KUMAR PHYSICS CLASSES E 281 BASEMENT M BLOCK MAIN ROAD GREATER KAILASH 2 NEW DELHI 9958461445,01141032244 www.kumarphysicsclasses.com www.kumarneetphysicsclasses.com IIT JEE PHYSICS PAPER SOLUTION-LAST PAPER OF PHASE 1 JEE-2022 29 JUNE 2022 EVENING SHIFT QUESTIONS BASED ON ELECTROSTATIC,MOLES, DISPLACEMENT CURRENT & DIODE ARE TRICKY
1. A small toy starts moving from the position of rest under a constant acceleration. If it travels a distance of 10 m in t s, the distance travelled by the toy in the next t s will be : (A) 10 m (B) 20 m (C) 30 m (D) 40 m 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. Choose the correct answer : ✓ U= 0 ← 10Mt ! a -1 1- see 1- see 5- Ucf) -1k at 10 = UCO) + 1- act ↳ = 1- at ① 2-110 = Uco) 1- % act> ↳ 2+10 = } a (4%-4)=2 at ② EQUATION ② EQUATION ① k = }÷¥= 4=>2+10--40 E- 30Mt
2. At what temperature a gold ring of diameter 6.230 cm be heated so that it can be fitted on a wooden bangle of diameter 6.241 cm? Both the diameters have been measured at room temperature (27°C). (Given: coefficient of linear thermal expansion of gold αL = 1.4 × 10–5 K–1) (A) 125.7°C (B) 91.7°C (C) 425.7°C (D) 152.7°C ilt-locr-t.AT ) " lt lo lo (a) Dt ) Al = to (a) ( se ) ← 6.230cm ← 6.241cm CST ) = §(I, bl= 6-241-6-230 Tf Ti = @ D= ¥ 152 = • 011 am G- 241×10-2×1 -4×10-5 Tf 27-1126.11 = 126 11 OR = 153-110 A
3. Two point charges Q each are placed at a distance d apart. A third point charge q is placed at a distance x from midpoint on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulombs force is ✗ 2 -Ñ fq§y £ c- ffn af find FNET = 2f Cos a ¥= 2 'kQ_Ñj÷%€ for man = { % Q d → Coulombs force dFdN÷=o 2kQd%{ÉdYa aka / Lakota) " -8-9 • Cñ+d% ) " )=o¢k4d4D%} (ñ+d÷)%_ sez-ca-i-d-YK.GR > = -0 Ge4d %( at-1% sat] 2ñ=dYy r=§
4. The speed of light in media ‘A’ and ‘B’ are 2.0 × 1010 cm/s and 1.5 × 1010 cm/s respectively. A ray of light enters from the medium B to A at an incident angle ‘θ’. If the ray suffers total internal reflection, then imma ñ→ VA = 271010cm/see DB = 1^5×10 " < m/see µA=%a ↳ = % ✓ =3 ✗ 1_ = ?¥%¥2×10/8 A = 1.5 = 2. 0 790° RARGR CDGNSER) I 2.0 Sin @ = 1.5 Sinew B since = ¥0 = } D= sin '( %) 0 > sin '( %)
✓ 1>182 178 74 zHe& -1 > 2D , ↓ 178 a 13° -1 D< → _ HI + Doth 73 7k 174 or ° + Dg7h bA⑤
6. The electric field at a point associated with a light wave is given by E = 200[sin(6 × 1015)t + sin(9 × 1015)t] Vm–1 Given : h = 4.14 × 10–15 eVs If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be (A) 1.90 eV (B) 3.27 eV (C) 3.60 eV (D) 3.42 eV F- = 200 ( sin (6×1015) t'Ñ•%y*g '☆ᵈ + sin (9×1015) f) Hmt FREQUENCY of f- MW W , # 6×10154 Wzt= 9×10151-1 217/-1=6× 1015 271-1<=9×1015 ✓ f , =6 5-2=9×-2%5 hf -=- Wo + KE KE = hf Wo ENERGY of F- MW① ENERGY of f- MW ② FOR ① WAVE E. = hf , E- = hfz = 4.14×1515×6×1015 = 4.14×1815×9×1015④ E) • =3 -95-2-50 get 27T = 1.45 EV = 3.95 ev 5- Belt for ② WAVE ( KE)< = 5-93-2-50 =3 -43 eV
i ✓ fAᵗᵗ=ᵗ• t=tz R Mr 1=(1%2) / or "=%o qzTake eogein ↳ 1£ / % = -2C efuatiow ! q=%é ᵗ4Rc-11%2 = lofee ③ % ' g- =F◦é ᵗYR$ take loge in eeuatim ② -4¢ 7¥ 8 =e ᵗYRc3 loge > = ¥ , lofee ④ a- %÷ (een③µn⑥ 9=90 eᵗYRc (2) 3 ⑤ ↳ = ¥ # ¥ ¥,=x◦é ᵗYrc I=eᵗ%=¥=éᵗYRc¥=% ①
8. Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways. The work done by the gas is W1 if the process is purely isothermal, W2, if the process is purely adiabatic and W3 if the process is purely isobaric. Then, choose the correct option.
(A) W1 < W2 < W3
(B) W2 < W3 < W1
(C) W3 < W1 < W2
(D) W2 < W1 < W3
60m paring the area under goapn
> At > Az
As
1/-13>1--1 , > Ha
9. Two long current carrying conductors are placed parallel to each other at a distance of 8 cm between them. The magnitude of magnetic field produced at mid-point between the two conductors due to current flowing in them is 30 μT. The equal current flowing in the two conductors is: (A) 30 A in the same direction (B) 30 A in the opposite direction (C) 60 A in the opposite direction (D) 300 A in the opposite direction ✓ BNEt= / Bit -1 / Bit ± a VI. < 4cm → ← 4cm → • n=%÷;÷ ×z ⑦ . /Bp 30×10-6 = 1z × % I = 30×156×21×152 = 30A ✗ ✗ 10-7 I 2
10. The time period of a satellite revolving around earth in a given orbit is 7 hours. If the radius of orbit is increased to three times its previous value, then approximate new time period of the satellite will be (A) 40 hours (B) 36 hours (C) 30 hours (D) 25 hours ^Fe mᵈfgT , > ✓ I. = !÷ IT (7)2 g. =L ( 3873 ei=É¥ʰ aY#=☒¥ 72%49×9×3 951=8 > w :-O -4¥? 72=7 × 3 × 53 F. = Tovsraet = 21×1-732 72283 = 36.372 I 36 hours
11. The TV transmission tower at a particular station has a height of 125 m. For doubling the coverage of its range, the height of the tower should be increased by (A) 125 m (B) 250 m (C) 375 m (D) 500 m A- CB A DERIVATION LhtR)_-d2+RL ↓ ✓ h4R+2Rh=d2tR/ d. f- height 9- tower chic > Rh ignore D= # B>eeR CASE I 1 CASE I R e di=F de # ÷=Fn÷ ¥=T÷ k=aa . Increase lnlengtn -- 4h , hi hi = 3×125 = 375Mt
12. The motion of a simple pendulum executing S.H.M. is represented by the following equation. y = A sin(πt + φ), where time is measured in second. The length of pendulum is (A) 97.23 cm (B) 25.3 cm (C) 99.4 cm (D) 406.1 cm J = A sin ( Ttt + 4.) ✓ wt = it w = TT 29 f = TT f- = I 7=>21--4%1 ¥ ¥ = 1g l = ¥-2 = a a. 4cm ( Approx ?
13. A vessel contains 16 g of hydrogen and 128 g of oxygen at standard temperature and pressure. The volume of the vessel in cm3 is: (A) 72 × 105 (B) 32 × 105 (C) 27 × 104 (D) 54 × 104 93 $5 f- Total number of moles are ✓ qt =NH< + no = ¥+128 -32 = 12 Moles PV= NRT V= NT- = 12 × 8-31 × 273.15 1 × 105 = 0.27Mt 3=27×104 Cm -3
Given below are two statements:
Statement I: The electric force changes the speed of the charged particle and hence changes its kinetic energy; whereas the magnetic force does not change the kinetic energy of the charged particle.
Statement II: The electric force accelerates the positively charged particle perpendicular to the direction of electric field. The magnetic force accelerates the moving charged particle along the direction of magnetic field.
In the light of the above statements, choose the most appropriate answer from the options given below:
(A) Both statement I and statement II are correct
(B) Both statement I and statement II are incorrect
(C) Statement I is correct but statement II is incorrect
(D) Statement I is incorrect but statement II is correct
ELECTRIC FIELD ACCELERATES
PARTICLE IN THE DIRECTION of FIELD
14.
THE
✓ F- =qI=ma → And magnetic field accelerates the tartrate perpendicular to the field f- = or ( Ñ ✗ E.) =ma →
15. A block of mass 40 kg slides over a surface, when a mass of 4 kg is suspended through an inextensible massless string passing over frictionless pulley as shown below.
coefficient of kinetic friction between the surface and block is
The acceleration of block
The
0.02.
is (Given g = 10 ms–2.) (A) 1 ms–2 (B) 1/5 ms–2 (C) 4/5 ms–2 (D) 8/11 ms–2 a TRY & ☒ FER MKR 4kg → @ LOCK) ( 4kg ) ~ 4g T= 4h -10 48 e- µ ,<R= 40h ✓ T (-02×408)--40 a ② ADD EQUATION ① & EQUATION ②4g (a) ( 4081=449 4 × 10 (4*(1/0)=449 40-8=449 a- 35¥? ,=m5 '
In the given figure, the block of mass m is dropped from the point ‘A’. The expression for kinetic energy of block when it reaches point
16.
‘B’ is (E)a = £ m ( o ) 2 + mg (g) ( TE) , = 1- m B) < -1mg ( 8- To ) ( TE) A = ( TE ) & Mg y = ± m ( Ors) -1mg ( g- 8D mg (8/-8/+80) = ± Mori Mf Yo = ( K E) is = 1- MOB "
17. A block of mass M placed inside a box descends vertically with acceleration ‘a’. The block exerts a force equal to one-fourth of its weight on the floor of the box.The value of ‘a’ will be
R = " % a Mg Mg Hgh = Ma 3%1 = Ma a = 384
19. The combination of two identical cells, whether connected in series or parallel combination provides the same current through an external resistance of 2 Ω. The value of internal resistance of each cell is (A) 2 Ω (B) 4 Ω (C) 6 Ω (D) 8 Ω ✓ B < K L N {% M A < ' % Is = ( p → given Mes h k L M N K Mesh A- B C D A. 2 Cp + 8 ( Gg ) E = -0 2 Is -1 If F- + If E = 0 ( b = E2 Is -1 2 If = 2 f2 -1% Is = f bIs = ¥÷r= Er f÷g=±21-81 2+1 = • + r 2 I = 81 8=2 hi
20. A person can throw a ball upto a maximum range of 100 m. How high above the ground he can throw the same ball? (A) 25 m (B) 50 m (C) 100 m (D) 200 m Answer (B) R = U2 51h20 For Mar g range ✓ too = Ugˢ 0=450 U= Hot v± uh agse IA- = 2gal 7450 a § 100/8=281 too wit 2=50 Mt ↑ "
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. The vernier constant of Vernier callipers is 0.1 mm and it has zero error of (–0.05) cm. While measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5. The corrected diameter will be ________× 10–2 cm. Answer (180) -q2181
ZERO ERROR
SINCE
is @ e.) CADD • 05cm ) Reading = 17cm -15 (0.1mm) + 05cm =/ 80×10-2 cm
2. A small spherical ball of radius 0.1 mm and density 104 kg m–3 falls freely under gravity through a distance h before entering a tank of water. If, after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value of h will be ______ m. (Given g = 10 ms–2, viscosity of water = 1.0 × 10–5 N-sm–2). Answer (20) i. / Mo 8=0^1×10-3 Mt 20 at D= 104181Mt ? away ↑ Velocity attaint A= # h }qAfter entering into water t.CA , → Ign Wg Fr Fu=0 _ → -F ↑ FF = ¥178 > dg 69m80 g- to > Sg * ng 43783g (d- 53=69480 • = ¥¥ f ᵈn = }•%{ e- sont Fn=¥¥¥×ae
In an experiment to determine the velocity of sound in
temperature using
at
is
column
Hz is used.
of
cm
tuning fork of frequency
of the sound at
temperature is
column
ms–1.
third resonance is observed when
of
cm.
3.
air
room
a resonance tube, the first resonance
observed when the air
has a length
20.0
for a
400
The velocity
room
336
The
the air
has a length
______
Answer (104) ↑3RD a. "" ← f- 400ha RESONANCE 20.0 D= 336 ME " Cm 5×-2 % @ + e)5×÷=k+e ✗ = 4( e. + e) ✗ 2=4 / late ] 5[ FIRST f- = 4%+5 = 400 f- = ) 4( late] | RESONANCE 4 (20×157-2)=34%0 (20×154-8)=34%-54400=3%1%-0 , , e. onmt lz = 1.04Mt = 104cm
4. Two resistors are connected in series across a battery as shown in figure. If a voltmeter of resistance 2000 Ω is used to measure the potential difference across 500 Ω resistor, the reading of the voltmeter will be _____ V.8 % ◦ B In v 2000 -0hm i vi. • = I (%;%-) 400 = # (57%+79.0) + soo # sooo 25-0051 ) 20 = ( Goo + soo ) ✗ 400=7*2×5--8 voet
5. A potential barrier of 0.4 V exists across a p-n junction. An electron enters the junction from the n-side with a speed of 6.0 × 105 ms–1. The speed with which electrons enters the p side will be . @ botentral barrier § P N ← ⑦ 14 to 6×105 his ' Abbey souse ration of energy e KE developed from electron = KE / e- enters tnep side + potential enereg at 1- m (6×105) > = ¥ mÑ-o4¥o% depletion region £ MO ' = 1- m (6×105)<-04×1.6×10-19 V2 = (6×105) 2- 2m ( O' 4) (1^6)×10-19 = 36×1010 × , ok / * = 14 = 14-3×105 MJ " = 7- ✗ 105
6. The displacement current of 4.425 μA is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 106 Vs–1. The area of each plate of the capacitor is 40 cm2. The distance between each plate of the capacitor x × 10–3 m. The value of x is, (Permittivity of free space, E0 = 8.85 × 10–12 C2 N–1 m–2) & BE -43 8 -1$ Id = d ᵈdt Id = 4^425×10-6 Amp dv Id = d_dt( < V ? q = 106 Volt/see A = 40×10-9 hit ≥ = < dug = ↳dA_.d£ D= :# ( %-) = 8.85×1512×40×154 4. 425 × 10-8 ✗ 10£ D= 8×10-3 Mt = xx 10-3 2=8
7. The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I1. The same rod is bent into a ring and its moment of inertia about a diameter is I2. If
as
.mg → M¥=I ,
← el¥÷=t¥¥¥ : ¥7 E- ME L=2RR = 73¥ R= 42ft 2 f- eE- 8 In Ga ) " = :÷
The half life of a radioactive substance is 5 years.
years,
gets
8.
After x
a given sample of the radioactive substance
reduced to 6.25% of its initial value. The value of x is ______. Answer (20) 20 years N = No (E) h G. 25=100 (f) h tFE¥;÷( n % (E) " = (f) 4 n= 4 ✗ = n ( tic ) = 4×5=20 TEARS
9. In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by 5 × 10–2 m towards the slits, the change in fringe width is 3 × 10–3 cm. If the distance between the slits is 1 mm, then the wavelength of the light will be _______ nm. Answer (600) *-*i ¥ ① Be ② Goo Da =D , Dz=(D 5×10-2 ) Mt Ba Ba = 3. ✗ 183cm = 3.✗ 155Mt EQUATION ① EQUATION ② D= 1×10-3 Mt ✗ =3 B , B , = Id [ D , Dr ] = ÷ / D- C. D- 5 × 157 } = [5×10-2] 3×10-5=6%-365×10--7 ✗ = 3×10-5 ✗ • ✗ 153 5×10-2 = 06×10-6 = 600mm
10. An inductor of 0.5 mH, a capacitor of 200 μF and a resistor of 2 Ω are connected in series with a 220 V ac source. If the current is in phase with the emf, the frequency of ac source will be ______ × 102 Hz. Answer (5) If current is In those with emf , then resonance £ & ✗ < = ✗ ↳ occurs f- = ¥gg 27ft = c. = ¥ J ×o-5×2001×10-6 = dZTT 5×154×2×10=4 = 27%-0×15-8=2*1%-4 5MHz = 5 × 18 Hz
ROUGH WORK
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