Tes Pttern Eglish REVIEW TEST CLAS ONT PROGRAMME (Aemi

READ THE INSTRUCTIONS CAREFULLY

GENERAL :

1. This sealed booklet is your Question Paper. Do not break the seal till you are toldto do so.

2. Use the Optical Response sheet (ORS) provided separately for answering the questions.

3. Blank spaces are provided within this booklet for rough work.

4. Write your name, form number and sign in the space provided on the back cover of thisbooklet.

5. After breaking the seal of the booklet, verify that the booklet contains 36 pages andthat all the 18 questions in each subject and along with the options are legible. If not,contact the invigilator for replacement of the booklet.

6. You are allowed to take away the Question Paper at the end of the examination.

OPTICAL RESPONSE SHEET :

7. The ORS will be collected by the invigilator at the end of the examination.

8. Do not tamper with or mutilate the ORS. Do not use the ORS for rough work.

9. Write your name, form number and sign with pen in the space provided for this purposeon the ORS. Do not write any of these details anywhere else on the ORS. Darkenthe appropriate bubble under each digit of your form number.

DARKENING THE BUBBLES ON THE ORS :

10. Use a BLACK BALL POINT PEN to darken the bubbles on the ORS.

11. Darken the bubble COMPLETELY.

12. The correct way of darkening a bubble is as :

13. The ORS is machine-gradable. Ensure that the bubbles are darkened in the correctway.

14. Darken the bubbles ONLY IF you are sure of the answer. There is NO WAY to eraseor "un-darken" a darkened bubble.

15. Take g = 10 m/s2 unless otherwise stated.

Please see the last page of this booklet for rest of the instructions

Time : 3 Hours Maximum Marks : 198

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PAPER-1

Test Pattern

CLASSROOM CONTACT PROGRAMME(Academic Session : 2020 - 2021)

SAMPLE PAPER-1

JEE(Advanced)

REVIEW TESTEngl

ish

E-2/36

Target : JEE (Main + Advanced) 2021/SAMPLE PAPER-1/Paper-1

Space for Rough Work

SOME USEFUL CONSTANTSAtomic No. : H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16,

Cl = 17, Br = 35, Xe = 54, Ce = 58Atomic masses : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,

Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,Xe = 131, Ba=137, Ce = 140,

· Boltzmann constant k = 1.38 × 10–23 JK–1

· Coulomb's law constant pe9

0

1 = 9×104

· Universal gravitational constant G = 6.67259 × 10–11 N–m2 kg–2

· Speed of light in vacuum c = 3 × 108 ms–1

· Stefan–Boltzmann constant s = 5.67 × 10–8 Wm–2–K–4

· Wien's displacement law constant b = 2.89 × 10–3 m–K· Permeability of vacuum µ0 = 4p × 10–7 NA–2

· Permittivity of vacuum Î0 = 20

1

cm· Planck constant h = 6.63 × 10–34 J–s

ALLEN

SAMPLE PAPER-1/Paper-1ALLEN

E-3/36

PART-1 : PHYSICSSECTION-I(i) : (Maximum Marks: 28)

� This section contains SEVEN questions.� Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of

these four option(s) is (are) correct option(s).� For each question, choose the correct option(s) to answer the question.� Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +4 If only (all) the correct option(s) is (are) chosen.Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.Partial Marks : +2 If three or more options are correct but ONLY two options are chosen,

both of which are correct options.Partial Marks : +1 If two or more options are correct but ONLY one option is chosen

and it is a correct option.Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).Negative Marks : –2 In all other cases.

� For Example : If first, third and fourth are the ONLY three correct options for a question withsecond option being an incorrect option; selecting only all the three correct options will resultin +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options),without selecting any incorrect option (second option in this case), will result in +2 marks.Selecting only one of the three correct options (either first or third or fourth option), withoutselecting any incorrect option (second option in this case), will result in +1 marks. Selectingany incorrect option(s) (second option in this case), with or without selection of any correctoption(s) will result in –2 marks.

1. The jockey of the rheostat is slided from far right to far left. During this time, comment onthe reading of different meters. Assume that the battery and all voltmeters and ammeter areideal.

V

A

V1

(A) Reading of V remains same (B) Reading of V1 will decrease(C) Reading of V1 will increase (D) Reading of A will increase

BEWARE OF NEGATIVE MARKINGHAVE CONTROL ¾® HAVE PATIENCE ¾® HAVE CONFIDENCE Þ 100% SUCCESS


Target : JEE (Main + Advanced) 2021/SAMPLE PAPER-1/Paper-1ALLEN

E-4/36

2. A uniform rod of mass m and length L is held at one of its ends such that the other end isvertically above it. Now holding the lower end, we release the rod so that it rotates in a verticalplane. When it reaches the lowermost position, we release the end which has been held in ourhands. It is seen that the end which was held in our hands strikes the ground first with rod beingvertical at that time. What could be the possible height of the point above the ground where wehad held the rod. (Take : p2 = 10 & neglect friction)

(A) 11L

6(B) 31L (C) 8L (D)

64L

3

3. In an organ pipe (which may be closed or open) of 99 cm length, standing wave is set up whoseequation is given by longitudinal displacement. x = 0.3 (mm) cos [(2p /0.8) (z + 0.01 m)] cos 400 t,where z is measured from the top of tube in metres and t is in seconds. Take end correction = 1 cm.Choose the CORRECT statements.

z = 0 Upper end

Lower end

z

(A) The upper end and lower end of tube are open and closed respectively.(B) The upper end and lower end of tube are closed and open respectively.(C) The air column is vibrating in third harmonic.(D) The air column is vibrating in second overtone.

4. In an AC circuit, we have an inductance L joined in series to two resistors of resistance Reach. In the first case, the resistors are in series with each other and in the second case,the resistors are parallel to each other. The power consumed by the circuit in the first case is810W and in the second case, it is 506.25 W. If the source voltage is 225 V,(A) Resistance R = 20 W(B) In the first case, the phase difference between current through the source and voltage across it

is 37°.(C) In the second case, the phase difference between current through the source and voltage

across it is tan-1(3)(D) Inductive reactance is 30 W



E-5/36


5. Light of wavelength l is incident on a spherical zinc ball of radius 10m. The ball has a chargeof 1 nC. The ball is in vacuum. The work function of zinc is 4 eV.(A) If l = 250 nm, electrons will come out of the ball but all of them will re-enter the ball.(B) If l = 350 nm, electrons will not come out of the ball.(C) If l = 300 nm, electrons will come out of the ball but all of them will reenter the ball.(D) If l = 155 nm, electrons will come out of the ball but some of them will never re-enter

the ball.6. One mole of monoatomic ideal gas is taken from an initial condition 1 to final 4 in the process

presented in figure. The temperature difference between state 4 and state 1 is 100 K.

1 2

3 4

V

T

(A) Change in internal energy of gas is same in modulus in each of the parts (1-2, 2-3, 3-4)(B) The work done by the gas from 2-3 is –100R(C) The total heat given to the gas during the series of processes is 50R(D) The final pressure is half of the initial pressure

7. A stationary observer in space observes a system of binary stars undergoing circular motionabout their centre of mass. Because the observer is in the plane of the motion and is quite faraway, he sees both undergoing SHM with same angular frequency. The velocity versus phase

angle f rotated for both A & B are shown below (Take : G = 1120 103

-´ Nm2/kg2)

MA

MB

RA

RB

f

Observer

A

B

60 km/s

40 km/s p 2pf

v

(A) A

B

M 3M 2

=

(B) If radius RA = 6 × 1010 m, MA = 2 × 1020 kg(C) Using data in option (B), MB = 9 × 1030 kg(D) Using data in option (B) time period T = 6.28 × 106 sec


E-6/36

SECTION–I(ii) : (Maximum Marks : 18)� This section contains THREE paragraphs.� Based on each paragraph, there are TWO questions.� Each question has FOUR options (A), (B), (C) and (D) ONLY ONE of these four options is

correct.� For each question, darken the bubble corresponding to the correct option in the ORS.� For each question, marks will be awarded in one of the following categories :

Full Marks : +3 If only the bubble corresponding to the correct answer is darkened.Zero Marks : 0 If none of the bubbles is darkened.Negative Marks : –1 In all other cases

Paragraph for Questions 8 and 9Fortin’s barometer is an improved form of mercury barometer. It is used for accurate measurementof atmospheric pressure. It consists of a thick walled glass tube of length 1m. The glass tubecontains a cistern of mercury. The bottom of the vessel is made of leather which can be raised orlowered by means of a screw provided below. The whole barometer is enclosed in a brass tube andglass windows are provided to take the readings. An ivory pointer is fixed to the lid of the vesselcontaining mercury. The tip of this index coincides with zero of the vertical scale on the glasstube. At the upper end of the glass tube a vernier scale is fixed which slides up and down withhelp of a screw. The height of the mercury column in the barometer is found using the vernierscale. This is done by making the zero of vernier coincide with the convex meniscus of mercurycolumn in the tube by adjusting the screw R given. However, the instrument has two possiblesources of systematic errors.

1. The scale is calibrated at 20o celcius. If the temperature is higher or lower, expansion of mercurywill change the reading. The expansion of glass tube and scale can be neglected.

2. The fortin barometer has to be exactly vertical. If it is not vertical, reading will not reflect thetrue pressure.Since the tube is wide, effect of surface tension can be neglected.



E-7/36

0

Glass tubeSlit

Screw RVernier scale

Main scale

Brass tube

Screw S

Grass vessel

Ivory pointer

Thermometer

Mercury

MercuryLeather cup

0

Assume that the vernier scale has 20 divisions which coincide with 19 divisions of main scale.Each main scale division is 1 mm.

8. Select the CORRECT statement:(A) If the scale is tilted slightly, the reading of fortin barometer will be more than the true value

and if the temperature of the mercury is more than 20o C, then also the reading of the barometerwill be more than the true value.

(B) If the scale is tilted slightly, the reading of fortin barometer will be less than the true valueand if the temperature of the mercury is more than 20o C, then the reading of the barometerwill be more than the true value.

(C) If the scale is tilted slightly, the reading of fortin barometer will be less than the true valueand if the temperature of the mercury is more than 20o C, then also the reading of the barometerwill be less than the true value.

(D) If the scale is tilted slightly, the reading of fortin barometer will be more than the true valueand if the temperature of the mercury is more than 20o C, then the reading of the barometerwill be less than the true value.

9. If the reading of vernier scale is such that 18th division is coinciding with a main scale, zero ofvernier reads 752 mm and there is no zero error, what is the percentage uncertainity in thepressure (approx) ?(A) 0.013% (B) 0.04% (C) 0.007% (D) 0.003%



E-8/36

Paragraph for Questions 10 and 11We know that if charge is distributed on a body, these charges interact to give a certain selfenergy to the body. This self energy can be calculated in many ways. One of the ways is to

integrate the energy spread out in entire space in form of electric field. U = ò Î dVE21 2

0 .

We can assume an electron to be like a uniformly charged thin shell of radius r0 having selfenergy E0. This energy has been estimated in nuclear reactions as 0.51 MeV. From this we canfind out an approximate radius of the electron.

10. What is the approximate radius of electron ?

(A) 00

2

E4eÎp (B)

00

2

E8eÎp (C)

00

2

E4eÎp (D)

00

2

E8eÎp

11. If we were to assume that electron is a uniformly charged sphere,(A) its radius would be more than estimated in previous question.(B) its radius would be less than estimated in previous question(C) its radius would be equal to that estimated in previous question(D) can't be said



E-9/36

Paragraph for Questions 12 and 13Bohm Aharonov effect : Consider a YDSE experiment done with electrons.

Source

d

W

1

S

Lines of B

L

2

IDy

The debroglie hypothesis says that electrons behave like waves and so in presence of double slit,they show interference pattern as shown by dotted line.But if we introduce a thin strip (w << L) of magnetic field just in front of slits, the electrontrajectory would bend by a small angle and so the central maxima shifts up by a distance Dy. Thisshift can be easily shown to depend on magnetic field B, charge q, width w, the momentum ofelectron p, slit distance d and L.The same result can be derived using principles of quantum mechanics. In quantum mechanicswe cannot talk about definite path. So we cannot talk about bending of electron trajectory. Thereit is found that on passing through the region of magnetic field, the electron waves undergo aphase change. So the electron wave coming from upper slit and that from lower slit have a phasedifference. Interestingly the path difference resulting from this phase difference predicts thesame shift of central maxima Dy as does the classical theory using magnetism.

12. The phase difference between the waves from upper slit and lower slit does not depend on(A) d and p (B) d and L (C) p and d (D) p and L

13. In a typical set up, electrons have a momentum of 10–25 kg m/s, B = 1T, w = 10–8 m, L = 1m. Atwhat minimum value of d will central point be a minimum? (Take h = 6.4 × 10–34 Jsec)(A) 200 nm (B) 300 nm (C) 400 nm (D) 500 nm



E-10/36

SECTION–II : (Maximum Marks : 20)MATCHING LIST TYPE QUESTION (ANSWER AS FOUR DIGIT INTEGER)

� This section contains FIVE questions.� The answer to each question is a FOUR DIGIT INTEGER ranging from 1111 to 4444, both

inclusive� Each question having two matching lists. Choices for the correct combination of elements form

List-I (P), (Q), (R), (S) and List-II (1), (2), (3), (4)� Given matching type questions in which List-I (P,Q,R,S) match with List-II (1,2,3,4) and represent

your answer as four digit integer for example correct match for P ® 2, Q ® 3, R ® 1,S ® 4 .Then you have fill the answer in Section-II as 2314.

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

Each matching in List-II may be/may not be correct & can be used more than one timefor matching in List-I. For example your answer can be 2233, 1333, 1433, 4321 etc...

� For each question, marks will be awarded in one of the following categories :Full Marks : +4 If only the bubble corresponding to the correct answer is darkened.Zero Marks : 0 In all other cases.


E-11/36

14. A sound source emits a sound wave of wavelength 2m when stationary. The velocity of soundwith respect to the medium is 330 m/s. List-I indicates a situation and List-II indicates thepossible values of wavelength and frequency recorded by recorder. Speeds of source recorderand wind are subsonic.

List-I List-II

(P)Source Recorder

Source & recorder stationary

wind blowing (1) l = 2.5 m, f = 165 Hz

(Q)Source RecorderSource moving as shown,

wind and recorder stationary(2) l = 2 m, f = 170 Hz

(R)Source Recorder

Source & wind stationary, recorder moving as shown.

(3) l = 1.5 m, f = 220 Hz

(S)Source Recorder

wind(4) l = 2 m, f = 160 Hz

Source & wind and recordermoving in direction shown.Source & wind have same speed.



E-12/36

15. Match the description of image in List–I with optical system in List–II.List-I List-II

(P)

30cm

f=10cm

(1) Real enlarged image/ of same size as the object.

(Q)

object

water

air

(2) Virtual enlarged image.

(R) n=1.51mair

R=3m

(3) Virtual image of same size as object/diminished.

(S) 20cm

air

n=1.5

(4) Real diminished image

equiconvex lens of focal length20cm in air whose rear side is silvered.



E-13/36

16. A block of mass 2 kg is connected to a vertical spring of spring constant 800 N/m as shown.At t = 0, the spring is relaxed. Now the block is released. List-I gives time t and List-II givesthe comments on the velocity and the position of block at that time.

List-I List-II

(P) 15p

sec. (1) Block is above mean position and moving towards mean.

(Q) 30p

sec. (2) Block is below mean position and moving towards mean.

(R) 60p

sec. (3) Block is above mean position and moving away from mean.

(S) 12p

sec. (4) Block is below mean position and moving away from mean.



E-14/36

17. Two men of mass 60 kg and 80 kg stand on a plank of mass 20 kg. Both of them can jump witha velocity of 1 m/s relative to the plank. In each event shown in List-I, find the velocity ofplank after the event. (Initially system is at rest)

Ram Shyam

60 kg 80 kg

Smooth level ground

+

List-I List-II

(P) Ram alone jumps to the left (1) – 4017

m/s

(Q) Shyam alone jumps to the right (2) – 21

m/s

(R) Ram jumps to left and shyam jumps to right simultaneously. (3) 83

m/s

(S) Ram jumps to left and after that shyam jumps to right (4) – 81

m/s



E-15/36

18. List–I List–II(P) A single dipole is kept on x axis at x = +r (1) Electric field increases as r

such that it is oriented in positive x direction. increases and has downwardWe find electric field at a point on +y axis or negative y component.(not origin)

(Q) A conductor of radius r & length l (2) Electric field decreases as r is connected to battery whose

l

r

y

increases and has downwardother end is earthed as shown. or negative y component.The volume of conductor is keptconstant as r increases or decreases.Electric field is inside the conductor.

(R) An ideal solenoid has current

x r

yÄ

Pi

(3) Electric field increases as ri and radius R. Current is increases and has upwardsdecreasing with time. Point P or positive y component.remains outside the solenoid.Electric field at point P isbeing investigated.

(S) Cylindrical capacitor is r

P(4) Electric field decreases as r

connected to a battery. increases and has upwardWe consider electric or positive y componentfield at P at a distance rfrom axis



E-16/36

PART-2 : CHEMISTRYSECTION-I(i) : (Maximum Marks: 28)







1. Potassium crystallizes in BCC lattice with cell length as 'a', then which are correct statement.

(A) Along one body diagonal effectively 1.25 atoms are present per unit cell

(B) Along one face diagonal effectively 1 atoms is present per unit cell

(C) If 'r' is radius of atom then 4a 3 r=(D) Co-ordination number of atoms in topmost layer is 8

2. For first order reaction

A ¾® B , Ea = 9.2 kcal / moleAt 500 K rate constant for reaction is 6.93 × 10–4 s–1. Which of the following option(s) is/arecorrect ? (Use : ln X = 2.3 log X, ln2 = 0.693)

(A) Value of rate constant for reaction at 1000 K is 6.93 × 10–2 s–1

(B) In 20 seconds , 75 % decomposition of A will occur at 1000 K(C) For 99.9 % completion at 500 K, 104 seconds are required.

(D) In 30 seconds, 87.5 % decomposition of A will occur at 1000 K



E-17/36

3. How many enantiomeric pairs are formed when isopentane undergoes monochlorinationwith Cl2, hn.(A) 1 (B) 2 (C) 3 (D) 4

4. Invalid resonating structure of following compound is/are

NH3

(A)

NH3

(B)

NH3

(C)

NH3

(D)

NH3

5. The true statement(s) regarding the brown ring test carried out in the laboratory for the

detection of 3NO- is/are?

(A) Brown ring is due to the formation of the iron nitrosyl complex(B) Concentrated nitric acid is used for the test(C) The complex formed in the reaction is [Fe(CN)5NO]2–

(D) The brown colored complex is paramangetic in nature



E-18/36

6. Select CORRECT statement(s) :-(A) When F– is co-ordinated to F3ClO hybridisation change from sp3d to sp3d2.(B) When F– is removed from F3ClO geometry changed from sea-saw to trigonal pyramidal(C) When F– is co-ordinated to FClO3 hybridisation change from sp3 to sp3d.(D) When F– is removed from F3ClO2 oxidation number of Cl is not changed

7. A sample of bauxite contain SiO2, TiO2 and iron oxides, treated with conc. NaOH at 500K &36 bar pressure.Which is/are not dissolved in this process ?(A) Al(OH)3 (B) Iron oxide (C) SiO2 (D) TiO2



E-19/36


SECTION–I(ii) : (Maximum Marks : 18)� This section contains THREE paragraphs.� Based on each paragraph, there are TWO questions.� Each question has FOUR options (A), (B), (C) and (D) ONLY ONE of these four options

is correct.� For each question, darken the bubble corresponding to the correct option in the ORS.� For each question, marks will be awarded in one of the following categories :


Paragraph for Questions 8 and 9A mixture of hydrogen gas and the theoretical amount of air at 25°C and a total pressure of1 atm, is exploded in a closed rigid vessel. If the process occurs under dq = 0 condition thenusing the given data answer the questions that follow :Given (i)

(N )2VC = 6.3 Cal deg–1 mol–1 ; (ii)

(H O,g )2VC = 9.3 Cal deg–1 mol–1 ; DHf[H2O(g)] = –57.8 Kcal

Take air as 80% N2, 20% O2 by volume.8. What will be maximum temperature (approx.) attained if the process occurs in above container.

(A) 298 K (B) 1900 K (C) 2665 K (D) 2940 K9. What will be the final pressure in atm (approx.) :

(A) 0.85 (B) 5.46 (C) 7.6 (D) 8.5


E-20/36


Paragraph for Questions 10 and 11Decarboxylation of sodium salt of carboxylic acid in presence of NaOH, CaO, D is calledsodalime process. It is also called OAKWOOD process.

R – C – OHO

NaOH, CaO R – C – ONaO

rdsD RNa

–CO2

NaOH

Na CO2 3

R – HAlkane

H – O – H

Rate of reaction µ stability of carbanion10. In which of the following rate of decarboxylation is maximum

(A) CH – CH – C – OH3 2

ONaOH,CaO

Δ¾¾¾¾¾® (B) CH = CH – C – OH2

ONaOH,CaO

Δ¾¾¾¾¾®

(C) H – C C – C – OHº

ONaOH,CaO

Δ¾¾¾¾¾® (D) C – OH

O

NaOH,CaOΔ¾¾¾¾¾®

11. How many ketal(s) is/are formed in following reaction

C – OH

OO

CH3

NaOH,CaOΔ¾¾¾¾¾® (A)

HOHO Dry HCl

(B)

(A) 0 (B) 1 (C) 2 (D) 3


E-21/36

Paragraph for Questions 12 and 13An inorganic white crystalline salt (X) which is freely dissolve in water and gives followingobservation.Pest of (X(s) + conc. HCl) produce light violet coloration on flame.

Acidic solution XAq.

very small amount ofCa(OCl) .CaCl .Ca(OH) .2H O2 2 2 2 dark brown

coloration (Y)

shaken withChloroform

(P)

(Q) Violet colorationin chloroform

Colourless solutionaq. solution

of hypo

12. Which is NOT a redox change ?(A) Preparation of "Y" from "X".(B) Reaction of Y with hypo solution(C) Process of changing Y to P(D) Reaction of "X"Aq. with K3[Fe(CN)6]

13. Select reaction which produce ppt. in final product?

(A) XAq. 3

4

(i ) AgNO Aq.(ii) Excess NH OH¾¾¾¾¾¾¾¾® (B) XAq.

3 2( i ) Hg(NO ) Aq.( ii ) Excess KI¾¾¾¾¾¾¾®

(C) XAq. 3 2(i) Pb(CH COO)

(ii) Heating¾¾¾¾¾¾¾® (D) XAq. 3BiCl (aq.)(i)

E(ii) xcess KI¾¾¾¾¾¾®



E-22/36




inclusive� Each question having two matching lists. Choices for the correct combination of elements

form List-I (P), (Q), (R), (S) and List-II (1), (2), (3), (4)� Given matching type questions in which List-I (P,Q,R,S) match with List-II (1,2,3,4) and

represent your answer as four digit integer for example correct match forP ® 2, Q ® 3, R ® 1, S ® 4 .Then you have fill the answer in Section-II as 2314.

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••



1. Component A and B are mixed to form a solution. Match correct nature of solution withcomponents mixed -List-I (Components) List-II (Nature)(P) Ethanol + cyclohexane (1) Negative deviation(Q) Water + HNO3 (2) Ideal solution(R) Ethyl bromide + ethyl iodide (3) Henry's Law(S) N2(g) + H2O (l) (4) Positive deviation


E-23/36


2. List-I (Components) List-II (Nature)

(P)

HÅN NO3

–

(1) 7 + 1 (pKa log c)2

+

(Q)

COO–

H H

NH3+

(2) 7 – +b1 (pK log c)2

(R)CH3

COONa

(3) + -a b17 (pK pK )2

(S) NH4CN (4) 1 21 (pKa pKa )2

+


E-24/36


3. Matching based integer List-I List-II

(P) b-napthol Ph N NClpH8to9

Å- º¾¾¾¾¾¾®

1 (1) Nucleophilic substitution in aromatic system

(Q) Cl

23

(i )KNH / Δ(ii) liq.NH¾¾¾¾¾® (2) Electrophilic substitution in aromatic system

(R)

N NClº

3 2CH CH OHΔ¾¾¾¾¾¾® (3) SNEA

(S)

ClNO2

NO2

(i)OH / Δ(ii )HÅ¾¾¾¾¾®1

(4) Redox reaction with N2 gas evolution

4. Matching list based integer. List-I List-II Compound Positive test(P) Valeric acid (1) Tollen's test as well as FeCl3 test(Q) D-glucose (2) Iodoform test(R) Vanillin (3) NaHCO3 test

(S) O

H3C CH3(4) Molish test


E-25/36

5. List-I List-II(Consider following conversion) (Reagent-1 & Reagent-2 respectively)

(P)

OH

OH

EtO

O

Et

Reagent-2

Reagent-1

(1) (Excess conc. HCl) & (ice cold water)

(Q) Reagent-2

Reagent-1

CoCl .6H O2 2 Aq. [CoCl ]42–Aq.

(2) (Excess of dihydrogen oxide) &

(excess of conc. HCl)

(R) Reagent-2

Reagent-1

K CO2 3 KHCO3 (3) (dioxygen gas) & (dihydrogen

gas with Pd catalyst)

(S) Reagent-2

Reagent-1

BiCl3Aq. BiO Cl+ – ¯ (4) (Carbonic acid in excess) &

(Ammonium hydroxide)



E-26/36

PART-3 : MATHEMATICSSECTION-I(i) : (Maximum Marks: 28)







1. Let ( )

3 2x x 10x ; 1 x 0

ƒ x sin x ; 0 x2

1 cosx ; x2

ìï + - - £ <ï pï= £ <íï

pï + £ £ pïî

then for ƒ(x) which of the following is/are NOT TRUE(A) ƒ(x) is continuous for x Î (–1,p)(B) ƒ(x) is differentiable for x Î (–1,p)(C) ƒ(x) is non-differentiable at only one point in (–1,p)(D) ƒ(x) is non-differentiable at two points in (–1,p)



E-27/36

2. Two sides of the rhombus PQRS are parallel to the lines 7x2 – 8xy + y2 + 17x – 5y + 6 = 0. If thediagonals of rhombus intersect at point (1, 2) and P lies on Y-axis, then(A) co-ordiantes of P can be (0,0)

(B) co-ordiantes of P can be æ öç ÷è ø

50,2

(C) sum of slope of diagonals of rhombus is 32

(D) sum of slope of diagonals of rhombus is 52

3. Let a1,a2,a3, ..., a100 be in Arithmetic Progression and h1,h2, ..., h100 be in Harmonic Progression.If a1 = h1= 2 and a100 = h100 = 23, then which of the following is/are correct(A) a5 + a96 + a10 + a91 = 50 (B) h5 + h10 + h91 + h96 = 50(C) a1h100 + a4h97 = 92 (D) a2h99 + a5h96 + a8h93 = 138



E-28/36

4. Consider 5 Boxes B1,B2,B3,B4,B5. Let the probability of selecting Box Br be r

15 and box Br

contains r balls numbered from 1 to r. If a box is randomly selected and a ball is taken outthen(Where E represents the event of drawing an even numbered ball and P(Br) denotes probabilityof selecting box Br )

(A) ( ) =2P E5 (B) æ ö =ç ÷

è ø4B 1P

E 3 (C) æ ö

=ç ÷è ø4

E 1PB 2 (D)

æ ö=ç ÷

è ø4B 2P

E 3

5. Let A and B are two non-singular matrices of order 3 with real entries such that adj(A) = 2Band adj(B) = A, then (where |A| = det A)(A) |A| + |B| = 6 (B) |A| + |B| = –6(C) adj(A2B) + adj(AB2) = 4(A + 2B) (D) adj(A2B) + adj(AB2) = 4(2A + B)

6. Let Tr (for r = 1, 2, ..., 13) be the rth term from beginning in the binomial expansion of

æ ö-+ç ÷

+ -è ø

12

5

1 cosxx1 4x 1

. If ®

= lrx 0limT exists and l ¹ 0, then

(A) r = 4 (B) r = 5 (C) [l] = 3 (D) [l] = 4(Where [.] denotes greatest integer function)



E-29/36


7. Let the angles A,B,C of a triangle ABC be in Arithmetic Progression and b : c = 3 : 2 . If theinternal angle bisector of ÐA meets BC in D, DE ^ AD meets AC extended in E and AB in F,then

(A) DABC is right angled (B) = +AD 1 sin BEF

(C) DAEF is isosceles (D) =+

2bc AAD sinb c 2


E-30/36

SECTION–I(ii) : (Maximum Marks : 18)� This section contains THREE paragraphs.� Based on each paragraph, there are TWO questions.� Each question has FOUR options (A), (B), (C) and (D) ONLY ONE of these four options

is correct.� For each question, darken the bubble corresponding to the correct option in the ORS.� For each question, marks will be awarded in one of the following categories :


Paragraph for Questions 8 and 9

Let = - +uuur ˆˆ ÂB 2i 3j 6k and = - -

uuur ˆˆ ÂC 3i 4j 3k . If D,E and F are the mid-points of BC,CA and AB

respectively. P,Q and R are three points such that =

uuuruuur 4ADAP

3, =uuur uuur4BQ BE

3 and =

uuur uuur4CR CF3

then

8. Area of polygon ARBPCQ is equal to

(A) 7 34 (B) 14 34 (C) 14 18 (D) 7 189. If the vector A is at (1,1,1), then the equation of median through B is

(A) ( )= - + + l - +r ˆ ˆˆ ˆ ˆ ˆr 3i 2j 7k 2i j 10k

(B) ( )= - + + l - +r ˆ ˆˆ ˆ ˆ ˆr 2i 3j 6k i 2j 5k

(C) ( )= - + + l - +r ˆ ˆˆ ˆ ˆ ˆr 3i 2j 7k i 2j 15k

(D) ( )= - + + l - +r ˆ ˆˆ ˆ ˆ ˆr 2i 3j 6k i j 15k



E-31/36


Let ƒ(x) be a cubic polynomial with ƒ(2) = 18 and ƒ(1) = –1. Also ƒ(x) has local maxima atx = –1 and ƒ'(x) has local minima at x = 0, then

10. Number of solutions of equation |ƒ(x)| – 18 = 0, is(A) 1 (B) 2 (C) 3 (D) 4

11. Distance between the tangents drawn at point of local maxima and local minima is

(A) 19 units (B) 2 units (C) 19

2 units (D) 19 2 units


Consider the system of equations in x,y and z

x sina + y sinb + z sing = 0

x cosa + y cosb + z cosg = 0

x + y + z = 012. If system of equations has a non-trivial solution and a,b,g are angles of a triangle then the

triangle must be(A) isosceles (B) equilateral (C) right angled (D) none of these

13. If a,b,g forms an arithmetric progression and system of equations has unique solution thenwhich of the following can be the common difference of arthmetic progression

(A) p2 (B) p (C) 3p (D) 2p



E-32/36



inclusive� Each question having two matching lists. Choices for the correct combination of elements

form List-I (P), (Q), (R), (S) and List-II (1), (2), (3), (4)� Given matching type questions in which List-I (P,Q,R,S) match with List-II (1,2,3,4) and

represent your answer as four digit integer for example correct match forP ® 2, Q ® 3, R ® 1, S ® 4 .Then you have fill the answer in Section-II as 2314.

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••



1. If all roots of the equation x3 – 3x = 0 satisfy the equation(a – sin–1(sin2))x2 – (b – tan–1(tan1))x + g2 – 2g + 1 = 0 then match list I with list II

List-I List-IIP. sina – sin(b + g) 1. 2 sin2Q. cosa + cos(b + g) 2. 0R. tana + tan(b + g) 3. 2 tan2S. cota – cot(b + g) 4. –2cot2



E-33/36

2. Let A = {z1 : z130 = 1}, B = {z2 : z2

42 = 1} andC = {z3 : 1 + z3 + z3

2 +.... + z369 = 0} be three sets. Match List I with List II

List-I List-IIP. n(A Ç B) 1. 1Q. n(B Ç C) 2. 6R. n(C Ç A) 3. 9S. n(A Ç B Ç C) 4. 13

3. Number of 5 letter words which can be formed using the letters of the word CORONAVIRUSaccording to the condition given in List-I, then match List-I with List-II

List-I List-IIP. Contain exactly two pairs of alike letters and 1. 8P4

the remaining distinct letter always have alikeletters adjacent to it

Q. Containing all different letters 2. 9P5

R. Containing exactly one pair of alike letters 3. 42S. Words starting with vowels containing all 4. 8P5

distinct letterSpace for Rough Work


E-34/36

4. If the pair of lines (x – 2)2 – (m1 + m2) (x – 2) (y + 2) + m1m2(y + 2)2 = 0 is tangent to circle(x – 1)2 + (y – 1)2 = 2 (where m1 > m2) then match List-I with List II

List-I List-II

P.1

1m is equal to 1. 1

Q. |m2| is equal to 2. 7

R. Square of radius of circumcircle of triangle 3.52

formed by tangents and chord of contact, isS. If O is centre of circle and A,B are point of 4. 4

contact of tangents, P is point of intersection oftangents then area of quadrilateral AOBP is

5. Let pæ öç ÷è ø

P6

and pæ öç ÷è ø

5Q6

be two points on a hyperbola x2 – y2 = 3 with centre 'C'. L1 is tangent

line at P and L2 is normal line at Q. If from 'C' a line parallel to L2 is drawn which meets L1

at M, then match List-I with List IIList-I List-II

P. Equation of L1 is 1. x + 2y + 4 = 0Q. Equaiton of L2 is 2. 2x – y – 3 = 0

R. CM is equal to 3.35

S. Distance of point C from normal line at Q 4.45


SAMPLE PAPER-1/Paper-1

E-35/36

ALLENSpace for Rough Work

Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005

+91-744-2757575 [email protected] www.allen.ac.in

QUESTION PAPER FORMAT AND MARKING SCHEME :16. The question paper has three parts : Physics, Chemistry and Mathematics.

17. Each part has two sections as detailed in the following table.

I have read all the instructionsand shall abide by them.

____________________________Signature of the Candidate

I have verified the identity, name and Formnumber of the candidate, and that questionpaper and ORS codes are the same.

____________________________Signature of the Invigilator

NAME OF THE CANDIDATE ................................................................................................

FORM NO. .............................................

Target : JEE (Main + Advanced) 2021/SAMPLE PAPER-1/Paper-1

E-36/36

ALLEN

Que. No. Category-wise Marks for Each Question MaximumSection Type of Full Partial Zero Negative Marks of the

Que. Marks Marks Marks Marks section+4 +1 0 –2

One or more If only the bubble(s) For darkening a bubble If none In all I(i) correct 7 corresponding corresponding to each of the other 28option(s) to all the correct correct option, provided bubbles is cases

option(s) is(are) NO incorrect option darkeneddarkened darkened

Paragraph +3 0 –1Based If only the bubble If none In all

I(ii) (Single 6 corresponding to — of the other 18correct the correct option bubbles is casesoption) is darkened darkened

+4 0 Matching List If only the bubble In all

II Based Integer 5 corresponding — other — 20(1111 to 4444) to correct answer cases

is darkened

Your Target is to secure Good Rank in JEE 2021

Ïi;k bu funsZ'kks a dks /;ku ls i<+ s a

lkekU; %

1. ;g eksgjcU/k iqfLrdk vkidk iz'ui= gSA bldh eqgj rc rd u rksM+s tc rd bldk funsZ'k u fn;k tk;sA

2. iz'uksa dk mÙkj nsus ds fy, vyx ls nh x;h vkWfIVdy fjLikal 'khV (vks- vkj- ,l-) (ORS) dk mi;ksx djsaA

3. dPps dk;Z ds fy, bl iqfLrdk esa [kkyh LFkku fn;s x;s gSaA

4. bl iqfLrdk ds fiNys i`"B ij fn, x, LFkku esa viuk uke o QkWeZ uEcj fyf[k, ,oa gLrk{kj cukb;sA

5. bl iqfLrdk dh eqgj rksM+us ds ckn Ïi;k tk¡p ysa fd blesa 36 i`"B gSa vkSj izR;sd fo"k; ds lHkh 18 iz'u vkSj muds mÙkj

fodYi Bhd ls i<+ s tk ldrs gSaA ;fn ugha] rks iz'ui= dks cnyus ds fy, fujh{kd ls lEidZ djsaA

6. ijh{kkFkhZ iz'ui= dks ijh{kk dh lekIrh ij ys tk ldrs gSaA

vkWfIVdy fjLikal 'khV (vks-vkj-,l-) %

7. vks- vkj- ,l- dks ijh{kk ds lekiu ij fujh{kd ds }kjk ,d= dj fy;k tk,xkA

8. vks- vkj- ,l- esa gsj&Qsj@foÏfr u djsaA vks-vkj-,l- dk dPps dke ds fy, iz;ksx u djs aA

9. viuk uke vkSj QkWeZ uEcj vks-vkj-,l- esa fn, x, [kkuksa esa dye ls fy[ksa vkSj vius gLrk{kj djsaA buesa ls dksbZ Hkh fooj.k

vks-vkj-,l- es a dgha vkSj u fy[ksaA QkWeZ uEcj ds gj vad ds uhps vuq:i cqycqys dks dkyk djsaA

vks-vkj-,l- ij cqycqyks a dks dkyk djus dh fof/k :

10. vks-vkj-,l- ds cqycqyksa dks dkys ckWy ikWbUV dye ls dkyk djsaA

11. cqycqys dks iw.k ± :i ls dkyk djsaA

12. cqycqys dks dkyk djus dk mi;qDr rjhdk gS :

13. vks-vkj-,l- e'khu tk¡P; gSA lqfuf'pr djsa dh cqycqys lgh fof/k ls dkys fd, x;sa gSaA

14. cqycqys dks rHkh dkyk djsa tc vki mÙkj ds ckjs esa fuf'pr gksA dkys fd, gq, cqycqys dks feVkus vFkok lkQ djus dk dksbZ

rjhdk ugha gSA

15. g = 10 m/s2 iz;qDr djsa] tc rd fd vU; dksbZ eku ugha fn;k x;k gksA

fujh

{kd

ds v

uqns'k

ksa d

s fcu

k eqgj

sa u r

ksM+s

Ïi;k 'ks"k funs Z'kks a ds fy, bl iqfLrdk ds vfUre i`"B dks i<+ s aA

Time : 3 Hours Maximum Marks : 198PAPER-1

CLASSROOM CONTACT PROGRAMME(Academic Session : 2020 - 2021)

SAMPLE PAPER-1

JEE(Advanced)

REVIEW TEST

Test Pattern

Hin

di

H-2/36

dPps dk;Z ds fy, LFkku

SOME USEFUL CONSTANTSAtomic No. : H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16,

Cl = 17, Br = 35, Xe = 54, Ce = 58Atomic masses : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,

Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,Xe = 131, Ba=137, Ce = 140,

· Boltzmann constant k = 1.38 × 10–23 JK–1

· Coulomb's law constant pe9

0

1 = 9×104

· Universal gravitational constant G = 6.67259 × 10–11 N–m2 kg–2

· Speed of light in vacuum c = 3 × 108 ms–1

· Stefan–Boltzmann constant s = 5.67 × 10–8 Wm–2–K–4

· Wien's displacement law constant b = 2.89 × 10–3 m–K· Permeability of vacuum µ0 = 4p × 10–7 NA–2

· Permittivity of vacuum Î0 = 20

1

cm· Planck constant h = 6.63 × 10–34 J–s



H-3/36

BEWARE OF NEGATIVE MARKINGHAVE CONTROL ¾® HAVE PATIENCE ¾® HAVE CONFIDENCE Þ 100% SUCCESS


Hkkx-1 : HkkSfrd foKku[kaM-I(i) (vf/kdre vad : 28)

� bl [kaM esa lkr iz'u gSaA� izR;sd iz'u ds lgh mÙkj (mÙkjksa) ds fy, pkj fodYi fn, x, gSaA bl pkj fodYiksa esa ls ,d ;k ,d ls vfèkd

fodYi lgh gS(gSa)A� izR;sd iz'u ds fy,] iz'u dk (ds) mÙkj nsus gsrq lgh fodYi (fodYiksa) dks pqusA� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstuk ds vuqlkj gksxk %

iw.kZ vad : +4 ;fn dsoy (lkjs) lgh fodYi (fodYiksa) dks pquk x;k gSAvkaf'kd vad : +3 ;fn pkjksa fodYi lgh gSa ijUrq dsoy rhu fodYiksa dks pquk x;k gSAvkaf'kd vad : +2 ;fn rhu ;k rhu ls vf/kd fodYi lgh gSa ijUrq dsoy nks fodYiksa dks pquk x;k gS vkSj

pqus gq, nksuks a fodYi lgh fodYi gSaAvkaf'kd vad : +1 ;fn nks ;k nks ls vf/kd fodYi lgh gSa ijUrq dsoy ,d fodYi dks pquk x;k gS vkSj

pquk gqvk fodYi lgh fodYi gSaA'kwU; vad : 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS (vFkkZr~ iz'u vuqÙkfjr gS)A½.k vad : –2 vU; lHkh ifjfLFkfr;ksa esaA

� mnkgj.k Lo:i % ;fn fdlh iz'u ds fy, dsoy igyk] rhljk ,oa pkSFkk lgh fodYi gSa vkSj nwljk fodYi xyrgS_ rks dsoy lHkh rhu lgh fodYiks a dk p;u djus ij gh +4 vad feys axs aA fcuk dksbZ xyr fodYi pqus(bl mnkgj.k esa nwljk fodYi) rhu lgh fodYiksa esa ls flQZ nks dks pquus ij (mnkgj.kr% igyk rFkk pkSFkk fodYi)+2 vad feysaxsA fcuk dksbZ xyr fodYi pqus (bl mnkgj.k esa nwljk fodYi)] rhu lgh fodYiksa esas ls flQZ ,ddks pquus ij (igyk ;k rhljk ;k pkSFkk fodYi) +1 vad feysaxsA dksbZ Hkh xyr fodYi pquus ij (bl mnkgj.kesa nwljk fodYi), –2 vad feysaxs] pkgs lgh fodYi (fodYiksa) dks pquk x;k gks ;k u pquk x;k gksA

1. iznf'kZr ifjiFk esa /kkjk fu;=ad dh tkWfd dks nwj nka;h vksj ls cka;h vksj f[kldk;k tkrk gSA bl le; ds nkSjku fofHkUu ehVjkas

ds ikB~;kad ij fopkj dhft;sA ekuk cSVjh] lHkh oksYVehVj rFkk vehVj vkn'kZ gSA

V

A

V1

(A) V dk ikB~;kad leku cuk jgrk gSA (B) V1 dk ikB~;kad ?kVsxkA

(C) V1 dk ikB~;kad c<+sxkA (D) A dk ikB~;kad c<+sxkA


H-4/36


2. æO;eku m o yEckbZ L okyh ,d le:i NM + dks blds fdlh ,d fljs ls bl izdkj idM+dj j[kk tkrk gS fd bldk nwljk

fljk blds Å/okZ/kj Åij gksrk gSA vc fupys fljs dks idM+s gq, NM+ dks bl izdkj fojkekoLFkk ls NksM+k tkrk gS fd ;g ,d

Å/okZ/kj ry esa ?kw.kZu djrh gSA tc ;g fuEure fLFkfr ij igq¡prh gS rks ml fljs dks NksM+ fn;k tkrk gS ftls gkFkksa esa idM+k

gqvk gSA ;g ns[kk tkrk gS fd tks fljk gkFkksa esa idM+k gqvk Fkk] og /kjkry ls igys Vdjkrk gS rFkk bl le; NM+ Å/okZ/kj jgrh

gSA ftl fcUnq ls NM + dks idM+k gqvk Fkk] ml fcUnq dh /kjkry ls Åij laHkkfor Å¡pkbZ D;k gks ldrh gS\

( p2 = 10 ysa o ?k"kZ.k ux.; ekusa)

(A) 11L

6(B) 31L (C) 8L (D)

64L

3

3. fdlh 99 cm yEckbZ ds ,d vkxZu ikbZi (tks can ;k [kqyk gks ldrk gS) esa vizxkeh rjax gS ftldh lehdj.k vuqnS/; Z foLFkkiu

x = 0.3 (mm) cos [(2p /0.8) (z + 0.01 m)] cos 400 t }kjk nh xbZ gS tgk¡ z (ehVj esa) ufydk ds 'kh"kZ ls ekik x;k gS ,oa

t lsd.M esa gSA fljk la'kks/ku = 1 cm yhft;sA lgh dFku@dFkuks dks pqfu,sA

z = 0 Upper end

Lower end

z

(A) ufydk dk Åijh fljk o fupyk fljk Øe'k% [kqyk ,oa can gSA

(B) ufydk dk Åijh fljk o fupyk fljk Øe'k% can o [kqyk gSA

(C) ok;q LrEHk r`rh; lauknh esa dEiUu dj jgk gSA

(D) ok;q LrEHk f}rh; vf/kLojd esa dEiUu dj jgk gSA

4. ,d AC ifjiFk esa ,d izsjd dq.Myh L] izR;sd R izfrjks/k okys nks izfrjks/kdks ls Js.kh Øe esa tqM+h gqbZ gSA izFke izdj.k esa

izfrjks/kd ,d&nwljs ds Js.kh Øe esa gS rFkk f}rh; izdj.k esa izfrjks/kd ,d&nwljs ds lekUrj Øe esa gSA izFke izdj.k esa ifjiFk

}kjk 'kfDr O;; 810 W o f}rh; izdj.k esa 506.25 W gSA ;fn L=ksr oksYVrk dk eku 225 V gks rks %&

(A) izfrjks/k R = 20 W gSA

(B) izFke izdj.k esa L=ksr ls fuxZr /kkjk o bl ij oksYVrk ds e/; dykUrj dk eku 37° gSA

(C) f}rh; izdj.k esa L=skr ls fuxZr /kkjk o bl ij oksYVrk ds e/; dykUrj dk eku tan-1(3) gSA

(D) izsjdh; izfr?kkr dk eku 30 W gSA


H-5/36


5. rjaxnS/;Z l dk izdk'k 10m f=T;k dh xksykdkj ftad xsan ij vkifrr gksrk gSA xsan ij 1 nC vkos'k gSA ;g xsan fuokZr~ esa gSA

ftad dk dk;ZQyu 4 eV gksrk gSA

(A) ;fn l = 250 nm gks rks xsan ls bysDVªkWu ckgj fudysxsa ijUrq ; s lHkh iqu% xsan esa izos'k dj tk;sxsaA

(B) ;fn l = 350 nm gks xsan ls bysDVªkWu ckgj ugh fudysxsaA

(C) ;fn l = 300 nm gks rks xsan ls bysDVªkWu ckgj fudysxsa ijUrq ; s lHkh iqu% xsan esa izos'k dj tk;sxsaA

(D) ;fn l = 155 nm gks rks xsan ls bysDVªkWu ckgj fudysxsa ijUrq muesa ls dqN dHkh Hkh iqu% xsan esa izos'k ugha djsxsaA

6. fdlh ,d eksy ,dijekf.od vkn'kZ xSl dks fp= esa iznf'kZr izØe esa izkjfEHkd voLFkk 1 ls vafre voLFkk 4 rd ys tk;k

tkrk gSA voLFkk 4 o voLFkk 1 ds e/; rkikUrj 100 K gSA

1 2

3 4

V

T

(A) izR;sd Hkkx (1-2, 2-3, 3-4) esa xSl dh vkarfjd ÅtkZ esa ifjorZu ifjek.k esa leku gSA

(B) 2-3 rd xSl }kjk fd;k x;k dk;Z –100R gSA

(C) izØeksa dh Jà[kyk ds nkSjku xSl dks nh x;h dqy Å"ek dk eku 50R gSA

(D) vafre nkc] izkjfEHkd nkc dk vk/kk gSA

7. varfj{k esa ,d fLFkj izs{kd] f}&rkjksa ds ,d fudk; dks muds æO;eku dsUæ ds lkis{k o`Ùkkdkj xfr djrk gqvk ns[krk gSA pwafd

izs{kd] xfr ds ry esa gS rFkk cgqr nwj gS vr% mls nksuksa gh leku dks.kh; vkofÙk ls ljy vkorZ xfr djrs gq, fn[kkbZ nsrs gaSA

A o B nksuksa ds fy;s r; dyk dks.k (f) o osx ds e/; vkjs[k uhps n'kkZ;k x;k gSA (G = 1120 103

-´ Nm2/kg2 ysa)

MA

MB

RA

RB

f

Observer

A

B

60 km/s

40 km/s p 2pf

v

(A) A

B

M 3M 2

=

(B) ;fn f=T;k RA = 6 × 1010 m gks rks MA = 2 × 1020 kg gksxkA

(C) fodYi (B) esa iz;qDr vk¡dM+ksa ds mi;ksx djus ij MB = 9 × 1030 kg gksxkA

(D) fodYi (B) esa iz;qDr vk¡dM+ksa ds mi;ksx djus ij vkorZdky T = 6.28 × 106 sec gksxkA


H-6/36


[k.M–I(ii) : (vf/kdre vad : 18)� bl [k.M esa rhu vuqPNsn gSa

� izR;sd vuqPNsn ij nks iz'u fn, x;sa gSaA

� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa dsoy ,d lgh gSaA� izR;sd iz'u ds fy, vks-vkj-,l- ij lgh mÙkj fodYi ds vuq:i cqycqys dks dkyk djsaA

� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,axs %

iw.kZ vad : +3 ;fn flQZ lgh fodYi ds vuq:i cqycqys dks dkyk fd;k gSA

'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA

½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA

iz'u 8 ,oa 9 ds fy;s vuqPNsn

QkWfVZu nkcekih (Fortin’s barometer)] ikjn~ nkcekih dk ,d mUur :i gSA bldk mi;ksx ok;qe.Myh; nkc ds ;FkkFkZ

ekiu ds fy;s fd;k tkrk gSA ;g 1 m yEckbZ okyh eksVh dk¡p dh nhokjksa okyh uyh ls cuk gksrk gSA dk¡p dh uyh esa ikjs dk

,d ik= gksrk gSA ik= dk iSank peM+s dk cuk gksrk gS ftls ,d fn;s x;s LØw }kjk Åij ;k uhps fd;k tk ldrk gSA bl laiw.kZ

nkcekih dks ihry dh uyh esa j[kk x;k gS rFkk blesa yxh gqbZ dk¡p dh f[kM+fd;ksa }kjk ikB~;kad dks i<+k tk ldrk gSA ikjs ls

Hkjs ik= ds <Ddu ij gkFkh nk¡r ls cuk gqvk ,d IokbUVj yxk gksrk gSA bl ladsr dh uksad] dk¡p dh uyh ij yxs Å/okZ/kj

iSekus ds 'kwU; ds laikrh gksrk gSA dk¡p dh uyh ds Åijh fljs ij ,d ofuZ;j iSekuk yxk gksrk gS] tks LØw dh lgk;rk ls

Åij&uhps xfr dj ldrk gSA ofuZ;j iSekus dh lgk;rk ls nkcekih esa ikjn~ LrEHk dh Å¡pkbZ Kkr dh tkrh gSA fn;s x;s LØw

R dks lek;ksftr dj uyh esa ikjn~ LrEHk ds mÙkykdkj uopUæd dks ofuZ;j ds 'kwU; ds lkFk laikrh dj mijksDr izfØ;k laiUu

dh tkrh gSA rFkkfi bl midj.k esa Øec¼ =qfV;kas ds nks laHkkfor L=ksr gksrs gSA

1. ;g iSekuk 20o lsfYl;l ij va'kkfdr fd;k x;k gSA rkieku ds c<+us ;k ?kVus ij ikjs ds izlkj ls ikB~;kad ifjofrZr gks tk;sxkA

dk¡p dh uyh o iSekus ds izlkj dks ux.; ekuk tk ldrk gSA

2. QkWfVZu nkcekih dks Bhd Å/okZ/kj gh j[kuk gksrk gSA ;fn ;g Å/okZ/kj fLFkfr esa ugh gks rks ikB~;kad] okLrfod nkc ugha

n'kkZ;sxkA

pwafd uyh pkSM+h gksrh gS] vr% i`"B ruko dk izHkko ux.; ekuk tk ldrk gSA


H-7/36

0

Glass tubeSlit

Screw RVernier scale

Main scale

Brass tube

Screw S

Grass vessel

Ivory pointer

Thermometer

Mercury

MercuryLeather cup

0

ekuk ofuZ;j iSekus ij 20 Hkkx gS tks eq[; iSekus ds 19 Hkkxksa ds laikrh gSA eq[; iSekus dk izR;sd Hkkx 1 mm gSA

8. lgh dFku pqfu;s%&

(A) ;fn iSekus dks gYdk lk > qdk;k tk;s rks QkWfVZu nkcekih dk ikB~;kad okLrfod eku ls vf/kd gksxk rFkk ;fn ikjs dk

rkieku 20o C ls vf/kd gks rks nkcekih dk ikB~;kad Hkh okLrfod eku ls vf/kd gksxkA

(B) ;fn iSekus dks gYdk lk > qdk;k tk;s rks QkWfVZu nkcekih dk ikB~;kad okLrfod eku ls de gksxk rFkk ;fn ikjs dk rkieku

20o C ls vf/kd gks rks nkcekih dk ikB~;kad okLrfod eku ls vf/kd gksxkA

(C) ;fn iSekus dks gYdk lk > qdk;k tk;s rks QkWfVZu nkcekih dk ikB~;kad okLrfod eku ls de gksxk rFkk ;fn ikjs dk rkieku

20o C ls vf/kd gks rks nkcekih dk ikB~;kad Hkh okLrfod eku ls de gksxkA

(D) ;fn iSekus dks gYdk lk > qdk;k tk;s rks QkWfVZu nkcekih dk ikB~;kad okLrfod eku ls vf/kd gksxk rFkk ;fn ikjs dk

rkieku 20o C ls vf/kd gks rks nkcekih dk ikB~;kad Hkh okLrfod eku ls de gksxkA

9. ;fn ofuZ;j iSekus dk ikB~;kad bl çdkj gS fd 18oka Hkkx eq[; iSekus ds lkFk laikrh gS rFkk ofuZ;j dk 'wkU; 752 mm i<+rk

gS rFkk ;gk¡ dksbZ 'kwU;kadh =qfV ugha gS] rks nkc esa çfr'kr vfuf'prrk yxHkx gS :(A) 0.013% (B) 0.04% (C) 0.007% (D) 0.003%



H-8/36


gesa irk gS fd ;fn vkos'k ,d oLrq ij forfjr fd;k tk;s] rks ; s vkos'k oLrq dks ,d fuf'pr Lo&ÅtkZ nsus ds fy, vkil

esa vU;ksU; fØ;k djrs gSaA bl Lo&ÅtkZ dks dbZ rjhdksa ls ifjdfyr fd;k tk ldrk gSA muesa ls ,d rjhdk fo|qr {kS= ds

:i esa lEiw.kZ LFkku esa QSyh gqbZ ÅtkZ dks lekdfyr djuk gSA U = ò Î dVE21 2

0 .

ge bysDVªkWu dks E0 Lo&ÅtkZ okys r0 f=T;k ds ,d le:i vkosf'kr irys dks'k ds leku eku ldrs gSaA ;g ÅtkZ ukfHkdh;

vfHkfØ;kvksa esa 0.51 MeV vuqekfur dh xbZ gSA blls ge bysDVªkWu dh yxHkx f=T;k Kkr dj ldrs gSaA

10. bysDVªkWu dh yxHkx f=T;k D;k gS ?

(A) 00

2

E4eÎp (B)

00

2

E8eÎp (C)

00

2

E4eÎp (D)

00

2

E8eÎp

11. ;fn ge ;g ekusa fd bysDVªkWu ,d le:i vkosf'kr xksyk gS] rks

(A) bldh f=T;k iwoZ iz'u esa vuqekfur f=T;k ls vf/kd gksxhA

(B) bldh f=T;k iwoZ iz'u esa vuqekfur f=T;k ls de gksxhA

(C) bldh f=T;k iwoZ iz'u esa vuqekfur f=T;k ds cjkcj gksxhA

(D) dgk ugha tk ldrkA



H-9/36


ckWge ,gjksukWo izHkko : bysDVªkWuksa ds lkFk fd;s x;s YDSE iz;ksx ij fopkj dhft;sA

Source

d

W

1

S

Lines of B

L

2

IDy

nh&czksXyh ifjdYiuk ds vuqlkj] bysDVªkWu rjaxksa dh Hkk¡fr O;ogkj djrs gSa rFkk blfy;s f}fLyV dh mifLFkfr esa] os fcUnqor~

js[kk }kjk n'kkZ;s vuqlkj O;frdj.k izfr:i iznf'kZr djrs gSaA

ysfdu ;fn ge fLyVksa ds Bhd lEeq[k pqEcdh; {kS= dh ,d iryh iV~Vh (w << L) yxkrs gSa] rks bysDVªkWu dk iz{ksi iFk ,d vYi

dks.k ls eqM+ tk;sxk rFkk blfy;s dsUnzh; mfPp"B Dy nwjh Åij dh vksj foLFkkfir gks tkrk gSA bl foLFkkiu dh pqEcdh; {kS=

B, vkos'k q, pkSM+kbZ w, bysDVªkWu ds laosx p, fLyV nwjh d o L ij fuHkZjrk dks ljyrk ls n'kkZ;k tk ldrk gSA

blh ifj.kke dks DokUVe ;kaf=dh ds fl¼kUrksa dk mi;ksx djds O;qRiUu fd;k tk ldrk gSA DokUVe ;kaf=dh esa ge fuf'pr

iFk ds ckjs esa ckr ugha dj ldrsA blfy;s ge bysDVªkWu ds iz{ksi iFk ds ?kqeko ds ckjs esa ckr ugha dj ldrs gSaA ;gk¡ ;g ik;k

tkrk gS fd pqEcdh; {kS= okys LFkku ls xqtjus ij bysDVªkWu rjaxksa dh dyk esa ifjorZu gks tkrk gSA blfy;s Åijh fLyV o fupyh

fLyV ls vkus okyh bysDVªkWu rjax esa dykUrj gksrk gSA bl dykUrj ls izkIr iFkkUrj }kjk dsUnzh; mfPp"B dk foLFkkiu Dy ogh

izkIr gksrk gS tks pqEcdRo ds iz;ksx ls fpjlEer fl¼kUr }kjk izkIr gksrk gSA

12. Åijh fLyV o fupyh fLyV ls vkus okyh rjaxksa ds e/; dykUrj fdl ij fuHkZj ugha djrk gS \

(A) d rFkk p (B) d rFkk L (C) p rFkk d (D) p rFkk L13. ,d fo'ks"k O;oLFkk esa] bysDVªkWuksa dk laosx 1025 kg m/s, B = 1T, w = 10–8 m, L = 1m gS] rks d ds fdl U;wure eku

ij dsUnzh; fcUnq fufEu"B gksxk \ (h = 6.4 × 10–34 Jsec yhft;s)(A) 200 nm (B) 300 nm (C) 400 nm (D) 500 nm



H-10/36

[k.M–II : (vf/kdre vad : 20)lqesyu&lwph izdkj iz'u (pkj vadh; iw.kk±d esa mÙkj)

� bl [k.M esa ik¡p iz'u gSaA� izR;sd iz'u dk mÙkj 1111 ls 4444 rd (nksuksa 'kkfey) ds chp dk ,d pkj vadh; iw.kk±d gSA� izR;sd iz'u esa nks lwesyu lwfp;k¡ gSaA lgh vo;oksa esa esy ls fodYiksa ds fy, lwph-I (P), (Q), (R) rFkk (S) vkSj lwph-

II (1), (2) (3) rFkk (4) gSaA� fn, x, lqesyu izdkj iz'uksa esa ls lwph-I (P,Q,R,S) dks lwph-II (1,2,3,4) ls esy dhft, vkSj vius mÙkj dks rhu

vadh; iw.kk±d esa n'kkZb, tSls fd mnkgj.k P ® 2; Q ® 3; R ® 1, S ® 4 gS] rks vki [k.M-II esa mÙkj dks 2314ds :i esa nsosaA

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

lwph-II esa gj ,d lqesyu tks fd ;k rks lgh gS @ lgh ugha gSa dk iz;ksx lwph-I ls lqesyu ds fy, ,d lsT;knk ckj fd;k tk ldrk gSA tSls fd vkidk mÙkj 2233, 1333, 1433, 4321 bR;kfn ... gks ldrk gSA

� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,axs %iw.kZ vad : +4 ;fn flQZ lgh fodYi ds vuq:i cqycqys dks dkyk fd;k gSA'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA


H-11/36

14. ,d /ofu lzksr fLFkjkoLFkk esa 2m rjaxnS/;Z dh /ofu rjax mRlftZr djrk gSA /ofu dk ek/;e ds lkis{k osx 330 m/s gSA

lwph&I, ,d fLFkfr dks iznf'kZr djrh gS rFkk lwph&II, izs{kd }kjk izsf{kr rjaxnS/; Z rFkk vkofr ds lEHkkfor eku iznf'kZr djrh

gSA L=ksr çs{kd rFkk ok;q dh pky vo/ofud (subsonic) gSA

lwph&I lwph&II

(P) L=ksr izs{kd

L=ksr o izs{kd fLFkj gSA

gok

cgrh gSA (1) l = 2.5 m, f = 165 Hz

(Q)L=ksr izs{kd

L=ksr fp=kuqlkj xfr djrk gSA

ok;q rFkk izs{kd fLFkj gSA

(2) l = 2 m, f = 170 Hz

(R)L=ksr izs{kd

L=ksr o ok;q fLFkj gSA

izs{kd fp=kuqlkj xfr djrk gSA

(3) l = 1.5 m, f = 220 Hz

(S) L=ksr] ok;q rFkk isz{kd fn[kkbZ

xbZ fn'kk esa xfr djrs gSA

ok;q

L=ksr izs{kd

L=ksr o ok;q dh pky leku gSA

(4) l = 2 m, f = 160 Hz



H-12/36

15. lwph–I esa çfrfcEc ds o.kZu dk lwph–II esa fn;s x;s çdkf'kd fudk; ds lkFk feyku dhft;sA

lwph–I lwph–I

(P)

30cm

f=10cm

(1) fcEc ds leku vkdkj @ ls cM+k okLrfod çfrfcEc gSA

(Q)

object

water

air

(2) vkHkklh cM+k çfrfcEcA

(R) n=1.51mair

R=3m

(3) fcEc ds leku vkdkj @ ls NksVk vkHkklh çfrfcEc gSA

(S) 20cm

air

n=1.5

(4) okLrfod NksVk çfrfcEcA

ok;q esa fLFkr leksÙky ySal dh Qksdl nwjh 20cm gS

ftldh ik'oZ lrg jtfrr gSA



H-13/36

16. 2 kg nzO;eku dk ,d CykWd n'kkZ;s vuqlkj 800 N/m fLçax fu;rkad dh ,d Å/okZ/kj fLçax ls tqM+k gSA t = 0 ij fLçax

foJkUr gSA vc CykWd eqDr NksM+k tkrk gS] lwph-I esa le; t fn;k x;k gS rFkk lwph-II esa ml le; ij CykWd ds osx o fLFkfr

dh lwpuk nh x;h gSA

lwph-I lwph-II

(P) 15p

sec. (1) CykWd ek/; fLFkfr ds Åij gS o ek/; fLFkfr dh vksj xfreku gSA

(Q) 30p

sec. (2) CykWd ek/; fLFkfr ds uhps gS o ek/; fLFkfr dh vksj xfreku gSA

(R) 60p

sec. (3) CykWd ek/; fLFkfr ds Åij gS o ek/; fLFkfr ls nwj xfreku gSA

(S) 12p

sec. (4) CykWd ek/; fLFkfr ds uhps gS o ek/; fLFkfr ls nwj xfreku gSA



H-14/36

17. nks O;fDr ftuds nzO;eku 60 kg vkSj 80 kg gSa] ,d 20 kg ds r[rs ij [kM+s gSA nksuksa 1 m/s ds osx ls r[rs ds lkis{k dwn

ldrs gSA lwph&I esa of.kZr çR;sd ?kVuk ds ckn r[rs dk osx Kkr dhft;sA (çkjEHk esa fudk; fojkekoLFkk esa gSA)

Ram Shyam

60 kg 80 kg

Smooth level ground

+

lwph&I lwph&II

(P) jke vdsyk ck;sa dwnrk gSA (1) – 4017

m/s

(Q) ';ke vdsyk nk;sa dwnrk gSA (2) – 21

m/s

(R) jke ck;sa ,oa ';ke nk;sa ,d lkFk dwnrs gSA (3) 83

m/s

(S) jke ck;sa dwnrk gS rFkk fQj ';ke nk;sa dwnrk gSA (4) – 81

m/s



H-15/36

18. lwph–I lwph–II(P) ,d ,dy /kzqo dks x v{k ij x = +r ij bl çdkj (1) fo|qr {kS= r ds c<+us ij c<+rk gS rFkk bldk

j[kk tkrk gS fd ;g /kukRed x fn'kk esa foU;kflr uhps dh vksj ;k ½.kkRed y ?kVd gSA

gksrk gSA ge +y-v{k ij fdlh fcUnq ij (ewyfcUnq ij ugha)

fo|qr {ks= çkIr djrs gSA

(Q) f=T;k r rFkk yEckbZ ldk ,d pkyd , slh cSVjh (2) fo|qr {kS= r ds c<+us ij ?kVrk gS rFkk bldk

ls tksM+k tkrk gS ftldk nwljk fljk fp=kuqlkj

l

r

y

uhps dh vksj ;k ½.kkRed y ?kVd gSA

HkwlaifdZr fd;k x;k gSA pkyd dk vk;ru

r ds c<+us ;k ?kVus ij fu;r j[kk tkrk gSA

pkyd ds vUnj fo|qr {ks= fo|eku gS

(R) ,d vkn'kZ ifjukfydk esa /kkjk i gS rFkk

x r

yÄ

Pi

(3) fo|qr {kS= r ds c<+us ij c<+rk gS rFkk bldk

bldh f=T;k R gSA /kkjk le; ds lkFk Åij dh vksj ;k /kukRed y ?kVd gSA

?kV jgh gSA fcUnq P ifjukfydk ds ckgj

jgrk gSA fcUnq P ij fo|qr {ks= Kkr

fd;k tkrk gSA

(S) csyukdkj la/kkfj= dks ,d cSVjh ls r

P(4) fo|qr {kS= r ds c<+us ij ?kVrk gS rFkk bldk

tksM+k tkrk gSA ge v{k ls r nwjh ij Åij dh vksj ;k /kukRed y ?kVd gSA

P ij fo|qr {ks= ij fopkj djrs gSA



H-16/36

Hkkx-2 : jlk;u foKku[kaM-I(i) (vf/kdre vad : 28)






� mnkgj.k Lo:i % ;fn fdlh iz'u ds fy, dsoy igyk ] rhljk ,oa pkSFkk lgh fodYi gSa vkSj nwljk fodYi xyrgS_ rks dsoy lHkh rhu lgh fodYiks a dk p;u djus ij gh +4 vad feys axs aA fcuk dksbZ xyr fodYi pqus(bl mnkgj.k esa nwljk fodYi) rhu lgh fodYiksa esa ls flQZ nks dks pquus ij (mnkgj.kr% igyk rFkk pkSFkk fodYi)+2 vad feysaxsA fcuk dksbZ xyr fodYi pqus (bl mnkgj.k esa nwljk fodYi)] rhu lgh fodYiksa esas ls flQZ ,ddks pquus ij (igyk ;k rhljk ;k pkSFkk fodYi) +1 vad feysaxsA dksbZ Hkh xyr fodYi pquus ij (bl mnkgj.kesa nwljk fodYi), –2 vad feysaxs ] pkgs lgh fodYi (fodYiksa) dks pquk x;k gks ;k u pquk x;k gksA

1. ikSVsf'k;e BCC tkyd esa fØLVyhd`r gksrk gS ftldh lSy yEckbZ 'a' gS] rks dkSuls dFku lgh gS

(A) ,d bdkbZ lSy esa ,d dk; fod.kZ ds lkis{k izHkkoh :i ls 1.25 ijek.kq mifLFkr gS

(B) ,d bdkbZ lSy esa ,d Qyd fod.kZ ds lkis{k izHkkoh :i ls 1 ijek.kq mifLFkr gS

(C) ;fn 'r' ijek.kq dh f=T;k gS rks 4a 3 r=

(D) lcls mijh ijr (topmost) esa ijek.kqvksa dh leUo; la[;k 8 gS

2. izFke dksfV vfHkfØ;k

A ¾® B , Ea = 9.2 kcal / mole

500 K ij vfHkfØ;k dk nj fu;rkad 6.93 × 10–4 s–1 gS fuEu esa ls dkSuls dFku lgh gS@gSa ?

(Use : ln X = 2.3 log X, ln2 = 0.693)

(A) 1000 K ij vfHkfØ;k ds nj fu;rkad dk eku 6.93 × 10–2 s–1 gS

(B) 1000 K ij 20 lSd.M esa, A dk 75 % fo?kVu gksxk

(C) 500K ij vfHkfØ;k ds 99.9 % iwjh gksus ds fy;s 104 lSd.M dh vko';drk gksrh gS

(D) 1000 K ij 30 lSd.M eas] A dk 87.5 % fo?kVu gksxk



H-17/36

3. tc vkblksisUVsu dk Cl2 ] hn ds lkFk eksuksDyksfjuhdj.k djk;k tkrk gS ] rks fdrus izfrfcEc:i leko;h ; qXe curs

gS(A) 1 (B) 2 (C) 3 (D) 4

4. fuEu ;kSfxd dh voS/k (invalid) vuquknh lajpuk gS@gSa

NH3

(A)

NH3

(B)

NH3

(C)

NH3

(D)

NH3

5. 3NO- dh igpku ds fy, iz;ksx'kkyk esa fd;s tkus okys Hkwjh oy; ijh{k.k ds lUnHkZ esa lgh dFku gS@gSa?

(A) Hkwjh oy; cuus dk dkj.k] vk;ju ukbVªksfly ladqy dk fuekZ.k gS(B) ijh{k.k ds fy, lkUæ ukbfVªd vEy dk iz;ksx fd;k tkrk gS(C) vfHkfØ;k esa fufeZr ladqy [Fe(CN)5NO]2– gS(D) Hkwjs jax ds ladqy dh izÏfr vuqpqEcdh; gS



H-18/36

6. lgh dFku pqfu, :-(A) tc F– , F3ClO ds lkFk milgla;ksftr gksrk gS rks ladj.k sp3d ls sp3d2 esa ifjofrZr gks tkrk gS

(B) tc F3ClO ls F– gVk;k tkrk gS rks T;kfefr] <sadqyh (sea-saw) ls ifjofrZr gksdj f=Hkqth; fijkfeMh; gks tkrh gS

(C) tc F– , FClO3 ds lkFk milgla;ksftr gksrk gS rks ladj.k sp3 ls sp3d esa ifjofrZr gks tkrk gS

(D) tc F3ClO2 ls F– dks gVk;k tkrk gS rks Cl dh vkWDlhdj.k la[;k ifjofrZr ugha gksrh gS

7. ckWDlkbV ds ,d uewus ftlesa SiO2, TiO2 rFkk vk;ju ds vkWDlkbM mifLFkr gSa] dks 500K rFkk 36 bar nkc ij

lkUæ NaOH ds lkFk mipkfjr fd;k x;k gS

bl izØe esa] fuEu esa ls dkSu] foys; ugha gksrk gS?(A) Al(OH)3 (B) vk;ju vkWDlkbM (C) SiO2 (D) TiO2



H-19/36


� izR;sd vuqPNsn ij nks iz'u fn, x;sa gSaA� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa dsoy ,d lgh gSaA

� izR;sd iz'u ds fy, vks-vkj-,l- ij lgh mÙkj fodYi ds vuq:i cqycqys dks dkyk djsaA

� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,axs %





25°C rFkk 1 atm ds dqy nkc ij gkbMªkstu xSl rFkk ok;q dh lS¼kfUrd ek=k ds feJ.k dks ,d can n`<+ ik= esa

foLQksfVr fd;k x;k gS ;fn izØe dq = 0 ifjfLFkfr ij gksrk gS] rks fn;s x;s vkadM+ks dk iz;ksx djrs gq, fuEu iz'uksa ds mÙkj

nhft;s:fn;k gS (i)

(N )2VC = 6.3 Cal deg–1 mol–1 ; (ii)

(H O,g )2VC = 9.3 Cal deg–1 mol–1 ; DHf[H2O(g)] = –57.8 Kcal

fyft;s ok;q esa vk;ru }kjk 80% N2, 20% O2

8. ;fn izØe mijksDr ik= esa djk;k tkrk gS] rks izkIr gksus okyk vf/kdre rkieku (yxHkx) D;k gksxk \(A) 298 K (B) 1900 K (C) 2665 K (D) 2940 K

9. vfUre nkc atm esa D;k gksxk (yXkHkx):(A) 0.85 (B) 5.46 (C) 7.6 (D) 8.5



H-20/36


NaOH, CaO, D dh mifLFkfr esa dkckZfDlfyd vEy ds lksfM;e yo.k ds fodkcksZfDlyhdj.k dks lksMkykbe izØe

dgk tkrk gS] bls vkdoqM (OAKWOOD) izØe Hkh dgk tkrk gS

R – C – OHO

NaOH, CaO R – C – ONaO

rdsD RNa

–CO2

NaOH

Na CO2 3

R – HAlkane

H – O – H

vfHkfØ;k dh nj µ dkcZ½.kk;u dk LFkkf;Ro

10. fuEu esa ls fdlesa fodkcksZfDlyhdj.k dh nj lokZf/kd gS

(A) CH – CH – C – OH3 2

ONaOH,CaO

Δ¾¾¾¾¾® (B) CH = CH – C – OH2

ONaOH,CaO

Δ¾¾¾¾¾®

(C) H – C C – C – OHº

ONaOH,CaO

Δ¾¾¾¾¾® (D) C – OH

O

NaOH,CaOΔ¾¾¾¾¾®

11. fuEu vfHkfØ;k esa fdrus dhVy cusxsa

C – OH

OO

CH3

NaOH,CaOΔ¾¾¾¾¾® (A)

HOHO Dry HCl

(B)

(A) 0 (B) 1 (C) 2 (D) 3



H-21/36

iz'u 12 ,oa 13 ds fy;s vuqPNsn,d vdkcZfud 'osr fØLVyh; yo.k (X) tks ty esa vklkuh ls ?kqy tkrk gS] fuEu izs{k.k nsrk gS(X(s) + lkUæ HCl) dk isLV Tokyk esa gYdk cSaxuh jax nsrk gS

Acidic solution XAq.

very small amount ofCa(OCl) .CaCl .Ca(OH) .2H O2 2 2 2 dark brown

coloration (Y)

shaken withChloroform

(P)

(Q) Violet colorationin chloroform

Colourless solutionaq. solution

of hypo

12. fuEu esa ls dkSu] ,d jsMkWDl ifjorZu ugha gS?(A) "X" ls "Y" dk fuekZ.k(B) Y dh gkbiks foy;u ds lkFk vfHkfØ;k(C) Y ds P esa ifjofrZr gksus dk izØe(D) "X"Aq. dh K3[Fe(CN)6] ds lkFk vfHkfØ;k

13. og vfHkfØ;k pqfu, ftlds vafre mRikn esa vo{ksi mRiUu gksrk gS?

(A) XAq. 3

4

(i ) AgNO Aq.(ii) Excess NH OH¾¾¾¾¾¾¾¾® (B) XAq.

3 2( i ) Hg(NO ) Aq.( ii ) Excess KI¾¾¾¾¾¾¾®

(C) XAq. 3 2(i) Pb(CH COO)

(ii) Heating¾¾¾¾¾¾¾® (D) XAq. 3BiCl (aq.)(i)

E(ii) xcess KI¾¾¾¾¾¾®



H-22/36

[k.M–II : (vf/kdre vad : 20)lqesyu&lwph izdkj iz'u (pkj vadh; iw.kk ±d es a mÙkj)

� bl [k.M esa ik¡p iz'u gSaA� izR;sd iz'u dk mÙkj 1111 ls 4444 rd (nksuksa 'kkfey) ds chp dk ,d pkj vadh; iw.kk ±d gSA� izR;sd iz'u esa nks lwesyu lwfp;k¡ gSaA lgh vo;oksa esa esy ls fodYiksa ds fy , lwph-I (P), (Q), (R) rFkk (S) vkSj

lwph-II (1), (2) (3) rFkk (4) gSaA� fn, x, lqesyu izdkj iz'uksa esa ls lwph -I (P,Q,R,S) dks lwph-II (1,2,3,4) ls esy dhft, vkSj vius mÙkj dks

rhu vadh; iw.kk±d esa n'kkZb, tSls fd mnkgj.k P ® 2; Q ® 3; R ® 1, S ® 4 gS] rks vki [k.M-II esa mÙkjdks 2314 ds :i esa nsosaA

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

lwph-II esa gj ,d lqesyu tks fd ;k rks lgh gS @ lgh ugha gS a dk iz;k sx lwph-I ls lqesyu ds fy , ,dls T;knk ckj fd;k tk ldrk gSA tSls fd vkidk mÙkj 2233, 1333, 1433, 4321 bR;kfn ... gks ldrk gSA


1. ?kVd A rFkk B dks fefJr djds ,d foy;u cuk;k x;k fefJr ?kVdks ds lkFk foy;u dh lgh izdfr dks lqesfyr

dhft;s -

lwph-I (?kVd) lwph-II (izd`fr)

(P) ,sFksukWy + lkbDyksgsDlsu (1) ½.kkRed fopyu

(Q) ty + HNO3 (2) vkn'kZ foy;u

(R) ,sfFky czksekbM + ,sfFky vk;ksMkbM (3) gSujh fu;e

(S) N2(g) + H2O (l) (4) /kukRed fopyu



H-23/36

2. lwph-I (?kVd) lwph-II (izd`fr)

(P)

HÅN NO3

–

(1) 7 + 1 (pKa log c)2

+

(Q)

COO–

H H

NH3+

(2) 7 – +b1 (pK log c)2

(R)CH3

COONa

(3) + -a b17 (pK pK )2

(S) NH4CN (4) 1 21 (pKa pKa )2

+



H-24/36

3. lwph-I lwph-II

(P) b-us¶FkkWy Ph N NClpH8to9

Å- º¾¾¾¾¾¾®

1 (1) ,sjksesfVd ra= esa ukfHkd Lusgh izfrLFkkiu

(Q) Cl

23

(i )KNH / Δ(ii) liq.NH¾¾¾¾¾® (2) ,sjksesfVd ra= esa bySDVªkWu Lusgh izfrLFkkiu

(R)

N NClº

3 2CH CH OHΔ¾¾¾¾¾¾® (3) SNEA

(S)

ClNO2

NO2

(i)OH / Δ(ii )HÅ¾¾¾¾¾®1

(4) N2 xSl mRltZu ds lkFk jsMkWDl vfHkfØ;k

4. lwph-I lwph-II

;k Sfxd /kukRed ijh{k.k

(P) osysfjd vEy (1) VkWysUl ds lkFk&lkFk FeCl3 ijh{k.k

(Q) D-Xywdksl (2) vk;ksMksQkeZ ijh{k.k

(R) osfufyu (3) NaHCO3 ijh{k.k

(S) O

H3C CH3(4) eksfy'k ijh{k.k



H-25/36

5. lwph-I lwph-II(fuEu :ikUrj.kk s a ij fopkj dhft,) (Øe'k% vfHkdeZd-1 rFkk vfHkdeZd-2)

(P)

OH

OH

EtO

O

Et

Reagent-2

Reagent-1

(1) (lkUæ HCl dk vkf/kD;) rFkk (cQZ tSlk BaMk ikuh)

(Q) Reagent-2

Reagent-1

CoCl .6H O2 2 Aq. [CoCl ]42–Aq.

(2) (MkbgkbMªkstu vkWDlkbM dk vkf/kD;) rFkk (lkUæ HCl

dk vkf/kD;)

(R) Reagent-2

Reagent-1

K CO2 3 KHCO3 (3) (MkbZvkWDlhtu xSl) rFkk (Pd mRiz sjd ds lkFk

MkbgkbMªk stu xSl)

(S) Reagent-2

Reagent-1

BiCl3Aq. BiO Cl+ – ¯ (4) (dkcksZfud vEy] vf/kdrk esa) rFkk (veksfu;e

gkbMªksDl kbM)dPps dk;Z ds fy, LFkku


H-26/36


Hkkx-3 : xf.kr[kaM-I(i) (vf/kdre vad : 28)






� mnkgj.k Lo:i % ;fn fdlh iz'u ds fy, dsoy igyk] rhljk ,oa pkSFkk lgh fodYi gSa vkSj nwljk fodYi xyrgS_ rks dsoy lHkh rhu lgh fodYiks a dk p;u djus ij gh +4 vad feys axs aA fcuk dksbZ xyr fodYi pqus(bl mnkgj.k esa nwljk fodYi) rhu lgh fodYiksa esa ls flQZ nks dks pquus ij (mnkgj.kr% igyk rFkk pkSFkk fodYi)+2 vad feysaxsA fcuk dksbZ xyr fodYi pqus (bl mnkgj.k esa nwljk fodYi)] rhu lgh fodYiksa esas ls flQZ ,ddks pquus ij (igyk ;k rhljk ;k pkSFkk fodYi) +1 vad feysaxsA dksbZ Hkh xyr fodYi pquus ij (bl mnkgj.kesa nwljk fodYi), –2 vad feysaxs] pkgs lgh fodYi (fodYiksa) dks pquk x;k gks ;k u pquk x;k gksA

1. ekuk ( )

3 2x x 10x ; 1 x 0

ƒ x sin x ; 0 x2

1 cosx ; x2

ìï + - - £ <ï pï= £ <íï

pï + £ £ pïî

gks] rks ƒ(x) ds fy;s fuEu esa ls dkSulk lR; ugha gksxk\

(A) x Î (–1,p) ds fy;s ƒ(x) lrr~ gksxk

(B) x Î (–1,p) ds fy;s ƒ(x) vodyuh; gksxk

(C) vUrjky (–1,p) esa dsoy ,d fcUnq ij ƒ(x) vuvodyuh; gksxk

(D) vUrjky (–1,p) esa nks fcUnq ij ƒ(x) vuvodyuh; gksxk


H-27/36


2. leprqHkqZt PQRS dh nks Hkqtk;sa] js[kkvksa 7x2 – 8xy + y2 + 17x – 5y + 6 = 0 ds lekUrj gSA ;fn leprqHkZqt ds

fod.kZ fcUnq (1, 2) ij izfrPNsn djrs gS rFkk P, Y-v{k ij fLFkr gks] rks

(A) P ds funsZ'kkad (0,0) gks ldrs gS

(B) P ds funsZ'kkad æ öç ÷è ø

50,2 gks ldrs gS

(C) leprqHkqZt ds fod.kks± dh izo.krk dk ;ksxQy 32 gksxk

(D) leprqHkqZt ds fod.kks± dh izo.krk dk ;ksxQy 52 gksxk

3. ekuk a1,a2,a3, ..., a100 lekUrj Js<+h esa rFkk h1,h2, ..., h100 gjkRed Js<+h esa gSA ;fn a1 = h1= 2 rFkk a100 = h100 = 23gks] rks fuEu esa ls dkSulk@dkSuls lgh gksxk@gksaxs\(A) a5 + a96 + a10 + a91 = 50 (B) h5 + h10 + h91 + h96 = 50(C) a1h100 + a4h97 = 92 (D) a2h99 + a5h96 + a8h93 = 138


H-28/36


4. ekuk 5 ckWDl B1,B2,B3,B4,B5 gSA ekuk p;u fd;s x;s ckWDl Br dh izkf;drk r

15 rFkk Br ckWDl esa 1 ls r rd

la[;kvksa okyh r xsans gSA ;fn ;kn`PN;k ,d ckWDl dk p;u fd;k tkrk gS rFkk ,d xsan fudkyh tkrh gS] rks

(tgk¡ E fudkyh x;h le la[;kvksa okyh xsanksa dh ?kVuk dks n'kkZrk gS rFkk P(Br) p;fur ckWDl Br dh izkf;drk

dks n'kkZrk gS)

(A) ( ) =2P E5 (B) æ ö

=ç ÷è ø

4B 1PE 3 (C)

æ ö=ç ÷

è ø4

E 1PB 2 (D)

æ ö=ç ÷

è ø4B 2P

E 3

5. ekuk A rFkk B dksfV 3 dh okLrfod izfof"V;k¡ okys nks O;qRØe.kh; vkO;wg bl izdkj gS fd adj(A) = 2B rFkk

adj(B) = A gks] rks (tgk¡ |A| = det A)(A) |A| + |B| = 6 (B) |A| + |B| = –6(C) adj(A2B) + adj(AB2) = 4(A + 2B) (D) adj(A2B) + adj(AB2) = 4(2A + B)

6. ekuk æ ö-

+ç ÷+ -è ø

12

5

1 cosxx1 4x 1

ds f}in izlkj esa izkjEHk ls rok¡ in Tr (r = 1, 2, ..., 13 ds fy;s) gSA ;fn ®

= lrx 0limT

fo|eku rFkk l ¹ 0 gks] rks(A) r = 4 (B) r = 5 (C) [l] = 3 (D) [l] = 4(tgk¡ [.] egÙke iw.kk±d Qyu dks n'kkZrk gS)


H-29/36


7. ekuk f=Hkqt ABC ds dks.k A,B,C lekUrj Js<+h esa gS rFkk b : c = 3 : 2 gSA ;fn ÐA dk vUr% dks.k v¼Zd Hkqtk BC

dks D ij feyrk gS] DE ^ AD tks AC dks vkxs c<+kus ij E rFkk AB dks F ij feyrk gS] rks

(A) f=Hkqt ABC ledks.k f=Hkqt gksxk (B) = +AD 1 sin BEF

(C) f=Hkqt AEF lef}ckgq f=Hkqt gksxk (D) =+

2bc AAD sinb c 2


H-30/36


� izR;sd vuqPNsn ij nks iz'u fn, x;sa gSaA

� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa dsoy ,d lgh gSaA

� izR;sd iz'u ds fy, vks-vkj-,l- ij lgh mÙkj fodYi ds vuq:i cqycqys dks dkyk djsaA� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,axs %





ekuk = - +uuur ˆˆ ÂB 2i 3j 6k rFkk = - -

uuur ˆˆ ÂC 3i 4j 3k gSA ;fn Hkqtk BC,CA rFkk AB ds e/; fcUnq Øe'k% D,E rFkk

F gSA rhu fcUnq P,Q rFkk R bl izdkj gS fd =

uuuruuur 4ADAP

3, =uuur uuur4BQ BE

3 rFkk =

uuur uuur4CR CF3

gks] rks

8. cgqHkqt ARBPCQ dk {ks=Qy gksxk

(A) 7 34 (B) 14 34 (C) 14 18 (D) 7 18

9. ;fn lfn'k A ds funsZ'kkad (1,1,1) gks] rks fcUnq B ls xqtjus okyh ekf/;dk dk lehdj.k gksxk

(A) ( )= - + + l - +r ˆ ˆˆ ˆ ˆ ˆr 3i 2j 7k 2i j 10k

(B) ( )= - + + l - +r ˆ ˆˆ ˆ ˆ ˆr 2i 3j 6k i 2j 5k

(C) ( )= - + + l - +r ˆ ˆˆ ˆ ˆ ˆr 3i 2j 7k i 2j 15k

(D) ( )= - + + l - +r ˆ ˆˆ ˆ ˆ ˆr 2i 3j 6k i j 15k



H-31/36


ekuk ƒ(x) ,d ?kuh; cgqin gS ftlesa ƒ(2) = 18 rFkk ƒ(1) = –1 gSA ƒ(x), x = –1 ij LFkkuh; mfPp"B rFkk ƒ'(x),x = 0 LFkkuh; fufEu"B gS] rks

10. lehdj.k |ƒ(x)| – 18 = 0 ds gyksa dh la[;k gksxh(A) 1 (B) 2 (C) 3 (D) 4

11. LFkkuh; mfPp"B rFkk LFkkuh; fufEu"B ds fcUnqvksa ij [khaph xbZ Li'kZ js[kkvksa ds e/; nwjh gksxk

(A) 19 bdkbZ (B) 2 bdkbZ (C) 19

2 bdkbZ (D) 19 2 bdkbZ


ekuk x, y rFkk z esa lehdj.k fudk; fuEu gSx sina + y sinb + z sing = 0x cosa + y cosb + z cosg = 0x + y + z = 0

12. ;fn lehdj.k fudk; ds vfujFkZd gy gS rFkk f=Hkqt ds dks.k a,b,g gks] rks f=Hkqt vo'; gh gksxk

(A) lef}ckgq (B) leckgq (C) ledks.k (D) buesa ls dksbZ ugha

13. ;fn a,b,g lekUrj Js<+h ds :i esa gS rFkk lehdj.k fudk; dk vf}rh; gy gks] rks fuEu esa ls dkSulk lekUrj Js<+h dk

lkoZvUrj gks ldrk gS

(A) p2 (B) p (C) 3p (D) 2p



H-32/36

[k.M–II : (vf/kdre vad : 20)lqesyu&lwph izdkj iz'u (pkj vadh; iw.kk ±d es a mÙkj)

� bl [k.M esa ik¡p iz'u gSaA� izR;sd iz'u dk mÙkj 1111 ls 4444 rd (nksuksa 'kkfey) ds chp dk ,d pkj vadh; iw.kk ±d gSA� izR;sd iz'u esa nks lwesyu lwfp;k¡ gSaA lgh vo;oksa esa esy ls fodYiksa ds fy , lwph-I (P), (Q), (R) rFkk (S) vkSj

lwph-II (1), (2) (3) rFkk (4) gSaA� fn, x, lqesyu izdkj iz'uksa esa ls lwph-I (P,Q,R,S) dks lwph-II (1,2,3,4) ls esy dhft, vkSj vius mÙkj dks

rhu vadh; iw.kk±d esa n'kkZb, tSls fd mnkgj.k P ® 2; Q ® 3; R ® 1, S ® 4 gS] rks vki [k.M-II esa mÙkjdks 2314 ds :i esa nsosaA

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

lwph-II esa gj ,d lqesyu tks fd ;k rks lgh gS @ lgh ugha gS a dk iz;k sx lwph-I ls lqesyu ds fy, ,dls T;knk ckj fd;k tk ldrk gSA tSls fd vkidk mÙkj 2233, 1333, 1433, 4321 bR;kfn ... gks ldrk gSA


1. ;fn lehdj.k x3 – 3x = 0 ds lHkh ewy lehdj.k

(a – sin–1(sin2))x2 – (b – tan–1(tan1))x + g2 – 2g + 1 = 0 dks larq"V djrs gS] rks lwph I dk lwph II ds lkFk

feyku dhft;sA

lwph-I lwph-IIP. sina – sin(b + g) 1. 2 sin2Q. cosa + cos(b + g) 2. 0R. tana + tan(b + g) 3. 2 tan2S. cota – cot(b + g) 4. –2cot2



H-33/36

2. ekuk A = {z1 : z130 = 1}, B = {z2 : z2

42 = 1} rFkk

C = {z3 : 1 + z3 + z32 +.... + z3

69 = 0} rhu leqPp; gSA lwph–I dk lwph–II ds lkFk feyku dhft;sA

lwph-I lwph-IIP. n(A Ç B) 1. 1Q. n(B Ç C) 2. 6R. n(C Ç A) 3. 9S. n(A Ç B Ç C) 4. 13

3. 'kCn CORONAVIRUS ds v{kjksa esa ls ikap v{kjksa dks ysdj lwph-I esa fn;s x;s izfrca/k ds vuqlkj cuk;s tk ldus

okys 'kCn rc lwph-I dk lwph-II ds lkFk feyku dhft;s

lwph-I lwph-II

P. le:i v{kjksa ds Bhd nks ;qXe fo|eku gS rFkk 1. 8P4

cps gq;s fHkUu v{kj ds layXu (vkl&ikl okys)

lnSo le:i v{kj gks

Q. lHkh fHkUu v{kj fo|eku gks 2. 9P5

R. le:i v{kjks a dk Bhd ,d ; qXe fo|eku gks 3. 42

S. Loj ls 'kq: gksus okys ftlesa lHkh fHkUu v{kj fo|eku gS 4. 8P5



H-34/36

4. ;fn js[kkvksa dk ;qXe (x – 2)2 – (m1 + m2) (x – 2) (y + 2) + m1m2(y + 2)2 = 0, o`Ùk (x – 1)2 + (y – 1)2

= 2 (tgk¡ m1 > m2) dh Li'kZ js[kk gks] rks lwph-I dk lwph-II ds lkFk feyku dhft;s

lwph-I lwph-II

P.1

1m gksxk 1. 1

Q. |m2| gksxk 2. 7

R. Li'kZ js[kk rFkk Li'kZ thok }kjk fufeZr f=Hkqt 3.52

ds ifjo`Ùk dh f=T;k dk oxZ gksxk

S. ;fn o`Ùk dk dsUæ O rFkk Li'kZ js[kk ds Li'kZ fcUnq A,B rFkk 4. 4Li'kZ js[kkvksa dk izfrPNsn fcUnq P gks] rks prqHkqZt AOBP dk

{ks=Qy Kkr dhft;s

5. ekuk pæ öç ÷è ø

P6

rFkk pæ öç ÷è ø

5Q6

, vfrijoy; x2 – y2 = 3 ij nks fcUnq gSA ftldk dsUæ 'C' gSA fcUnq P ij Li'kZ js[kk

L1 rFkk fcUnq Q ij vfHkyEc js[kk L2 gSA ;fn fcUnq 'C' ls L2 ds lekUrj js[kk [khaph xbZ gS tks L1 dks M ij feyrhgS] rks lwph-I dk lwph-II ds lkFk feyku dhft;s

lwph-I lwph-IIP. L1 dk lehdj.k gksxk 1. x + 2y + 4 = 0Q. L2 dk lehd.k gksxk 2. 2x – y – 3 = 0

R. CM gksxk 3.35

S. fcUnq Q ij [khaph xbZ vfHkyEc js[kk ls fcUnq C dh nwjh gksxh 4.45


H-35/36

SAMPLE PAPER-1/Paper-1ALLENdPps dk;Z ds fy, LFkku

Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005

+91-744-2757575 [email protected] www.allen.ac.in

iz'ui= dk izk:i vkSj vadu ;kstuk %16. bl iz'ui= esa rhu Hkkx gSa % HkkSfrd foKku] jlk;u foKku ,oa xf.krA17. izR;sd Hkkx esa nks [k.M gSa ftudk fooj.k fuEufyf[kr rkfydk esa fn;k x;k gSA

eSaus lHkh funsZ'kks a dks i<+ fy;k gS vkSj eS mudk

vo'; ikyu d:¡xk@d:¡xhA

____________________________

ijh{kkFkhZ ds gLrk{kj

eSaus ijh{kkFkhZ dk ifjp;] uke vkSj QkWeZ uEcj dks iwjh rjgtk¡p fy;k gS ,oa iz'u i= vkSj vks- vkj- ,l- dksM nksuks aleku gSaA

____________________________

fujh{kd ds gLrk{kj

ijh{kkFkh Z dk uke ................................................................................................

QkWeZ uEcj .............................................

H-36/36


[k.M iz'u dk iz'uks a oxkZuqlkj izR;sd iz'u ds vad [k.M es aizdkj dh iw.kZ vad vkaf'kd vad 'kwU; vad ½.k vad vf/kdÙke

la[;k vad,d ;k ,d ls +4 +1 0 –2

I(i) vf/kd lgh ;fn flQZ lkjs lgh fodYi izR;sd lgh fodYi ds vuq:i ;fn fdlh Hkh vU; lHkhfodYi 7 (fodYiksa) ds vuq:Ik cqycqys dks dkyk djus ij] cqycqys dks ifjfLFkfr;ksa 28

cqycqys (cqycqyksa) dks ;fn dksbZ xyr fodYi dkyk dkyk ugha esa dkyk fd;k x;k gS ugha fd;k gS fd;k gS

vuqPNsn ij +3 0 –1 I(ii) vk/kkfjr ;fn flQZ lgh fodYi ds ;fn fdlh Hkh vU; lHkh

(,dy lgh 6 vuq:Ik cqycqys dks — cqycqys dks dkyk ifjfLFkfr;ksa 18

fodYi) dkyk fd;k gS ugha fd;k gS esa

lqpha lqesyu +4 0 II ij vk/kkfjr ;fn flQZ lgh mÙkj ds vU; lHkh

la[;kRed eku 5 vuq:Ik cqycqys dks — ifjfLFkfr;ksa — 20

izdkj dkyk fd;k gS esa(111 ls 4444)

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