# Jee Main 2014 Solution Code E English

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JEE MAIN EXAMINATION - 2014

QUESTIONS WITH SOLUTIONS

PAPER CODE - E

Fastest Growing Institute of Kota (Raj.)

FOR JEE Advanced (IIT-JEE) | JEE Main (AIEEE) | CBSE | SAT | NTSE | OLYMPIADS

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[PHYSICS] Code - E1. The current voltage relation of diode is givenby I = (e1000V/T – 1) mA, where the applied voltage Vis in volts and the temperature T is in degree Kelvin.If a student makes an error measuring 0.01 V whilemeasuring the current of 5 mA at 300 K, what willbe the error in the value of current in mA?(1) 0.2 mA (2) 0.02 mA(3) 0.5 mA (4) 0.05 mASol. 1

I =

1e T

V1000

mA

TV1000

e = I + 1

log (I + 1) = T

V1000

1I1

× dI = 1000

T × dv

dI = 3001000

× (5 + 1) mA × 0.01

= 0.2 mA

2. From a tower of height H, a particle is thrownvertically upwards with a speed u. The time takenby the particle, to hit the ground, is n times thattaken by it to reach the highest point of its path.The relation between H, u and n is :(1) 2 gH = n2u2 (2) g H = (n – 2)2u2

(3) 2 g H = nu2(n – 2) (4) g H = (n – 2)u2

Sol. 3–H = ut – 1/2 g t2

t' = u/gGiven t = nt'

–H = u

gnu

– 1/2 g 2

gnu

–H = gnu2

– 1/2 gun 22

u

H

–H = gnu2

(1 – 21

n)

–2Hg = nu2 (2 – n) 2Hg = nu2 (n – 2)

3. A mass 'm' is supported by a massless stringwound around a uniform hollow cylinder of mass mand radius R. If the string does not slip on thecylinder, with what acceleration will the mass fall onrelease ?

(1) 3g2

(2) 2g

(3) 6g5

(4) g

Sol. 2mg – T = ma .....(1)TR = I

TR = mR2 aR

T = ma ......(2) T

m

mg – ma = maa = g/2

4. A block of mass m is placed on a surface

with a vertical cross section given by y = 3x

6. If the

coefficient of friction is 0.5, the maximum heightabove the ground at which the block can be placedwithout slipping is :

(1) 16 m (2)

23 m

(3) 13 m (4)

12

m

Sol. 1

m

f

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y = 6x3

dxdy

= 6x3 2

= 2x2

f = mg cos = mg sin = tan

0.5 = 2x2

x = 1

y = 6x3

= 61

m

5. When a rubber-band is stretched by adistance x, it exerts a restoring force of magnitudeF = ax + bx2 where a and b are constants. The workdone in stretching the unstretched rubber-band byL is :

(1) aL2 + bL3 (2) 12

(aL2 + bL3)

(3) 2 3aL bL

2 3 (4)

2 31 aL bL2 2 3

Sol. 3

W = Fdx

W = Fdx(ax + bx2) dx

W =

L

0

32

3bx

2ax

W =

3bL

2aL 32

6. A bob of mass m attached to an inextensiblestring of length is suspended from a verticalsupport. The bob rotates in a horizontal circle withan angular speed rad/s about the vertical. Aboutthe point of suspension :(1) Angular momentum is conserved.(2) Angular momentum changes in magnitude butnot in direction.(3) Angular momentum changes in direction but notin magnitude.(4) Angular momentum changes both in directionand magnitude.

Sol. 3

L

L

L vector rotates in direction but magnitude

remains constant.

7. Four partic les, each of mass M andequidistant from each other, move along a circle ofradius R under the action of their mutual gravitiationalattraction. The speed of each particle is :

(1) GMR

(2) GM2 2R

(3) GM 1 2 2R

(4) 1 GM 1 2 22 R

Sol. 4

(2)(1)

(3) (4)

r

rmv2

= F14 + F24 + F34

= 2

2

r4Gm

+ 2

2

r2Gm

2

= 2

2

r2Gm

221

2mvr

= 2

2

r2Gm

2

221

v = Gm(1 2 2)4r

v = 21

)221(r

Gm

8. The pressure that has to be applied to theends of a steel wire of length 10 cm to keep its lengthconstant when its temperature is raised by 100ºC is(For steel Young's modulus is 2 × 1011 N m–2 and coef-ficient of thermal expansion is 1.1 × 10–5 K–1)(1) 2.2 × 108 Pa (2) 2.2 × 109 Pa(3) 2.2 × 107 Pa (4) 2.2 × 106 Pa

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Sol. 1y = 2 × 1011

lf = l0 (1 + t) t = 100° C = 1.1 × 10–5 k–1

lf – l0 = t l = 1.1 × 10–5 × 100= 1.1 × 10–3

y = 3

p1.1 10

2 × 1011 = 3

p1.1 10

p = 2 × 1.1 × 1011 × 10–3

p = 2.2 × 108 Pa

9. There is a circular tube in a vertical plane.Two liquids which do not mix and of densities d1 andd2 are filled in the tube. Each liquid subtends 90ºangle at centre. Radius joining their interface makes

an angle with vertical. Ratio 1

2

dd is :

(1) 1 sin1 sin

(2) 1 cos1 cos

(3) 1 tan1 tan

(4) 1 sin1 cos

Sol. 3

R

Rsin

Rcos

R–RcosR-Rsin

RR

R – R sin = d1 (R – R sin ) + d2 (R sin + R cos )d1 (1 – sin ) = d1 (1 – cos ) + d2 (sin + cos )d1 – d1 sin = d1 – d1 cos + d2 (sin + cod )d1 (cos – sin ) = d2 (sin + cos )

2

1

dd

=

sincoscossin

=

tan1tan1

10. On heating water, bubbles being formed atthe bottom of the vessel detatch and rise. Take thebubbles to be spheres of radius R and making a cir-cular contact of radius r with the bottom of thevessel. If r << R, and the surface tension of thewater is T, value of r just before bubbles detatch is:(density of water is W)

(1) 2 WgR

3T

(2) 2 WgR

6T

(3) 2 WgR

T

(4) 2 W3 gR

T

Sol.

R T

r

34

R3 w g = P (r2) + T (2r) sin

34

R2 w g = RT2

(r2) + T (2r) Rr

34

R3 w g = RT4

r2

r = R2 T3gw

11. Three rods of Copper, Brass and Steel arewelded together to form a Y - shaped structure.Area of cross - section of each rod = 4 cm2. End ofcopper rod is maintained at 100ºC where as ends ofbrass and steel are kept at 0ºC. Lengths of thecopper, brass and steel rods are 46, 13 and 12 cmsrespectively. The rods are thermally insulated fromsurroundings except at ends. Thermal conductivitiesof copper, brass and steel are 0.92, 0.26 and 0.12CGS units respectively. Rate of heat flow throughcopper rod is :(1) 1.2 cal/s (2) 2.4 cal/s(3) 4.8 cal/s (4) 6.0 cal/s

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Sol. 3

100° 0°C

Cu46cmk=0.92

Br13 cmk=0.26T

st, 12 cmk = 0.12

0°C

By Kirchoff law

1R100T

+ 2

T 0R

+ 3

T 0R

= 0

)46(100T

(0.92) + 13T

(0.26) + 12T

(0.12) = 0

50100T

+ 50T

+ 100T

= 0

2T – 200 + 2T + T = 0T = 40

Cu

dQdt

= 46

)40100( (0.92)(4) = 4.8 cal/s

12. One mole of diatomic ideal gas undergoes acyclic process ABC as shown in figure. The processBC is adiabatic. The temperature at A, B and C are400 K, 800 K and 600 K respectively. Choose thecorrect statement :

(1) The change in internal energy in whole cyclicprocess is 250 R.(2) The change in internal energy in the process CAis 700 R.(3) The change in internal energy in the process ABis –350 R.(4) The change in internal energy in the process BCis –500 R.

Sol. 4

C600

800B

A400

P

V(1) dUAB = n cv dT

= 1 × 2R5

(400)

= 1000 R(2) dUBC = n cv dT

= 1 × 2R5

(–200)

= – 500 R(3) dUABCA = 0

(4) dUCA = 1 × 2R5

(–200)

= – 500 R

13. An open glass tube is immersed in mercury insuch a way that a length of 8 cm extends abovethe mercury level. The open end of the tube is thenclosed and sealed and the tube is raised verticallyup by additional 46 cm. What will be length of theair column above mercury in the tube now?(Atmospheric pressure = 76 cm of Hg)(1) 16 cm (2) 22 cm(3) 38 cm (4) 6 cmSol. 1

8cm

P0

76 × 8 = × pgas

76 8

= pgas

76 8

+ (54 – ) = 76

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54 – = 7681

= 76 8

54 – 2 = 76 – 6082 + 22 – 608 = 0

= 222 (22) 4 808

2

= 2

5422

= 16 cm

14. A particle moves with simple harmonicmotion in a straight line. In first s, after startingfrom rest it travels a distance a, and in nest s ittravels 2a, in same direction, then :(1) Amplitude of motion is 3a(2) Time period of oscillations is 8(3) Amplitude of motion is 4a(4) Time period of oscillations is 6Sol. 4

a

3a

A – A cos () = a ....(1)A – A cos (2) = 3a2A (1 – cos2 ) = 3a ...(2)Divide (2) By (1)2 (1 + cos ) = 3cos = 1/2= /3

T2

= /3

T = 6

15. A pipe of length 85 cm is closed from oneend. Find the number of possible natural oscillationsof air column in the pipe whose frequencies lie below1250 Hz. The velocity of sound in air is 340 m/s.(1) 12 (2) 8 (3) 6 (4) 4

Sol. 3

f = L4V

= 210854340

= 100 Hz

100 H, 300 Hz, 500 Hz, 700 Hz, 900 Hz, 1100 Hz = 6

16. Assume that an electric field 2E 30x i

exists in space. Then the potential differenceVA – VO, where VO is the potential at the origin andVA the potential at x = 2 m is :(1) 120 J (2) – 120 J(3) – 80 J (4) 80 JSol. 3

dv = dr.E

VA – V0 = 2

0

2dxx30

= –

2

0

3

3x30

= – [10x3]= – 10 [8 – 0]VA – V0 = –80 J

17. A parallel plate capacitor is made of twocircular plates separated by a distance of 5 mm andwith a dielectric of dielectric constant 2.2 betweenthem. When the electric field in the dielectric is3 × 104 V/m, the change density of the positiveplate will be close to :(1) 6 × 10–7 C/m2

(2) 3 × 10–7 C/m2

(3) 3 × 104 C/m2

(4) 6 × 104 C/m2

Sol. 1Q = cv

Q = d

A. r0 V

= AQ

= 0 r.d

V

= 0 r .E= 8.85 × 10–12 × 2.2 × 3 × 104

= 58.41 × 10–8 _~ 6 × 10–7 c/m2

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18. In a large building, there are 15 bulbs of 40W, 5 bulbs 100 W, 5 fans of 80 W and 1 heater of 1kW. The voltage of the electric mains is 220 V. Theminimum capacity of the main fuse of the buildingwill be :(1) 8 A (2) 10 A(3) 12 A (4) 14 ASol. 3

P = RV2

R1 = 40

)200( 2

Total current

= 1R

V × 15 +

2RV

× 5 + 3R

V × 5 +

4RV

× 1

R = V2/P

= 220

1540 +

2205100

+ 220

580 +

2201000

= 11.3626 AMinimum value is 12 A.

19. A conducor lies along the z-axis at– 1.5 z < 1.5 m and carries a fixed current of 10.0

A in – za direction (see figure). For a field

B

= 3.0 × 10–4 e–0.2x ya T, find the power required tomove the conductor at constant speed to x = 2.0m, y = 0 m in 5 × 10–3 s. Assume parallel motionalong the x-axis.

(1) 1.57 W (2) 2.97 W(3) 14.85 W (4) 29.7 WSol. 2

w = dxe103310 x2.04

= 4. 5 × 10–2 (1 – e–0.4)

P = Tw

= 2.97

20. The coercivity of a small magent where theferromagnet gets demagnetized is 3 × 103 A m–1.The current required to be passed in a solenoid oflength 10 cm and number of turns 100, so that themagnet gets demagnetized when inside the sole-noid, is :(1) 30 mA (2) 60 mA(3) 3 A (4) 6 ASol. 3

Coercivity 0

B = 3 × 103 A/m

B = 0 (3 × 103) = 0 nI

= 0

1.0

100 I

I = 100

1.0103 3 = 3 A

21. In the circuit shown here, the point 'C' iskept connected to point 'A' till the current flowingthrough the circuit becomes constant. Afterward,suddenly, point 'C' is disconnected from point 'A' andconnected to point 'B' at time t = 0. Ratio of thevoltage across resistance and the inductor att = L/R will be equal to :

(1) e

1 e(2) 1 (3) –1 (4)

1 ee

Sol. 3When BC are joined onlyL & R are in circuit

R

– +

+

–L

Applying KVL, VR + VL = 0 VR = –VL

L

R

VV

= – 1

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22. During the propagation of electromagneticwaves in a medium :(1) Electric energy density is double of themagnetic energy density.(2) Electric energy density is half of the magneticenergy density.(3) Electric energy density is equal to the magneticenergy density.(4) Both electric and magnetic energy densities arezero.Sol. 3

In an EM wave

1/2 0 E2 =

0

2

2B

Both are equal.

23. A thin convex lens made from crown grass

23

has focal length f. When it is measured in

two different liquids having refractive indices 34

and

35

, it has the focal length f1 and f2 respectively..

The correct relation between the focal lengths is :(1) f1 = f2 < f(2) f1 > f and f2 becomes negative(3) f2 > f and f1 becomes negative(4) f1 and f2 both become negativeSol. 2

Since 4/3 < 3/2 thus f1 > fsince 5/3 > 3/2 thus f2 changes sign

24. A green light is incident from the water tothe air - water interface at the critical angle ().Select the correct statement.(3) The spectrum of visible light whose frequency ismore than that of green light will come out to theair medium.(4) The entire spectrum of visible light will come outof the water at various angles to the normal.(1) The entire spectrum of visible light will come outof the water at an angle of 90º to the normal.(2) The spectrum of visible light whose frequency isless than that of green light will come out to the airmedium.

Sol. 2

Green

1

f > fgreen

< green

> green T.I.R

25. Two beams, A and B, of plane polarized lightwith mutually perpendicular planes of polarizationare seen through a polaroid. From the position whenthe beam A has maximum intensity (and beam B haszero intensity), a rotation of polaroid through 30ºmakes two beams appear equally bright. If the initialintensities of the two beams are IA and IB

respectively, then A

B

II equals :

(1) 3 (2) 32

(3) 1 (4) 13

Sol. 4IA cos2 30° = IB cos2 60°

B

A

II

=

30cos60cos

2

2

= 2

2

)2/3()2/1(

= 4/34/1

= 1/3

26. The radiation corresponding to 3 2transition of hydrogen atom falls on a metal surfaceto produce photoelectrons. These electrons are madeto enter a magnetic field of 3 × 10–4 T. If the radiusof the largest circular path followed by theseelectrons is 10.0 mm, the work function of the metalis close to :(1) 1.8 eV (2) 1.1 eV(3) 0.8 eV (4) 1.6 eVSol. 2

r = qBmv

v = m

qBr =

19 4 3

31

1.6 10 3 10 10 109.1 10

= 5.27 × 105

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KEmax = 1/2 mv2

= 1/2 × 9.1 × 10–31 × (5.27 × 105)2

= 19

21

106.11036.126

= 78.975 × 10–2

= 0.78 eV

E = 13.6

2

22

1 n1

n1

= 13.6

91

41

= 1.88 ev1.88 = 0.78 + = 1.1 eV

27. Hydrogen (1H1), Deuterium (1H2), singlyionised Helium (2He4)+ and doubly ionised lithium(3Li6)++ all have one electron around the nucleus.Consider an electron transition from n = 2 to n = 1.If the wave lengths of emitted radiation are 1, 2,3 and 4 respectively then approximately which oneof the following is correct?(1) 41 = 22 = 23 = 4(2) 1 = 22 = 23 = 4(3) 1 = 2 = 43 = 94(4) 1 = 22 = 33 = 44Sol. 3

hc

= Rc z2

2

22

1 n1

n1

1

z2

1

1 (1)2,

2

1 (1)2,

3

1 (2)2,

4

1 (3)2

1 : 2 : 3 : 4 = 1 : 1 : 41

: 91

28. The forward biased diode connection is :

(1)

(2)

(3)

(4)

Sol. 1

+2V

(High) (Low)

–2V

Forward biased

29. Match List-I (Electromagnetic wave type)with List-II (Its association/application) and selectthe correct option from the choices given below thelists :

(a) Infrared waves (i)To treat muscular strain

(b) Radio waves (ii) For broadcasting

(c) X-rays (iii)To detect fracture of bones

(d) Ultraviolet rays (iv)Absorbed by the ozone layer of the atmosphere

List-I List-II

(a) (b) (c) (d)(1) (iv) (iii) (ii) (i)(2) (i) (ii) (iv) (iii)(3) (iii) (ii) (i) (iv)(4) (i) (ii) (iii) (iv)Sol. 4

Knowledge based

30. A student measured the length of a rod andwrote it as 3.50 cm. Which instrument did he use tomeasure it?(1) A meter scale(2) A vernier calliper where the 10 divisions in ver-nier scale matches with 9 division in main scale andmain scale has 10 divisions in 1 cm.(3) A screw gauge having 100 divisions in the circu-lar scale and pitch as 1 mm.(4) A screw gauge having 50 divisions in the circularscale and pitch as 1 mm.Sol. 2

The vernier calliper has least count 10

cm1.0

= 0.01 cmAlso 3.50 cm has smallest reading0.01 cm

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31. The correct set of four quantum numbers forthe valence electrons of rubidium atom (Z =37) is ?(1) 5, 0, 0, + 1/2(2) 5, 0, 1, + 1/2(3) 5, 1, 1 + 1/2(4) 5, 1, 0, + 1/2

Sol. (1)

32. If Z is a compressibility factor, van der waalsequation at low pressure can be written as ?

(1) Z = 1 – RTPb

(2) Z = 1 – VRT

a

(3) Z = 1 + PbRT

(4) Z = 1 + RTPb

Sol. (2)According to vander waal’s equation forone mole gas

2

aP (V b) RTV

at low pressure volume will be high b can be neglected,so,

2

aP (V) RTV

aPV RTV

Z = PVRT

= 1 – a

VRT

33. CsCl crystallises in body centred cubic lat-tice. If 'a' is its edge length then which ofthe following expressions is correct ?

(1) Cs Clr r 3a

(2) Cs Clr r 3a

(3) Cs Cl3r r a2

(4) Cs Cl3ar r2

Sol. 3Cl– forms SCCs+ goes to centre of the cube

–3ar r2

34. For the estimation of nitrogen, 1.4 g of anorganic compound was digested by kjeldahlmethod and the evolved ammonia was ab-

sorbed in 60 mL of 10M

sulphuric acid. The

unreacted acid required 20 mL of 10M

sodium

hydroxide for complete neutralization. Thepercentage of nitrogen in the compound is ?(1) 3 % (2) 10 % (3) 6 % (4) 5 %

Sol. (2)1.4 gm NH3

2 4

M M60mL H SO 20mL NaOH10 10

Let’s Assume x % N

x1.4 1 20 1 60100 1 214 10 1000 10 1000

14 100x (0.012 – 0.002)1.4

= 10 %

35. Resistance of 0.2 M solution of an electro-lyte is 50 . The specific conductance ofthe solution is 1.4 S m–1. The resistance of0.5 M solution of the same electrolyte is 280. The molar conductivity of 0.5 M solutionof the electrolyte in S m2 mol–1 is ?(1) 5 × 10–4

(2) 5 × 102

(3) 5 × 103

(4) 5 × 10–3

Sol. 1

R = a

50 = 1

1.4 a

170ma

R = a

[CHEMISTRY]

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280 = 1 70

= 0.25 Sm–1

= 0.25 × 10–2 –1 - cm–1

m1000

c

=

–21000 0.25 100.5

= 5 –1-cm2 mole–1

= 5 × 10–4 S-m2 mol–1

36. For complete combustion of ethanol,C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l),the amount of heat produced as measured inbomb calorimeter, is 1364.47 kJ mol–1 at 25ºC.Assuming ideality the Enthalpy of combustion,CH, for the reaction will be :(R = 8.314 kJ mol–1)(1) –1366.95 kJ mol–1

(2) –1350.50 kJ mol–1

(3) –1460.50 kJ mol–1

(4) –1361.95 kJ mol–1

Sol. (1)H = U + ngRT= – 1364.47 + (–1) × 8.314 × 10–3 × 298= –1364.47 – 2.47= – 1366.95 kJ mol–1

37. the equivalent conductance of NaCl at con-centration C and at infinite dilution are C

and respectively. The correct relationship

between C and is given as ?(where the constant B is positive)(1) C = + (B) C

(2) C = + (B) C

(3) C = – (B) C

(4) C = – (B) CSol. (3)

c = — CB

38. Consider separate solutions of 0.500 MC2H5OH(aq), 0.100 M Mg3(PO4)2(aq), 0.250M KBr(aq) and 0.125 M Na3PO4(aq) at 25ºC.Which statement is true about these solu-tions, assuming all salts to be strong elec-trolytes?(1) they all the have the same osmotic pres-sure

(2) 0.500 M C2H5OH(aq) has the highest os-motic pressure.(3) 0.125 M Na3PO4(aq) has the highest os-motic pressure.(4) 0.100 M Mg3(PO4)2(aq) has the highestosmotic pressure.

Sol. 1i ic

0.5 M C2H5OH (aq.) 1 0.50.10 M Mg3 (PO4)2 5 0.50.25 M KBr 2 0.50.125 M Na3PO4 4 0.5All have same colligative prop.

39. For the reaction SO2(s) + 21

O2(g) SO3(g),

if Kp = KC (RT)x where the symbols have usualmeaning then the value of x is: (assumingideality) ?(1) 1/2(2) – 1/2(3) –1(4) 1

Sol. (2)

SO2(g) 21 O (g)2

SO3(g)

Kp = Kc (RT)ng

ng = 1 – (1 + 21

) = 21–

Kp = Kc (RT)–1/2

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40. For the non-stoichiometre reaction 2A + B C + D, the following kinetic data were ob-tained in three separate expreiments, all at298 K

3

3

3

104.2M1.0M2.0102.1M2.0M1.0102.1M1.0M1.0

)SLmol(

CofformationofrateInitial

)B(ionConcentrat

Initial

)A(ionConcentrat

Initial

the rate law for the formation of C is

(1) dtdc

= k[A] [B]2

(2) dtdc

= k[A]2 [B]

(3) dtdc

= k[A] [B]

(4) dtdc

= k[A]

Sol. (4)2A + B C + D(1) & (3) [A]1

(1) & (2) [B]º r = k[A]1

41. Among the following oxoacids, the correctdecreasing order of acid strength is :(1) HOCl > HClO2 > HClO3 > HClO4(2) HClO2 > HClO4 > HClO3 > HOCl(3) HClO4 > HClO3 > HClO2 > HOCl(4) HClO4 > HOCl > HClO2 > HClO3

Sol. (3)

42. The metal that connnot be obtained by elec-trolysis of an aqueous solution of its salts is(1) Cu(2) Ca(3) Ag(4) Cr

Sol. 2

22 2H o /H Ca / Ca

Eº Eº

Ca is more electro +ve so aqueous solu-tions cannot be use.

43. The octahedral complex of a metal ion M3+

with four monodentate ligands L1, L2, L3 andL4 absorb wavelengths in the region of red,green, yellow and blue, respectively. The in-creasing order of ligand strength of the fourligands is ?(1) L3 < L2 < L4 < L1(2) L1 < L3 < L2 < L4(3) L4 < L3 < L2< L1(4) L1 < L2 < L4 < L3

Sol. (2)(a) According to spectro chemical seriesmore absorption frequency stronger theligand

V I B G Y O R

EL1 < L3 < L2 < L4

44. Which one of the following properties is notshown by NO?(1) It is diamagnetic in gaseous state(2) It's bond order is 2.5(3) It combines with oxygen to form nitrogendioxide(4) It is a neutral oxide

Sol. (1)(1) NO + 1/2 O2

NO2(2) bond Order of NO is 2.5(3) It is paramagnetic in gaseous state(4) It is neutral oxide

45. In which of the following reactions H2O2 actsas a reducing agent?(a) H2O2 + 2H+ + 2e– 2H2O(b) H2O2 – 2e– O2 + 2H+

(c) H2O2 + 2e– 2OH–

(d) H2O2 + 2OH– –2e– O2 + 2H2O(1) (a), (c)(2) (c), (d)(3) (a), (b)(4) (b), (d)

Sol. 4As a reducing agent

01 –

22 2H O O 2e

(b) & (d)

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46. The correct statement for the molecule, CsI3,is ?(1) It contains Cs3+ and I– ions.(2) It contains Cs+ and I3

– ions.(3) It is a covalent molecule.(4) It contains Cs+, I– and lattice I2 mol-ecule.

Sol. (2)

47. The ratio of masses of oxygen and nitrogenin a particular gaseous mixture is 1 : 4. Theratio of number of their molecule is :(1) 1 : 8 (2) 7 : 32(3) 1 : 4 (4) 3 : 16

Sol. (2)O2 : N2

mass 1 : 4

Moles321

:284

7 : 32

48. Given below are the half - cell reactionsMn2+ + 2e– Mn : Eº = – 1.18 V2(Mn3 + e– Mn2+) : Eº = + 1.51 VThe Eº for 3Mn2+ Mn + 2Mn3+ will be(1) – 2.69 V ; the reaction will not occur(2) – 0.33 V ; the reaction will occur(3) – 0.33 V ; the reaction will not occur(4) – 2.69 V ; the reaction will occur

Sol. (1)Mn2+ + 2e– Mn ; Eº = – 1.18 V2Mn3+ + 2e– 2Mn2+ ; Eº = +1.51 V(1) – (2)Mn2+ + 2e– + 2Mn2+ 2Mn3+ + 2e–1 + MnEº = – 1.18 – 1.51 ( n is same) Eº = – 2.69 V(3) The reaction will nor occur

49. Which series of reactions correctly representschemical relations related to iron and its com-pound ?(1) Fe Cl ,heat2 FeCl3 heat, air FeCl2

Zn Fe

(2) Fe dil H SO2 4 FeSO4 H SO2 4, 2O

Fe(SO4)3 heat Fe

(3) Fe O heat2, FeO dil H SO2 4 FeSO4

heat Fe

(4) Fe O heat2, Fe3O4 CO, 600ºC FeOCO, 700ºC Fe

Sol. (4)This reaction is used in blast furnace.

50. The equation which is balanced and repre-sents the correct product(s) is?(1)[Mg(H2O)6]

2+ + (EDTA)4– excess NaOH

[Mg(EDTA)]2+ + 6H2O(2) [CoCl(NH3)5]

+ + 5H+ CO2+ + 5NH4+ + Cl–

(3) Li2O + 2KCl 2LiCl + K2O(4) CuSO4 + 4KCN K2[Cu(CN)4] + K2SO4

Sol. (2)In this reaction H+ ion attacks ammonia mol-ecule and hence decomposes the complexwhich results information of amonia

51. In SN2 reactions, the correct order of reac-tivity for the following compounds ?CH3Cl, CH3CH2Cl, (CH3)2CHCl and (CH3)3CCl is(1) CH3CH2Cl > CH3Cl > (CH3)2CHCl > (CH3)3CCl(2) CH3Cl > (CH3)2CHCl > CH3CH2Cl > (CH3)3CCl(3) (CH3)2CHCl > CH3CH2Cl > CH3Cl > (CH3)3CCl(4) CH3Cl> CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl

Sol. (2)SN2

Reactivity of SN2 1

Steric hindranceOrder of reactivity towards SN2.CH3Cl > CH3–CH2–Cl > (CH3)2CH–Cl > (CH3)3C–Cl

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52. On heating an aliphatic primary amine withchloroform and ethanolic potassium hydrox-ide, the organic compound formed is :(1) An alkyl cyanide(2) An alkanediol(3) An alkanol(4) An alkyl isocyanide

Sol. (4)

R—NH2 + CHCl3 +KOH R–NC1º-amine chloroform Alkyl isocyanideit is carbyl amine reaction or isocyanide test.

53. The most suitable reagent for the conversionof R – CH2 – OH R – CHO is :(1) CrO3(2) K2Cr2O7(3) KMnO4(4) PCC (Pyridininum Chlorochromate)

Sol. (4)

R—CH2—OH

PCC (pyridinium chloro chromate) is mildoxidising agent.

54. The major organic compound formed by thereaction of 1,1,1-trichloroethane with silverpowder is ?(1) Acetylene (2) 2 - Butene(3) 2 Butyne (4) Ethene

Sol. (3)

CH3—CC—CH3 + 6AgCl2-butyne

55. Sodium phenoxide when heated with CO2 un-der pressure at 125º C yields a product whichon acetylation produces C ?

ONa

+ CO2 B C125º H+

5 Atm Ac O2

The major product C would be;

(1) COOH

OCOCH3

(2)

OCOCH3

COOH

(3) COOCH3

OH

(4)

OH

COCH3

COCH3

Sol. (1)

56. Considering the basic strength of amines inaqueous solution, which one has the small-est pKb value?(1) (CH3)2NH (2) C6H5NH2(3) (CH3)3N (4) CH3NH2

Sol. (1)

Delocalised lone pair electrons.

(least Basic)Order of basic strength in aqueous solution 2º > 1º >3º amine > aniline.

Basic strength Kb bpk

1

(CH3)2NH is most basic so it has smallest pKb.

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57. For which of the following molecule signifi-cant 0 ?

(a)

Cl

Cl

(b)

CN

CN

(c)

OH

OH

(d)

SH

SH(1) Only (c)(2) (a) and (b)(3) Only (a)(4) (c) and (d)

Sol. (4)

(a) (b)

(c) (d)

58. Which one is classified as a condensationpolymer ?(1) Dacron (2) Acrylonitrile(3) Teflon (4) Neoprene

Sol. (1)Terylene or Dacron

(Dacron)

Other than dacron are addition polymers.

59. Which one of the following bases is notpresent in DNA?(1) Quinoline (2) Thymine(3) Cytosine (4) Adenine

Sol. (1)

(Quinoline)

Quinoline is not present into DNA.Cytosin, Thymine and Adenine are present inDNA.

60. In the reaction,CH3COOH LiAlH4 A PCl5 B Alc. KOH C,the product C is ?(1) Acetaldehyde(2) Acetyl chloride(3) Ethylene(4) Acetylene

Sol. (3)

CH3–CH2–OH

CH3–CH2–Cl

CH2=CH2 (Ethylene)

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[MATHEMATICS]

SECTION – ASingle Correct

61. If X = {4n – 3n – 1) : n N} andY = {9 (n – 1) : n N}, where N is the set ofnatural numbers, then X Y is equal to :(1) X (2) Y(3) N (4) Y – X

Sol. [2]x : 4n – 3n –1x : (1 + 3)n – 3n – 1x : 1 + 3n + nC2 3

2 + nC3 33 + ........ –3n–1

x : 9 {nC2 + nC3 3 + ......} ...(1)& y : 9 (n – 1) ...(2)From (1) & (2) multiply of 9 Hence X Y Y

62. If z is a complex number such that |z| 2,

then the minimum value of 21z :

(1) is strictly greater than 25

(2) is strictly greater than23

but less than25

(3) is equal to 25

(4) lies in the interval (1, 2)

Sol. [4]|z| 2

1 1 1| z | z | z |2 2 2

min

1 3Z2 2

63. If a R and the equation– 3(x – [x])2 + 2(x – [x]) + a2 = 0(where [x] denotes the greatest integer x)has no integral solution, then all possiblevalues of a lie in the interval :(1) (–2, –1)(2) (–, –2) (2, )(3) (–1, 0) (0, 1) (4) (1, 2)

Sol. [3]

x – [x] = {x} = t [0, 1)–3t2+ 2t + a2 = 0 a2= 3t2 – 2t [0, 1)Since eqn cannot have integralroot : t 0 a2 (0, 1) a (–1, 0) (0, 1)

64. Let and be the roots of equationpx2 + qx + r = 0, p 0. If p, q, r are in A.P.

and 1

+ 1

= 4, then the value of | – | is:

(1) 934

(2) 9132

(3) 961

(4) 9172

Sol. [2]

Given = –qp =

rp

p, q, r A. P. , then

q – p = r – q 2q r p ...(1)

again 1 1 4

= 4

q 4r ...(2)

(1) & (2) gives– 8r = r + p

p 9r ...(3)

Now, | – | = 2( ) 4

= 2q 4pr|p |

= 2 216r 36r

|9r |

= 52 2 139 9

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65. If 0, and f(n) = n + n and

)4(f1)3(f1)2(f1)3(f1)2(f1)1(f1)2(f1)1(f13

= K(1– )2 (1 – )2 ( – )2, then K is equalto :(1) 1 (2) –1

(3) (4) 1

Sol. [1]f(n) = n + n

3

1 1 1 1 11 1 1

1 1 1

= 2 2

1 1 1 1 1 11 11 1

= (1 – )2 ( – )2 ( – 1)2

k 1

66. If A is an 3 × 3 non - singular matrix suchthat AA' = A'A and B = A–1 A', then BB' equals:(1) B–1 (2) (B–1)'(3) I + B (4) I

Sol. [4]AA' = A'AB = A–1A'B' = A(A–1)'B . B' = A–1 (A' A) (A–1)'= A–1 (A A') (A–1)1

= (A–1 A) (A') (A')–1

= I

67. If the coefficients of x3 and x4 in theexpansion of (1 + ax + bx2) (1 – 2x)18 inpowers of x are both zero, then (a, b) isequal to :

(1)

3

272,14 (2)

3

272,16

(3)

3

251,16 (4)

3

251,14

Sol. [2](1 + ax + bx2) (1 – 2x)18

coeff. of x3 = 18C3 (–2)3 + a. 18C2 (–2)2

+ b . 18C1(–2)1

= 6528 a612 36b 0

coeff. of x4 = 18C4 (–2)4 + a . 18C3 (–2)3 + b 18C2 (–2)2

= 48960 a.( 6528) b612 0

We get (16, 2723

)

68. If (10)9 + 2(11)1 (10)8 + 3(11)2 (10)7 +........ + 10(11)9 = k(10)9, then k is equal to(1) 100 (2) 110

(3) 10121

(4) 100441

Sol. [1]109 + 2.111 108 + 3.112 107 ---- + 10.119

S = 109 (1+2 111

10

+ 3. 211

10

+ ..... + 10911

10

)

S111

10

= 109 (111

10

+ 2211

10

+ 3311

10

+.......101011

10

)

11s 110

=109 1 2 9 1 01 1 1 1 1 1 1 1

1 1 01 0 1 0 1 0 1 0

= 109

10

10111 1

1110 1011 10110

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S

10 1091 11 1110 1 10 10

10 10 10

= 109 1010 1011 111

10 10

S . 1

10

= 109 (–10) S = 100. 109

So K = 100

69. Three positive numbers form an increasingG.P. If the middle term in this G.P. is doubled,the new numbers are in A.P. Then the commonratio of the G.P. is :

(1) 2 – 3 (2) 2 + 3

(3) 2 + 3 (4) 3 + 2

Sol. [2]a, ar, ar2 in GPa, 2ar, ar2 in AP

2ar = 2a ar

2

4r = 1 + r2

r2 – 4r + 1 = 0

r = 4 16 4

2

r = 4 2 3

2

r =2 3

so r = 2 + 3 as In GP

70. 2

2

0x x)xcossin(Lim

is equal to :

(1) – (2)

(3) 2

(4) 1

Sol. [2]

2x 0

sin( c'x)limx

2

2 2x 0

sin( c x) (1 cx)lim (1 cx)( c x) x

= 1 . . 12

. 2

=

71. If g is the inverse of a function f and

f'(x) = 5x11

, then g'(x) is equal to :

(1) 5)}x(g{11

(2) 1 + {g(x)}5

(3) 1 + x5 (4) 5x4

Sol. [2]f(g(x)) = xf'(g(x)) g'(x) = 1

g'(x) = 1

f '(g(x))

g'(x) = 5

11

1 (g(x))

1 + (g(x))5

72. If f and g are differentiable functions in [0,1] satisfying f(0) = 2 = g(1), g(0) = 0 andf(1) = 6, then for some c ]0, 1[ :(1) f'(c) = g'(c)(2) f'(c) = 2g'(c)(3) 2f'(c) = g'(c)(4) 2f'(c) = 3g'(c)

Sol. [2]2g'(c) = f'(c)

= 2 g(1) g(0) f(1) f(0)

1 0 1 0

= 22 0 6 2

1 1

4 = 4

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73. If x = – 1 and x = 2 are extreme points off(x) = log |x| + x2 + x then :

(1) = 2, = –21

(2) = 2, = 21

(3) = – 6, = 21

(4) = – 6, = –21

Sol. [1]

f'(x) = 2 x 1x

at x = –1, 2 f'(x) = 0 – – 2+ 1 = 0 ....(1)

2

+ 4 + 1 = 0 .... (2)

= 2, = 12

74. The integral dxex1x1 x

1x

is equal

to:

(1) (x + 1) ce x1x

(2) –x ce x

1x

(3) (x – 1) ce x1x

(4) x ce x

1x

Sol. [4]

1 1x xx x1e x e dx

x

...(1)

1xxe

= f(x)

1xxe

2

11x

dx = f'(x)

1 1x+ x+x x

2f(x)

xf '(x)

1e x.e 1 dxx

x f(x) = x . 1x+xe + C

75. The integral

0

2 dx2xsin4

2xsin41

equals :

(1) 4 3 – 4 (2) 4 3 – 4 – 3

(3) – 4 (4) 32

– 4 – 4 3

Sol. [2]

0

x1 2sin dx2

= 3

0

x1 2sin dx2

3

x1 2sin dx2

= 3

03

x xx 4cos x 4cos2 2

= 4cos 0 43 6 4cos 4cos

2 3 6

= 4 3 43

76. The area of the region described byA = {(x, y) : x2 + y2 1 and y2 1 – x} is :

(1) 2

– 32

(2) 2

+ 32

(3) 2

+ 34

(4) 2

– 34

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Sol. [3]

A1 = 21

01 x dx

A1 = 2 0 2

12t dt

A1 = 4 . 13

4Area2 3

77. Let the population of rabbits surviving at atime t be governed by the differential equation

dt)t(dp

= 21

p(t) – 200.

If p(0) = 100, then p(t) equals :(1) 600 – 500 et/2 (2) 400 – 300 e–t/2

(3) 400 – 300 et/2 (4) 300 – 200 e–t/2

Sol. [3]2dp(t) dt

p(t) 400

2 log|p(t) – 400| = t + c ...(1)

t = 0, p = 1002 log (300) = C

From (1)2log|p(t) – 400| = t + 2 log (300)|p(t) – 400| = et/2 . elog(300)

t /2p(t) 400 e (300)

78. Let PS be the median of the triangle withvertices P(2, 2), Q(6, –1) and R(7, 3). Theequation of the line passing through (1, –1)and parallel to PS is :(1) 4x + 7y + 3 = 0 (2) 2x – 9y – 11 = 0(3) 4x – 7y – 11 = 0 (4) 2x + 9y + 7 = 0

Sol. [4]Slope of PS

m = 2 1 1

13 922 2

m = 29

Eqn of line

y + 1 = 29

(x – 1)

9y + 9 = 2x + 22x + 9y + 7 = 0

79. Let a, b, c and d be non-zero numbers. Ifthe point of intersection of the lines4ax + 2ay + c = 0 and 5bx + 2by + d = 0 liesin the fourth quadrant and is equidistant fromthe two axes then :(1) 3bc – 2ad = 0 (2) 3bc + 2ad = 0(3) 2bc – 3ad = 0 (4) 2bc + 3ad = 0

Sol. [1]4ax + 2ay + c = 05bx + 2by + d = 0let intersectionpoint be (, – )4a – 2a + c = 0

= – c2a

....(1)

5b – 2b + d = 0

= – d3b

....(2)

– c d2a 3b

3bc 2ad

3bc – 2ad = 0

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80. The locus of the foot of perpendicular drawnfrom the centre of the ellipse x2 + 3y2 = 6 onany tangent to it is :(1) (x2 + y2)2 = 6x2 + 2y2

(2) (x2 + y2)2 = 6x2 – 2y2

(3) (x2 – y2)2 = 6x2 + 2y2

(4) (x2 – y2)2 = 6x2 – 2y2

Sol. [1]

Ellipse2 2x y 16 2

If foot of r is (h,k) then equationof line xh + ky = k2 + h2

is y = hk

x + 2 2k h

k

Now if this tangent then condition of tan-gencyc2 = a2m2 + b2

2 2k hk

= 2h6

k

+ 2

22 2

2

k h

k

=

2

26h 2k

(k2 + h2)2 = 6h2 + 2k2

(x2 + y2)2 = 6x2 + 2y2

81. Let C be the circle with centre at (1, 1) andradius = 1. If T is the circle centred at (0,y), passing through origin and touching thecircle C externally, then the radius of T isequal to :

(1) 21

(2) 41

(3) 23

(4) 23

Sol. [2]

C1C = 1 + y(1 – 0)2 + (1 – y)2 = (1 + y)2

1 + 1 – 2y + y = 1 + y2 + 2y4y = 1

1y4

82. The slope of the line touching both theparabolas y2 = 4x and x2 = – 32y is :

(1) 81

(2) 32

(3) 21

(4) 23

Sol. [3]y2 = 4x, x2 = –32yEquation of Tangent, Equation of Tangent

y = mx + 1m

, y = mx + 8m2

1m

= 8m2

m = 12

83. The image of the line 3

1x =

13y

= 54z

in the plane 2x – y + z + 3 = 0 is the line :

(1) 3

3x =

15y

= 52z

(2) 33x

= 15y

= 5

2z

(3) 3

3x =

15y

= 52z

(4) 33x

= 15y

= 5

2z

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Sol. [3]

x 1 y 3 z 43 1 5

2 3 4 3x 1 y 3 z 4 22 1 1 4 1 1

x 1 y 3 z 4 22 1 1

x= 1 – 4, y = 3 + 2, z = 4 – 2P' (–3, 5 ,2)Line is || to the planeEquation of image line

x 3 y 5 z 23 1 5

84. The angle between the lines whose directioncosines satisfy the equations + m + n = 0and 2 = m2 + n2 is :

(1) 6

(2) 2

(3) 3

(4) 4

Sol. [3]l + m + n = 0l2 – m2 – n2 = 0(m + n)2 – m2 – n2 = 0mn = 0m = 0 or n = 01.l + 1.m + 1.n = 0, 1.l + 1.m + 1.n = 00.l + 1.m + 0.n = 0, 0.l + 0.m + 1.n = 0

l m n1 1 1 1 1 11 0 0 0 0 1

,

l m n1 1 1 1 1 10 1 1 0 0 0

l m n1 0 1

or

l m n1 1 0

Cos = 1 0 02 2

Cos = 12

= 3

85. If a b b c c a

=

2a b c

then is

equal to :(1) 0 (2) 1(3) 2 (4) 3

Sol. [2]Clearly = 1

86. Let A and B be two events such that

)BA(P = 61

, P (A B) = 41

and

P( A ) = 41

, where A stands for the

complement of the event A. Then the eventsA and B are :(1) independent but not equally likely(2) independent and equally likely(3) mutually exclusive and independent.(4) equally likely but not independent

Sol. [1]

1P(A B)6

1P(A B)4

1P(A)4

clearly P(A)= 34

and P(A B) = 56

P(A) + P(B) – P(A B) = 56

34

+ P(B) – 1 54 6

P(B) = 5 3 16 4 4 =

13

P(A B) = 14

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P(A) . P(B) = 3 1 14 3 4

so independent

P(A) P(B) so not equally likelyso option (3)

87. The variance of first 50 even natural numbersis :

(1) 437 (2) 4

437

(3) 4

833(4) 833

Sol. [4]

50 51 101V 4

6 50

50 51 50 514

50 50 4

= 17 [202 – 153]= 17 × 49= 833

88. Let fk(x) = k1

(sink x + cosk x) where x R

and k 1. Then f4(x) – f6(x) equals :

(1) 41

(2) 121

(3) 61

(4) 31

Sol. [2]

f4(x) – f6(x) = 14

(sin4x + cos4x) – 16

(sin6 x + cos6 x)

= 14

(1 – 2sin2x cos2x) – 16 (1 – 3 sin2x cos2x)

= 1 1 3 2 14 6 12 12

89. A bird is sitting on the top of a vertical pole20 m high and its elevation from a point O onthe ground is 45º. It flies off horizontallystraight away from the point O. After onesecond, the elevation of the bird from O isreduced to 30º. Then the speed (in m/s) ofthe bird is :

(1) 20 2 (2) 20( 3 – 1)

(3) 40( 2 – 1) (4) 40( 3 – 2 )

Sol. [2]

tan30° = 20 1

x 20 3

x + 20 = 20 3

x = 20( 3 1)

90. The statement ~ (P ~q) is :(1) a tautology(2) a fallacy(3) equivalent to p q(4) equivalent to ~p q

Sol. [3]

p q ~ q p ~ q ~ p ~ qT T F F TT F T T FF T F T FF F T F T

p qTFFT