READ THE INSTRUCTIONS CAREFULLY GENERAL : 1. This sealed booklet is your Question Paper. Do not break the seal till you are told to 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 this booklet. 5. After breaking the seal of the booklet, verify that the booklet contains 36 pages and that 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 purpose on the ORS. Do not write any of these details anywhere else on the ORS. Darken the 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 correct way. 14. Darken the bubbles ONLY IF you are sure of the answer. There is NO WAY to erase or "un-darken" a darkened bubble. 15. Take g = 10 m/s 2 unless otherwise stated. Please see the last page of this booklet for rest of the instructions Time : 3 Hours Maximum Marks : 198 DO NOT BREAK THE SEALS WITHOUT BEING INSTRUCTED TO DO SO BY THE INVIGILATOR PAPER-1 Test Pattern CLASSROOM CONTACT PROGRAMME (Academic Session : 2020 - 2021) SAMPLE PAPER-1 JEE(Advanced) REVIEW TEST English
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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
� 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
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
Space for Rough Work
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Space for Rough Work
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
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.
Space for Rough Work
SAMPLE PAPER-1/Paper-1ALLEN
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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%
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
Space for Rough Work
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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
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.
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.
SAMPLE PAPER-1/Paper-1ALLEN
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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.
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.
Space for Rough Work
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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.
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
Space for Rough Work
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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
� 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. 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
Space for Rough Work
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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
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
Space for Rough Work
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Space for Rough Work
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 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)
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
SAMPLE PAPER-1/Paper-1ALLEN
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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?
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 forP ® 2, Q ® 3, R ® 1, S ® 4 .Then you have fill the answer in Section-II as 2314.
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.
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
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 )Hž¾¾¾¾®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
SAMPLE PAPER-1/Paper-1ALLEN
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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) &
� 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. 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)
Space for Rough Work
SAMPLE PAPER-1/Paper-1ALLEN
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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
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
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
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 9
Let = - +uuur ˆˆ ˆAB 2i 3j 6k and = - -
uuur ˆˆ ˆAC 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
(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
Space for Rough Work
SAMPLE PAPER-1/Paper-1ALLEN
E-31/36
Paragraph for Questions 10 and 11
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
Paragraph for Questions 12 and 13
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
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 forP ® 2, Q ® 3, R ® 1, S ® 4 .Then you have fill the answer in Section-II as 2314.
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.
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
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
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
Space for Rough Work
SAMPLE PAPER-1/Paper-1
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ALLENSpace for Rough Work
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
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
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
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
dPps dk;Z ds fy, LFkku
SAMPLE PAPER-1/Paper-1ALLEN
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?