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PHYSICS TEAM
IIT-JEE 2012 PAPER-I WITH SOLUTION [CODE - 7]
Supported by a pool of 20 Faculty Members
Nitin VijayEx-Sr. Faculty Bansal Classes
B.Tech IT-BHU
Amit Verma Ex-Sr. Faculty Bansal Classes
M.Sc. Gold Medalist
Nirbhay Pandey Ex-Faculty Bansal Classes
Ravi Vijay Sr. Faculty Physics
New
Kumar Sourabh Ex-Sr. Faculty Vibrant Academy
Ex-HOD Career point
New
Bhupendra Sudan Ex-Sr. Faculty Vibrant Academy
Ex-Sr. Faculty Bansal Classes
SECTION – A
Single Correct
1. Young's double slit experiment is carried out by
using green, red and blue light, one color at a
time. The fringe widths recorded are βG,βR andβB, respectively. Then
(A) βG > βB > βR (B) βB > βG > βR
(C) βR > βB > βG (D) βR > βG > βB
Sol. Fringe width
d
Pλ=β
as (λR > λ
G > λ
b)
so βR > β
G > β
B
2. Two large vertical and parallel metal plates having
a separation of 1 cm are connected to a DC
voltage source of potential difference X. A proton
is released at rest midway between the two plates.
It is found to move at 45° to the vertical JUSTafter release. Then X is nearly
(A) 1 × 10–5 V (B) 1 × 10–7 V(C) 1 × 10–9 V (D) 1 × 10–10 V
Sol. Fx = Fy ⇒ qE = mg
qv
d = mg
45°
Fx
Fy
45°
⇒ V = mgd
q= 10–9 v
3. A mixture of 2 moles of helium gas (atomic mass= 4 amu) and 1 mole of argon gas (atomic mass
= 40 amu) is kept at 300 K in a container. The
ratio of the rms speeds ( )( )
rms
rms
v helium
v argon
is
(A) 0.32 (B) 0.45 (C) 2.24 (D) 3.16
Sol.M
RT3vrms =
He Ar
Ar He
v M 4010 3.16
v M 4= = = =
4. A small block is connected to one end of a
massless spring of un-stretched length 4.9 m.
The other end of the spring (see the figure) isfixed. The system lies on a horizontal frictionless
surface. The block is stretched by 0.2 m and
released from rest at t = 0. It then executessimple harmonic motion with angular frequency
ω = 3
πrad/s .Simultaneously at t = 0, a small
pebble is projected with speed ν from point P atan angle of 45° as shown in the figure. Point P is
at a horizontal distance of 10 m from O. If the
pebble hits the block at t = 1s, the value of ν is(take g = 10 m/s2)
45°
10m PxO
Z
(A) 50 m/s (B) 51 m/s
(C) 52 m/s (D) 53m/s
Sol. T = 1 S = g
sinu2 θ
u = g
2sinθ =
10
12
2× = 5025 =
5. Three very large plates of same area are keptparallel and close to each other. They areconsidered as ideal black surfaces and have veryhigh thermal conductivity. The first and thirdplates are maintained at temperatures 2T and3T respectively. The temperature of the middle(i.e. second) plate under steady state conditionis
(A)
1
465
2
T (B)
1
497
4
T
(C)
1
497
2
T (D) ( )1
497 T
PART - I [PHYSICS]
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Sol.
2T T = ?0 3T
for middle flate (in unit time)Heat absorbed/ Area = Heat emitted/Areaσ (2T)4 + σ(3T)4 = σ(T
0)4×2
T0 = T
2
974/1
6. A thin uniform rod, pivoted at O is rotating in thehorizontal plane with constant angular speed ω,as shown in the figure. At time t = 0, small insectstarts from O and moves with constant speed νwith respect to the rod towards the other end.it reaches the end of the rod at t = T and stops.The angular speed of the system remains ω
throughout. The magnitude of the torque ( )τ
on
the system about O, as a function of time isbest represented by which plot?
O
Z
(A)
0 tT
(B)
0 tT
(C)
0 tT
(D)
0 tT
Sol. τ = dt
dL = )I(
dt
dω
τ = dI
dtω = ω )II(
dt
dmrod +
as Irod
= com ⇒ τ = ( )insec t
wdI
dt
= )mr(dt
d 2ω = mω dr
2rdt
= 2m rωv
= 2m(vt)ωv ⇒ τ ∝ t
7. In the determination of young's modulus
4MLgY
d2
=
π by using Searle's method, a wire of
length L= 2 m and diameter d = 0.5 mm is used.For a load M = 2.5 kg, an extension l = 0.25 mmin the length of the wire is observed. Quantitiesd and l are measured using a screw gauge and amicrometer, respectively. The have the samepitch of 0.5 mm. The number of divisions on theircircular scale is 100. The contributions to themaximum probable error of the Y measurement(A) due to the errors in the measurements of dand l are the same(B) due to the error in the measurement of d istwice that due to the error in the measurementof l.(C) due to the error in the measurement of l istwice that due to the error in the measurementof d.(D) due to the error in the measurement of d isfour times that due to the error in themeasurement of l.
Sol. 2d
MgL4Y
π=
d
d2
L
L
g
g
M
M
Y
Y ∆+
∆+
∆+
∆+
∆=
∆
d∆=∆ (same instrment)
so d
d2∆=
∆
hence both contribute same in
error (Hence)8. Consider a thin spherical shell of radius R with its
centre at the origin, carrying uniform positive
surface charge density. The variation of the
magnitude of the electric field ( )E r
and the
electric potential V(r) with the distance r fromthe centre, is best represented by which graph?
(A)
( )rE
V(r)
0 R r
(B)
( )rE
V(r)
0 R r
(C)
( )rE
V(r)
0 R r
(D)
( )rE
V(r)
0 R r
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Sol. Inside shell E = 0& V = const.and outside shell
E ∝ 2r
1 & v ∝
r
1
9. A bi-convex lens is formed with two thin plano-
convex lenses as shown in the figure. Refractiveindex n of the first lens is 1.5 and that of the
second lens is 1.2. Both the curved surfaces areof the same radius of curvature R = 14 cm. For
this bi-convex lens, for an object distance of 40
cm, the image distance will be
n=1.5 n=1.2
R =14cm(A) –280.0 cm (B) 40.0 cm
(C) 21.5 cm (D) 13.3 cmSol. Left lens
1
1 1 1 1(1.5 1) –
f 14 28
= − =
∞
Right lens
2
1 1 1 1(1.2 1)
f 14 70
= − + =
∞
1 2
1 1 1 1 1 1
feq f f 28 70 20= + = + =
feq = 20 cm
1 1 1
v u f− = ⇒
20
1
40
1
v
1=+ ⇒ v = 40 cm
10. A small mass m is attached to a massless string
whose other end is fixed at P as shown in the
figure. The mass is undergoing circular motion inthe x-y plane with centre at O and constant
angular speed ω.If the angular momentum of thesystem, calculated about O and P are denoted
by 0L
and PL
respectively, then.
m
P
O
z
(A) 0L
and PL
do not vary with time
(B) 0L
varies with time while PL
remains constant
(C) 0L
remains constant while PL
varies with time
(D) 0L
and PL
both vary with time.
Sol. L0 remains cons. in magnitude and direction but
LP changes its direction continously hence L
P is
variable
L0
v x
L (varies direction)P
SECTION – BMultiple Correct
11. A person blows into open-end of a long pipe. Asa result, a high-pressure pulse of air travels downthe pipe. When this pulse reaches the other endof the pipe,(A) a high-pressure pulse starts travelling up thepipe, if the other end of the pipe is open.(B) a low-pressure pulse starts travelling up thepipe, if the other end of the pipe is open.(C) a low-pressure pulse starts travelling up thepipe, if the other end of the pipe is closed.(D) a high-pressure pulse starts travelling up thepipe, if the other end of the pipe is closed.
Sol. At closed end prssure doesn't change phase andat open end the phase is reversed.
12. A small block of mass of 0.1 kg lies on a fixed
inclined plane PQ which makes an angle θ with
the horizontal. A horizontal force of 1 N acts onthe block through its center of mass as shown inthe figure. The block remains stationary if (takeg = 10 m/s2)
(A) θ = 45o
(B) θ > 45o and a frictional force acts on the
block towards P.
(C) θ > 45o and a frictional force acts on the
block towards Q.
(D) θ < 45o and a frictional force acts on the
block towards Q.
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Sol.
Q
P
at θ = 45° No friction is requiredat θ > 45° , sin θ > cos 45°hence friction is required towards Qat θ < 45°, cos θ > sin θhence friction acts towards P
13. A cubical region of side a has its centre at theorigin. It encloses three fixed point charges, -qat (0,-a/4,0), +3q at (0,0,0) and -q at (0,+a/4,0). Choose the correct option(s).
(A) The net electric flux crossing the plane x =+a/2 is equal to the net electric flux crossing
the plane x = -a/2.(B) The net electric flux crossing the plane y =+a/2 is equal to the net electric flux crossingthe plane y = -a/2.(C) The net electric flux crossing the entire region
is 0
q
ε
(D) The net electric flux crossing the plane z =+a/2 is equal to the net electric flux crossingthe plane x = +a/2.
Sol. For the charge distribution
for x = +a/2 and x = -a/2flux is symmetric & same
also flux is same throughy = +a/2 and y = -a/2
for net flux 0
00
in /qqqq3q
∈=∈
−−=
∈=φ
z = +a/2 and x = +a/2 aresimilar so same flux.
14. For the resistance network shown in the figure,
choose the correct option(s).
Ω
Ω
Ω
Ω Ω
ΩΩ
Ω
2I
1I
(A) The current through PQ is zero
(B) l1 = 3A(C) The potential at S is less than that at Q(D) l2 = 2A
Sol. By symmetry the 1Ω resistances don't get any
current so circuit reduces to
126
126
12I1
+
×=
= 3A
15. Consider the motion of a positive point charge in
a region where there are simultaneous uniform
electric and magnetic fields E= E0 j and B
= B0
j . At time t = 0. At time t = 0, this charge has
velocity v in the x - y plane, making and angle
θ with the x-axis. Which of the following option(s)
is(are) correct for time t > 0 ?
( A ) I f θ = 0o, the charge moves in a circular
path in the x-z plane.
(B) If θ = 0o, the charge undergoes helical motion
with constant pitch along the y-axis.
(C) If If θ = 10o, the charge undergoes helical
motion with its pitch increasing with time, along
the y-axis.
(D) If If θ = 90o, the charge undergoes linear
but acclerated motion along the y-axis.Sol. (i) The path of the particle will be helical with
increasing pitch in x-z plane .
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B0
YE0
x
(ii) if Q = 0 no effect due to B. but due to E there is a
force along E which Accelerates if along +y direction so
speed will increase in +y direction
SECTION – CInteger Answer Type
16. A proton is fired from very far away towards a
nucleus with charge Q = 120 e, where e is theelectronic charge. It makes a closest approach
of 10 fm to the nucleus. The de Broglie wave-
length (in units of fm) of the proton at its startis: (take the proton mass, mp = (5/3) x 10
-27
kg; h/e = 4.2 x 10-15
J.s/C; o4
1
πε= 9 X 10
9 m/F;
1 fm = 10-15
m)
Sol. KE = PE ⇒r
qkqmv
2
1 212 =
15
92
1010
e120e109
m2
p−×
×××=
& p
h=λ (from debroglie)
solving λ = 7 x 10-15 = 7 fm
17. A lamina is made by removing a small disc ofdiameter 2R from a bigger disc of uniform mass
density and radius 2R, as shown in the figure.The moment of inertia of this lamina about axes
passing through O and P is Io and IP, respec-tively. Both these axes are perpendicular to the
plane of the lamina. The ratio o
P
I
I to the nearest
integer is
Sol. 0
M/4
MP
Let σ be the density off disc.
∴ 22 R4
M
)R2(
M
π=
π=σ
Here M → Mass of disc without cavity
∴ Mass of cavity = σ x πR2 = M/4πR2 x πR2 = M/4
Io = MI of disc with non cavity
- MI of cavity (About O)
+
−
= 222
o R4
MR
4
M
2
1)R2(M
2
1I
222
o R4
MMR
8
1
2
MR4I −−= =
8
MR3
8
MR16 22
−
= 8
MR13 2
Ip = M.I. of Disc without
cavity about P - M.I. of 0 2R
5 R
R
cavity(aboutP)
= ( )
+
−
+
2222 R5
4
MR
4
M
2
1)R2(M)R2(M
2
1
8
MR37I
2
P =
38.213
37
I
I
o
P ≈≈=∴
18. An infinitely long solid cylinder of radius R has a
uniform volume charge density ρ . It has a spheri-
cal cavity of radius R/2 with its centre on the
axis of the cylinder, as shown in the figure. The
magnitude of the electric field at the point P,which is at a distance 2R from the axis of the
cylinder, is given by the expression o16
R23
κε
ρ. The
value of κ is
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Sol. For cylinder λ = 2M Rπ ρ
=
= πR2ρ
at E = Ecylinder
+ Esphere
λ = ρ × πR2 = ( )
2
2
2k kQ
2R 2R
λ+
= ( )
3
2
2
4 Rk
2k R 3 8–2R 2R
× π × ρπ ρ
= π kRρ 1
1–24
= 0 0
R 23 23R
4 24 16 6
π ρ ρ× =
πε ε ×
so k = 6
19. A cylindrical cavity of diameter a exists inside a
cylinder of diamter 2a as shown in the figure.Both the cylinder and the cavity are infinitely
long. A uniform current density J flows along thelength. If the magnitude of the magnetic field at
the point P is given by 012
Nµ aJ , then the value
of N is
Sol. B = B1( 2a dimeter) – B
2(a diameter)
a2
)aJ( 2
π
π×µ –
2
a32
)4/aJ( 2
×π
π×µ
= aJ.12
5µ = N = 5
20. A circular wire loop of radius R is placed in the x-y plane centered at the origin O. A square loop
of side a(a<<R) having two turns is placed with
its center at z = 3 R along the axis of the
circular wire loop, as shown in figure. The plane
of the square loop makes an angle of 45o with
respect to the z-axis. If the mutual inductance
between the loops is given by R2
a2/p
20µ
, then value
of p is
R3
o45
Sol. At the position of square loop
B =
20
2 2 3/2
iR
2(R 2R )
µ
+ =
0i
16R
µ
α = BA cos θ
= 20i 1
2a16R 2
µ× ×
=
20 ia
8 2R
µ
M = i
θ =
20a
8 2R
µ =
20
7/2
a
2 R
µ
P =7
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Exp. 5 yrs
New
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CHEMISTRY TEAMSupported by a pool of 28 Faculty Members
PART - II [CHEMISTRY]
SECTION – A
Single Correct
21.The colour of light absorbed by an aqueous solutionof CuSO4 is
(A) orange-red (B) blue-green
(C) yellow (D) violet
Sol. A
B
G
V
R
O Y
munshel wheel
⇒ Since CuSO4 appears blue-green it would have
absorbed organe-red.
22.Which ordering of compounds is according to thedecreasing order of the oxidation state of nitrogen ?
(A) HNO3, NO, NH4Cl, N2
(B) HNO3, NO, N2, NH4Cl
(C) HNO3, NH4Cl, NO, N2
(D) NO, HNO3, NH4Cl, N2
Sol. B
HNO3 O.S. of N = +5
NO O.S. of N = +2
N2 O.S. of N = 0
NH4Cl O.S. of N = –3
Decreasing order
HNO3, NO, N2, NH4Cl
23.The kinetic energy of an electron in the second Bohrorbit of a hydrogen atom is [a0 is Bohr radius]
(A) 2
0
2
2
ma4
h
π (B) 2
0
2
2
ma16
h
π
(C) 2
0
2
2
ma32
h
π (D) 2
0
2
2
ma64
h
π
Sol. C
mvr = π2
nh ⇒ m2v2r2 = 2
22
4
hn
π
21mv2 =
2 2 2
2 2 2
0
n h Z
4 2 m(r n )
×
π × × = 2
0
2
2
ma32
h
π
24.The number of aldol reaction(s) that occurs in thegiven transformation is
CH3CHO + 4HCHO conc.aq. NaOH
OHOH
OHOH
(A) 1 (B) 2 (C) 3 (D) 4
Sol. C
OH–
–H O2CH CHO3 CH – C – H2
H–C–H
O
O
CH – CH – C – H2 2
O–
H OH
O
CH –CH–C–H2
OH
H
O
2 times aldol
CH2–C – C – H
CH2OH
CH2OH
CH2OH
O
OH,HCHO
CH2HO
HO
HO–CH –C–CH –OH2 2
H – C – O
+
Ocannizaro
25.For one mole of a van der Waals gas when b = 0 andT = 300 K, the PV vs. l/V plot is shown below. The valueof the van der Waals constant a
(atm. liter2 mol–2) is
20.121.6
23.1
24.6
pV (lite
r-atm
mol
)–1
0 2.0 3.01v (mol liter )
–1
(A) 1.0 (B) 4.5 (C) 1.5 (D) 3.0
Sol. C
Since b = 0
2
2
anP
v
+
(v) = nRT
PV = nRT – 2an
v
∴ slope = – an2 = – an2 = 15.1
a = 1.5
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26.In Allen (C3H4), the type (s) of hybridisation of thecarbon atoms is (are)
(A) sp and sp3 (B) sp and sp2
(C) only sp2 (D) sp2 and sp3
Sol. B
In allene (C3H4)
H C = C = C H 2 2
sp2
spsp2
27.A compound MpXq has cubic close packing (ccp)arrangement of X. Its unit cell structure is shown below.The empirical formula of the compound is
M =
X =
(A) MX (B) MX2 (C) M2X (D) M5X14
Sol. B
X = 8 × 81
+ 6 + 21 = 4
M = 4 × 41
+ 1 = 2
M =
X =
∴ formula of unit cell
= M2X4
& expirical formula of compound = MX2
Ans. – (B)
28.The number of optically active product obtained fromthe complete ozonolysis of the given compound is
CH –CH=CH–C–CH=CH–C–CH=CH–CH33
CH3
CH3
H
H
(A) 0 (B) 1 (C) 2 (D) 4
Sol. A
CH –CH=CH–C–CH=CH–C–CH=CH–CH33
H
CH3
CH3
H
ozo.
2CH3CH=O + O = CH – C – CH = O
CH3
H
+
O = CH – C – CH = O
CH3
H
All products are optical inactive.
29.As per IUPAC nomenclature, the name of the complex[Co(H2O)4(NH3)2]Cl3 is
(A) Tetraaquadiaminecobalt (III) chloride
(B) Tetraaquadiamminecobalt (III) chloride
(C) Diaminetetraaquacobalt (III) chloride
(D) Diamminetetraaquacobalt (III) chloride
Sol. D
[Co(H2O)4(NH3)2]Cl3Diamminetetraaquacobalt (III) chloride
30.The carboxy functional group (–COOH) is present in
(A) picric acid (B) barbituric acid
(C) ascorbic acid (D) aspirin
Sol. D
(A)Picric acid
OH
NO2NO2
NO2
(B) Barbituric acid
OO
O
NH NH
(C) Ascorbic acid
O O
OHHO
HOHO
(D) Aspirin
O – C – CH3
O
C – OH
O
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SECTION – A
Multiple Correct
31.For an ideal gas, consider only P-V work in goingfrom a initial state X to the final state Z. The final stateZ can be reached by either of the two paths shown inthe figure. Which of the following choice (s) is (are)correct ? [take ∆S as change in entropy and w as workdone]
Z
YX
P(a
tmosp
here
)
V(liter)
(A) x z x y y zS S S
→ → →∆ = ∆ + ∆
(B) x z x y y zw w w
→ → →= +
(C) yxzyxww
→→→=
(D) yxzyxSS
→→→∆=∆
Sol. A,C
Explanation → Entropy is a state function andindependent of path taken.
A l s o , W y–z = 0 (Isochoric)
32.Which of the following molecules in pure from is (are)unstable at room temperature
(A) (B)
(C)
O
(D)
O
Sol. B
It is anti-aromatic compound. It dimerise at roomtemperature.
+dimerise room temperature
33.Identify the binary mixture(s) that can be separatedinto individual compounds, by differential extraction, asshown in the given scheme.
Binary mixture containing Compound 1 and compound 2
Compound 1 Compound 2+
+Compound 1 Compound 2
NaOH(aq)
NaHCO (aq)3
(A) C6H5OH and C6H5COOH
(B) C6H5COOH and C6H5CH2OH
(C) C6H5CH2OH and C6H5OH
(D) C6H5CH2OH and C6H5CH2COOH
Sol. B,D
NaOH NaHCO3
Ph–OH
Ph–COOH
Ph–CH –OH2
Ph–CH –COOH2
+ve
–ve
=
=
34.Choose the correct reason(s) for the stability of thelyophobic colloidal particles.
(A) Preferential adsorption of ions on their surface fromthe solution
(B) Preferential adsorption of solvent on their surfacefrom the solution
(C) Attraction between different particles having oppositecharges on their surface
(D) Potential difference between the fixed layer andthe diffused layer of opposite charges around the colloidalparticles.
Sol. A,D
Option A – Due to like charges on DP. Sol. particles arestabiltsed.
Option D – If magnitude of electrokinetic potential ishigh then stability of sol is more .
35.Which of the following hydrogen halides react(s) withAgNO3(aq) to give a precipitate that dissolves inNa2S2O3(aq)?
(A) HCl (B) HF (C) HBr (D) HI
Sol. ACD
SECTION – C
Integer Answer Type
36.An organic compound undergoes first-orderdecomposition. The time taken for its decomposition of1/8 and 1/10 of its initial concentration are t1/8 and t1/10 respectively.
What is the value of 10]t[
]t[
10/1
8/1× ? (take log102 = 0.3)
Sol. 0009
t1/8 = 2.303 1
logk 1 / 8
t1/10 = 2.303 1
logk 1 /10
10]t[
]t[
10/1
8/1× = 3 log 2 × 10 = 3 × 0.3 × 10 = 9
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37.When the following aldohexose exists in its D-configuration, the total number stereoisomers in itspyranose form is
CHO
CH2
CHOH
CHOH
CHOH
CH OH2
Sol. 0008
C–H
CH2
CHOH
CHOH
H–C–OH
CH OH2
O
CH2OH
HH
OHCH2
OH
H
*
*
*
O
Chiral centre is fix for D-configuration
HO
H
Total number of stereoisomers in pyranose form ofD-configuration.
∴ 23 = 8
38.The substituents R1 and R2 for nine peptides arelisted in the table given blow. How many of thesepeptides are positively charged at pH = 7.0?
H N–CH–CO–NH–CH–CO–NH–CH–CO–NH–CH–COO3
R1 R2H H
Peptide
I H HH CH3
CH COOH2 H
CH CONH2 2 (CH )2 4NH2
CH CONH2 2 CH CONH2 2
(CH NH2 2)4 (CH NH2 2)4
(CH NH2 2)4 CH3
CH COOH2 CH CO2 NH2
CH OH2 (CH ) NH2 4 2
II
III
IVV
VI
VIIVIII
IX
R1 R2
Sol. 0004
Peptide no. IV, VI, VIII & IX under goes protonation atpH = 7 because side chain alkyl group are basic.
39.The periodic table consists of 18 groups. An isotopeof copper, on bombardment with protons, undergoes anuclear reaction yielding element X as shown below. Towhich group, element X belongs in the periodic table?
63 1 1 129 1 0 1Cu H 6 n 2 H+ → + α + + X
Sol. 0008
63 1 1 4 1 A29 1 0 2 1 zCu H 6 n 2 H X+ → + α + +
30 = 2 + 2 + z
z = 26 (Fe)
1 2 3 4 5 6 7 8
K Ca Sc Ti V Cr Mn Fe
Sol. Ans. 8
40.29.2% (w/w) HCl stock solution has a density of1.25 g mL
–1. The molecular weight of HCl is 36.5 g
mol–1. The volume (mL) of stock solution required to
prepare a 200 mL solution of 0.4 M HCl is.
Sol. 0008
Molarity = 1000100
251
536
229××
.
.
. = 10 M
From question,
MV = 200 × 0.4
V = 10
40200 .× = 8
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PART - III [MATHEMATICS]
SECTION – A
Single Correct
41. The ellipse E1 : 14
y
9
x 22
=+ is inscribed in a
rectangle R whose sides are parallel to the coordinateaxes. Another ellipse E2 passing through the point(0, 4) circumscribes the rectangle R. The eccentricityof the ellipse E2 is
(A) 2
2(B)
2
3(C)
2
1(D)
4
3
Sol. C
116
y
a
x 2
2
2
=+
116
4
a
92
=+
2
(0,4)
(3,2)
a2 = 12
e2 = 1 – 16
121 − =
4
1
e = 2
1
42. The point P is the intersection of the straightline joining the points Q(2, 3, 5) and R(1, –1, 4) withthe plane 5x – 4y – z = 1. If S is the foot of theperpendicular drawn from the point T(2, 1, 4) to QR,then the length of the line segment PS is
(A) 2
1(B) 2 (C) 2 (D) 2 2
Sol. A
Line )Let(1
5z
4
3y
1
2xλ=
−=
−=
−
dso (λ + 2, 4λ + 3, λ + 5)Line on plane 5x – 4y – z = 15λ + 10 – 16λ – 12 – λ – 5 = 1–12λ = 8
λ = – 2/3 so P
3
13,
3
1,
3
4
for foot of perpendiuclar of T(2, 1, 4)(λ 4λ + 2, λ + 1) . (1, 4, 1) = 0
λ + 16λ + 8 + λ + 1 = 0λ = – 9/18 ⇒ λ = – 1/2
So R(3/2, 1, 9/2), distance a = 1/ 2
43. The integral ∫+ 2/9
2
)xtanx(sec
xsec dx equals
(for some arbitrary constant K)
(A) – 2/11)xtanx(sec
1
+
+− 2)xtanx(sec7
1
11
1+K
(B) 2/11)xtanx(sec
1
+
+− 2)xtanx(sec7
1
11
1+K
(C) – 2/11)xtanx(sec
1
+
++ 2)xtanx(sec7
1
11
1+K
(D) 2/11)xtanx(sec
1
+
++ 2)xtanx(sec7
1
11
1+K
Sol. C
tan x = t ⇒ sec2x dx=dt
2 9/2
dt
(t 1 t )+ +∫ put 2t 1 t z+ + =
2
13/2
1 z 1dz
2 z
+= ∫ 7/2 11/2
1 1 1 1c
7 11z z
−= − +
2
11/2
1 1 1(sec x tanx) c
11 7(sec x tanx)
= − + + +
+
44. Let z be a complex number such that theimaginary part of z is nonzero and a = z2 + z + 1 is real.Then a cannot take the value
(A) –1 (B) 3
1(C)
2
1(D)
4
3
Sol. D
Im(z) ≠ 0 a = z2 + z + 1z = α + iβ β ≠ 0a = α2 – β2 + 2iαβ + α + iβ + 1= α2 – β2 + α + 1 + iβ(2α + 1)α = –1/2 = 3/4 – β2
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45. Let f(x) =
≠
≠π
0x,0
0x,x
cosx2
, x ∈ IR, then f is
(A) differentiable both at x = 0 and at x = 2(B) differentiable at x = 0 but not differentiable at x = 2(C) not differentiable at x = 0 but differentiable at x = 2(D) differentiable neither at x = 0 nor at x = 2
Sol. B
46. The total number of ways in which 5 balls ofdifferent colours can be distributed among 3 persons sothat each person gets at least one ball is
(A) 75 (B) 150 (C) 210 (D) 243
Sol. B
47. If ∞→xlim
−−
+
++bax
1x
1xx2
= 4, then
(A) a = 1, b = 4 (B) a = 1, b = –4(C) a = 2, b = –3 (D) a = 2, b = 3
Sol. B
∞→xlim
+
−−−−+
1x
baxbxaxxHx 22
= 4
∞→xlim
+
−+−−+−
1x
b1x)ab1(x)a1( 2
= 4
1 – a = 0 ⇒ a = 11 – b – a = 4b = –4
48. The function f : [0, 3] → [1, 29], defined byf(x) = 2x3 – 15x2 + 36x + 1, is
(A) one-one and onto.(B) onto but not one-one.(C) one-one but not onto.(D) neither one-one nor onto
Sol. B
f(X) = 2X3 – 15X2 + 36X + 1f'(X) = 6X2 – 30X + 36= 6 (X2 – 5X + 6)= 6(x – 3) (x – 2)
(0,1)
(0,29)(0,28)
(2,0) (3,0)f(0) = 1
f(2) = 2 × 8 – 15 × 4 + 36 × 2 +1 = 29f(3) = 2 × 27 – 15 × 9 + 36 × 3 + 1 = 28onto but not one one so 'B'.
49. The locus of the mid-point of the chord ofcontact of tangents drawn from points lying on thestraight line 4x – 5y = 20 to the circle x2 + y2 = 9 is
(A) 20(x2 + y2) – 36x + 45y = 0(B) 20(x2 + y2) + 36x – 45y = 0(C) 36(x2 + y2) – 20x + 45y = 0
(D) 36(x2 + y2) + 20x – 45y = 0Sol. A
P(h,k)
−
5
20a4,a x + y = 9
2 2
Equation of chord of contact
ax +
−
5
20a4y – 9 = 0
5ax + 4ay – 20y – 45 = 05ax + (4a – 20) y – 45 = 0 .....(i)equation of chord of mid point
hx + ky = h2 + k2 .....(ii)
22 kh
45
k
20a4
h
a5
+=
−= ⇒ a = 22 kh
h9
+
4a – 20 = 22 kh
k45
+ put the value of a
22 kh
h36
+ 20 = 22 kh
k45
+
36x – 20(x2 + y2) = 45y20(x2 + y2) + 45y – 36x = 0
50. Let P = [aij] be a 3 × 3 matrix and let Q = [bij],where bij = 2i + jaij for 1 ≤ i, j ≤ 3. If the determinant of Pis 2, then the determinant of the matrix Q is
(A) 210 (B) 211 (C) 212 (D) 213
Sol. D
11 12 13
21 22 23
31 32 33
a a a
P a a a
a a a
=
,
11 12 13
21 22 23
31 32 33
4a 8a 16a
Q 8a 16a 32a
16a 32a 64a
=
Q = 11 12 13
2 3 4 121 22 23
31 32 33
a a a
2 .2 .2 .2 a a a
a a a
Q = 122 P ⇒ |Q| = 213
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Multiple Correct51. If. y (x) satisfies the differential equation
y' – y tan x = 2x sex x and y (0) = 0, then
(A)
2
y4 8 2
π π =
(B)
2
y'4 18
π π =
(C)
2
y3 9
π π =
(D)
24 2y'
3 3 3 3
π π π = +
Sol. A,D
I.F. = cos x
y . cos x = ∫ dx.xcos.xsecx2
y . cos x = x2 + c, c = 0y = x2 sec x
52. A ship is fitted with three engines E1, E2 and E3.
The engines function independently of each other
with respective probabilities 1
2,
1
4 and
1
4. For
the ship to be operational at least two of its
engines must function. Let X denote the eventthat the ship is operational and let X1, X2 and
X3 denote respectively the events that theengines E1 , E2 and E3 are functioning. Which of
the following is (are) true ?
(A) c1
3P X | X
16 =
(B) P [Exactly two engines of the ship are functioning|X] 7
8=
(C) 2
5P X | X
16= (D) 1
7P X | X
16=
Sol. B,D
1
1P(E )
2= 2
1P(E )
4= 3
1P(E )
4=
1 1 1 1 1 3 1 1 1P(X) . . . . 2 . .
2 4 4 2 4 4 2 4 4
= + +
= 1 3 1 6 1
32 16 32 4
++ = =
(A) 1
1 1 1. .P(E X) 12 4 4
P(x) 1 / 4 8
∩= =
(B) 4/1
32/7 = 7/8
(C)
1 1 1 1 1 3 1 1 1. . . . . .
2 4 4 2 4 4 2 4 41
4
+ +=
1 3 5516 32 32
11 / 4 8
4
+= = =
(D) 1
1
1 1 1 1 1 3. . . . 2
P(X X ) 2 4 4 2 4 4 7 / 32 7
1P(X ) 1 / 2 16
2
+ ∩ = = =
53. Let , [0,2 ]θ ϕ∈ π be such that
22cos (1 sin ) sin tan cot cos 12 2
θ θ θ − ϕ = θ + ϕ −
3tan(2 ) 0and 1 sin
2π − θ > − < θ < −
Then ϕ cannot satisfy
(A) 02
π< ϕ < (B)
4
2 3
π π< ϕ <
(C) 4 3
3 2
π π< ϕ < (D)
32
2
π< ϕ < π
Sol. A,C,D
2cos (1 sin ) 2sin cos 1θ − ϕ = θ φ −
2 cos θ + 1 = 2 sin (θ + φ)
2
1 = sin(θ + φ) – cos θ
tan(2 ) 0π − θ > , tan θ < 0 ⇒ –1 < sin θ < –2
3
270º < θ < 300º
54. Let S be the area of the region enclosed by
2xy e ,y 0,x 0,−= = = and x = 1. Then
(A) 1
se
≥ (B) 1
s 1e
≥ −
(C) 1 1
s 14 e
≤ +
(D)
1 1 1s 1
2 e 2
≤ + −
Sol. A,B,D
(B) x ≥ x2
∫∫ −− ≤1
0
x1
0
x dxedxe2
0
1
12/1
1/eA D
EBC
F
(D) S ≤ area OABC + area AFDE
1 1 1s 1
2 e 2
≤ + −
55. Tangents are drawn to the hyperbola2 2x y
19 4
− = ,
parallel to the straight line 2x y 1− = . The points
of contact of the tangents on the hyperbola are
(A) 9 1
,2 2 2
(B) 9 1
,2 2 2
− −
(C) ( )3 3, 2 2− (D) ( )3 3,2 2−
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Sol. A,B
4x1 x –9y1y = 36
Slope ; 1
1
4x2
9y=
11
9yx
2= .............(1)
21 1x y 2
19 4
− =
1y 1 / 2= ± & 1x 9 / 2 2= ±
A & B
SECTION – CInteger Answer Type
56. Let S be the focus of the parabola y2 = 8x and let
PQ be the common chord of the circle
x2 + y
2 – 2x – 4y = 0 and the given parabola. The area
of the triangle PQS is
Sol. 0004
S(2,0)P
Q(2,4)
Q is end point of latus rectum
Area is ( ) 4242
1=×
57. Let p(x) be a real polynomial of least degree which
has a local maximum at x = 1 and a local minimum at x =
3. If p(1) = 6 and p(3) = 2, then p'(0) isSol. 0009
p'(x) = a(x – 1) (x – 3) = ax2 – 4ax + 3a
p(x) = 3
ax3
2ax2 + 3ax + c
3
a – 2a + 3a + c = 6 & 9a – 18a + 9a = 2
a = 3
58. Let f : IR → IR be defined as f(x) = |x|+|x2 – 1|.
The total number of points at which f attains either alocal maximum or a local minimum is
Sol. 0005
f(x) =
−≤−−
≤≤−+−−
≤≤++−
=−+
1x;1xx
0x1;1xx
1x0;1xx
1x;1xx
2
2
2
2
–1 –1/2 1/2 10
59. The value of 6+
−−− ......23
14
23
14
23
14
23
1log
2
3 is
Sol. 0004
3/2
16 log 4
3 2
+
1y 4 y
3 2= −
2 1y 4 y
3 2= −
23 2y 12 2 y= −
23 2y y 12 2 0+ − =
1 17y
6 2
− ±= =
17 1 16 8
6 2 6 2 3 2
−= =
3/2
1 8log
3 2 3 2×
3/2
4log
9
2
3/2
26 log
3
+
= 4
60. If candb,a are unit vectors satisfying
9accbba222
=−+−+−
, then c5b5a2
++ is
Sol. 0003
Given is maximum value of expression & hence
angle b/w any two is 120°
1a.b b.c c.a
2
−= = =
so |2a 5b 5c |+ +
is 3
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