TU IOST | Entrance paper
Answer (b): 'I am sick' has no fixed truth value because it depends on who the speaker is and when. The other three have definite truth values (even the false sun statement is a logical statement).
Revise: Algebra (Class 11 notes)
Answer (d): $|x-2|=5$ gives $x-2=5$ or $x-2=-5$, so $x=7$ or $x=-3$.
Revise: Algebra (Class 11 notes)
Answer (c): $A-B=A\cap \bar B$, and $\bar B-\bar A=\bar B\cap A$, which is the same set.
Revise: Algebra (Class 11 notes)
Answer (a): Put $y=2x-3$ and solve for $x$: $x=\frac{y+3}{2}$, so $f^{-1}(x)=\frac{x+3}{2}$.
Revise: Algebra (Class 11 notes)
Answer (c): Period of $\sin(bx+c)$ is $\frac{2\pi}{|b|}=\frac{2\pi}{2}=\pi$.
Revise: Trigonometry (Class 11 notes)
Answer (a): $l=ar^3$, $m=ar^6$, $n=ar^9$, so $ln=a^2r^{12}=m^2$.
Revise: Algebra (Class 11 notes)
Answer (c): The $n$th term is $(2n-1)(2n+1)=4n^2-1$.
Revise: Algebra (Class 11 notes)
Answer (d): Let $\theta=\cos^{-1}x$, so $\cos\theta=x$ and $\sin\theta=\sqrt{1-x^2}$.
Revise: Trigonometry (Class 11 notes)
Answer (c): $\csc^2x=2$ gives $\sin^2x=\frac12$, so $\sin x=\pm\frac{1}{\sqrt2}$ and $x=n\pi\pm\frac{\pi}{4}$.
Revise: Trigonometry (Class 11 notes)
Answer (d): Expanding along row 1: $k(0-2k)-1(-2-0)=-2k^2+2=0$, so $k^2=1$ and $k=\pm1$.
Revise: Algebra (Class 11 notes)
Answer (c): No inverse means $|A|=0$: $6x-3(x-2)=3x+6=0$, so $x=-2$.
Revise: Algebra (Class 11 notes)
Answer (b): A homogeneous system whose only solution is the zero solution has the trivial solution.
Revise: Algebra (Class 11 notes)
Answer (a): $(1+i)^2=2i$, so $(1+i)^6=(2i)^3=-8i$; similarly $(1-i)^6=(-2i)^3=8i$. The sum is $0$.
Revise: Algebra (Class 11 notes)
Answer (c): The square is $\frac{-4}{(1+i)^2}=\frac{-4}{2i}=\frac{-2}{i}=2i$.
Revise: Algebra (Class 11 notes)
Answer (a): Product of roots $=\frac{-k}{5}=1$, so $k=-5$.
Revise: Algebra (Class 11 notes)
Answer (b): Here $a=1,b=1,h=-1$, so $h^2=ab$, which is the condition for a parabola.
Revise: Analytic geometry (Class 11 notes)
Answer (c): Area $=\frac12|1(1+1)+(-1)(-1+1)+(-1)(-1-1)|=\frac12(2+0+2)=2$.
Revise: Analytic geometry (Class 11 notes)
Answer (a): Sum is $\frac{n(n+1)}{2}$, so the limit is $\lim\frac{n+1}{2n}=\frac12$.
Revise: Calculus (Class 12 notes)
Answer (a): Divide by $x^2$: $5+4\frac{\tan x}{x}\to5+4=9$.
Revise: Calculus (Class 12 notes)
Answer (d): Squaring, $y^2=x+y$. Differentiate: $2yy'=1+y'$, so $y'=\frac{1}{2y-1}$.
Revise: Calculus (Class 12 notes)
Answer (b): By the chain rule, $y'=e^{\sqrt{ax^2+b}}\cdot\frac{2ax}{2\sqrt{ax^2+b}}=\frac{ax}{\sqrt{ax^2+b}}e^{\sqrt{ax^2+b}}$.
Revise: Calculus (Class 12 notes)
Answer (d): Standard result: $\int\frac{dx}{1-x^2}=\frac12\ln\left|\frac{1+x}{1-x}\right|+c$.
Revise: Calculus (Class 12 notes)
Answer (a): Put $u=\ln t$: $\int_0^{\ln2}\sin u,du=1-\cos(\ln2)$.
Revise: Calculus (Class 12 notes)
Answer (c): $f'(x)=3x^2-3=0$ gives $x=\pm1$.
Revise: Calculus (Class 12 notes)
Answer (a): Area $=\int_2^4x^3dx=\frac{4^4-2^4}{4}=\frac{256-16}{4}=60$.
Revise: Calculus (Class 12 notes)
Answer (b): "Answer for" means to be responsible for something.
Revise: Grammar and Usage (Class 11 notes)
Answer (c): 'Ran through' means used up or squandered, so he inherited a fortune and ran through it in a year.
Revise: Grammar and Usage (Class 11 notes)
Answer (d): 'Stand up for' means to defend or support someone, which matches the mother saying he acted sensibly.
Revise: Grammar and Usage (Class 11 notes)
Answer (a): The phrase is 'take someone for' meaning to regard as, as in 'Do you take me for a fool?'.
Revise: Grammar and Usage (Class 11 notes)
Answer (a): "Turned in" means went to bed, which fits the captain unaware of danger.
Revise: Grammar and Usage (Class 11 notes)
Answer (d): "As fit as a fiddle" is the idiom for being in very good health.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (c): Squeamish means easily nauseated or upset.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (b): A bramble is a prickly, thorny wild shrub such as the blackberry bush.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (a): Congruous means in agreement or harmony, i.e. consistent.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (b): A savior rescues; the opposite is a destroyer.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (c): Scabrous means rough or indecent; its opposite among the options is decent.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (c): Unavailing means ineffective or unsuccessful, so the opposite is successful.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (d): We say 'died of cancer' for the disease and 'died in a train accident' for the accident.
Revise: Grammar and Usage (Class 11 notes)
Answer (d): We divide a number 'by' another and share things 'between' two people.
Revise: Grammar and Usage (Class 11 notes)
Answer (a): The infinitive "to live" takes the preposition "in" at the end.
Revise: Grammar and Usage (Class 11 notes)
Answer (c): The correct collocations are "angry with" and "indifferent to".
Revise: Grammar and Usage (Class 11 notes)
Answer (a): "Relieve someone from" a worry is the usual form.
Revise: Grammar and Usage (Class 11 notes)
Answer (c): "Lest" introduces a feared result and fits the warning.
Revise: Grammar and Usage (Class 11 notes)
Answer (c): "On the bottle" means drinking heavily.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (b): "On the rocks" means in difficulty or close to failure.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (a): It means to stop what you are doing for the night.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (c): 'Over the hill' can mean past the hardest or worst part, which fits option c best among the choices.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (d): plea-sur-a-ble has four syllables.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (d): Hematophobia is the fear of blood.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (b): $a/V^2$ has the dimensions of pressure, so $[a]=[p][V^2]=ML^{-1}T^{-2}\cdot L^6=ML^5T^{-2}$.
Revise: Physical Quantities (Class 11 notes)
Answer (c): Trailing zeros without a decimal point are not significant, so only 1 and 4 count: 2.
Revise: Physical Quantities (Class 11 notes)
Answer (d): The two legs are perpendicular, so displacement is $\sqrt{8^2+6^2}=10$ m.
Revise: Vectors (Class 11 notes)
Answer (b): Range $R=\frac{u^2\sin2\theta}{g}\propto u^2$, so doubling $u$ makes it 4 times.
Revise: Kinematics (Class 11 notes)
Answer (d): Limiting static friction is greater than kinetic and rolling friction.
Revise: Dynamics (Class 11 notes)
Answer (a): Effective $g'=g-\omega^2R\cos^2\lambda$ falls as $\omega$ rises, so weight decreases.
Revise: Gravitation (Class 11 notes)
Answer (d): The balloon displaces its own weight of water, so buoyancy cancels weight and apparent weight is zero.
Revise: Fluid Statics (Class 12 notes)
Answer (c): A change of 1 degree C equals a change of 1 K, so an increase of 270 C is an increase of 270 K.
Revise: Heat and Temperature (Class 11 notes)
Answer (a): $v_{rms}=\sqrt{3RT/M}$ is largest for the smallest molar mass; $H_2$ has $M=2$.
Revise: Ideal Gas (Class 11 notes)
Answer (c): Black body radiation covers all wavelengths, a continuous spectrum.
Revise: Quantization of Energy (Class 12 notes)
Answer (c): A convex mirror forms a diminished virtual image and gives the widest field of view.
Revise: Reflection at Curved Mirrors (Class 11 notes)
Answer (a): Short wavelengths scatter more (Rayleigh scattering), so blue light reaches us from all directions.
Revise: Dispersion (Class 11 notes)
Answer (d): Needed lens is concave with $f=-0.5$ m, so $P=\frac{1}{f}=-2$ D.
Revise: Lenses (Class 11 notes)
Answer (b): Positive charging means electrons are removed, so the mass decreases slightly.
Revise: Electric Charges (Class 11 notes)
Answer (a): $\frac1C=\frac13+\frac13+\frac13=1$, so $C=1,\mu$F.
Revise: Capacitor (Class 11 notes)
Answer (b): The loop rule states that the net energy gained per unit charge around a loop is zero.
Revise: DC Circuit (Class 11 notes)
Answer (c): The magnetic force is perpendicular to velocity and does no work, so kinetic energy is constant.
Revise: Magnetic Field (Class 12 notes)
Answer (a): A straight wire has no loop and links no flux with itself, so its self-inductance is taken as zero.
Revise: Electromagnetic Induction (Class 12 notes)
Answer (a): Einstein's equation $h\nu=\phi+K_{max}$ expresses energy conservation.
Revise: Photons (Class 12 notes)
Answer (c): In beta decay a neutron becomes a proton, so the atomic number rises by 1 while the mass number is unchanged.
Revise: Radioactivity and Nuclear Reaction (Class 12 notes)
Answer (d): At the breakdown voltage the reverse current rises sharply.
Revise: Semiconductor Devices (Class 12 notes)
Answer (b): Wave number $k=62.4$, so $\lambda=\frac{2\pi}{k}=\frac{6.283}{62.4}\approx0.1$ unit.
Revise: Wave Motion (Class 12 notes)
Answer (d): Reflection at a rigid (denser) boundary inverts the wave, a phase change of $\pi$.
Revise: Mechanical Waves (Class 12 notes)
Answer (a): A source moving toward the observer raises the apparent frequency (Doppler effect), so the pitch is higher.
Revise: Acoustic Phenomena (Class 12 notes)
Answer (a): Lyman series ends on $n=1$ and gives wavelengths of about 91 to 122 nm, in the ultraviolet.
Revise: Quantization of Energy (Class 12 notes)
Answer (b): $3(+1)+x+6(-1)=0$ gives $x=+3$.
Revise: Oxidation and Reduction (Class 11 notes)
Answer (b): 8 g of O is $\frac{8}{16}=0.5$ mol of atoms, so $0.5\times6.023\times10^{23}=3.0115\times10^{23}$.
Revise: Stoichiometry (Class 11 notes)
Answer (c): Filling orbitals in order of increasing energy is the Aufbau principle.
Revise: Atomic Structure (Class 11 notes)
Answer (c): Lyman lines come from transitions to $n=1$ and lie in the ultraviolet.
Revise: Atomic Structure (Class 11 notes)
Answer (d): NaCl has Cl$^-$ ions in a fcc lattice with Na$^+$ in the octahedral holes.
Revise: States of Matter (Class 11 notes)
Answer (d): Chlorine has the highest electron affinity; fluorine's is lower because of small size and electron repulsion.
Revise: Classification of Elements (Class 11 notes)
Answer (a): $\ce{N2 + 3H2 <=> 2NH3}$ is the Haber process.
Revise: Chemistry of Non-Metals (Class 11 notes)
Answer (a): $\ce{SO2}$ reacts with water to give an acid and is an acidic oxide; $\ce{N2O}$ and CO are neutral and ZnO is amphoteric.
Revise: Chemistry of Non-Metals (Class 11 notes)
Answer (b): Be has two bond pairs and no lone pair, giving a linear molecule.
Revise: Chemical Bonding (Class 11 notes)
Answer (d): Potassium permanganate is not a primary standard because it is not stable and is hard to obtain pure. Oxalic acid, potassium dichromate and potassium bromate are primary standards (option c is a misspelling of bromate).
Revise: Volumetric Analysis (Class 12 notes)
Answer (d): $\ce{NH3}$ has a lone pair to donate; the others are electron-deficient Lewis acids.
Revise: Ionic Equilibrium (Class 12 notes)
Answer (a): Equivalent conductivity is $\Lambda_{eq}=\kappa\times1000/N$, with units $S,cm^2,equivalent^{-1}$. The mole unit applies to molar conductivity.
Revise: Electrochemistry (Class 12 notes)
Answer (a): By definition, 1 cal $=4.184$ J.
Revise: Chemical Thermodynamic (Class 12 notes)
Answer (b): The Gibbs-Helmholtz relation is $\Delta G=\Delta H-T\Delta S$.
Revise: Chemical Thermodynamic (Class 12 notes)
Answer (b): $k$ has unit $(\text{mol L}^{-1})^{1-n},\text{s}^{-1}$; for $n=3$ this is $\text{mol}^{-2}\text{L}^2\text{s}^{-1}$.
Revise: Chemical Kinetics (Class 12 notes)
Answer (a): Alkylation of benzene with an alkyl halide and $\ce{AlCl3}$ is the Friedel-Crafts reaction.
Revise: Aromatic Hydrocarbons (Class 11 notes)
Answer (a): Victor Meyer's test gives different colours (red, blue, colourless) for 1°, 2° and 3° alcohols.
Revise: Alcohols (Class 12 notes)
Answer (d): $\ce{R-ONa + R'-X -> R-O-R'}$ gives ethers.
Revise: Ethers (Class 12 notes)
Answer (d): Rosenmund reduction of acid chlorides to aldehydes uses Pd on $\ce{BaSO4}$ (poisoned with quinoline).
Revise: Aldehydes and Ketones (Class 12 notes)
Answer (a): Conc. $\ce{HNO3}$ with $\ce{H2SO4}$ nitrates phenol fully to 2,4,6-trinitrophenol (picric acid).
Revise: Phenols (Class 12 notes)
Answer (a): This is Hofmann bromamide degradation: $\ce{C6H5CONH2 -> C6H5NH2}$.
Revise: Amines (Class 12 notes)
Answer (c): Amino acids carry both an amino group $-NH_2$ and a carboxyl group $-COOH$.
Revise: Amines (Class 12 notes)
Answer (b): Martius yellow (a nitro naphthol dye) is classed as an acid dye.
Answer (b): Green vitriol is hydrated ferrous sulphate, $\ce{FeSO4.7H2O}$.
Revise: Transition Metals (Class 12 notes)
Answer (b): Zinc blende (ZnS, sphalerite) is the chief ore of zinc.
Revise: Chemistry of Metals (Class 11 notes)