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Answers & Solutions for 26 JULY EVENING SHIFT JEE (Main)-2022 (Online) Phase-2 (Physics) ✓R = U2 51ns O g B = Bo Sin 2%(01--2) FK NITA 3 M • = ¥gis=TR% 7=29 K

SECTION - A Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct. 1. Two projectiles are thrown with same initial velocity making an angle of 45° and 30° with the horizontal respectively. The ratio of their respective ranges will be E- = # , ✓ ¥;o , = " n - o , Sin 2 @ L = 51h 2 ( 45 ) sin 2 ( 30° ) = 51h90 Sin 60 = ¥ , = %}

2. In a Vernier Calipers, 10 divisions of Vernier scale is equal to the 9 divisions of main scale. When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and 4th Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to 1 mm. While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and 6th Vernier scale division exactly coincides with the main scale reading. The diameter of the spherical body will be A) 3.02 cm (B) 3.06 cm (C) 3.10 cm (D) 3.20 cm✓ to V5 D= 9 MSD L C = |-O TT SD = 0.01 cm Negative error = (01 • ◦ 4) am = i 06cm Reading = 3.0cm -16 Go 1) cm -1.06cm = 3.12 Cnn I 3- 10cm

3. A ball of mass 0.15 kg hits the wall with its initial speed of 12 ms–1 and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is 100 N, calculate the time duration of the contact of ball with the wall. (A) 0.018 s (B) 0.036 s (C) 0.009 s (D) 0.072 s ✓ , ,wael m I → U / ve + re U I m I f. At m ( Vf Vi ? / If. At = m C- U U ? / 3 At = -2hfˢ = I ; ¥ m = 21 × 0*12-6 1%-00×10%-10 = 036sec

4) A body of mass 8kg and another of mass 2kg are moving with equal kinetic energy. The ratio of their respective momenta will be A) 1:1. (C) 1:4 (B) 2:1 (D) 4:1 kE= { mvkfm-Y.tn✓ = ≤ m¥=±Fn ( KE) , = ( KE ) ≥ * Em , %÷ !÷=÷ ÷=1÷=F? ÷= }

¥5 E.

5) Two uniformly charged spherical conductors A and B of radii 5mm and 10mm are separated by a distance of 2 cm. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitude of the electric ﬁelds at the surface of the sphere A and B will be (A) 1:2 (B) 2:1 (C) 1:1 (D) 1:4

V , = V2

ear . e. r ,

Bg= Bo Sin 2¥ (It a) = 5×10-6 Tesla Compare with D= Ee Bf= 5×10-6 5in (50009k Bo 4000×1081-11-7 f. ◦ =D Bo 500074=2%4 ✗ = ÷wmt =4¥¥×§¥ 4000×108*4=2*01-1 = 54-50×102 0=4000×108×1*0 = 2 × 102

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7. Light travels in two media M1 and M2 with speeds 1.5 × 108 ms–1 and 2.0 × 108 ms–1 respectively. The critical angle between them is:

☆ •8 ✓ µ , sink = lls 51h90 since UI = 0%9--1 , µ , 9%17 , = !- = " 5×1081 2×1081 = ÷ < = sin " %) or • = rai ' ,¥

8. A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be: (Take radius of earth = 6400 km and g = 10 ms–2) (A) 800km (B) 1600km (C) 2133 km (D) 4800 km Abbey conservation of energy • B between point A and taint Bµ( TE ) A = ( TE ) @ ( KE) At E)a = ( KE )B 1- ( PE).rs A 7 t.int#-E5-I.m < = 9Mm Rth am=*q4 ¥ ˢ÷ -4*7%-7 ↳ = ¥ n h=%=%ᵈ= ooo km

9) The maximum and minimum voltage of an amplitude modulated signal are 60 V and 20 V respectively. The percentage modulation index will be: A) 0.5% (C) 2% (B) 50% (D) 30% % modulation man men man mln (60-20)×100 60 1- 20 = 50%

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10) A nucleus of mass M at rest splits into two parts having masses M'/3and 2M'/3 (M'<M).The ratio of de Broglie wavelength of two parts will be: A) 1 : 2 (B) 2 : 1 (C) 1 : 1 (D) 2 : 3

Linear momentum is conserved ✓ 0 = ᵗ¥ I , 2¥ ' V2 1¥ I , = 2¥ V2 ] since ✗ • = ¥ ¥: "(+1%4) ✗ z = ↑A-

11. A nice cube of dimensions 60cm×50cm×20cm is placed in an insulation box of wall thickness 1 cm. The box keeping the ice cube at 0°C of temperature is brought to a room of temperature 40°C. The rate of melting of ice is approximately. (Latent heat of fusion of ice is 3.4 × 105 J kg–1 and thermal conducting of insulation wall is 0.05 Wm–1°C–1) (A) 61 × 10–3 kg s–1 (B) 61 × 10–5 kg s–1 (C) 208 kg s–1 (D) 30 × 10–5 kg s–1 ✗ $5 + →• ✓ × Mi ✗ ima DO = KA ( % %) e. Mad = KA ( Ta E) e. h = K ACT. -72 > = • 05 ✗ 2 (06×0.5-105×0 -2-10-2×06 ) L ( l ) 3.4 × 105 × 1 × 152 = 61×10-5 kglseo

12. A gas has n degrees of freedom. The ratio of speciﬁc heat of gas at constant volume to the speciﬁc heat of gas at constant pressure will be Cv = NRI ep = n¥ +R = R (1-+1) ÷=¥÷=¥

13. A transverse wave is represented by y = 2sin(ωt – kx) cm. The value of wavelength (in cm) for which the wave velocity becomes equal to the maximum particle velocity, will be A) 4π (C) π (B) 2π (D) 2 ✓ 8=2 5in ( wt KR) compare Mtn eF- a sin ( ¥ not a.) 2¥vt=wt v=w¥ -10 Differential g f- w.o.tt ) > 4- = 2 ( w) cos ( wt Kae] ↳ maximum velocity ② EQUATE ① &② 1×7=2 @X ANSWER✗ = 41T

14) A battery of 6 V is connected to the circuit as shown below. The current I drawn from the battery is → Thts Is balanced wheat stone ✓ Bridge 3. = % ¥r gw Hence 5- ohm is openmÉÉ \ circuited 12h ⇐ 6Th How 6 v 2h The GVᵗ=÷i¥=¥¥-;y= / Amb

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capacitors.

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total energy stored

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is connected to the combination of two identical capacitors as shown in

combination

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15) A source of potential difference V is connected to the combination of two identical capacitors as shown in the ﬁgure. When

‘K’ is closed, the total

stored across the combination is E1. Now

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is E . Now

is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now E . The ratio will be :E / E You

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in HL and SL physics CASE I E , = § (2C) V2 CASE -2 F< = § ( Kc) V24 ¥ ( KE) ( ¥ ) ≥ = IV. Ike -1k¥ ] = ± v2 ( Kc + & ]✓ = { < v2 / 5-1%-1--1=6 " (¥ ) ¥643K" Yea = ☒µ(zgj= " ¥513 = %

Q16: Two concentric circular loops of radii and are placed in x-y plane as shown in the figure. A current is flowing through them in the direction as shown in figure. The net magnetic moment of this system of two circular loops is approximately: Ma = I -1T ( O 5) 2 ( E) M2 = I TT (0^3) _ ( É ) IF = TT , + Ma = * I (¥oo %) É ✓ = Ey CI ) ( Foo ) É IT = 3. 52 É Am ? = § I Ampmt -2

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physics E- = F- k^ Ñ = By " 13=12× 10-37 F- = 728×1-6×18 '9J=§m = 2×728×1 -6×10-17 9. 1×10-31 = 16×106 mtlsec 9VB= ¢ E E=DB= 12×153×16×106 = 192×10 > Mm = 192kV/m

Mig F- Mia / / / T Mzg=Mzh ADI" Mm , f- a gem , my _- alms -1ms ✓ me ↓ my t.mg a=8%÷h% CASE I CASE -2 h= 2m18 MiG hz= 3mg Mig • 3mi / 4Mt h , = % 92=8/2 he ñ=%÷=%

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Q19: Mass numbers of two nuclei are in the ratio of 4: 3. Their nuclear densities will be in the ratio of

R= Ro A- % mass of ✓ Density of nucleus = nucleus volume of nucleus g=%→É→ = m(AX ¥7 Ros # = {7-2} f. → Independent of mass number

Q20: The area of cross section of the rope used to lift a load by a crane is . The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross section of the rope should be : (take 8=10 MIZ ✓ MAX LAFFIN 4 BREAKING STRESS = CAPACITY Area of cross-section of rope I 2-5×10-4 = 25 AA- = 625×10-6 Mt = 6-25×10-4 m2

SECTION - B 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. Q21: If and . The magnitude of component of vector A along vector B will be ______ zi -13J I ↑-12J -12kt 2 II. B- = AB Los ② component of A- along B = A→ É = A→ BTo = A→;%- = (2) (e) + (3) G) + c- 1) (2) ¥ +2T = 21+63-21--63 = 2

Q22: The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through the center will be ______. Given, the length of the rod is5 1055M XIE = ☒ K ≥ → k= ¥ L = = 1%5-1=5

Q23: In the given ﬁgure, the face AC of the equilateral prism is immersed in a liquid of refractive index 'n'. For incident angle at the side AC, the refracted light beam just grazes along face AC. The refractive index of the liquid . The value of x is ______. (Given refractive index of glass =1.5 )% \ Abbey snail 's Law at surface AC 1. 5 51h60 = U 51h90 A" 5 (E) =3÷=¥. E. 4 2=27

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4. Two lighter nuclei combine to from a comparatively heavier nucleus by the relation given below: 2 X + 2 X = 4Y The binding energies per nucleon for 12 X and 24Y are 1.1 MeV and 7.6 MeV respectively. The energy released in the process is ______ MeV

2*2 •i •i BAT $41 26 4 * + i' = ¥ Energy released = change in BE = (76×4) (4×1^1) = 26 MeV

25. A uniform heavy rod of mass 20 kg, cross sectional area 0.4 m2 and length 20 m is hanging from a ﬁxed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is x×10–9 m.The value of x is _____ (Given Young's modulus Y = 2 × 1011 Nm–2 and g = 10 ms–2) z , mama 25 + ☒ _u2 4= F/A TAVG ( L ) ALIL = AY = MGL 20×10×20 ZAY = 2×04×2×1071 = 4×103×18 " 4 × 04 = 2^5×10-8 = 25×10-9=2×10-9 2=25

26. The typical transfer characteristics of a transistor in CE conﬁguration is shown in ﬁgure. A load resistor of 2 kΩ is connected in the collector branch of the circuit used. The input resistance of the transistor is 0.50 kΩ. The voltage gain of the transistor is ________. W gain = current gain ✗ ¥ ' =L :÷ .)( E.) 5×10-3 = @ ✗ a- g) ( 2×103 05×153 ) = ' ° ) = 200

27) Three point charges of magnitude 5 μC, 0.16 μC and 0.3 μC are located at the vertices A, B, C of a right angled triangle whose sides are AB = 3 cm,BC= 3 2 cm and CA=3cm and point A Is the right angle corner. Charge at point A, experiences ______N of electrostatic force due to the other two charges. 1IT C 03h6 35cm 3cm FAB A B 016ha g- µ , Fao 3cm FN£e= (F*B ) 't ( Fao ) " (5×03)×10-12FAC = 4T¥ 9×10-4 = 9×109×10-12 9×10-9 (1.55+0.0)-3 FAB = z , (5×016)×10-12 = 17N 9 × 10-4

value of total heat produced in the coil, till the

becomes zero, will be ______ J.

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28. In a coil of resistance 8 Ω, the magnetic ﬂux due to an external magnetic ﬁeld varies with time as φ = 3 9 t . The

ﬂux

qEBM-t-E.IR = 8- 0hm ☆ = } ( 9- C- 2.) 14--1%-1=31%-69-+4 ] =/ ¥ for at ] = 4¥ let = 41 3. µ .gg, , , .gg, , , , , ≥ = 3,1¥ ] ? = 3,1%3=4

Q29: A potentiometer wire of length 300 cm is connected in series with a resistance 780 Ω and a standard cell of emf 4V. A constant current ﬂows through potentiometer wire. The length of the null point for cell of emf 20 mV is found to be 60 cm. The resistance of the potentiometer wire is____ Ω20 R Resistance of potentiometer were i = ¥780 300cm is having a resistance R 60 cm ✗ , c. , = RX 60 300 Under Balance condition 20×10-3 = > go) ✗ 8×60 (300 ) ¥-- = 2-48%40 ☒Rto ) UoR=R -1780 39k = 780 R=}%- = 20 -0hm

Q30: As per given figures, two springs of spring constants K and 2K are connected to mass m. If the period of oscillation in figure (a) is 3s, then the period of oscillation in figure (b) will be . The value of x is _____. 2.00 ÷ 2 BE CASE -1 keg = (K ) ( 2K ) = 2¥ 3K F- ZTT Jg= 27%-2 CASE -2 keg = 3K E- 27T¥ 2. 00 Tp E- F¥m¥=É ¥ = ¥ 72=5 = Foe 2=2^0

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After connection the potential will remain cons rent

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