TU IOST | Entrance paper
Answer (b): Reciprocal means mutual, shared by both sides. Its opposite is unilateral (one-sided).
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (a): Nimble means quick and light in movement. The opposite is lazy.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (b): To swelter is to suffer from oppressive heat, which matches stifling.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (d): A parody is a humorous imitation that mocks something, so lampoon is the closest.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (c): To swill means to wash or rinse something with plenty of water.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (a): An antiquarian studies or collects antiques and ancient objects.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (b): Sophism is a clever but misleading argument meant to deceive.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (c): "Go about" one's business means to carry on with one's daily activities.
Revise: Grammar and Usage (Class 11 notes)
Answer (c): "Set to" means to begin a task energetically.
Revise: Grammar and Usage (Class 11 notes)
Answer (d): A rough patch is a difficult period in life, closest to time of trouble.
Revise: Grammar and Usage (Class 11 notes)
Answer (a): Wildfire spreads rapidly, so the idiom means to spread very quickly and widely.
Revise: Grammar and Usage (Class 11 notes)
Answer (d): The idiom is a good ear for music, so the first blank is 'for'. 'Stick with grammar' completes it, which is option d.
Revise: Grammar and Usage (Class 11 notes)
Answer (c): A good ear for music, and sick of grammar exercises.
Revise: Grammar and Usage (Class 11 notes)
Answer (d): Interested in a thing, good at a skill.
Revise: Grammar and Usage (Class 11 notes)
Answer (a): Xylophone is pronounced /ˈzaɪləfəʊn/, so the primary stress falls on the first syllable.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (c): Angle is a ratio of arc to radius, so it has no dimension.
Revise: Physical Quantities (Class 11 notes)
Answer (d): Multiplying a vector by a positive scalar 2 keeps its direction and doubles its magnitude. Its unit does not change.
Revise: Vectors (Class 11 notes)
Answer (b): $R/H = 4\cot\theta = 4\sqrt{3}$, so $\cot\theta=\sqrt3$ and $\theta=30^\circ$.
Revise: Kinematics (Class 11 notes)
Answer (b): The slope of kinetic energy against position is $dK/dx = F$, the net force. Force is the rate of change of momentum, so momentum is the only option that fits; this is the least bad of four loose options.
Revise: Work, Energy and Power (Class 11 notes)
Answer (c): Work $=T\times 2\times4\pi r^2$, so it scales as $r^2$. At $3R$ the total is $9W$, so the extra work is $9W-W=8W$.
Revise: Elasticity (Class 11 notes)
Answer (a): A floating body displaces water equal to its weight, so the total weight the boy carries is unchanged.
Revise: Elasticity (Class 11 notes)
Answer (d): A bimetallic strip bends on heating because the two metals expand unequally, closing the alarm circuit.
Revise: Thermal Expansion (Class 11 notes)
Answer (b): $\sin C = 1/\mu$. Violet has the largest $\mu$, so the smallest critical angle.
Revise: Refraction at Plane Surface (Class 11 notes)
Answer (c): Fall time $t_1=\sqrt{2h/g}=\sqrt{2(78.4)/9.8}=4$ s. Sound time $=78.4/350\approx0.22$ s. Total about 4.2 s.
Revise: Kinematics (Class 11 notes)
Answer (c): At the critical temperature liquid and vapour become indistinguishable, so latent heat vanishes.
Revise: Quantity of Heat (Class 11 notes)
Answer (a): For zero net force on the third charge at distance $x$ from $q$: $q/x^2=4q/(r-x)^2$, so $r-x=2x$ and $x=r/3$ from $q$, i.e. $2r/3$ from $4q$.
Revise: Electric Charges (Class 11 notes)
Answer (b): Powers add in contact: $P = +3 - 7 = -4$ D, so $f = 1/P = -0.25$ m $= -25$ cm. The printed powers are garbled but the values +3 D and -7 D are clear.
Revise: Lenses (Class 11 notes)
Answer (d): Induced emf $=-d\Phi/dt$, which exists only while the flux is changing.
Revise: Electromagnetic Induction (Class 12 notes)
Answer (b): $Q=X_L/R=100/5=20$.
Revise: Alternating Currents (Class 12 notes)
Answer (c): On the axial line both $E=2kp/r^3$ and $V=kp/r^2$ are non-zero.
Revise: Electric Field (Class 11 notes)
Answer (b): Resonance gives $L = \frac{1}{4\pi^2 f^2 C}$. With $f = 1.02\times10^6$ Hz and $C = 420\times10^{-12}$ F, $4\pi^2 f^2 C \approx 1.725\times10^4$, so $L \approx 5.8\times10^{-5}$ H. The printed 420 H is a misprint for 420 pF.
Revise: Alternating Currents (Class 12 notes)
Answer (d): Thermoelectric power is $dE/dT$, which is zero at the neutral temperature where emf is maximum.
Revise: Thermoelectric Effects (Class 12 notes)
Answer (a): The photoelectric effect shows light energy comes in quanta (photons).
Revise: Photons (Class 12 notes)
Answer (a): $K_{max}=hf-\phi$, so it depends on the frequency of light, not intensity.
Revise: Photons (Class 12 notes)
Answer (d): Energy per fission is $200\times1.6\times10^{-13} = 3.2\times10^{-11}$ J. Fissions per second $= 10^4/3.2\times10^{-11} \approx 3.1\times10^{14}$, closest to $5\times10^{14}$ among the options.
Revise: Radioactivity and Nuclear Reaction (Class 12 notes)
Answer (b): Mean life $=T_{1/2}/0.693=10/0.693=14.43$ years.
Revise: Radioactivity and Nuclear Reaction (Class 12 notes)
Answer (d): The majority carriers in NPN are electrons, which are more mobile than holes, giving faster operation.
Revise: Semiconductor Devices (Class 12 notes)
Answer (a): The emf is $B_v l v$. The vertical component of the Earth's field is greatest at the poles.
Revise: Electromagnetic Induction (Class 12 notes)
Answer (d): $E=Pt=200\times10^6\times86400=1.728\times10^{13}$ J.
Revise: Work, Energy and Power (Class 11 notes)
Answer (c): A donor impurity supplies an extra electron, so it must have five valence electrons, i.e. pentavalent (such as P or As).
Revise: Semiconductor Devices (Class 12 notes)
Answer (a): All six carbons are $sp^2$ hybridised and lie in one plane.
Revise: Aromatic Hydrocarbons (Class 11 notes)
Answer (d): Ethanol and its isomer dimethyl ether share the formula $C_2H_6O$ but have different functional groups (-OH versus -O-), so they are functional isomers. The printed 'diethyl ether' is a misprint for dimethyl ether.
Revise: Fundamental Principles (Class 11 notes)
Answer (a): Myoglobin is the oxygen-storing protein of muscle; transport of oxygen is the job of haemoglobin.
Revise: Bio-Inorganic Chemistry (Class 11 notes)
Answer (b): Bronze is copper with tin.
Revise: Chemistry of Metals (Class 11 notes)
Answer (c): Iron is rendered passive by conc. nitric acid through a protective oxide layer, commonly given as $Fe_3O_4$.
Revise: Chemistry of Metals (Class 11 notes)
Answer (a): PbO reacts with both acids and bases, so it is amphoteric.
Revise: Chemistry of Non-Metals (Class 11 notes)
Answer (c): $+1+x+3(-2)=0$, so $x=+5$.
Revise: Oxidation and Reduction (Class 11 notes)
Answer (b): Five bond pairs and no lone pairs give a trigonal bipyramidal shape.
Revise: Chemical Bonding (Class 11 notes)
Answer (a): Noble gases form group 18.
Revise: Classification of Elements (Class 11 notes)
Answer (b): fixes the shell and its size and energy.
Revise: Atomic Structure (Class 11 notes)
Answer (a): 32 g of $O_2$ is 1 mole, which contains $6.023\times10^{23}$ molecules.
Revise: Stoichiometry (Class 11 notes)
Answer (d): Camphor sublimes on heating while sand does not.
Revise: Fundamentals of Chemistry (Class 11 notes)
Answer (b): $AgNO_3$ and $NaCl$ form insoluble $AgCl$, so it is a precipitation titration.
Revise: Volumetric Analysis (Class 12 notes)
Answer (c): The law applies to weak electrolytes, and only $NH_4OH$ is weak here.
Revise: Ionic Equilibrium (Class 12 notes)
Answer (d): Rate $=k[A]$, so $k=$ rate$/$conc $=$ time$^{-1}$.
Revise: Chemical Kinetics (Class 12 notes)
Answer (a): A calorimeter measures heat changes.
Revise: Chemical Thermodynamic (Class 12 notes)
Answer (c): The Nernst equation gives cell emf at non-standard concentrations.
Revise: Electrochemistry (Class 12 notes)
Answer (d): Boiling points of carboxylic acids rise with molar mass, so butanoic acid is highest.
Revise: Carboxylic Acids (Class 12 notes)
Answer (a): Tollens' reagent is ammoniacal silver nitrate.
Revise: Aldehydes and Ketones (Class 12 notes)
Answer (c): Nitrating phenol with conc. $HNO_3$/$H_2SO_4$ gives 2,4,6-trinitrophenol (picric acid).
Revise: Phenols (Class 12 notes)
Answer (a): Zinc and water reduce $CHCl_3$ fully to methane.
Revise: Haloalkanes (Class 12 notes)
Answer (c): Cinnabar is $HgS$, the ore of mercury.
Revise: Chemistry of Metals (Class 11 notes)
Answer (b): Copper with dilute nitric acid is the standard answer for the lower oxide of nitrogen listed in textbooks, while concentrated acid gives $NO_2$. Among the options dilute $HNO_3$ is the only defensible choice.
Revise: Chemistry of Non-Metals (Class 11 notes)
Answer (c): $Fe^{3+}$ is $3d^5$, so all five $d$ orbitals hold one electron each.
Revise: Atomic Structure (Class 11 notes)
Answer (a): Gypsum slows setting of cement, so it is a retarder.
Revise: Cement (Class 12 notes)
Answer (b): $ax+by+c=0$ is a straight line.
Revise: Analytic geometry (Class 11 notes)
Answer (a): $-7\le y-3\le7$ gives $-4\le y\le10$.
Revise: Algebra (Class 11 notes)
Answer (c): $A-B$ is the elements of $A$ not in $B$, i.e. $A\cap B'$.
Revise: Algebra (Class 11 notes)
Answer (d): $f(f(x))=1-(1-x)=x$.
Revise: Algebra (Class 11 notes)
Answer (b): $f(-x)=-x-2x^3=-f(x)$, so it is odd.
Revise: Algebra (Class 11 notes)
Answer (c): Geometric series with $a=1$, $r=-1/2$: $S=1/(1+1/2)=2/3$.
Revise: Algebra (Class 11 notes)
Answer (c): A symmetric matrix needs $a_{12}=a_{21}$, so $x+2 = 2x+2$, giving $x=0$.
Revise: Algebra (Class 11 notes)
Answer (c): Expanding along row 1: $1(w^3-1) - w(w^2-w^2) + w^2(w - w^4)$. Using $w^3=1$ each bracket is 0, so the determinant is 0.
Revise: Algebra (Class 11 notes)
Answer (d): $(1+i)^2 = 2i$ and $2i^2=-2$, so the expression is $-2/2i = -1/i = i$. Its imaginary part is 1.
Revise: Algebra (Class 11 notes)
Answer (c): Product of roots $=k/p=1$, so $k=p$.
Revise: Algebra (Class 11 notes)
Answer (a): $\cos^2x=1/4=\cos^2(\pi/3)$, so $x=n\pi\pm\pi/3$.
Revise: Trigonometry (Class 11 notes)
Answer (a): Put $x=\cos\theta$: $\cos2\theta=2x^2-1$.
Revise: Trigonometry (Class 11 notes)
Answer (b): For $(a,0)$: $|3a-8|/5=5$, so $3a-8=25$ and $a=11$.
Revise: Analytic geometry (Class 11 notes)
Answer (a): Lines are perpendicular when coefficients of $x^2$ and $y^2$ sum to zero: $k-1=0$.
Revise: Analytic geometry (Class 11 notes)
Answer (c): Distance from the yz plane is $|x|=3$.
Revise: Analytic geometry (Class 11 notes)
Answer (d): Vectors $(1,1,3)$ and $(3,3,n)$ are parallel with ratio 3, so $n=9$.
Revise: Vectors (Class 11 notes)
Answer (c): In a symmetric distribution mean, median and mode coincide. (Option d is also true, but c is the full condition.)
Revise: Statistics and Probability (Class 11 notes)
Answer (b): One outcome (TT) out of four equally likely outcomes.
Revise: Statistics and Probability (Class 11 notes)
Answer (a): $\frac{\sin2x}{x}\to2$ and $4x/x=4$, giving 6.
Revise: Calculus (Class 11 notes)
Answer (b): $y=e^{\log \sin x^2}=\sin x^2$, so $y' = 2x\cos x^2 = 2x,\frac{\cos x^2}{\sin x^2}\sin x^2 = 2xy\cot x^2$.
Revise: Calculus (Class 11 notes)
Answer (a): $f'=3x^2-12=0$ gives $x=\pm2$. $f''=6x<0$ at $x=-2$, so it is a maximum.
Revise: Calculus (Class 11 notes)
Answer (c): Put $u=\log x$: $\int_0^{\log2}\cos u,du=\sin(\log2)$.
Revise: Calculus (Class 11 notes)
Answer (b): By the intermediate value theorem, a continuous $f$ with $f(a)f(b)<0$ changes sign on the interval and so has a root there.
Revise: Computational methods (Class 11 notes)
Answer (d): $R^2=P^2+Q^2+2PQ\cos\theta$ is largest at $\theta=0$.
Revise: Mechanics (Class 11 notes)
Answer (c): $s=ut-\tfrac12gt^2=400-500=-100$ m, so the tower is 100 m high.
Revise: Mechanics (Class 11 notes)
Answer (b): Deliberately scrambling program code so it is hard to read is called obfuscation. Encryption scrambles data for secrecy, not code readability.
Revise: Information Security and Cyber Law (Class 11 notes)
Answer (b): Melissa spread through infected e-mail attachments.
Revise: Information Security and Cyber Law (Class 11 notes)
Answer (b): Topology is the physical or logical layout of network nodes.
Revise: Concept of Network and Data Communication (Class 12 notes)
Answer (c): Windows is the personal computer OS built around a graphical desktop.
Revise: Computer Software and Operating System (Class 11 notes)
Answer (b): DMA is Direct Memory Access.
Revise: Computer System (Class 11 notes)
Answer (c): Spyware (here Spybot) secretly collects user data and sends it out.
Revise: Information Security and Cyber Law (Class 11 notes)
Answer (d): Excel saves its files as workbooks.
Revise: Application Package (Class 11 notes)
Answer (a): Copied items are held in the clipboard.
Revise: Computer System (Class 11 notes)
Answer (b): A simulator reproduces a real environment for training.
Revise: Computer System (Class 11 notes)