JEE Main 2021 — 26 Aug Shift 1 (Previous Year Paper)

JEE Main · 2021 · 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

In a Screw Gauge, fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are 50 divisions on the circular scale, and the main scale moves by 0.5 mm on a complete rotation. For a particular observation the reading on the main scale is 5 mm and the 20th^{th} division of the circular scale coincides with reference line. Calculate the true reading.

  1. A.5.20 mm
  2. B.5.25 mm
  3. C.5.15 mm
  4. D.5.00 mm

Correct answer: C

Explanation: LC = 0.01 mm; zero error +0.05 mm; reading 5+0.20−0.05=5.155+0.20-0.05=5.15 mm.

Question 2

Physics · 4 marks · −1

The rms speeds of the molecules of Hydrogen, Oxygen and Carbondioxide at the same temperature are VHV_H, VOV_O and VCV_C respectively then:

  1. A.VH=VO=VCV_H=V_O=V_C
  2. B.VH=VO>VCV_H=V_O>V_C
  3. C.VH>VO>VCV_H>V_O>V_C
  4. D.VC>VO>VHV_C>V_O>V_H

Correct answer: C

Explanation: vrms∝1/Mv_{rms}\propto1/\sqrt M: 2 < 32 < 44.

Question 3

Physics · 4 marks · −1

Identify the logic operation carried out by the given circuit:

Figure for question 3
  1. A.AND
  2. B.OR
  3. C.NAND
  4. D.NOR

Correct answer: D

Explanation: Z=Aˉ⋅Bˉ=A+B‾Z=\bar A\cdot\bar B=\overline{A+B}: NOR.

Question 4

Physics · 4 marks · −1

An electric appliance supplies 6000 J/min heat to the system. If the system delivers a power of 90 W. How long it would take to increase the internal energy by 2.5×1032.5\times10^3 J?

  1. A.2.5×1012.5\times10^1 s
  2. B.2.5×1022.5\times10^2 s
  3. C.2.4×1032.4\times10^3 s
  4. D.4.1×1014.1\times10^1 s

Correct answer: B

Explanation: Net rate 100−90=10100-90=10 W; t=250t=250 s.

Question 5

Physics · 4 marks · −1

A particular hydrogen like ion emits radiation of frequency 2.92×10152.92\times10^{15} Hz when it makes transition from n=3n=3 to n=1n=1. The frequency in Hz of radiation emitted in transition from n=2n=2 to n=1n=1 will be:

  1. A.4.38×10154.38\times10^{15}
  2. B.6.57×10156.57\times10^{15}
  3. C.0.44×10150.44\times10^{15}
  4. D.2.46×10152.46\times10^{15}

Correct answer: D

Explanation: ν∝1−1n2\nu\propto1-\frac1{n^2}: 2.92×3/48/9=2.46×10152.92\times\frac{3/4}{8/9}=2.46\times10^{15} Hz.

Question 6

Physics · 4 marks · −1

Two narrow bores of diameter 5.0 mm and 8.0 mm are joined together to form a U-shaped tube open at both ends. If this U-tube contains water, what is the difference in the level of two limbs of the tube. [Take surface tension of water T=7.3×10−2T=7.3\times10^{-2} Nm−1^{-1}, angle of contact = 0, g=10g=10 ms−2^{-2} and density of water =1.0×103=1.0\times10^3 kg m−3^{-3}]

  1. A.5.34 mm
  2. B.3.62 mm
  3. C.4.97 mm
  4. D.2.19 mm

Correct answer: D

Explanation: Δh=2Tρg(1r1−1r2)=0.0146(10.0025−10.004)≈2.19\Delta h=\frac{2T}{\rho g}\left(\frac1{r_1}-\frac1{r_2}\right)=0.0146\left(\frac1{0.0025}-\frac1{0.004}\right)\approx2.19 mm.

Question 7

Physics · 4 marks · −1

In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function ϕ=2.5\phi=2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. (h=6.63×10−34h=6.63\times10^{-34} Js, c=3×108c=3\times10^8 ms−1^{-1})

  1. A.1.3 V
  2. B.1.9 V
  3. C.0.6 V
  4. D.1.1 V

Correct answer: A

Explanation: ΔV=hc(1280−1400)/e≈4.44−3.11=1.3\Delta V=hc\left(\frac1{280}-\frac1{400}\right)/e\approx4.44-3.11=1.3 V.

Question 8

Physics · 4 marks · −1

Car B overtakes another car A at a relative speed of 40 ms−1^{-1}. How fast will the image of car B appear to move in the mirror of focal length 10 cm fitted in car A, when the car B is 1.9 m away from the car A?

  1. A.0.1 ms−1^{-1}
  2. B.0.2 ms−1^{-1}
  3. C.40 ms−1^{-1}
  4. D.4 ms−1^{-1}

Correct answer: A

Explanation: Convex rear-view mirror: vi=(fu−f)2vo=(0.12)2⋅40=0.1v_i=\left(\frac{f}{u-f}\right)^2v_o=\left(\frac{0.1}{2}\right)^2\cdot40=0.1 m/s.

Question 9

Physics · 4 marks · −1

An inductor coil stores 64 J of magnetic field energy and dissipates energy at the rate of 640 W when a current of 8 A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit in seconds:

  1. A.0.2
  2. B.0.4
  3. C.0.8
  4. D.0.125

Correct answer: A

Explanation: L=2U/I2=2L=2U/I^2=2 H, R=P/I2=10 ΩR=P/I^2=10\ \Omega, τ=0.2\tau=0.2 s.

Question 10

Physics · 4 marks · −1

A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance R=3R=3 kΩ\Omega, an inductor of inductive reactance XL=250π ΩX_L=250\pi\ \Omega and an unknown capacitor. The value of capacitance to maximize the average power should be: (take π2=10\pi^2=10)

  1. A.400 μ\muF
  2. B.40 μ\muF
  3. C.25 μ\muF
  4. D.4 μ\muF

Correct answer: D

Explanation: Resonance: 1ωC=250π⇒C=1100π⋅250π=4 μ\frac{1}{\omega C}=250\pi\Rightarrow C=\frac{1}{100\pi\cdot250\pi}=4\ \muF.

Question 11

Physics · 4 marks · −1

The fractional change in the magnetic field intensity at a distance 'r' from centre on the axis of current carrying coil of radius 'a' to the magnetic field intensity at the centre of the same coil is: (Take r < a).

  1. A.32a2r2\frac32\frac{a^2}{r^2}
  2. B.23a2r2\frac23\frac{a^2}{r^2}
  3. C.23r2a2\frac23\frac{r^2}{a^2}
  4. D.32r2a2\frac32\frac{r^2}{a^2}

Correct answer: D

Explanation: B=B0(1+r2/a2)−3/2≈B0(1−32r2a2)B=B_0(1+r^2/a^2)^{-3/2}\approx B_0(1-\frac32\frac{r^2}{a^2}).

Question 12

Physics · 4 marks · −1

In the given figure, the emf of the cell is 2.2 V and if internal resistance is 0.6 Ω\Omega. Calculate the power dissipated in the whole circuit:

Figure for question 12
  1. A.2.2 W
  2. B.4.4 W
  3. C.0.65 W
  4. D.1.32 W

Correct answer: A

Explanation: The resistor network between A and B reduces to 1.6 Ω1.6\ \Omega; with r=0.6 Ωr=0.6\ \Omega the total is 2.2 Ω2.2\ \Omega, so I=1I=1 A and the power in the whole circuit is EI=2.2EI=2.2 W.

Question 13

Physics · 4 marks · −1

A solid metal sphere of radius R having charge q is enclosed inside the concentric spherical shell of inner radius a and outer radius b as shown in figure. The approximate variation electric field E⃗\vec E as a function of distance r from centre O is given by:

Figure for question 13
  1. A.E rises from r = 0 up to R, is zero between R and a... (graph 1: rising curve up to R, gap, then decaying curve beyond b)
  2. B.E is zero up to b, then decays as 1/r^2 beyond b
  3. C.E is zero for r < R, decays as 1/r^2 from R to a, is zero from a to b, and decays as 1/r^2 beyond b
  4. D.E is zero for r < R and then decays continuously as 1/r^2 from R onwards

Correct answer: C

Explanation: Inside the solid conductor E = 0; between R and a, E=kq/r2E=kq/r^2; inside the shell material E = 0; outside b, E=kq/r2E=kq/r^2.

Question 14

Physics · 4 marks · −1

The material filled between the plates of a parallel plate capacitor has resistivity 200 Ω\Omegam. The value of capacitance of the capacitor is 2 pF. If a potential difference of 40 V is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is: (given the value of relative permitivity of material is 50)

  1. A.9.0 mA
  2. B.0.9 mA
  3. C.0.9 μ\muA
  4. D.9.0 μ\muA

Correct answer: B

Explanation: RC=ρε0εrRC=\rho\varepsilon_0\varepsilon_r; I=VCρε0εr=40×2×10−12200×8.85×10−12×50≈0.9I=\frac{VC}{\rho\varepsilon_0\varepsilon_r}=\frac{40\times2\times10^{-12}}{200\times8.85\times10^{-12}\times50}\approx0.9 mA.

Question 15

Physics · 4 marks · −1

Inside a uniform spherical shell: (a) the gravitational field is zero. (b) the gravitational potential is zero. (c) the gravitational field is same everywhere. (d) the gravitation potential is same everywhere. (e) all of the above. Choose the most appropriate answer from the options given below:

  1. A.(b), (c) and (d) only
  2. B.(a), (b) and (c) only
  3. C.(a), (c) and (d) only
  4. D.(e) only

Correct answer: C

Question 16

Physics · 4 marks · −1

The initial mass of a rocket is 1000 kg. Calculate at what rate the fuel should be burnt so that the rocket is given an acceleration of 20 ms−2^{-2}. The gases come out at a relative speed of 500 ms−1^{-1} with respect to the rocket: [Use g = 10 m/s2^2]

  1. A.60 kg s−1^{-1}
  2. B.6.0×1026.0\times10^2 kg s−1^{-1}
  3. C.500 kg s−1^{-1}
  4. D.10 kg s−1^{-1}

Correct answer: A

Explanation: udmdt=m(g+a)⇒dmdt=1000×30500=60u\frac{dm}{dt}=m(g+a)\Rightarrow\frac{dm}{dt}=\frac{1000\times30}{500}=60 kg/s.

Question 17

Physics · 4 marks · −1

What equal length of an iron wire and a copper-nickel alloy wire, each of 2 mm diameter connected parallel to give an equivalent resistance of 3 Ω\Omega? (Given resistivities of iron and copper-nickel alloy wire are 12 μΩ\mu\Omega cm and 51 μΩ\mu\Omega cm respectively)

  1. A.110 m
  2. B.97 m
  3. C.90 m
  4. D.82 m

Correct answer: B

Explanation: LA⋅ρ1ρ2ρ1+ρ2=3\frac{L}{A}\cdot\frac{\rho_1\rho_2}{\rho_1+\rho_2}=3 with A=π×10−6A=\pi\times10^{-6}: L≈97L\approx97 m.

Question 18

Physics · 4 marks · −1

The magnitude of vectors OA⃗\vec{OA}, OB⃗\vec{OB} and OC⃗\vec{OC} in the given figure are equal. The direction of OA⃗+OB⃗−OC⃗\vec{OA}+\vec{OB}-\vec{OC} with xx-axis will be:

Figure for question 18
  1. A.tan⁡−1(1−3−2)(1+3+2)\tan^{-1}\frac{(1-\sqrt3-\sqrt2)}{(1+\sqrt3+\sqrt2)}
  2. B.tan⁡−1(3−1+2)(1+3−2)\tan^{-1}\frac{(\sqrt3-1+\sqrt2)}{(1+\sqrt3-\sqrt2)}
  3. C.tan⁡−1(1+3−2)(1−3−2)\tan^{-1}\frac{(1+\sqrt3-\sqrt2)}{(1-\sqrt3-\sqrt2)}
  4. D.tan⁡−1(3−1+2)(1−3+2)\tan^{-1}\frac{(\sqrt3-1+\sqrt2)}{(1-\sqrt3+\sqrt2)}

Correct answer: A

Explanation: Components (unit magnitude): x=32+12+12x=\frac{\sqrt3}{2}+\frac12+\frac{1}{\sqrt2}, y=12−32−12y=\frac12-\frac{\sqrt3}{2}-\frac1{\sqrt2}; ratio 1−3−21+3+2\frac{1-\sqrt3-\sqrt2}{1+\sqrt3+\sqrt2}.

Question 19

Physics · 4 marks · −1

Statement I: By doping silicon semiconductor with pentavalent material, the electrons density increases. Statement II: The n-type semiconductor has net negative charge. In the light of the above statements, choose the most appropriate 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: n-type material is electrically neutral.

Question 20

Physics · 4 marks · −1

If E, L, M and G denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of P in the formula P=EL2M−5G−2P=EL^2M^{-5}G^{-2} are:

  1. A.[M1L1T−2][M^1L^1T^{-2}]
  2. B.[M−1L−1T2][M^{-1}L^{-1}T^2]
  3. C.[M0L1T0][M^0L^1T^0]
  4. D.[M0L0T0][M^0L^0T^0]

Correct answer: D

Explanation: ML2T−2⋅M2L4T−2⋅M−5⋅M2L−6T4=M0L0T0ML^2T^{-2}\cdot M^2L^4T^{-2}\cdot M^{-5}\cdot M^2L^{-6}T^4=M^0L^0T^0.

Question 21

Physics · 4 marks

A uniform chain of length 3 meter and mass 3 kg overhangs a smooth table with 2 meter laying on the table. If k is the kinetic energy of the chain in joule as it completely slips off the table, then the value of k is ________. (Take g = 10 m/s2^2)

Correct answer: 40

Explanation: ΔU\Delta U: hanging 1 kg at −0.5-0.5 m to 3 kg at −1.5-1.5 m: K=45−5=40K=45-5=40 J.

Question 22

Physics · 4 marks

Consider a badminton racket with length scales as shown in the figure. If the mass of the linear and circular portions of the badminton racket are same (M) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, r2\frac r2 distance from the end A of the handle will be ________ Mr2Mr^2.

Figure for question 22

Correct answer: 52

Explanation: Rod: M(6r)212+M(2.5r)2=9.25Mr2\frac{M(6r)^2}{12}+M(2.5r)^2=9.25Mr^2; ring: Mr22+M(6.5r)2=42.75Mr2\frac{Mr^2}{2}+M(6.5r)^2=42.75Mr^2; total 52.

Question 23

Physics · 4 marks

A soap bubble of radius 3 cm is formed inside the another soap bubble of radius 6 cm. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is ________ cm.

Correct answer: 2

Explanation: 1r=13+16⇒r=2\frac1r=\frac13+\frac16\Rightarrow r=2 cm.

Question 24

Physics · 4 marks

Two travelling waves produces a standing wave represented by equation. y=1.0 mmcos⁡(1.57 cm−1)xsin⁡(78.5 s−1)ty=1.0\text{ mm}\cos(1.57\text{ cm}^{-1})x\sin(78.5\text{ s}^{-1})t. The node closest to the origin in the region x>0x>0 will be at xx = ________ cm.

Correct answer: 1

Explanation: 1.57x=π2⇒x=11.57x=\frac\pi2\Rightarrow x=1 cm.

Question 25

Physics · 4 marks

Two short magnetic dipoles m1m_1 and m2m_2 each having magnetic moment of 1 Am2^2 are placed at point O and P respectively. The distance between OP is 1 meter. The torque experienced by the magnetic dipole m2m_2 due to the presence of m1m_1 is ________ ×10−7\times10^{-7} Nm.

Figure for question 25

Correct answer: 1

Explanation: P is on the equatorial line of m1m_1: B=μ0m4πr3=10−7B=\frac{\mu_0m}{4\pi r^3}=10^{-7} T (antiparallel to m1m_1), perpendicular to m2m_2: τ=10−7\tau=10^{-7} N m.

Question 26

Physics · 4 marks

The electric field in a plane electromagnetic wave is given by E⃗=200cos⁡[(0.5×103m)x−(1.5×1011rads×t)]Vmj^\vec E=200\cos\left[\left(\frac{0.5\times10^3}{m}\right)x-\left(1.5\times10^{11}\frac{rad}{s}\times t\right)\right]\frac{V}{m}\hat j. If this wave falls normally on a perfectly reflecting surface having an area of 100 cm2^2. If the radiation pressure exerted by the E.M. wave on the surface during a 10 minute exposure is x109Nm2\frac{x}{10^9}\frac{N}{m^2}. Find the value of xx.

Correct answer: 354

Explanation: I=12cε0E02I=\frac12c\varepsilon_0E_0^2; P=2Ic=ε0E02=8.85×10−12×4×104≈354×10−9P=\frac{2I}{c}=\varepsilon_0E_0^2=8.85\times10^{-12}\times4\times10^4\approx354\times10^{-9} N m−2^{-2}.

Question 27

Physics · 4 marks

White light is passed through a double slit and interference is observed on a screen 1.5 m away. The separation between the slits is 0.3 mm. The first violet and red fringes are formed 2.0 mm and 3.5 mm away from the central white fringes. The difference in wavelengths of red and voilet light is ________ nm.

Correct answer: 300

Explanation: Δλ=Δy dD=1.5×10−3×3×10−41.5=300\Delta\lambda=\frac{\Delta y\,d}{D}=\frac{1.5\times10^{-3}\times3\times10^{-4}}{1.5}=300 nm.

Question 28

Physics · 4 marks

An amplitude modulated wave is represented by Cm(t)=10(1+0.2cos⁡12560t)sin⁡(111×104t)C_m(t)=10(1+0.2\cos12560t)\sin(111\times10^4t) volts. The modulating frequency in kHz will be ________.

Correct answer: 2

Explanation: f=125602π=2000f=\frac{12560}{2\pi}=2000 Hz =2=2 kHz.

Question 29

Physics · 4 marks

Two spherical balls having equal masses with radius of 5 cm each are thrown upwards along the same vertical direction at an interval of 3 s with the same initial velocity of 35 m/s, then these balls collide at a height of ________ m. (take g = 10 m/s2^2)

Correct answer: 50

Explanation: They meet when 35t−5t2=35(t−3)−5(t−3)2⇒t=535t-5t^2=35(t-3)-5(t-3)^2\Rightarrow t=5 s: height 50 m.

Question 30

Physics · 4 marks

A source and a detector move away from each other in absence of wind with a speed of 20 m/s with respect to the ground. If the detector detects a frequency of 1800 Hz of the sound coming from the source, then the original frequency of source considering speed of sound in air 340 m/s will be ________ Hz.

Correct answer: 2025

Explanation: 1800=f320360⇒f=20251800=f\frac{320}{360}\Rightarrow f=2025 Hz.

Question 31

Chemistry · 4 marks · −1

Given below are two statements: Statement I: Frenkel defects are vacancy as well as interstitial defects. Statement II: Frenkel defect leads to colour in ionic solids due to presence of F-centres. Choose the most appropriate answer for the statements 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 32

Chemistry · 4 marks · −1

Given below are two statements: Statement I: According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in positive charges on the nucleus as there is no strong hold on the electron by the nucleus. Statement II: According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in principal quantum number. In the light of the above statements, choose the most appropriate 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: D

Explanation: v∝Z/nv\propto Z/n.

Question 33

Chemistry · 4 marks · −1

Given below are two statements: Statement I: The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of CH3_3COOH (weak electrolyte). Statement II: Molar conductivity decreases with decrease in concentration of electrolyte. In the light of the above statements, choose the most appropriate 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: B

Explanation: Λ∘\Lambda^\circ(CH3_3COOH) ≈390.5\approx390.5 S cm2^2 mol−1^{-1} exceeds Λ∘\Lambda^\circ(KCl) ≈149.9\approx149.9, so I is false; molar conductivity increases on dilution, so II is false.

Question 34

Chemistry · 4 marks · −1

Which one of the following is correct for the adsorption of a gas at a given temperature on a solid surface?

  1. A.ΔH>0,ΔS>0\Delta H>0, \Delta S>0
  2. B.ΔH<0,ΔS>0\Delta H<0, \Delta S>0
  3. C.ΔH>0,ΔS<0\Delta H>0, \Delta S<0
  4. D.ΔH<0,ΔS<0\Delta H<0, \Delta S<0

Correct answer: D

Question 35

Chemistry · 4 marks · −1

Given below are two statements. Statement I: The choice of reducing agents for metals extraction can be made by using Ellingham diagram, a plot of ΔG\Delta G vs temperature. Statement II: The value of ΔS\Delta S increases from left to right in Ellingham diagram. In the light of the above statements, choose the most appropriate 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 36

Chemistry · 4 marks · −1

Which one of the following methods is most suitable for preparing deionized water?

  1. A.Clark's method
  2. B.Synthetic resin method
  3. C.Calgon's method
  4. D.Permutit method

Correct answer: B

Question 37

Chemistry · 4 marks · −1

What are the products formed in sequence when excess of CO2_2 is passed in slaked lime?

  1. A.CaCO3_3, Ca(HCO3_3)2_2
  2. B.Ca(HCO3_3)2_2, CaCO3_3
  3. C.CaO, CaCO3_3
  4. D.CaO, Ca(HCO3_3)2_2

Correct answer: A

Question 38

Chemistry · 4 marks · −1

The incorrect statement is:

  1. A.Cl2_2 is more reactive than ClF.
  2. B.F2_2 is more reactive than ClF.
  3. C.On hydrolysis ClF forms HOCl and HF.
  4. D.F2_2 is a stronger oxidizing agent than Cl2_2 in aqueous solution.

Correct answer: A

Explanation: Interhalogens (except F2) are more reactive than the parent halogens: Cl2 is less reactive than ClF.

Question 39

Chemistry · 4 marks · −1

Which one of the following when dissolved in water gives coloured solution in nitrogen atmosphere?

  1. A.ZnCl2_2
  2. B.CuCl2_2
  3. C.Cu2_2Cl2_2
  4. D.AgCl

Correct answer: B

Explanation: Cu2+^{2+} (d9^9) is coloured.

Question 40

Chemistry · 4 marks · −1

The conversion of hydroxyapatite occurs due to presence of F−^- ions in water. The correct formula of hydroxyapatite is:

  1. A.[3Ca3_3(PO4_4)2_2·CaF2_2]
  2. B.[Ca3_3(PO4_4)2_2·CaF2_2]
  3. C.[3Ca3_3(PO4_4)2_2·Ca(OH)2_2]
  4. D.[3Ca(OH)2_2·CaF2_2]

Correct answer: C

Question 41

Chemistry · 4 marks · −1

Which one of the following complexes is violet in colour?

  1. A.[Fe(SCN)6_6]4−^{4-}
  2. B.[Fe(CN)5_5NOS]4−^{4-}
  3. C.Fe4_4[Fe(CN)6_6]3_3·H2_2O
  4. D.[Fe(CN)6_6]4−^{4-}

Correct answer: B

Explanation: Sodium nitroprusside test for sulphide gives the violet [Fe(CN)5NOS]4- complex.

Question 42

Chemistry · 4 marks · −1

Excess of isobutane on reaction with Br2_2 in presence of light at 125°C gives which one of the following, as the major product?

  1. A.(CH3)2CH−CH2Br\mathrm{(CH_3)_2CH-CH_2Br}
  2. B.(CH3)3C−Br\mathrm{(CH_3)_3C-Br}
  3. C.(CH3)2C(Br)−CH2Br\mathrm{(CH_3)_2C(Br)-CH_2Br}
  4. D.CH3−CH(CH2Br)−CH2Br\mathrm{CH_3-CH(CH_2Br)-CH_2Br}

Correct answer: B

Explanation: Bromination is highly selective for the tertiary C–H.

Question 43

Chemistry · 4 marks · −1

The major product formed in the following reaction is: isoprene (2-methylbuta-1,3-diene) + HBr (excess) →\rightarrow Major Product

Figure for question 43
  1. A.1,4-dibromo-2-methylbutane (Br-CH2-CH(CH3)-CH2-CH2-Br)
  2. B.1-bromo-3-methylbut-2-ene ((CH3)2C=CH-CH2Br)
  3. C.2,3-dibromo-2-methylbutane ((CH3)2CBr-CHBr-CH3)
  4. D.3-bromo-3-methylbut-1-ene ((CH3)2CBr-CH=CH2)

Correct answer: B

Explanation: Protonation of isoprene gives the allylic tertiary cation; 1,4-addition (thermodynamic product) gives 1-bromo-3-methylbut-2-ene (official key).

Question 44

Chemistry · 4 marks · −1

Among the following compounds I–IV, which one forms a yellow precipitate on reacting sequentially with (i) NaOH (ii) dil. HNO3_3 (iii) AgNO3_3? I: 4-chlorotoluene; II: 3-chlorotoluene; III: 2-bromotoluene; IV: 2-chlorobenzyl iodide (Cl on ring, CH2_2I side chain)

Figure for question 44
  1. A.I
  2. B.II
  3. C.III
  4. D.IV

Correct answer: D

Explanation: Only the benzylic iodide hydrolyses to give I- (yellow AgI).

Question 45

Chemistry · 4 marks · −1

The major products formed in the following reaction sequence A and B are: acetophenone →Br2, KOH\xrightarrow{\text{Br}_2\text{, KOH}} A + B

  1. A.A = C6_6H5_5COOK, B = CHBr3_3
  2. B.A = 3,5-dibromoacetophenone, B = 3,5-dihydroxyacetophenone
  3. C.A = 2-bromo-1-(3-bromophenyl)ethanone, B = 1-(3-bromophenyl)-2-hydroxyethanone
  4. D.A = C6_6H5_5COCBr3_3, B = C6_6H5_5CHO

Correct answer: A

Explanation: Haloform reaction: potassium benzoate and bromoform.

Question 46

Chemistry · 4 marks · −1

The correct options for the products A and B of the following reactions are: phenol →Br2 (Excess), H2O\xrightarrow{\text{Br}_2\text{ (Excess), H}_2\text{O}} A; phenol →Br2, CS2, <5∘C\xrightarrow{\text{Br}_2\text{, CS}_2\text{, }<5^\circ\text{C}} B

  1. A.A = 4-bromophenol, B = 4-bromophenol
  2. B.A = 2,3,4-tribromophenol (as drawn), B = 2,6-dibromophenol
  3. C.A = 2,4,6-tribromophenol, B = 4-bromophenol
  4. D.A = 2,4,6-tribromophenol, B = 2,4-dibromophenol

Correct answer: C

Question 47

Chemistry · 4 marks · −1

The major product formed in the following reaction is: tryptophan →SOCl2, CH3OH\xrightarrow{\text{SOCl}_2\text{, CH}_3\text{OH}} Major Product

Figure for question 47
  1. A.tryptophanyl chloride (–COCl, free NH2)
  2. B.tryptophan methyl ester hydrochloride (–CO2CH3, –NH2·HCl)
  3. C.methyl ester with both the alpha-amino group and indole N-H as hydrochlorides
  4. D.5-chlorotryptophan methyl ester (Cl on the benzene ring)

Correct answer: B

Explanation: SOCl2/MeOH generates HCl in situ: esterification with the amine isolated as its hydrochloride; the indole N is not basic.

Question 48

Chemistry · 4 marks · −1

The correct sequential addition of reagents in the preparation of 3-nitrobenzoic acid from benzene is:

  1. A.HNO3_3/H2_2SO4_4, Br2_2/AlBr3_3, Mg/ether, CO2_2, H3_3O+^+
  2. B.Br2_2/AlBr3_3, NaCN, H3_3O+^+, HNO3_3/H2_2SO4_4
  3. C.Br2_2/AlBr3_3, HNO3_3/H2_2SO4_4, NaCN, H3_3O+^+
  4. D.Br2_2/AlBr3_3, HNO3_3/H2_2SO4_4, Mg/ether, CO2_2, H3_3O+^+

Correct answer: A

Explanation: Nitration first makes the ring meta-directing, so bromination goes meta; the Br is then converted to COOH via the Grignard and CO2 (official key; the textbook caveat that a nitro group is incompatible with Grignard formation is ignored by the question).

Question 49

Chemistry · 4 marks · −1

The polymer formed on heating Novolac with formaldehyde is:

  1. A.Melamine
  2. B.Bakelite
  3. C.Polyester
  4. D.Nylon 6,6

Correct answer: B

Question 50

Chemistry · 4 marks · −1

Given below are two statements: Statement I: In the titration between strong acid and weak base methyl orange is suitable as an indicator. Statement II: For titration of acetic acid with NaOH phenolphthalein is not a suitable indicator. In the light of the above statements, choose the most appropriate 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 51

Chemistry · 4 marks

An aqueous KCl solution of density 1.20 g mL−1^{-1} has a molality of 3.30 mol kg−1^{-1}. The molarity of the solution in mol L−1^{-1} is ________. (Nearest integer) [Molar mass of KCl = 74.5]

Correct answer: 3

Explanation: 1245.9 g solution per 3.30 mol: V=1.038V=1.038 L, M≈3.18≈3M\approx3.18\approx3.

Question 52

Chemistry · 4 marks

AB3_3 is an interhalogen T-shaped molecule. The number of lone pairs of electrons on A is ________. (Integer answer)

Correct answer: 2

Explanation: T-shaped (sp3d): 2 lone pairs.

Question 53

Chemistry · 4 marks

The Born-Haber cycle for KCl is evaluated with the following data: ΔfH⊖\Delta_fH^\ominus for KCl = −436.7-436.7 kJ mol−1^{-1}; ΔsubH⊖\Delta_{sub}H^\ominus for K = 89.2 kJ mol−1^{-1}; ΔionizationH⊖\Delta_{ionization}H^\ominus for K = 419.0 kJ mol−1^{-1}; Δelectron gainH⊖\Delta_{electron\ gain}H^\ominus for Cl(g)_{(g)} = −348.6-348.6 kJ mol−1^{-1}; ΔbondH⊖\Delta_{bond}H^\ominus for Cl2_2 = 243.0 kJ mol−1^{-1}. The magnitude of lattice enthalpy of KCl in kJ mol−1^{-1} is ________. (Nearest integer)

Correct answer: 718

Explanation: U=436.7+89.2+419+121.5−348.6=717.8≈718U=436.7+89.2+419+121.5-348.6=717.8\approx718.

Question 54

Chemistry · 4 marks

Of the following four aqueous solutions, total number of those solutions whose freezing point is lower than that of 0.10 M C2_2H5_5OH is ________. (Integer answer) (i) 0.10 M Ba3_3(PO4_4)2_2 (ii) 0.10 M Na2_2SO4_4 (iii) 0.10 M KCl (iv) 0.10 M Li3_3PO4_4

Correct answer: 4

Explanation: All four electrolytes have i > 1: 4.

Question 55

Chemistry · 4 marks

The OH−^- concentration in a mixture of 5.0 mL of 0.0504 M NH4_4Cl and 2 mL of 0.0210 M NH3_3 solution is x×10−6x\times10^{-6} M. The value of xx is ________. (Nearest integer) [Given Kw=1×10−14K_w=1\times10^{-14} and Kb=1.8×10−5K_b=1.8\times10^{-5}]

Correct answer: 3

Explanation: [OH−]=Kb0.0420.252=3×10−6[OH^-]=K_b\frac{0.042}{0.252}=3\times10^{-6}.

Question 56

Chemistry · 4 marks

The following data was obtained for chemical reaction given below at 975 K. 2NO(g)+2H2(g)→N2(g)+2H2O(g)\mathrm{2NO_{(g)}+2H_{2(g)}\rightarrow N_{2(g)}+2H_2O_{(g)}}. (A) [NO] 8×10−58\times10^{-5}, [H2_2] 8×10−58\times10^{-5}, Rate 7×10−97\times10^{-9}; (B) 24×10−524\times10^{-5}, 8×10−58\times10^{-5}, 2.1×10−82.1\times10^{-8}; (C) 24×10−524\times10^{-5}, 32×10−532\times10^{-5}, 8.4×10−88.4\times10^{-8} (concentrations in mol L−1^{-1}, rate in mol L−1^{-1}s−1^{-1}). The order of the reaction with respect to NO is ________. [Integer answer]

Correct answer: 1

Explanation: Tripling [NO] triples the rate: order 1.

Question 57

Chemistry · 4 marks

These are physical properties of an element (A) Sublimation enthalpy (B) Ionisation enthalpy (C) Hydration enthalpy (D) Electron gain enthalpy. The total number of above properties that affect the reduction potential is ________. (Integer answer)

Correct answer: 3

Question 58

Chemistry · 4 marks

The number of 4f electrons in the ground state electronic configuration of Gd2+^{2+} is ________. [Atomic number of Gd = 64]

Correct answer: 7

Explanation: Gd: [Xe]4f$^75d5d^16s6s^2;Gd; Gd^{2+}:4f: 4f^75d5d^1$: 7.

Question 59

Chemistry · 4 marks

The ratio of number of water molecules in Mohr's salt and potash alum is ________ ×10−1\times10^{-1}. (Integer answer)

Correct answer: 5

Explanation: Mohr's salt FeSO4⋅_4\cdot(NH4_4)2_2SO$_4\cdot6H6H_2OandpotashalumwrittenasKAl(SOO and potash alum written as KAl(SO_4))_2\cdot12H12H_2$O: 6/12=0.5=5×10−16/12=0.5=5\times10^{-1}.

Question 60

Chemistry · 4 marks

The total number of negative charge in the tetrapeptide, Gly-Glu-Asp-Tyr, at pH 12.5 will be ________. (Integer answer)

Correct answer: 4

Explanation: C-terminal COO-, Glu and Asp side-chain COO-, and Tyr phenoxide: 4.

Question 61

Mathematics · 4 marks · −1

Out of all the patients in a hospital 89% are found to be suffering from heart ailment and 98% are suffering from lungs infection. If K% of them are suffering from both ailments, then K can not belong to the set:

  1. A.{84,86,88,90}\{84, 86, 88, 90\}
  2. B.{80,83,86,89}\{80, 83, 86, 89\}
  3. C.{79,81,83,85}\{79, 81, 83, 85\}
  4. D.{84,87,90,93}\{84, 87, 90, 93\}

Correct answer: C

Explanation: 87≤K≤8987\le K\le89; the set {79,81,83,85}\{79,81,83,85\} has no admissible value.

Question 62

Mathematics · 4 marks · −1

The equation arg⁡(z−1z+1)=π4\arg\left(\frac{z-1}{z+1}\right)=\frac\pi4 represents a circle with:

  1. A.centre at (0,0)(0,0) and radius 2\sqrt2
  2. B.centre at (0,1)(0,1) and radius 2
  3. C.centre at (0,1)(0,1) and radius 2\sqrt2
  4. D.centre at (0,−1)(0,-1) and radius 2\sqrt2

Correct answer: C

Explanation: x2+y2−2y−1=0x^2+y^2-2y-1=0 (arc): centre (0,1)(0,1), radius 2\sqrt2.

Question 63

Mathematics · 4 marks · −1

If A=(1525−2515)A=\begin{pmatrix}\frac1{\sqrt5}&\frac2{\sqrt5}\\ \frac{-2}{\sqrt5}&\frac1{\sqrt5}\end{pmatrix}, B=(10i1)B=\begin{pmatrix}1&0\\i&1\end{pmatrix}, i=−1i=\sqrt{-1}, and Q=ATBAQ=A^TBA, then the inverse of the matrix AQ2021ATAQ^{2021}A^T is equal to:

  1. A.(10−2021i1)\begin{pmatrix}1&0\\-2021i&1\end{pmatrix}
  2. B.(15−2021202115)\begin{pmatrix}\frac1{\sqrt5}&-2021\\2021&\frac1{\sqrt5}\end{pmatrix}
  3. C.(1−2021i01)\begin{pmatrix}1&-2021i\\0&1\end{pmatrix}
  4. D.(102021i1)\begin{pmatrix}1&0\\2021i&1\end{pmatrix}

Correct answer: A

Explanation: AA is orthogonal, so AQ2021AT=B2021=(102021i1)AQ^{2021}A^T=B^{2021}=\begin{pmatrix}1&0\\2021i&1\end{pmatrix}, whose inverse has −2021i-2021i.

Question 64

Mathematics · 4 marks · −1

Let θ∈(0,π2)\theta\in\left(0,\frac\pi2\right). If the system of linear equations (1+cos⁡2θ)x+sin⁡2θ y+4sin⁡3θ z=0(1+\cos^2\theta)x+\sin^2\theta\,y+4\sin3\theta\,z=0, cos⁡2θ x+(1+sin⁡2θ)y+4sin⁡3θ z=0\cos^2\theta\,x+(1+\sin^2\theta)y+4\sin3\theta\,z=0, cos⁡2θ x+sin⁡2θ y+(1+4sin⁡3θ)z=0\cos^2\theta\,x+\sin^2\theta\,y+(1+4\sin3\theta)z=0 has a non-trivial solution, then the value of θ\theta is:

  1. A.π18\frac{\pi}{18}
  2. B.7π18\frac{7\pi}{18}
  3. C.5π18\frac{5\pi}{18}
  4. D.4π9\frac{4\pi}{9}

Correct answer: B

Explanation: Δ=2+4sin⁡3θ=0⇒sin⁡3θ=−12⇒θ=7π18\Delta=2+4\sin3\theta=0\Rightarrow\sin3\theta=-\frac12\Rightarrow\theta=\frac{7\pi}{18}.

Question 65

Mathematics · 4 marks · −1

The sum of the series 1x+1+2x2+1+22x4+1+⋯+2100x2100+1\frac{1}{x+1}+\frac{2}{x^2+1}+\frac{2^2}{x^4+1}+\dots+\frac{2^{100}}{x^{2^{100}}+1} when x=2x=2 is:

  1. A.1−21014101−11-\frac{2^{101}}{4^{101}-1}
  2. B.1−21004100−11-\frac{2^{100}}{4^{100}-1}
  3. C.1+21014101−11+\frac{2^{101}}{4^{101}-1}
  4. D.1+21004101−11+\frac{2^{100}}{4^{101}-1}

Correct answer: A

Explanation: Telescoping with 2kx2k+1=2kx2k−1−2k+1x2k+1−1\frac{2^k}{x^{2^k}+1}=\frac{2^k}{x^{2^k}-1}-\frac{2^{k+1}}{x^{2^{k+1}}-1}: sum =1x−1−2101x2101−1=\frac{1}{x-1}-\frac{2^{101}}{x^{2^{101}}-1} at x=2x=2.

Question 66

Mathematics · 4 marks · −1

If 20Cr{}^{20}C_r is the co-efficient of xrx^r in the expansion of (1+x)20(1+x)^{20}, then the value of ∑r=020r2 20Cr\sum_{r=0}^{20}r^2\,{}^{20}C_r is equal to:

  1. A.420×219420\times2^{19}
  2. B.380×219380\times2^{19}
  3. C.380×218380\times2^{18}
  4. D.420×218420\times2^{18}

Correct answer: D

Explanation: n(n+1)2n−2=420×218n(n+1)2^{n-2}=420\times2^{18}.

Question 67

Mathematics · 4 marks · −1

If the sum of an infinite GP a,ar,ar2,ar3,…a, ar, ar^2, ar^3,\dots is 15 and the sum of the squares of its each term is 150, then the sum of ar2,ar4,ar6,…ar^2, ar^4, ar^6,\dots is:

  1. A.52\frac52
  2. B.12\frac12
  3. C.92\frac92
  4. D.252\frac{25}{2}

Correct answer: B

Explanation: a1−r=15\frac{a}{1-r}=15 and a21−r2=150\frac{a^2}{1-r^2}=150 give a1+r=10\frac{a}{1+r}=10, so r=15r=\frac15, a=12a=12; ar21−r2=12/2524/25=12\frac{ar^2}{1-r^2}=\frac{12/25}{24/25}=\frac12.

Question 68

Mathematics · 4 marks · −1

Let f(x)=cos⁡(2tan⁡−1sin⁡(cot⁡−11−xx))f(x)=\cos\left(2\tan^{-1}\sin\left(\cot^{-1}\sqrt{\frac{1-x}{x}}\right)\right), 0<x<10<x<1. Then:

  1. A.(1−x)2f′(x)−2(f(x))2=0(1-x)^2f'(x)-2(f(x))^2=0
  2. B.(1+x)2f′(x)+2(f(x))2=0(1+x)^2f'(x)+2(f(x))^2=0
  3. C.(1−x)2f′(x)+2(f(x))2=0(1-x)^2f'(x)+2(f(x))^2=0
  4. D.(1+x)2f′(x)−2(f(x))2=0(1+x)^2f'(x)-2(f(x))^2=0

Correct answer: C

Explanation: sin⁡(cot⁡−1(1−x)/x)=x\sin(\cot^{-1}\sqrt{(1-x)/x})=\sqrt x, so f=1−x1+xf=\frac{1-x}{1+x} and f=ˊ−2(1+x)2f\'=\frac{-2}{(1+x)^2}: (1+x)2f+ˊ2f2≠0(1+x)^2f\'+2f^2\ne0; (1−x)2f+ˊ2f2=0(1-x)^2f\'+2f^2=0.

Question 69

Mathematics · 4 marks · −1

The value of ∫−1/21/2((x+1x−1)2+(x−1x+1)2−2)1/2dx\int_{-1/\sqrt2}^{1/\sqrt2}\left(\left(\frac{x+1}{x-1}\right)^2+\left(\frac{x-1}{x+1}\right)^2-2\right)^{1/2}dx is:

  1. A.4log⁡e(3+22)4\log_e(3+2\sqrt2)
  2. B.log⁡e4\log_e4
  3. C.log⁡e16\log_e16
  4. D.2log⁡e162\log_e16

Correct answer: C

Explanation: Integrand =∣x+1x−1−x−1x+1∣=4∣x∣1−x2=\left|\frac{x+1}{x-1}-\frac{x-1}{x+1}\right|=\frac{4|x|}{1-x^2}; integral =2⋅2ln⁡2=ln⁡16=2\cdot2\ln2=\ln16.

Question 70

Mathematics · 4 marks · −1

The value of lim⁡n→∞1n∑r=02n−1n2n2+4r2\lim_{n\to\infty}\frac1n\sum_{r=0}^{2n-1}\frac{n^2}{n^2+4r^2} is:

  1. A.14tan⁡−1(4)\frac14\tan^{-1}(4)
  2. B.12tan⁡−1(4)\frac12\tan^{-1}(4)
  3. C.tan⁡−1(4)\tan^{-1}(4)
  4. D.12tan⁡−1(2)\frac12\tan^{-1}(2)

Correct answer: B

Explanation: ∫02dx1+4x2=12tan⁡−14\int_0^2\frac{dx}{1+4x^2}=\frac12\tan^{-1}4.

Question 71

Mathematics · 4 marks · −1

The sum of solutions of the equation cos⁡x1+sin⁡x=∣tan⁡2x∣\frac{\cos x}{1+\sin x}=|\tan2x|, x∈(−π2,π2)−{π4,−π4}x\in\left(-\frac\pi2,\frac\pi2\right)-\left\{\frac\pi4,-\frac\pi4\right\} is:

  1. A.π10\frac{\pi}{10}
  2. B.−11π30-\frac{11\pi}{30}
  3. C.−7π30-\frac{7\pi}{30}
  4. D.−π15-\frac{\pi}{15}

Correct answer: B

Question 72

Mathematics · 4 marks · −1

Let ABC be a triangle with A(−3,1)A(-3,1) and ∠ACB=θ\angle ACB=\theta, 0<θ<π20<\theta<\frac\pi2. If the equation of the median through B is 2x+y−3=02x+y-3=0 and the equation of angle bisector of C is 7x−4y−1=07x-4y-1=0, then tan⁡θ\tan\theta is equal to:

  1. A.12\frac12
  2. B.34\frac34
  3. C.43\frac43
  4. D.2

Correct answer: C

Question 73

Mathematics · 4 marks · −1

If a line along a chord of the circle 4x2+4y2+120x+675=04x^2+4y^2+120x+675=0, passes through the point (−30,0)(-30,0) and is tangent to the parabola y2=30xy^2=30x, then the length of this chord is:

  1. A.7
  2. B.353\sqrt5
  3. C.5
  4. D.535\sqrt3

Correct answer: B

Explanation: Tangent y=mx+152my=mx+\frac{15}{2m} through (−30,0)(-30,0): m2=14m^2=\frac14. Circle centre (−15,0)(-15,0), r2=2254r^2=\frac{225}{4}; distance of centre from x−2y+30=0x-2y+30=0 is 155=35\frac{15}{\sqrt5}=3\sqrt5; chord =22254−45=35=2\sqrt{\frac{225}4-45}=3\sqrt5.

Question 74

Mathematics · 4 marks · −1

On the ellipse x28+y24=1\frac{x^2}{8}+\frac{y^2}{4}=1 let P be a point in the second quadrant such that the tangent at P to the ellipse is perpendicular to the line x+2y=0x+2y=0. Let S and S' be the foci of the ellipse and e be its eccentricity. If A is the area of the triangle SPS' then, the value of (5−e2)⋅A(5-e^2)\cdot A is:

  1. A.24
  2. B.14
  3. C.12
  4. D.6

Correct answer: D

Explanation: Slope 2 tangent at P(−83,23)P(-\frac{8}{3},\frac23); e2=12e^2=\frac12, foci (±2,0)(\pm2,0); A=12⋅4⋅23=43A=\frac12\cdot4\cdot\frac23=\frac43; (4.5)43=6(4.5)\frac43=6.

Question 75

Mathematics · 4 marks · −1

A plane P contains the line x+2y+3z+1=0=x−y−z−6x+2y+3z+1=0=x-y-z-6, and is perpendicular to the plane −2x+y+z+8=0-2x+y+z+8=0. Then which of the following points lies on P?

  1. A.(2,−1,1)(2,-1,1)
  2. B.(0,1,1)(0,1,1)
  3. C.(1,0,1)(1,0,1)
  4. D.(−1,1,2)(-1,1,2)

Correct answer: B

Question 76

Mathematics · 4 marks · −1

Let y=y(x)y=y(x) be a solution curve of the differential equation (y+1)tan⁡2x dx+tan⁡x dy+y dx=0(y+1)\tan^2x\,dx+\tan x\,dy+y\,dx=0, x∈(0,π2)x\in\left(0,\frac\pi2\right). If lim⁡x→0+xy(x)=1\lim_{x\to0+}xy(x)=1, then the value of y(π4)y\left(\frac\pi4\right) is:

  1. A.π4−1\frac\pi4-1
  2. B.π4+1\frac\pi4+1
  3. C.−π4-\frac\pi4
  4. D.π4\frac\pi4

Correct answer: D

Question 77

Mathematics · 4 marks · −1

Let a⃗=i^+j^+k^\vec a=\hat i+\hat j+\hat k and b⃗=j^−k^\vec b=\hat j-\hat k. If c⃗\vec c is a vector such that a⃗×c⃗=b⃗\vec a\times\vec c=\vec b and a⃗⋅c⃗=3\vec a\cdot\vec c=3, then a⃗⋅(b⃗×c⃗)\vec a\cdot(\vec b\times\vec c) is equal to:

  1. A.−2-2
  2. B.2
  3. C.−6-6
  4. D.6

Correct answer: A

Explanation: a⃗⋅(b⃗×c⃗)=b⃗⋅(c⃗×a⃗)=−∣b⃗∣2=−2\vec a\cdot(\vec b\times\vec c)=\vec b\cdot(\vec c\times\vec a)=-|\vec b|^2=-2.

Question 78

Mathematics · 4 marks · −1

If the truth value of the Boolean expression ((p∨q)∧(q→r)∧(∼r))→(p∧q)((p\vee q)\wedge(q\to r)\wedge(\sim r))\to(p\wedge q) is false, then the truth values of the statements p, q, r respectively can be:

  1. A.T F T
  2. B.F F T
  3. C.F T F
  4. D.T F F

Correct answer: D

Explanation: Need antecedent true and p∧qp\wedge q false: r=Fr=F, q=Fq=F (else q→rq\to r fails), p=Tp=T.

Question 79

Mathematics · 4 marks · −1

Let A and B be independent events such that P(A)=pP(A)=p, P(B)=2pP(B)=2p. The largest value of p, for which P (exactly one of A, B occurs) =59=\frac59, is:

  1. A.13\frac13
  2. B.49\frac49
  3. C.512\frac{5}{12}
  4. D.29\frac29

Correct answer: C

Explanation: 3p−4p2=59⇒p=133p-4p^2=\frac59\Rightarrow p=\frac13 or 512\frac{5}{12}; largest 512\frac{5}{12}.

Question 80

Mathematics · 4 marks · −1

The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. If α\alpha and β\sqrt\beta are the mean and standard deviation respectively for correct data, then (α,β)(\alpha,\beta) is:

  1. A.(11, 25)
  2. B.(11, 26)
  3. C.(10.5, 26)
  4. D.(10.5, 25)

Correct answer: C

Explanation: α=10.5\alpha=10.5; ∑x2=20(106.25)−625+1225=2725\sum x^2=20(106.25)-625+1225=2725; β=136.25−110.25=26\beta=136.25-110.25=26.

Question 81

Mathematics · 4 marks

The number of three-digit even numbers, formed by the digits 0, 1, 3, 4, 6, 7 if the repetition of digits is not allowed, is ________.

Correct answer: 52

Explanation: Last 0: 5⋅4=205\cdot4=20; last 4 or 6: 2⋅4⋅4=322\cdot4\cdot4=32; total 52.

Question 82

Mathematics · 4 marks

Let a,b∈Ra,b\in\mathbb{R}, b≠0b\ne0. Define a function f(x)={asin⁡π2(x−1),x≤0tan⁡2x−sin⁡2xbx3,x>0f(x)=\begin{cases}a\sin\frac\pi2(x-1), & x\le0\\ \frac{\tan2x-\sin2x}{bx^3}, & x>0\end{cases}. If ff is continuous at x=0x=0, then 10−ab10-ab is equal to ________.

Correct answer: 14

Explanation: −a=4b⇒ab=−4-a=\frac{4}{b}\Rightarrow ab=-4; 10−ab=1410-ab=14.

Question 83

Mathematics · 4 marks

The sum of all integral values of kk (k≠0k\ne0) for which the equation 2x−1−1x−2=2k\frac{2}{x-1}-\frac{1}{x-2}=\frac2k in xx has no real roots, is ________.

Correct answer: 66

Question 84

Mathematics · 4 marks

A wire of length 36 m is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is k (meter), then (4π+1)k\left(\frac4\pi+1\right)k is equal to ________.

Correct answer: 36

Explanation: Minimum when square side == circle diameter: k=36ππ+4k=\frac{36\pi}{\pi+4}, so (4+ππ)k=36\left(\frac{4+\pi}{\pi}\right)k=36.

Question 85

Mathematics · 4 marks

The area of the region S={(x,y):3x2≤4y≤6x+24}S=\{(x,y): 3x^2\le4y\le6x+24\} is ________.

Correct answer: 27

Explanation: ∫−24(6x+24−3x24)dx=27\int_{-2}^{4}\left(\frac{6x+24-3x^2}{4}\right)dx=27.

Question 86

Mathematics · 4 marks

If y=y(x)y=y(x) is an implicit function of xx such that log⁡e(x+y)=4xy\log_e(x+y)=4xy, then d2ydx2\frac{d^2y}{dx^2} at x=0x=0 is equal to ________.

Correct answer: 40

Question 87

Mathematics · 4 marks

If 1P1+2⋅2P2+3⋅3P3+⋯+15⋅15P15=qPr−s{}^1P_1+2\cdot{}^2P_2+3\cdot{}^3P_3+\dots+15\cdot{}^{15}P_{15}={}^qP_r-s, 0≤s≤10\le s\le1, then q+sCr−s{}^{q+s}C_{r-s} is equal to ________.

Correct answer: 136

Explanation: ∑k⋅k!=16!−1=16P16−1\sum k\cdot k!=16!-1={}^{16}P_{16}-1: q=r=16,s=1q=r=16,s=1; 17C15=136{}^{17}C_{15}=136.

Question 88

Mathematics · 4 marks

Let the line L be the projection of the line x−12=y−31=z−42\frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2} in the plane x−2y−z=3x-2y-z=3. If d is the distance of the point (0,0,6)(0,0,6) from L, then d2d^2 is equal to ________.

Correct answer: 26

Question 89

Mathematics · 4 marks

Let z=1−i32z=\frac{1-i\sqrt3}{2}, i=−1i=\sqrt{-1}. Then the value of 21+(z+1z)3+(z2+1z2)3+(z3+1z3)3+⋯+(z21+1z21)321+\left(z+\frac1z\right)^3+\left(z^2+\frac1{z^2}\right)^3+\left(z^3+\frac1{z^3}\right)^3+\dots+\left(z^{21}+\frac1{z^{21}}\right)^3 is ________.

Correct answer: 13

Explanation: z=e−iπ/3z=e^{-i\pi/3}: zk+z−k=2cos⁡kπ3z^k+z^{-k}=2\cos\frac{k\pi}{3}; the cubes over one period of 6 sum to 1−1−8−1+1+8=01-1-8-1+1+8=0; k=19..21k=19..21 add 1−1−81-1-8; total 21−8=1321-8=13.

Question 90

Mathematics · 4 marks

The locus of a point, which moves such that the sum of squares of its distances from the points (0,0)(0,0), (1,0)(1,0), (0,1)(0,1), (1,1)(1,1) is 18 units, is a circle of diameter d. Then d2d^2 is equal to ________.

Correct answer: 16

Explanation: 4x2+4y2−4x−4y+4=18⇒r2=44x^2+4y^2-4x-4y+4=18\Rightarrow r^2=4, d2=16d^2=16.

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