Online Physics Tutor In Gurugram-IIT JEE PHYSICS PAPER SOLUTION 25 JULY 2022 MORNING SHIFT

Page 1

JEE Main 2022 Question Paper with Solutions for July 25 Morning Session (Shift 1) ✓ € Asf ⑤ * & D= C-0% die%:<c• one
Q61: If momentum [P], area [A] and time [T] are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is : (A) (B) (C) (D) ✓ 7--1<10 ] " / A ]b ft ]f- Mit = IYLT " Mili "=k / Met ' ] ʰ fi ]b / e) & =k fit / a feet " ] /eate ] A- → L2 h=9 , ht2b= M , ate = 1 T T 1+26=-1 21-1-45 'T " 2b = -2 -1+1=-9 b = p 6=0 7- k / b) ' (A) → IT ] °

Which of the following physical quantities have the same dimensions?

(B) Displacement current and electric field

Current density and surface charge density

Electric potential and energy

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Q62:
(A)
(C)
(D)
✓ D= EOE = 40%-1 D= 6.
Q63: A person moved from A to B on a circular path as shown in figure. If the distance travelled by him is 60 m, then the magnitude of displacement would be : (Given Displacement ✓ = R4R<_2R2cos 135 Angle = ?Éadeus= zp2( d- Cos / 357 7- are = 2R2( 1*0-707-5 RT ARE --RG3% ) = 2. R2 (1-707) R = &- Mt = 1.84k = 38.9-4×1.84 = 47Mt
Q64: A body of mass 0.5 kg travels on straight line path with velocity . The net work done by the force during its displacement from x =0 to x = 2m is: (A) 64 J (B) 60 J (C) 120 J (D) 128 J mtsea ✓ WORK done = ( kf Kc ) a- 322+4 Ii =3 ( o) -14 = 4m51 If = 3 (2) 2+4 = 16m51 AW = Im { ( 1632 (4) 2 ] = 1- ✗ (0-5) (16+4) (16-4) = 605
Q65: A solid cylinder and a solid sphere, having same mass M and radius R, roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be : (A) mghi-l-zmo-l-l-IW-mgn-t.in/o---Lmk2Y-I ↑ 2gn=Ñ- ¥02 h✓ D= 21 • + % ↳ 1. cylinder = ( • + ¥ ) sbhere ntsbhere @ +1¥ ) cylinder = ::÷-=¥÷ =F%s
Q66: Three identical particle A, B and C of mass 100 kg each are placed in a straight line with AB = BC = 13 m. The gravitational force on a fourth particle P of the same mass is F, when placed at a distance 13 m from the particle B on the perpendicular bisector of the line AC. The value of F will be approximately: (A) 21 G (B) 100 G (C) 59 G (D) 42 G f- Cosas-0 P ✓ # fastest V2Fcd450 L vf , > f JE f 13Mt r A 14813Mt B 401 c Iookg • twig 13Mt 100kg FNET = 28-66450 1- Fp = -4%7761-7+917 = a ;: ˢ( • + %) = 4,3%-4-1%-7=1004
Q67: A certain amount of gas of volume V at. temperature and pressure expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (Use ) 27°C P , = 2×107 Nlmt ✓ Pill , = Pz V2 Vz=2V Hence B. = P% Pz ( Vz ) ✗ = % ( 2k ) 't 83=74%-5--3.536×106 Pa
Q68: Following statements are given: (1) The average kinetic energy of a gas molecule decreases when the temperature is reduced. (2) The average kinetic energy of a gas molecule increases with increase in pressure at constant temperature. (3) The average kinetic energy of a gas molecule decreases with increases in volume. (4) Pressure of a gas increases with increase in temperature at constant pressure. (5) The volume of gas decreases with increase in temperature. Choose the correct answer from the options given below: (A) (1) and (4) only (B) (1), (2) and (4) only (C) (2) and (4) only (D) (1), (2) and (5) only ✓ (KE ) avg =3 KT 2. P = § S Vrms ④ CORRECT consider constant volume end not constant pressure C. But most appropriate as we ✓ Is option A)
Q69: In figure (A), mass ‘2 m’ is fixed on mass ‘m’ which is attached to two springs of spring constant k. In figure (B), mass ‘m’ is attached to two spring of spring constant ‘k’ and ‘2k’. If mass ‘m’ in (A) and (B) are displaced by distance ‘x’ horizontally and then released, then time period and corresponding to (A) and (B) respectively follow the relation. % 27T ' 3m -2K ✓ F- 27T M2 2K 27T 3m To 2K a= = ÷ 251 M 2k
Q70: A condenser of capacitance is charged steadily from 0 to 5C. Which of the following graph represents correctly the variation of potential difference (V) across it’s plates with respect to the charge (Q) on the condenser? Q = CV✓ V = stringent line V= (f) Q meth slope % slope = 1- = 2×1-1,6=5×105

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Q71: Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of radii of their circular paths is 6 : 5 and their respective masses ratio is 9 : 4. Then, the ratio of their charges will be : (A) 8 : 5 (B) 5 : 4 (C) 5 : 3 (D) 8 : 7
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✓ KE = 1- ME D= + 2 C. KE ) MEDB = Mob → r=m¥• D= JB 2 CKE ) in = ↓ m @ E) 9 B %= ✗ ¥ , = 9- ✗ ÷=¥

Q72: To increase the resonant frequency in series LCR circuit,

(D) The source frequency should be decreased

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(A) Source frequency should be increased
(B) Another resistance should be added in series with the first resistance.
(C) Another capacitor should be added in series with the first capacitor
✓ af- = Fatso to increase f- ↑ , Ltt , 6¥ For capacity to decrease another insert capacitor is in series with the circuit
Q73: A small square loop of wire of side l is placed inside a large square loop of wire . Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is: (A) (B) 4 = B A M # = 41 µ÷ ¥1 , { sin 45 + tna 5 3) × l ≥ " = 1÷(;Éje ✓ = 2. F. hole ? LTT
Q74: The rms value of conduction current in a parallel plate capacitor is 6.9 The capacity of this capacitor, if it is connected to 230 V ac supply with an angular frequency of 600 rad/s, will be : MA ✓ I= VI. = ¥w , , = VCWC ? c. = = 6-9×10-6 230 ✗ 600 = 50 Pf

Q75: Which of the following statement is correct?

(B) In primary rainbow, observer sees violet colour on the top and red on the bottom

(C) In primary rainbow, light wave suffers total internal reflection twice before coming out of water drops

(D) Primary rainbow is less bright than secondary rainbow

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(A) In primary rainbow, observer sees red colour on the top and violet on the bottom
www.kumarneetphysicsclasses.com ✓ i. Red color C.at top ) Violet ( bottom ?
of
saints Is less than primary oath bow
✓ %A = % , =%ˢa= £ Mrs Let the thickness be K % Bga = 5×10-10 x= 5×10-10 ( DAUB →A- Irs ) ÑA=2DB R= 5×10-10×209 = 5×10-1 ◦ VA

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Q77: A metal exposed to light of wavelength 800 nm and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500 nm is used. The work function of the metal is (Take hc = 1230 eV-nm). (A) 1.537 eV (B) 2.46 eV (C) 0.615 eV (D) 1.23 eV
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men 7¥

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Q78: The momentum of an electron revolving in usual meanings) दिली
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Q79: The magnetic moment of an electron (e) revolving in an orbit around nucleus with an orbital angular momentum is given by :
Ratio of magnetic moment ✓ Angular moment a a= Fm for ea = Emi
Q80: In the circuit, the logical value of A = 1 or B = 1 when potential at A or B is 5V and the logical value of A = 0 or B = 0 when potential at A or B is 0 V. Pr AND ✓ GATE

with

the same car is moving with

distance.

km/h

after applying the brake it will move 27 m before it stops.

speed of one third the reported speed then it will stop after

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Q81: A car is moving
speed of 150
and
If
a
travelling ________ m
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Q82: Four forces are acting at a point P in equilibrium as shown in figure. The ratio of force to is 1 : x Where x = ______. 3 2 Cosh 5° , ↑ Coste 5° ^ > 15th 450 4 2 Sin 450 2 Cos 450+1 Cos 45° = Fz fz = 3 Cos 45° = ¥ N 2 51h45 ◦ = f , + 1 Sin 450 tf , = I 51h45 ◦ = ↳ N &÷=¥¥j=l_=tg * =3

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Q83: A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force F, its length increases by 5 cm. Another wire of the same material of length 4L and radius 4r is pulled by a force 4F under same conditions. The increase in length of this wire is ______ cm.
IN
5 += % e Y = , ¥ % Be , = != 5cm • e. = 4*19%7=5 'm
Q84: A unit scale is to be prepared whose length does not change with temperature and remains 20 cm, using a bimetallic strip made of brass and iron each of different length. The length of both components would change in such a way that difference between their lengths remains constant. If length of brass is 40 cm and length of iron will be ______cm. . 60.00cm It = to ( at ✗ At ) It do = to (d) ( be ? Al = to (d) ( At ) ✗ BIB = di le di -7%-8×109×40 2/6142×10/-5 = 12% = 60cm
Q85: An observer is riding on a bicycle and moving towards a hill at . He hears a sound from a source at some distance behind him directly as well as after its reflection from the hill. If the original frequency of the sound as emitted by source is 640 Hz and velocity of the sound in air is 320 m/s, the beat frequency between the two sounds heard by observer will be _____ Hz.20 Ña= 320mF " 64011-2 SOURCE of SOUND f-reflected = t farce e=(%÷s)f = (33%5)×640 = (332%5)×640=6301-12 = 650 Hz f.BE#e-- 650-630=201+2
Q86: The volume charge density of a sphere of radius 6 m is . The number of lines of force per unit surface area coming out from the surface of the sphere is ______ . [Given: Permittivity of vacuum ] 45 ✗ 1010% 01 = EA lo f- = :& % ☒ = E ↳ number # lines 6 (4-3%31) of force Mtsu = ¥Ñ ☒area = ELECTRIC FIELD = :& = 2×6 3×8.85×10--12 = 045 × 1012 NIG = 45×1010 %
Q87: In the given figure, the value of V0 will be ____ V.8 2 V → time =;÷nÉix÷ It Izt Is = 0 :# + Y÷u+Y÷ 4 V 3 V0 12=0 No = 4 Volt

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Q88: Eight copper wire of length and diameter d are joined in parallel to form a single composite conductor of resistance R. If a single copper wire of length 2L have the same resistance (R) then its diameter will be _________ d.
ffalo
4 R=÷ " ? Eight copper wire in parallel Reg =%= # (4) 871dL = £¥d2 = G- ( 4.) -114,2 ¥ñ=•¥¥ i. Resistance dF= 16dL ofsenple were of d) =4d length #
Q89: The energy band gap of semiconducting material to produce violet (wavelength ) 4000 A LED is _____ eV. (Round off to the nearest integer)3 105 Eg=¥=÷m,=%÷ =3 105 ev
Q90: The required height of a TV tower which can cover the population of 6.03 lakh is h. If the average population density is 100 per square km and the radius of earth is 6400 km, then the value of h will be ______ m.II. 0 d = 2 R h = 2 ✗ 6 4 00 ✗ h ✗ l Area = IT d 2 = (TT ✗ 2 ✗ 64 oo ✗ h ✗ 10-3 ) km = 6- 03 ✗ 100000 = I 00 Tl ✗ 2 ✗ 6 4 00 × 10-3 9 6 03 ✗ 105h = 10 ✗ A ✗I 28 h = I 50Mt
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