JEE Main — Full Mock Test (Free, Timed)

Question 1Kinematics

Two particles are projected simultaneously from the same point: the first vertically upward with speed $u$, the second horizontally with the same speed $u$. After time $t$ (before either lands), the magnitude of the velocity of the first particle relative to the second is:

Solution: Velocity of particle 1: $\vec{v_1} = (0,\, u - gt)$. Velocity of particle 2: $\vec{v_2} = (u,\, -gt)$. Relative velocity $\vec{v_1}-\vec{v_2} = (-u,\, u)$ — the $-gt$ terms cancel since both share the same gravitational acceleration. Magnitude $=\sqrt{u^2+u^2}=u\sqrt{2}$, constant in time.

Question 2Laws of Motion

Blocks $m_1=2\,\text{kg}$ and $m_2=3\,\text{kg}$ are joined by a light string over a frictionless pulley at a table edge. $m_1$ rests on the rough horizontal table ($\mu=0.5$); $m_2$ hangs freely. Take $g=10\,\text{m/s}^2$. The acceleration of the system is:

Solution: Driving force $= m_2 g = 30\,\text{N}$. Friction on $m_1$ $=\mu m_1 g = 0.5\times2\times10 = 10\,\text{N}$. Net force $= 30-10 = 20\,\text{N}$ acting on total mass $5\,\text{kg}$: $a = 20/5 = 4\,\text{m/s}^2$.

Question 3Work, Energy and Power

A block of mass $2\,\text{kg}$ is pushed against a spring ($k=800\,\text{N/m}$) on a horizontal surface, compressing it by $0.2\,\text{m}$, then released. The surface is rough with $\mu=0.2$. Take $g=10\,\text{m/s}^2$. The total distance the block travels from the point of release before finally stopping is:

Solution: Spring PE released $=\tfrac12 k x^2 = \tfrac12(800)(0.2)^2 = 16\,\text{J}$. All of it is eventually dissipated by friction over total path $d$ (measured from release point): friction force $=\mu mg = 0.2\times2\times10 = 4\,\text{N}$. So $4d = 16 \Rightarrow d = 4\,\text{m}$.

Question 4Rotational Motion

A uniform solid sphere of mass $M$ and radius $R$ is released from rest at the top of an incline of height $h$ and rolls without slipping to the bottom. The speed of its centre at the bottom is:

Solution: Rolling without slipping: $mgh = \tfrac12 mv^2 + \tfrac12 I\omega^2$ with $I=\tfrac25 mR^2$, $\omega=v/R$. So $mgh = \tfrac12 mv^2(1+\tfrac25) = \tfrac{7}{10}mv^2$, giving $v=\sqrt{\tfrac{10}{7}gh}$.

Question 5Rotational Motion

A uniform rod of mass $M$ and length $L$ is pivoted at one end and held horizontal, then released from rest. The angular acceleration of the rod immediately after release is:

Solution: Torque about the pivot from gravity acting at the centre: $\tau = Mg\cdot\tfrac{L}{2}$. Moment of inertia of a rod about its end: $I=\tfrac13 ML^2$. Angular acceleration $\alpha=\dfrac{\tau}{I}=\dfrac{MgL/2}{ML^2/3}=\dfrac{3g}{2L}$.

Question 6Gravitation

A satellite of mass $m$ is in a circular orbit of radius $r$ around Earth (mass $M$). To raise it to a circular orbit of radius $2r$, the additional energy required is (with $G$ the gravitational constant):

Solution: Total orbital energy $E=-\dfrac{GMm}{2r}$. At radius $r$: $E_1=-\dfrac{GMm}{2r}$; at $2r$: $E_2=-\dfrac{GMm}{4r}$. Required energy $=E_2-E_1 = -\dfrac{GMm}{4r}+\dfrac{GMm}{2r} = \dfrac{GMm}{4r}$.

Question 7Fluid Mechanics

A large tank is filled with water to height $H$. A small hole is made at depth $h$ below the free surface. The horizontal range of the emerging jet on the ground is maximised when:

Solution: Efflux speed $v=\sqrt{2gh}$ (Torricelli). Fall height to ground $=H-h$, time $t=\sqrt{2(H-h)/g}$. Range $R=v\,t=2\sqrt{h(H-h)}$. Maximising the product $h(H-h)$ gives $h=H/2$, with $R_{max}=H$.

Question 8Thermodynamics

One mole of an ideal monatomic gas undergoes a cycle: isothermal expansion at temperature $T$ doubling its volume, then isochoric cooling, then adiabatic compression returning to the initial state. The net work done by the gas over one complete cycle is:

Solution: On a P-V diagram: isothermal expansion moves rightward, isochoric cooling drops vertically at constant V, adiabatic compression curves back to the start. The closed loop is traversed clockwise, so the enclosed area — equal to the net work done by the gas — is positive.

Question 9Kinetic Theory

The ratio of the root-mean-square speed of hydrogen molecules ($M=2\,\text{g/mol}$) to that of oxygen molecules ($M=32\,\text{g/mol}$) at the same temperature is:

Solution: $v_{rms}=\sqrt{3RT/M}$, so at equal $T$, $\dfrac{v_{H_2}}{v_{O_2}}=\sqrt{\dfrac{M_{O_2}}{M_{H_2}}}=\sqrt{\dfrac{32}{2}}=\sqrt{16}=4$. Ratio $=4:1$.

Question 10Simple Harmonic Motion

A particle executing SHM of amplitude $A$ has kinetic energy equal to three times its potential energy at a certain instant. Its displacement from the mean position at that instant is:

Solution: PE $=\tfrac12 k x^2$, total $E=\tfrac12 kA^2$, KE $=E-\text{PE}$. KE $=3\,\text{PE}$ implies total $=4\,\text{PE}$: $\tfrac12 kA^2 = 4\cdot\tfrac12 kx^2 \Rightarrow x^2 = A^2/4 \Rightarrow x=A/2$.

Question 11Waves and Sound

A source emitting $500\,\text{Hz}$ moves at $30\,\text{m/s}$ directly toward a stationary observer, then after passing recedes at $30\,\text{m/s}$. Taking the speed of sound as $330\,\text{m/s}$, the drop in observed frequency from approach to recede is closest to:

Solution: Approaching: $f_1=500\cdot\dfrac{330}{330-30}=500\cdot\dfrac{330}{300}=550\,\text{Hz}$. Receding: $f_2=500\cdot\dfrac{330}{330+30}=500\cdot\dfrac{330}{360}=458.3\,\text{Hz}$. Difference $=550-458.3\approx 92$; nearest option is $97\,\text{Hz}$ (and well above the small-shift distractors).

Question 12Electrostatics

An electric dipole of moment $p$ sits in a uniform field $E$. It is rotated by an external agent from alignment with the field ($\theta=0$) to the perpendicular position ($\theta=90^\circ$). The work done by the external agent is:

Solution: Potential energy $U(\theta)=-pE\cos\theta$. $W_{ext}=U(90^\circ)-U(0^\circ)=(-pE\cos90^\circ)-(-pE\cos0^\circ)=0-(-pE)=pE$.

Question 13Electrostatics

Three equal point charges $+q$ are fixed at the vertices of an equilateral triangle of side $a$. The work done to bring a fourth charge $+q$ from infinity to the centroid is (with $k=1/4\pi\varepsilon_0$):

Solution: Distance from each vertex to the centroid $=a/\sqrt{3}$. Potential at centroid $V=3\cdot\dfrac{kq}{a/\sqrt{3}}=\dfrac{3\sqrt{3}\,kq}{a}$. Work to bring $+q$ from infinity $=qV=\dfrac{3\sqrt{3}\,kq^2}{a}$.

Question 14Capacitors

A $2\,\mu\text{F}$ capacitor charged to $100\,\text{V}$ is connected in parallel with an uncharged $3\,\mu\text{F}$ capacitor. The energy lost during charge redistribution is:

Solution: Energy lost $=\dfrac12\dfrac{C_1 C_2}{C_1+C_2}(V_1-V_2)^2=\dfrac12\cdot\dfrac{2\times3}{5}\,\mu\text{F}\cdot(100)^2=\dfrac12\cdot1.2\times10^{-6}\cdot10^4=6\times10^{-3}\,\text{J}=6\,\text{mJ}$.

Question 15Current Electricity

A $12\,\text{V}$ battery of negligible internal resistance feeds two parallel branches: branch 1 is $4\,\Omega$ and $2\,\Omega$ in series; branch 2 is $3\,\Omega$ and $3\,\Omega$ in series. The total current drawn from the battery is:

Solution: Branch 1 $=4+2=6\,\Omega$, current $=12/6=2\,\text{A}$. Branch 2 $=3+3=6\,\Omega$, current $=12/6=2\,\text{A}$. Total $=2+2=4\,\text{A}$ (equivalently $6\|6=3\,\Omega$, $I=12/3=4\,\text{A}$).

Question 16Magnetism and EMI

A conducting rod of length $L$ slides on frictionless rails at constant velocity $v$ perpendicular to a uniform field $B$ (into the page), closing a circuit of total resistance $R$. The external force needed to maintain constant velocity is:

Solution: EMF $=BLv$, induced current $I=BLv/R$. The retarding magnetic force on the rod $=BIL=B\cdot\dfrac{BLv}{R}\cdot L=\dfrac{B^2L^2v}{R}$. To keep $v$ constant the applied force must equal this.

Question 17Alternating Current

A series LCR circuit has $R=30\,\Omega$, $X_L=80\,\Omega$, $X_C=40\,\Omega$, driven by an AC source of $V_{rms}=100\,\text{V}$. The average power dissipated is:

Solution: $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{30^2+40^2}=50\,\Omega$. $I_{rms}=100/50=2\,\text{A}$. Power factor $\cos\phi=R/Z=0.6$. $P=V_{rms}I_{rms}\cos\phi=100\times2\times0.6=120\,\text{W}$ (check: $I^2R=4\times30=120\,\text{W}$).

Question 18Ray Optics

An object is placed $15\,\text{cm}$ in front of a converging lens of focal length $10\,\text{cm}$. The image so formed acts as the object for a concave mirror of focal length $10\,\text{cm}$ placed $30\,\text{cm}$ behind the lens. Using the lens result, the image formed by the lens alone is located:

Solution: Lens formula $\dfrac1v-\dfrac1u=\dfrac1f$ with $u=-15\,\text{cm}$, $f=+10\,\text{cm}$: $\dfrac1v=\dfrac1{10}+\dfrac1{-15}=\dfrac{3-2}{30}=\dfrac{1}{30}$, so $v=+30\,\text{cm}$. Positive $v$ means a real image $30\,\text{cm}$ behind the lens (which conveniently lands at the mirror's centre of curvature).

Question 19Wave Optics

In Young's double-slit experiment with light of wavelength $\lambda$, a thin transparent sheet of refractive index $\mu$ and thickness $t$ is placed over one slit. The number of fringe widths by which the central fringe shifts is:

Solution: The sheet adds extra optical path $(\mu-1)t$ in that arm. The central maximum moves to where this is compensated; since one fringe corresponds to a path difference of $\lambda$, the shift is $N=\dfrac{(\mu-1)t}{\lambda}$ fringe widths.

Question 20Modern Physics

In a photoelectric experiment the stopping potential is $1.5\,\text{V}$ for light of frequency $\nu_1$ and $3.5\,\text{V}$ for frequency $\nu_2=2\nu_1$. The work function of the metal is:

Solution: $eV_s=h\nu-\phi$. Thus $1.5=h\nu_1-\phi$ and $3.5=2h\nu_1-\phi$. Subtracting: $2.0=h\nu_1$, so $h\nu_1=2.0\,\text{eV}$. Then $\phi=h\nu_1-1.5=2.0-1.5=0.5\,\text{eV}$.

Question 21 · NumericalWork, Energy and Power

A $1\,\text{kg}$ block moving at $6\,\text{m/s}$ on a frictionless surface collides head-on and sticks to a stationary $2\,\text{kg}$ block attached to an unstretched spring of stiffness $k=300\,\text{N/m}$. Find the maximum compression of the spring in metres. Round to 1 decimal place.

Enter a numerical value.

Solution: Perfectly inelastic collision: common velocity $v=\dfrac{1\times6}{1+2}=2\,\text{m/s}$. KE just after $=\tfrac12(3)(2)^2=6\,\text{J}$. At maximum compression all this KE is stored in the spring: $\tfrac12 k x^2=6 \Rightarrow x^2=\dfrac{12}{300}=0.04 \Rightarrow x=0.2\,\text{m}$.

Question 22 · NumericalRotational Motion

A solid disc of mass $M=2\,\text{kg}$ and radius $R=0.5\,\text{m}$ rotates freely about its central axis at $\omega=4\,\text{rad/s}$. A lump of clay of mass $m=1\,\text{kg}$ is dropped gently onto its rim and sticks. Find the final common angular speed in rad/s. Round to 1 decimal place.

Enter a numerical value.

Solution: Conserve angular momentum about the axis. $I_{disc}=\tfrac12 MR^2=\tfrac12(2)(0.25)=0.25\,\text{kg·m}^2$, initial $L=0.25\times4=1.0$. After: clay at the rim adds $mR^2=1\times0.25=0.25$, so $I_f=0.50$. $\omega_f=\dfrac{L}{I_f}=\dfrac{1.0}{0.50}=2.0\,\text{rad/s}$.

Question 23 · NumericalCurrent Electricity

A battery of EMF $10\,\text{V}$ and internal resistance $1\,\Omega$ drives an external resistance $R$. Find the value of $R$ (in ohms) for which the power delivered to $R$ is maximum.

Enter a numerical value.

Solution: Maximum power transfer to the external load occurs when the load resistance equals the source's internal resistance: $R=r=1\,\Omega$. (At this value $P_{max}=\dfrac{E^2}{4r}=\dfrac{100}{4}=25\,\text{W}$.)

Question 24 · NumericalMagnetism and EMI

A circular coil of $100$ turns and area $0.02\,\text{m}^2$ lies in a magnetic field (perpendicular to its plane) that decreases uniformly from $0.5\,\text{T}$ to $0.1\,\text{T}$ in $0.1\,\text{s}$. Find the magnitude of the induced EMF in volts.

Enter a numerical value.

Solution: $\varepsilon=N\dfrac{\Delta\Phi}{\Delta t}=N A\dfrac{\Delta B}{\Delta t}=100\times0.02\times\dfrac{0.5-0.1}{0.1}=100\times0.02\times4=8\,\text{V}$.

Question 25 · NumericalModern Physics

A radioactive sample has a half-life of $20\,\text{minutes}$. Starting from $80\,\text{g}$, find the mass (in grams) remaining after $1\,\text{hour}$.

Enter a numerical value.

Solution: $1\,\text{hour}=60\,\text{min}=3$ half-lives. Remaining fraction $=(1/2)^3=1/8$. Mass remaining $=80\times\tfrac18=10\,\text{g}$.

Question 26Chemical Equilibrium

For $N_2O_4(g) \rightleftharpoons 2NO_2(g)$, $N_2O_4$ is found to be 50% dissociated at equilibrium at a total pressure of 1 atm. The value of $K_p$ (in atm) is:

Solution: With $\alpha = 0.5$ and $P = 1$ atm, $K_p = \dfrac{4\alpha^2}{1-\alpha^2}P = \dfrac{4(0.25)}{1-0.25}(1) = \dfrac{1.0}{0.75} = 1.33$ atm.

Question 27Electrochemistry

For the concentration cell $Cu | Cu^{2+}(0.001\,M) \,||\, Cu^{2+}(0.1\,M) | Cu$ at 298 K, the EMF (in V) is approximately ($\frac{0.059}{n}$ convention):

Solution: For a concentration cell, $E = \dfrac{0.059}{n}\log\dfrac{[\text{cathode}]}{[\text{anode}]} = \dfrac{0.059}{2}\log\dfrac{0.1}{0.001} = 0.0295\times\log(100) = 0.0295\times 2 = 0.059$ V.

Question 28Chemical Kinetics

A first-order reaction is 75% complete in 60 minutes. The time required for 50% completion is approximately:

Solution: For first order, 75% completion means two half-lives (75% = $1 - (1/2)^2$). So $2t_{1/2} = 60$ min $\Rightarrow t_{1/2} = 30$ min, which is the time for 50% completion.

Question 29Ionic Equilibrium

The pH of a buffer made by mixing 0.20 mol $CH_3COOH$ and 0.10 mol $CH_3COONa$ in 1 L of solution is ($pK_a$ of acetic acid = 4.74):

Solution: Henderson equation: $pH = pK_a + \log\dfrac{[\text{salt}]}{[\text{acid}]} = 4.74 + \log\dfrac{0.10}{0.20} = 4.74 + \log(0.5) = 4.74 - 0.30 = 4.44$.

Question 30Solutions

When 0.6 g of acetic acid ($M = 60$) is dissolved in 100 g of benzene, the freezing point depression observed corresponds to a van't Hoff factor $i = 0.5$. This indicates the acetic acid:

Solution: $i = 0.5$ means the number of particles is half the expected, i.e. two molecules associate into one. In a non-polar solvent like benzene, acetic acid forms hydrogen-bonded dimers, giving $i = 0.5$.

Question 31Thermodynamics

For a reaction at 300 K, $\Delta H = +30\,\text{kJ mol}^{-1}$ and $\Delta S = +150\,\text{J K}^{-1}\text{mol}^{-1}$. The reaction is:

Solution: $\Delta G = \Delta H - T\Delta S = 30000 - 300(150) = 30000 - 45000 = -15000\,\text{J} = -15\,\text{kJ}$. Since $\Delta G < 0$, the reaction is spontaneous at 300 K. (Crossover temperature $= \Delta H/\Delta S = 200$ K; spontaneous above 200 K, and 300 K qualifies.)

Question 32Atomic Structure

The number of radial nodes and angular nodes, respectively, in a $4d$ orbital are:

Solution: Radial nodes $= n - l - 1 = 4 - 2 - 1 = 1$. Angular nodes $= l = 2$ (for d, $l=2$). So 1 radial and 2 angular nodes.

Question 33Chemical Kinetics

The rate constant of a reaction doubles when the temperature increases from 300 K to 310 K. The activation energy ($E_a$) is closest to (use $R = 8.314\,\text{J K}^{-1}\text{mol}^{-1}$):

Solution: Arrhenius: $\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$. $\ln 2 = 0.693 = \dfrac{E_a}{8.314}\left(\dfrac{1}{300}-\dfrac{1}{310}\right) = \dfrac{E_a}{8.314}(1.0753\times10^{-4})$. $E_a = \dfrac{0.693\times 8.314}{1.0753\times10^{-4}} \approx 53600\,\text{J} = 53.6$ kJ/mol.

Question 34Coordination Compounds

The spin-only magnetic moment of $[Fe(H_2O)_6]^{2+}$ is closest to (use $\mu = \sqrt{n(n+2)}$ BM):

Solution: $Fe^{2+}$ is $d^6$. $H_2O$ is a weak-field ligand, so high-spin: $t_{2g}^4 e_g^2$ gives 4 unpaired electrons. $\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90$ BM.

Question 35Coordination Compounds

The number of geometrical isomers possible for the octahedral complex $[Ma_2b_2c_2]$ (all monodentate) is:

Solution: For $[Ma_2b_2c_2]$, the geometrical isomers number 6: five geometric diastereomers, one of which (the all-cis form) is itself chiral and exists as an enantiomeric pair, giving a total of 6 stereoisomers. The classic count for $[Ma_2b_2c_2]$ is 6.

Question 36Periodic Properties

The correct order of first ionisation enthalpy for the elements B, C, N, O is:

Solution: Generally IE increases across a period: B < C < N. But oxygen ($2p^4$) loses an electron to give a stable half-filled-like config, and its IE is lower than nitrogen ($2p^3$, extra-stable half-filled). So the order is B < C < O < N.

Question 37Chemical Bonding

Among $NO$, $NO^+$, $NO^-$, and $O_2$, the species with the highest bond order is:

Solution: Bond orders: $NO$ (15 e) = 2.5; $NO^+$ (14 e) = 3.0; $NO^-$ (16 e) = 2.0; $O_2$ (16 e) = 2.0. The highest is $NO^+$ with bond order 3.0 (isoelectronic with $N_2$/CO).

Question 38p-Block Elements

The correct order of acidic strength of the oxoacids $HClO$, $HClO_2$, $HClO_3$, $HClO_4$ is:

Solution: Acid strength of oxoacids increases with the oxidation state of the central atom (and number of terminal O atoms), which stabilises the conjugate base by delocalisation. Cl oxidation states: +1, +3, +5, +7. So $HClO_4 > HClO_3 > HClO_2 > HClO$.

Question 39Organic Reaction Mechanism

The major product when 2-bromo-2-methylbutane is treated with hot alcoholic KOH is:

Solution: Alcoholic KOH promotes E2/E1 elimination. By Saytzeff's rule the more substituted (more stable) alkene predominates. Eliminating to form the tetrasubstituted double bond gives 2-methylbut-2-ene as the major product.

Question 40Carbonyl Chemistry

When a mixture of benzaldehyde and formaldehyde is treated with concentrated NaOH (crossed Cannizzaro), the major organic products are:

Solution: In crossed Cannizzaro, formaldehyde (more readily oxidised, having no $\alpha$-H) acts as the reducing agent and is oxidised to formate, while benzaldehyde is reduced to benzyl alcohol. Products: benzyl alcohol + sodium formate.

Question 41Aromatic Chemistry

Which of the following is the most reactive towards electrophilic aromatic substitution?

Solution: Anisole ($C_6H_5OCH_3$) has the strongly activating $-OCH_3$ group (+M effect, strong electron donation by resonance), making it the most reactive. Toluene is activating but weaker; chlorobenzene is deactivating; nitrobenzene is strongly deactivating.

Question 42Stereochemistry

The number of stereoisomers possible for 2,3-dichlorobutane ($CH_3CHClCHClCH_3$) is:

Solution: The molecule has two similar stereocentres. It gives a pair of enantiomers (d and l) plus one achiral meso form. Total = 3 stereoisomers (the meso compound has an internal plane of symmetry).

Question 43Organic Acidity

The correct order of acidic strength is:

Solution: $pK_a$ values: acetic acid (~4.76) < carbonic acid (~6.35) < phenol (~10) < ethanol (~16). Lower $pK_a$ means stronger acid. Order: acetic acid > carbonic acid > phenol > ethanol.

Question 44Carbocation Stability

The correct order of stability of the following carbocations is: (I) $CH_3^+$ (II) $CH_3CH_2^+$ (III) $(CH_3)_2CH^+$ (IV) $C_6H_5CH_2^+$ (benzyl)

Solution: Benzyl cation (IV) is most stable due to resonance delocalisation into the ring. Among alkyl cations stability increases with substitution: $3°$-like isopropyl (III, secondary) > ethyl (II) > methyl (I). Order: IV > III > II > I.

Question 45Solid State

An element crystallises in a face-centred cubic (FCC) lattice with edge length $a = 400$ pm. The radius of the atom is approximately:

Solution: For FCC, atoms touch along the face diagonal: $4r = \sqrt{2}\,a$, so $r = \dfrac{\sqrt{2}\,a}{4} = \dfrac{1.414\times 400}{4} = \dfrac{565.6}{4} \approx 141$ pm.

Question 46 · NumericalElectrochemistry

A current of 9.65 A is passed through molten $AlCl_3$ for 1000 seconds. Calculate the mass of aluminium (atomic mass 27) deposited at the cathode, in grams. Round to 2 decimal places. (Faraday constant $F = 96500\,\text{C mol}^{-1}$)

Enter a numerical value.

Solution: Charge $Q = It = 9.65 \times 1000 = 9650$ C. Moles of electrons $= 9650/96500 = 0.1$ mol. $Al^{3+} + 3e^- \to Al$, so moles of Al $= 0.1/3 = 0.0333$ mol. Mass $= 0.0333 \times 27 = 0.90$ g.

Question 47 · NumericalIonic Equilibrium

Calculate the pH of a 0.01 M solution of a weak monobasic acid HA whose dissociation constant $K_a = 1.0\times10^{-5}$. Give the answer to 1 decimal place.

Enter a numerical value.

Solution: $[H^+] = \sqrt{K_a \cdot C} = \sqrt{(1.0\times10^{-5})(0.01)} = \sqrt{1.0\times10^{-7}} = 3.16\times10^{-4}$ M. $pH = -\log(3.16\times10^{-4}) = 4 - \log 3.16 = 4 - 0.5 = 3.5$.

Question 48 · NumericalChemical Kinetics

The half-life of a first-order reaction is 138.6 seconds. Calculate the time (in seconds) required for the concentration of the reactant to fall to 1/8 of its initial value. Give a whole number.

Enter a numerical value.

Solution: Falling to 1/8 means three half-lives ($(1/2)^3 = 1/8$). Time $= 3 \times t_{1/2} = 3 \times 138.6 = 415.8$ s.

Question 49 · NumericalRedox / Oxidation State

Determine the oxidation state of chromium in the dichromate ion $Cr_2O_7^{2-}$. Give the answer as a signed integer (e.g. +6).

Enter a numerical value.

Solution: Let oxidation state of Cr be $x$. Oxygen is $-2$. $2x + 7(-2) = -2 \Rightarrow 2x - 14 = -2 \Rightarrow 2x = 12 \Rightarrow x = +6$.

Question 50 · NumericalMole Concept / Empirical Formula

An organic compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its molar mass is 180 g/mol. Calculate the number of carbon atoms in one molecule (i.e. subscript of C in the molecular formula).

Enter a numerical value.

Solution: Moles per 100 g: C $= 40/12 = 3.33$; H $= 6.7/1 = 6.7$; O $= 53.3/16 = 3.33$. Ratio C:H:O $= 1:2:1$, empirical formula $CH_2O$ (mass 30). $n = 180/30 = 6$, so molecular formula $C_6H_{12}O_6$. Carbon atoms $= 6$.

Question 51Combinatorial Identities

$\binom n{r+1}+\binom n{r-1}+2\binom nr$ equals

Solution: $(\binom n{r+1}+\binom nr)+(\binom nr+\binom n{r-1})=\binom{n+1}{r+1}+\binom{n+1}r=\binom{n+2}{r+1}$.

Question 52Trig Identities

$\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}$ can be written as

Solution: The expression simplifies to $1+\tan A+\cot A=1+\frac{\sin^2A+\cos^2A}{\sin A\cos A}=1+\sec A\operatorname{cosec}A$.

Question 53Standard Form

A value of $\theta$ for which $\dfrac{2+3i\sin\theta}{1-2i\sin\theta}$ is purely imaginary is

Solution: Rationalising, the real part is $\frac{2-6\sin^2\theta}{1+4\sin^2\theta}$; setting it to $0$ gives $\sin^2\theta=\frac13$, i.e. $\theta=\sin^{-1}\frac1{\sqrt3}$.

Question 54Location Of Roots

All values of $m$ for which both roots of $x^2-2mx+m^2-1=0$ are greater than $-2$ but less than $4$ lie in

Solution: Roots are $m\pm1$; $m-1>-2\Rightarrow m>-1$ and $m+1<4\Rightarrow m<3$, so $-1<m<3$.

Question 55Coefficients

For $r>1,n>2$, if the coefficients of the $(r+2)$th and $(3r)$th terms in $(1+x)^{2n}$ are equal, then $n$ equals

Solution: $\binom{2n}{r+1}=\binom{2n}{3r-1}\Rightarrow(r+1)+(3r-1)=2n\Rightarrow n=2r$.

Question 56Hyperbola

$(at^2,2bt)$ lies on $\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}=1$ for

Solution: $t^4-4t^2-1=0\Rightarrow t^2=2\pm\sqrt5$; the positive value is $2+\sqrt5$.

Question 57Integration

$\displaystyle\int f'(ax+b)\,[f(ax+b)]^n\,dx$ equals

Solution: With $u=f(ax+b)$, $du=a f'(ax+b)dx$, giving $\frac{1}{a(n+1)}[f(ax+b)]^{n+1}+C$.

Question 58Trigonometric Identities

$\tan\left(\tfrac\pi4\sin^2x\right)$, $-\infty<x<\infty$, lies between

Solution: $\sin^2x\in[0,1]$ so the argument $\in[0,\tfrac\pi4]$ and $\tan(\cdot)\in[0,1]$.

Question 59Coordinate Geometry Advanced

$AB,AC$ are tangents to $y^2=4ax$. If $l_1,l_2,l_3$ are the perpendicular distances from $A,B,C$ to any tangent to the parabola, then $l_1,l_2,l_3$ are in

Solution: For the parabola these perpendiculars satisfy $l_1^2=l_2l_3$, so they are in G.P.

Question 60Parabola

$ax^2+2hxy+by^2+2gx+2fy+c=0$ (non-degenerate) represents a parabola if

Solution: A conic is a parabola when $h^2=ab$.

Question 61Circles

$C_1$ has centre $O$ radius $r$; $C_2$ has centre $(3r,0)$ radius $2r$. The number of common tangents is

Solution: Distance between centres $=3r=r+2r$, so the circles touch externally — exactly $3$ common tangents.

Question 62Ellipse

$P$ is variable on $\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1$ with major axis $AA'$. The maximum area of $\triangle APA'$ is

Solution: Base $AA'=2a$, maximum height $=b$ (at the minor-axis end), so max area $=\tfrac12(2a)(b)=ab$.

Question 63Probability

6 boys and 6 girls sit in a row randomly. The probability that all 6 girls sit together is

Solution: $\frac{7!\cdot6!}{12!}=\frac{6!}{12\cdot11\cdot10\cdot9\cdot8}=\frac{720}{95040}=\frac{1}{132}$.

Question 64Straight Lines

A line $L$ has intercepts $a,b$ on the axes. After rotating the axes (origin fixed), $L$ has intercepts $p,q$. Then

Solution: The perpendicular distance from the origin to $L$ is unchanged: $\tfrac1{d^2}=\tfrac1{a^2}+\tfrac1{b^2}=\tfrac1{p^2}+\tfrac1{q^2}$.

Question 65Permutations And Combinations

A number greater than $1000$ but less than $4000$ is formed using the digits $0,1,2,3,4$ (repetition allowed). The number of such numbers is

Solution: Four-digit numbers with first digit $\in\{1,2,3\}$ (3 ways) and the other three digits free ($5^3=125$): $3\times125=375$.

Question 66Sequence And Series

An A.P. has $2n$ terms. The sum of terms in even positions is $P$ and in odd positions is $Q$. The $n$th term of the A.P. is

Solution: The $n$ odd-position terms $a_1,a_3,\dots,a_{2n-1}$ have average $a_n$ (mean of first and last), so $Q=n\,a_n\Rightarrow a_n=Q/n$.

Question 67Matrices And Determinants

For a square matrix $A$, which is NOT symmetric?

Solution: $A-A'$ is skew-symmetric, not symmetric.

Question 68Logarithm

For positive reals $x,y,z$, the minimum of $\tfrac{y+z}{x}+\tfrac{z+x}{y}+\tfrac{x+y}{z}$ is

Solution: Group into three pairs $\left(\tfrac xy+\tfrac yx\right)+\cdots$, each $\ge2$ by AM–GM, so the minimum is $6$.

Question 69Trigonometric Equations

If $\cos2\theta=\sqrt2(\cos\theta-\sin\theta)$, then $\tan\theta$ is

Solution: $\cos2\theta=(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)=\sqrt2(\cos\theta-\sin\theta)$ gives $\cos\theta+\sin\theta=\sqrt2$, i.e. $\theta=\tfrac\pi4$, so $\tan\theta=1$.

Question 70Inverse Trigonometric Functions

If $\sum_{i=1}^{2n}\sin^{-1}x_i=n\pi$, then $\sum_{i=1}^{2n}x_i$ equals

Solution: Each $\sin^{-1}x_i\le\frac\pi2$, so the sum of $2n$ terms is at most $n\pi$; equality forces every $x_i=1$, hence $\sum x_i=2n$.

Question 71 · NumericalNth Roots Unity

For integer $k$, $\alpha_k=\cos\frac{k\pi}7+i\sin\frac{k\pi}7$. The value of $\dfrac{\sum_{k=1}^{12}|\alpha_{k+1}-\alpha_k|}{\sum_{k=1}^{3}|\alpha_{4k-1}-\alpha_{4k-2}|}$ is

Enter a numerical value.

Solution: Each $|\alpha_{k+1}-\alpha_k|=|e^{i\pi/7}-1|=2\sin\frac\pi{14}$ — a constant, since consecutive indices differ by $1$. Numerator $=12c$, denominator $=3c$, ratio $=4$.

Question 72 · NumericalSystem Of Equations

$(x,y,z)$ are integer points with $3x-y-z=0$ and $-3x+2y+z=0$. The number of such points with $x^2+y^2+z^2\le100$ is

Enter a numerical value.

Solution: Adding the equations gives $y=0$, then $z=3x$. So points are $(x,0,3x)$ with $10x^2\le100\Rightarrow|x|\le3$ — seven integer values.

Question 73 · NumericalDistribution And Partition

$n_1<n_2<n_3<n_4<n_5$ are positive integers with sum $20$. The number of such $(n_1,\dots,n_5)$ is

Enter a numerical value.

Solution: Partitions of $20$ into $5$ distinct positive parts $=$ partitions of $20-15=5$, which number $7$.

Question 74 · NumericalCoefficients

The sum of the coefficients of $(1+x-3x^2)^{2163}$ is

Enter a numerical value.

Solution: Put $x=1$: $(1+1-3)^{2163}=(-1)^{2163}=-1$.

Question 75 · NumericalModulus Triangle Inequality

If $|z-3-2i|\le2$, then the minimum value of $|2z-6+5i|$ is

Enter a numerical value.

Solution: $|2z-6+5i|=2\left|z-\left(3-\tfrac52 i\right)\right|$. The centre $3+2i$ is at distance $4.5$ from $3-\tfrac52 i$; minus radius $2$ gives $2.5$, doubled $=5$.