JEE Main 2023 — 24 Jan Shift 1 (Previous Year Paper)

JEE Main · 2023 · with answers and explanations

questions
90
Total marks
300
Duration
180 minutes
Negative marking
−1

Subjects: Physics · Chemistry · Mathematics

Question 1

Physics · 4 marks · −1

Match List I with List II. List I: A. Planck's constant (h); B. Stopping potential (Vs); C. Work function (ϕ\phi); D. Momentum (p). List II: I. [M1L2T−2][M^1L^2T^{-2}]; II. [M1L1T−1][M^1L^1T^{-1}]; III. [M1L2T−1][M^1L^2T^{-1}]; IV. [M1L2T−3A−1][M^1L^2T^{-3}A^{-1}]. Choose the correct answer from the options given below:

  1. A.A-I, B-III, C-IV, D-II
  2. B.A-II, B-IV, C-III, D-I
  3. C.A-III, B-I, C-II, D-IV
  4. D.A-III, B-IV, C-I, D-II

Correct answer: D

Question 2

Physics · 4 marks · −1

A travelling wave is described by the equation y(x,t)=[0.05sin⁡(8x−4t)]y(x,t)=[0.05\sin(8x-4t)] m. The velocity of the wave is: [all the quantities are in SI unit]

  1. A.8 ms−1^{-1}
  2. B.4 ms−1^{-1}
  3. C.0.5 ms−1^{-1}
  4. D.2 ms−1^{-1}

Correct answer: C

Explanation: v=ω/k=4/8=0.5v=\omega/k=4/8=0.5 m/s.

Question 3

Physics · 4 marks · −1

The maximum vertical height to which a man can throw a ball is 136 m. The maximum horizontal distance upto which he can throw the same ball is:

  1. A.136 m
  2. B.68 m
  3. C.272 m
  4. D.192 m

Correct answer: C

Explanation: u22g=136⇒Rmax=u2g=272\frac{u^2}{2g}=136\Rightarrow R_{max}=\frac{u^2}{g}=272 m.

Question 4

Physics · 4 marks · −1

Given below are two statements: Statement I: An elevator can go up or down with uniform speed when its weight is balanced with the tension of its cable. Statement II: Force exerted by the floor of an elevator on the foot of a person standing on it is more than his/her weight when the elevator goes down with increasing speed. In the light of the above statements, choose the correct answer from the options given below:

  1. A.Both Statement I and Statement II are true
  2. B.Both Statement I and Statement II are false
  3. C.Statement I is true but Statement II is false
  4. D.Statement I is false but Statement II is true

Correct answer: C

Question 5

Physics · 4 marks · −1

As per given figure, a weightless pulley P is attached on a double inclined frictionless surfaces. The tension in the string (massless) will be (if g = 10 m/s2^2)

Figure for question 5
  1. A.(43−1)(4\sqrt3-1) N
  2. B.(43+1)(4\sqrt3+1) N
  3. C.4(3+1)4(\sqrt3+1) N
  4. D.4(3−1)4(\sqrt3-1) N

Correct answer: C

Explanation: a=4gsin⁡60∘−gsin⁡30∘5=43−1a=\frac{4g\sin60^\circ-g\sin30^\circ}{5}=4\sqrt3-1; T=1⋅(gsin⁡30∘+a)=43+4=4(3+1)T=1\cdot(g\sin30^\circ+a)=4\sqrt3+4=4(\sqrt3+1) N.

Question 6

Physics · 4 marks · −1

The weight of a body at the surface of earth is 18 N. The weight of the body at an altitude of 3200 km above the earth's surface is (given, radius of earth ReR_e = 6400 km):

  1. A.4.9 N
  2. B.19.6 N
  3. C.8 N
  4. D.9.8 N

Correct answer: C

Explanation: W=18(R1.5R)2=8W=18\left(\frac{R}{1.5R}\right)^2=8 N.

Question 7

Physics · 4 marks · −1

A 100 m long wire having cross-sectional area 6.25×10−46.25\times10^{-4} m2^2 and Young's modulus is 101010^{10} Nm−2^{-2} is subjected to a load of 250 N, then the elongation in the wire will be:

  1. A.6.25×10−66.25\times10^{-6} m
  2. B.6.25×10−36.25\times10^{-3} m
  3. C.4×10−44\times10^{-4} m
  4. D.4×10−34\times10^{-3} m

Correct answer: D

Explanation: ΔL=FLAY=250×1006.25×10−4×1010=4×10−3\Delta L=\frac{FL}{AY}=\frac{250\times100}{6.25\times10^{-4}\times10^{10}}=4\times10^{-3} m.

Question 8

Physics · 4 marks · −1

1 g of a liquid is converted to vapour at 3×1053\times10^5 Pa pressure. If 10% of the heat supplied is used for increasing the volume by 1600 cm3^3 during this phase change, then the increase in internal energy in the process will be:

  1. A.4800 J
  2. B.432000 J
  3. C.4.32×1084.32\times10^8 J
  4. D.4320 J

Correct answer: D

Explanation: W=PΔV=480W=P\Delta V=480 J =0.1Q⇒Q=4800=0.1Q\Rightarrow Q=4800 J; ΔU=4320\Delta U=4320 J.

Question 9

Physics · 4 marks · −1

Given below are two statements: Statement I: The temperature of a gas is −73∘-73^\circC. When the gas is heated to 527∘527^\circC, the root mean square speed of the molecules is doubled. Statement II: The product of pressure and volume of an ideal gas will be equal to translational kinetic energy of the molecules. In the light of the above statements, choose the correct answer from the options given below:

  1. A.Both Statement I and Statement II are true
  2. B.Both Statement I and Statement II are false
  3. C.Statement I is true but Statement II is false
  4. D.Statement I is false but Statement II is true

Correct answer: C

Explanation: 200 K to 800 K quadruples T, doubling vrmsv_{rms} (I true). PV=23KEtransPV=\frac23KE_{trans} (II false).

Question 10

Physics · 4 marks · −1

A modulating signal is a square wave, as shown in the figure. If the carrier wave is given as c(t)=2sin⁡(8πt)c(t)=2\sin(8\pi t) volts, the modulation index is:

Figure for question 10
  1. A.14\frac14
  2. B.1
  3. C.12\frac12
  4. D.13\frac13

Correct answer: C

Explanation: μ=AmAc=12\mu=\frac{A_m}{A_c}=\frac12.

Question 11

Physics · 4 marks · −1

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Photodiodes are preferably operated in reverse bias condition for light intensity measurement. Reason R: The current in the forward bias is more than the current in the reverse bias for a p−np-n junction diode. In the light of the above statements, choose the correct answer from the options given below:

  1. A.Both A and R are true and R is the correct explanation of A
  2. B.Both A and R are true but R is NOT the correct explanation of A
  3. C.A is true but R is false
  4. D.A is false but R is true

Correct answer: B

Question 12

Physics · 4 marks · −1

Consider the following radioactive decay process 84218A→αA1→β−A2→γA3→αA4→β+A5→γA6{}^{218}_{84}A\xrightarrow{\alpha}A_1\xrightarrow{\beta^-}A_2\xrightarrow{\gamma}A_3\xrightarrow{\alpha}A_4\xrightarrow{\beta^+}A_5\xrightarrow{\gamma}A_6. The mass number and the atomic number of A6A_6 are given by:

  1. A.210 and 82
  2. B.210 and 80
  3. C.210 and 84
  4. D.211 and 80

Correct answer: B

Explanation: Mass: 218−8=210218-8=210. Charge: 84−2+1−2−1=8084-2+1-2-1=80.

Question 13

Physics · 4 marks · −1

From the photoelectric effect experiment, following observations are made. Identify which of these are correct. A. The stopping potential depends only on the work function of the metal. B. The saturation current increases as the intensity of incident light increases. C. The maximum kinetic energy of a photo electron depends on the intensity of the incident light. D. Photoelectric effect can be explained using wave theory of light. Choose the correct answer from the options given below:

  1. A.B only
  2. B.A, B, D only
  3. C.B, C only
  4. D.A, C, D only

Correct answer: A

Explanation: Stopping potential depends on frequency too; KE_max is independent of intensity; wave theory fails. Only B is correct.

Question 14

Physics · 4 marks · −1

Given below are two statements: Statement I: If the Brewster's angle for the light propagating from air to glass is θB\theta_B, then Brewster's angle for the light propagating from glass to air is π2−θB\frac{\pi}{2}-\theta_B. Statement II: The Brewster's angle for the light propagating from glass to air is tan⁡−1(μg)\tan^{-1}(\mu_g) where μg\mu_g is the refractive index of glass. In the light of the above statements, choose the correct answer from the options given below:

  1. A.Both Statement I and Statement II are true
  2. B.Both Statement I and Statement II are false
  3. C.Statement I is true but Statement II is false
  4. D.Statement I is false but Statement II is true

Correct answer: C

Explanation: Glass to air: tan⁡θ=1/μg=cot⁡θB\tan\theta=1/\mu_g=\cot\theta_B, so θ=π2−θB\theta=\frac{\pi}{2}-\theta_B (I true); II gives the air-to-glass angle (false).

Question 15

Physics · 4 marks · −1

If E⃗\vec E and K⃗\vec K represent electric field and propagation vectors of the EM waves in vacuum, then magnetic field vector is given by: (ω\omega - angular frequency):

  1. A.ω(K⃗×E⃗)\omega(\vec K\times\vec E)
  2. B.ω(E⃗×K⃗)\omega(\vec E\times\vec K)
  3. C.K⃗×E⃗\vec K\times\vec E
  4. D.1ω(K⃗×E⃗)\frac{1}{\omega}(\vec K\times\vec E)

Correct answer: D

Explanation: From Faraday\'s law for a plane wave, K⃗×E⃗=ωB⃗\vec K\times\vec E=\omega\vec B.

Question 16

Physics · 4 marks · −1

A conducting circular loop of radius 10π\frac{10}{\sqrt\pi} cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is:

  1. A.emf = 100 mV
  2. B.emf = 1 mV
  3. C.emf = 10 mV
  4. D.emf = 5 mV

Correct answer: C

Explanation: A=100A=100 cm2=10−2^2=10^{-2} m2^2, dB/dt=1dB/dt=1 T/s: emf =10=10 mV (constant).

Question 17

Physics · 4 marks · −1

Two long straight wires P and Q carrying equal current 10A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F1F_1. If distance between wires is halved and currents on them are doubled, force F2F_2 on 10 cm length of wire P will be:

  1. A.8F18F_1
  2. B.10F110F_1
  3. C.F18\frac{F_1}{8}
  4. D.F110\frac{F_1}{10}

Correct answer: A

Explanation: F∝I1I2dF\propto\frac{I_1I_2}{d}: ×4×2=8F1\times4\times2=8F_1.

Question 18

Physics · 4 marks · −1

A circular loop of radius rr is carrying current I A. The ratio of magnetic field at the center of circular loop and at a distance r from the center of the loop on its axis is:

  1. A.1:321:3\sqrt2
  2. B.22:12\sqrt2:1
  3. C.1:21:\sqrt2
  4. D.32:23\sqrt2:2

Correct answer: B

Explanation: Baxis=μ0Ir22(2r2)3/2=Bc22B_{axis}=\frac{\mu_0Ir^2}{2(2r^2)^{3/2}}=\frac{B_c}{2\sqrt2}, so the ratio is 22:12\sqrt2:1.

Question 19

Physics · 4 marks · −1

As shown in the figure, a network of resistors is connected to a battery of 24V with an internal resistance of 3Ω\Omega. The currents through the resistors R4R_4 and R5R_5 are I4I_4 and I5I_5 respectively. The values of I4I_4 and I5I_5 are:

Figure for question 19
  1. A.I4=25I_4=\frac25 A and I5=85I_5=\frac85 A
  2. B.I4=85I_4=\frac85 A and I5=25I_5=\frac25 A
  3. C.I4=65I_4=\frac65 A and I5=245I_5=\frac{24}{5} A
  4. D.I4=245I_4=\frac{24}{5} A and I5=65I_5=\frac65 A

Correct answer: A

Explanation: External =1+2+4+2=9 Ω=1+2+4+2=9\ \Omega, total 12 Ω\Omega, I=2I=2 A. Voltage across R4∥R5R_4\|R_5 is 8 V: I4=820=25I_4=\frac{8}{20}=\frac25 A, I5=85I_5=\frac85 A.

Question 20

Physics · 4 marks · −1

If two charges q1q_1 and q2q_2 are separated with distance 'd' and placed in a medium of dielectric constant K. What will be the equivalent distance between charges in air for the same electrostatic force?

  1. A.dkd\sqrt k
  2. B.1.5dk1.5d\sqrt k
  3. C.2dk2d\sqrt k
  4. D.kdk\sqrt d

Correct answer: A

Explanation: q1q24πε0Kd2=q1q24πε0r2⇒r=dK\frac{q_1q_2}{4\pi\varepsilon_0Kd^2}=\frac{q_1q_2}{4\pi\varepsilon_0r^2}\Rightarrow r=d\sqrt K.

Question 21

Physics · 4 marks · −1

A block of a mass 2 kg is attached with two identical springs of spring constant 20 N/m each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is πx\frac{\pi}{\sqrt x} in SI unit. The value of xx is ________.

Figure for question 21

Correct answer: 5

Explanation: keq=40k_{eq}=40 N/m, T=2π2/40=π/5T=2\pi\sqrt{2/40}=\pi/\sqrt5, so x=5x=5.

Question 22

Physics · 4 marks · −1

Vectors ai^+bj^+k^a\hat i+b\hat j+\hat k and 2i^−3j^+4k^2\hat i-3\hat j+4\hat k are perpendicular to each other when 3a+2b=73a+2b=7, the ratio of aa to bb is x2\frac x2. The value of xx is ________.

Correct answer: 1

Explanation: 2a−3b+4=02a-3b+4=0 and 3a+2b=73a+2b=7 give a=1,b=2a=1,b=2; a/b=1/2⇒x=1a/b=1/2\Rightarrow x=1.

Question 23

Physics · 4 marks · −1

A spherical body of mass 2 kg starting from rest acquires a kinetic energy of 10000 J at the end of 5th^{th} second. The force acted on the body is ________ N.

Correct answer: 40

Explanation: v=100v=100 m/s after 5 s, a=20a=20 m/s2^2, F=40F=40 N.

Question 24

Physics · 4 marks · −1

Solid sphere A is rotating about an axis PQ. If the radius of the sphere is 5 cm then its radius of gyration about PQ will be x\sqrt x cm. The value of xx is ________.

Figure for question 24

Correct answer: 110

Explanation: k2=25(25)+100=110k^2=\frac25(25)+100=110.

Question 25

Physics · 4 marks · −1

A hole is drilled in a metal sheet. At 27∘27^\circC, the diameter of hole is 5 cm. When the sheet is heated to 177∘177^\circC, the change in the diameter of hole is d×10−3d\times10^{-3} cm. The value of d will be ________ if coefficient of linear expansion of the metal is 1.6×10−5/∘1.6\times10^{-5}/^\circC.

Correct answer: 12

Explanation: Δd=5×1.6×10−5×150=12×10−3\Delta d=5\times1.6\times10^{-5}\times150=12\times10^{-3} cm.

Question 26

Physics · 4 marks · −1

Assume that protons and neutrons have equal masses. Mass of a nucleon is 1.6×10−271.6\times10^{-27} kg and radius of nucleus is 1.5×10−15A1/31.5\times10^{-15}A^{1/3} m. The approximate ratio of the nuclear density and water density is n×1013n\times10^{13}. The value of nn is ________.

Correct answer: 11

Explanation: ρ=1.6×10−2743π(1.5×10−15)3≈1.13×1017\rho=\frac{1.6\times10^{-27}}{\frac43\pi(1.5\times10^{-15})^3}\approx1.13\times10^{17} kg m−3^{-3}; ratio ≈11×1013\approx11\times10^{13}.

Question 27

Physics · 4 marks · −1

As shown in the figure, a combination of a thin plano concave lens and a thin plano convex lens is used to image an object placed at infinity. The radius of curvature of both the lenses is 30 cm and refraction index of the material for both the lenses is 1.75. Both the lenses are placed at distance of 40 cm from each other. Due to the combination, the image of the object is formed at distance xx = ________ cm, from concave lens.

Figure for question 27

Correct answer: 120

Explanation: f=±300.75=±40f=\pm\frac{30}{0.75}=\pm40 cm. Concave lens forms a virtual image 40 cm to its left, i.e. 80 cm from the convex lens; 1v=140−180⇒v=80\frac1v=\frac1{40}-\frac1{80}\Rightarrow v=80 cm, i.e. x=120x=120 cm from the concave lens.

Question 28

Physics · 4 marks · −1

In the circuit shown in the figure, the ratio of the quality factor and the band width is ________ s.

Figure for question 28

Correct answer: 10

Explanation: Q=1RLCQ=\frac1R\sqrt{\frac LC}, bandwidth =R/L=R/L; QBW=L3/2R2C=33100×33×10−3=10\frac{Q}{BW}=\frac{L^{3/2}}{R^2\sqrt C}=\frac{3\sqrt3}{100\times3\sqrt3\times10^{-3}}=10 s.

Question 29

Physics · 4 marks · −1

A hollow cylindrical conductor has length of 3.14 m, while its inner and outer diameters are 4 mm and 8 mm respectively. The resistance of the conductor is n×10−3Ωn\times10^{-3}\Omega. If the resistivity of the material is 2.4×10−8Ω2.4\times10^{-8}\Omegam. The value of nn is ________.

Correct answer: 2

Explanation: A=π(42−22)×10−6=12π×10−6A=\pi(4^2-2^2)\times10^{-6}=12\pi\times10^{-6} m2^2; R=2.4×10−8×3.1412π×10−6=2×10−3 ΩR=\frac{2.4\times10^{-8}\times3.14}{12\pi\times10^{-6}}=2\times10^{-3}\ \Omega.

Question 30

Physics · 4 marks · −1

A stream of a positively charged particles having qm=2×1011Ckg\frac qm=2\times10^{11}\frac{C}{kg} and velocity v⃗0=3×107i^\vec v_0=3\times10^7\hat i m/s is deflected by an electric field 1.8j^1.8\hat j kV/m. The electric field exists in a region of 10 cm along xx direction. Due to the electric field, the deflection of the charge particles in the yy direction is ________ mm.

Correct answer: 2

Explanation: y=12qEm(Lv)2=12(3.6×1014)(0.13×107)2=2×10−3y=\frac12\frac{qE}{m}\left(\frac{L}{v}\right)^2=\frac12(3.6\times10^{14})\left(\frac{0.1}{3\times10^7}\right)^2=2\times10^{-3} m.

Question 31

Chemistry · 4 marks · −1

Order of Covalent bond; A. KF > KI; LiF > KF. B. KF < KI; LiF > KF. C. SnCl4_4 > SnCl2_2; CuCl > NaCl. D. LiF > KF; CuCl < NaCl. E. KF < KI; CuCl > NaCl. Choose the correct answer from the options given below:

  1. A.A, B only
  2. B.B, C only
  3. C.C, E only
  4. D.B, C, E only

Correct answer: D

Question 32

Chemistry · 4 marks · −1

In the depression of freezing point experiment A. Vapour pressure of the solution is less than that of pure solvent B. Vapour pressure of the solution is more than that of pure solvent C. Only solute molecules solidify at the freezing point D. Only solvent molecules solidify at the freezing point. Choose the most appropriate answer from the options given below:

  1. A.A only
  2. B.A and C only
  3. C.B and C only
  4. D.A and D only

Correct answer: D

Question 33

Chemistry · 4 marks · −1

Statement I: For colloidal particles, the values of colligative properties are of small order as compared to values shown by true solutions at same concentration. Statement II: For colloidal particles, the potential difference between the fixed layer and the diffused layer of same charges is called the electrokinetic potential or zeta potential. In the light of the above statements, choose the correct answer from the options given below

  1. A.Both Statement I and Statement II are true
  2. B.Both Statement I and Statement II are false
  3. C.Statement I is true but Statement II is false
  4. D.Statement I is false but Statement II is true

Correct answer: C

Question 34

Chemistry · 4 marks · −1

It is observed that characteristic X-ray spectra of elements show regularity. When frequency to the power "n" i.e. νn\nu^n of X-rays emitted is plotted against atomic number "Z", following graph is obtained. The value of "n" is

Figure for question 34
  1. A.1
  2. B.2
  3. C.3
  4. D.12\frac12

Correct answer: D

Explanation: Moseley: ν∝(Z−b)\sqrt\nu\propto(Z-b), so n=12n=\frac12.

Question 35

Chemistry · 4 marks · −1

Match List I with List II. List I: A. Reverberatory furnace; B. Electrolytic cell; C. Blast furnace; D. Zone Refining furnace. List II: I. Pig Iron; II. Aluminum; III. Silicon; IV. Copper. Choose the correct answer from the options given below:

  1. A.A-IV, B-II, C-I, D-III
  2. B.A-I, B-IV, C-II, D-III
  3. C.A-I, B-III, C-II, D-IV
  4. D.A-III, B-IV, C-I, D-II

Correct answer: A

Question 36

Chemistry · 4 marks · −1

Decreasing order of the hydrogen bonding in following forms of water is correctly represented by A. Liquid water B. Ice C. Impure water. Choose the correct answer from the options given below:

  1. A.A > B > C
  2. B.B > A > C
  3. C.C > B > A
  4. D.A = B > C

Correct answer: B

Question 37

Chemistry · 4 marks · −1

Reaction of BeO with ammonia and hydrogen fluoride gives A which on thermal decomposition gives BeF2_2 and NH4_4F. What is 'A'?

  1. A.(NH4_4)BeF3_3
  2. B.(NH4_4)2_2BeF4_4
  3. C.H3_3NBeF3_3
  4. D.(NH4_4)Be2_2F5_5

Correct answer: B

Question 38

Chemistry · 4 marks · −1

Match List I with List II. List I: A. Chlorophyll; B. Soda ash; C. Dentistry, Ornamental work; D. Used in white washing. List II: I. Na2_2CO3_3; II. CaSO4_4; III. Mg2+^{2+}; IV. Ca(OH)2_2. Choose the correct answer from the options given below:

  1. A.A-II, B-I, C-III, D-IV
  2. B.A-II, B-III, C-IV, D-I
  3. C.A-III, B-IV, C-I, D-II
  4. D.A-III, B-I, C-II, D-IV

Correct answer: D

Question 39

Chemistry · 4 marks · −1

Which of the Phosphorus oxoacid can create silver mirror from AgNO3_3 solution?

  1. A.H4_4P2_2O5_5
  2. B.(HPO3_3)n_n
  3. C.H4_4P2_2O6_6
  4. D.H4_4P2_2O7_7

Correct answer: A

Explanation: Only pyrophosphorous acid H4_4P2_2O5_5 has P–H bonds (reducing).

Question 40

Chemistry · 4 marks · −1

The magnetic moment of a transition metal compound has been calculated to be 3.87 B.M. The metal ion is

  1. A.Ti2+^{2+}
  2. B.V2+^{2+}
  3. C.Cr2+^{2+}
  4. D.Mn2+^{2+}

Correct answer: B

Explanation: 3.87 BM =15=\sqrt{15}: 3 unpaired electrons, V2+^{2+} (d3^3).

Question 41

Chemistry · 4 marks · −1

Which of the following is true about freons?

  1. A.These are chlorofluorocarbon compounds
  2. B.These are radicals of chlorine and chlorine monoxide
  3. C.These are chemicals causing skin cancer
  4. D.All radicals are called freons

Correct answer: A

Question 42

Chemistry · 4 marks · −1

The primary and secondary valencies of cobalt respectively in [Co(NH3)5Cl]Cl2[\mathrm{Co(NH_3)_5Cl}]\mathrm{Cl_2} are:

  1. A.2 and 6
  2. B.3 and 5
  3. C.2 and 8
  4. D.3 and 6

Correct answer: D

Question 43

Chemistry · 4 marks · −1

Increasing order of stability of the resonance structures is: A. OHC−C+H−CH=CH−C−H−NH2\mathrm{OHC-\overset{+}{C}H-CH=CH-\overset{-}{C}H-NH_2}; B. OHC−C−H−CH=CH−C+H−NH2\mathrm{OHC-\overset{-}{C}H-CH=CH-\overset{+}{C}H-NH_2}; C. H−C(O+)=CH−CH=CH−C−H−NH2\mathrm{H-C(\overset{+}{O})=CH-CH=CH-\overset{-}{C}H-NH_2} (oxygen drawn with a single bond and a positive charge); D. OHC−C−H−CH=CH−CH=N+H2\mathrm{OHC-\overset{-}{C}H-CH=CH-CH=\overset{+}{N}H_2}. Choose the correct answer from the options given below:

Figure for question 43
  1. A.D, C, B, A
  2. B.C, D, B, A
  3. C.C, D, A, B
  4. D.D, C, A, B

Correct answer: C

Question 44

Chemistry · 4 marks · −1

In the following given reaction, 'A' is: (cyclobutyl)C(CH3_3)=CH2_2 + HBr →\rightarrow 'A' (major product)

Figure for question 44
  1. A.2-bromo-1-methylcyclohexane with CH3 and Br on adjacent ring carbons (six-membered ring)
  2. B.1-bromo-1-methylcyclohexane
  3. C.2-bromo-1,1-dimethylcyclopentane
  4. D.1-bromo-1,2-dimethylcyclopentane

Correct answer: D

Explanation: Markovnikov protonation gives a tertiary cation next to the strained ring; ring expansion to a cyclopentyl cation followed by a methyl shift gives the tertiary cation, trapped as 1-bromo-1,2-dimethylcyclopentane.

Question 45

Chemistry · 4 marks · −1

Assertion A: Hydrolysis of an alkyl chloride is a slow reaction but in the presence of NaI, the rate of the hydrolysis increases. Reason R: I−^- is a good nucleophile as well as a good leaving group. In the light of the above statements, choose the correct answer from the options given below

  1. A.Both A and R are true and R is the correct explanation of A
  2. B.Both A and R are true but R is NOT the correct explanation of A
  3. C.A is true but R is false
  4. D.A is false but R is true

Correct answer: A

Question 46

Chemistry · 4 marks · −1

'A' and 'B' formed in the following set of reactions are: 3-(hydroxymethyl)phenol →HBr,Δ\xrightarrow{\text{HBr},\Delta} A; 3-methoxyphenol →HBr,Δ\xrightarrow{\text{HBr},\Delta} B

Figure for question 46
  1. A.A = 3-(bromomethyl)phenol, B = benzene-1,3-diol (resorcinol)
  2. B.A = 3-(bromomethyl)phenol, B = 3-bromophenol
  3. C.A = 3-bromobenzyl alcohol (Br replaces OH on ring), B = 1,3-dibromobenzene
  4. D.A = 3-(bromomethyl)phenol, B = 2-bromo-5-methoxyphenol-type (Br on ring, OCH3 retained)

Correct answer: A

Explanation: HBr substitutes the benzylic OH (A = 3-(bromomethyl)phenol) and cleaves the aryl methyl ether to the phenol (B = resorcinol); phenolic C-OH is not replaced.

Question 47

Chemistry · 4 marks · −1

Compound (X) undergoes following sequence of reactions to give the Lactone (Y): Compound (X) →(i) HCHO, KOH; (ii) KCN(alc); (iii) H3O+\xrightarrow{\text{(i) HCHO, KOH; (ii) KCN(alc); (iii) H}_3\text{O}^+} Lactone (Y) [a five-membered lactone (γ-butyrolactone ring) with OH on the carbon alpha to C=O and two CH3_3 groups on the next (β) carbon, i.e. pantolactone]. Compound (X) is

Figure for question 47
  1. A.HOH2C−C(CH3)2−CHO\mathrm{HOH_2C-C(CH_3)_2-CHO}
  2. B.HO−CH(CH3)−CH2−CH2−CH2−CHO\mathrm{HO-CH(CH_3)-CH_2-CH_2-CH_2-CHO} (drawn as H2_2C(OH)–CH2_2–CH2_2–CH2_2–CHO)
  3. C.(CH3)2CH−CHO\mathrm{(CH_3)_2CH-CHO}
  4. D.CH3−CH2−CH2−CHO\mathrm{CH_3-CH_2-CH_2-CHO} (drawn as H2_2C(CH3_3)–CH2_2–CHO)

Correct answer: C

Explanation: Isobutyraldehyde undergoes crossed aldol with HCHO to HOCH2C(CH3)2CHO; cyanohydrin formation and acid hydrolysis give the hydroxy acid that lactonises to pantolactone.

Question 48

Chemistry · 4 marks · −1

'R' formed in the following sequence of reactions is: 4-chloroacetophenone →NaCN, HOAc\xrightarrow{\text{NaCN, HOAc}} 'P' →EtOH, H+\xrightarrow{\text{EtOH, H}^+} 'Q' →(i) 2MeMgBr, (ii) H3O+\xrightarrow{\text{(i) 2MeMgBr, (ii) H}_3\text{O}^+} 'R' (major product)

Figure for question 48
  1. A.2-(4-chlorophenyl)-3-methylbutane-2,3-diol: p-Cl-C6H4-C(OH)(Me)-C(OH)Me2
  2. B.2-(4-cyanophenyl)propan-2-ol: p-NC-C6H4-C(OH)Me2
  3. C.2-(4-chlorophenyl)propan-2-ol: p-Cl-C6H4-C(OH)Me2
  4. D.2-(4-cyanophenyl)-3,3-dimethylbutan-2-ol: p-NC-C6H4-C(OH)(CH3)-C(CH3)3

Correct answer: A

Explanation: Cyanohydrin, then the nitrile is converted to the ethyl ester; two MeMgBr convert the ester into a tertiary alcohol, giving the vicinal diol ArC(OH)(Me)C(OH)Me2.

Question 49

Chemistry · 4 marks · −1

An ammoniacal metal salt solution gives a brilliant red precipitate on addition of dimethylglyoxime. The metal ion is:

  1. A.Fe2+^{2+}
  2. B.Co2+^{2+}
  3. C.Ni2+^{2+}
  4. D.Cu2+^{2+}

Correct answer: C

Question 50

Chemistry · 4 marks · −1

Given below are two statements: Statement I: Noradrenaline is a neurotransmitter. Statement II: Low level of noradrenaline is not the cause of depression in human. In the light of the above statements, choose the correct answer from the options given below

  1. A.Both Statement I and Statement II are correct
  2. B.Both Statement I and Statement II are incorrect
  3. C.Statement I is correct but Statement II is incorrect
  4. D.Statement I is incorrect but Statement II is correct

Correct answer: C

Question 51

Chemistry · 4 marks · −1

5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution? ________ Given: Molar Mass of Na, O and H is 23, 16 and 1 g mol−1^{-1} respectively

Correct answer: 180

Explanation: Stock =0.1250.45=\frac{0.125}{0.45} M; need 0.05 mol: V=0.05×4500.125=180V=0.05\times\frac{450}{0.125}=180 mL.

Question 52

Chemistry · 4 marks · −1

When Fe0.93_{0.93}O is heated in presence of oxygen, it converts to Fe2_2O3_3. The number of correct statement/s from the following is ________ A. The equivalent weight of Fe0.93_{0.93}O is Molecular weight0.79\frac{\text{Molecular weight}}{0.79} B. The number of moles of Fe2+^{2+} and Fe3+^{3+} in 1 mole of Fe0.93_{0.93}O is 0.79 and 0.14 respectively C. Fe0.93_{0.93}O is metal deficient with lattice comprising of cubic closed packed arrangement of O2−^{2-} ions D. The % composition of Fe2+^{2+} and Fe3+^{3+} in Fe0.93_{0.93}O is 85% and 15% respectively

Correct answer: 4

Question 53

Chemistry · 4 marks · −1

If wavelength of the first line of the Paschen series of hydrogen atom is 720 nm, then the wavelength of the second line of this series is ________ nm. (Nearest integer)

Correct answer: 492

Explanation: λ2=720×1/9−1/161/9−1/25=720×7/14416/225≈492\lambda_2=720\times\frac{1/9-1/16}{1/9-1/25}=720\times\frac{7/144}{16/225}\approx492 nm.

Question 54

Chemistry · 4 marks · −1

For independent processes at 300 K: Process A: ΔH=−25\Delta H=-25 kJ mol−1^{-1}, ΔS=−80\Delta S=-80 J K−1^{-1}; B: −22-22, +40+40; C: +25+25, −50-50; D: +22+22, +20+20. The number of non-spontaneous processes from the following is ________

Correct answer: 2

Explanation: ΔG\Delta G (kJ): A −1-1, B −34-34, C +40+40, D +16+16: two non-spontaneous.

Question 55

Chemistry · 4 marks · −1

The dissociation constant of acetic acid is x×10−5x\times10^{-5}. When 25 mL of 0.2 M CH3_3COONa solution is mixed with 25 mL of 0.02 M CH3_3COOH solution, the pH of the resultant solution is found to be equal to 5. The value of xx is ________

Correct answer: 10

Explanation: 5=pKa+log⁡10⇒pKa=45=pK_a+\log10\Rightarrow pK_a=4, Ka=10×10−5K_a=10\times10^{-5}.

Question 56

Chemistry · 4 marks · −1

At 298 K, a 1 litre solution containing 10 mmol of Cr2_2O72−_7^{2-} and 100 mmol of Cr3+^{3+} shows a pH of 3.0. Given: Cr2_2O72−→_7^{2-}\rightarrow Cr3+^{3+}; E∘E^\circ = 1.330V and 2.303RTF\frac{2.303RT}{F} = 0.059 V. The potential for the half cell reaction is x×10−3x\times10^{-3} V. The value of xx is ________

Correct answer: 917

Explanation: E=1.33−0.0596log⁡(0.1)2(0.01)(10−3)14=1.33−0.0596×42≈0.917E=1.33-\frac{0.059}{6}\log\frac{(0.1)^2}{(0.01)(10^{-3})^{14}}=1.33-\frac{0.059}{6}\times42\approx0.917 V, so x=917x=917.

Question 57

Chemistry · 4 marks · −1

The number of correct statement/s from the following is ________ A. Larger the activation energy, smaller is the value of the rate constant. B. The higher is the activation energy, higher is the value of the temperature coefficient. C. At lower temperatures, increase in temperature causes more change in the value of k than at higher temperature D. A plot of ln k vs 1T\frac1T is a straight line with slope equal to −EaR-\frac{E_a}{R}

Correct answer: 3

Question 58

Chemistry · 4 marks · −1

The d-electronic configuration of [CoCl4]2−[\mathrm{CoCl_4}]^{2-} in tetrahedral crystal field is emt2ne^mt_2^n. Sum of "m" and "number of unpaired electrons" is ________

Correct answer: 7

Explanation: Co2+^{2+} d7^7: e4t23e^4t_2^3, 3 unpaired; 4+3=74+3=7.

Question 59

Chemistry · 4 marks · −1

Uracil is a base present in RNA with the following structure. % of N in uracil is ________ Given: Molar mass N = 14 g mol−1^{-1}, O = 16 g mol−1^{-1}, C = 12 g mol−1^{-1}, H = 1 g mol−1^{-1}

Figure for question 59

Correct answer: 25

Explanation: C4_4H4_4N2_2O2_2, M = 112: %N =28/112=25%=28/112=25\%.

Question 60

Chemistry · 4 marks · −1

Number of moles of AgCl formed in the following reaction is ________. [Substrate: a benzene ring bearing (i) a vinylic chloride side chain –CH=CH–Cl, (ii) an aryl Cl, and (iii) a side chain –CH(Cl)–CH2_2–C(CH3_3)2_2–Cl (one benzylic secondary Cl and one tertiary Cl)] →AgNO3\xrightarrow{\text{AgNO}_3} (A) + X AgCl↓\downarrow

Figure for question 60

Correct answer: 2

Explanation: Only the benzylic and tertiary chlorides ionise; vinylic and aryl chlorides do not: 2 mol AgCl.

Question 61

Mathematics · 4 marks · −1

The equation x2−4x+[x]+3=x[x]x^2-4x+[x]+3=x[x], where [x][x] denotes the greatest integer function, has:

  1. A.exactly two solutions in (−∞,∞)(-\infty,\infty)
  2. B.a unique solution in (−∞,∞)(-\infty,\infty)
  3. C.a unique solution in (−∞,1)(-\infty,1)
  4. D.no solution

Correct answer: B

Explanation: (x−1)(x−3)=[x](x−1)(x-1)(x-3)=[x](x-1): x=1x=1, or x−[x]=3x-[x]=3 (impossible). Unique solution x=1x=1.

Question 62

Mathematics · 4 marks · −1

The relation R={(a,b):gcd⁡(a,b)=1,2a≠b,a,b∈Z}R=\{(a,b): \gcd(a,b)=1, 2a\ne b, a,b\in\mathbb{Z}\} is:

  1. A.reflexive but not symmetric
  2. B.symmetric but not transitive
  3. C.transitive but not reflexive
  4. D.neither symmetric nor transitive

Correct answer: D

Explanation: (2,1)∈R(2,1)\in R but (1,2)∉R(1,2)\notin R (since 2⋅1=22\cdot1=2); (2,3),(3,4)∈R(2,3),(3,4)\in R... transitivity also fails, e.g. (2,3),(3,2)(2,3),(3,2) but (2,2)∉R(2,2)\notin R.

Question 63

Mathematics · 4 marks · −1

Let p,q∈Rp,q\in\mathbb{R} and (1−3i)200=2199(p+iq)(1-\sqrt3i)^{200}=2^{199}(p+iq), i=−1i=\sqrt{-1}. Then p+q+q2p+q+q^2 and p−q+q2p-q+q^2 are roots of the equation.

  1. A.x2+4x+1=0x^2+4x+1=0
  2. B.x2−4x−1=0x^2-4x-1=0
  3. C.x2+4x−1=0x^2+4x-1=0
  4. D.x2−4x+1=0x^2-4x+1=0

Correct answer: D

Explanation: (1−3i)200=2200e−i200π/3=2199(−1−3i)(1-\sqrt3i)^{200}=2^{200}e^{-i200\pi/3}=2^{199}(-1-\sqrt3i): p=−1,q=−3p=-1,q=-\sqrt3. Roots 2∓32\mp\sqrt3: x2−4x+1=0x^2-4x+1=0.

Question 64

Mathematics · 4 marks · −1

Let N denote the number that turns up when a fair die is rolled. If the probability that the system of equations x+y+z=1x+y+z=1, 2x+Ny+2z=22x+Ny+2z=2, 3x+3y+Nz=33x+3y+Nz=3 has unique solution is k6\frac k6, then the sum of value of k and all possible values of N is

  1. A.21
  2. B.18
  3. C.19
  4. D.20

Correct answer: D

Explanation: Δ=(N−2)(N−3)≠0\Delta=(N-2)(N-3)\ne0 for N∈{1,4,5,6}N\in\{1,4,5,6\}: k=4k=4, sum =4+16=20=4+16=20.

Question 65

Mathematics · 4 marks · −1

If A and B are two non-zero n×nn\times n matrices such that A2+B=A2BA^2+B=A^2B, then

  1. A.A2B=IA^2B=I
  2. B.A2=IA^2=I or B=IB=I
  3. C.A2B=BA2A^2B=BA^2
  4. D.AB=IAB=I

Correct answer: C

Explanation: (A2−I)(B−I)=I(A^2-I)(B-I)=I implies (B−I)(A2−I)=I(B-I)(A^2-I)=I, i.e. BA2=A2+B=A2BBA^2=A^2+B=A^2B.

Question 66

Mathematics · 4 marks · −1

Let α\alpha be a root of the equation (a−c)x2+(b−a)x+(c−b)=0(a-c)x^2+(b-a)x+(c-b)=0 where a, b, c are distinct real numbers such that the matrix [α2α1111abc]\begin{bmatrix}\alpha^2&\alpha&1\\1&1&1\\a&b&c\end{bmatrix} is singular. Then, the value of (a−c)2(b−a)(c−b)+(b−a)2(a−c)(c−b)+(c−b)2(a−c)(b−a)\frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)} is

  1. A.3
  2. B.6
  3. C.9
  4. D.12

Correct answer: A

Explanation: With p=a−c,q=b−a,r=c−bp=a-c,q=b-a,r=c-b, p+q+r=0p+q+r=0, so the expression p3+q3+r3pqr=3\frac{p^3+q^3+r^3}{pqr}=3.

Question 67

Mathematics · 4 marks · −1

tan⁡−1(1+33+3)+sec⁡−1(8+436+33)\tan^{-1}\left(\frac{1+\sqrt3}{3+\sqrt3}\right)+\sec^{-1}\left(\sqrt{\frac{8+4\sqrt3}{6+3\sqrt3}}\right) is equal to:

  1. A.π2\frac{\pi}{2}
  2. B.π3\frac{\pi}{3}
  3. C.π4\frac{\pi}{4}
  4. D.π6\frac{\pi}{6}

Correct answer: B

Explanation: tan⁡−113+sec⁡−123=π6+π6=π3\tan^{-1}\frac1{\sqrt3}+\sec^{-1}\frac{2}{\sqrt3}=\frac\pi6+\frac\pi6=\frac\pi3.

Question 68

Mathematics · 4 marks · −1

Let y=y(x)y=y(x) be the solution of the differential equation x3dy+(xy−1)dx=0x^3dy+(xy-1)dx=0, x>0x>0, y(12)=3−ey\left(\frac12\right)=3-e. Then y(1)y(1) is equal to

  1. A.2−e2-e
  2. B.e
  3. C.3
  4. D.1

Correct answer: D

Explanation: Linear with IF e−1/xe^{-1/x}: y=1+1x+Ce1/xy=1+\frac1x+Ce^{1/x}; y(12)=3+Ce2=3−e⇒C=−e−1y(\frac12)=3+Ce^2=3-e\Rightarrow C=-e^{-1}; y(1)=2−1=1y(1)=2-1=1.

Question 69

Mathematics · 4 marks · −1

The value of ∑r=02222Cr 23Cr\sum_{r=0}^{22}{}^{22}C_r\,{}^{23}C_r is

  1. A.44C22{}^{44}C_{22}
  2. B.44C23{}^{44}C_{23}
  3. C.45C23{}^{45}C_{23}
  4. D.45C24{}^{45}C_{24}

Correct answer: C

Explanation: ∑22Cr 23C23−r=45C23\sum{}^{22}C_r\,{}^{23}C_{23-r}={}^{45}C_{23} (Vandermonde).

Question 70

Mathematics · 4 marks · −1

Let a tangent to the curve y2=24xy^2=24x meet the curve xy=2xy=2 at the points A and B. Then the mid points of such line segments AB lie on a parabola with the

  1. A.directrix 4x=−34x=-3
  2. B.directrix 4x=34x=3
  3. C.length of latus rectum 32\frac32
  4. D.length of latus rectum 2

Correct answer: B

Question 71

Mathematics · 4 marks · −1

The distance of the point (7,−3,−4)(7,-3,-4) from the plane passing through the points (2,−3,1)(2,-3,1), (−1,1,−2)(-1,1,-2) and (3,−4,2)(3,-4,2) is:

  1. A.4
  2. B.5
  3. C.424\sqrt2
  4. D.525\sqrt2

Correct answer: D

Explanation: Plane x−z=1x-z=1; distance ∣7+4−1∣2=52\frac{|7+4-1|}{\sqrt2}=5\sqrt2.

Question 72

Mathematics · 4 marks · −1

For three positive integers p, q, r, xpq2=yqr=zp2rx^{pq^2}=y^{qr}=z^{p^2r} and r=pq+1r=pq+1 such that 3, 3log⁡yx3\log_yx, 3log⁡zy3\log_zy, 7log⁡xz7\log_xz are in A.P. with common difference 12\frac12. Then r−p−qr-p-q is equal to

  1. A.2
  2. B.12
  3. C.−6
  4. D.6

Correct answer: A

Question 73

Mathematics · 4 marks · −1

Let f(x)={x2sin⁡(1x),x≠00,x=0f(x)=\begin{cases}x^2\sin\left(\frac1x\right), & x\ne0\\ 0, & x=0\end{cases}. Then at x=0x=0

  1. A.ff is continuous but not differentiable
  2. B.f′f' is continuous but not differentiable
  3. C.ff and f′f' both are continuous
  4. D.ff is continuous but f′f' is not continuous

Correct answer: D

Explanation: f′(0)=0f'(0)=0 but f′(x)=2xsin⁡1x−cos⁡1xf'(x)=2x\sin\frac1x-\cos\frac1x has no limit at 0.

Question 74

Mathematics · 4 marks · −1

lim⁡t→0(11sin⁡2t+21sin⁡2t+⋯+n1sin⁡2t)sin⁡2t\lim_{t\to0}\left(1^{\frac{1}{\sin^2t}}+2^{\frac{1}{\sin^2t}}+\dots+n^{\frac{1}{\sin^2t}}\right)^{\sin^2t} is equal to

  1. A.n(n+1)2\frac{n(n+1)}{2}
  2. B.n2+nn^2+n
  3. C.n2n^2
  4. D.n

Correct answer: D

Explanation: Factor n1/sin⁡2tn^{1/\sin^2t}: the bracket tends to 1, limit =n=n.

Question 75

Mathematics · 4 marks · −1

The area enclosed by the curves y2+4x=4y^2+4x=4 and y−2x=2y-2x=2 is:

  1. A.223\frac{22}{3}
  2. B.233\frac{23}{3}
  3. C.9
  4. D.253\frac{25}{3}

Correct answer: C

Explanation: Intersections at y=2,−4y=2,-4; area =∫−42(2−y2−y24)dy=9=\int_{-4}^{2}\left(2-\frac y2-\frac{y^2}{4}\right)dy=9.

Question 76

Mathematics · 4 marks · −1

The distance of the point (−1,9,−16)(-1,9,-16) from the plane 2x+3y−z=52x+3y-z=5 measured parallel to the line x+43=2−y4=z−312\frac{x+4}{3}=\frac{2-y}{4}=\frac{z-3}{12} is

  1. A.13213\sqrt2
  2. B.20220\sqrt2
  3. C.26
  4. D.31

Correct answer: C

Explanation: Point (−1+3t,9−4t,−16+12t)(-1+3t,9-4t,-16+12t) on plane: 41−18t=5⇒t=241-18t=5\Rightarrow t=2; distance =2×13=26=2\times13=26.

Question 77

Mathematics · 4 marks · −1

Let u⃗=i^−j^−2k^\vec u=\hat i-\hat j-2\hat k, v⃗=2i^+j^−k^\vec v=2\hat i+\hat j-\hat k, v⃗⋅w⃗=2\vec v\cdot\vec w=2 and v⃗×w⃗=u⃗+λv⃗\vec v\times\vec w=\vec u+\lambda\vec v. Then u⃗⋅w⃗\vec u\cdot\vec w is equal to

  1. A.−23-\frac23
  2. B.1
  3. C.32\frac32
  4. D.2

Correct answer: B

Explanation: Dotting with v⃗\vec v: λ=−12\lambda=-\frac12, so v⃗×w⃗=(0,−32,−32)\vec v\times\vec w=(0,-\frac32,-\frac32); with v⃗⋅w⃗=2\vec v\cdot\vec w=2 this gives w⃗=(76,−16,16)\vec w=(\frac76,-\frac16,\frac16) and u⃗⋅w⃗=1\vec u\cdot\vec w=1.

Question 78

Mathematics · 4 marks · −1

Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that QAAR=RBBP=PCCQ=12\frac{QA}{AR}=\frac{RB}{BP}=\frac{PC}{CQ}=\frac12. Then Area(ΔPQR)Area(ΔABC)\frac{\text{Area}(\Delta PQR)}{\text{Area}(\Delta ABC)} is equal to

  1. A.2
  2. B.52\frac52
  3. C.3
  4. D.4

Correct answer: C

Question 79

Mathematics · 4 marks · −1

Let Ω\Omega be the sample space and A⊆ΩA\subseteq\Omega be an event. Given below are two statements: (S1): If P(A)=0P(A)=0, then A=∅A=\emptyset (S2): If P(A)=1P(A)=1, then A=ΩA=\Omega. Then

  1. A.both (S1) and (S2) are true
  2. B.both (S1) and (S2) are false
  3. C.only (S1) is true
  4. D.only (S2) is true

Correct answer: B

Question 80

Mathematics · 4 marks · −1

The compound statement (∼(P∧Q))∨((∼P)∧Q)⇒((∼P)∧(∼Q))(\sim(P\wedge Q))\vee((\sim P)\wedge Q)\Rightarrow((\sim P)\wedge(\sim Q)) is equivalent to

  1. A.((∼P)∨Q)∧(∼Q)((\sim P)\vee Q)\wedge(\sim Q)
  2. B.(∼P)∨Q(\sim P)\vee Q
  3. C.(∼Q)∨P(\sim Q)\vee P
  4. D.((∼P)∨Q)∧((∼Q)∨P)((\sim P)\vee Q)\wedge((\sim Q)\vee P)

Correct answer: D

Question 81

Mathematics · 4 marks · −1

Let a tangent to the curve 9x2+16y2=1449x^2+16y^2=144 intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is ________

Correct answer: 7

Explanation: For x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, min intercept =a+b=4+3=7=a+b=4+3=7.

Question 82

Mathematics · 4 marks · −1

The number of 9 digit numbers, that can be formed using all the digits of the number 123412341 so that the even digits occupy only even places, is ________

Correct answer: 60

Explanation: Even digits 2,2,4,4 in 4 even places: 4!2!2!=6\frac{4!}{2!2!}=6; odd digits 1,1,1,3,3 in 5 odd places: 5!3!2!=10\frac{5!}{3!2!}=10; total 60.

Question 83

Mathematics · 4 marks · −1

The 4th^{th} term of GP is 500 and its common ratio is 1m\frac1m, m∈Nm\in\mathbb{N}. Let SnS_n denote the sum of the first n terms of this GP. If S6>S5+1S_6>S_5+1 and S7<S6+12S_7<S_6+\frac12, then the number of possible values of m is ________

Correct answer: 12

Explanation: T6=500/m2>1⇒m≤22T_6=500/m^2>1\Rightarrow m\le22; T7=500/m3<12⇒m≥11T_7=500/m^3<\frac12\Rightarrow m\ge11: 12 values.

Question 84

Mathematics · 4 marks · −1

Suppose ∑r=02023r2 2023Cr=2023×α×22022\sum_{r=0}^{2023}r^2\,{}^{2023}C_r=2023\times\alpha\times2^{2022}. Then the value of α\alpha is ________

Correct answer: 1012

Explanation: ∑r2nCr=n(n+1)2n−2=2023⋅2024⋅22021⇒α=1012\sum r^2{}^nC_r=n(n+1)2^{n-2}=2023\cdot2024\cdot2^{2021}\Rightarrow\alpha=1012.

Question 85

Mathematics · 4 marks · −1

A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is ________

Correct answer: 546

Explanation: 7C5+5C17C4+5C27C3=21+175+350=546{}^7C_5+{}^5C_1{}^7C_4+{}^5C_2{}^7C_3=21+175+350=546.

Question 86

Mathematics · 4 marks · −1

Let λ∈R\lambda\in\mathbb{R} and let the equation E be ∣x∣2−2∣x∣+∣λ−3∣=0|x|^2-2|x|+|\lambda-3|=0. Then the largest element in the set S={x+λ:xS=\{x+\lambda: x is an integer solution of E}\} is ________

Correct answer: 5

Explanation: (∣x∣−1)2=1−∣λ−3∣(|x|-1)^2=1-|\lambda-3|; integer xx needs ∣λ−3∣∈{0,1}|\lambda-3|\in\{0,1\}: λ=3,x=±2,0\lambda=3,x=\pm2,0 or λ=2,4,x=±1\lambda=2,4,x=\pm1. Max x+λ=5x+\lambda=5.

Question 87

Mathematics · 4 marks · −1

The value of 12∫03∣x2−3x+2∣dx12\int_0^3\left|x^2-3x+2\right|dx is ________

Correct answer: 22

Explanation: 56+16+56=116\frac56+\frac16+\frac56=\frac{11}6; times 12 =22=22.

Question 88

Mathematics · 4 marks · −1

The value of 8π∫0π/2(cos⁡x)2023(sin⁡x)2023+(cos⁡x)2023dx\frac8\pi\int_0^{\pi/2}\frac{(\cos x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}}dx is ________

Correct answer: 2

Explanation: The integral is π4\frac\pi4 (King property), so the value is 2.

Question 89

Mathematics · 4 marks · −1

Let C be the largest circle centred at (2,0)(2,0) and inscribed in the ellipse x236+y216=1\frac{x^2}{36}+\frac{y^2}{16}=1. If (1,α)(1,\alpha) lies on C, then 10α210\alpha^2 is equal to ________

Correct answer: 118

Explanation: Min of (6cos⁡θ−2)2+16sin⁡2θ=20cos⁡2θ−24cos⁡θ+20(6\cos\theta-2)^2+16\sin^2\theta=20\cos^2\theta-24\cos\theta+20 is 12.8 at cos⁡θ=0.6\cos\theta=0.6; 1+α2=12.8⇒10α2=1181+\alpha^2=12.8\Rightarrow10\alpha^2=118.

Question 90

Mathematics · 4 marks · −1

The shortest distance between the lines x−23=y+12=z−62\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2} and x−63=1−y2=z+80\frac{x-6}{3}=\frac{1-y}{2}=\frac{z+8}{0} is equal to ________

Correct answer: 14

Explanation: n⃗=(3,2,2)×(3,−2,0)=(4,6,−12)\vec n=(3,2,2)\times(3,-2,0)=(4,6,-12), ∣n⃗∣=14|\vec n|=14; (4,2,−14)⋅n⃗=196(4,2,-14)\cdot\vec n=196; distance 14.

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