JEE (Main) · 2020
JEE (Main) 2020 — 5 September, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Physics
25 Q
- Oscillations and Waves3
- Electrostatics3
- Properties of Bulk Matter2
13 easy · 12 medium · 0 hard
Chemistry
25 Q
- Hydrocarbons2
- Hydrogen2
- Coordination Compounds2
15 easy · 10 medium · 0 hard
Mathematics
25 Q
- Applications of Derivatives3
- Matrices and Determinants2
- Sequences and Series2
10 easy · 15 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $x=\frac{1}{\sqrt{\mu_0\epsilon_0}}=c$, the speed of light: dimension $LT^{-1}$.
$y=\frac EB$: from $F=qE=qvB$, $\frac EB$ is a speed, $LT^{-1}$.
$CR$ is a time constant, so $z=\frac{l}{CR}$ is length/time, $LT^{-1}$.
All three have the dimension of velocity.
Answer: (D). | ||||
| 2 | KinematicsMotion graphs | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on motion graphs, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Distance = total area between the graph and the time axis (taking the part below the axis as positive).
0 to 2 s (O to A, rising to 4 m/s): $\frac12\times2\times4=4$ m.
2 to 3 s (A to B, 4 m/s): $1\times4=4$ m.
3 to 4.333 s (B to S): $\frac12\times\frac43\times4=\frac83$ m. The slope of BC is $-\frac{4}{4/3}=-3$ m/s², so at C ($t=5$ s) $v=-2$ m/s, as drawn.
4.333 to 6 s (S to C to D, below the axis): $\frac12\times\frac53\times2=\frac53$ m.
Total $=4+4+\frac83+\frac53=8+\frac{13}{3}=\frac{37}{3}$ m.
Answer: (A). | ||||
| 3 | Laws of MotionConservation of linear momentum | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on conservation of linear momentum, from Laws of Motion. NCERT: Class 11: Laws of Motion No external force acts, and the dust swept up is at rest, so the momentum $Mv$ of ship plus collected dust is constant:
$\frac{d(Mv)}{dt}=M\frac{dv}{dt}+v\frac{dM}{dt}=0$.
$\frac{dv}{dt}=-\frac{v}{M}\cdot bv^2=-\frac{bv^3}{M(t)}$.
Answer: (B). | ||||
| 4 | Oscillations and WavesSimple pendulum and spring systems | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on simple pendulum and spring systems, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Physical pendulum: $T=2\pi\sqrt{\frac{I}{mgR}}$, the centre being a distance $R$ below the nail.
(i) In its plane, the axis is perpendicular to the ring through a point on the rim: $I_1=mR^2+mR^2=2mR^2$.
(ii) Perpendicular to its plane, the axis is a tangent in the plane of the ring: $I_2=\frac{mR^2}{2}+mR^2=\frac32mR^2$.
$\frac{T_1}{T_2}=\sqrt{\frac{I_1}{I_2}}=\sqrt{\frac{2}{3/2}}=\frac{2}{\sqrt3}$.
Answer: (C). | ||||
| 5 | GravitationAcceleration due to gravity | Class 11: Gravitation | Single correctMedium | |
A medium single correct question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation At the equator the effective gravity is $g-\omega^2R$.
At height $h$ above a pole (no rotation effect): $g\left(1-\frac{2h}{R}\right)$ for $h\ll R$.
Equal weights: $g-\frac{2gh}{R}=g-\omega^2R \Rightarrow h=\frac{R^2\omega^2}{2g}$.
Answer: (D). | ||||
| 6 | Properties of Bulk MatterThermal expansion and calorimetry | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on thermal expansion and calorimetry, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter For a temperature rise $\Delta T$ the combined length increases by $\Delta L=L_1\alpha_1\Delta T+L_2\alpha_2\Delta T$.
Writing $\Delta L=(L_1+L_2)\alpha_{\text{eff}}\Delta T$: $\alpha_{\text{eff}}=\frac{\alpha_1L_1+\alpha_2L_2}{L_1+L_2}$.
Answer: (C). | ||||
| 7 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Speed after falling $h$ freely: $v=\sqrt{2gh}$.
Terminal velocity in water: $v_t=\frac{2r^2(\rho-\sigma)g}{9\eta}\propto r^2$.
$\sqrt{2gh}\propto r^2 \Rightarrow h\propto r^4$.
Answer: (D). | ||||
| 8 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Adiabatic: $PV^\gamma=$ constant, and $V\propto\frac1\rho$, so $P\propto\rho^\gamma$. For a diatomic gas $\gamma=\frac75$.
$n=32^{7/5}=(2^5)^{7/5}=2^7=128$.
Answer: (A). | ||||
| 9 | Oscillations and WavesDoppler effect | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on doppler effect, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves The wall receives and re-emits; the driver (moving observer) hears $f'=f\,\frac{v+u}{v-u}$.
$\frac{480}{440}=\frac{12}{11}=\frac{345+u}{345-u} \Rightarrow 12(345-u)=11(345+u) \Rightarrow 23u=345 \Rightarrow u=15$ m/s.
$15$ m/s $=15\times3.6=54$ km/hr.
Answer: (A). | ||||
| 10 | ElectrostaticsElectric field and potential | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on electric field and potential, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Potential: all charges are at distance R, and the total charge is $5q-5q=0$, so $V=0$.
Field: the charges are $36°$ apart. Charge $k$ and charge $k+5$ are diametrically opposite and of opposite sign, so each such pair gives a field $\frac{2q}{4\pi\epsilon_0R^2}$ along its diameter, pointing away from its + charge.
Taking charge $k$ at angle $36°(k-1)$, the five pair fields point at $180°$ (pair 1-6), $36°$ (2-7), $252°$ (3-8), $108°$ (4-9), $324°$ (5-10): five equal vectors $72°$ apart, whose sum is zero.
So $E=0$.
Answer: (A). | ||||
| 11 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance With the battery connected, $U=\frac12CV^2\propto C$, so $C$ must double.
With length $x$ of slab inside: $C=\frac{\epsilon_0w}{d}\left[kx+(l-x)\right]=\frac{\epsilon_0w}{d}(l+3x)$.
$l+3x=2l \Rightarrow x=\frac l3$.
Answer: (A). | ||||
| 12 | Current ElectricityKirchhoff's rules | Class 12: Current Electricity | Single correctMedium | |
A medium single correct question on kirchhoff's rules, from Current Electricity. NCERT: Class 12: Current Electricity At junction C, 1 A arrives from A (the current in EA flows E→A→C) and 2 A leaves along CF, so 1 A flows from D to C through the 2 Ω resistor.
Go from A to B along A→C→D→B.
A→C through the 1 V cell (from its negative to its positive terminal): $+1$ V.
C→D through 2 Ω against the 1 A current (D is at the higher potential): $+1\times2=+2$ V.
D→B through the 2 V cell (from its positive to its negative terminal): $-2$ V.
$V_B-V_A=1+2-2=+1$ V.
Answer: (A). | ||||
| 13 | ElectrostaticsCapacitance | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on capacitance, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Take O (the lower end of the 5 μF capacitor) at 0 V. Reading the cells (long plate = positive): the left 6 V cell has its positive plate on the left, so the bottom-left corner is at $+6$ V; the right 6 V cell also has its positive plate on its left (towards O), so the bottom-right corner is at $-6$ V.
Let the upper junction X be at potential $x$. Charge conservation at the isolated node X:
$2(x-6)+4(x+6)+5(x-0)=0 \Rightarrow 11x=-12 \Rightarrow x=-\frac{12}{11}$ V.
Charge on 5 μF: $5\times\frac{12}{11}=\frac{60}{11}\approx5.45$ μC.
Answer: (A); with both cells' positive plates on the outer side the charge is 16.36 μC, (B). | ||||
| 14 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current At distance r the wire's field is $B=\frac{\mu_0I}{2\pi r}$, perpendicular to the plane, the same along the whole connector (which is parallel to the wire).
Motional emf: $\varepsilon=Blv=\frac{\mu_0Ilv}{2\pi r}$; current $i=\frac{\varepsilon}{R}=\frac{\mu_0}{2\pi}\frac{Ivl}{Rr}$.
Answer: (B). | ||||
| 15 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctMedium | |
A medium single correct question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter $H=nI=1000\ \text{m}^{-1}\times0.5=500$ A/m.
Magnetisation $M=\chi H=(\mu_r-1)H=999\times500\approx5\times10^5$ A/m.
Magnetic moment $=M\times V=5\times10^5\times10^{-3}=5\times10^2\ \text{Am}^2$.
Answer: (A). | ||||
| 16 | Electromagnetic WavesEM spectrum | Class 12: Electromagnetic Waves | Single correctEasy | |
A easy single correct question on em spectrum, from Electromagnetic Waves. NCERT: Class 12: Electromagnetic Waves Microwaves: mm to cm range, $10^{-3}$ m (iv). Gamma rays: shortest, $10^{-15}$ m (ii). AM radio: hundreds of metres, 100 m (i). X-rays: about $10^{-10}$ m (iii).
Answer: (A). | ||||
| 17 | Oscillations and WavesSuperposition | Class 11: Oscillations; Class 11: Waves | Single correctMedium | |
A medium single correct question on superposition, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves At L: $S_1L=\sqrt{1.5^2+2^2}=2.5$ m, $S_2L=2$ m, path difference $0.5$ m $=\frac\lambda2$, a minimum.
Moving away from $S_1$ on the circle of radius 2 m about $S_2$ increases $S_1$'s distance, so the path difference grows; the next maximum is at path difference $\lambda=1$ m.
$d-2=1 \Rightarrow d=3$ m (possible, since $d$ can reach $2+1.5=3.5$ m).
Answer: (C). | ||||
| 18 | Atoms and NucleiRadioactivity | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on radioactivity, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Decay constants add: $\lambda=\lambda_1+\lambda_2$, so $\frac1T=\frac1{T_1}+\frac1{T_2}=\frac1{10}+\frac1{100}=\frac{11}{100}$.
$T=\frac{100}{11}\approx9.09$ s, close to 9 s.
Answer: (D). | ||||
| 19 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | Single correctMedium | |
A medium single correct question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits In the circuit all branches (Zener A; Zener B in series with 100 Ω; the load $R_L$) are connected directly across the input, and $V_o$ is taken across $R_L$, so $V_o=V_{in}$ as long as the input can rise.
When $V_{in}$ passes 4 V, Zener B conducts, but its 100 Ω resistor takes up the extra voltage; $V_o$ still equals $V_{in}$ and keeps rising at the same rate.
When $V_{in}$ reaches 6 V, Zener A breaks down and holds the voltage across the parallel branches at 6 V, so $V_o$ stays at 6 V.
Graph: $V_o$ rises linearly to 6 V and then remains constant.
Answer: (C); if the source is taken to have some internal resistance, the slope changes at 4 V, (A). | ||||
| 20 | Magnetic Effects of Current and MagnetismMoving coil galvanometer | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on moving coil galvanometer, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Figure of merit = current per unit deflection $=\frac{I}{\theta}=\frac{6\times10^{-3}\ \text{A}}{2\ \text{div}}=3\times10^{-3}$ A/div (one division taken as 1°).
Answer: (C). | ||||
| 21 | Work, Energy and PowerPower | Class 11: Work, Energy and Power | NumericalMedium | |
A medium numerical question on power, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power Work-energy: $Pt=\frac12mv^2 \Rightarrow v=\sqrt{\frac{2Pt}{m}}=\sqrt{t}$ (in SI units with $P=1$, $m=2$).
$x=\int_0^9\sqrt t\,dt=\frac23t^{3/2}\Big|_0^9=\frac23\times27=18$ m.
Answer: 18. | ||||
| 22 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | NumericalMedium | |
A medium numerical question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion Angular momentum about the pivot is conserved (the pivot force has no torque about it).
Before: $mvL=0.1\times80\times1=8$ kg m²/s.
After: $I=\frac{ML^2}{3}+mL^2=\frac{0.9}{3}+0.1=0.4$ kg m².
$\omega=\frac{8}{0.4}=20$ rad/s.
Answer: 20. | ||||
| 23 | Behaviour of Perfect Gas and Kinetic TheoryKinetic theory | Class 11: Kinetic Theory | NumericalEasy | |
A easy numerical question on kinetic theory, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory $v_{rms}=\sqrt{\frac{3RT}{M}}$; equal speeds need $\frac{T_{H_2}}{2}=\frac{T_{N_2}}{28}$.
$T_{N_2}=300+273=573$ K, so $T_{H_2}=\frac{573\times2}{28}=40.93$ K.
Answer: 40.93. | ||||
| 24 | OpticsTotal internal reflection and prisms | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalEasy | |
A easy numerical question on total internal reflection and prisms, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics For a thin prism $\delta_{min}=(\mu-1)A=0.5\times1°=0.5°=\frac{5}{10}$.
Answer: 5. | ||||
| 25 | Dual Nature of Matter and RadiationPhotoelectric effect | Class 12: Dual Nature of Radiation and Matter | NumericalEasy | |
A easy numerical question on photoelectric effect, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\frac12mv^2=E-\phi$. Speed ratio 2 means kinetic-energy ratio 4:
$4-\phi=4(2.5-\phi) \Rightarrow 3\phi=6 \Rightarrow \phi=2$ eV.
Answer: 2. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques An exocyclic C=CHCl shows geometrical (E/Z) isomerism only if the two ring bonds leaving the ring carbon of the double bond lead to different groups, i.e. the ring is not symmetric about that carbon.
(A) (chloromethylene)cyclohexane: the two ring arms are identical - no geometrical isomers.
(B) 4-chloro-1-methylenecyclohexane: $=CH_2$ carries two H atoms - no geometrical isomers.
(C) 1-(chloromethylene)-3-methylcyclohexane: one ring arm reaches the CH(CH₃) carbon in one step (C2→C3), the other only via C6-C5-C4, so the arms differ and Cl can lie on the side of the methyl-bearing arm or the other side - geometrical isomers exist.
(D) the 3,5-dimethyl compound: the ring is symmetric about C1 (both arms reach a CH(CH₃) carbon in the same way), so the two arrangements of H and Cl are identical - no geometrical isomers.
Answer: (C). | ||||
| 27 | HydrocarbonsElectrophilic substitution and directive influence | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on electrophilic substitution and directive influence, from Hydrocarbons. NCERT: Class 11: Hydrocarbons The starting material is 3-methylaniline (m-toluidine).
(i) $Ac_2O$/pyridine acetylates the amine to $-NHCOCH_3$, a moderately activating ortho/para director that is bulky and stops polybromination.
(ii) $Br_2/FeCl_3$: the acetamido group is the stronger director. Its para position (C4, also ortho to CH₃) is free and least hindered, so Br enters para to $-NHCOCH_3$.
(iii) $OH^-/\Delta$ hydrolyses the amide back to $-NH_2$.
Product: 4-bromo-3-methylaniline (Br para to NH₂ and ortho to Me).
Answer: (A). | ||||
| 28 | PolymersNylon, Teflon, Bakelite and PVC | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on nylon, teflon, bakelite and pvc, from Polymers. NCERT: No matching NCERT chapter Nylon 6,6 (hexamethylenediamine + adipic acid), Nylon 6 (from caprolactam, a condensation/step-growth polyamide) and Bakelite (phenol + formaldehyde) are condensation polymers.
Buna-N is a copolymer of 1,3-butadiene and acrylonitrile formed by addition (chain-growth) polymerisation.
Answer: (D). | ||||
| 29 | Chemistry in Everyday LifeDrugs and their classification | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on drugs and their classification, from Chemistry in Everyday Life. NCERT: No matching NCERT chapter Brompheniramine (like chlorpheniramine) blocks histamine receptors; it is an antihistamine used against allergies.
Answer: (D). | ||||
| 30 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids The substrate, 3-ethyl-3-hydroxycyclohexan-1-one, is a β-hydroxy ketone (the tertiary OH is on C3, β to the C=O).
With $H_2SO_4$ the tertiary alcohol dehydrates; losing the H on C2 gives the C2=C3 double bond conjugated with the carbonyl, the most stable (conjugated, trisubstituted) alkene, as in the dehydration step of an aldol condensation.
Product: 3-ethylcyclohex-2-en-1-one.
Answer: (B). | ||||
| 31 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | Single correctMedium | |
A medium single correct question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Protonation of the C=C gives a secondary carbocation. Protonating C2 gives $CH_3CH_2\overset{+}{C}HCH(CH_3)_2$, which is next to the CH(CH₃)₂ carbon.
A 1,2-hydride shift turns it into the tertiary cation $CH_3CH_2CH_2\overset{+}{C}(CH_3)_2$.
$Br^-$ attacks the tertiary cation: $CH_3CH_2CH_2C(Br)(CH_3)_2$ (2-bromo-2-methylpentane).
Answer: (C). | ||||
| 32 | Alcohols, Phenols and EthersPreparation and properties | Class 12: Alcohols, Phenols and Ethers | Single correctMedium | |
A medium single correct question on preparation and properties, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers All four are para-substituted phenols, so each forms intermolecular hydrogen bonds through OH; the boiling point then rises with the polarity and extra intermolecular attraction contributed by the para group.
I (–CH₃): non-polar, least extra attraction - lowest.
IV (–OCH₃): polar ether oxygen, a hydrogen-bond acceptor - higher than I.
III (–NH₂): itself forms intermolecular hydrogen bonds (donor and acceptor) - higher still.
II (–NO₂): strongly polar (large dipole) and a good acceptor for the phenolic OH, giving the strongest intermolecular association - highest.
Order: I < IV < III < II.
Answer: (B). | ||||
| 33 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties All six ions are isoelectronic (10 electrons). The radius falls as the nuclear charge rises: N (7) > O (8) > F (9) > Na (11) > Mg (12) > Al (13) in size.
Increasing order: $Al^{3+} < Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}$.
Answer: (B). | ||||
| 34 | Isolation of MetalsSmelting and refining | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on smelting and refining, from Isolation of Metals. NCERT: No matching NCERT chapter Zone refining (a heater moved along the rod; impurities stay in the melt) gives semiconductors and other elements of very high purity, such as germanium, silicon, boron, gallium and indium.
Answer: (D). | ||||
| 35 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter $H_2O_2$ has an open-book (non-planar) structure: dihedral angle about 111.5° in the gas phase (90.2° in the solid).
Pure $H_2O_2$ is an almost colourless (very pale blue) liquid.
Answer: (C). | ||||
| 36 | HydrogenHydrogen and its compounds | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on hydrogen and its compounds, from Hydrogen. NCERT: No matching NCERT chapter Permanent hardness (chlorides and sulphates of Ca and Mg) is removed by washing soda ($Na_2CO_3$), Calgon (sodium hexametaphosphate) or ion exchange.
Clark's method (adding slaked lime) removes only temporary hardness, by precipitating the bicarbonates as $CaCO_3$ and $Mg(OH)_2$.
Answer: (D). | ||||
| 37 | p-Block ElementsGroup 17 and 18 elements | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on group 17 and 18 elements, from p-Block Elements. NCERT: No matching NCERT chapter With excess ammonia: $8NH_3+3Cl_2\to6NH_4Cl+N_2$.
With excess chlorine: $NH_3+3Cl_2\to NCl_3+3HCl$.
Answer: (C). | ||||
| 38 | Chemical Bonding and Molecular StructureVSEPR theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on vsepr theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure $CH_4$: no lone pairs, tetrahedral, 109.5°. $NH_3$: one lone pair, 107°. $H_2O$: two lone pairs, 104.5°. $H_2S$: about 92°.
Largest: $CH_4$.
Answer: (D). | ||||
| 39 | Coordination CompoundsIsomerism in complexes | Class 12: Coordination Compounds | Single correctEasy | |
A easy single correct question on isomerism in complexes, from Coordination Compounds. NCERT: Class 12: Coordination Compounds trans-$[Co(en)_2Cl_2]^+$ has a mirror plane (the plane of Co and the four N atoms of the two en ligands, perpendicular to the Cl-Co-Cl axis), so it is achiral.
cis-$[Co(en)_2Cl_2]^+$ has no mirror plane; it exists as a non-superimposable pair of enantiomers ($\Delta$ and $\Lambda$), so it is optically active.
Answer: (D). | ||||
| 40 | Solid StateClose packed structures | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on close packed structures, from Solid State. NCERT: No matching NCERT chapter In fcc the octahedral voids are at the body centre $(\frac a2,\frac a2,\frac a2)$ and the edge centres, e.g. $(\frac a2,0,0)$.
Body centre to edge centre: $\sqrt{0+\frac{a^2}{4}+\frac{a^2}{4}}=\frac{a}{\sqrt2}$; two adjacent edge centres such as $(\frac a2,0,0)$ and $(0,\frac a2,0)$ are also $\frac a{\sqrt2}$ apart. (The voids themselves form an fcc array, whose nearest-neighbour distance is $\frac{a}{\sqrt2}$.)
Answer: (C). | ||||
| 41 | Structure of AtomAtomic orbitals | Class 11: Structure of Atom | Single correctEasy | |
A easy single correct question on atomic orbitals, from Structure of Atom. NCERT: Class 11: Structure of Atom Probability density $\psi^2$ is never negative, so (C) is false.
1s has no node, so $\psi^2$ is never zero at finite distance: (B) false.
2s has one radial node where $\psi^2=0$: (A) false.
3p has a radial node and an angular nodal plane, where $\psi^2=0$: (D) true.
Answer: (D). | ||||
| 42 | ThermodynamicsHess law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics Dissolution = break the lattice + hydrate the ions: $\Delta_{sol}H=\Delta_{lattice}H+\Delta_{hyd}H$.
$4=788+\Delta_{hyd}H \Rightarrow \Delta_{hyd}H=-784$ kJ mol$^{-1}$.
Answer: (D). | ||||
| 43 | ElectrochemistryConductance | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on conductance, from Electrochemistry. NCERT: Class 12: Electrochemistry The molar conductivity falls steeply at low concentration and stays low at higher $\sqrt c$ - the curve of a weak electrolyte, whose degree of dissociation rises sharply on dilution. Strong electrolytes (HCl, NaCl, $KNO_3$) give a nearly linear, gentle fall against $\sqrt c$.
X is the weak acid $CH_3COOH$.
Answer: (D). | ||||
| 44 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | Single correctMedium | |
A medium single correct question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics $\ln k=\ln A-\frac{E_a}{RT}=\ln A-\frac{E_a}{1000R}\cdot\frac{10^3}{T}$.
The line runs from $\ln k=10$ at $\frac{10^3}{T}=0$ to $\ln k=0$ at $\frac{10^3}{T}=5$: slope $=-2$.
$\frac{E_a}{1000R}=2 \Rightarrow E_a=2000R$ J mol$^{-1}$ $=2R$ kJ mol$^{-1}$ (R in J K$^{-1}$ mol$^{-1}$).
Answer: (D). | ||||
| 45 | Surface ChemistryFreundlich adsorption isotherm | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on freundlich adsorption isotherm, from Surface Chemistry. NCERT: No matching NCERT chapter Freundlich: $\frac xm=kp^{1/n}$ with $n>1$, so $\frac xm$ rises with p along a curve that bends towards the p-axis (not a straight line).
Physical adsorption is exothermic, so at a given pressure less gas is adsorbed at higher temperature: the 200 K curve lies highest and the 270 K curve lowest.
That is plot (A).
Answer: (A). | ||||
| 46 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $Ru^{2+}$ is $4d^6$. With $\Delta_0>P$ the electrons pair in $t_{2g}$: $t_{2g}^6e_g^0$, no unpaired electrons.
$\mu=\sqrt{n(n+2)}=0$ BM.
Answer: 0. | ||||
| 47 | Redox ReactionsBalancing redox equations | Class 11: Redox Reactions | NumericalMedium | |
A medium numerical question on balancing redox equations, from Redox Reactions. NCERT: Class 11: Redox Reactions $FeC_2O_4$: $M=56+2(12)+4(16)=144$ g mol$^{-1}$; moles $=\frac{0.288}{144}=2\times10^{-3}$.
Each $FeC_2O_4$ loses 3 electrons ($Fe^{2+}\to Fe^{3+}$: 1; $C_2O_4^{2-}\to2CO_2$: 2), so electrons $=6\times10^{-3}$ mol.
$Cr_2O_7^{2-}\to2Cr^{3+}$ takes 6 electrons: moles of $K_2Cr_2O_7=10^{-3}$.
Volume $=\frac{10^{-3}}{0.02}$ L $=0.05$ L $=50$ mL.
Answer: 50. | ||||
| 48 | ThermodynamicsEntropy and Gibbs energy | Class 11: Thermodynamics | NumericalMedium | |
A medium numerical question on entropy and gibbs energy, from Thermodynamics. NCERT: Class 11: Thermodynamics $\Delta n_g=1-2=-1$, so $\Delta H^\ominus=\Delta U^\ominus+\Delta n_gRT=-20000-8.314\times298=-20000-2477.57=-22477.57$ J mol$^{-1}$.
$\Delta G^\ominus=\Delta H^\ominus-T\Delta S^\ominus=-22477.57-298(-30)=-22477.57+8940=-13537.57$ J.
Answer: −13537.57. | ||||
| 49 | EquilibriumChemical equilibrium | Class 11: Equilibrium | NumericalEasy | |
A easy numerical question on chemical equilibrium, from Equilibrium. NCERT: Class 11: Equilibrium Z rises from 0.5 to 1.0 M, so 0.25 M each of X and Y reacted: $[X]=0.75$, $[Y]=1.25$, $[Z]=1.0$.
$K=\frac{[Z]^2}{[X][Y]}=\frac{1}{0.75\times1.25}=\frac{1}{0.9375}=\frac{16}{15}$.
$x=16$.
Answer: 16. | ||||
| 50 | BiomoleculesCarbohydrates | Class 12: Biomolecules | NumericalMedium | |
A medium numerical question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules Sucrose = α-D-glucopyranose (C1) linked to β-D-fructofuranose (C2).
Glucose unit: C1, C2, C3, C4, C5 are chiral (C6 is $CH_2OH$): 5.
Fructose unit: C2, C3, C4, C5 are chiral (C1 and C6 are $CH_2OH$): 4.
Total $=9$.
Answer: 9. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 51 | Complex NumbersModulus and argument | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on modulus and argument, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations $-1+i\sqrt3=2e^{i2\pi/3}$ and $1-i=\sqrt2e^{-i\pi/4}$, so the ratio is $\sqrt2\,e^{i(2\pi/3+\pi/4)}=\sqrt2\,e^{i\,11\pi/12}$.
Raised to 30: $(\sqrt2)^{30}e^{i\,330\pi/12}=2^{15}e^{i\,55\pi/2}$.
$\frac{55\pi}{2}=26\pi+\frac{3\pi}{2}$, and $e^{i3\pi/2}=-i$.
Value $=-2^{15}i$.
Answer: (D). | ||||
| 52 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations $\alpha+\beta=\frac37$, $\alpha\beta=-\frac27$.
$\frac{\alpha}{1-\alpha^2}+\frac{\beta}{1-\beta^2}=\frac{\alpha+\beta-\alpha\beta(\alpha+\beta)}{1-(\alpha^2+\beta^2)+\alpha^2\beta^2}=\frac{(\alpha+\beta)(1-\alpha\beta)}{1-(\alpha^2+\beta^2)+(\alpha\beta)^2}$.
Numerator: $\frac37\cdot\frac97=\frac{27}{49}$.
$\alpha^2+\beta^2=\frac{9}{49}+\frac47=\frac{37}{49}$, so the denominator is $1-\frac{37}{49}+\frac{4}{49}=\frac{16}{49}$.
Value $=\frac{27}{16}$.
Answer: (B). | ||||
| 53 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $C_3\to C_3-C_1$: third column becomes $(a,b,c)$. Then $C_2\to C_2-C_3$: second column becomes $(y,y,y)$.
$\Delta=y\begin{vmatrix}x&1&a\\y&1&b\\z&1&c\end{vmatrix}$. Now $R_1\to R_1-R_2$, $R_3\to R_3-R_2$:
$\Delta=y\begin{vmatrix}x-y&0&a-b\\y&1&b\\z-y&0&c-b\end{vmatrix}=y\left[(x-y)(c-b)-(a-b)(z-y)\right]$.
From the condition: $x-y=b-a$ and $z-y=b-c-1$. So
$(x-y)(c-b)-(a-b)(z-y)=(b-a)(c-b)-(a-b)(b-c-1)=(a-b)\left[(b-c)-(b-c-1)\right]=a-b$.
$\Delta=y(a-b)$.
Answer: (B). | ||||
| 54 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Subtracting the first equation from the third: $2x=0$, so $x=0$. Then $y+3z=0$, i.e. $y=-3z$ (with $z\neq0$ for a non-zero solution; the second equation then needs $k^2=9$).
$x+\frac yz=0-3=-3$.
Answer: (B). | ||||
| 55 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations The numbers chosen from the three sections are a permutation of (3,1,1) or (2,2,1).
(3,1,1): $3\times\binom53\binom51\binom51=3\times250=750$.
(2,2,1): $3\times\binom52\binom52\binom51=3\times500=1500$.
Total $=2250$.
Answer: (D). | ||||
| 56 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $ar(1+r+r^2)=3$ and $ar^5(1+r+r^2)=243$; dividing, $r^4=81$, so $r=3$ (positive terms).
$a\cdot3\cdot13=3 \Rightarrow a=\frac1{13}$.
$S_{50}=\frac{a(r^{50}-1)}{r-1}=\frac{1}{13}\cdot\frac{3^{50}-1}{2}=\frac{1}{26}(3^{50}-1)$.
Answer: (A). | ||||
| 57 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $\log_{7^{1/n}}x=n\log_7x$. The 20 terms have $n=2,3,\ldots,21$, and $\sum_{n=2}^{21}n=\frac{21\cdot22}{2}-1=230$.
$230\log_7x=460 \Rightarrow \log_7x=2 \Rightarrow x=7^2$.
Answer: (C). | ||||
| 58 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Let $t=\frac{\sqrt{1+x^2+x^4}-1}{x}=\frac{x+x^3}{\sqrt{1+x^2+x^4}+1}\to0$ as $x\to0$.
The expression is $\frac{x(e^t-1)}{xt}=\frac{e^t-1}{t}\to1$.
Answer: (B). | ||||
| 59 | Limits, Continuity and DifferentiabilityDifferentiation of composite and implicit functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on differentiation of composite and implicit functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $u$: put $x=\tan\theta$: $\frac{\sec\theta-1}{\tan\theta}=\frac{1-\cos\theta}{\sin\theta}=\tan\frac\theta2$, so $u=\frac12\tan^{-1}x$ and $\frac{du}{dx}=\frac{1}{2(1+x^2)}$.
$v$: put $x=\sin\phi$ (near $x=\frac12$, $\phi=\frac\pi6<\frac\pi4$): $\frac{2\sin\phi\cos\phi}{\cos2\phi}=\tan2\phi$, so $v=2\sin^{-1}x$ and $\frac{dv}{dx}=\frac{2}{\sqrt{1-x^2}}$.
At $x=\frac12$: $\frac{du}{dx}=\frac{1}{2\cdot\frac54}=\frac25$, $\frac{dv}{dx}=\frac{2}{\sqrt3/2}=\frac{4}{\sqrt3}$.
$\frac{du}{dv}=\frac25\cdot\frac{\sqrt3}{4}=\frac{\sqrt3}{10}$.
Answer: (C). | ||||
| 60 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives Differentiating: $4x^3e^y+x^4e^yy'+\frac{y'}{\sqrt{y+1}}=0$. At (1, 0): $4+y'+y'=0 \Rightarrow y'=-2$.
Tangent: $y=-2(x-1)$. At $x=-2$: $y=6$, so (−2, 6) lies on it; (−2, 4), (2, 6), (2, 2) do not (at $x=2$, $y=-2$).
Answer: (D). | ||||
| 61 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x)=e^x\left[3x^2+(6+a)x-2\right]$. $f'(1)=0 \Rightarrow 3+6+a-2=0 \Rightarrow a=-7$.
$f'(x)=e^x(3x^2-x-2)=e^x(3x+2)(x-1)$.
Sign of $f'$: $+$ for $x<-\frac23$, $-$ on $\left(-\frac23,1\right)$, $+$ for $x>1$. So $x=-\frac23$ is a local maximum and $x=1$ a local minimum.
Answer: (D). | ||||
| 62 | Integral CalculusPartial fractions | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on partial fractions, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $s=\sin\theta$: the denominator is $5+7s-2(1-s^2)=2s^2+7s+3=(2s+1)(s+3)$.
$\frac{1}{(2s+1)(s+3)}=\frac15\left(\frac{2}{2s+1}-\frac{1}{s+3}\right)$, so the integral is $\frac15\ln\left|\frac{2\sin\theta+1}{\sin\theta+3}\right|+C$.
$A=\frac15$, $B(\theta)=\frac{2\sin\theta+1}{\sin\theta+3}$, $\frac{B}{A}=\frac{5(2\sin\theta+1)}{\sin\theta+3}$.
Answer: (B). | ||||
| 63 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Lower boundary: $(x-1)[x]=0$ on $[0,1)$ and $=x-1$ on $[1,2)$; it stays below $2\sqrt x$ throughout.
Area $=\int_0^22\sqrt x\,dx-\int_1^2(x-1)\,dx=\frac43\cdot2\sqrt2-\frac12=\frac83\sqrt2-\frac12$.
Answer: (D). | ||||
| 64 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Divide by $\cos x$: $y'+2\tan x\,y=2\sin x$. Integrating factor $e^{\int2\tan x\,dx}=\sec^2x$.
$(y\sec^2x)'=2\sin x\sec^2x=2\sec x\tan x \Rightarrow y\sec^2x=2\sec x+C$.
$y(\pi/3)=0$: $0=4+C \Rightarrow C=-4$.
At $\pi/4$: $2y=2\sqrt2-4 \Rightarrow y=\sqrt2-2$.
Answer: (B). | ||||
| 65 | CirclesIntersection of a line and a circle | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on intersection of a line and a circle, from Circles. NCERT: Class 11: Conic Sections Distance of the centre from the line $2x-y+3=0$: $d=\frac{3}{\sqrt5}$.
Half-chord $\frac r2$: $r^2=d^2+\frac{r^2}{4} \Rightarrow \frac34r^2=\frac95 \Rightarrow r^2=\frac{12}{5}$.
Answer: (B). | ||||
| 66 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Tangency to the circle: $c^2=36(1+m^2)$. Tangency to the hyperbola: $c^2=100m^2-64$.
$36+36m^2=100m^2-64 \Rightarrow m^2=\frac{25}{16}$, so $m=\pm\frac54$ (neither $8m+5=0$ nor $5m=4$).
$c^2=36\cdot\frac{41}{16}=\frac{369}{4}$, i.e. $4c^2=369$.
Answer: (C). | ||||
| 67 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $L_1$ passes through $(-1,2,1)$ with direction $(2,-1,1)$; $L_2$ through $(-2,-1,-1)$ with direction $(\alpha,5-\alpha,1)$.
Coplanar $\iff\begin{vmatrix}-1&-3&-2\\2&-1&1\\\alpha&5-\alpha&1\end{vmatrix}=0$.
$=-1(-1-5+\alpha)+3(2-\alpha)-2(10-2\alpha+\alpha)=6-\alpha+6-3\alpha-20+2\alpha=-8-2\alpha=0 \Rightarrow \alpha=-4$.
$L_2:\frac{x+2}{-4}=\frac{y+1}{9}=\frac{z+1}{1}$. With parameter $-1$: $(2,-10,-2)$.
Answer: (C). | ||||
| 68 | StatisticsStandard deviation and variance | Class 11: Statistics | Single correctEasy | |
A easy single correct question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Mean: $15+a+b=25 \Rightarrow a+b=10$.
Variance: $\frac{9+25+49+a^2+b^2}{5}-25=4 \Rightarrow a^2+b^2=62$.
$ab=\frac{(a+b)^2-(a^2+b^2)}{2}=\frac{100-62}{2}=19$. Equation: $x^2-10x+19=0$.
Answer: (B). | ||||
| 69 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Using $\sin^2\theta=\frac{1-\cos2\theta}{2}$, $\cos^2\theta=\frac{1+\cos2\theta}{2}$:
$L=\frac{1-\cos\frac\pi8}{2}-\frac{1-\cos\frac\pi4}{2}=\frac{1}{2\sqrt2}-\frac12\cos\frac\pi8$ (neither (A) nor (B)).
$M=\frac{1+\cos\frac\pi8}{2}-\frac{1-\cos\frac\pi4}{2}=\frac{1}{2\sqrt2}+\frac12\cos\frac\pi8$, which is (C).
Answer: (C). | ||||
| 70 | Mathematical ReasoningStatements and logical connectives | No matching NCERT chapter | Single correctEasy | |
A easy single correct question on statements and logical connectives, from Mathematical Reasoning. NCERT: No matching NCERT chapter $p\to(p\vee q)$ is always true (if p is true, $p\vee q$ is true). An implication whose conclusion is always true is always true.
So the statement is a tautology.
Answer: (B). | ||||
| 71 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Functions with 2 in the range: $4^3-3^3=64-27=37$.
One-one functions with 2 in the range: all one-one $=4\cdot3\cdot2=24$, minus those avoiding 2 $=3!=6$: 18.
Not one-one with 2 in the range: $37-18=19$.
Answer: 19. | ||||
| 72 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $(1+x+x^2+x^3)^6=(1+x)^6(1+x^2)^6$.
Coefficient of $x^4$: take $x^{2j}$ from $(1+x^2)^6$ and $x^{4-2j}$ from $(1+x)^6$:
$j=0$: $\binom64=15$; $j=1$: $\binom61\binom62=6\times15=90$; $j=2$: $\binom62=15$.
Total $=15+90+15=120$.
Answer: 120. | ||||
| 73 | Applications of DerivativesTangents and normals | Class 12: Application of Derivatives | NumericalEasy | |
A easy numerical question on tangents and normals, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives The curve meets the x-axis at $x=1$ and $x=2$; $y'=2x-3$ gives slopes $-1$ at (1, 0) and $+1$ at (2, 0).
$x+y=a$ has slope $-1$, so it touches at (1, 0): $a=1$. $x-y=b$ has slope 1, so it touches at (2, 0): $b=2$.
$\frac ab=\frac12=0.5$.
Answer: 0.5. | ||||
| 74 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | NumericalEasy | |
A easy numerical question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra Equal projections on $\vec a$: $\vec a\cdot\vec b=\vec a\cdot\vec c$. Perpendicular: $\vec b\cdot\vec c=0$.
$|\vec a+\vec b-\vec c|^2=4+16+16+2\vec a\cdot\vec b-2\vec a\cdot\vec c-2\vec b\cdot\vec c=36$.
Value $=6$.
Answer: 6. | ||||
| 75 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | NumericalMedium | |
A medium numerical question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability With n bombs, $P(\text{at least 2 hits})=1-\frac{1}{2^n}-\frac{n}{2^n}=1-\frac{n+1}{2^n}$.
Need $\frac{n+1}{2^n}\le0.01$. $n=10$: $\frac{11}{1024}\approx0.0107$ (too large); $n=11$: $\frac{12}{2048}\approx0.0059$ (enough).
Minimum $n=11$.
Answer: 11. | ||||
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