TU IOST | Entrance paper
Answer (d): The right side has dimensions $L^{2}\cdot L=L^{3}$, and $\pi/3$ is a pure number, so $x$ has the dimension of volume.
Revise: Physical Quantities (Class 11 notes)
Answer (b): A negative scalar flips the direction, and the magnitude is multiplied by $|-2|=2$. The unit stays the same.
Revise: Vectors (Class 11 notes)
Answer (c): Fall time: $45=\frac12(10)t^{2}$ gives $t=3$ s. Distance in the first 2 s is $\frac12(10)(4)=20$ m, so the last second covers $45-20=25$ m.
Revise: Kinematics (Class 11 notes)
Answer (c): Internal forces of the break-up do not change total momentum, but the explosion energy is converted into extra kinetic energy of the pieces, so kinetic energy increases.
Revise: Dynamics (Class 11 notes)
Answer (c): A jet expels gas backward at high speed, and the plane gains an equal forward momentum, which is conservation of linear momentum.
Revise: Dynamics (Class 11 notes)
Answer (d): Honey is much more viscous than water, blood or glycerin (honey is of the order of several Pa s, glycerin about 1.4 Pa s).
Revise: Fluid Statics (Class 12 notes)
Answer (c): $KE=\frac12 mV^{2}$, so KE versus V is a parabola through the origin, curving upward.
Revise: Work, Energy and Power (Class 11 notes)
Answer (d): With both springs, $k_{eff}=2k$ and $f=\frac{1}{2\pi}\sqrt{2k/m}$. With one spring, $k_{eff}=k$, so the frequency becomes $f/\sqrt{2}$.
Revise: Periodic Motion (Class 12 notes)
Answer (a): Thermal expansion acts on the hole as if it were made of the same metal, so the hole expands and its diameter increases.
Revise: Thermal Expansion (Class 11 notes)
Answer (b): $\lambda=\frac{kT}{\sqrt2,\pi d^{2}p}$, so $\lambda\propto p^{-1}$.
Revise: Ideal Gas (Class 11 notes)
Answer (c): For minima $a\sin\theta=m\lambda$: $\sin\theta=\frac{2(550\times10^{-9})}{22\times10^{-7}}=0.5$, so $\theta=30^\circ$.
Revise: Diffraction of Light (Class 12 notes)
Answer (b): By Wien's law $\lambda_m T$ is constant, so the curve whose peak is at the shortest wavelength has the highest temperature: T1 first, then T3, then T2.
Revise: Rate of Heat Flow (Class 11 notes)
Answer (a): An object is seen because of refraction or reflection at its boundary. With $\mu=1$ there is no change from vacuum, so it is invisible.
Revise: Refraction at Plane Surface (Class 11 notes)
Answer (a): For a convex mirror $v=\frac{fu}{f+u}$ (magnitudes), which rises with $u$ and levels off at $f$, matching graph (a).
Revise: Reflection at Curved Mirrors (Class 11 notes)
Answer (c): Musical (sweet) sound is made of regular, periodic vibrations, while noise is irregular.
Revise: Acoustic Phenomena (Class 12 notes)
Answer (b): Charge stays constant. Moving the plates apart lowers $C=\varepsilon_0A/d$, so $V=Q/C$ rises (and the stored energy also rises).
Revise: Capacitor (Class 11 notes)
Answer (a): Solving the network by node potentials, the two junctions across the 20 $\Omega$ resistor sit at equal potential, so it carries no current. The rest is $(10+10)\parallel(10+10)=10\ \Omega$.
Revise: DC Circuit (Class 11 notes)
Answer (c): The two poles feel unequal forces in a non-uniform field, so there is a net force, and the unequal forces also give a torque.
Revise: Magnetic Field (Class 12 notes)
Answer (d): $V^{2}=V_R^{2}+V_L^{2}$: $V_L=\sqrt{34^{2}-10^{2}}=\sqrt{1056}\approx32.5$ V, so about 32 V.
Revise: Alternating Currents (Class 12 notes)
Answer (c): Deflection in electric and magnetic fields shows the rays carry charge and mass, i.e. they are a stream of particles.
Revise: Electrons (Class 12 notes)
Answer (a): X-rays lie outside the visible range, so objects exposed to them do not appear to have any colour; they are invisible to the eye.
Revise: Electrons (Class 12 notes)
Answer (c): Lyman is ultraviolet, Balmer is visible, and Paschen, Brackett and Pfund are infrared. Pfund is the one listed.
Revise: Quantization of Energy (Class 12 notes)
Answer (b): ${}_5^{10}B+{}_0^{1}n\to{}_3^{7}Li+{}_2^{4}He$. The mass number of the resulting nucleus is $10+1-4=7$.
Revise: Nuclear Physics (Class 11 notes)
Answer (d): Doping with trivalent atoms creates holes, which become the majority carriers, while thermally generated electrons are the minority carriers.
Revise: Semiconductor Devices (Class 12 notes)
Answer (b): Decinormal means $0.1,N$. The equivalent weight of $Na_2CO_3$ is $106/2=53$, so 1000 mL needs $0.1\times53=5.3$ g.
Revise: Volumetric Analysis (Class 12 notes)
Answer (d): The pH scale was introduced by the Danish chemist S. P. L. Sorensen in 1909.
Revise: Ionic Equilibrium (Class 12 notes)
Answer (b): A higher temperature gives more molecules with energy above the activation energy, so the rate rises.
Revise: Chemical Kinetics (Class 12 notes)
Answer (c): A process is spontaneous when $\Delta G<0$, i.e. the free energy decreases.
Revise: Chemical Thermodynamic (Class 12 notes)
Answer (a): Cobalt is a d-block element (3d7 4s2). Potassium and magnesium are s-block, silicon is a metalloid.
Revise: Transition Metals (Class 12 notes)
Answer (b): Hydrated copper sulphate, CuSO4.5H2O, is blue and is called blue vitriol.
Revise: Copper (Class 12 notes)
Answer (c): Argentite (Ag2S) is leached with dilute NaCN solution (MacArthur-Forrest cyanide process) to give a soluble complex, then silver is displaced with zinc.
Revise: Chemistry of Metals (Class 11 notes)
Answer (a): 2,4-D is a selective weed killer, so it is a herbicide.
Revise: Chemistry in the Service to Mankind (Class 12 notes)
Answer (d): Alkyl magnesium halides RMgX are Grignard reagents.
Revise: Organometallic Compounds (Class 12 notes)
Answer (a): Aldehydes reduce Fehling's solution to a red precipitate of Cu2O, while ketones do not.
Revise: Aldehydes and Ketones (Class 12 notes)
Answer (b): Nitrating phenol with concentrated HNO3 and H2SO4 gives 2,4,6-trinitrophenol, which is picric acid.
Revise: Phenols (Class 12 notes)
Answer (c): Chloroform (CHCl3) was a classic general anaesthetic.
Revise: Haloalkanes (Class 12 notes)
Answer (d): Cinnamic acid, $C_6H_5CH=CHCOOH$, has a benzene ring. Formic, acetic and stearic acids are aliphatic.
Revise: Carboxylic Acids (Class 12 notes)
Answer (a): Urea is the diamide of carbonic acid, $\ce{NH2CONH2}$.
Revise: Chemistry in the Service to Mankind (Class 12 notes)
Answer (b): Strong oxidation of the triple bond by alkaline KMnO4 gives oxalic acid (as its salt, then acidified).
Revise: Hydrocarbons (Class 11 notes)
Answer (c): Acetaldehyde is CH3CHO, a two-carbon aldehyde, so ethanal.
Revise: Aldehydes and Ketones (Class 12 notes)
Answer (d): Methylmercury is lipid soluble, accumulates in the food chain and attacks the nervous system, making it the most toxic form.
Revise: Study of Heavy Metals (Class 12 notes)
Answer (a): Molar mass of $O_2$ is 32 g, so 32 g is 1 mole, which has $6.02\times10^{23}$ molecules.
Revise: Stoichiometry (Class 11 notes)
Answer (b): Half-filled d subshell is extra stable, so Cr is [Ar] 3d5 4s1.
Revise: Atomic Structure (Class 11 notes)
Answer (c): These ions are isoelectronic with 10 electrons. A larger nuclear charge pulls them in more, so $Al^{3+}$ with $Z=13$ has the smallest radius.
Revise: Classification of Elements (Class 11 notes)
Answer (d): Dipole moment falls with electronegativity of the halogen, which falls down the group, so HI is lowest.
Revise: Chemical Bonding (Class 11 notes)
Answer (a): In SI, R = 8.314 J K^-1 mol^-1. Option d is the same value in CGS (erg), not SI.
Revise: States of Matter (Class 11 notes)
Answer (c): Fluorine is the most electronegative halogen, so the H-F bond has the largest ionic character.
Revise: Chemical Bonding (Class 11 notes)
Answer (d): Magnesia is the name linked to magnesium oxide/hydroxide (milk of magnesia is Mg(OH)2). MgSO4.7H2O is Epsom salt.
Revise: Chemistry of Metals (Class 11 notes)
Answer (a): Compound nouns such as 'headland' carry the main stress on the first element: HEAD-land.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (c): A headland is a narrow piece of land jutting into the sea, i.e. a promontory.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (d): 'Remarkable' means worthy of notice, and 'notable' is the nearest synonym. 'Astonishing' is stronger and 'good' is too general.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (d): Pompous means self-important and showy, the opposite of humble.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (b): Reciprocal means done by both sides, so the opposite is one-sided, unilateral.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (a): Orthography means the conventional spelling system, which fits 'writing with correct spelling'.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (c): Pantheism identifies God with nature and the universe.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (b): 'Live on' a diet, and 'believe in' something.
Revise: Grammar and Usage (Class 11 notes)
Answer (d): 'Scared of', 'loss in' business, and 'keen on'.
Revise: Grammar and Usage (Class 11 notes)
Answer (a): 'Reared on' means brought up with (stories) from childhood.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (b): 'Break up' is the phrasal verb used when a school closes for holidays.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (c): 'Bright and breezy' means cheerful, lively and brisk.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (a): 'Go by the board' means to be abandoned or ignored.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (d): 'Whether ... or not' is the correct pairing.
Revise: Grammar and Usage (Class 11 notes)
Answer (c): MS-Word is a word processor, which is application software.
Revise: Application Package (Class 11 notes)
Answer (c): Booting loads the operating system into memory, so without one the computer cannot boot.
Revise: Computer Software and Operating System (Class 11 notes)
Answer (b): The default page orientation in word processors is portrait.
Revise: Application Package (Class 11 notes)
Answer (a): These keys modify the action of other keys, so they are modifier keys.
Revise: Computer System (Class 11 notes)
Answer (b): Non-volatile storage keeps data without power.
Revise: Computer System (Class 11 notes)
Answer (a): The usual textbook answer to this question is a telephone line, a modem, a computer and an ISP.
Revise: Web Technology I (Class 11 notes)
Answer (a): Booting starts the computer and loads the OS into RAM.
Revise: Computer Software and Operating System (Class 11 notes)
Answer (d): The gutter margin is extra space reserved for binding.
Revise: Application Package (Class 11 notes)
Answer (d): Permanent programs held in ROM are firmware.
Revise: Computer System (Class 11 notes)
Answer (d): A, B and D are true. C is false because columns use letters and rows use numbers.
Revise: Application Package (Class 11 notes)
Answer (c): $x+7=5$ contains an unknown, so it is an open sentence whose truth depends on $x$. It is therefore not a statement.
Revise: Algebra (Class 11 notes)
Answer (b): If every element of P is in Q, then anything outside Q is outside P, so the complement of Q is a subset of the complement of P.
Revise: Algebra (Class 11 notes)
Answer (a): $x-3=\pm7$ gives $x=10$ or $x=-4$.
Revise: Algebra (Class 11 notes)
Answer (d): $\sqrt{-1}=i$ is imaginary, which is a complex number.
Revise: Algebra (Class 11 notes)
Answer (b): Domain is $-3\le x\le3$, and $9-x^{2}$ goes from 0 to 9, so $f$ takes values from 0 to 3.
Revise: Algebra (Class 11 notes)
Answer (c): $f(-x)=-\sin x-x=-f(x)$, so it is odd.
Revise: Trigonometry (Class 11 notes)
Answer (d): In an A.P. the middle term is the mean of the other two, so $x+z=2y$.
Revise: Algebra (Class 11 notes)
Answer (c): The determinant of a 1 by 1 matrix is its single entry, $-3$.
Revise: Algebra (Class 11 notes)
Answer (b): $(5+2i)^{2}=25+20i-4=21+20i$. So $x=21$ and $-y=20$, giving $y=-20$.
Revise: Algebra (Class 11 notes)
Answer (d): Angles $B-d$, $B$, $B+d$ sum to $3B=180^\circ$, so $B=60^\circ$.
Revise: Trigonometry (Class 11 notes)
Answer (c): Area $=\frac12\times$ two sides $\times$ sine of the included angle $=\frac12 bc\sin A$.
Revise: Trigonometry (Class 11 notes)
Answer (a): Swapping the coefficients and changing one sign gives $3x-2y+k=0$, which matches option a.
Revise: Analytic geometry (Class 11 notes)
Answer (b): Lines are perpendicular when the coefficients of $x^{2}$ and $y^{2}$ sum to zero: $a+2=0$, so $a=-2$.
Revise: Analytic geometry (Class 11 notes)
Answer (d): Tangent at $(x_1,y_1)$ is $xx_1+yy_1=r^{2}$: $3x+4y=25$.
Revise: Analytic geometry (Class 11 notes)
Answer (c): Magnitude is $\sqrt{9+16}=5$, so the unit vector is $\frac15(3i+4j)$.
Revise: Vectors (Class 11 notes)
Answer (b): $(3i-4j)\cdot(3i+4j)=3\cdot3+(-4)(4)=9-16=-7$.
Revise: Vectors (Class 11 notes)
Answer (d): Even numbers 2, 4, 6 are 3 of 6 outcomes, so $3/6=1/2$.
Revise: Statistics and Probability (Class 11 notes)
Answer (c): Factor $x^{3}-27=(x-3)(x^{2}+3x+9)$, so the limit is $9+9+9=27$.
Revise: Calculus (Class 11 notes)
Answer (b): $y'=3x^{2}-6x=3x(x-2)=0$ gives $x=0,2$.
Revise: Calculus (Class 11 notes)
Answer (a): $e^{\log x^{2}}=x^{2}$, so $dy/dx=2x$.
Revise: Calculus (Class 11 notes)
Answer (c): $y''=6x-2$ changes sign at $6x-2=0$, so $x=\frac13$ is the point of inflection.
Revise: Calculus (Class 11 notes)
Answer (a): Put $u=x^{3}$: $\int_0^1e^{u}du=e-1$.
Revise: Calculus (Class 11 notes)
Answer (d): $\int_0^4x^{2}dx=\frac{4^{3}}{3}=\frac{64}{3}$.
Revise: Calculus (Class 11 notes)
Answer (b): $f'(2)\approx\frac{f(2.1)-f(2)}{0.1}=\frac{9.41-9}{0.1}=4.1$. (The exact derivative is $4$, and the option 4.1 is the difference quotient value.)
Revise: Calculus (Class 11 notes)
Answer (b): Forces along the same line and direction add: $P+Q$.
Revise: Mechanics (Class 11 notes)