JIIPMEERR MBBBBSS ENTRANCC E TEST 2001199 ......JIPMER. Paper has 200 questions but we have included...

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JJIIPPMMEERR MMBBBBSS EENNTTRRAANNCCEE TTEESSTT 22001199

EEXXAAMMIINNAATTIIOONN PPAAPPEERR

((BBAASSEEDD OONN MMEEMMOORRYY RREETTEENNTTIIOONN))

Date : 02-06-2019 (Sunday) | Time : 09.00 pm - 12.30 pm | Morning Session

NOTE:-

1. Questions are collected from the appeared students.

2. The solutions are prepared by the expert faculty team of Resonance Pre-Medical division,

Kota.

3. Questions may not be in the order or sequence as asked in the actual examination paper.

4. The questions collected may not have all the options similar to the actual paper. Students

are advised to see the question and answer / solutions.

5. Actual JIPMER Paper has 200 questions but we have included only those many questions which have been collected from the students as per following table :-

Subject No. of Question in Actual

JIPMER-2019 Paper No. of Question in this Paper

Chemistry 60 34

Physics 60 37

Biology 60 52

English & Comprehension +

Logical & Quantitative

Reasoning

20 06

Total 200 129

JIPMER : 1st Shift



PAGE # 1

PART - A (CHEMISTRY)

Total Questions (34) 1. 19F–1, 16O–2, 20Ne choose the correct statement - [XI] [PTB] (1) Both O–2 and F– are isoelectronic (2) All given have equal no of e– (3) F– and Ne have equal number of e– (4) All Ans. (4) Sol. O–2, F–, Ne contain same no. of e– = 10 2. Hund's Rule state that : [XI] [ATS] (1) Number of two e– can be in two separate orbitals (2) Number of two e– can be present with similar spin in a orbital. (3) no one e– can exist in 'd' orbitals (4) None of these Ans. (4) 3. If two atoms have equal number of electron it is called : [XI] [ATS] (1) iso electronic (2) isotone (3) isobar (4) None of these Ans. (1) Sol. 4. In an ideal gas equation which is constant : [XI] [GST] (1) Temperature (2) pressure (3) volume (4) universal Gas constant Ans. (4) Sol. PV = nRT (R = universal gas constant) 5. Formula for Half life of a zero order reaction is : [XII] [Chemical kinetics]

(1) K

C0 (2) K

C

2

0 (3) K

C02 (4)

K

C

2

2 0

Ans. (2)

Sol. K

Ct

2

0

2

1

6. What is the value of for Mono atomic gas (ideal gas) : [XI] [GST]

(1) 5

7 (2)

3

4 (3)

2

5 (4) None

Ans. (4)

Sol. = 1.66 for monoatomic gas 7. What is the oxidation number of Cr in Na2Cr2O7 : [XI] [CBO] (1) 2 (2) 6 (3) 10 (4) 16 Ans. (2) Sol. Na2Cr2O7 2(+1) + 2x + 7 (–2) = 0 = x = +6

8. CH3–CH2–CH = CH2 ''X CH3CH2–CH2–CH2OH [XI] [Hydrocarbon]

What is suitable reagent 'x' is ? (1) H2 | Pt (2) B2H6 | THF, H2O2|OH– (3) Br2 | HCl (4) HBr | H2O2 Ans. (2)

JIPMER : 1st Shift



PAGE # 2

9. If

react with Cl2 in presence of light and than react with Na metal in dry ether, final product is

(1)

Br

(2)

Cl

(3)

(4)

Ans. (4) [XI] [Hydrocarbon] 10. In which of the following oxidation state of phosphorous is +3? [XII] [PBC] (1) Orthophosphoric acid (2) Pyrophosphoric acid (3) Orthophosphorous acid (4) Meta phosphoric acid Ans. (3)

Sol. Orthophosphoric acid H3

5

P O4

Pyrophosphoric acid H4

5

P 2O7

Orthophosphorous acid H3

3

P O3

Meta phosphoric acid H5

P O3 11. Which of the following amino acid optically Inactive : [XII] [Biomolecule] (1) Glycine (2) Valine (3) Alanine (4) Histidine Ans. (1) 12. Which of the following can form H-bond ? [XI] [CBO] (1) NH3 (2) R–CN (3) R–O–R (4) R–Br Ans. (1) Sol. NH3 form hydrogen bond

13. CH3–CH2–CH2–Br KOH.alc

Final product is : [XII] [Alkyl halide]

(1) Propene (2) Propanol (3) Cyclopropane (4) propane -1, 2-diol Ans. (1) 14. Which can not behave as a Nucleophile for SN2 reaction : [XII] [Alkyl halide]

(1) H2O (2) CN– (3)

2NH (4) I–

Ans. (1) 15. Which of the following act as epimeric pair ? [XII] [Biomolecule] (1) Glucose + Fructose (2) Fructose and Mannose (3) Glucose and Mannose (4) Glucose and Sucrose Ans. (3)

16. What is hybridisation of 'O' in H2O [XI] [CBO] (1) sp (2) sp3 (3) sp2 (4) No hybridisation Ans. (2)

Sol.

O

H H

….

bp = 2 lp = 2 total bp + lp = 4 sp3

JIPMER : 1st Shift



PAGE # 3

17. CH2=CH–CHO ?CH2=CH–CH2–OH [XII] [Reduction]

Suitable reagent for conversion of following reaction ? (1) NaBH4 (2) Ni/H2 (3) Zn/Hg/HCl (4) Red P + HI

Ans. (1)

18. Moles of N2 = 0.4 and moles of O2 = 0.1 find 2NP (Partial pressure) N2 = ? at atmospherim pressure

(1) 0.2 atm (2) 0.8 atm (3) 0.6 atm (4) 0.4 atm Ans. (2) [XI] [MOL]

Sol. PN2 = 15.0

4.0 = 0.8 atm

19. If Half life of a substance is 36 minutes. Find amount left after 2 hrs. Initial amount is 10 gm ? (1) 1 gm (2) 2gm (3) 3 gm (4) 4 gm Ans. (1) [XII] [Chemical kinetics] Sol. t1/2

= 36 min. t = 2 hrs = 120 min.

so, 36

120= 3.33

2log

log 0

2/1

tC

C

t

t

log 23.33 =

tC

C0log

tC

C0

Ct = 1 gram 20. Most common isotopes of hydrogen (Non radioactive) [XI] [ATS] (1) Protium (2) Deutorium (3) Titrium (4) All Ans. (1) Sol. most common isotope protium 21. Write IUPAC name of the following [XI] [IUPAC]

CHO

Br (1)2-Bromo-2-ethyl hexanal (2) 3-Bromo-2-methyl hexanal (3) 2-Methyl-3-Bromo hexanal (4) 3-Bromo-2-formyl hexane.

Ans. (2) 22. Which of the following statement correct for isotope of carbon [XI] [Carbon family] (1) graphite is conductor of electricity (2) diamond have all sp3 carbon (3) graphite is more stable thermodynamically than diamond (4) All are correct Ans. (4) Sol. Fact

JIPMER : 1st Shift



PAGE # 4

23.

Br

Br

Write the IUPAC name of given structure [XI] [IUPAC] (1) Para Bromo Benzene (2) 1,4-di bromo benzene (3) Both (1) and (2) are correct (4) Meta bromo benzene Ans. (3) 24. Bordeaux mixture consist of ? [XII] [DBC] (1) CuSO4 + Ca(OH)2 (2) CuSO4 + CaCl2

(3) ZnSO4 + Mg(OH)2 (4) FeSO4 + Ba(OH)2 Ans. (1) Sol. Bordeaux mixture CuSO4 + Ca(OH)2

25. –

3BrO Changes into Br2 in an acidic medium of a unbalanced equation how many electron should be

present on the balanced electron. [XII] [Halogen family] (1) 10 electron in left (2) 6 electron in left (3) 3 electron in left (4) 3 electron in right Ans. (1)

Sol. 12H+ + 10e– + –

3BrO — Br2 + 6H2O

26. Glucose HCN x Hydrolysis

y HIPdRe Z IUPAC name of 'y' and 'z'

(1) Hexa hydroxyl heptanoic acid, heptane (2) Hepta hydroxyl hexanoic acid, hexane (3) Penta hydroxyl hexanoic acid, hexane (4) Hepta hydroxyl hexanoic acid, heptane Ans. (4) [XII] [CBC]

27. Azimuthal quantum number () defined [XI] [ATS] (1) Shape of orbitals (2) Orientation of orbitals (3) Energy of orbitals (4) Size of orbitals. Ans. (1)

Sol. Azimuthal quantum number () tell about Shape of orbitals

28. CHCH Hq+/H+

H2O A 4LiAlH

B 2/ BrPC final product 'c' is [XI] [Hydrocarbon]

(1) CH3CH2OH (2) CH3CH2–Br (3) CH3–CH2–I (4) CH2–CH2–OH

Br

Ans. (2)

29. 2KHCO3 ..........+ CO2 + H2O find amount of gases formed (in lit.) when amount of KHCO3 is 33 gm. (1) 5.6 (2) 11.2 (3) 7.46 (4) 22.4 Ans. (3) [XI] [MOL]

Sol. 2KHCO3 K2CO3 + CO2 + H2O 33/100 = 1/3 mole 1/6 1/6 mole total moles of gas = 1/6 + 1/6 = 1/3 mole so total volume of gas is = 1/3 × 22.6 = 7.46 30. Frenkal defect is present in which of the following: [SST-XII] (1) NaOH (2) NaI (3) AgBr (4) None Ans. (3) 31. Which one follow 18 electron octet rule : [XII] [COR]

(1) 5COMn (2) 5COCr (3) [Fe(CO)5] (4) None

Ans. (3)

JIPMER : 1st Shift



PAGE # 5

32. How many ions obtain after dissociation of this complex 363 ClNHCo [XII] [COR]

(1) 3 (2) 2 (3) 5 (4) 4 Ans. (4)

Sol. 363 ClNHCo [Co(NH3)6]+2 + 3Cl–

total ion = 4

33. CH2=CH-CHO ?CH2=CH-CH2-OH [XII] [Reduction]

(1) NaBH4 (2) Ni/H2 (3) Zn/Hg conc. HCl (4) None Ans. (1) 34. Enthalpy of neutralisation of strong acid and strong base is –57.1 KJ/mole and for H2C2O4 with strong

base is –53.4 KJ/eqi. What is H2C2O4

2H+ + –2

42OC enthalpy change is :

(1) 7.4 KJ (2) –7.4 KJ (3) 3.7KJ (4) –3.7 KJ Ans. (1)

Sol. Difference = 57.1 –53.4 = 3.7 KJ for ½ mole H2C2O4 or 1 equivalent so for 1 mole H2C2O4 H = 7.4 KJ



PAGE # 1

PART - B (PHYSICS)

Total Questions (37)

1. If same charge ‘q’ is placed inside a sphere and cube having radius 1 m and side 2 m respectively.

What will be the ratio of flux passing through them:

(1) 1 : 1 (2) 1 : 8 (3) 8 : 1 (4) 1 : 2

Ans. (1)

Sol. Flux = o

enclosedq

Since enclosed charge is same, hence flux will be same

1 = 2

1 : 1

2. Find amount of charge flown from Y to X when switch S is closed.

3µF 6µF

( ) S

3 6

18V

X

Y

(1) 72µC (2) 0µC (3) 54µC (4) 36µC

Ans. (3)

Sol. When switch is open:

3µF 6µF

3 6

18V

V1

V2

18V 0V

In series capacitors, charge is same

CV = constant

B

A

V

V =

A

B

C

C (CA = 3µF, CB = 6µF)

B

A

V

V =

3

6 =

1

2

also VA + VB = 18V

VA = 12V & VB = 6V

V1 = 18 – VA = 6V

In series resistance, (RC = 3, RD = 6)



PAGE # 2

D

C

V

V =

D

C

R

R

D

C

V

V =

6

3 =

2

1

& VC + VD = 18V

VC = 6V & VD = 12D

so V2 = 18 – VC = 12V

so, charge, qA = 3 × 12µC = 36µC

qB = 6 × 6 µC = 36µC

On closing switch, V1 will become equal to V2

3µF 6µF

3 6

18V

V

V

18V 0V

VC VD

At steady state,

D

C

V

V =

6

3

VC + VD = 18V

VC = 6V, VD = 12V

Final charge, on A

qA = 6 × 3 = 18µC

qB = 6 × 12 = 72µC

+36µC

3 6

18V

VC VD

–36µC

+36µC –36µC

Initially

+18µC

3 6

18V

VC VD

–18µC

+72µC –72µC

Finally

54µC

3. Psychrometer is used to measure

(1) Relative humidity (2) Pressure (3) Temperature (4) Density

Ans. (1)

4. Induction furnace uses ______ to produce heat

(1) Eddy Current (2) Resistance (3) Capacitor (4) None

Ans. (1)

Sol. When AC current flowing through a coil creates reversing magnetic field that penetrates metal.

This induces eddy current flowing through electrical resistance of bulk metal producing heat.



PAGE # 3

5. The given combination will work as:

A

B Y

(1) NAND Gate (2) OR Gate (3) AND Gate (4) NOR Gate

Ans. (3)

Sol. Y = )B.A).(B.A(

= )B.A( + )B.A(

= A.B + A.B = A.B

6. For CE configuration NPN transistor, which of the following statement is correct:

(1) IC = IE + IB (2) IB = IE + IC (3) IE = IC + IB (4) All of these

Ans. (3)

Sol. For CE IE = IC + IB

7. When a light ray enters from oil to glass on oil-glass interface. Velocity of light changes by a factor of:

(µoil = 2, µglass = 3/2)

(1) 3

4 (2)

4

3 (3) 3 (4) 1

Ans. (1)

Sol. Voil = oilµ

C

Vglass = glassµ

C

oil

glass

V

V =

glass

oil

µ

µ =

2

3

2 =

3

4

8. A light ray of wavelength 600nm is incident on young’s double slit experiment. Distance between slits is

2 mm and 5th bright fringe is at a distance of 3 mm from central maxima, find the distance between slits

and screen:

(1) 20 m (2) 2 m (3) 3 m (4) 6 m

Ans. (2)

Sol. y = 5 = 5 d

D = 3 mm

3–

9–

102

D106005

= 3 × 10–3

D = 6–

6–

103

106

= 2 m



PAGE # 4

9. Huygen’s principle does not use:

(1) Reflection (2) Refraction (3) Diffraction (4) Point of spectra origin

Ans. (4)

Sol. Fact

10. A stationary wave equation is given as

y = 20sin(20x)cos(1000t)

What will be the speed of stationary wave:

(1) 20 m/s (2) 50 m/s (3) 2 cm/s (4) Not defined

Ans. (4)

Sol. For stationary wave, wave speed is not defined.

11. A particle of mass 7 kg is executing circular motion with time period of 11 sec. Find out centripetal force

if radius of circle is 10 m.

(1) 7

30N (2)

7

40N (3) 30 N (4)

7

160N

Ans. (4)

Sol. T = V

r2 = 11

V = 11

)10(2 =

117

10222

=

7

40

FC = r

mv2

= 10

7×

7

40×

7

40 =

7

160N

12. A particle is moving with 10 m/s in a circle of radius 5m, find out magnitude of average velocity if

particle moved by 60º in 1 sec.

(1) 5 m/s (2) 10 m/s (3) 35 m/s (4) 20 m/s

Ans. (1)

Sol.

A

B 60º

60º

Average velocity = time total

ntdisplaceme total

= 1

5 = 5 m/s

13. A particle is projected at an angle of 30º from the horizontal with a speed of 10 m/s on the earth. It’s

time of flight, range, velocity of impact are respectively T, R, V. If this projectile is projected on the moon

then its time of flight, range and velocity of impact is: (Assume gravitational field on moon is g/6 m/s2)

(1) 6T, 6R, 6V (2) 6T, 6R, V (3) T, R, V (4) T, 6R, 6V

Ans. (2)



PAGE # 5

Sol. T = g

sinu2

T’ = 6/g

sinu2 = 6T

R = g

2sinu2

R’ = 6R

V’ = V

14. Two objects moving with speed u m/s collide at 90º then final momentum is: (Assume collision is

inelastic)

(1) mu (2) 2 mu (3) 2 mu (4) 22 mu

Ans. (3)

Sol. pi = pf

mu i + mu j = pf

pf = 22 )mu()mu( = 2 mu

15. One horse power (1 HP) is equal to

(1) 700 W (2) 373 W (3) 746 W (4) 750 W

Ans. (3)

16. Coercivity and retentivity of soft iron is:

(1) high coercivity, high retentivity

(2) low coercivity, high retentivity

(3) low coercivity, low retentivity

(4) high coercivity, low retentivity

Ans. (1)

17. Van De graff generator is used to:

(1) Create a high potential of range of few million volts

(2) Create a low potential of range of few million volts

(3) To de-accelerate projectile like protons, deutrons etc.

(4) It can not be used to study collision experiments in physics

Ans. (1)

18. If resistivity of copper is 1.72 × 10–8 -m and number of free electrons in copper is 8.5 × 1028 /m3. Find

mobility.

(1) 4.25 × 10–3 m2/C (2) 6.8 × 10–3 m2/C (3) 8.5 × 10–3 m2/C (4) 3.4 × 10–3 m2/C

Ans. (1)



PAGE # 6

Sol. Mobility, µ = ne

1

= 19–3288– 106.1m/105.8m1072.1

1

= 4.25 × 10–3 m2/C

19. A toy truck of mass 2m elastically collides with a toy car of mass m. speed of truck is V and car is at

rest. Find the velocity of car after collision

(1) 3

4v (2)

3

v (3) v (4)

3

2v

Ans. (1)

Sol. Initial momentum = 2mv + O

2m m V

Final momentum = 2mv1 + mv2

V1 m 2m V2

By momentum conservation

2mV = 2mV1 + mv2

2v = 2v1 + v2 …… (i)

e = 1 = approachofvelocity

separationofvelocity

V = V2 – V1 …… (ii)

By solving (i) and (ii), we get

V2 = 3

4v

20. Two blocks of mass m1 and m2 are connected at the both ends of a light string which is passing over a

pulley. Radius of pulley is R and mass M, then find the acceleration of block 1?

Radius = R

Mass=M

//////////////////

m1 m2

(1)

2

Mmm

g)m–m(

21

21 (2) 21

1

mm

gm

(3)

21

21

mm

g)mm(

(4)

21

2

mm

gm



PAGE # 7

Sol.

////////////////

//

m1

M, R

T1

m2

T2

a a

m1g – T1 = m1a …… (i)

T2 –m2g = m2a …… (ii)

T1 . R – T2.R = = R

a ( =

2

MR2

) …… (ii)

T1 – T2 = 2

MR2

. 2R

a=

2

Ma ……. (iii)

By adding (i), (ii) and (iii) we get

m1 g – m2g = (m1 + m2 + 2

M)a a =

)Mm2m2(

g)m–m(2

21

21

21. A protons enters in a uniform magnetic field (in x direction) of strength which is 2

times of its

q

m

ratio. At an angle of 30º from magnetic field find velocity of proton after 4 sec, if initial velocity was

)ji3(2

(1) )ji3(2 (2) )ji3(2– (3) )j–i3(2 (4) )ji3(–2

Ans. (1)

Sol. T = qB

m2 = 2.

q

m.

2

m

q

= 4 sec

So after 4 sec, one cycle will be completed so velocity will be same as initial.

22. Find out the rise of column of liquid with in a fine capillary tube (if density of liquid = 103, angle of

contact = 60º, acceleration due to gravity = 10 m/s2), surface tension = 70 × 10–3 N/m r = 0.2 mm

(1) 2

5cm (2) 3.6 cm (3) 7 cm (4*) 3.5 cm

Ans. (4)

Sol. H = gr

cosT2

= 1010102

60cos1070234–

3–



PAGE # 8

= 2

7cm = 3.5

23. A planet of radius R has a time period of revolution 'T'. Find time period of a planet of radius 9R?

(1) T33 (2) 9T (3) 27T (4) T39

Ans. (3)

Sol. T2 R3

27

1

R9

R

T

T

R

R

T

T 2

3

2

2

3

2

1

2

1

T2 = 27T

24. Potential difference is given as V(x) = – x2y volt. Find electric field at a point (1,2)?

(1) j4i V/m (2) j–i4– V/m (3) ji4 V/m (4) j–i4 V/m

Ans. (3)

Sol. V–E

jxixy2 2

ji4E

V/m

25. In the given circuit find voltage across 12 resistance.

12A

12

4 8

4

8

(1) 12 Volt (2) 36 Volt (3) 72 Volt (4) 48 Volt

Ans. (4)

Sol. Req of

8

4

8

is

2

1

4

1

8

1

8

1

R

1

eq

Req = 2

12A

12

4 2



PAGE # 9

Current through 12 resistance A43

ii

126

6 00

V = iR = 4 × 12 = 48 volt

26. Ratio of charge on positron to mass of positron is approximately.

(1) +2 × 1011 (2) + 5 × 1012 (3) – 2 × 1011 (4) – 5 × 1011

Ans. (1)

Sol. 11

31–

19–

1075.1101.9

106.1

m

e

m

Q

27. What is dimensions of energy in terms of linear momentum (P), area (A) and Time (T)

(1) [P1A1T1] (2) [P2A2T–1] (3) [P1A1/2T–1] (4) [P1/2A1/2T–1]

Ans. (3)

Sol. Energy = (P)x (A)y (T)z

M1L2T–2 = (M1L1T–1)x (L2)y (T)z

M1L2T–2 = MxLx+2y T–x+z

Comparing powers

of M x = 1

of L x + 2y = 2

2

1y

or T – x + z = – 2

z = x – 2

z = – 1

[P1A1/2T–1]

28. A man applying force of 20 N on an object at an angle of 60º from the horizontal. If object moved by 20

m in horizontal direction then work done by the force on the object is -

(1) 300 J (2) 400 J (3) 100 J (4) 200 J

Ans. (4)

Sol.

20N

60º

20m

S.FW

= FS cos

= 20 × 20 × cos 60º

= 200 J



PAGE # 10

29. A galvanometer of 50 resistance has 25 divisions. A current of 4 × 10–4 A gives a deflection of one

division. To convert this galvanometer into a voltmeter having a range of 25V, it should be connected

with a resistance of: [AIPMT 2004][DPP JR 2013-14]

(1) 2500 as a shunt (2) 245 as a shunt (3) 2550 in series (4*) 2450 in series

50 izfrjks/k ds xsYouksehVj esa 25 va'k gSA 4 × 10–4 A dh /kkjk ,d va'k dk fopyu nsrh gSA bl xsYouksehVj dks

25V, ijkl ds oksYVehVj esa cnyus ds fy, izfrjks/k tksMuk pkfg,A [AIPMT 2004]

(1) 2500 'kaV :i esa (2) 245 'kaV :i esa (3) 2550 Js.kh Øe esa (4*) 2450 Js.kh Øe esa

Sol. To convert a galvanometer into voltmeter, high resitance should be connected in series with it. Let R is

the resistance connected in series with the galvanometer.

,d xsYosuksehVj dks oksYVehVj esa ifjofrZr djus ds fy, blds Js.kh Øe esa mPp izfrjks/k dk rkj tksM+k tkrk gSA

ekuk xsYosuksehVj ds Js.kh Øe esa R izfrjks/k tksM+k x;k gSA

xsYosuksehVj /kkjk

ig = V

G R or R =

g

V

i – G

fn;k x;k gS Given, G = 50,

ig = 25 × 4 × 10–4 = 10–2 A, V = 25V

R = –2

25

10 – 50 = 2500 – 50 = 2450

30. If a ball of mass 0.1 kg hits the ground from the height of 20m and bounce back to the same height then

find out the force exerted on the ball if the time of impact is 0.04 sec. (g = 10 m/s2)

(1) ˆ100N( j) (2) ˆ200N( j) (3) ˆ100 ( j)N (4) ˆ1000N( j)

Sol. The velocity attained by the ball before hitting the ground = ˆ2gh 2 10 20 20m/s( j)

As ball bounce back to the same height it means collision is elastic so velocity just after hitting the

ground = ˆ20m/s( j)

So, change in velocity = vf – vi = 20 – (– 20) = ˆ40jm/s

Force =

P

t =

f imv mv

t =

f im(v v )

t =

0.1 40

0.04 = ˆ100N j

31. Find resonance frequency in the given circuit:

~

C

C

L

L

(1)

1

LC (2)

2

LC (3)

1

2 LC (4)

4

LC

Ans. (1)



PAGE # 11

Sol. Leq =

L.L

L L = L

2

Ceq = C + C = 2C

W = eq eq

1

L C

W = eq eq

1

L 2C =

1

LC

32. If frequency of a photon is 6 × 1014 Hz, then find its wavelength [speed of light, C = 3 × 108 m/s]

(1) 500 Å (2) 500 nm (3) 200 Å (4) 200 nm

Ans. (2)

Sol. C = f

=

8

14

3 10

6 10 = 5 × 10–7 m = 500 nm

33. 2 wire of same material having radius in ratio 2 : 1 and lengths in ratio 1 : 2. If same force is applied on

them, then ratio of their change in length will be:

(1) 1 : 1 (2) 1 : 2 (3) 1 : 4 (4) 1 : 8

Ans. (4)

Sol. y =

F

A

=

F

Ay

2

1 1 2

2

2 21

F r .y

Fr .y

2

1 1 2

2 2 1

r

r

=

21 1

2 2

1

2

1

8

34. A projectile of man 1 kg is projected with a speed of 10 m/s at an angle of 60º from the horizontal when

projectile is at its highest point, its magnitude of angular momentum (about point of projection) :

(1) 75

Nm / sec2 (2)

75Nm / sec

4 (3) 75 N-m/sec (4) 150 N-m/sec

Ans. (2)

Sol. L = r p

r p



PAGE # 12

=

2 2u sinmucos60º

2g

=

10 10 3 11 10

4 2 10 2

= 75

N m/ sec4

35. A man moving with 2m/s of 75 kg in west and another man of 50 kg is moving at 4m/s in North collides

in elastically and move together. Find final direction of motion:

(1) 53º N-W (2) 37º N-E (3) 53º N-E (4) 37º N-E

Ans. (1)

Sol.

N (j)

S

W E (i)

1P = 2 × 75 kg m/s

ˆ( i)

2P =

ˆ4 50kgm/s( j)

By momentum conservation

f iP P

f i 2P P P

fP =

ˆ ˆ(–150i 200j)kgm/s

200

S

150

tan = 200

150

= 53º 53º North of West

36. If balls collides elastically with fixed wedge and travels horizontally. Find h/H for maximum horizontal

range:

h

H

m

(1) 1

2 (2) 1

4 (3) 1

8 (4) 1

16



PAGE # 13

Sol. Speed gain before collision

u = 2g(H h)

usin

vsin

ucos

u

vcos

vsin

v

u sin = v cos

v = u tan

v = 2g(H h) tan

Range, R =

2hV

g

=

2h

2g(H h). tang

R = 2 h(H h) tan

Range will be maximum when h (H – h) = P is maximum

P = hH – h2

dP

dh = H – 2h = 0

h =

H

2

h 1

H 2

37. Lenz law is based in principle of conservation of:

(1) Linear momentum (2) Energy (3) Charge (4) Mass

Ans. (2)



PAGE # 1

PART – C (BIOLOGY)

Total Number of Questions (52)

1. Vernalisation is

(1) Low pH treatment (2) Low temp treatment

(3) High temp treatment (4) High pH treatment

Ans (2)

2. Synaptonemal Complex is formed

(1) During Anaphase (2) During Metaphase

(3) During Prophase II (4) During Prophase I of meiosis

Ans (4)

3. Some organisms are poisonous or colored (red or black) in order to protect themselves from predator

is called as

(1) Mutualistic mimicry (2) Batesian mimicry

(3) Mullerian mimicry (4) Predation mimicry

Ans (2)

4. Which of the following is similar in DNA of elephant and mango tree

(1) Size of DNA (2) Type of nucleotide

(3) Sequence of nucleotide (4) Total bases

Ans (2)

5. Which of the following is not a micronutrient?

(1) B (2) zn (3) Mn (4) Mg

Ans (4)

6. Inhibitory effect of O2 on photosynthesis is

(1) CAM pathway (2) Photorespiration

(3) Fermentation (4) Krebs cycle

Ans (2)

7. Eutrophication is caused due to

(1) accumulation of minerals (2) effect of UV-C

(3) accumulation of metals only (4) accumulation of Zooplankton

Ans (1)



PAGE # 2

8. Cell placed in hypertonic solution is shown by

(1) (2)

(3) (4) Both (1) and (3)

Ans (1)

9. Identify the diagram of heterocyst.

(1) (2)

(3) (4)

Ans (1)

10. tRNA binds to mRNA through

(1) Anticodon loop (2) TC loop

(3) Amino acid binding loop (4) D loop

Ans (1)

11. Chaperones' function is

(1) Hydrogen bonding (2) Folding of protein

(3) modification of protein (4) Proteolytic cleavage

Ans (2)



PAGE # 3

12. Chl b differs from chl a by

(1) – CH3 group (2) CHO group

(3) COOH group (4) CH2OH group

Ans (2)

13. What can be same in a mango/ Mangifera indica cell and human / Homo sapiens cells

(1) Nucleosides used in DNA (2) DNA base sequence

(3) DNA length (4) mRNA sequence

Ans (1)

14. Derdrochronometer is used to measure / count

(1) Ring in tree for age measurement (2) width of tree

(3) length of trees (4) types of wood

Ans (1)

15. Find odd one out

(1) stamen (2) stigma

(3) style (4) ovary

Ans (1)

16. Diagram is of

(1) Sporic meiosis (2) Zygotic meiosis (3) Gametic meiosis (4) Brachymeiosis

Ans (1)

17. Gynaecomastia is present in which syndrome?

(1) Turner (2) Klinefelter (3) Down (4) None

Ans (2)

18. Which one is the oldest in following?

(1) Gymnosperms (2) Angiosperms (3) Bryophytes (4) None

Ans (3)



PAGE # 4

19. Which is called the amphibians of plant kingdom?

(1) Angiosperms (2) Gymnosperms (3) Bryophytes (4) Pteridophytes

Ans (3)

20. Hardest substance in plant kingdom is

(1) Saple (2) Corolla (3) Sporopollenin (4) Anther

Ans (3)

21. What it is called when thousands of aquatic species are killed over a wide range in sea.

(1) Bateson – effect (2) El Nino effect (3) Warburg effect (4) None

Ans (2)

22. Dark reaction –

(1) In light (2) In dark

(3) Requires product of light reaction (4) All of these

Ans (4)

23. Which one of the labelled part utilizes fructose as a source of energy-

(1) Head (2) Tail (3) Acrosome (4) Middle piece

Ans (4)

24. Among the following choose the correct epimers-

(1) Glucose, galactose (2) Glucose, Fructose

(3) Glucose, ribose (4) Gluconic acid, glucose

Ans (1)

25. In gluconeogenesis, which of the following cannot acts as substrate

(1) Fructose (2) Palmitic acid

(3) Glycine (4) Glycerol

Ans (1)



PAGE # 5

26. Choose correct option for larva of housefly

(1) Grub (2) Cypris (3) Maggot (4) Wriggler

Ans (3)

27. Traditional anesthesia is obtained from

(1) Datura (2) Poppy (3) Cannabis (4) Erythroxylum

Ans (4)

28. Band 3 protein in RBC has the function of

(1) It mediates the exchange of cellular –3HCO with Cl– in plasma

(2) It is a cytoplasmic protein

(3) It mediates the exchange of cations between cell and plasma

(4) Both (1) and (2)

Ans (1)

29. Taut form of Haemoglobin is favoured by

(1) High partial pressure of O2 (2) High pH

(3) High PCO2 (4) Low level of 2,3-BPG

Ans (3)

30. Number of chromosomes present in secondary spermatocyte is

(1) 22 (2) 23 (3) 24 (4) 25

Ans (2)

31. Choose incorrect statement

(1) Deuterostomes show radial and indeterminate cleavage

(2) Deuterostomes show spiral and indeterminate cleavage

(3) Protostomes show radial and determinate cleavage

(4) Protostomes show spiral and indeterminate cleavage

Ans (2)

32. Flame cells are present in

(1) Aschelminthes (2) Platyhelminthes

(3) Annelida (4) Cephalochordate

Ans (2)

33. Reduced coelom is found in

(1) Annelida (2) Echinodermata

(3) Mollusca (4) Platyhelminthes

Ans (3)



PAGE # 6

34. Level of progesterone is highest

(1) Just before the ovulation (2) Just after the ovulation

(3) 7 days after ovulation (4) During menstrual bleeding phase

Ans (3)

35. Which of the following is not considered as a secondary messenger

(1) Acetyl choline (2) cAMP

(3) Ca2+ (4) Diacylglycerol

Ans (1)

36. Coacervates were discovered by

(1) Huxley (2) Fox (3) Oparin (4) Miller

Ans (3)

37. Length of the spinal cord is

(1) 30 cm (2) 25 cm (3) 55 cm (4) 45 cm

Ans (4)

38. Atlas is

(1) 1st cervical vertebrae (2) 2nd cervical vertebrae

(3) 1st thoracic vertebrae (4) 2nd Lumbar vertebrae

Ans (1)

39. Which embryonic layer is responsible for gaseous exchange

(1) Amnion (2) Yolk sac (3) Chorion (4) All of the above

Ans (3)

40. Choose the correct regarding antibodies

(1) IgA – Helps in allergic reaction (2) IgG – Cross placenta

(3) IgE – Found in secretions (4) IgM – exist as Dimer

Ans (2)

41. Which of the following is uricotelic

(1) Insects (2) Birds (3) Lizards (4) All

Ans (4)

42. In Annealing

(1) Primers anneal to the DNA template

(2) Taq polymerase adds nucleotides to the annealed primer

(3) Two strands of the DNA separate

(4) Temperature is 92ºC

Ans (1)



PAGE # 7

43. Which is the first step in PCR cycle

(1) Denaturation (2) Extension (3) Annealing (4) None

Ans (1)

44. Hypothalamus can regulate

(1) Anger (2) Patience (3) Thermoregulation (4) Pleasure

Ans (3)

45. In DNA 20% bases are adenine. What percentage of bases are pyrimidines?

(1) 30% (2) 60% (3) 50% (4) 20%

Ans (3)

46. Typhoid is caused by :

(1) Pneumonia (2) filarial worm (3) Salmonella typhi (4) None

Ans (3)

47. Reproduction without actual fertilization is called as :

(1) Parthenogenesis (2) Parthenocarpy

(3) Pseudo-reproduction (4) All of the above

Ans (1)

48. Radula is a part of which animal :

(1) Mollusca (2) Poriferans (3) Coelenterata (4) Annelida

Ans (1)

49. Length of fallopian tube in female is:

(1) 40 cm (2) 30 cm (3) 12 cm (4) 15 cm

Ans (3)

50. Which is also called molecular glue:

(1) DNA Gyrase (2) DNA Helicase (3) DNA Ligase (4) DNA Polymerase

Ans (3)

51. Which one of them is odd with respect to Human male :

(1) Absence of one X chromosome (2) Addition of one X chromosome

(3) Presence of XY chromosome (4) Absence of barr body

Ans (2)

52.

Should be species

Species now at present

Which type of evolution is shown :

(1) Disruptive (2) Destructive (3) Competitive (4) Stabilizing

Ans (4)



PAGE # 1

PART – D

(ENGLISH & COMPREHENSION + LOGICAL & QUANTITATIVE REASONING)

Total Number of Questions (5)

1. Which statement is grammatically correct?

(1) He have tried to control his anger despite bully nature

(2) He had tried to control his anger despite bully nature

(3) He would tried to control his anger despite bully nature

(4) He could tried to control his anger despite bully nature

Ans (2)

2. Harsh is jovial. Jovial in this context means:

(1) Bully (2) Pessimistic (3) Optimistic (4) Angersome

Ans (3)

3. M is taller than Q and L. Q is only taller there N. L is taller than Q but shorter than M. then who is the

tallest and the shortest

(1) L & Q (2) Q & N (3) M & N (4) None of these

Ans (3)

4. If HYDROGEN is coded as YHCQPHFM. Then DRY HEN will be coded as what

(1) RDXGFO (2) DRXGFO (3) ODRXGF (4) None of these

Ans (1)

5. Bateson sees a gentleman and says "He is my brother's daughter's father". Then what is Bateson to

that gentleman.

(1) Grandfather (2) Uncle (3) Son (4) Brother

Ans (4)

JIIPMEERR MBBBBSS ENTRANCC E TEST 2001199 ......JIPMER. Paper has 200 questions but we have included...

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Transcript of JIIPMEERR MBBBBSS ENTRANCC E TEST 2001199 ......JIPMER. Paper has 200 questions but we have included...