BSc CSIT Entrance Exam 2069 Questions with Answers

1. a$0$
2. b$4$
3. c$2$
4. d$-2$

Answer (a): Put $x=2$: $f(2)=\sqrt{4-4}=0$.

Revise: Algebra (Class 11 notes)

1. a${2}$
2. b${-2,2}$
3. c$(-2,2)$
4. d$[-2,2]$

Answer (b): $|x|=2$ gives $x=2$ or $x=-2$, so the solution set is ${-2,2}$. Option a prints ${2,2}$, which is just ${2}$, and the interval options are not solution sets of an equation.

Revise: Algebra (Class 11 notes)

1. a$B$
2. b$A$
3. cempty set
4. d$C$

Answer (b): Every element of $A$ is already in $B$, so the common part is $A$.

Revise: Algebra (Class 11 notes)

1. a1
2. b0
3. c2
4. d-1

Answer (c): 1 is not prime and 0 and -1 are not natural primes; the smallest prime is 2.

Revise: Algebra (Class 11 notes)

1. aY axis
2. bX axis
3. cboth axes
4. dorigin

Answer (b): Replacing $y$ by $-y$ leaves the equation unchanged, so it is symmetric about the X axis.

Revise: Analytic geometry (Class 11 notes)

1. a8
2. b7
3. c13
4. d-7

Answer (b): Sum $=(x-4)+(2x-4)=3x-8=13$, so $x=7$.

Revise: Algebra (Class 11 notes)

1. asymmetric
2. bskew symmetric
3. cdiagonal
4. dunit

Answer (a): $A^T=A$ since the off-diagonal entries are both 3, so it is symmetric.

Revise: Algebra (Class 11 notes)

1. a0
2. b1
3. c-1
4. d2

Answer (a): Two rows (R1 and R2) are identical, so the determinant is 0.

Revise: Algebra (Class 11 notes)

1. a0
2. b1
3. c2
4. d3

Answer (d): Expand: $1-\omega-\omega^2+\omega^3=1-(\omega+\omega^2)+1=1+1+1=3$.

Revise: Algebra (Class 11 notes)

1. a$\sqrt{13}$
2. b$\sqrt{5}$
3. c1
4. d13

Answer (a): $|2-3i|=\sqrt{2^2+3^2}=\sqrt{13}$.

Revise: Algebra (Class 11 notes)

1. a$x^2-3x+2=0$
2. b$x^2+3x+2=0$
3. c$x^2+x+2=0$
4. d$x^2+3x+1=0$

Answer (a): Sum of roots 3 and product 2 give $x^2-3x+2=0$.

Revise: Algebra (Class 11 notes)

1. a0
2. b1
3. c-1
4. d$\sqrt{3}$

Answer (b): $\tan(\tan^{-1}x)=x$, so the value is 1.

Revise: Trigonometry (Class 11 notes)

1. a$2n\pi$
2. b$\frac{\pi}{3}$
3. c$2n\pi\pm\frac{\pi}{3}$
4. d$2n\pi+\frac{\pi}{3}$

Answer (c): $\cos x=\cos\frac{\pi}{3}$, and cosine is even, so $x=2n\pi\pm\frac{\pi}{3}$.

Revise: Trigonometry (Class 11 notes)

1. a$\frac{1}{\sqrt2}$
2. b$\sqrt2$
3. c$-\frac{1}{\sqrt2}$
4. d$-\sqrt2$

Answer (a): Distance $=\frac{|0+0+1|}{\sqrt{1+1}}=\frac{1}{\sqrt2}$.

Revise: Analytic geometry (Class 11 notes)

1. a$0^0$
2. b$45^0$
3. c$135^0$
4. d$90^0$

Answer (d): Here $a+b=2-2=0$, so the lines are perpendicular and the angle is $90^0$.

Revise: Analytic geometry (Class 11 notes)

1. a$(1,2)$
2. b$(0,0)$
3. c$(2,1)$
4. d$(2,2)$

Answer (a): Centre is $(-g,-f)$ with $2g=-2$, $2f=-4$, giving $(1,2)$.

Revise: Analytic geometry (Class 11 notes)

1. a$2x+3y=13$
2. b$x+2y=13$
3. c$2x+3y=10$
4. d$2x+y=13$

Answer (a): Tangent at $(x_1,y_1)$ is $xx_1+yy_1=r^2$: $2x+3y=13$.

Revise: Analytic geometry (Class 11 notes)

1. a12
2. b4
3. c9
4. d0/0

Answer (a): This is the derivative of $x^3$ at 2: $3\cdot2^2=12$.

Revise: Calculus (Class 11 notes)

1. a6
2. b-4
3. c0
4. d0/0

Answer (b): Split: $\frac{\tan x}{x}-5\to1-5=-4$.

Revise: Calculus (Class 11 notes)

1. a$x=0$
2. b$x=2$
3. c$x=-2$
4. d$x=1$

Answer (b): At $x=2$: left limit $2(2)=4$, right limit and value $2+2=4$, so it is continuous there.

Revise: Calculus (Class 11 notes)

1. a0
2. b-1
3. c2
4. d5

Answer (b): $2x+3=1$ gives $x=-1$.

Revise: Calculus (Class 11 notes)

1. a$t$
2. b$1/t$
3. c$2t$
4. d0

Answer (c): Differentiate $y=t^2$ directly with respect to $t$: $2t$.

Revise: Calculus (Class 11 notes)

1. a0
2. b1
3. c6
4. d-1

Answer (a): $f''(x)=6x=0$ at $x=0$, and $f''$ changes sign there.

Revise: Calculus (Class 11 notes)

1. a0
2. b1
3. c-1
4. d2

Answer (b): $2\cdot\left[\frac{x^2}{2}\right]_0^1=2\cdot\frac12=1$.

Revise: Calculus (Class 11 notes)

1. a$\sin x$
2. b$-\frac{\cos x}{2}$
3. c$\sin\frac{2x}{2}$
4. d$2\sin 2x$

Answer (a): The integral of $\cos x$ is $\sin x$ (+ C).

Revise: Calculus (Class 11 notes)

1. aalthough
2. bin spite of
3. cas it
4. ddespite

Answer (a): A clause follows, so a conjunction is needed: 'Although'. 'In spite of' and 'despite' take a noun phrase.

Revise: Grammar and Usage (Class 11 notes)

1. abut also
2. bboth
3. cas well as
4. dnot only

Answer (b): 'Both ... and' is the correlative pair; 'friendly' follows 'and'.

Revise: Grammar and Usage (Class 11 notes)

1. ahave come
2. bhas come
3. cis coming
4. dwas coming

Answer (a): 'People' is plural and the present perfect fits, so 'have come'.

Revise: Grammar and Usage (Class 11 notes)

1. aas well as
2. band also
3. cand
4. dwith

Answer (c): 'Both' pairs with 'and'.

Revise: Grammar and Usage (Class 11 notes)

1. adiscussion
2. bdiscussing
3. cto discuss
4. ddiscusses

Answer (b): The list is of gerunds (reading, observing, teaching), so parallel form 'discussing'.

Revise: Grammar and Usage (Class 11 notes)

1. aan
2. ba
3. cthe
4. dno article

Answer (d): Names of towns and places take no article.

Revise: Grammar and Usage (Class 11 notes)

1. awith
2. bfor
3. cabout
4. dof

Answer (c): The adjective 'concerned' takes 'about' (concerned about something).

Revise: Grammar and Usage (Class 11 notes)

1. aby ; for
2. bby ; with
3. cin ; for
4. dwith ; by

Answer (b): 'By' marks the agent (robber); 'with' marks the instrument (gun).

Revise: Grammar and Usage (Class 11 notes)

1. ahas
2. bhave
3. cwas
4. dhas been

Answer (a): The subject is the singular 'fear', so the present perfect 'has caused'.

Revise: Grammar and Usage (Class 11 notes)

1. acutting
2. bto cut
3. cto be cut
4. dboth a and b

Answer (a): 'Need' followed by a gerund has a passive sense: 'needs cutting'.

Revise: Grammar and Usage (Class 11 notes)

1. aat
2. bby
3. cfrom
4. din

Answer (c): Here 'from this evidence' means 'judging from' it, the usual collocation for drawing a conclusion.

Revise: Grammar and Usage (Class 11 notes)

1. aat
2. babout
3. cin
4. dto

Answer (d): The verb 'object' takes the preposition 'to'.

Revise: Grammar and Usage (Class 11 notes)

1. aAbout
2. bover
3. camong
4. dout

Answer (d): 'Work out' means to solve or calculate, so 'worked out the answer' is correct. 'Worked over' means revised or beaten up.

Revise: Vocabulary and Word Formation (Class 11 notes)

1. ahistory
2. bathlete
3. cartist
4. dsuccess

Answer (d): We form historic, athletic and artistic, but 'success' takes -ful or -ive, not -ic.

Revise: Vocabulary and Word Formation (Class 11 notes)

1. aspectacular
2. bmarvel
3. ccomparable
4. dpeculiar

Answer (b): 'Marvel' (to be amazed) is a verb; the others are adjectives.

Revise: Vocabulary and Word Formation (Class 11 notes)

1. aprovided
2. bpoisonous
3. cprotected
4. ddetected

Answer (b): We have unprovided, unprotected and undetected, but 'poisonous' does not take a negative prefix.

Revise: Vocabulary and Word Formation (Class 11 notes)

1. awill come
2. bwould come
3. ccould come
4. dcomes

Answer (d): In a time clause with 'before' the simple present is used for future: 'comes'.

Revise: Grammar and Usage (Class 11 notes)

1. ais
2. bwas
3. chas been
4. dhave been

Answer (d): 'Memoranda' is the plural of memorandum, so 'have been'.

Revise: Grammar and Usage (Class 11 notes)

1. aamnesia
2. binsomnia
3. canemia
4. depicurean

Answer (c): Anemia is the condition of too little blood or haemoglobin.

Revise: Vocabulary and Word Formation (Class 11 notes)

1. alawyer
2. bcosmopolitan
3. cveteran
4. dteacher

Answer (c): A veteran is a person with long experience in a field. A cosmopolitan is a person at home in many countries.

Revise: Vocabulary and Word Formation (Class 11 notes)

1. athat
2. bif
3. cwhen
4. dwhy

Answer (b): 'Doubt if' expresses uncertainty about whether she will come.

Revise: Grammar and Usage (Class 11 notes)

1. aadverb
2. badjective
3. cpreposition
4. ddemonstrative adjective

Answer (a): In 'They walk very fast', fast modifies the verb walk and is itself modified by very, so it is an adverb.

Revise: Grammar and Usage (Class 11 notes)

1. aHe said that congratulations.
2. bHe congratulated me.
3. cHe congratulated that I passed.
4. dHe said that I deserved congratulations.

Answer (b): An exclamation of congratulation is reported with the verb 'congratulate'.

Revise: Grammar and Usage (Class 11 notes)

1. a6 sec
2. b3 sec
3. c2 sec
4. d1 sec

Answer (d): From the mean position $y=A\sin\omega t=A/2$ gives $\omega t=\pi/6$, so $t=T/12=1$ s.

Revise: Periodic Motion (Class 12 notes)

1. a1.5 sec
2. b1.5 min
3. c15 sec
4. d15 min

Answer (b): Distance $=1.6$ km, so $t=1.6/64=0.025$ h $=1.5$ min.

Revise: Kinematics (Class 11 notes)

1. aWork and power
2. bEnergy and power
3. cWork and energy
4. dAngular momentum and linear momentum

Answer (c): Both work and energy have dimension $[ML^2T^{-2}]$.

Revise: Physical Quantities (Class 11 notes)

1. a$v=\sqrt F$
2. b$v=\sqrt{2F}$
3. c$v=\frac12\sqrt F$
4. d$v=\sqrt{\frac F2}$

Answer (d): $F=\frac{mv^2}{r}=\frac{1\cdot v^2}{0.5}=2v^2$, so $v=\sqrt{F/2}$.

Revise: Circular Motion (Class 11 notes)

1. a$2,km,s^{-1}$
2. b$1,km,s^{-1}$
3. c$0.5,km,s^{-1}$
4. d$4,km,s^{-1}$

Answer (b): $v\propto\sqrt{T/M}$: $v=2\sqrt{\frac{1083}{273}\cdot\frac{2}{32}}\approx2\times0.5=1$ km/s.

Revise: Ideal Gas (Class 11 notes)

1. acoefficient of thermal conductivity
2. bthermo emf
3. cSpecific heat
4. dheat radiation

Answer (d): A bolometer detects the intensity of heat (thermal) radiation.

Revise: Rate of Heat Flow (Class 11 notes)

1. a296 K
2. b263 K
3. c250 K
4. d$1.5^0$ C

Answer (b): $C=\frac{5}{9}(14-32)=-10^0$C, which is $263$ K.

Revise: Heat and Temperature (Class 11 notes)

1. a1.5 D
2. b0.66 D
3. c-1.5 D
4. d-0.66 D

Answer (d): A diverging lens of $f=-1.5$ m is needed: $P=1/f=-0.66$ D.

Revise: Lenses (Class 11 notes)

1. a$\sin^{-1}\left(\frac89\right)$
2. b$\sin^{-1}\left(\frac45\right)$
3. c$\sin^{-1}\left(\frac34\right)$
4. d$\sin^{-1}(2/3)$

Answer (a): $\sin C=\frac{n_w}{n_g}=\frac{4/3}{3/2}=\frac89$.

Revise: Refraction at Plane Surface (Class 11 notes)

1. a2
2. b3
3. c4
4. d0

Answer (b): $d=\frac{10^{-2}}{5000}=2\times10^{-6}$ m; $n_{max}\le\frac{d}{\lambda}=3.3$, so $n=3$.

Revise: Diffraction of Light (Class 12 notes)

1. a$\frac{t}{nc}$
2. b$cnt$
3. c$\frac{nt}{c}$
4. d$\frac{tc}{n}$

Answer (c): Speed in glass is $c/n$, so $time=\frac{t}{c/n}=\frac{nt}{c}$.

Revise: Refraction at Plane Surface (Class 11 notes)

1. a50 cm from $2,\mu C$ charge
2. b59 cm from $2,\mu C$ charge
3. c41 cm from $2,\mu C$ charge
4. d66 cm from $2,\mu C$ charge

Answer (c): $\frac{2}{x^2}=\frac{4}{(1-x)^2}$ gives $\frac{1-x}{x}=\sqrt2$, so $x=\frac{1}{1+\sqrt2}\approx0.41$ m.

Revise: Electric Field (Class 11 notes)

1. aonly odd harmonics
2. bonly even harmonics
3. cboth odd and even harmonics
4. ddepends upon the length of pipe

Answer (c): An open pipe has antinodes at both ends and supports all harmonics.

Revise: Wave in Pipes and Strings (Class 12 notes)

1. aelectric potential
2. belectric field intensity
3. celectric power
4. delectric energy

Answer (a): $P=VI$, so $\frac{W}{A}=V$, the unit of potential.

Revise: DC Circuit (Class 11 notes)

1. aless than that of B
2. bequal to that of B
3. cmore than that of B
4. ddepends upon the amount of current flowing in wires

Answer (b): Resistivity depends only on the material, not on length, so it is equal.

Revise: DC Circuit (Class 11 notes)

1. aincrease
2. bdecrease
3. cremains unchanged
4. dincreases or decreases depending upon the frequency of a.c

Answer (d): Adding a capacitor in series lowers the net capacitance and raises the capacitive reactance. $Z=\sqrt{R^2+(X_L-X_C)^2}$ then falls if $X_L>X_C$ and rises if $X_C>X_L$, so it depends on the frequency.

Revise: Alternating Currents (Class 12 notes)

1. a$X-(Y/R)$
2. b$(X-Y)R$
3. c$(X-Y)R/X$
4. d$(X-Y)R/Y$

Answer (d): $Y=IR$ and $X=I(R+r)$, so $r=\frac{(X-Y)R}{Y}$.

Revise: DC Circuit (Class 11 notes)

1. a5:27
2. b20:27
3. c27:20
4. d1:4

Answer (a): $\frac1{\lambda_L}=R\left(1-\frac14\right)=\frac{3R}{4}$, $\frac1{\lambda_B}=R\left(\frac14-\frac19\right)=\frac{5R}{36}$; ratio $=\frac{5/36}{3/4}=\frac{5}{27}$.

Revise: Quantization of Energy (Class 12 notes)

1. a$2\times10^{15}$ Hz
2. b$8\times10^{14}$ Hz
3. c$4\times10^{16}$ Hz
4. d$6\times10^{17}$ Hz

Answer (b): $\nu_0=\frac{\phi}{h}=\frac{3.3\times1.6\times10^{-19}}{6.63\times10^{-34}}\approx8\times10^{14}$ Hz.

Revise: Photons (Class 12 notes)

1. a125 g
2. b250 g
3. c550 g
4. d937.5 g

Answer (d): 20 days is 4 half lives, so remaining $=\frac{1000}{16}=62.5$ g and decayed $=937.5$ g.

Revise: Radioactivity and Nuclear Reaction (Class 12 notes)

1. aspeed
2. bmass
3. ccharge
4. dmagnetic induction A

Answer (a): $T=\frac{2\pi m}{qB}$ has no speed term.

Revise: Magnetic Field (Class 12 notes)

1. a$250,ms^{-1}$
2. b$300,ms^{-1}$
3. c$360,ms^{-1}$
4. d$450,ms^{-1}$

Answer (c): $v\propto\sqrt T$: $v=330\sqrt{300/273}\approx346$, nearest option is 360.

Revise: Mechanical Waves (Class 12 notes)

1. a$90^0$
2. b$60^0$
3. c$45^0$
4. d$30^0$

Answer (c): $T=\frac{2u\sin\theta}{g}$, $R=\frac{u^2\sin2\theta}{g}$; $gT^2=2R$ gives $\tan\theta=1$, so $45^0$.

Revise: Kinematics (Class 11 notes)

1. a$1.7\times10^{11},ms^{-2}$
2. b$1\times10^{6},ms^{-2}$
3. c$1\times10^{12},ms^{-2}$
4. d$1\times10^{7},ms^{-2}$

Answer (a): $a=\frac{eE}{m}=\frac{1.6\times10^{-19}}{9.1\times10^{-31}}\approx1.7\times10^{11}$.

Revise: Electric Field (Class 11 notes)

1. asame
2. b2 times
3. c4 times
4. d16 times

Answer (c): Heat produced equals kinetic energy $\frac12mv^2$, which becomes 4 times when $v$ doubles.

Revise: Quantity of Heat (Class 11 notes)

1. aalkanes
2. balkenes
3. calkynes
4. dethylenes

Answer (a): Saturated hydrocarbons with little affinity for reagents are called paraffins, i.e. alkanes.

Revise: Hydrocarbons (Class 11 notes)

1. a$\ce{Fe4[Fe(CN)6]3}$
2. b$\ce{Fe(CNS)}$
3. c$\ce{Na4[Fe(CN)5NOS]}$
4. d$\ce{Na[Fe(CN)6]}$

Answer (a): Nitrogen gives NaCN, then sodium ferrocyanide, which with Fe(III) forms ferric ferrocyanide (Prussian blue), seen as a blue-green colour.

Revise: Fundamental Principles (Class 11 notes)

1. a$\ce{C2H4}$
2. b$\ce{C2H2}$
3. c$\ce{H2}$
4. d$\ce{CH4}$

Answer (d): $\ce{Al4C3 + 12H2O -> 4Al(OH)3 + 3CH4}$.

Revise: Hydrocarbons (Class 11 notes)

1. aacetanilide
2. bbenzalinide
3. cSchiff's base
4. dphenol

Answer (c): An amine condenses with an aldehyde to form an imine, called a Schiff's base.

Revise: Amines (Class 12 notes)

1. aTetraethyl lead
2. bmethane
3. cmethoxy propane
4. dmethyl cyanide

Answer (a): Tetraethyl lead (TEL) was the classic antiknock additive in petrol.

Revise: Hydrocarbons (Class 11 notes)

1. achloretone
2. bchloropicrin
3. ccarbon tetrachloride
4. dcarbonyl chloride

Answer (b): $\ce{CHCl3 + HNO3 -> CCl3NO2 + H2O}$, giving chloropicrin (tear gas).

Revise: Haloalkanes (Class 12 notes)

1. aNinhydrin reagent
2. bTollen's reagent
3. cGrignard reagent
4. dBaeyer's reagent

Answer (b): Aldehydes reduce Tollen's reagent to give a silver mirror.

Revise: Aldehydes and Ketones (Class 12 notes)

1. aaddition polymer
2. bcondensation polymer
3. cpolyamide
4. dpolyimide

Answer (b): Terylene is a polyester formed from ethylene glycol and terephthalic acid with loss of water.

Revise: Chemistry in the Service to Mankind (Class 12 notes)

1. a$\ce{ZnCO3}$
2. b$\ce{ZnS}$
3. cZnO
4. d$\ce{ZnO.Fe2O3}$

Answer (b): Zinc blende is the sulphide ore ZnS.

Revise: Chemistry of Metals (Class 11 notes)

1. a+2
2. b+3
3. c+1
4. d0

Answer (a): Mohr's salt is $\ce{FeSO4.(NH4)2SO4.6H2O}$, which contains Fe(II).

Revise: Oxidation and Reduction (Class 11 notes)

1. aFeO
2. b$\ce{CO2}$
3. cCO
4. dCaO

Answer (b): $\ce{CO2}$ is a non-metal oxide that gives carbonic acid with water; CaO and FeO are basic and CO is neutral.

Revise: Chemistry of Non-Metals (Class 11 notes)

1. aV-shaped
2. btetrahedral
3. clinear
4. dtrigonal pyramidal

Answer (d): $\ce{NH3}$ has three bond pairs and one lone pair on N, giving a trigonal pyramidal shape.

Revise: Chemical Bonding (Class 11 notes)

1. amonobasic acid
2. bdibasic acid
3. cTribasic acid
4. dtetrabasic acid

Answer (c): $\ce{H3PO4}$ has three replaceable hydrogens, so it is tribasic.

Revise: Chemistry of Non-Metals (Class 11 notes)

1. a$\ce{MgSO4.7H2O}$
2. bMgO
3. c$\ce{CaSO4.2H2O}$
4. dCaO

Answer (a): Epsom salt is hydrated magnesium sulphate, MgSO4.7H2O.

Revise: Chemistry of Metals (Class 11 notes)

1. aSulphuric acid
2. bnitric acid
3. cammonia
4. dwhite phosphorus

Answer (b): Ostwald's process oxidises ammonia over Pt catalyst to make $\ce{HNO3}$.

Revise: Chemistry of Non-Metals (Class 11 notes)

1. achlorine
2. bfluorine
3. ciodine
4. dbromine

Answer (a): Chlorine has the highest electron affinity (more than fluorine, which has a small, crowded 2p).

Revise: Classification of Elements (Class 11 notes)

1. a$6.023\times10^{23}$ no of $CO_2$ molecule
2. b$6.023\times10^{21}$ no of $CO_2$ molecule
3. c$6.023\times10^{19}$ no of $CO_2$ molecule
4. d$6.023\times10^{18}$ no of $CO_2$ molecule

Answer (a): Molar mass of $\ce{CO2}$ is 44 g, so 44 g is one mole, i.e. $6.023\times10^{23}$ molecules.

Revise: Stoichiometry (Class 11 notes)

1. aMol. wt/1
2. bMol. wt/6
3. cMol. wt/3
4. dMol. wt/5

Answer (b): $\ce{Cr2O7^2-}$ gains 6 electrons (Cr from +6 to +3 in two atoms), so $E=M/6$.

Revise: Volumetric Analysis (Class 12 notes)

1. a0.53 gm
2. b5.3 gm
3. c0.353 gm
4. d5.6 gm

Answer (a): Equivalent weight of $Na_2CO_3$ is $106/2=53$. Mass $=0.1\times53\times0.1\text{ L}=0.53$ g.

Revise: Volumetric Analysis (Class 12 notes)

1. adynes.cm
2. b$dynes.cm^{-1}$
3. c$dynes.cm^{2}$
4. d$dynes.cm^{-2}$

Answer (b): Surface tension is force per unit length: dyne per cm.

Revise: States of Matter (Class 11 notes)

1. aHund's rule
2. bAufbau principle
3. cUncertainty principle
4. dPaul's exclusion principle

Answer (a): Filling degenerate orbitals singly with parallel spins is Hund's rule of maximum multiplicity.

Revise: Atomic Structure (Class 11 notes)

1. a$[Ar]4s^23d^8$
2. b$[Ar]4s^13d^{10}$
3. c$[Ar]4s^23d^{10}$
4. d$[Ar]4s^23d^8$

Answer (b): A filled 3d shell is extra stable, so one 4s electron moves to 3d: $[Ar]4s^13d^{10}$.

Revise: Atomic Structure (Class 11 notes)

1. a$6.3\times10^{-5}$ M
2. b$6.3\times10^{-6}$ M
3. c$6.3\times10^{-7}$ M
4. d$6.3\times10^{-4}$ M

Answer (b): $[H^+]=10^{-5.2}=10^{0.8}\times10^{-6}=6.3\times10^{-6}$ M.

Revise: Ionic Equilibrium (Class 12 notes)

1. a$sec^{-1}$
2. b$mol,sec^{-1}$
3. c$mol^{-1}sec^{-1}$
4. d$litre^2mol^{-2}s^{-1}$

Answer (a): Rate $=k[A]$ so $k=\frac{mol,L^{-1}s^{-1}}{mol,L^{-1}}=s^{-1}$.

Revise: Chemical Kinetics (Class 12 notes)

1. aalways positive
2. balways negative
3. ccan be positive or negative
4. dunpredictable

Answer (c): Formation can be exothermic (negative, e.g. water) or endothermic (positive, e.g. NO).

Revise: Chemical Thermodynamic (Class 12 notes)