IOE Entrance Model Set 3 with Answers

1. a5
2. b7
3. c6
4. d8

Answer (c): A set with $n$ elements has $2^n$ subsets, and $2^n=64=2^6$ gives $n=6$.

Revise: Algebra (Class 11 notes)

1. aif $\neg p$ then $\neg q$
2. bif $q$ then $p$
3. cif $p$ then $\neg q$
4. dif $\neg q$ then $\neg p$

Answer (d): The contrapositive of $p\to q$ swaps and negates both parts: $\neg q\to\neg p$. It is always equivalent to the original statement.

Revise: Algebra (Class 11 notes)

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

Answer (b): The vertex is at $x=-b/2a=2$, and $f(2)=4-8+3=-1$.

Revise: Algebra (Class 11 notes)

1. a$\begin{pmatrix}1&6\0&1\end{pmatrix}$
2. b$\begin{pmatrix}1&3\0&1\end{pmatrix}$
3. c$\begin{pmatrix}1&8\0&1\end{pmatrix}$
4. d$\begin{pmatrix}1&4\0&1\end{pmatrix}$

Answer (a): $A^2=\begin{pmatrix}1&4\0&1\end{pmatrix}$ and $A^3=A^2A=\begin{pmatrix}1&6\0&1\end{pmatrix}$. In general $A^n$ has $2n$ in the corner.

Revise: Algebra (Class 11 notes)

1. a$24$
2. b$15$
3. c$12$
4. d$30$

Answer (a): For a triangular matrix the determinant is the product of the diagonal entries: $1\cdot4\cdot6=24$.

Revise: Algebra (Class 11 notes)

1. a$5\sqrt2$
2. b$\dfrac52$
3. c$\dfrac{25}{2}$
4. d$\dfrac{5}{\sqrt2}$

Answer (d): $|z_1/z_2|=|z_1|/|z_2|=\sqrt{9+16}/\sqrt{1+1}=5/\sqrt2$.

Revise: Algebra (Class 11 notes)

1. a$\dfrac23$
2. b$-\dfrac32$
3. c$-\dfrac23$
4. d$\dfrac32$

Answer (c): Rewrite as $y=-\tfrac23x+2$, so the slope is $-2/3$.

Revise: Analytic geometry (Class 11 notes)

1. a$192$
2. b$384$
3. c$768$
4. d$256$

Answer (b): Here $a=3$, $r=2$, so $T_8=ar^7=3\cdot128=384$.

Revise: Algebra (Class 11 notes)

1. a$60$
2. b$125$
3. c$120$
4. d$10$

Answer (a): Choose and arrange 3 of 5 digits: ${}^5P_3=5\cdot4\cdot3=60$.

Revise: Algebra (Class 11 notes)

1. a$56$
2. b$70$
3. c$28$
4. d$35$

Answer (b): The middle term has the largest coefficient: ${}^8C_4=70$.

Revise: Algebra (Class 11 notes)

1. a$\dfrac12$
2. b$\dfrac{\sqrt3}{4}$
3. c$\dfrac14$
4. d$\dfrac18$

Answer (c): $\sin A\cos A=\tfrac12\sin2A$, so the value is $\tfrac12\sin30^\circ=\tfrac14$.

Revise: Trigonometry (Class 11 notes)

1. a$\dfrac{5\pi}{6}$
2. b$-\dfrac{\pi}{6}$
3. c$\dfrac{\pi}{3}$
4. d$\dfrac{\pi}{6}$

Answer (d): The principal value of $\sin^{-1}$ lies in $[-\pi/2,\pi/2]$. Since $\sin\tfrac{5\pi}{6}=\tfrac12$, the answer is $\pi/6$.

Revise: Trigonometry (Class 12 notes)

1. a$x+2y+3z=1$
2. b$6x+3y+2z=1$
3. c$2x+3y+6z=6$
4. d$6x+3y+2z=6$

Answer (d): The intercept form is $\dfrac x1+\dfrac y2+\dfrac z3=1$. Multiplying by 6 gives $6x+3y+2z=6$.

Revise: Analytic geometry (Class 12 notes)

1. a$x^2+y^2-4x+2y+4=0$
2. b$x^2+y^2+4x-2y-4=0$
3. c$x^2+y^2-4x+2y-4=0$
4. d$x^2+y^2-4x+2y-9=0$

Answer (c): $(x-2)^2+(y+1)^2=9$ expands to $x^2+y^2-4x+2y+5-9=0$, i.e. $x^2+y^2-4x+2y-4=0$.

Revise: Analytic geometry (Class 11 notes)

1. a$(2,0)$
2. b$(4,0)$
3. c$(0,2)$
4. d$(8,0)$

Answer (a): For $y^2=4ax$, $4a=8$ so $a=2$ and the focus is $(a,0)=(2,0)$.

Revise: Analytic geometry (Class 12 notes)

1. a$0$
2. b$6$
3. c$3$
4. d$9$

Answer (b): Cancel the common factor: $\dfrac{(x-3)(x+3)}{x-3}=x+3\to6$.

Revise: Calculus (Class 11 notes)

1. a$\tan x$
2. b$-\cot x$
3. c$\cot x$
4. d$\sec x$

Answer (c): By the chain rule, $\dfrac{1}{\sin x}\cdot\cos x=\cot x$.

Revise: Calculus (Class 11 notes)

1. a$1$
2. b$e^2$
3. c$e^2-1$
4. d$2$

Answer (d): The integral is $[\ln x]_1^{e^2}=\ln e^2-\ln1=2$.

Revise: Calculus (Class 12 notes)

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

Answer (b): Perpendicular means $\vec a\cdot\vec b=0$: $2-2\lambda+4=0$, so $\lambda=3$.

Revise: Vectors (Class 11 notes)

1. a$6$
2. b$5$
3. c$7$
4. d$6.5$

Answer (a): Sorted: $1,3,5,7,8,9$. With 6 values the median is the mean of the 3rd and 4th: $(5+7)/2=6$.

Revise: Statistics and Probability (Class 11 notes)

1. a$20$
2. b$25$
3. c$15$
4. d$10$

Answer (c): $n(A\cup B)=60-10=50$. Then $n(A\cap B)=35+30-50=15$.

Revise: Algebra (Class 11 notes)

1. a$I$
2. b$-I$
3. c$A$
4. d$-A$

Answer (b): $A^2=-I$, so $A^4=I$. Since $2026=4\cdot506+2$, $A^{2026}=A^2=-I$.

Revise: Algebra (Class 11 notes)

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

Answer (a): $\dfrac{1+i\sqrt3}{2}=\cos60^\circ+i\sin60^\circ$. By De Moivre, the 6th power is $\cos360^\circ+i\sin360^\circ=1$.

Revise: Algebra (Class 11 notes)

1. a$\dfrac{10}{11}$
2. b$1$
3. c$\dfrac89$
4. d$\dfrac9{10}$

Answer (d): Each term is $\dfrac1k-\dfrac1{k+1}$, so the sum telescopes to $1-\dfrac1{10}=\dfrac9{10}$.

Revise: Algebra (Class 11 notes)

1. a$24$
2. b$48$
3. c$6$
4. d$12$

Answer (d): Treat the pair as one unit: 4 units in a circle give $3!=6$ ways, and the pair can swap in 2 ways, so $6\cdot2=12$.

Revise: Algebra (Class 11 notes)

1. a$120$
2. b$160$
3. c$80$
4. d$20$

Answer (b): The term is ${}^6C_3(2x)^3=20\cdot8x^3=160x^3$.

Revise: Algebra (Class 11 notes)

1. a$2$
2. b$3$
3. c$2.5$
4. d$4$

Answer (a): Area $=\tfrac12\cdot5\cdot12=30$ and $s=15$, so $r=\Delta/s=30/15=2$.

Revise: Trigonometry (Class 11 notes)

1. a$48$
2. b$14$
3. c$24$
4. d$12$

Answer (c): The intercepts are $x=6$ and $y=8$, so the area is $\tfrac12\cdot6\cdot8=24$.

Revise: Analytic geometry (Class 11 notes)

1. a$\sqrt{13}$
2. b$13$
3. c$\sqrt{10}$
4. d$5$

Answer (a): The circle through the origin is $x^2+y^2+2gx+2fy=0$. Using the other points, $g=-2$ and $f=-3$, so $r=\sqrt{g^2+f^2}=\sqrt{13}$.

Revise: Analytic geometry (Class 11 notes)

1. a$3$
2. b$2\sqrt7$
3. c$9$
4. d$6$

Answer (d): $c^2=a^2-b^2=16-7=9$, so $c=3$ and the foci are $2c=6$ apart.

Revise: Analytic geometry (Class 12 notes)

1. a$1$
2. b$\dfrac13$
3. c$3$
4. d$\dfrac23$

Answer (b): Distance $=\dfrac{|2+2-6+3|}{\sqrt{4+1+4}}=\dfrac{1}{3}$.

Revise: Analytic geometry (Class 12 notes)

1. a$3$
2. b$\dfrac23$
3. c$6$
4. d$9$

Answer (c): Separating, $\ln y=\ln x+\ln c$, so $y=cx$. With $y(1)=2$, $c=2$ and $y(3)=6$.

Revise: Calculus (Class 12 notes)

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

Answer (b): Taking logs, $y'=x^x(\ln x+1)$. At $x=1$ this is $1\cdot(0+1)=1$.

Revise: Calculus (Class 11 notes)

1. a$\dfrac12$
2. b$1$
3. c$\dfrac14$
4. d$2$

Answer (a): $f'(x)=\dfrac{1-x^2}{(1+x^2)^2}=0$ gives $x=1$ (a maximum), and $f(1)=\dfrac12$.

Revise: Calculus (Class 12 notes)

1. a$0.4$
2. b$0.2$
3. c$0.5$
4. d$0.6$

Answer (c): $P(A\cap B)=0.5+0.4-0.7=0.2$, so $P(A\mid B)=0.2/0.4=0.5$.

Revise: Statistics and Probability (Class 11 notes)

1. a$MLT^{-2}$
2. b$ML^{2}T^{-2}$
3. c$ML^{-2}T^{-2}$
4. d$ML^{-1}T^{-2}$

Answer (d): Pressure is force per area: $MLT^{-2}/L^2=ML^{-1}T^{-2}$.

Revise: Physical Quantities (Class 11 notes)

1. a20 m
2. b80 m
3. c100 m
4. d40 m

Answer (d): $v^2=u^2-2as$ gives $0=400-10s$, so $s=40$ m.

Revise: Kinematics (Class 11 notes)

1. a3 m/s
2. b6 m/s
3. c15 m/s
4. d2 m/s

Answer (b): Impulse equals change in momentum: $Ft=mv$, so $v=10\cdot3/5=6$ m/s.

Revise: Dynamics (Class 11 notes)

1. a1000 W
2. b100 W
3. c10000 W
4. d10 W

Answer (a): $P=mgh/t=100\cdot10\cdot10/10=1000$ W.

Revise: Work, Energy and Power (Class 11 notes)

1. a9 m/s
2. b4.5 m/s
3. c3 m/s
4. d1.5 m/s

Answer (c): At the top, $mg=mv^2/r$, so $v=\sqrt{gr}=\sqrt{9}=3$ m/s.

Revise: Circular Motion (Class 11 notes)

1. a600 N
2. b10 N
3. c360 N
4. d100 N

Answer (d): Weight $=m g_{moon}=60\cdot\dfrac{10}{6}=100$ N. The mass stays 60 kg.

Revise: Gravitation (Class 11 notes)

1. a3 N m
2. b30 N m
3. c0.3 N m
4. d13 N m

Answer (a): $\tau=Fr=10\cdot0.3=3$ N m.

Revise: Rotational Dynamic (Class 12 notes)

1. a$2\times10^{5}$ Pa
2. b$2\times10^{11}$ Pa
3. c$2\times10^{8}$ Pa
4. d$2\times10^{-5}$ Pa

Answer (b): $Y=\text{stress}/\text{strain}=2\times10^8/10^{-3}=2\times10^{11}$ Pa.

Revise: Elasticity (Class 11 notes)

1. a$\dfrac{2T}{r}$
2. b$\dfrac{T}{r}$
3. c$\dfrac{4T}{r}$
4. d$\dfrac{8T}{r}$

Answer (c): A soap bubble has two surfaces, so the excess pressure is twice the $2T/r$ of a single drop: $4T/r$.

Revise: Fluid Statics (Class 12 notes)

1. a$3.36\times10^{5}$ J
2. b$6.72\times10^{5}$ J
3. c$1.68\times10^{4}$ J
4. d$1.68\times10^{5}$ J

Answer (d): $Q=mL=0.5\cdot3.36\times10^5=1.68\times10^5$ J.

Revise: Quantity of Heat (Class 11 notes)

1. a0.5 atm
2. b2 atm
3. c1 atm
4. d4 atm

Answer (b): At constant volume $P\propto T$, so the pressure doubles to 2 atm.

Revise: Heat and Temperature (Class 11 notes)

1. a$2\times10^{8}$ m/s
2. b$4.5\times10^{8}$ m/s
3. c$1.5\times10^{8}$ m/s
4. d$3\times10^{8}$ m/s

Answer (a): $v=c/n=3\times10^8/1.5=2\times10^8$ m/s.

Revise: Refraction at Plane Surface (Class 11 notes)

1. a516 Hz
2. b258 Hz
3. c4 Hz
4. d2 Hz

Answer (c): Beat frequency is the difference of the two frequencies: $260-256=4$ Hz.

Revise: Acoustic Phenomena (Class 12 notes)

1. a$3\ \Omega$
2. b$16\ \Omega$
3. c$8\ \Omega$
4. d$6\ \Omega$

Answer (a): $R=\dfrac{4\cdot12}{4+12}=3\ \Omega$.

Revise: Electrical Circuit One (Class 12 notes)

1. amaximum
2. b$evB$
3. chalf of the maximum
4. dzero

Answer (d): The force is $F=evB\sin\theta$. For motion parallel to the field $\theta=0$, so the force is zero.

Revise: Magnetic Field (Class 12 notes)

1. a$^{234}_{92}$U
2. b$^{234}_{90}$Th
3. c$^{238}_{90}$Th
4. d$^{236}_{90}$Th

Answer (b): An alpha particle ($^4_2$He) removes 4 from the mass number and 2 from the atomic number: $^{234}_{90}$Th.

Revise: Radioactivity and Nuclear Reaction (Class 12 notes)

1. a10 m
2. b20 m
3. c5 m
4. d2.5 m

Answer (c): $H=\dfrac{u^2\sin^2\theta}{2g}=\dfrac{400\cdot0.25}{20}=5$ m.

Revise: Kinematics (Class 11 notes)

1. a12 J
2. b36 J
3. c6 J
4. d24 J

Answer (d): Momentum: $2\cdot6=6v$ gives $v=2$ m/s. KE before $=36$ J, after $=\tfrac12\cdot6\cdot4=12$ J, so the loss is 24 J.

Revise: Work, Energy and Power (Class 11 notes)

1. a$4\ \text{m/s}^2$
2. b$3\ \text{m/s}^2$
3. c$2\ \text{m/s}^2$
4. d$1\ \text{m/s}^2$

Answer (c): Friction $=\mu mg=0.2\cdot50=10$ N. Net force $=20-10=10$ N, so $a=10/5=2\ \text{m/s}^2$.

Revise: Dynamics (Class 11 notes)

1. a11200 m/s
2. b8000 m/s
3. c800 m/s
4. d6400 m/s

Answer (b): $v=\sqrt{gR}=\sqrt{10\cdot6.4\times10^6}=\sqrt{6.4\times10^7}=8000$ m/s.

Revise: Gravitation (Class 11 notes)

1. a$0.8\ \text{kg m}^2/\text{s}$
2. b$0.4\ \text{kg m}^2/\text{s}$
3. c$8\ \text{kg m}^2/\text{s}$
4. d$0.08\ \text{kg m}^2/\text{s}$

Answer (a): $I=\tfrac25MR^2=\tfrac25\cdot5\cdot0.04=0.08$ kg m$^2$, so $L=I\omega=0.08\cdot10=0.8$.

Revise: Rotational Dynamic (Class 12 notes)

1. a$50^\circ$C
2. b$30^\circ$C
3. c$40^\circ$C
4. d$35^\circ$C

Answer (d): Heat lost equals heat gained: $100(80-T)=300(T-20)$, so $400T=14000$ and $T=35^\circ$C.

Revise: Quantity of Heat (Class 11 notes)

1. a12 cm
2. b50 cm
3. c60 cm
4. d10 cm

Answer (c): $\dfrac1f=\dfrac1{20}-\dfrac1{30}=\dfrac1{60}$, so $f=60$ cm.

Revise: Lenses (Class 11 notes)

1. a480 Hz
2. b500 Hz
3. c320 Hz
4. d400 Hz

Answer (b): $f'=f\dfrac{v}{v-v_s}=400\cdot\dfrac{340}{272}=500$ Hz.

Revise: Acoustic Phenomena (Class 12 notes)

1. a0.05 J
2. b0.5 J
3. c0.1 J
4. d0.005 J

Answer (a): $U=\tfrac12CV^2=\tfrac12\cdot10^{-5}\cdot10^4=0.05$ J.

Revise: Capacitor (Class 11 notes)

1. a$1\ \Omega$
2. b$0.5\ \Omega$
3. c$0.25\ \Omega$
4. d$2\ \Omega$

Answer (b): Terminal voltage $V=E-Ir$, so $r=(E-V)/I$. The external resistor gives $V=IR=2\cdot4=8$ V, hence $r=(9-8)/2=0.5\ \Omega$.

Revise: Electrical Circuit One (Class 12 notes)

1. a$4\pi\times10^{-4}$ T
2. b$8\pi\times10^{-4}$ T
3. c$8\pi\times10^{-7}$ T
4. d$2\pi\times10^{-3}$ T

Answer (b): $B=\mu_0nI=4\pi\times10^{-7}\cdot1000\cdot2=8\pi\times10^{-4}$ T, about $2.5\times10^{-3}$ T.

Revise: Magnetic Field (Class 12 notes)

1. a10.2 eV
2. b12.09 eV
3. c1.89 eV
4. d0.66 eV

Answer (c): $\Delta E=13.6\left(\dfrac14-\dfrac19\right)=13.6\cdot\dfrac5{36}\approx1.89$ eV.

Revise: Quantization of Energy (Class 12 notes)

1. a98 g
2. b196 g
3. c24.5 g
4. d49 g

Answer (d): Mass $=n\times M=0.5\times98=49$ g.

Revise: Stoichiometry (Class 11 notes)

1. a$\ce{Na+}$
2. b$\ce{Cl-}$
3. c$\ce{K+}$
4. d$\ce{Ca^2+}$

Answer (a): Neon has 10 electrons. $\ce{Na+}$ has $11-1=10$, while $\ce{Cl-}$, $\ce{K+}$ and $\ce{Ca^2+}$ each have 18.

Revise: Atomic Structure (Class 11 notes)

1. achlorine
2. bfluorine
3. coxygen
4. dnitrogen

Answer (b): Electronegativity increases across a period and up a group, so small fluorine is highest (4.0 on the Pauling scale).

Revise: Classification of Elements (Class 11 notes)

1. a$sp^3$
2. b$sp$
3. c$sp^2$
4. d$sp^3d$

Answer (c): Each carbon forms three sigma bonds and one pi bond, which needs three hybrid orbitals: $sp^2$.

Revise: Chemical Bonding (Class 11 notes)

1. a22.4 L
2. b11.2 L
3. c67.2 L
4. d44.8 L

Answer (d): One mole occupies 22.4 L at STP, so 2 mol occupy $2\times22.4=44.8$ L.

Revise: States of Matter (Class 11 notes)

1. a$\ce{NH2-}$
2. b$\ce{NH4+}$
3. c$\ce{OH-}$
4. d$\ce{N2H4}$

Answer (b): A conjugate acid forms when the base gains a proton: $\ce{NH3 + H+ -> NH4+}$.

Revise: Ionic Equilibrium (Class 12 notes)

1. a3
2. b10
3. c12
4. d11

Answer (d): $[\ce{OH-}]=10^{-3}$, so pOH $=3$ and pH $=14-3=11$.

Revise: Ionic Equilibrium (Class 12 notes)

1. a$+6$
2. b$+5$
3. c$+7$
4. d$+4$

Answer (c): $+1+x+4(-2)=0$ gives $x=+7$.

Revise: Oxidation and Reduction (Class 11 notes)

1. asodium
2. bchlorine
3. chydrogen
4. dsodium hydroxide

Answer (a): $\ce{Na+}$ ions are reduced at the cathode to sodium metal, while chlorine gas forms at the anode.

Revise: Electrochemistry (Class 12 notes)

1. afreezing of water
2. bcondensation of steam
3. cformation of crystals from solution
4. dmelting of ice

Answer (d): Melting turns an ordered solid into a more disordered liquid, so $\Delta S>0$. The other processes increase order.

Revise: Chemical Thermodynamic (Class 12 notes)

1. a$\ce{CaCl2}$
2. b$\ce{Ca(OH)2}$
3. c$\ce{CaOCl2}$
4. d$\ce{CaCO3}$

Answer (c): Bleaching powder is calcium oxychloride, $\ce{CaOCl2}$, made by passing chlorine over slaked lime.

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

1. acopper and zinc
2. bcopper and tin
3. ciron and carbon
4. dcopper and nickel

Answer (a): Brass is copper with zinc. Copper with tin is bronze.

Revise: Chemistry of Metals (Class 11 notes)

1. amethanoic acid
2. bethanoic acid
3. cpropanoic acid
4. dethanal

Answer (b): It has two carbons and a carboxylic acid group, so the name is ethanoic acid.

Revise: Fundamental Principles (Class 11 notes)

1. adiethyl ether
2. bethane
3. cethene
4. dethyne

Answer (c): At $170^\circ$C the acid dehydrates ethanol by removing water: $\ce{C2H5OH -> C2H4 + H2O}$. Ether forms at about $140^\circ$C.

Revise: Alcohols (Class 12 notes)

1. a1.12 L
2. b2.24 L
3. c0.56 L
4. d11.2 L

Answer (a): $n=5.3/106=0.05$ mol. One mole of $\ce{Na2CO3}$ gives one mole of $\ce{CO2}$, so $V=0.05\times22.4=1.12$ L.

Revise: Stoichiometry (Class 11 notes)

1. a0.1 M
2. b0.2 M
3. c0.4 M
4. d2.0 M

Answer (b): Moles $=4/40=0.1$. Molarity $=0.1/0.5\ \text{L}=0.2$ M.

Revise: Volumetric Analysis (Class 12 notes)

1. a1
2. b10
3. c0.1
4. d100

Answer (b): $K_c=\dfrac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]}=\dfrac{0.04}{0.04\cdot0.1}=10$.

Revise: Chemical Equilibrium (Class 11 notes)

1. a4
2. b8
3. c6
4. d16

Answer (b): The overall order is 3, so the rate increases by $2^3=8$ times.

Revise: Chemical Kinetics (Class 12 notes)

1. a12
2. b1
3. c13
4. d7

Answer (c): HCl gives 0.01 mol and NaOH gives 0.03 mol, so 0.02 mol of $\ce{OH-}$ remains in 0.2 L: 0.1 M. pOH $=1$ and pH $=13$.

Revise: Ionic Equilibrium (Class 12 notes)

1. a1930 s
2. b482.5 s
3. c965 s
4. d96.5 s

Answer (c): Moles of silver $=2.16/108=0.02$, and each needs one electron, so $Q=0.02\times96500=1930$ C. Then $t=Q/I=1930/2=965$ s.

Revise: Electrochemistry (Class 12 notes)

1. aethyl methanoate
2. bdiethyl ether
3. cethanal
4. dethyl ethanoate

Answer (d): Excess oxidant converts ethanol to ethanoic acid (X). Ethanoic acid and ethanol undergo esterification to give ethyl ethanoate (Y).

Revise: Carboxylic Acids (Class 12 notes)

1. a6 sigma and 2 pi
2. b7 sigma and 1 pi
3. c5 sigma and 3 pi
4. d6 sigma and 3 pi

Answer (a): Four C-H bonds and two C-C bonds (one single, one sigma of the triple bond) give 6 sigma. The triple bond also has 2 pi bonds.

Revise: Chemical Bonding (Class 11 notes)

1. awill complete
2. bwill have completed
3. chas completed
4. dwould have completed

Answer (b): 'By the end of next year' points to an action finished before a future time, which takes the future perfect: 'will have completed'.

Revise: Grammar and Usage (Class 11 notes)

1. aare
2. bwere
3. cis
4. dhave been

Answer (c): 'News' is an uncountable noun that takes a singular verb, so 'is' is correct.

Revise: Grammar and Usage (Class 11 notes)

1. aa, an
2. ban, an
3. ca, a
4. dan, a

Answer (d): 'Umbrella' starts with a vowel sound, so it takes 'an'. 'Hat' starts with a consonant sound, so it takes 'a'.

Revise: Grammar and Usage (Class 11 notes)

1. asince
2. bfor
3. cfrom
4. dby

Answer (a): 'Since' is used with a point in time (Monday). 'For' is used with a period of time.

Revise: Grammar and Usage (Class 11 notes)

1. aneedn't
2. bmustn't
3. cmight
4. dought

Answer (b): 'Mustn't' expresses prohibition. 'Needn't' means it is not necessary.

Revise: Grammar and Usage (Class 11 notes)

1. aHe is given a prize by the teacher.
2. bHe had given a prize by the teacher.
3. cHe gave a prize by the teacher.
4. dHe was given a prize by the teacher.

Answer (d): The active verb is simple past, so the passive uses 'was' plus the past participle: 'was given'.

Revise: Grammar and Usage (Class 11 notes)

1. aShe asked me where do I live.
2. bShe told me where I lived.
3. cShe asked me where I lived.
4. dShe asked me where you lived.

Answer (c): A question is reported with 'asked', uses statement word order and shifts the tense back: 'where I lived'.

Revise: Grammar and Usage (Class 11 notes)

1. afrank
2. bsecretive
3. cpolite
4. dcautious

Answer (a): 'Candid' means open and honest, which is 'frank'.

Revise: Vocabulary and Word Formation (Class 11 notes)

1. abibliographer
2. bbibliophile
3. clibrarian
4. dphilatelist

Answer (b): 'Biblio' means book and 'phile' means lover. A philatelist collects stamps.

Revise: Vocabulary and Word Formation (Class 11 notes)

1. aphotograph
2. bbeautiful
3. cfamily
4. dphotography

Answer (d): Photography is stressed as pho-TOG-ra-phy. The others are stressed on the first syllable.

Revise: Grammar and Usage (Class 11 notes)

1. aneed very little labour
2. bstop rain completely
3. creduce soil erosion
4. dreplace rice with other crops

Answer (c): The passage says the steps slow the rain water so less soil is washed away, which is reduced erosion.

Revise: Reading Comprehension (Class 11 notes)

1. aSunita always forgets her umbrella.
2. bIt rained heavily by noon.
3. cIt did not rain that morning.
4. dSunita did not go out.

Answer (c): The dry roads at noon and her pleasure at the choice show that rain did not come.

Revise: Reading Comprehension (Class 11 notes)

1. aThe number of applicants has increased this year.
2. bThe number of applicants have increased this year.
3. cA number of applicant has increased this year.
4. dThe number of applicants are increasing this year.

Answer (a): 'The number of' takes a singular verb, so 'has increased' is correct.

Revise: Grammar and Usage (Class 11 notes)

1. aam
2. bwere
3. cwould be
4. dhad been

Answer (b): A wish about an unreal present uses the past subjunctive, so 'were' is used for all persons.

Revise: Grammar and Usage (Class 11 notes)

1. aa prayer said in secret
2. ba gift that is wrapped well
3. ca person who hides their feelings
4. dsomething that seems bad at first but turns out good

Answer (d): The phrase describes an apparent misfortune that later brings a benefit.

Revise: Vocabulary and Word Formation (Class 11 notes)