JEE (Main)-2022 (Online) JEE Main July 25th 2022 Shift 2 Physics Question Paper ✓ 2 = N • Cb , 1- Nz Cop , Ni Cv , 1- NL ( V2 a = V date • = :-B 1 main twmusFin this paper
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modulation index is
1. In AM modulation, a signal is modulated on a carrier wave such that maximum and minimum amplitudes are found to be 6 V and 2 V respectively. The
Amar = 6 Voet Amin = 2 Volt µ = Amar Amin Amaro 1- Amen = %÷z = 05 µ = 50%
2. The electric current in a circular coil of 2 turns produces a magnetic induction B1 at its centre. The coil is unwound and is rewound into a circular coil of 5 turns and the same current produces a magnetic induction B2 at its centre. The ratio of B2 B1 is (A) 5 (B) 25 (C) 5 (D) 25 µ Bz a/ • • 12 14 14 Be = tf÷ I. 2629A 12 8¥ 1=2 ( 29th ) Betty , I 5 ( 29k ) = 5 ( 29827 ri = M¥,*%¥ %≠*% =É×¥ =÷÷%÷=ˢ×¥=÷
3) A drop of liquid of density ρ is floating half immersed in a liquid of density σ and surface tension 7.5 × 10–4 N cm–1. The radius of drop in cm will be (g = 10 ms–2)* y UPTHRUST % TIR > • g. z☒Rt+§HRˢs8 = } -HRˢog 271-4-3 Neg = 3- R' 6g t.it t.it , ↓ ZTTRT * = 3% (6-25) × 10 3×75×154×102 £ TIR > Pg R = %IÑ= (6-25) × 10 = I %: *a R=%÷
6. Capacitance of an isolated conducting sphere of radius R1 becomes n times when it is enclosed by a concentric conducting sphere of radius R2 connected to earth.& . • : Itiitiallg 60=4-112-0 Re ✓ finally 4TÑRR=nco Ra Ri = Hon r=n R-p=n a- %÷=n ¥=n ¥=÷ ,
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7) The ratio of wavelengths of proton and deuteron accelerated by potential Vp and Vd is 1 : the ratio of Vp to Vd will be:
. ev = % met ✓ V= 2¥ ✗ = E. = = ÷ ÷= ¥ 20¥ = % ÷ = ;
8. For an object placed at a distance 2.4 m from a lens, a sharp focused image is observed on a screen placed at a distance 12 cm form the lens. A glass plate of refractive index 1.5 and thickness 1 cm is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen? A) 0.8 m (C) 1.2 m (B) 3.2 m (D) 5.6 m shift glass slab second D= 1- ( r ʰµ)= • ✗ ( a- ¥ )✓ case / = Is Cm so final image must be too deuced • at ( ez f) cm= 3¥ cm from c- 2.4Mt → the lens so that glass plate 12cm Must 5 heft to trochee image at screen so ¥ ◦ = ¥ = ¥3 te Te = 355 % ◦ U = 560cm Sheff = 5-6-24 = 3- 2Mt
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← ✓ c. = 1¥ 540 Bo = ¥ = 3 × 108 = 18×10-7-1
10.
you walk through
phenomenon works on:
(A) Electromagnetic induction
(B) Resonance in ac circuits
metal
(C) Mutual induction in ac circuits
(D) Interference of electromagnetic waves
metal object in your pocket,
raises an alarm.
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¥ works on principle of Resonance
In A. C. CIRCUIT
When
a
detector carrying a
it
This
1
11. An electron with energy 0.1 keV moves at right angle to the earth’s magnetic field of 1 × 10–4 Wbm–2. The frequency of revolution of the electron will be Take mass of electron = 9.0 × 10–31 kg) (A) 1.6×105 Hz (C) 2.8×106 Hz (B) 5.6×105 Hz (D) 1.8×106 Hz https://kumarphysicsclasses.com/ Physics Tutor in Ahmedabad Cantt Physics Tutor in Central Ahmedabad Physics Tutor in East Ahmedabad Physics Tutor in North Ahmedabad Physics Tutor in South Ahmedabad Physics Tutor in West Ahmedabad -4g 31 ✗ naan mjᵈ=9vB5 • ✓ 6*4 D= MB 5%5 6Mt D= 2%1=298 f 1 5- = ¥r=¥µ% f- = _m= 154×1.6×10-19 29×9×10-31 = 2-8 ✗ 10GHz
12 ) A current of 15 mA flows in the circuit as shown in figure. The value of potential difference between the points A and B will be I , = I ✗ 10 -15 = 15M¥ = 10mA ^ I I = I¥5= 15,4=5 > In = 5mA I= UAB = 5×15-110×5 +10×15 = 75-150-1150 ✓ = 275 Volt
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13. The length of a seconds pendulum at a height h = 2R from earth surface will be (Given R = Radius of earth and acceleration due to gravity at the surface of earth, g = π2 ms–2)
GIGIat surface of the earth' (3RD " ✓ 7=24 lg , & * " % L T '= 27T¥ it ' Fsecond pendulum length I = 8%4 , e- e'= > sea = 'ⁿᵗ 8 'l=ge ' F- (e) =8XeY é=%=%mt
14. Sound travels in a mixture of two moles of helium and n moles of hydrogen. If rms speed of gas molecules in the mixture is 2 times the speed of sound, then the value of n will be A) 1. (B) 2 (C) 3 (D) 4✓ 2×4+2×1 Molar mass = ztn V= N ' Cb , -1N , 6ps NICV , 1- NL ( V2 given that = 2. ( SR ) -1N ( TR ) Urms = JI V sound 2 ( SR) -1h ( SR ) 3kt z.ro/Rp D= 10 -17N 6 -15h 8=3/2 , 2- = 10-171 6 -15h 18+1521=20 -114h 2=2
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Vadodara Ma = d- 420 720 = 720-420 > 20 = 30C 720 1--1-7%-0 = 1220-320 900 1220 = I 220 720 ✗ / 220M¥ = 300 900 M¥ = 056
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16 ) An object is taken to a height above the surface of earth at a distance 54 R from the centre of the earth.Where radius of earth, R = 6400 km. The percentage decrease in the weight of the object will be
/ It ¥=E,R mg=¥÷ Mg : 41 = GI ( Rtn) (Ear ) " mg ' → = ÷ mg '=¥.mg tool:( 1- mImg)=(- E) ✗ too I. decrease = 36J in weight
17. A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of 200 g travelling with speed 10 ms–1 gets embedded in it, then loss of kinetic energy will be A) 4.9 J (B) 9.8 J (C) 14.7 J (D) 19.6 J ✓ Mi LV ) = ( M , 1- MDV ' V' = M , ( V ) ( mi -1mL ) Loss of KE = Ma V 1- Milk 1- ( M , -1ms V ' " ± / mm cm+ms(mm÷m .] } 1- ( V ) 1m1cm -1mi) mi]Cm , -1M ) = 1" ( mml.tt#m)v2=l- ✗ 98×(0-2) ( logs 10 = -9.8J
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18. A ball is projected from the ground with a speed 15 ms–1 at an angle θ with horizontal so that its range and maximum height are equal, then ‘tan θ’ will be equal to
R= It ✓ * 51h20 = ↳ 1h20 18 I 251*050 = 51nF 1- ano=4
error
and time for which current
in
electrical circuit are
2%
respectively. The maximum percentage error in the detection of the dissipated heat will be
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19. The maximum
in the measurement of resistance, current
flows
an
1%,
and 3%
H = I < Rt ✓ ↑¥- ✗ too % = 2 B¥ ✗ too % + ARI ✗ too % -1b¥ ✗ too % = 2×2 % + I % + 3 % = 8%
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20. Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength λ. The value of principal quantum number ‘n’ of the excited state will be, (R: Rydberg constant)
✓ ÷=r( ÷ ÷. ) ⇐ = ' ÷ ÷ = a- ÷ , ÷=×÷ a- ÷ .
Numerical Value Type Questions: This section contains 10 questions. In Section B, attempt any five questions out of 10. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. 1. A particle is moving in a straight line such that its velocity is increasing at 5 ms–1 per meter. The acceleration of the particle is _______ms–2 at a point where its velocity is 20 ms–1. SECTION B Ge = 5M 5 Ym Acceleration of particle When V = 20M 5 " oh = V. d÷• = @ ☐ (5) = 100m52
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2) Three identical spheres each of mass M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 3 m each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be x m.The value of x is_______
5 2 14 ..É¥:-. MEI Sin 450 = Ig K= 3 Sin 450 & 450 ( pg 3Mt a- = MIO.jp?MC&)-=ZzxÑ = } In 45°) / = # = I = Fe , 2=2
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3. A block of ice of mass 120 g at temperature 0°C is putin300gofwaterat25°C.Thexgoficemelts as the temperature of the water reaches 0°C. The value of x is _______. [Use specific heat capacity of water = 4200 Jkg–1K–1, Latent heat of ice = 3.5 ×105 Jkg–1]
90 Heat lost by water = Heat gained by foe 0-3×4200 ✗ 25 = & (3-5) ✗ 105 ✗ = 03×4200×25 35 ✗ 105 = 90×100 ✗ 105 ✗ 103 = 90 from
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En = 13n ev ÷ 13 ≥ = & set 4 5 9-1 = ¥+49×4×41 2=5
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5. In a potentiometer arrangement, a cell of emf 1.20 V gives a balance point at 36 cm length of wire. This cell is now replaced by another cell of emf 1.80 V. The difference in balancing length of potentiometer wire in above conditions will be _______ cm
18 ¥÷=÷ :¥= :-. l , = 3%18--9-3 12-6/4 = 54 cm Diffence in length = 54 36 = 18cm
6)Two ideal diodes are connected in the network as shown is figure. The equivalent resistance between A and B is ___ Ω. 25 D , Dre D , forward bias B- Reverse bias 0 0 } zow 0 B. If or 15W 20h Inform 25 -0hm 20W 15W
7. Two waves executing simple harmonic motions travelling in the same direction with same amplitude and frequency are superimposed. The resultant amplitude is equal to the 3 times of amplitude of individual motions. The phase difference between the two motions is____ (degree).6JANET = AT + At -12A , A > Cos of Fs A = A- + A2 -12 A- cost 3 # = * ( ✗ + cos d) 3- ✗ = cos of ¥ = Coscf D= 600
8. Two parallel plate capacitors of capacity C and 3C are connected in parallel combination and charged to a potential difference 18 V. The battery is then disconnected and the space between the plates of the capacitor of capacity C is completely filled with a material of dielectric constant 9. The final potential difference across the combination of capacitors will be ____ V.6 -1 ¥ 8. =/ 8C V. %= 54 CV 3C ← 18W 9 , -19<=9 , ' + % " 9k 18W -15401=(96+36) V ' 1- 1<=9 > & -1 V' = 7%1=6 Voet , ' 1- V '
lens of focal length 20 cm is placed in front of a convex mirror with principal axis coinciding each other. The distance between the lens and mirror is 10 cm. A point object is placed on principal axis at a distance of 60 cm from the convex lens. The image formed by combination coincides the object itself. The focal length of the convex mirror is_______cm.
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9. A convex
I 0 1- + te = ¥ ↓ ◦ = Fo U = 30 cm Radius of curvature of mirror = 30 I 0 = 20 Con fmirror = 20-2=10 Can
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10) Magnetic flux (in weber) in a closed circuit of resistance 20 Ω varies with time t(s) as φ = 8t2 –9t + 5. The magnitude of the induced current at t = 0.25 s will be _____ mA.
BG 250 01=81-2-91-+5 • = ¥ = 1161--91 htÉ=O -25sec = / 1660-257-9 / = 5- volt 1- = & = = 0-25 A = 0%57×103=250 MA
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