravitational forced experienced by m. 2GMm Ans. L2 L Sol. r=  Gdm dg = sin  r2 GM = rd sin  r2 L  G G M = =. g r L  sin d 0 GM g= (2) rL F = mg GM = m2 rL 2GMm  = L L 2GMm = L2 2. 5 convex lens are kept together each having power of 25 D. Find the focal length. Ans. 0.8 cm Sol. Peq = P  5 = 25 × 5 = 125D 1 = 125 m feq 100 = cm 125 4 = cm 5 = 0.8 cm 3. Position of a particle is related to time as given equation x = t4 + 6t2 + 2t Find its acceleration at t = 5 sec. Ans. 480 m/s2 dx Sol. V= dt V = 4t 3 + 18t 2 + 2 dV a= dt = 12t2 + 36 t At t = 5 sec a = 12 × 25 + 36 × 5 = 300 + 180 = 480 m/s2 JEE-Main-04-04-2024 Morning Shift www.esaral.com 11  JEE | NEET | Class 8 - 10 Download eSaral App 4. A body moving with constant acceleration covers 102.5 m in nth second of its motion and covers 115.0 m in (n + 2)th second then find its acceleration. Ans. 6.25 m/s2 Sol. Let, acceleration = a (constant) a Snth = u + [2n – 1] = 102.5 …(i) 2 a S(n+ 2)th = u + [2(n + 2) – 1] = 115 2 a  u + [2n + 3] = 115 …(ii) 2 by using (i) and (ii) a a 102.5 – [2n – 1] + [2n + 3] = 115 2 2 a 3a  102.5 + + = 115 2 2  2a = 115 – 102.5 12.5 a= = 6.25m / s2 2 5. A particles of mass m dropped from height h above the ground. After collision, rises to height h/2, Then loss in energy during collision and speed of particle just before collision respectively are. (1) 50%, 2gh (2) 40%, 2gh (3) 50%, gh (4) 40%, gh Ans. (1) h h Sol. E = mg – mgh = –mg 2 2 i.e. 50% loss in energy v= 2gh 6. If the electric field vector at a point in an electromagnetic wave is given by  z E = 40cos   t −  ˆi then corresponding B will be:  c  z Sol. E = 40cos   t –  ˆi  c  z | E |= 40cos   t –   c |E | =C |B| 40  z | B |= cos   t –  ; also E.B = 0 C  C 7. Infinite charge sheet in xy plane of surface charge density  and infinite long wire of linear charge density  placed at (0, 0, 4) and  = 2. Then net electric field (0, 0, 2).   2r − 1  Ans. Enet    N/C 0  2r  Sol. Given :  = 2 JEE-Main-04-04-2024 Morning Shift www.esaral.com 12  JEE | NEET | Class 8 - 10 Download eSaral App  2K Enet = − 20 r 2 2 Enet = − 20 40r 2 2 Enet = = − 20 40r   2r − 1     N/C 0  2r  8. A hollow cylinder and solid sphere of same mass and radius are rolling with same initial velocity v on a rough inclined plane. Find the ratios of their kinetic energies and maximum height reached by them. 10 Ans. 7 2 1 1 1 1 V Sol. Kcylinder = MV2 + Icm2 = MV2 + (MR2 )   2 2 2 2 R  = MV2 1 1 Ksphere = Icm2 + MV 2 2 2 2 1 2 2  V  1 2 =  MR    + MV 25 R   2 1 1 = MV2 + MV2 5 2 7 = MV2 10 Kcylinder 10  = ksphere 7 At top point kinetic energy will convert into potential energy Mghcylinder 10 = Mghsphere 7 hcylinder 10  = hsphere 7  2nt   2x  9. In given equation y = 2A sin   cos  . Find the dimension of n.       Ans. [n] = [L1T–1] [2nt] Sol. [n]  + M0L0 T0 [ ] [n][T 1 ] = M0L0 T0 1 [L ] [n] = [L1T–1] JEE-Main-04-04-2024 Morning Shift www.esaral.com 13  JEE | NEET | Class 8 - 10 Download eSaral App 10. When a conducting platinum wire is placed in ice, its resistance is 8 and when placed in steam it is 10. Find the resistance of wire at 400°C. Ans. 8.8 Sol. RT = R0 (1 + T) R0 at 0°  8 RT at 100°C → 10 10 = 8(1 + (100)) 10 = 1 + 100 8  10  1  − 1  = 8  100 2 1 =  8 100 1 = 400 R at 40° R = R0 (1 + T)  1  = 8 1+  40   400   1  = 8 1+   10  11  8 = 10 R = 8.8 v u 11. Fractional error in image distance and object distance are and then find the v u fractional error in focal length of the given spherical mirror. df uv  dv du  Ans.  =  +  f u + v  v 2 u2  1 1 1 Sol. = + f v u 1 u+ v = f uv uv f= u+ v 1 dv du − df = − − 2 2 f v u2 df  1 dv 1 du   = f +  f v v u u  df uv  dv du   =  +  f u + v  v 2 u2  12. Instantaneous current in a circuit is zero. In which of the options voltage will be maximum. (a) L (b) C (c) R (d) LC (1) ABD (2) B (3) BC (4) D JEE-Main-04-04-2024 Morning Shift www.esaral.com 14  JEE | NEET | Class 8 - 10 Download eSaral App Ans. (1) Sol. Phase difference between current and voltage is 90°. So, possible circuit are (A), (B) and (D). 13. x and y coordinates of a body performing some motion is given as: x = 3 + 4t y = 3t2 + 4t Identify the trajectory of motion. (1) Parabola (2) Circular (3) Straight line (4) Hyperbola Ans. (1) x−3 Sol. x = 3 + 4t  t = ….(1) 4 y = 3t2 + 4t ….(2) equation (1) in (2) (x − 3)2 (x − 3) y= 3 +4 16 4 3 2  y= (x + 9 − 6x) + (x − 3) 16 1  2 y= 3x + 27 − 18x + 16x − 48 16   1  2 y= 3x − 2x − 21 16    it is quadratic in x  its trajectory is parabola. 14. Choose the correct graph for kinetic energy vs r for an electron revolving around a infinite line of charge. Ans. Theoretical mv 2 Sol. Net force acting towards centre = r F = qE KE  F = e × 2k r mv 2  2k  2ke  = (e)   r  r  mv 2 = 2k  e r  KE = 2ke 15. Pressure vs temperature graph is given for gas of different density. Compare 1, 2 and 3? Ans. 1 > 2 > 3 Sol. PM = RT PM = RT P  T   slope Hence 1 > 2 > 3 JEE-Main-04-04-2024 Morning Shift www.esaral.com 15  JEE | NEET | Class 8 - 10 Download eSaral App 16. Work done to expand the bubble of diameter 7 cm and surface tension 40 dyne/cm is 36960 erg. Find the radius of the expanded bubble? Ans. 14 cm Sol. Surface energy = T (area) Bubble has tw surface of interface Ei = 2TSi Ef = 2TSf  Work done = Ef – Ei  36960 = 2[TSf -TSi]  3690 = TS ×2 36960  S = 40  2   S = 462 cm2 Sf – Si = 462  4 rf2 = 462 + 4 ri2 1  2 7  rf2 = 462 + 4     4  2    1 49 rf2 = [462 + 4  ] 4 4 462  7 49 = + 4  22 4 196 = rf2 = = 49 4 rf = 7 cm diameter = 7 × 2 = 14 cm 17. De-Broglie wavelength of electron moving from n = 4 to n = 3 of a hydrogen is b(a); Where a is bohr radius of the hydrogen atom. Find the value of b. Ans. b=2 hc nh Sol. E= , mvr =  2 h r = = mv n (2)(a0n2 ) ( 1 )n=4 = n (1 )n=4 = (2)(a0n) = 8a0 (2 )n=3 = 6a0  = 1 – 2 = 8a0 – 6a0  = 2a0 Therefore b = 2 18. An elastic string under tension of 3N has a length of ‘a’. If length is ‘b’ then tension is 2N. Find tension when length is (3a – 2b). 5F Ans. K Sol. F = kx 3F 3F = Ka  a = K 2F 2F = Kb  b = K JEE-Main-04-04-2024 Morning Shift www.esaral.com 16  JEE | NEET | Class 8 - 10 Download eSaral App 9F 4F 5F Now, 3a–2b = − = K K K 19. An electron projected inside the solenoid along its axis which carries constant current, then its trajectory would be: Ans. Straight line V B Sol. e– F = q(V  B) B and V are parallel at axis of solenoid so, their cross product will be zero i.e. F = 0 So, electron will move with constant velocity in a straight line.   20. Current as a function of time is given as i = 6 + 56 sin  100t +  A. Find rms value of current.  3 Ans. 8A ( 56)2 Sol. irms = 62 + 2 = 36 + 28 = 64 =8A 21. In Celsius the temperature of a body increases by 40°C. The increasing temperature on Fahrenheit scale is: Ans. 72°F 9 Sol. TF = T + 32 5 c 9 Tf = T 5 c 9  TF =  40 5  TF = 72 F 22. Force on a particle varies linearly with time(t) (F  t). Then select correct acceleration vs time graph. a a (1) (2) t t a a (3) (4) t t Ans.  at JEE-Main-04-04-2024 Morning Shift www.esaral.com 17  JEE | NEET | Class 8 - 10 Download eSaral App F Sol. F = ma  a = m  at 23. Which graph correctly represents the photo current (i) vs stopping potential (Vs) for the same frequency but different intensity? (Here I1 > I2) i i (1) I1 (2) I1 I2 I2 Vs Vs i i I1 I2 (3) (4) I2 I1 Vs Vs Ans. Theoretical 24. A cubical arrangement of 12 resistors each having resistance R is shown. Find  shown in the given circuit. R T R R N R I R R R R B R S R R A R M V0 V0 Ans. i 6R R R 1 1 1 4 Sol. = + = i Req 3R R 3R R R 3R R v/3R R Req = 4  V0 = IRe q 4V0 R R I= 3R So, I1 + I2 = I R R  in parallel combination, current is divided into inverse ratio of resistance V0 JEE-Main-04-04-2024 Morning Shift www.esaral.com 18  JEE | NEET | Class 8 - 10 Download eSaral App 2R I R 2R 2R R I1 I2 2R R I1 = V0/3R R V0 I2 R I I1 I  = I2 3 1 V  I1 + 3I1 = I  I1 = I= 0 4 3R V0 Now, I1 gets divided equally in both branches I1 V0 1 V i= =  i= 0 2 3R 2 6R 25. A wooden block is initially at rest on at rest a smooth surface. Now a horizontal force is applied on the block which increases linearly with time. The acceleration time (a-t) graph for the block would be: F a a (1) (2) t t a a (3) (4) t t k Ans. F= t m Sol. This horizontal force increases linearly with time F t F = kt + c ( F = ma) k c a= t+ m m c k if, =0 tan =  m m then :  F = kt k a= F = t m JEE-Main-04-04-2024 Morning Shift www.esaral.com 19  JEE | NEET | Class 8 - 10 Download eSaral App 26. Find Req ? 15 10 –6V –8V 5 Sol. Below diode is in reverse bias so no current flow through it circuit looks like. 5 Re q = = 2.5 2 5 5 JEE-Main-04-04-2024 Morning Shift www.esaral.com 20  JEE | NEET | Class 8 - 10 Download eSaral App 9 structural isomers are possible of All are octahedral complex JEE-Main-04-04-2024 Morning Shift www.esaral.com 21  JEE | NEET | Class 8 - 10 Download eSaral App ⎯⎯→ H2  ( ) − ( ) = − − =     = =           P=      − − , NO − − → JEE-Main-04-04-2024 Morning Shift www.esaral.com 22  JEE | NEET | Class 8 - 10 Download eSaral App = = = =   = = = + ⎯⎯⎯→  8 =   =  =  JEE-Main-04-04-2024 Morning Shift www.esaral.com 23  JEE | NEET | Class 8 - 10 Download eSaral App JEE-Main-04-04-2024 Morning Shift www.esaral.com 24  JEE | NEET | Class 8 - 10 Download eSaral App ⎯⎯⎯⎯⎯⎯ → + + →  = + + →  =      − ⎯⎯⎯⎯⎯  → CHO OH O OH OH Zn–Hg HCl JEE-Main-04-04-2024 Morning Shift www.esaral.com 25