A FEW RESULTS OF JEE-MAIN NEET 2017 AMONG 766 … (UG) 201… · Solution to NEET (UG)-2018 (Paper Code-AA) Held on: 6th May, Sunday 3 3k T B 125.44 106 m = × 6 B m T 125.44 10 3k

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A FEW RESULTS OF JEE-MAIN NEET 2017AMONG 766 SELECTIONS

&

Solution to NEET (UG)-2018 (Paper Code-AA)Held on: 6th May, Sunday 2

1. The volume (V) of a monatomic gas varies withits temperature (T), as shown in the graph. The ratioof work done by the gas, to the heat absorbed by it,when it undergoes a change from state A to stateB, is

V

AB

TO

(1) 31

(2) 32

(3) 52

(4) 72

Solution: (3)

AB

PV nRT=

V T P constantα ⇒ =

Q U W= ∆ +

U W1Q Q

∆= +

W U1Q Q

∆= −

V

P

nC T1nC T

∆= −

∆

V

P

C1C

= −

11= −γ

1 3 21 15 / 3 5 5

= − = − =

2. The fundamental frequency in an open organ pipeis equal to the third harmonic of a closed organpipe. If the length of the closed organ pipe is 20cm, the length of the open organ pipe is(1) 12.5 cm (2) 8 cm(3) 13.2 cm (4) 16 cm

Solution: (3)

l

11 P2

γν =

ρ

l =1 20cm

31

3 P4

γ′ν =ρ

Now 3 1′ν = ν

1

3 P 1 P4 2

γ γ=

ρ ρ

⇒ 12 2 20 13.3cm3 3

= = × =

3. At what temperature will the rms speed of oxygenmolecules become just sufficient for escaping fromthe Earth’s atmosphere ?(Given:Mass of oxygen molecule (m) = 2.76 × 10–26 kgBoltzmann’s constant kB = 1.38 × 10–23 JK–1)(1) 5.016 × 104 K (2) 8.360 × 104K(3) 2.508 × 104K (4) 1.254 × 104K

Solution: (2)

Brms

3k T3RTVM m

= =

rms escapeV V=

3B3k T 11.2 10 m/sm

= ×

on squaring both side


6B3k T 125.44 10m

= ×

6

B

mT 125.44 103k

= × ×

266

23

2.76 10 125.44 103 1.38 10

−

−

×= × ×

× ×

3 4346.2 10 83623K 8.36 10 K4.14

= × = = ×

4. The efficiency of an ideal heat engine workingbetween the freezing point and boiling point ofwater, is:(1) 6.25% (2) 20%(3) 26.8% (4) 12.5%

Solution: (3)

2

1

T% 1 100T

η = − ×

2731 100 26.8%373

= − × =

5. A carbon resistor of(47 ± 4.7) k is to be markedwith rings of different colours for its identification.The colour code sequence will be(1) Yellow-Green-Violet-Gold(2) Yellow -Violet-Orange -Silver(3) Violet-Yellow-Orange-Silver(4) Green-Orange-Violet-Gold

Solution: (2)

(47 4.7)k± Ω

⇒ % error 4.7 100 10%47

= × =

347k 47 10Ω = × ΩColour code is

4 → yellow7 →Violet3 →Orange10% →Silver

6. A set of ‘n’ equal resistors, ofvalue ‘R’ each, areconnected in series to a battery of emf ‘E’ andinternal resistance ‘R’. The Current drawn is I.Now, the ‘n’ resistors are connected in parallel tothe same battery. Then the current drawn frombattery becomes 10 I. The value of ‘n’ is:(1) 20 (2) 11(3) 10 (4) 9

Solution: (3)

R R Rn

R I

I(n 1)R

ε=

+ ...(1)

R

10I

RR

R

10I RRn

n10I(n 1)R

ε=

+

ε=

+

...(2)

for (1) and (2)

10I nI

ε=

ε

n = 107. A battery consists of a variable number identical

cells (having internal resistance ‘r’ each) which areconnected in series. The terminals of the batteryare short-circuited and the current I is measured.Which of the graphs shows the correct relationshipbetween I and n ?

(1)

n

I

O

(2)

n

I

O


(3)

n

I

O

(4)

n

I

O

Solution: (3)r r r

nInr rε ε

= =

⇒ ‘I’ is independent of ‘n’I

n8. Unpolarised light is incident from air on a plane

surface of a material of refractive index ‘µ’. At aparticular angle of incidence ‘i’, it is found that thereflected and refracted rays are perpendicular toeach other. Which of the following options iscorrect for this situation ?

(1) 1 1i sinµ

− =

(2) Reflected light is poldrised with its electric

vector perpendicular to the plane of incidence(3) Reflected light is polarised with its electric

vector parallel to the plane of incidence

(4) 1 1i tanµ

− =

Solution: (2)

From Brewster’s Law Ptan I µ=

⇒ 1PI tan (µ)−=

9. In Young’s double slit experiment the separation dbetween the slits is 2 mm, the wavelength λ of thelight used is 5896 Å and distance D between thescreen and slits is 100 cm. It is found that theangular width of the fringes is 0.20°. To increasethe fringe angular width to 0.21° (with sane λ andD) the separation between the slits needs to bechanged to(1) 2.1 mm (2) 1.9 mm(3) 1.8 mm (4) 1.7 mmSolution: (2)

3d 2mm 2 10 m−= = ×65896Å 0.5896 10 m−λ = = ×

D 100cm 1m= =

1 0.20θ = °

2 0.21θ = °

Angular width of fungeswe know that

Dd

λβ =

D dβ λ

θ = =

1 2

2 1

d1d d

θθα ⇒ =

θ

12 1

2

0.20d d 2mm 1.9mm0.21

θ °⇒ = × = × =

θ °

10. An astronomical refracting telescope will havelarge angular magnification and high angularresolution, when it has an objective lens of(1) large focal length and large diameter(2) large focal length and small diameter(3) small focal length and large diameter(4) small focal length and small diameterSolution: (1)

0

e

fMPf

=


11. The ratio of kinetic energy to the total energy of anelectron in a Bohr orbit of the hydrogen atom, is(1) 2: – 1 (2) 1: –1(3) 1 : 1 (4) 1: – 2Solution: (2)

KE 1TE 1

=−

12. An electron of mass m with an initial velocity

0 0ˆV V i (V 0)= >

enters an electric field 0

Ê E i= −

(E0 = constant > 0) at t = 0. If 0λ is its de-Brogliewavelength initially, then its de-Broglie wavelengthat time t is

(1) 0tλ (2) 00

0

eE1 tmV

λ +

(3) 0

0

0

eE1 tmV

λ

+

(4) 0λ

Solution: (3)Velocity of electron at t = 0sec is V0

⇒ de-Broglie wavelength 00

hmV

λ = ...(1)

Velocity of electron at ‘t’ sec.V = U + at

00

eEV V tm

= + 0

0

F eEêE i

eEF â im m

= −

= + ⇒ = =

⇒ de-Broglie wavelength of electron at ‘t’ sec

0

000

0

h heEeEmV 1 tm V tmVm

λλ = = =

++

13. For a radioactive material, half-life is10 minutes.If initially there are 600 number of nuclei, the timetaken (in minutes) for the disintegration of 450nuclei is

(1) 30 (2) 10(3) 20 (4) 15Solution: (3)

0N 600=

No. of Nuclei Disintegrated = 450No. of Nuclei leftN = 600 – 450 = 150

We know that 0n

NN2

=

⇒ n 20N 6002 4 2N 150

= = = =

⇒ n = 2

1/2

tn 2T

= =

1/2t 2T 2 10 min 20min= = × =

14. When the light of frequency 2v0 (where v0 isthreshold frequency), is incident on a metal plate,the maximum velocity of electrons emitted is v1.When the fequency of the incident radiation isincreased to 5v0, the maximum velocity ofelectrons emitted from the same plate is v2. Theratio of v1 to v2 is(1) 4: 1 (2) 1: 4(3) 1 : 2 (4) 2 : 1Solution: (3)

20

1h h mv2

ν = ν +

02ν = ν let velocity = VV1

20 0 1

1h2 h mV2

ν = ν + ...(1)

05ν = ν let velocity = V2

20 0 2

1h5 h mV2

ν = ν +


⇒ 20 2

14h mV2

ν = ...(2)

for (2) and (1)

22

0

201

1 mV4h 21h mV2

ν=

ν

22

2 121

V4 V 2VV

= ⇒ = ⇒ 1

2

V 1V 2

=

15. In the circuit shown in the figure, the input voltage Vi

is 20V, VBE = 0 and VCE = 0. The values of IB, IC and βare given by

RB

RC

500k Vi

4k

20V

B

C

E

(1) B CI 20µA, I 5mA, 250= = β =

(2) B CI 25µA, I 5mA, 200= = β =

(3) B CI 40µA, I 10mA, 250= = β =

(4) B CI 40µA, I 5mA, 125= = β =

Solution: (4)

RB

RB

RC

IB 500k

Vi

20V

4k 20V

0V

0V

0V

E

C

BEV 0V=

CEV 0V=

6

B 3 5

20V 20 10I µA 40µA500 10 5 10

×= = =

× ×

3C 3

20VI 5 10 A 5mA 5000µA4 10

−= = × = =×

C

B

I 5000µA 125I 40µA

β = = =

16. In a p-n junction diode, change in temperature dueto heating(1) does not affect resistance of p-n junction(2) affects only forward resistance(3) affects only reverse resistance(4) affects the overall V - I characteristics of p-n

junctionSolution: (4)

17. In the combination of the following gates theoutput Y can be written in terms of inputs A and Bas

Y

AB

(1) A B A B⋅ + ⋅ (2) A B A B⋅ + ⋅(3) A B⋅ (4) A B+Solution: (2)

y A B A B= ⋅ + ⋅

18. An em wave is propagating in a medium with avelocity ˆV Vi.=

The instantaneous oscillating

electric field of this em wave is along +y axis. Thenthe direction of oscillating magnetic field of theem wave will be along(1) –y direction (2) + z direction(3) –z direction (4) – x directionSolution: (2)

ˆV Vi=

Ê Ej=

Velocity of e.m. wave is along E B×

This imply magnetic field intensity is along k+

⇒ +ve z direction.


19. The refractive index of the material of a prism is2 and the angle of the prism is 30°. One of the

two refracting surfaces of the prism is made amirror inwards, by silver coating. A beam ofmonochromatic light entering the prism from theother face will retrace its path (after reflection fromthe silvered surface) if its angle of incidence onthe prism is(1) 30° (2) 45°(3) 60° (4) zeroSolution: (2)

60°30°i

sin i µsin r

=

sin i 2sin 30

=

1 1sin i 2 sin 30 22 2

= = × =

i = 45°20. An object is placed at a distance of 40 cm from a

concave mirror of focal length 15 cm. If the objectis displaced through a distance of 20 cm towardsthe mirror, the displacement of the image will be(1) 30 cm towards the mirror(2) 36 cm away from the mirror(3) 30 cm away from the mirror(4) 36 cm towards the mirrorSolution: (4)

O C F15cm40cm

u = –40cmf = –15cm

1 1 1v u f

+ =

1 1 1v 40 15

− = −

1 1 1v 15 40

= − = −

8 3 5 1120 120 24

− + − −= = =

v = –24cm

C O F15cm20cm

u = –20cmf = –15cm

1 1 1v u f

+ =

1 1 1v 20 15

− =−

1 1 1v 20 15

= −

1 3 4 1 1 v 60cmv 60 v 60

−= ⇒ = ⇒ = −

Displacement of Image = 60–24 = 36cm away frommirror

21. The magnetic potential energy stored in a certaininductor is 25 mJ, when the cunent in the inductoris 60 mA. This inductor is of inductance(1) 1.389 H (2) 138.88 H(3) 0.138 H (4) 13.89 HSolution: (4)

2L

1U LI2

=

3 3 2125 10 L (60 10 )2

− −× = × × ×

3

6

50 10 L3600 10

−

−

×=

×L = 13.89 H


22. An electron falls from rest through a vertical/distance h in a uniform and vertically upwarddirected electric field E. The direction of electricfield is now reversed, keeping its magnitude thesame. A proton is allowed to fall from rest in itthrough the same vertical distance h. The time offall of the electron, in comparison to the time offall of the proton is(1) 10 times greater (2) 5 times greater(3) smaller (4) equalSolution: (3)

21S Ut at2

= +

21S at2

= ⇒2sta

=

2h 2mht qE qEm

= =

p eq q ,E constant, h constant= = =

e ee p

p p

t mt m t tt m

α ⇒ = ⇒ <

23. The electrostatic force between the metal plates ofan isolated parallel plate capacitor C having acharge Q and area A, is(1) proportional to the square root of the distance

between the plates.(2) linearly proportional to the distance between

the plates.(3) independent of the distance between the plates.(4) inversely proportional to the distance between

the plates.Solution: (3)

A

–Q +Q

0

1 1 QF QE Q2 2 A

= = ∈

2

0

QF2 A

=∈

24. A tuning fork is used to produce resonance in aglass tube. The length of the air column in this tubecan be adjusted by a variable piston. At roomtemperature of 27°C two succeessive resonancesare produced at 20 cm and 73 cm of column length.If the frequency of the tuning fork is 320 Hz, thevelocity of sound in air at 27.°C is(1) 350 m/s (2) 33.9 m/s(3) 330 m/s (4) 300 m/sSolution: (2)

1 x4λ

= +l1

23 x4λ

= + l2

2 134 4λ λ

− = −

2 12λ

= −

2 12( )λ = −

2(73 20)λ = − = 2 × 53

106cmλ =

320Hzν =

106cm 1.06mλ = =

V 320 1.06 339m/s= νλ = × =25. A pendulum is hung from the roof of a sufficiently

‘high building and is moving freely to and fro hikea simple harmonic oscillator. The acceleration ofthe bob of the pendulum is 20 m/s2 at a distance of5 m from the mean position. The time period ofoscillation is(1) 2 s (2) sπ

(3) 2 sπ (4) 1 s


Solution: (2)2a x= −ω2| a | x= ω

2

2

4| a | xTπ

=2Tπ ω =

2 2 xT 4a

= π

x 5T 2 2a 20

= π = π

2T sec2π

= = π

26. A metallic rod of mass per unit length 0.5 kg m–1 islying horizontally on a smooth inclined planewhich makes an angle of 30° with the horizontal.The rod is not allowed to slide down by flowinga current through it when a magnetic field ofinduction 0.25 T is acting on it in the verticaldirection. The current flowing in the rod to keep itstationary is(1) 14.76 A (2) 5.98 A(3) 7.14 A (4) 11.32 ASolution: (4)

10.5kgm−λ =

30θ = °

B 0.25T=

×

I Bl

mg

N

Rod at rest ⇒

I Bcos mg sinθ = θ

m gI tan= θβ

9.80.5 tan 300.25

= × °

I = 11.32A27. A thin diamagnetic rod is placed vertically between

the poles of an electromagnet. When the current inthe electromagnet is switched on, then thediamagnetic rod is pushed up, out of the horizontalmagnetic field. Hence the rod gains gravitationalpotential energy. The work required to do thiscomes from(1) the lattice structure of the material of the rod(2) the magnetic field(3) the curent source(4) the induced electric field due to the changing

magnetic fieldSolution: (3)

28. An inductor 20 mH, a capacitor 100 µF and aresistor 50Ω are connected in series across asource of emf, V = 10 sin 314 t. The power loss inthe circuit is(1) 2.74 W (2) 0.43 W(3) 0.79 W (4) 1.13 WSolution: (3)V=10 sin 314t ...(1)

3L 20mH 20 10 H−= = ×

4C 100µF 10 F−= =

R 50= Ω3

LX L 314 20 10 6.28−= ω = × × = Ω

C 4

1 1X 31.85C 314 10−= = = Ω

ω ×

2 2 2 2C L| Z | R (X X ) 50 (31.85 6.28)= + − = + −


| Z | 56= Ω

rr

10V 2I 126A| Z | 56

= = =

2 2loss rP I R (.126) 50 0.8W= = × =

29. Current sensitivity of a moving coil galvanometeris 5 div/mA and its voltage sensitivity (angulardeflection per unit voltage applied) is 20 div/V.The resistance of the galvanometer is(1) 250 Ω (2) 25Ω

(3) 40Ω (4) 500Ω

Solution: (1)

Current sensitivity = 5div div5000mA Amp

=

Voltage sensitivity =div20volt

iNAB

Cα =

iV

NABCR R

αα = =

⇒ i

v

R α=

α5000

20= ⇒ R 250= Ω

30. A body initially at rest and sliding along africtionless track from a height h (as shown in thefigure) just completes a vertical circle of diameterAB = D. The height h is equal to

B

A

h

(1)7 D5 (2) D

(3)3 D2 (4)

5 D4

Solution: (4)

h r

B

A

2r DDr2

=

=

To complete vertical circle

Velocity at ‘A’ is AV 5gr=

from conservation of M.E.

AV 2gh=

Now 2gh 5gr=

2gh = 5gr

5h r2

=

5h D4

=

31. Three objects, A : (a solid sphere), B : (a thincircular disk) and C : (a circular ring), each havethe same mass M and radius R. They all spin withthe same angular speed ω about their own symmetryaxes. The amounts of work (W) required to bringthem to rest, would satisfy the relation(1) B A CW W W> > (2) A B CW W W> >

(3) C B AW W W> > (4) A C BW W W> >

Solution: (3)

A → solid sphere 2A

2I MR5

=

B → Disc 2B

1I MR2

=

C → Ring 2CI 1MR=

Now C B AI I I> >


⇒ C B A(KE) (KB) (KE)> >

⇒ C B AW W W> >

32. A moving block having mass m, collides withanother stationary block having mass 4m. Thelighter block comes to rest after collision. Whenthe initial velocity of the lighter block is v, then thevalue of coefficient of restitution (e) will be(1) 0.8 (2) 0.25(3) 0.5 (4) 0.4Solution: (2)

m v 4m m 4m v2

i fP P=

2mv 0 0 4mv+ = +

⇒ 2vv4

=

coefficient of restitution

2 1

1 2

u 0v v 14eu u u 0 4

−−= = =

− −

e = 0.2533. Which one of the following statements is

incorrect ?(1) Frictional force opposes the relative motion.(2) Limiting value of static friction is directly

proportional to normal reaction.(3) Rolling friction is smaller than sliding

friction.(4) Coefficient of sliding friction has dimensions

of length.Solution (4)

34. A toy car with charge q moves on a frictionlesshorizontal plane surface under the influence of auniform electric field E

. Due to the force q E , its

velocity increase from 0 to 6 m/s in one secondduration. At that instant the direction of the field isreversed. The car continues to move for two moreseconds under the influence of this field. Theaverage velocity and the average speed of the toy

car between 0 to 3 seconds are respectively.(1) 1 m/s, 3.5 m/s (2) 1 m/s, 3 m/s(3) 2 m/s, 4 m/s (4) 1.5 m/s, 3 m/sSolution: (2)

x x 0

2s1s

6m/s

0s0

3s x 2s

0s to 1s acceleration qEam

=

1s to 2s retardation qEam

= −

2s to 3s acceleration qEam

=

0s to 1s Distance covered

U Vs t2+ =

0 6x (1) 3m2+ = =

0s to 3s Av. velocity 3m 1m3s sec

= =

0s to 3s Av - speed 9m m33s s

= =

35. A block of mass m is placed on a smooth inclinedwedge ABC of inclination θ as shown in the figure.The wedge is given an acceleration ‘a’ towards theright. The relation between a and θ for the block toremain stationary on the wedge is

a

m

BC

A

(1) a g cosθ= (2) asin

gθ

=


(3) acosec

gθ

= (4) a g tanθ=

Solution: (4)Reference: Incline Plane

macos

mgsin

N

ma

mgcos masin

mg

Mass At rest ma cos mgsin⇒ θ = θ

a g tan= θ

36. The moment of the force, F 4i 5j 6k= + −

at (2, 0,–3), about the point (2, –2, –2), is given by

(1) 7i 8j 4k− − − (2) 4i j 8k− − −

(3) 8i 4 j 7k− − − (4) 7i 4 j 8k− − −

Solution: (4)ˆ ˆ ˆF 4i 5j 6k= + −

Position vector of point of application of force

1ˆ ˆ ˆr 2i 0j 3k= + −

Position vector of refrence

2ˆ ˆ ˆr 2i 2 j 2k= + −

1 2r r r= −

ˆ ˆ ˆr 0i 2 j 1k= + −

Torque r Fτ = ×

ˆ ˆ î j k0 2 14 5 6

τ = −−

2 1 0 1 0 2ˆ ˆ î j k5 6 4 6 4 5

− −τ = − +

− −

ˆ ˆ ˆ7i 4 j 8kτ = − − −

37. A student measured the diameter of a small steelball using a screw gauge of least count 0.001 cm.The main scale reading is 5 mm and zero of circularscale division coincides with 25 divisions abovethe reference level. If screw gauge has a zero errorof –0.004 cm, the correct diameter of the ball is

(1) 0.053 cm (2) 0.525 cm(3) 0.521 cm (4) 0.529 cmSolution: (4)L.C. = 0.001cm = 0.01mmMSR = 5mmVSR = 25divDiameter of ball = zero error + MSD + LC × VSR

= 0.04 + 5 + 0.01 × 25= 0.04 + 5 + 0.25= 5.29 mm= 0.529 cm

38. A solid sphere is rotating freely about its symmetryaxis in free space. The radius of the sphere isincreased keeping its mass same. Which of thefollowing physical quantities would remainconstant for the sphere ?(1) Rotational kinetic energy(2) Moment of inertia(3) Angular velocity(4) Angular monentumSolution: (4)Angular Momentum L = constant

MOI 22I MR5

= increase

221 LKE I

2 2I= ω = decreases

LL I constantI

= ω = ⇒ ω = decreases


39. The kinetic energies of a planet in an elliptical orbitabout the Sun, at positiona A, B and C are AK ,

BK and CK , respectively. AC is the major axisand SB is perpendicular to AC at the position ofthe Sun S as shown in the figure. Then

B

CA S

(1) B A CK K K< < (2) A B CK K K> >

(3) A B CK K K< < (4) B A CK K K> >

Solution: (2)

BsunA C

KA > KB > KC

40. If the mass of the Sun were ten times smaller andthe universal gravitational constant were ten timeslarger in magnitude, which of the following is notcorrect ?(1) Time period of a simple pendulum on the Earth

would decrease.(2) Walking on the ground would become more

difficult.(3) Raindrops will fall faster.(4) ‘g’ on the Earth will not change.Solution: (4)

2

GMgR

=

earth

sunsun

M M(Same)MM10

G 10G

=

′ =

′ =

2

10GMg 10gR

′ = =

(1) Time period of Pendulum

T 2g

= π

TT 210g 10

′ = π =

Time period decreases(2) Gravity of Earth increases

Walking becomes difficult(3) As gravity increases

Velocity of rain drops increases(4) g changes

41. A solid sphere is in rolling motion. In rolling motiona body possesses translational kinetic energy ( )tKas well as rotational kinetic energy ( )rKsimultaneously. The ratio ( )t t rK : K K+ for thesphere is(1) 10 : 7 (2) 5 : 7(3) 7 : 10 (4) 2 : 5Solution: (4)

2rot

2trans

K K 2K R 5

− = trans

Total

K 5K 7

⇒ =

42. A small sphere of radius ‘r’ falls from rest in aviscous liquid. As a result, heat is produced due toviscous force. The rate of production of head whenthe sphere attains its terminal velocity, isproportional to

(1) 5r (2) 2r

(3) 3r (4) 4rSolution: (1)The forces acting on the sphere are its weight

34 r g3

π ρ downwards, buoyancy force 34 r g3

π σ

upwards, and viscous force 6 rvπη upwards. The

sphere attains the terminal velocity tv when theresultant force on it is zero i.e.,

3 3t

4 4r g r g 6 rv3 3

π ρ = π σ + πη


Solve above equation to get the terminal velocity,2

t2r g( )v

9ρ − σ

=η

The rate of heat generation is equal to the rate ofwork done by the viscous force which, in turn, isequal to its power. Thus,

2 2 5

t tdQ 8 ( ) g r(6 r v )(v )dt 27

π ρ − σ= π η =

η

43. The power radiated by a black body is P and itradiates maximum energy at wavelength, 0λ . If thetemperature of the black body is now changed sothat it radiates maximum energy at wavelength

03 ,4

λ the power radiated by it becomes nP. Thevalue of n is

(1)25681 (2)

43

(3)34 (4)

81256

Solution: (1)4Power (T)α ...(1)

From Wein’s Displacement Law

mT b (constant )λ =

00 1 2

3T T4λ

λ =

⇒1

2

T 3T 4

= ...(2)

4 41 1

2 2

(Power) T 3 81(Power) T 4 256

= = =

P 81nP 256

=

256n81

=

44. Two wires are made of the same material and havethe same volume. The first wire has cross-sectionalarea A and the second wire has cross-sectional area3A. If the length of the first wire is increased by

l∆ on applying a force F, how much force is neededto stretch the second wire by the same amount ?(1) 4 F (2) 6 F(3) 9 F (4) FSolution: (3)

lA

V V′ =

3A A′ =l l

3′ =

ll

Same material Y Y′⇒ =

Wen know that F / AY( ) /

=∆l l

YA( )F ∆⇒ =

ll

Y = Same, ∆l = Same

AF ∝l

F A 3A 9F A A / 3 1′ ′

= × = × =′l ll l

F 9F′ =45. A sample of 0.1 g of water at 100°C and normal

pressure ( )5 21.013 10 Nm−× requires 54 cal of heat

energy of convert to steam at 100°C. If the volumeof the steam produced is 167.1 cc, the change ininternal energy of the sample, isSolution: (2)(1) 42.2 J (2) 208.7 J(3) 104.3 J (4) 84.5 J


m 0.1gm=

Q 54cal 54 4.18J 225.72= = × =

fV 167.1cc=

iV 0.1cc=

6f iV V V 167cc 167 10 m−∆ = − = = ×

5 2P 1.013 10 Nm−= ×

5 6W P( V) 1.013 10 167 10−= ∆ = × × ×

1.013 16.7= ×

16.92 Joule=

Q U W= ∆ +

225.72 U 16.92= ∆ +

U 208.8J∆ =

46. The correct order of N-compounds in decreasingorder of oxidation states is:(1) 243 N,NO,ClNH,HNO

(2) 243 N,ClNH,NO,HNO

(3) ClNH,N,NO,HNO 423

(4) 324 HNO,NO,N,ClNH

Solution : (3)HNO3 = +5NO = +2N2 = 0NH4Cl = –3

47. Which one of the following elements is unable to

form 36MF ion?

(1) B (2) Al(3) Ga (4) InSolution : (1)

36MF−

Boron doesn’t have vacant d-orbital so it can’t have6 covalury.

48. Considering Ellingham diagram, which of thefollowing metals can be used to reduce alumia ?(1) Mg (2) Zn(3) Fe (4) CuSolution : (1)Ellingham diagram the metal which lie below canreduce metal which lie above to it.

49. The correct order of atomic radii in group 13elements is:(1) B < Ga < Al < Tl < In(2) B < Al < Ga < In < Tl(3) B < Al < In < Ga < Tl(4) B < Ga < Al < In < TlSolution : (4)B < Ga < Al < In < Tl

↓due to poor shriding of d-orbitals

50. Which of the following statements is not true forhalogens ?(1) All but fluorine show positive oxidation states.(2) All are oxidizing agents.(3) All form monobasic oxyacids.(4) Chlorine has the highest electron-gain enthalpy.Solution : (2)

51. In the structure of ,ClF3 the number of lone pairsof electrons on central atom ‘Cl’ is(1) four (2) two(3) one (4) threeSolution : (2)

F |Cl – F | F

:

:

52. Identify the major products P, Q and R in thefollowing sequence of reactions:

+ CH CH CH Cl3 2 23

AnhydrousAlCl→

23

(i)O(ii)H O+/

P Q+R∆

→


(1)CH(CH )3 2 OH

CH CH(OH)CH3 3, ,

(P) (Q) ( )R

(2)

CH CH2 3CH2

,

CHO

,

COOH

(3)

CH CH2 3CH2

,

CHO

, CH CH – OH3 2

(4)CH(CH )3 2

,

OH

, CH – CO – 3 CH3

Solution: (4)

+ CH – CH – CH – Cl3 2 23

./

AnhydA Cl

→

(F.C. Alkylation)with Rearrangementof carbonation

CHCH3 CH3

2

3

(i)O(ii)H O /+ ∆

→

OH O ||+ CH – e – eH3 3

53. Which of the following compounds can form azwitterion ?(1) Benzoic acid (2) Acetanlide(3) Aniline (4) GlycineSolution : (4)

Nh3

+o

o

–

54. Regarding cross-linked or network polymers,which of the following statements is incorrect ?

(1) Examples are bakelite and melainine.(2) They are formed from bi-and tri-functional

monomers.(3) They contain covalent bonds between various

linear polymer chains.(4) They contain strong covalent bonds in their

polymer chains.Solution : (3)Regarding cross linked polyer.

55. Nitration of aniline in strong acidic medi’um alsogives in-nitroaniline because(1) In absence of substituents nitro group always

goes to m-position.(2) In electrophilic substitution reactions amino

group is meta directive.(3) In spite of substituents nitro group always goes

to only m-position.(4) In acidic (strong) medium aniline is present as

aniliniun ion.Solution : (4)In acidic medium aniline is present as anilium ion.

56. The difference between anylose and amylopectinis:(1) Amylopectin have 1 4 -linkage and 1 6

-linkage.(2) Amylose have 1 4 -linkage and 1 6

-linkage.(3) Amylopectin have 1 4 -linkage and 1 6

-linkage.(4) Amylose is made up of glucose and galactoseSolution : (3)

57. A mixture of 2.3 g formic acid and 4.5 g oxalicacid is treated with conc. 42SOH . The evolvedgaseous mixture is passed through KOH pellets.Weight (in g) of the repeining product at STP willbe(1) 2.8 (2) 3.0(3) 1.4 (4) 4.4Solution: (1) 2.8 gm

58. Which of the following oxides is most acidic innature ?


(1) BaO (2) BeO(3) MgO (4) CaOSolution: (2) BeO

59. Which oxide of nitrogen is not a common pollutantintroduced into the atmosphere both due to naturaland human activity ?(1) ON2 (2) 2NO

(3) 52ON (4) NO

Solution: (3) 2 5N O

60. The compound A on treatment with Na gives B,and with 5PCl gives C, B and C react together togive diethyl ether. A, B and C are in the order(1) OHHC,HC,ClHC 526252

(2) ONaHC,ClHC,OHHC 525252

(3) ClHC,HC,OHHC 526252

(4) ClHC,ONaHC,OHHC 525252

Solution : (4) 2 5 2 5 2 5C H OH,C H ONa,C H Cl

61. The compound 87HC undergoes the followingreactions:

C H7 8

3 C / l2 ∆A

Br / Fe2 B Zn / HCl CThe product ‘C’ is(1) 3-bromo-2, 4, 6-trichlorotoluene(2) o-bromotoluene(3) m-bromotoluene (4) p-bromotolueneSolution: (3)

CH3

23 /3Cl

HCl∆

−→

CCl3

2 /Br Fe→

CCl3

Br

CH3

Br

Zn/HCl

62. Hydrocarbon (A) reacts with bromine bysubstitution to form an alkyl bromide which byWurtz reaction is converted to gaseous hydrocarboncontaining less than four carbon atoms. (A) is:(1) 33 CHCH (2) 22 CHCH(3) CHCH (4) 4CHSolution : (4)

24 2 3 2 6CH Br CH Br C HNa

ether+ → − →

63. Which of the following molecules represents theorder of hybridisation sp2, sp2, sp, sp from left toright atoms ?(1) 22 CHCHCHCH(2) CHCCHCH2(3) CHCCHC(4) 33 CHCHCHCH

Solution : (2)

CH = CH – C CH2 ≡

Sp Sp Sp Sp2 2

64. Which of the following carbocations is expectedto be most stable ?

(1)

NO2

HY ⊕

(2)

NO2

HY

⊕

(3)

NO2

HY⊕

(4)

NO2

HY ⊕

Solution : (1)NO2

⇒ 3 Resonatic structureH

Y ⊕


65. Which of the following is correct with respect to–I effect of the substituents ? (R = alkyl)

(1) 2NH OR F− > − > −

(2) 2NR OR F<− − < −

(3) 2NH OR F<− − < −

(4) 2NR OR F>− − > −

Solution : (3)

2NH OR F<− − < −

66. In the reaction

OH

+ CHCl + NaOH3

O Na– +

CHO

the electrophile involved is:

(1) dichloromethyl anion (CHCl )2

(2) formyl cation (CHO)

(3) dichloromethyl cation (CHCl )2

(4) dichlorocarbene )CCl(: 2

Solution : (4)Carbylamine reactionIntermediate = carbone (: CCl2)

67. Carboxylic acids have higher boiling points thanaldehydes, ketones and even alcohols ofcomparable molecular mass. It is due to their(1) more extensive association of carboxylic acid

via van der Waals force of attraction(2) formation of carboxylate ion(3) formation of intramolecular H-bonding(4) formation of intermolecular H-bondingSolution : (4)Formation of intermolecular H-Bonding

68. Cumpqund A, ,OHC 108 is found to react withNaOI (produced by reacting Y with NaOH) andyields a yellow precipitate with characteristic smell.A and Y are respectively

(1) CH – CH and I3 2

OH

(2) CH – CH – OH and I2 2 2

(3) CH – OH and I2 2H C3

(4) OH and I2CH3

CH3

Solution : (1)Haloform Reaction

CH – |

OH

3

Ch and I2

69. Match the metal ions given in Column I with thespin magnetic moments of the ions given inColumn II and assign the corret code:Column I Column II

a. 3Co i. 8 B.M.

b. 3Cr ii. 35 B.M.

c. 3Fe iii. 3 B.M.

d. 2Ni iv. 24 B.M.

v. 15 B.M.

a b c d(1) iv i ii iii(2) i ii iii iv(3) iv v ii i(4) iii v i iiSolution : (3)

63 24 .Co d Bm+ ⇒ =33 15 .Cr d Bm+ ⇒ =53 35 .Fe d Bm+ ⇒ =82 8 .Ni d Bm+ ⇒ =


70. Which one of the following ions exhibits d-dtransition and paramagnetism as well ?

(1) 4MnO (2) 272OCr

(3) 24CrO (4) 2

4MnO

Solution: (4) 24MnO

71. Iron carbonyl, 5)CO(Fe is:(1) trinuclear (2) mononuclear(3) tetranuclear (4) dinuclearSolution : (2)Iron Carboxy Fe(CO)s isMononuclear

72. The type of isomerism shown by the complex])en(CoCl[ 22 is:

(1) Ionization isomerism(2) Coordination isomerism(3) Geometrical isomerism(4) Linkage isomerismSolution : (3)The type of Isomerism.geometrical isomerism

73. The geometry and magnetic behaviour of thecomplex ])CO(Ni[ 4 are:(1) square planar geometry and paramagnetic(2) tetrahedral geometry and diamagnetic(3) square planar geometry and diamagnetic(4) tetrahedral geometry and paramagneticSolution : (2)tetrahedral geometry and diamegnetic.

74. Following solutions were, prepared by mixingdifferent volumes of NaOH and HCl of differentconcentrations:

a. NaOH10MmL40HCl

10MmL60

b. NaOH10MmL45HCl

10MmL55

c. NaOH5MmL25HCl

5MmL75

d. NaOH10MmL100HCl

10MmL100

pH of which one of them will be equal to 1 ?(1) d (2) a(3) b (4) cSolution : (4)

M M75mL HCl 25mL5 5

NaOH+

75 15

× 25−15

×10

100 100=

[ ]110 H−= = +

PH = 175. On which of the following properties does the

coagulating puwer of an ion depend ?(1) Both magnitude and sign of the charge on the

ion.(2) Size of the ion alone(3) The magnitude of the charge on the ion alone(4) The sign of charge on the ion aloneSolution : (3)The magnitude of the charge on the Ion alone.

76. Given van der Waals constant for 3NH , 2H , 2Oand 2CO are respectively 4.17, 0.244, 1.36 and3.59, which one of the following gases is mosteasily liquefied ?(1) 2O (2) 2H

(3) 3NH (4) 2CO

Solution : (3)liquification of agas α, ‘a’a → vander waal constant for pressureGas a

3NH 4.17

2H 0.244

2O 1.36

2CO 3.59

So, 3NH can be easily liquified.


77. The solubility of 4BaSO in water is3 12.42 10 gL− −× at 298 K. The value of its

solubility product ( )spK will be

(Given molar mass of 14BaSO 233g mol−= )

(1) 14 2 21.08 10 mol L− −×

(2) 12 2 21.08 10 mol L− −×

(3) 10 2 21.08 10 mol L− −×

(4) 8 2 21.08 10 mol L− −×Solution : (3)solubility of BaSO4, s = 2.42 × 10–3 g/L.

32.42 10 mol/L.233

−= ×

4 2BaSO (s) H O( )l 2 24Ba (aq.) SO (aq.)

t = 0 – –s s

t = teq.2 2

4Ba SOspK + − =

22 2.42s s s 10 3

233 = × = = × −

10 2 21.08 10 /mol L−= ×78. In which case is the number of molecules of water

maximum ?(1) 0.00224 L of water vapours at 1 atm and

273K(2) 0.18 g of water(3) 18 mL of water

(4) 310− mol of waterSolution : (3)In which case18 ml of water ⇒ m = v × d

m = 18 ml × 1 gml–1

= 18g = 1 mol.79. The correct difference between first-and second-

order reaction is that

(1) a first-order reaction can be catalyzed; asecond-order reaction cannot be catalyzed

(2) the half-life of a first-order reaction does not

depend on [ ]0A ; the half-life of a second-order

reaction does depend on [ ]0A .(3) the rate of a first-order reaction does not

depend on reactant concentration; the rate ofa second-order reaction does depend onreactant concentration.

(4) the rate of a first-order reaction does dependon reactant concentration; the rate of a second-order reaction does not depend on reactantconcentrations

Solution : (2)

( )1

2 11

ntAo −∝

80. Among 2CaH , 2BeH , 2BaH , the order of ioniccharacter is(1) 2 2 2BeH BaH CaH< <

(2) 2 2 2CaH BeH BaH< <

(3) 2 2 2BeH CaH BaH< <

(4) 2 2 2BaH BeH CaH< <

Solution : (3)

2 2 2BeH CaH BaH< <

81. Consider the charge in oxidation state of Brominecoresponding to different emf values as shown inthe diagram below :

1.82 V 1.5 V4BrO−

3BrO− HBrO

1.595 VBr21.0652 V

Br–

Then the species undergoing disproportionation is

(1) 2Br (2) 4BrO−

(3) 3BrO− (4) HBrOSolution : (4)


1 02HBrO Br .

+→ 2

0HBrO/Br 1.595vE =

5 13BrO HBrO.

+ +− → 3

0BrO /HBrO 1.5v

E − =

[oxidation]

[Reduction]

Hence, HBrO has equal tendency for oxidation aswell as reduction.HBrO undergo disproposition reaction.

82. For the redox reaction2 2

4 2 4 2 2MnO C O H Mn CO H O− − + ++ + → + +the correct coefficients of the reactants for thebalanced equation are

4MnO− 22 4C O −

H+

(1) 2 16 5(2) 2 5 16(3) 16 5 2(4) 5 16 2Solution : (2)

2 24 2 4 2 2MnO C O H Mn CO H O.− − + ++ + → + +

+3

5+7

+41

n = 1 2×= 2

n = 5

So, 24 2 4MnO 5C O 16H− −+ + +

22 22Mn 10CO 8H O.+→ + +

83. Which one of the following conditions will favourmaximum formation of the product in the reaction.

2 2A ( ) B ( )g g 2 rX ( ) H= X kJ ?g(1) High temperature and high pressure(2) Low temperature and low pressure(3) Low temperature and high pressure(4) High temperature and low pressureSolution : (3)

2 2A g B g 2 , H x kJX g .For formation of product, reaction should proceedin forward direction.So, For exothermic reaction,

so reaction proceeds in forward direction.

T K↓ ⋅ ↑ ⋅

For the reaction, 2 2

2pQ

.T

A B T

nx nn n P

=

On P Pand Q < K so reaction,

proceeds in forward direction.PP Q↑ ↓

84. When initial concentration of the reactant isdoubled, the half-life period of a zero orderreaction.(1) is tripled (2) is doubled(3) is halved(4) remains unchangedSolution : (2)For zero order reaction,

( )12

0initial

=2atk

( ) ( )1 12 2

0final

2= 2 initial.2at tk

=

85. The bond dissociation energies of 2X , 2Y and XYare in the ratio of 1 : 0.5 : 1. ∆H for the formation

of XY is 1200 kJ mol .−− The bond dissociationenergy of 2X will be

(1) 1800 kJ mol− (2) 1100 kJ mol−

(3) 1200 kJ mol− (4) 1400 kJ mol−

Solution : (1)∆H

2X 2X→ x. kJ.

2Y 2Y→ 0.5 x. kJ.

XY X Y→ + x. kJ.


X2 Y2 2 4× ∆ ×H = 2 (– 200)

+ x kJ

2x 2y+

– 2x kJ = – 400 kJ+ 0.5x kJ

Apply Hess’s law → ax + 0.5x – 2x = – 400

2 400x− = −

x = 800 kJ.

86. The correction factor ‘a’ to the ideal gas equationcorresponds to(1) electric field present between the gas molecules(2) volume of the gas molecules(3) density of the gas molecules(4) forces of attraction between the gas moleculesSolution : (4)‘a’ denotes, forces of attraction between gasmolecules.

87. Consider the following species:

CN ,CN ,NO and CN+ −

Which one of these will have the highest bond order ?

(1) CN+ (2) CN− (3) NO (4) CNSolution : (2)

No. of electrons Electronic configuation

CN+

CN–

NO

CN

6 + 7 – 1 = 12

6 + 7 + 1 = 14

7 + 8 = 15

6 + 7 = 13

2 2 2 2 2 1 11 1 2 2 2 2 2 2s s s s p px pyσ σ σ σ σ π π=

2 2 2 2 2 1 11 1 2 2 2 2 2 2s s s s p px pyσ σ σ σ σ π π=

2 2 2 2 2 1 1 *1 *1 1 2 2 2 2 2 2 2 2s s s s p px py px pyσ σ σ σ σ π π π π= =

2 *2 2 *2 2 2 11 1 2 2 2 2 2 2s s s s p px pyσ σ σ σ σ π π=

So, B.O. for 8 4 22 2

b aN NCN + − −= = =

B.O. for 10 4 32

CN − −= =

B.O. for 10 5 2.52

NO −= =

B.O. for 9 4 2.52

CN −= =

88. Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration

of (X) is 2 2 31s 2s 3p , the simplest formula for this compound is

(1) 2Mg X (2) 2MgX (3) 2 3Mg X (4) 3 2Mg X


Solution : (4)Electronic Configuration Valency

Mg. 2 21s 2s 6 22p 3s 2

X. 2 21s 2s 32p 3

So, Mg + 2 & x – 3 will form E formula will be

3 2Mg X .

89. Iron exhibits bcc structure at room temperature.Above 900°C, it transforms to fcc structure. Theratio of density of iron at room temperature to thatat 900°C (assuming molar mass and atomic radiiof iron remains constant with temperature) is

(1)3 34 2

(2)4 33 2

(3)32

(4)12

Solution : (1)

3A

ZMdensityN a

=

For BCC, ( )3

4RA 3

2.2 Mdensity ,3 4R.N

z

a

=×=

=

For FCC, ( )3

4RA 2

4.4 Mdensity ,2 4R.N

z

a

=×=

=×

( )

( )

34RA 3

34RA 2

density/BCC 2M 3 3density/FCC 4 2N

4M

N

= =

90. Which one is a wrong statements ?(1) The electronic configuration of N atom is

1s2 2s 1y2p 1

z2p1x2p

(2) An orbital is designated by three quantumnumbers while an electron in an atom isdesignated by four quantum numbers.

(3) Total orbitals angular momentum of electronin ‘s’ orbital is equal to zero.

(4) The value of m for zd 2 is zero.

Solution : (1)In option (1), Electronic configuration of N is

1s2 2s 2p

Hund’s Rule is voilatedbecause acc. to Hund’s Rule, when electrons arefilled in degenerate orbitals they have some spinquantum number.

91. Oxygen is nof produced during photosynthesis by(1) Cycas(2) Nostoc(3) Green sulphur bacteria(4) Chara

Ans: (3)

92. Double fertilization is:(1) Fusion of two male gametes with one egg(2) Fusion of one male gamete with two polar

nuclei(3) Fusion of two male gametes of a pollen tube

with two different eggs(4) Syngamy and triple fusion

Ans: (4)

93. Which one of the following plants shows a veryclose relationship with a species of moth, wherenone of the two can complete its life cycle withoutthe other ?(1) Banana (2) Yucca(3) Hydrilla (4) Viola

Ans: (2)

94. Pollen grains can be stored for Several years inliquid nitrogen having a temperatuie of(1) – 196ºC (2) – 80ºC(3) – 120ºC (4) – 160ºC

Ans: (1)


95. Which of the following elements is responsible formaintaining turgor in cells ?(1) Potassium (2) Sodium(3) Magnesiun (4) Calcium

Ans: (1)

96. What is the role of NAD+ in cellular respiration ?(1) It is a nucleotide source for ATP synthesis.(2) It functions as an electron carrier(3) It functions as an enzyme.(4) It is the final electron acceptor for anaerobic

respiration. Ans: (2)

97. In which of the following forms is iron absorbedby plants ?(1) Free element(2) Ferrous(3) Ferric(4) Both ferric and ferrous Ans: (3)

98. Which of the following is commonly used as avector for introducing a DNA fragment in humanlymphocytes ?(1) phage (2) Ti plasmid(3) Retrovirus (4) pBR 322

Ans: (3)

99. Use of bioresources by multinational companiesand organisations without authorisation from theconcerned country and its peoople is called(1) Biodegradation (2) Biopiracy(3) Bio-infringement (4) Bioexploitation

Ans: (2)

100. In India, the organisation responsible for assessingthe safety of introducing genetically modifiedorganisms for public use is:(1) Research Committee on Genetic Manipulation

(RCGM)(2) Council for Scientific and Industrial Research

(CSIR)(3) Indian Council of Medical Research (ICMR)(4) Genetic Engineering Appraisal Committee

(GEAC) Ans: (4)

101. The correct order of steps in Polymerase ChainReaction (PCR) is:

(1) Denaturation, Extension, Annealing(2) Annealing, Extension, Denaturation(3) Extension, Denaturation, Annealing(4) Denaturation, Annealing, Extension

Ans: (4)

102. Select the correct Match:(1) T.H. Morgan – Transduction(2) F2 × Recessive parent – Dihybrid cross(3) Ribozyme – Nucleic acid(4) G. Mendel – Transformation

Ans: (3)

103. A ‘new variety of rice was patented by a foreigncompany though such varieties have been presentin India for a long time. This is related to:(1) Lerma Rojo (2) Sharbati Sonora(3) Co-667 (4) Basmati

Ans: (4)

104. Which of the following pairs in wrongly matched?(1) XO type sex : Grasshopper

determination(2) ABO blood grouping : Co-dominance(3) Starch synthesis in pea : Multiple alleles(4) T.H. Morgan : Linkage

Ans: (3)

105. Select the correct statement:(1) Spliceosomes take part in translation.(2) Punnett square was developed by a British

scientist.(3) Franklin Stahl coined the term “linkage”.(4) Transduction was discovered by S. Altman.

Ans: (2)

106. The experimental proof for semiconservativereplication of DNA was first shown in a(1) Plant (2) Bacterium(3) Fungus (4) Virus

Ans: (2)

107. Which of the following flowers only once in itslife-time ?(1) Mango (2) Jackfruit(3) Bamboo species (4) Papaya

Ans: (3)


108. Offsets are produced by:(1) Parthenocarpy (2) Mitotic divisions(3) Meiotic divisions (4) Parthenogenesis

Ans: (2)

109. Select the correct match:(1) Matthew Meselson – Pisum sativum

and F. Stahl

(2) Alfred Hershey and – TMVMartha Chase

(3) Alec Jeffreys – Streptococcuspneumoniae

(4) Francois Jacob and – Lac operonJacques Monod

Ans: (4)

110. Which of the following has proved helpful inpreserving pollen as fossils ?(1) Oil content (2) Cellulosic intine(3) Pollenkitt (4) Sporopollenin

Ans: (4)

111. Natality refers to:(1) Number of individuals leaving the habitat(2) Birth rate(3) Death rate(4) Number of individuals entering a habitat

Ans: (2)

112. World Ozone Day is celebrated on:(1) 16th September (2) 21st April(3) 5th June (4) 22nd April

Ans: (1)

113. Which of the following is a secondary pollutant(1) SO2 (2) CO2(3) CO (4) O3 Ans: (4)

114. Niche is:(1) the range of temperature that the organism

needs to live(2) the physical space where an organism live(3) all the biological factors in the organism

environment(4) the functional role played by the organism

where it livesAns: (4)

115. What type of ecological pyramid would obtainedwith the following data ?

Secondary consumer : 120 gPrimary consumer : 60 gPrimary producer : 10 g

(1) Upright pyramid of numbers(2) Pyramid of energy(3) Inverted pyramid of biomass(4) Upright pyramid of biomass Ans: (3)

116. In stratosphere, which of the following element actsas a catalyst in degradation of ozone a release ofmolecular oxygen ?(1) Fe (2) Cl(3) Carbon (4) Oxygen

Ans: (2)

117. The two functional groups characteristic of sugarsare:(1) carbonyl and phosphate(2) carbonyl and methyl(3) hydroxyl and methyl(4) carbonyl and hydroxyl Ans: (4)

118. Which among the following is not a prokaryote ?(1) Nostoc (2) Mycobacterium(3) Saccharomyces (4) Oscillatoria

Ans: (3)

119. The Golgi complex participates in:(1) Respiration in bacteria(2) Formation of secretory vesicles(3) Fatty acid breakdown(4) Activation of amino acid

Ans: (2)

120. Which of the following is not a product of lightreaction of photosynthesis ?(1) NADPH (2) NADH(3) ATP (4) Oxygen

Ans: (2)

121. Which of the following is true for nucleolus ?(1) It takes part in spindle formation.(2) It is a membrane-bound structure.(3) Larger nucleoli are present in dividing cells.(4) It is a site for active ribosomal RNA synthesis.


Ans: (4)

122. Stomatal movement is not affected by:(1) O2 concentration (2) Light(3) Temperature (4) CO2 concentration

Ans: (1)

123. The stage during which separation of the pairedhomologous chromosomes begins is:(1) Diakinesis (2) Diplotene(3) Pachytene (4) Zygotene

Ans: (2)

124. Stomata in grass leaf are:(1) Rectangular (2) Kidney shaped(3) Dumb-bell shaped (4) Barrel shaped

Ans: (3)

125. Secondary xylem and phloem in dicot stem areproduced by:(1) Phellogen (2) Vascular cambium(3) Apical meristems (4) Axillary meristems

Ans: (2)

126. Pneumatophores occur in:(1) Carnivorous plants(2) Free-floating hydrophytes(3) Halophytes(4) Submerged hydrophytes Ans: (3)

127. Casparian strips occur in:(1) Cortex (2) Pericycle(3) Epidermis (4) Endodermis

Ans: (4)

128. Plants having little or no secondary growth are:(1) Conifers(2) Deciduous angiosperms(3) Grasses(4) Cycads Ans: (3)

129. Sweet potato is a modified:(1) Tap root (2) Adventitious root(3) Stem (4) Rhizome

Ans: (2)

130. Which of the following statement is correct ?(1) Horsetails are gymnosperms(2) Selaginella is heterosporous, while Salvinia is

homosporous

(3) Ovules are not enclosed by ovary wall ingymnosperms

(4) Stems are usually unbranched in both Cycasand Cedrus Ans: (4)

131. Select the wrong statement:(1) Pseudopodia are locomotory and feeding

structures in Sporozoans(2) Mushrooms belong to Basidiomycetes(3) Cell wall is present in members of Fungi and

Plantae(4) Mitochondria are the powerhouse of the cell in

all kingdoms except Monera Ans: (1)

132. After karyogamy followed by meiosis, spores areproduced exogenously in:(1) Agaricus (2) Alternaria(3) Neurospora (4) Saccharomyces

Ans: (1)

133. Match the items given in Column I with those inColumn II and select the correct option givenbelow:

Column-I Column-IIa. Herbarium i. It is a place having a collection

of preserved plants and animals.

b. Key ii. A list that enumeratesmethodically all the species foundin an area with brief descriptionaiding identification.

c. Museum iii. Is a place where dried and pressedplant specimens mounted onsheets are kept.

d. Catalogue iv. A booklet containing a list ofcharacters and their alternateswhich are helpful in identificationof various taxa.

a b c d

(1) ii iv iii i(2) iii ii i iv(3) i iv iii ii(4) iii iv i ii

Ans: (4)

134. Winged pollen grains are present in:


(1) Mango (2) Cycas(3) Mustard (4) Pinus

Ans: (4)

135. Which one is wrongly matched ?(1) Gemma cups – Marchantia(2) Biflagellate zoospores – Brown algae(3) Uniflagellate gametes – Polysiphonia(4) Unicellular organism – Chlorella

Ans: (3)

136. Which of the following options correctly representsthe lung conditions in asthma and emphysema,respectively ?(1) Increased respiratory surface; Inflammation of

bronchioles(2) Increased number of bronchioles; Increased

respiratory surface(3) Inflammation of bronchioles; Decreased

respiratory surface(4) Decreased respiratory surface; Inflammation of

bronchiolesAns: (3)


Column I Column II

a. Tricuspid valve i. Between left atriumand left ventricle

b. Bicuspid valve ii. Between right ventricleand pulmonary artery

c. Semilunar valve iii. Between right atriumand right ventricle

a b c(1) i ii iii(2) i iii ii(3) iii i ii(4) ii i iii Ans: (3)

138. Match the items given Column I with those inColumn II and select the correct option givenbelow:

Column I Column II

a. Tidal volume i. 2500-3000 mL

b. Inspiratory Reserve ii. 1100-1200 mLvolume

c. Expiratory Reserve iii. 500-550 mLvolume

d. Residual volume iv. 1000-1100 mL

a b c d

(1) i iv ii iii(2) iii i iv ii(3) iii ii i iv(4) iv iii ii i

Ans: (2)

139. The transparent lens in the human eye is held in itsplace by:(1) smooth muscles attached to the iris(2) ligaments attached to the iris(3) ligaments attached to the ciliary body(4) smooth muscles attached to the ciliary body

Ans: (3)

140. Which of the following is an amino acid derivedhormone ?(1) Estradiol (2) Ecdysone(3) Epinephrine (4) Estriol

Ans: (3)

141. Which of the following hormones can play asignificant role in osteoporesis ?(1) Estrogen and Parathyroid hormone(2) Progesterone and Aldosterone(3) Aldosterone and Prolactin(4) Parathyroid hormone and Prolactin

Ans: (1)

142. Which of the following structures or regions isincorrectly paired with its function ?

(1) Hypothalamus: production of releasinghormones and regulation oftemperature, hunger andthirst.

(2) Limbic system: consists of fibre tracts thatinterconnect different regionsof brain; controls movement.


(3) Medulla controls respiration andoblongata: cardiovascular reflexes.

(4) Corpus band of fibers connecting leftcallosum: and right cerebral hemispheres.

Ans: (2)

143. The amnion of mammalian embryo is derived from(1) mesoderm and trophoblast(2) endoderm and mesoderm(3) ectoderm and mesoderm(4) ectoderm and endoderm

Ans: (3)

144. Hormones secreted by the placenta to maintainpregnancy are:(1) hCG, hPL, progestogens, estrogens(2) hCG, hPL, estrogens, relaxin, oxytocin(3) hCG, hPL, progestogens, prolactin(4) hCG, progestogens, estrogens, glucocorticoids

Ans: (1)

145. The difference between spermiogenesis andspermiation is:(1) In spermiogenesis spermatozoa from Sertoli

cells are released into the cavity of seminiferoustubules, while in spermiation spermatozoa areformed.

(2) In spermiogenesis spermatozoa are formed,while in spermiation spermatids are formed.

(3) In spermiogenesis spermatids are formed, whilein spermiation spermatozoa are formed.

(4) In spermiogenesis spermatozoa are formed,while in spermiation spermatozoa are releasedfrom Sertoli cells into the cavity ofseminiferous tubules.

Ans: (4)

146. The contraceptive ‘SAHELI’(1) is an IUD.(2) increases the concentration of estrogen and

prevents ovulation in females.(3) blocks estrogen receptors in the uterus,

preventing eggs from getting implanted.(4) is a post-coital contraceptive.

Ans: (3)

147. Ciliates differ from all other protozoans in:(1) using pseudopodia for capturing prey(2) having a contractile vacuole for removing

excess water(3) using flagella for locomotion(4) having two types of nuclei

Ans: (4)

148. Identify the vertebrate group of animalscharacterized by crop and gizzard in its digestivesytstem.(1) Aves (2) Reptilia(3) Amphibia (4) Osteichthyes

Ans: (1)

149. Which of the following features is used to identifya male cockroach from a female cockroach ?(1) Forewings with darker tegmina(2) Presence of caudal styles(3) Presence of a boat shaped sternum on the 9th

abdominal segment(4) Presence of anal cerci

Ans: (4)

150. Which one of these animals is not a homeotherm?(1) Camelus (2) Chelone(3) Macropus (4) Psittacula

Ans: (2)

151. Which of the following animals does not undergometamorphosis ?(1) Moth (2) Tunicate(3) Earthworm (4) Starfish

Ans: (3)

152. Which of the following organisms are known aschief producers in the oceans ?(1) Cyanobacteria (2) Diatoms(3) Dinoflagellates (4) Euglenoids

Ans: (2)

153. Which one of the following population interactionsis widely used in medical science for the productionof antibiotics ?(1) Parasitism (2) Mutualism(3) Commensalism (4) Amensalism

Ans: (4)


154. All of the following are included in ‘Ex-situconservation’ except(1) Botanical gardens(2) Sacred groves(3) Wildlife safari parks(4) Seed banks

Ans: (2)


Column I Column II

a. Eutrophication i. UV-B radiation

b. Sanitary landfill ii. Deforestation

c. Snow blindness iii. Nutrient enrichment

d. Jhum cultivation iv. Waste disposal

a b c d

(1) iii iv i ii(2) i iii iv ii(3) ii i iii iv(4) i ii iv iii

Ans: (1)

156. In a growing population of a country:(1) reproductive and pre-reproductive individuals

are equal in number.(2) reproductive individuals are less than the

post-reproductive individuals.(3) pre-reproductive individuals are more than the

reproductive individuals.(4) pre-reproductive individuals are less than the

reproductive individuals.Ans: (3)

157. Which part of poppy plant is used to obtain thedrug. “Smack” ?(1) Roots (2) Latex(3) Flowers (4) Leaves

Ans: (2)

158. All of the following are part of an operon except:(1) an enhancer (2) structural genes(3) an operator (4) a promoter

Ans: (1)

159. A woman has an X-linked condition on one of herX-chromosomes. This chromosome can beinherited by:

(1) Only grandchildren(2) Only sons(3) Only daughters(4) Both sons and daughters

Ans: (4)

160. According to Hugo de Vries, the mechanism ofevolution is:(1) Phenotypic variations(2) Saltation(3) Multiple step mutations(4) Minor mutations

Ans: (2)

161. AGGTATCGCAT is a sequence from the codingstrand of a gene. What will be the correspondingsequence of the transcribed mRNA ?

(1) ACCUAUGCGAU

(2) UGGTUTCGCAT

(3) AGGUAUCGCAU

(4) UCCAUAGCGUAAns: (3)


Column I Column II

a. Proliferative Phase i. Breakdown ofendometrial lining

b. Secretory Phase ii. Follicular Phase

c. Menstruation iii. Luteal Phase

a b c

(1) ii iii i

(2) i iii ii

(3) iii ii i

(4) iii i iiAns: (1)



Column I Column II

a. Glycosuria i. Accumulation of uricacid in joints

b. Gout ii. Mass of crystallisedsalts within thekidney

c. Renal calculi iii. Inflammation inglomeruli

d. Glomerular iv. Presence of glucose innephritis urine

a b c d

(1) ii iii i iv(2) i ii iii iv(3) iii ii iv i(4) iv i ii iii

Ans: (4)

164. Match the items given in Column I with thoseColumn II and select the correct option givenbelow:

Column I Column II(Function) (Part of Excretory System)

a. Ultrafiltration i. Henle’s loop

b. Concentration ii. Ureterof urine

c. Transport of iii. Urinary bladderurine

d. Storage of urine iv. Malpighian corpuscle

v. Proximal convolutedtubule

a b c d

(1) v iv i ii(2) iv i ii iii(3) iv v ii iii(4) v iv i iii

Ans: (2)

165. Which of the following gastric cells indirectly helpin erythropoiesis ?(1) Goblet cells (2) Mucous cells(3) Chief cells (4) Parietal cells

Ans: (4)


Column I Column II

a. Fibrinogen i. Osmotic balance

b. Globulin ii. Blood clotting

c. Albumin iii. Defence mechanism

a b c

(1) i iii ii(2) i ii iii(3) iii ii i(4) ii iii i

Ans: (4)

167. Which of the following is an occupationalrespiratory disorder ?(1) Botulism (2) Silicosis(3) Anthracis (4) Emphysema

Ans: (2)

168. Calcium is important in skeletal muscle contractionbecause it:(1) detaches the myosin head from the actin

filament.(2) activates the myosin ATPase by binding to it.(3) binds to troponin to remove the masking of

active sites on actin for myosin.(4) prevents the formation of bonds between the

myosin cross bridges and the actin filament.Ans: (3)

169. Nissl bodies are mainly composed of:(1) Nucleic acids and SER(2) DNA and RNA(3) Proteins and lipids(4) Free ribosomes and RER

Ans: (4)

170. Which of these statements is incorrect ?


(1) Glycolysis operates as long as it is supplied withNAD that can pick up hydrogen atoms.

(2) Glycolysis occurs in cytosol.(3) Enzymes of TCA cycle are present in

mitochondrial matrix.(4) Oxidative phosphorylation takes place in outer

mitochondrial membrane.Ans: (4)

171. Select the incorrect match:(1) Submetacentric – L-shaped chromososmes

chromosomes

(2) Allosomes – Sex chromosomes

(3) Lampbrush – Diplotene bivalentschromosomes

(4) Polytene – Oocytes of amphibianschromosomes

Ans: (4)

172. Which of the following terms describe humandentition ?(1) Pleurodont, Monophyodont, Homodont(2) Thecodont, Diphyodont, Heterodont(3) Thecodont, Diphyodont, Homodont(4) Pleurodont, Diphyodont, Heterodont

Ans: (2)

173. Which of the following events does not occur inrough endoplasmic reticulum ?

(1) Cleavage of signal peptide(2) Protein glycosylation(3) Protein folding(4) Phospholipid synthesis

Ans: (4)

174. Many ribosomes may associate with a singlemRNA to for m mu ltiple copies of a polypeptidesimultaneously. Such strings of ribosomes aretermed as:(1) Plastidome (2) Polyhedral bodies(3) Polysome (4) Nucleosome

Ans: (3)

175. In which disease does mosquito transmittedpathogen cause chronic inflammation of lymphaticvessels?

(1) Ringworm disease (2) Ascariasis(3) Elephantiasis (4) Amoebiasis

Ans: (3)

176. Which of the following is not at autoimmunedisease?(1) Alzheimer’s disease(2) Rheumatoid arthritis(3) Psoriasis(4) Vitiligo

Ans: (1)

177. Among the following sets of examples for divergentevolution, select the incorrect option:(1) Brain of bat, man and cheetah(2) Heart of bat, man and cheetah(3) Forelimbs of man, bat and cheetah(4) Eye of octopus, bat and man

Ans: (3)

178. Conversion of milk to curd improves its nutritionalvalue by increasing the amount of(1) Vitamin B12 (2) Vitamin A(3) Vitamin D (4) Vitamin E

Ans: (1)

179. The similarity of bone structure in the forelimbs ofmany vertebrates is an example of:(1) Convergent evolution(2) Analogy(3) Homology(4) Adaptive radiation

Ans: (3)

180. Which of the following characteristics represent‘Inheritance of blood groups’ in humans ?

a. Dominanceb. Co-dominancec. Multiple dominanced. Incomplete dominancee. Polygenic inheritance

(1) b, d and e (2) a, b and c(3) b, c and e (4) a, c and e

Ans: (2)

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A FEW RESULTS OF JEE-MAIN NEET 2017 AMONG 766 … (UG) 201… · Solution to NEET (UG)-2018 (Paper Code-AA) Held on: 6th May, Sunday 3 3k T B 125.44 106 m = × 6 B m T 125.44 10 3k