Answers & Solutions for 26 JULY MORNING SHIFT JEE (Main)-2022 (Online) Phase-2 (Physics) V. aw cos wt cost = ¥ I = R4( ✗ i. Ñ ✓ R= US In 20 ' g F- 2ft ¥-
Find current through cell? I B A- K L K an • } " " Rl Wr MmR }÷=:÷%:::;WHEATSTONE A " 2W • B 4Th %wgw% Yiu en t mug4)Naw 4W Zovolt I= # = 5A
2. Assume all surfaces are friction less. Find value of force required such that 20 kg block moves with acceleration 2 m/s2qq* ii ñi E. ± T 20g = 20 (2) 20g 7=40+200 = 240N f-=p f= (100+10+20) a " = 130h " 130h ' = 240 a " = 2,4% Ms ≥ f= 130 ✗ 2,4% = 240N
3. A charged particle moving in a uniform magnetic field B = 2i 3+j has acceleration a = (αi 4j). The value of α is equal to 73kWh % I. B- = 0 , ( di 4J ) (21^+35)=0 2. (2) -12--0 2 (2) = 12 2=6
4. In S.H.M. v-x graph will be X = A sin wt Ñ= hw cos wt ✓ D= hw 1- Sind wt D= hw a 2%2 = W AL 22 WE w ( a :& ) Ñ=a ʰ w2 abet , -1%1=1 rehearse = a- w " t 1- ÷.+¥÷= . enervation of ellipse
5 ) In an LR circuit if XL = R then power factor is P1. In another LCR series circuit if XL = XC then power factor is P2. Then value of 1 is equal is a & Me BE % , MM ✓ Cos & , =P . Rz = R_ = RR + ✗ L " R2-1 R2 = ÷ ÷ ↳ % = Pa = Rz % =p at resonance cos Q= t ÷=÷
6. A coil of 200 turns and another coil of 400 turns have same length 20 cm. Find ratio of magnetic field at centre. ✓ B=ho_N•¢Ñ•)±=M÷";÷ • 49 H2 B. =¥Hj÷ l=yoo ¥ ) = 400 8D 2 ¥ ¥ "¥¥µ:* a. = ra = 14182 200 = Too :# = 'z
7 )A monkey climbs rope with 4 m/s2 acceleration and when it climbs down his acceleration is 5 m/s2. Weight of monkey is 50 kg and maximum tension is 350 N. Find correct option. (1) T = 700 N, when climbs upwards (2) T = 350 N, when climbs downwards (3) Rope will break when climbs upward (4) Rope will break when climbs downward ✓ at T mg = ma goinggÑwpM = 50 (10+4) down /¥ .mg " ! !! ↑ = 700N t.mg mg 7- ma F- Mlg a) = 50 (10-5) = 250N
✓ f= 2×10-8 sin Ckxtwt -10 ) compare with f. = a sin ( Kae + wt -10) h= 2X 188cm
9) In YDSE experiment fringe width β = 12 cm is given, if the setup is dipped in medium having refractive index μ= find new fringe width413 ✓ B = % when dipped B ' = D¥ P F. = ¥ h • • = He = ¥ , → an
10) With spring at its natural length two blocks are given velocity v = 1 m/s. The maximum extension in the spring is equal to ✓ Sods¥ 2 ( Hmv ) = # kN 2 × 25 ✗ (1) 2 = 200 x2 ñ= ÷g= & a- % Mt -05Mt
_ All capacitors are in parallel Leg = ( 11-2-13 + 4) Mf = 10 µ F 9= Eeg ) ( V7 = doll F) ( 20) = 200 MG
✓ T , = TI U•&gl = Us 51h02 / F I. = :*,=:¥÷=¥÷=⇐
13) The decrease in weight of a rocket when it in 32 km above surface of earth. ✓ 8=8 ( a- ¥ ) g ' g= 1- ¥ 2¥ a- % = 8 g ' g= %mgⁿˢ= ÷! %ˢ=%?→ %ˢ= Foo = •
14. If velocity of electron is x times than neutron and de-Broglie wavelengths are same then find x.he = ✗ n Merle = Mn On oe=Cʰme ) " k= Mm ˢe = 1835
Q15: A source of potential difference V is connected to the combination of two identical capacitors as shown in the figure. When key K is closed, the total energy stored across the combination is E . Now key K 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 a 2 CASE I E , = § (2C) V2 I 2 CASE -2 F< = § ( Kc) V4 § ( KE) ( ¥ ) ? = IV. Ike -1k¥ ] = ± v2 ( Kc -1€ ] = { < v2 / 5-1%-1=1-2 6ñ(¥ ) ✓ EYE , = ¥ = ¥+5,3 § # ( 2% ) = %
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
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=%÷=%
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. In the circuit shown the potential drop across the diode is 60 V then current through diode is ____ mA. 21 I=g µ = 6mA ☒ IF ✗ 103← >6•v_V ↑ ⑤ ^ 60 Voet = 1mA I → current through the Zener MEET diode =
22 )A drop breaks in 729 smaller identical droplets. It T is the surface tension and R is the radius of bigger drop then change in the surface potential energy is nπR2T. The value of n is _____.at F- i = 4 IT R T Ff = 729 ✗ 4982 ( t ) 729 (4-3/983) = 4-3/9 R ? R = -95 8 = Rlq DE = ff- f- i = 729 × 47 ✗ (F) ¥ T 49 RT = 329 R2 T
23. In an EM wave if amplitude of magnetic field component is 2 × 10–8 T then the value amplitude of electric field component is _____ V/m. me F- o = C Bo = 3×1 ✗ 2×10-81 = 6 Volt/mt
In a meter bridge experiment balance point is l1 = 40 cm away from point A. Now if an unknown resistance of x Ω is added to 4 Ω resistance in series then balance point is 80 cm from point A. Then value of x is ______. M case I 4 To = % 0=6 ~ Case -2 4%ˢ = §g 4+x= 24 2=20 oh in
25 )Temperature of 7 moles of a monoatomic gas is raised by 40 K. The change in internal energy of the sample is equal to ______ R. (R is universal gas constant)420 AU = § n R ( AT ) = 3-2×7×40 ✗ R = 42 OR
26 )Find the number of photons coming out per unit time of a source that emits a light of wavelength 900 nm of intensity 100 W/m2 through its surface area of 1 m2. (In multiple of 1019)$2 HEY a = I A Energy of ohe photon = h÷ Number of photon coming out per unit time = 1%1=100×9×10-7 6- 625 ✗ 10-34×3×108 = 45×1019
f- serano ( a- F) Y= a. ( a- %) the 0=1 ,R=1O- 0=45-0 U251gn = 10 , A- = 100 U= 10hr5 "
In a biconvex lens graph between v1 and u1 is as shown. The focal length of lens is equal to _______ cm ⑧ Ya Ya ↓ tu = ¥ from gratin to = ≠ f- = 10cm
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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