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NEET - 2 EXAMINATION - 2016
QUESTION WITH SOLUTION
PAPER CODE - _XX_
NEET - 2 Examination (2016) (Code - _)
[CHEMISTRY]
(Page # 2)
1.
A given nitrogen-containing aromatic compound A reacts with Sn/HCl, followed by HNO2 to give
an unstable compound B. B, on treatment with phenol, forms a beautiful coloured compound C
with the molecular formula C12H10N2O. The structure of compound A is
(1)
(2)
(3)
(4)
, d fn; k x; k ukbVªkst u ; q
ä , sjkseS
fVd ; kS
fxd A, Sn/HCl r Fkk ckn esaHNO2 l sfØ; k dj ds, d vLFkk; h; kS
fxd B nsrk gS
A
B fQukW
y dsl kFk fØ; k dj ds, d l q
Unj j a
xhu ; kS
fxd C cukr k gS
] ft l dk v.kq
&l w
=k C12H10N2O gS
A ; kS
fxd A dhl a
j puk gS
(1)
(2)
(3)
(4)
Sol.
3
2.
Consider the reaction
CH3CH2CH2Br + NaCN  CH3CH2CH2CN + NaBr
This reaction will be the fastest in
(1) water
(2) ethanol
(3) methanol
(4) N, N'-dimethylformamide (DMF)
fuEu vfHkfØ; k i j fopkj dhft ; s%
CH3CH2CH2Br + NaCN  CH3CH2CH2CN + NaBr
; g vfHkfØ; k fdl esavfr ' kh?kzgksxh\
(1) t y
(2) , s
FksukW
y
(3)
esFksukW
y
Sol.
4
SN2 reaction
Polar aprotic solvent
DMF
3.
The correct structure of the product A formed in the reaction
(4) N, N'-Mkbes
fFky QkW
ekZ
ekbM(DMF)
H2  gas,1atmosphere

A is
Pd / carbon, ethanol
(1)
(3)
(3)
(4)
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NEET - 2 Examination (2016) (Code - _)
(Page # 3)
vfHkfØ; k
H2  gas,1atmosphere

A
Pd / carbon, ethanol
esacuusoky smRikn A dhl gh l a
j puk gS
(1)
(2)
(3)
Sol.
3
C = O Multiple bond is not affected by pd
4.
Which among the given molecules can exhibit tautomerism ?
(1) Both II and III
(2) III only
(3) Both I and III
(4)
(4) Both I and II
fn; sx; sv.kq
v ksaesal sfdl esapy ko; or k gksxh\
Sol.
5.
(1) II vkS
j III nksuska (2) dsoy III
(3) I vkS
j III nksuksa
(4) I
2
Bridge carbon atom does not take participate in tauetomerism
Acidic hydrogen must be present
vkS
j II nksuksa
The correct order of strengths of the carboxylic acids
is
(1) II > I > III
(2) I > II > III
(3) II > III > I
(4) III > II > I
(3) II > III > I
(4) III > II > I
dkcksZ
fDl fy d vEy
dsl keF; Zdk l ghØe gS%
Sol.
(1) II > I > III
3
(2) I > II > III
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NEET - 2 Examination (2016) (Code - _)
(Page # 4)
6.
The compound that will react most readily with gaseous bromine has the formula
(1) C2H4
(2) C3H6
(3) C2H2
(4) C4H10
ml ; kS
fxd dk] t ksfd xS
l h; czksehu l svR; f/kd vkl kuhl sfØ; k dj r k gS
]lw
=k gS
Sol.
7.
(1) C2H4
(2) C3H6
(3) C2H2
2
C3H6 propene  CnH2n
more reactive
(Nucleophilic strength )
(4) C4H10
Which one of the following compounds shows the presence of intramolecular hydrogen bond ?
(1) Concentrated acetic acid
(2) H2O2
(3) HCN
(4) Cellulose
fuEu esal sfdl ; kS
fxd esavUr %vkf.od gkbMªkst u vkca
/k mi fLFkr gS\
(1) l ka
nz, sl hfVd vEy
(2) H2O2
(3) HCN
(4) l s
yq
y ksl
Sol.
4
8.
The molar conductivity of a 0.5 mol/dm3 solution of AgNO3 with electrolytic conductivity of 5.76
× 10–3 S cm–1 at 298 K is
(1) 28.8 S cm2/mol (2) 2.88 S cm2/mol (3) 11.52 S cm2/mol (4) 0.086 S cm2/mol
0.5 mol/dm3 AgNO3 dsfoy; u] ft l dhfo| q
r vi ?kVuhpkydr k5.76 × 10–3 S cm–1 gS
] dh298 K i j eksy j pkydr k
gS
Sol.
(1) 28.8 S cm2/mol (2) 2.88 S cm2/mol (3) 11.52 S cm2/mol (4) 0.086 S cm2/mol
2
m = K ×
1000
M
1000
5.76
576
57.6
=
=
=
0.5
0.50
50
5
= 11.52 S cm2/mole
= 5.76 × 10–3 ×
9.
Sol.
10.
The decomposition of phosphine (PH3) on tungsten at low pressure is a first order reaction. It is
because the
(1) rate of decomposition is very slow
(2) rate is proportional to the surface coverage
(3) rate is inversely proportional to the surface coverage
(4) rate is independent of the surface coverage
Va
XLVu i j QkW
LQhu (PH3) dk U; w
u nkc i j vi ?kVu , d i zFke dksfV dhvfHkfØ; k gS
] D; ksa
fd
(1) vi ?kVu dk os
x cgq
r /khek gS
(2) os
x] i `"B ds?ksjko dsl ekuq
i kr hgS
(3) os
x] i `"B ds?ksjko dsO
;q
RØekuq
i kr hgS
(4) os
x] i `"B ds?ksjko l sLor a
=k gS
2
The coagulation values in millimoles per litre of the electrolytes used for the coagulation of As2S3
are given below :
I. (NaCl) = 52,
II. (BaCl2) = 0.69, III. (MgSO4) = 0.22
The correct order of their coagulating power is
(1) III > I > II
(2) I > II > III
(3) II > I > III
(4) III > II > I
As2S3 dsLda
nu esai z;q
ä fo| q
r~
&vi ?kV~
; ksadsLda
nu eku fey heksy i zfr y hVj esauhpsfn; sx; gS%
I. (NaCl) = 52,
II. (BaCl2) = 0.69, III. (MgSO4) = 0.22
budsLda
nu ' kfä dk l ghØe gS
Sol.
(1) III > I > II
4
As2S3 = (–)
(2) I > II > III
(3) II > I > III
(4) III > II > I
1
Coagulating power  charge  coagulation value III > II > I
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NEET - 2 Examination (2016) (Code - _)
11.
Sol.
During the electrolysis of molten sodium chloride, the time required to produce 0.10 mole of
chlorine gas using a current of 3 amperes is
(1) 330 minutes
(2) 55 minutes
(3) 110 minutes
(4) 220 minutes
xfyr l ksfM; e DyksjkbMdsfo| q
r~
&vi ?kVu dsnkS
j ku 3 , sfEi; j /kkjkl s0.10 eksy Dyksjhu xS
l dscuusesafdr ukl e; yxr kgS\
(1) 330 feuV
3
gm E Cl2 = F
moles × V.F. =
0.1 × 2 =
t=
12.
Sol.
13.
(Page # 5)
(2) 55
feuV
(3) 110
feuV
(4) 220
feuV
i t
96500
3 t
96500
0.1  2  96500
3  60
min = 107.22  110 min
How many electrons can fit in the orbital for which n = 3 and  = 1 ?
(1) 14
(2) 2
(3) 6
(4) 10
n = 3 , oa = 1 dsd{kd es
afdr usby sDVªkW
u vk l dr sgS?
(1) 14
(2) 2
(3) 6
(4) 10
2
For a sample of perfect gas when its pressure is changed isothermally from pi to pf, the entropy
change is given by
 pi 
 pf 
 pi 
(1) S = RT ln  p  (2) S = nR ln  p  (3) S = nR ln  p 
 f
 i
 f
 pf 
(4) S = nRT in  p 
 i
, d vkn' kZxS
l dsuew
usdk nkc esai fj or Z
u pi l spf l er ki i j gksrk gS
A bl dh, UVªkW
i hesai fj or Z
u gksxk
 pi 
 pf 
 pi 
(1) S = RT ln  p  (2) S = nR ln  p  (3) S = nR ln  p 
 f
 i
 f
Sol.
 pf 
(4) S = nRT in  p 
 i
3
p
nRT loge i
q
E  w
w
p
S =
=
=
=
f
T
T
T
T
pi
= n R log p
f
14.
Sol.
15.
The van’t Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is
(1) 3
(2)0
(3) 1
(4) 2
i zcy fo| q
r~
&vi ?kV; csfj ; e gkbMªkW
Dl kbM dsr uqt y h; foy ; u dsfy , okUV gkW
Q xq
. kd (i) gS
(1) 3
(2)0
(3) 1
(4) 2
1
Ba(OH)2  Ba+2 + 2OH–
i=3
The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5N+H) in a 0.10 M aqueous
pyridine solution (Kb for C5H5N= 1.7 × 10–9) is
(1) 1.6%
(2) 0.0060%
(3) 0.013%
(4) 0.77%
0.10 M fi fj Mhu (C5H5N) dst y h; foy ; u (C5H5N+ dsfy , Kb = 1.7 × 10–9) es
afi fj Mhfu; e vk; u (C5H5N+H)
cukusdsfy , fi fj Mhu dh i zfr ' kr r k gS
(1) 1.6%
(2) 0.0060%
(3) 0.013%
(4) 0.77%
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NEET - 2 Examination (2016) (Code - _)
(Page # 6)
Sol.
3
C5H5N + HOH  C5H5NH+ + OH–
1
0
0
(1 – ) C
c
c
% = 
k eq
c
= 1.3 × 10–4
= 0.013 %
16.
=
1.7  109
=
0.1
–4
1.7 × 10
ln calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion (Ca2+)
and fluoride ion (F–) are
(1) 4 and 8
(2) 4 and 2
(3) 6 and 6
(4) 8 and 4
dS
fYl ; e ¶y q
v ksjkbM esa
] ft l dh ¶y q
v ksjkbV l a
j puk gS
] dS
fYl ; e vk; u (Ca2+) , oa¶y q
v ksjkbM vk; u (F–) ds fy ,
mi l gl a
; kst u l a
[ ; k, ¡ gS
(1) 4 vkS
j 8
(2) 4 vkS
j 2
(3) 6
vkS
j 6
(4) 8
vkS
j 4
Sol.
4
CaF2
8:4
17.
If the E°cell for a given reaction has a negative value, which of the following gives the correct
relationships for the values of G° and Keq ?
(1) G° < 0 ; Keq < 1
(2) G°> 0 ; Keq < 1
(3) G° > 0 ; Keq > 1
(4) G° < 0 ; Keq > 1
; fn fdl hnhx; hvfHkfØ; k dsfy , E°cell dk eku _ .kkRed gS
] r ksG° , oaKeq dsekuksadsfy , l ghl EcU/k gS?
(1) G° < 0 ; Keq < 1
(2) G°> 0 ; Keq < 1
(3) G° > 0 ; Keq > 1
(4) G° < 0 ; Keq > 1
2
E°cell = (–)
G°cell = –nf E°cell = 
Sol.
E°cell =
18.
0.06
log k
n
k<1
Which one of the following is incorrect for ideal solution ?
(1) Gmix  0
(2) Hmix  0
(3) Umix  0
(4) P  Pobs  Pcalculated by Rault's law  0
vkn' kZfoy ; u dsfy , fuEu esal sdkS
u l k , d xy r gS\
Sol.
19.
Sol.
(1) Gmix  0
(2) Hmix  0
(3) Umix  0
(4) P  Pobs  Pj kmYV fu; e } kj k i fj dfyr  0
1
Gmix = (–)
The solubility of AgCl(s) with solubility product 1.6 × 10–10 in 0.1 M NaCl solution would be
(1) Zero
(2) 1.26 × 10–5 M
(3) 1.6 × 10–9 M
(4) 1.6 × 10–11 M
–10
0.1 M NCl foy ; u es
a1.6 × 10 foy s;r k xq
. kuQy oky sAgCl dhfoy s;r k gskxh
(1) Zero
(2) 1.26 × 10–5 M
(3) 1.6 × 10–9 M
(4) 1.6 × 10–11 M
3
S=
20.
K sp
c
=
1.6  1010
= 1.6 × 10–9
0.1
Suppose the elements X and Y combine to form two compounds XY2 and X3Y2. When 0.1 mole of
XY2 weights 10 g and 0.05 mole of X3Y2 weight 9 g, the atomic weights of X and Y are
(1) 30 , 20
(2) 40 , 30
(3) 60 , 40
(4) 20 , 30
eku y safd nksr Ro X vkS
j Y fey dj nks; kS
fxd XY2 , oaX3Y2 nsrsgS
At c 0.1 eksy XY2 dkHkkj 10 g r Fkk0.05 eksy X3Y2
dk Hkkj 9 g gS
] r ksX vkS
j Y dsi j ek.kqHkkj gS
&
(1) 30 , 20
(2) 40 , 30
(3) 60 , 40
(4) 20 , 30
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NEET - 2 Examination (2016) (Code - _)
Sol.
2
1 mole
x y2 = 100
x + 2y = 100
0.05 –––––9gm
(Page # 7)
....(i)
9
 100 =180
0.05
3x + 2y = 180
....(ii)
x  2y = 100
....(i)
– –
–
2x = 80
x = 40
y = 30
1 ––-----
21.
Sol.
The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in
60 seconds is (Charge on electron = 1.60 × 10–19 C)
(1) 7.48 × 1023
(2) 6 × 1023
(3) 6 × 1020
(4) 3.75 × 1020
1 ,s
fEi ; j /kkj ki j fo| q
r ~vi ?kVu dsnkS
j ku 60 l sd .MesadS
FkksMi j by sDVªkW
uksadheq
ä la
[ ; kgS(by sDVªkW
u dkvkos'k= 1.60
× 10–19 C)
(1) 7.48 × 1023
(2) 6 × 1023
(3) 6 × 1020
(4) 3.75 × 1020
4
Q=Q
i×t=n×e
1 × 60 = n × 1.6 ×10–19
n=
=
22.
6
1.6  1019
60
60
× 10+19 =
× 1020 = 3.75 × 1020
1.6
16
Boric acid is an acid because its molecule
(1) combines with proton from water molecule
(2) contains replaceable H+ ion
(3) gives up a proton
(4) accepts OH– from water releasing proton
cksfj d , d vEy gS
] D; ksa
fd bl dsv.kq
(1) t y dsv.kqdsi z
ksVkW
u dsl kFk l fEefy r gksrsgS
(2) es
ai zfr LFkkI; H+ vk; u gS
(3) i z
ksVkW
u nsrsgS
(4) t y es
aOH– xzg.k dj dsi zksVkW
u nsrsgS
Sol.
23.
Sol.
24.
Sol.
4
AlF3 is soluble in HF only in presence of KF. It is due to the formation of
(1) K[AlF3H]
(2) K3[AlF3H3]
(3) K3[AlF6]
(4) AlH3
AlF3 dk HF es
afoy s; dsoy KF dhmi fLFkfr esagksrk gS
A , sl k fdl dscuusdsdkj .k gksrk gS\
(1) K[AlF3H]
(2) K3[AlF3H3]
(3) K3[AlF6]
(4) AlH3
3
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is
because
(1) zinc has higher negative electrode potential than iron
(2) zinc is lighter than iron
(3) zinc has lower melting point than iron
(4) zinc has lower negative electrode potential than iron
ft a
d dksvk; j u i j y sfi r dj usl st Lr snkj y ksgk (vk; j u) cur k gS
] t cfd bl dk foi j hr l a
Hko ughagS
A bl dk dkj .k gS
(1) ft a
d dk _ .kkRed by sDVªksM foHko vk; j u l sT; knk gS
(2) ft a
d ] vk; j u l sgYdk gksrk gS
(3) ft a
d dk xy uka
d vk; j u l sde gS
(4) ft a
d dk _ .kkRed by sDVªkW
M foHko vk; j u l sde gS
1
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NEET - 2 Examination (2016) (Code - _)
(Page # 8)
25.
The suspension of slaked lime in water is known as
(1) aqueous solution of slaked lime
(2) limewater
(3) quicklime
(4) milk of lime
cq
>spw
usdk i kuhesafuy a
cu dgy kr k gS
(1) t q
>spw
usdk t y h; foy ; u
(3) vucq
>k pw
uk
(2)
(4)
pw
usdk i kuh
nq
f/k; k pw
uk
Sol.
4
26.
The hybridizations of atomic orbitals of nitrogen in NO2 , NO–3 and NH4 respectively are
(1) sp2, sp and sp3 (2) sp, sp3 and sp2 (3) sp2, sp3 and sp (4) sp, sp2 and sp3
vkS
j NH4 desaukbVªkst u dsi j ekf.od d{kdksadsl a
d j .k gS
] Øe' k%
2
3
3
2
2
3
(1) sp , sp vkS
j sp
(2) sp, sp vkS
j sp
(3) sp , sp vkS
j sp (4) sp, sp2 vkS
j sp3
NO2 , NO–3
Sol.
4
27.
Which of the following fluoro-compounds is most likely to behave as a Lewis base?
(1) SiF4
(2) BF4
(3) PF3
(4) CF4
fuEu esal sdkS
u&l k ¶y q
v ksjks; kS
fxd l okZ
f/kd : i l sy w
bl {kkj dhr j g O
; ogkj dj r k gS\
Sol.
(1) SiF4
3
28.
Which of the following pairs of ions is isoelectronic and isostructural?
(1) ClO–3 ,SO2–
3
(2) BF4
–
(2) CO2–
3 ,NO3
(3) PF3
(3) ClO–3 ,CO2–
3
(4) CF4
–
(4) SO2–
3 ,NO3
fuEu esal svk; uksadk dkS
u&l k ; q
Xe l eby sDVªkW
fud , oal el a
j pukRed gS\
Sol.
29.
(1) ClO–3 ,SO2–
3
1
–
(2) CO2–
3 ,NO3
(3) ClO–3 ,CO2–
3
–
(4) SO2–
3 ,NO3
In context with beryllium, which one of the following statements is incorrect?
(1) Its hydride is electron-deficient and polymeric.
(2) It is rendered passive by nitric acid.
(3) It forms Be2C.
(4) Its salts rarely hydrolyze.
csfj fy ; e dsl a
nHkZesafuEu dFkuksaesal sdkS
u&l k xy r gS\
(1) bl dk gkbMªkbM by s
DVªkW
u&U; w
u , oacgq
y d gS
(2) bl dksukbfVªd vEy } kj k fuf"Ø; cuk fn; k t kr k gS
A
(3) ; g Be2C cukr k gS
(4) bl dsy o.k nq
yZ
Hkr k l st y &vi ?kfVr gksrsgS
A
Sol.
30.
4
Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following
reactions does not show oxidizing behaviour?
(1) CaF2 + H2SO4  CaSO4 + 2HF
(2) Cu + 2H2SO4  CuSO4 + SO2 + 2H2O
(3) 3S + 2H28O4  3SO2 + 2H2O
(4) C + 2H2SO4  CO2 + 2SO2 + 2H20
xeZl ka
nzl Y¶; w
fj d vEy , d e/; e i zcy vkW
Dl hdkj d gS
A fuEu esal sdkS
u&l hvfHkfØ; k vkW
Dl hdj .k O
; ogkj ughan' kkZ
r hgS\
Sol.
(1)
(2)
(3)
(4)
2
CaF2 + H2SO4  CaSO4 + 2HF
Cu + 2H2SO4  CuSO4 + SO2 + 2H2O
3S + 2H28O4  3SO2 + 2H2O
C + 2H2SO4  CO2 + 2SO2 + 2H20
31.
Which of the following pairs of d-orbitals will have electron density along the axes?
(1) dxy , dx2 –y2
(2) dx2 , dxy
(3) dxz, dyz
(4) dz2 , dx2 –y2
fuEu esal sd-d{kdksdsfdl ; q
Xe esaby sDVªkW
u ?kuRo v{kksadsvuq
fn' k gS\
(1) dxy , dx2 –y2
Sol.
(2) dx2 , dxy
(3) dxz, dyz
(4) dz2 , dx2 –y2
4
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NEET - 2 Examination (2016) (Code - _)
32.
(Page # 9)
Sol.
The correct geometry and hybridization for XeF4 are
(1) square planar, sp3 d2
(2) octahedral, sp3d2
3
(3) trigonal bipyramidal, sp d
(4) planar triangle, sp3d3
XeF4 dsfy , l gh T; kfefr , oal ghl a
d j .k gS
3
2
(1) oxZl er y h; ] sp d
(2) v"VQy dh; , sp3d2
3
(3) f=kdks
. kh; f} fi j S
feMh] sp d
(4) l er y h; f=kdks
. k, sp3d3
2
33.
Among the following, which one is a wrong statement?
(1) I3 has bent geometry
(3) p-d bonds are present in SO2
(2) PH5 and BiCl5 do not exist
(4) SeF4 and CH4 have same shape
fuEu esal sdkS
Uk l k dFku xy r gS\
(1) I3 dh T; kfefr
ca
fdr gS
A
(3) SO2 es
ap-d vkca
/k gksrk gS
A
vkS
j BiCl5 dk vfLr Ro ughagS
(4) SeF4 vkS
j CH4 dk vkdkj l eku gS
(2) PH5
Sol.
4
34.
The correct increasing order of trans-effect of the following species is
(1) CN– > Br– > C6H–5 > NH3
(2) NH3 > CN– > Br– > C6H–5
–
–
–
(3) CN > C6H 5 > Br > NH3
(4) Br– > CN– > NH3 > C6H–5
fuEu Li h'kht dsfy , Vªka
l &i zHkko dk c<+
r k gq
v k l ghØe gS
Sol.
35.
(1) CN– > Br– > C6H–5 > NH3
(3) CN– > C6H–5 > Br– > NH3
3
(2) NH3 > CN– > Br– > C6H–5
(4) Br– > CN– > NH3 > C6H–5
Which one of the following statements related to lanthanons is incorrect?
(1) Ce (+ 4) solutions are widely used as oxidizing agent in volumetric analysis.
(2) Europium shows +2 oxidation state.
(3) The basicity decreases as the ionic radius decreases from Pr to Lu.
(4) All the lanthanons are much more reactive than aluminiulm.
fuEu dFkuksaesal sdkS
Uk&l k y S
FksukW
u dsl a
nHkZesaxy r gS\
(1) vk; r uh fo' y s
"k.k esavkW
Dl hdkj d ds: i esaCe (+ 4) foy ; uksadk og̀n~: i esami ; ksx fd; k t kr k gS
A
(2) ; w
j ksfi ; e +2 vkW
Dl hdj .k voLFkk n' kkZ
r k gS
A
(3) Pr l sLu r d vk; fud f=kT; k ds?kVusdsl kFk {kkj dr k ?kVr h gS
(4) l H
kh y S
FksukW
u] , sY; q
fefu; e dhvi s{kk vR; f/kd fØ; k' khy gS
A
Sol.
36.
4
Jahn-Teller effect is not observed in high spin complexes of
(1) d9
(2) d7
(3) d8
(4) d4
fuEu esal sfdl dsmPp pØ.k l a
dq
y ksaesa; ku&Vsy j i zHkko n`' ; ughagS\
Sol.
37.
(1) d9
3
(2) d7
(3) d8
(4) d4
Which of the following can be used as the halide component for Friedel-Crafts reaction?
(1) Isopropyl chloride
(2) Chlorobenzene
(3) Bromobenzene
(4) Chloroethene
fuEu esal sfdl ÝhMsy &Øk¶V~
l vfHkfØ; k esagS
y kbM ?kVd ds: i esami ; ksx esay k; k t k l dr k gS\
(1) vkbl ks
i zksfi y Dy kj ksbM
(2) Dy ks
j kscsa
t hu
(3) cz
kseksca
st hu
(4) Dy ks
j ks,sFkhu
37.
1
Isopropylchloride
(Friedel craft alkylation)
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NEET - 2 Examination (2016) (Code - _)
(Page # 10)
38.
In which of the following molecules, all atoms are coplanar?
(1)
(2)
(3)
(4)
(3)
(4)
fuEu esal sfdl v.kqesal Hkhi j ek.kql er y h; gS\
(1)
(2)
Sol.
2
Due to resonance
39.
Which one of the following structures represents nylon 6, 6 polymer?
H2
C
(1)
(2)
H
C
H2
NH 2
(3)
H2
C H
C
H2
NH2
H
C
(4)
H2
C H
C
NH 2 66
H
C
CH3 66
H2
NH 2
H2
C H
C
H
C
H2
6
COOH
CH3
Cl
H
C
6
fuEu esal sdkS
u&l hl j a
puk ukby kW
u 6, 6 cgq
y d dksi znf' kZ
r dj r hgS
\
H2
C
(1)
(2)
H
C
H2
NH 2
(3)
H2
C H
C
NH2
Sol.
H2
H
C
(4)
NH 2 66
H2
C H
C
H
C
CH3 66
H2
NH 2
H2
C H
C
H
C
CH3
Cl
H2
H
C
6
COOH
6
1
n(Hexamethylene diamine) + n(adipic acid)
NYLN–6,6
(Condensation polymer)
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NEET - 2 Examination (2016) (Code - _)
40.
(Page # 11)
In pyrrole
4
3
5
2
··
N1
H
The electron density is maximum on
(1) 2 and 5
(2) 2 and 3
(3) 3 and 4
(4) 2 and 4
i kbj ksy
4
3
5
2
··
N1
H
esaby sDVªkW
u ?kuRo vf/kdr e gS
(1) 2 vkS
j 5 ij
(2) 2 vkS
j 3 ij
(3) 3
vkS
j 4 ij
(4) 2
vkS
j 4 ij
Sol.
1
41.
Which of the following compounds shall not produce propene by reaction with HBr followed by
elimination or direct only elimination reaction?
H2C
H2
(1) H C—C—CH
3
2Br
(2)
CH 2
H2
(3) H C—C—CH
3
2OH
C
H2
(4) H2C
C
O
fuEu esal sdkS
u&l k; kS
fxd HBr l sfØ; kdj dsr Fkkckn esafoy ksiu vfHkfØ; k; kl h/khdsoy foy ksiu vfHkfØ; kl si zksihu ugha
nsrk gS\
H2C
H2
(1) H C—C—CH
3
2Br
Sol.
42.
4
H2C C
Ketone
(2)
CH 2
H2
(3) H C—C—CH
3
2OH
C
H2
(4) H2C
C
O
O
Which one of the following nitro-compounds does not react with nitrous acid?
CH3
(1)
H3C
C
H
O
H2
C
H3C
NO2
(2)
C
H2
H2
C
H3C
CH
NO2 (3)
H 3C
NO2
H3C
(4)
H3C
C
NO2
H3C
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NEET - 2 Examination (2016) (Code - _)
(Page # 12)
fuEu esal sdkS
Uk&l k ukbVªks; kS
fxd] ukbVªl vEy l sfØ; k ughadj r k gS\
CH3
(1)
H3C
C
H
H3C
NO2
(2)
O
Sol.
H2
C
C
H2
H2
C
H3C
CH
NO2 (3)
H 3C
NO2
(4)
H3C
C
NO2
H3C
H3C
4
Due to absence of –H
H 3C
H3C
C
NO2
H3C
43.
The central dogma of molecular genetics states that the genetic information flows from
(1) DNA  RNA  Carbohydrates
(2) Amino acids  Proteins  DNA
(3) DNA  Carbohydrates  Proteins
(4) DNA  RNA  Proteins
vkf.od vkuq
oa
f' kdr k dk dsUnzh; fl ) ka
r dgr k gSfd vkuq
oa
f' kd l w
puk dk i zokg gksrk gS
(1) DNA  RNA  dkcks
Z
gkbMªsV
(2) , s
ehuksvEy  i zksVhu DNA
(3) DNA  dkcks
Z
gkbMªsV  i zksVhu
(4) DNA  RNA  i z
ksVhu
Sol.
4
DNA  RNA  Proteins
44.
The correct corresponding order of names of four aldoses with configuration given below :
respectively, is
(1) D-erythrose, D-threose, L-erythrose, L-threose
(2) L-erythrose, L-threose, L-erythrose, D-threose
(3) D-threose, D-erythrose, L-threose, L-erythrose
(4) L-erythrose, L-threose, D-erythrose, D-threose
foU; kl l g fn; sx; s, sYMksl
dsukeksadsl ghl a
xr Øe gS
] Øe' k%
(1) D-, fj Fkz
ksl , D-fFkzv ksl , L-, fj Fkzksl , L-fFkzv ksl
(3) D-fFkz
v ksl , D-, fj Fkzksl , L-fFkzv ksl , L-, fj Fkzksl
Sol.
(2) L-, fj Fkz
ksl , L-fFkzv ksl , L-, fj Fkzksl , D-fFkzv ksl
(4) L-, fj Fkz
ksl , L-fFkzv ksl , D-, fj Fkzksl , D-fFkzv ksl
1
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NEET - 2 Examination (2016) (Code - _)
45.
(Page # 13)
In the given reaction
the product p is
(1)
(2)
(3)
(4)
(2)
(3)
(4)
nh x; h vfHkfØ; k
esamRikn p gS
(1)
Sol.
4
followed by formation of carbocation
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NEET - 2 Examination (2016) (Code - _)
[BIOLOGY]
(Page # 14)
46.
A foreign DNA and plasmid cut by the same restriction endonuclease can be joined to form a
recombinant plasmid using
(1) ligase
(2) Eco Rl
(3) Taq polymerase
(4) polymerase III
,d gh izfrca/ku ,aMksU;wfDy,t ls dkVs x;s ,d fotkrh; DNA vkSj IykfTem dks iqu;ksZxt IykfTeM cukus ds fy, fdldk
mi;ksx djds bUgsa tksM+k tk ldrk gSA
(1)
ykbxst
(2) Eco Rl
(3)
VSd ikWfyejst
(4)
Ans.
1
47.
Which of the following is not a component of downstream processing?
(1) Expression
(2) Separation
(3) Purification
ikWfyejst III
(4) Preservation
fuEufyf[kr esa ls dkSulk vuqizokg izØe.k dk ,d vo;u ugha gS\
(1)
vfHkO;fDr
(2)
i`FkDdj.k
(3)
'kqf)dj.k
(4)
Ans.
1
48.
Which of the following restriction enzymes produces blunt ends?
(1) Hind III
(2) Sal I
(3) Eco RV
ifjj{k.k
(4) Xho I
fuEufyf[kr esa ls dkSulk izfrca/ku ,atkbe dqafBr fljs mRiUu djrk gS\
(1) Hind III
(2) Sal I
(3) Eco RV
(4) Xho I
Ans.
3
49.
Which kind of therapy was given in 1990 to a four-year-old girl with adenosine deaminase (ADA)
deficiency?
(1) Radiation therapy
(2) Gene therapy
(3) Chemotherapy
(4) Immunotherapy
o"kZ 1990 esa ,fMukslhu Mh,sfeust (,-Mh-,-) dh deh ls ihfM+r pkj o"kZ dh ckfydk dks fuEufyf[kr essa ls dkSulh fpfdRlk nh
x;h\
(1)
fofdj.k fpfdRlk
(2)
thu fpfdRlk
(3)
jlk;u fpfdRlk
(4)
izfrj{kk fpfdRlk
Ans.
2
50.
How many hot spots of biodiversity in the world have been identified till date by Norman Myers?
(1) 43
(2) 17
(3) 25
(4) 34
ukWeZu ek;lZ }kjk vc rd fo'o esa fdrus tSo fofo/krk okys gkWV LikWV igpkus x;sa gS\
(1) 43
Ans.
(2) 17
(3) 25
(4) 34
4
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NEET - 2 Examination (2016) (Code - _)
51.
(Page # 15)
The primary producers of the deep-sea hydrothermal vent ecosystem are
(1) coral reefs
(2) green algae
(3) chemosynthetic bacteria
(4) blue-green algae
xgjs leqnz ds m".ktyh; osUV ifjra= esa izkFkfe mRiknd dkSu gS\
(1)
izoky&fHkfÙk
(2)
gfjr 'kSoky
(3)
jlk;u&la'ys"k.k thok.kq
(4)
uhy&gfjr 'kSoky
Ans.
3
52.
Which of the following is correct for r-selected species?
(1) Small number of progeny with large size
(2) Large number of progeny with small size
(3) Large number of progeny with large size
(4) Small number of progeny with small size
r-p;fur
tkfr;ksa ds ckjs esa fuEufyf[kr esa ls dkSulk lgh gS\
(1)
de la[;k esa cM+s vkeki okyh larfr
(2)
cM+h la[;k esa NksVs vkeki okyh larfr
(3)
cM+h la[;k esa cM+s vkeki okyh larfr
(4)
de la[;k esa NksVs vkeki okyh larfr
Ans.
2
53.
If '+' sign is assigned to beneficial interaction, '–' sign to detrimental and '0' sign to neutral
interaction, then the population interaction represented by '+' '–' refers to
(1) parasitism
(2) mutualism
(3) amensalism
(4) commensalism
;fn '+' fpà dks ykHknk;h ifjLifjfØ;k ds fy,] '–' fpà dks gkfudkjd ds fy, vkSj '0' fpà dks mnklhu ijLijfØ;k ds fy,
fn;k tkrk gS] rks '+' '–' }kjk nf'kZr lef"V ijLijfØ;k fdls lanfHkZr djrh gS\
(1)
ijthfork
(2)
lgksidkfjrk
(3)
Ans.
1
54.
Which of the following is correctly matched?
varjtkrh; ijthfork (4) lgHkksftrk
(1) Stratification-Population
(2) Aerenchyma-Opuntia
(3) Age pyramid-Biome
(4) Parthenium hysterophorus–Threat to biodiversity
fuEufyf[kr esa ls dkSulk lgh lqesfyr gS\
Ans.
(1)
Lrj.k&lef"V
(2)
ok;wrd&vksiaf'k;k
(3)
vk;q fijSfeM&thokse
(4)
ikFkhZfu;e fgLVsjksQksjl&tso fofo/krk ds fy, ladV
4
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NEET - 2 Examination (2016) (Code - _)
(Page # 16)
55.
Red List contains data or information on
(1) marine vertebrates only
(2) all economically important plants
(3) plants whose products are in international trade
(4) threatened species
yky lwph esa fduds ckjs esa vk¡dM+sa ;k lwpuk gksrh gS\
(1)
dsoy leqnzh d'ks:dh izk.kh
(2)
vkfFkZd :i ls egRoiw.kZ lHkh ikni
(3)
os ikni ftuds mRikn varjkZ"Vªjh; O;kikj esa gS
(4)
ladVkiUu tkfr;k¡
Ans.
4
56.
Which one of the following is wrong for fungi?
(1) They are both unicellular and multicellular.
(2) They are eukaryotic.
(3) All fungi possess a purely cellulosic cell wall.
(4) They are heterotrophic.
dodksa ds fy, fuEufyf[kr esa ls dkSulk&xyr gS\
(1)
;s ,ddksf'kdh; vkSj cgqdksf'kdh; nksuksa izdkj ds gksrs gSA
(2)
;s llhedsUnzdh gSaA
(3)
lHkh dodksa esa 'kq) lsyqyksl okyh dksf'kdk&fHkfÙk gksrh gSA
(4)
;s fo"keiks"kh gksrs gSaA
Ans.
3
57.
Methanogens belong to
(1) Slime moulds
(2) Eubacteria
(3) Archaebacteria
(4) Dinoflagellates
feFksukstu fdlesa lEcfU/kr gksrs gSa\
(1)
voiad QQ¡wnh
(2)
lqthok.kq
Ans.
3
58.
Select the wrong statement.
(3)
vk|thok.kq
(4)
Mkbuks¶yStsysV
(1) Diatoms are microscopic and float passively in water.
(2) The walls of diatoms are easily destructible.
(3) 'Diatomaceous earth' is formed by the cell walls of diatoms.
(4) Diatoms are chief producers in the oceans.
xyr dFku pqfu;sA
Ans.
(1)
Mk;Ve lwPen'khZ; gksrs gSa vkSj ty esa fu'ps"V rSjrs gSaA
(2)
Mk;Ve dh fHkfÙk;k¡ vklkuh ls ?oal gksrh gSA
(3)
^Mk;Veh e`fÙkdk^ dk fuekZ.k Mk;Veksa dh dksf'kdk fHkfÙk;ksa ls gksrk gSA
(4)
egklkxjksa esa Mk;Ve izeq[k mRiknd gSA
2
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NEET - 2 Examination (2016) (Code - _)
59.
(Page # 17)
The label of a herbarium sheet does not carry information on
(1) height of the plant
(2) date of collection
(3) name of collector
(4) local names
ikniky; i= ds ukei= esa fuEufyf[kr esa ls dkSulh lwpuk vafdr ugha gksrh\
(1)
ikS/ks dh Å¡pkbZ
(2)
laxzg dh rkjh[k
(3)
laxzgdrkZ dk uke
(4)
LFkkuh; uke
Ans.
1
60.
Conifers are adapted to tolerate extreme environmental conditions because of
(1) presence of vessels
(2) broad hardy leaves
(3) superficial stomata
(4) thick cuticle
'kadq/kkjh ikni i;kZoj.k dh pje n'kkvksa dks lgu djus ds fy, vuqdwfyr gksrs gSa] D;ksafd muesa
(1)
okfgdkvksa dh mifLFkfr gksrh gS
(2)
pkSM+h dBksj ifÙk;k¡ gksrh gS
(3)
ja/kz lrg ij gksrs gS
(4)
eksVh miRopk gksrk gksrh gS
Ans.
4
61.
Which one of the following statements is wrong?
(1) Laminaria and Sargassum are used as food .
(2) Algae increase the level of dissolved oxygen in the immediate environment.
(3) Algin is obtained from red algae, and carrageenan from brown algae.
(4) Agar-agar is obtained from Gelidium and Gracilaria.
fuEufyf[kr esa ls dkSulk dFku xyr gS\
(1)
ysfeusfj;k vkSj lkxSZle [kk| ds :i esa iz;qä fd;s tkrs gSA
(2)
'kSokysa vius lfUudV i;kZoj.k esa ?kqfyr vkWDlhtu ds Lrj dks c<+krh gSA
(3)
,fYtu yky 'kSokyksa ls rFkk dSjkfxuu Hkwj 'kSokyksa ls izkIr fd;k tkrk gSA
(4)
,sxkj&,sxkj] ftyhfM;e vkSj xzsflysfj;k ls izkIr fd;k tkrk gSA
Ans.
3
62.
The term 'polyadelphous' is related to
(1) calyx
(2) gynoecium
(3) androecium
(4) corolla
^cgqla?kh^ in fdlls lEcfU/kr gS\
(1)
ckányiqat
(2)
tk;kax
(3)
iqeax
(4)
nyiqat
Ans.
3
63.
How many plants among Indigofera, Sesbania, Salvia, Allium, Aloe, mustard, groundnut, radish,
gram and turnip have stamens with different lengths in their flowers?
(1) Six
(2) Three
(3) Four
(4) Five
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NEET - 2 Examination (2016) (Code - _)
(Page # 18)
bafMxksQsjk] lsLcsfu;k] lSfYo;k] ,sfy;e] ,syks] ljlksa] ew¡xQyh] ewyh] puk] vkSj 'kyte esa ls fdrus ikS/kksa ds iq"iksa esa iqadsljksa dh
yEckbZ fHkUu&fHkUu gksrh gS\
(1)
N^
(2)
rhu
Ans.
3
64.
Radial symmetry is found in the flowers of
(1) Cassia
(2) Brassica
(3)
pkj
(4)
(3) Trifolium
ik¡p
(4) Pisum
vjh; lefefr fdlds iq"iksa esa ik;h tkrh gS\
(1)
dSfl;k
(2)
czSfldk
Ans.
2
65.
Free-central placentation is found in
(1) Citrus
(2) Dianthus
(3)
VªkbQksfy;e
(4)
(3) Argemone
ikble
(4) Brassica
eqä&v{kh; chtk.MU;kl fdlesa ik;k tkrk gS\
(1)
flVªl
(2)
Mkb,sUFkl
Ans.
2
66.
Cortex is the region found between
(3)
vkftZeksu
(4)
(1) endodermis and vascular bundle
(2) epidermis and stele
(3) pericycle and endodermis
(4) endodermis and pith
czSfldk
oYdqV {ks= fduds chp esa ik;k tkrk gS\
(1)
vUrLropk vkSj laogu cUMy
(2)
ckáropk vkSj jEHk
(3)
ifjjEHk vkSj vUrLRopk
(4)
vUrLRopk vkSj eTtk
Ans.
2
67.
The balloon-shaped structures called tyloses
(1) are linked to the ascent of sap through xylem vessels
(2) originate in the lumen of vessels
(3) characterize the sapwood
(4) are extensions of xylem parenchyma cells into vessels
xqCckjsuqek lajpuk;sa] tks Vkbyksfll dgykrh gS] os
Ans.
(1)
tkbye okfgdkvksa ls gksdj jlkjksg.k ls lEcfU/kr gksrh gS
(2)
okfgdkvksa dh vodkf'kdk ls mRiUu gksrh gS
(3)
jldk"B dks vfHkyf{kr djrh gS
(4)
okfgdkvksa esa tkbye e`nwrd dksf'kdkvksa dh izlkj gksrh gS
4
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68.
(Page # 19)
A non-proteinaceous enzyme is
(1) deoxyribonuclease
(2) lysozyme
(3) ribozyme
(4) ligase
fuEufyf[kr esa ls dkSulk ,d xSj izksVhu okyk a,atkbe gS\
(1)
fMvkWDlhjkbcksU;wfDy,t
(2)
ykblkstkbe
(3)
jkbckstkbe
(4)
ykbxst
Ans.
3
69.
Select the mismatch .
(1) Methanogens-Prokaryotes
(2) Gas vacuoles-Green bacteria
(3) Large central vacuoles-Animal cells
(4) Protists-Eukaryotes
csesy pqfu;sA
(1)
feFksukstu&izkd~dsUnzdh
(2)
xsl jl/kkfu;k¡ &gfjr thok.kq
(3)
cM+h dsUnzh; jl/kkfu;k¡ &tarq dksf'kdk;sa
(4)
izksfVLV&llhedsUnzdh
Ans.
3
70.
Select the wrong statement.
(1) Mycoplasma is a wall-less microorganism.
(2) Bacterial cell wall is made up of peptidoglycan.
(3) Pili and fimbriae are mainly involved in motility of bacterial cells.
(4) Cyanobacteria lack flagellated cells.
xyr dFku pqfu;sA
(1)
ekbdksIykTek ,d fHkfÙk&jfgr lw{etho gSA
(2)
thok.kq dksf'kdk&fHkfÙk isIVhMksXykbdSu dh cuh gksrh gSA
(3)
jksed vkSj >kyj eq[; :i ls thok.kq dksf'kdkvks dh xfr'khyrk ds fy, gksrs gSA
(4)
lk;ukscSDVhfj;k esa d'kkHkh dksf'kdkvksa dk vHkko gksrk gSA
Ans.
3
71.
A cell organelle containing hydrolytic enzymes is
(1) mesosome
(2) lysosome
(3) microsome
(4) ribosome
fdl dksf'kdh; vaxd esa ty&vi?kVuh ,atkbe gksrk gS\
(1)
e/;dk;
(2)
y;udk;
(3)
lw{edk;
Ans.
2
72.
During cell growth, DNA synthesis takes place in
(1) M phase
(2) S phase
(3) G1 phase
(4)
jkbckstkse
(4) G2 phase
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dksf'kdk o`f) ds nkSjku] DNA dk la'ys"k.k fdl izkoLFkk esa gksrk gS\
(1) M
izkoLFkk
(2) S
izkoLFkk
(3) G1
izkoLFkk
(4) G2 izkoLFkk
Ans.
2
73.
Which of the following biomolecules is common to respiration-mediated breakdown of fats, carbohydrates and proteins?
(1) Acetyl CoA
(2) Glucose-6-phosphate
(3) Fructose 1,6-bisphosphate
(4) Pyruvic acid
fuEufyf[kr esa ls dkSulk tso v.kq olk] dkcksgkbMªsV vkSj izksVhu ds 'olu &ekf/;r Hkatu esa loZfu"B gS\
(1)
,slhfVy CoA
(2)
Xywdksl -6-QkWLQsV
(3)
ÝDVksl 1,6-fclQkWLQsV
(4)
ikb:fod vEy
Ans.
1
74.
A few drops of sap were collected by cutting across a plant stem by a suitable method. The sap
was tested chemically. Which one of the following test results indicates that it is phloem sap?
(1) Absence of sugar
(2) Acidic
(3) Alkaline
(4) Low refractive index
,d mi;qä fof/k }kjk ,d ikni ds rus dh vkjikj dkVdj jl dh dqN cw¡nsa ,df=r dh x;haA jl dk jklk;fud ijh{k.k fd;k
x;kA fuEufyf[kr esa ls dkSulk ijh{k.k ifj.kke ;g n'kkZ;sxk fd ;g ¶yks,e jl gS\
(1)
'kdZjk dh vuqifLFkfr
(2)
vEyh;
(3)
{kkjh;
(4)
fuEu viorZukad
Ans.
3
75.
You are given a tissue with its potential for differentiation in an artificial culture. Which of the
following pairs of hormones would you add to the medium to secure shoots as well as roots?
(1) Gibberellin and abscisic acid
(2) IAA and gibberellin
(3) Auxin and cytokinin
(4) Auxin and abscisic acid
vkidks ,d d`f=e ek/;e esa foHksnu dh {kerk okyk ,d Ård fn;k x;k gSA izjksgksa vkSj nksuska dks izkIr djus ds fy, vki ek/
;e esa fuEufyf[kr esa ls gkWeksZauksa ds fdl ;qXe dks feyk;saxs\
(1)
ftCcjsfyu vkSj ,sfClfld vEy
(2) IAA
(3)
vkWfDlu vkSj lkbVksdkbfuu
(4)
Ans.
3
76.
Phytochrome is a
(1) chromoprotein
(2) flavoprotein
vkSj ftCcjsfyu
vkWfDlu vksj ,sfClfld vey
(3) glycoprotein
(4) lipoprotein
iknio.kZd D;k gS\
(1)
Ans.
ØkseksizksVhu
(2)
¶ySoksizksVhu
(3)
XykbdksizksVhu
(4)
ykbiksizksVhu
1
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77.
(Page # 21)
Which is essential for the growth of root tip?
(1) Mn
(2) Zn
(3) Fe
(4) Ca
ewykxzksa dh o`f) ds fy, fuEufyf[kr esa ls dkSulk vko';d gS\
(1) Mn
(2) Zn
(3) Fe
(4) Ca
Ans.
4
78.
The process which makes major difference between C3 and C4 plants is
(1) respiration
(2) glycolysis
(3) Calvin cycle
(4) photorespiration
fuEufyf[kr esa ls og dkSulh izfØ;k gS] tks C3 vkSj C4 ikniksa ds chp eq[; :i ls foHksn djrh gS\
(1)
'olu
(2)
Xykbdksykbfll
(3)
dSfYou pØ
Ans.
4
79.
Which one of the following statements is not correct?
(4)
izdk'k'olu
(1) Water hyacinth, growing in the standing water, drains oxygen from water that leads to the
death of fishes.
(2) Offspring produced by the asexual reproduction are called clone.
(3) Microscopic, motile a sexual reproductive structures are called zoospores.
(4) In potato, banana and ginger, the plantlets arise from the internodes present in the modified
stem.
fuEufyf[kr esa ls dkSulk dFku lgh ugha gS\
(1)
:ds gq, ty esa mxyh gqbZ ty gk;kflUFk ty ls vkWDlhtu [khap ysrh gS ftlls eNfy;ksa dh e`R;q gks tkrh gSA
(2)
vySafxd iztuu }kjk mRiUu larkuksa dks Dyksu dgk tkrk gSA
(3)
lw{en'khZ;] py vySafxd iztuu lajpuk;sa py chtk.kq dgykrh gSA
(4)
vkyw] dsyk vkSj vnjd esa iknid] :ikarfjr rus esa mifLFkr ioksZ ls mRiUu gksrs gSA
Ans.
4
80.
Which one of the following generates new genetic combinations leading to variation?
(1) Nucellar polyembryony
(2) Vegetative reproduction
(3) Parthenogenesis
(4) Sexual reproduction
fuEufyf[kr esa ls dkSulk fofHkUurk ykus okys u;s vkuqoaf'kd la;kstu dks mRiUu djrk gS\
(1)
chtk.Mdkf;d cgqHkwz.krk
(2)
dkf;d tuu
(3)
vfu"ksdtuu
(4)
ySafxd tuu
Ans.
4
81.
Match Column-I with Column-II and select the correct option using the codes given below :
Column-I
Column-II
a.
Pistils fused together
(i)
Gametogenesis
b.
Formation of gametes
(ii)
Pistillate
c.
Hyphae of higher Ascomycetes
(iii)
Syncarpous
d.
Unisexual female flower
(iv)
Dikaryotic
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Codes:
a
b
c
d
(1 )
(iii)
(i)
(iv)
(ii)
(2)
(iv)
(iii)
(i)
(ii)
(3)
(ii)
(i)
(iv)
(iii)
(4)
(i)
(ii)
(iv)
(iii)
dkWye -I dks dkWye -II ls lqesfyr dhft;s rFkk uhps fn;s x;s dwV dk iz;ksx dj lgh fodYi dks pqfu;s %
dkW ye -I
dkWy e -II
a.
vkil esa tqM+s L=hdslj
(i)
;qXedtuu
b.
;qXedksa dk cuuk
(ii)
L=hdsljh
c.
mPprj ,sLdksekbflfVt ds dod gsrq
(iii)
;qäk.Mih
d.
,dfyaxh eknk iq"i
(iv)
f}dsUnzdh
dwV :
a
b
c
d
(1 )
(iii)
(i)
(iv)
(ii)
(2)
(iv)
(iii)
(i)
(ii)
(3)
(ii)
(i)
(iv)
(iii)
(4)
(i)
(ii)
(iv)
(iii)
Ans.
1
82.
In majority of angiosperms
(1) a small central cell is present in the embryo sac
(2) egg has a filiform apparatus
(3) there are numerous antipodal cells
(4) reduction division occurs in the megaspore mother cells
vf/kdka'k vko`rchth ikniksa esa
(1)
Hkwz.k dks"k esa ,d y?kq dsUnzh; dksf'kdk gksrh gS
(2)
v.M esa rarq"i leqPp; gksrk gS
(3)
cgqr lh izfrO;klkar dksf'kdk;sa gksrh gS
(4)
v)Zlw=h foHkktu] xq:fctk.kq ekr` dksf'kdkvksa esa gksrk gS
Ans.
4
83.
Pollination in water hyacinth and water lily is brought about by the agency of
(1) bats
(2) water
(3) insects or wind
(4) birds
ty gk;kflU/k vkSj ty dqeqfnuh esa ijkx.k fdlds }kjk gksrk gS\
(1)
Ans.
pexknM+
(2)
ty
(3)
dhV ;k iou
(4)
i{kh
3
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84.
(Page # 23)
The ovule of an angiosperm is technically equivalent to
(1) megaspore
(2) megasporangium
(3) megasporophyll
(4) megaspore mother cell
vko`rchth ikniksa dk chtk.M rduhdh :i esa fdlds led{k gksrk gS\
(1)
xq:chtk.kq
(2)
xq:chtk.kq/kkuh
(3)
xq:chtk.kqi.kZ
(4)
xq:chtk.kq ekr` dksf'kdk
Ans.
2
85.
Taylor conducted the experiments to prove semiconservative mode of chromosome replication on
(1) E. coli
(2) Vinca rosea
(3) Vida faba
(4) Drosophila melanogaster
Vsyj us fdl tho dk xq.klw= izfrd`fr;u dh v)Z&laj{kh fof/k dks izekf.kr djus ds fy, iz;ksx fd;k Fkk\
(1)
bZ- dksykbZ
(2)
foUdk jksft;k
(3)
fofl;k QSck
(4)
MªkslksfQyk esykusxSLVj
Ans.
3
86.
The mechanism that causes a gene to move from one linkage group to another is called
(1) crossing-over
(2) inversion
(3) duplication
(4) translocation
ml fØ;kfof/k dks] ftlds dkj.k ,d thu ,d lgyXurk lewg ls nwljs lgyXurk lewg dks pyk tkrk gS] D;k dgykrk gSA
(1)
thu&fofue;
(2)
izfrykseu
Ans.
4
87.
The equivalent of a structural gene is
(1) recon
(2) muton
(3)
f}xq.ku
(3) cistron
(4)
LFkkukUrj.k
(4) operon
fuEufyf[kr esa ls dkSu lajpukRed thu ds leku gS\
(1)
iqujk.kq
(2)
Ans.
3
88.
A true breeding plant is
mRik.kq
(3)
leikj
(4)
izpkysd
(1) always homozygous recessive in its genetic constitution
(2) one that is able to breed on its own
(3) produced due to cross-pollination among unrelated plants
(4) near homozygous and produces offspring of its own kind
,d okLrfod iztuu ikni og gS] tks fd
Ans.
(1)
vius vkuqoaf'kd xBu esa ges'kk letkr vizHkkoh gks
(2)
vius vki iztuu dj lds
(3)
vlEc) ikniksa ds chp ij&ijkx.k ls mRiUu fd;k x;k gks
(4)
yxHkx letkr gks vkSj viuh rjg dh larku mRiUu djrk gks
4
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NEET - 2 Examination (2016) (Code - _)
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89.
Which of the following rRNAs acts as structural RNA as well as ribozyme in bacteria?
(1) 5·8 S rRNA
(2) 5 S rRNA
(3) 18 S rRNA
(4) 23 S rRNA
thok.kq esa] fuEufyf[kr esa ls dkSulk rRNA, lajpukRed RNA vkSj jkbckstkbe] nksuksa dh rjg dk;Z djrk gS\
(1) 5·8 S rRNA
(2) 5 S rRNA
(3) 18 S rRNA
Ans.
4
90.
Stirred-tank bioreactors have been designed for
(4) 23 S rRNA
(1) ensuring anaerobic conditions in the culture vessel
(2) purification of product
(3) addition of preservatives to the product
(4) availability of oxygen throughout the process
foyksfM+r VSad tSo fj,sdVj fdl fy, vfHkdfYir fd;s x;s gSa\
(1)
izo/kZu ufydk esa vok;oh; n'kkvksa dks cuk;s j[kus ds fy,
(2)
mRiknksa ds 'kqf)dj.k ds fy,
(3)
mRiknksa esa ifjj{kdksa dks feykus ds fy,
(4)
lkjh izfØ;k ds nkSjku vkWDlhtu dh izkI;rk cuk;s j[kus ds fy,
Ans.
4
91.
A molecule that can act as a genetic material must fulfill the traits given below, except
(1) it should provide the scope for slow changes that are required for evolution
(2) it should be able to express itself in the form of 'Mendelian characters'
(3) it should be able to generate its replica
(4) it should be unstable structurally and chemically
91.
fdlh v.kq esa] tks vkuqoaf'kd inkZFk ds :i esa dk;Z dj ldrk gS] uhpsa fn;s x;s fdl fo'ks"kd ds vfrfjDr] vU; lHkh fo'ks"kd
vo'; gksus pkfg;s\
(1) blesa fodkl ds fy, vko';d ean ifjorZuksa ds fy, volj gksuk pkfg;s
(2) besa ^esUMsyhl y{k.kksa^ ds :i esa Lo;a dks vfHkO;Dr djus ;ksX; gksuk pkfg;s
(3) besa viuh izfrd`fr mRiUu djus ;ksX; gksuk pkfg;s
(4) bls lajpukRed :i ls vkSj jklk;fud :i ls vfLFkj gksuk pkfg;sA
Ans.
4
92.
DNA-dependent RNA polymerase catalyzes transcription on one strand of the DNA which is called
the
(1) antistrand
(2) template strand (3) coding strand
(4) alpha strand
92.
DNA-vk/kkfjr RNA
Ans.
(1)
2
93.
izfrjTtqd
ikWsfyejst DNA ds fdl ,d jTtqd ij vuqys[ku dk mRizsj.k djrk gS\
(2) VsEIysV jTtqd
(3) dksMu jTtqd
(4) ,sYQk jTtqd
lnterspecilic hybridization is the mating of
(1) more closely related individuals within same breed for 4-6 generations
(2) animals within same breed without having common ancestors
(3) two different related species
(4) superior males and females of different breeds
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93.
(Page # 25)
varjkLih'khth ladj.k esa fdlds chp lekxe djk;k tkrk gS\
(1) 4-6
ihf<+;ksa rd ,d gh uLy okys rFkk utnhd ls lEcfU/kr O;f"V;ksa ds chp
(2)
leku iwoZt u gksrs gq, Hkh ,d gh uLy ds tarqvksa ds chp
(3)
nks vyx&vyx lEcfU/kr Lih'khtksa ds chp
(4)
fofHkUu uLyksa okys csgrj ujksa rFkk eknkvksa ds chp
Ans.
3
94.
Which of t.he following is correct regarding AIDS causative agent HIV?
(1) HIV does not escape but attacks the acquired immune response.
(2) HIV is enveloped virus containing one molecule of single-stranded RNA and one molecule of
reverse transcriptase.
(3) HIV is enveloped virus that contains two identical molecules of single-stranded RNA and two
molecules of revers transcriptase.
(4) HIV is unenveloped retrovirus.
94.
AIDS
ds jksxtud dkjd HIV ds ckjs esa fuEufyf[kr esa ls dkSu&lk lgh gS?
(1) HIV
ckgj ugha fudy ikrk ij miftZr izfrj{kh vuqfØ;k ij vkØe.k djrk gSA
(2) HIV
,d vko`r okbjl gS] ftlds Hkhrj ,dy jTtqd okys RNA dk ,d v.kq vkSj mRØe VªaklfØIVst dk ,d v.kq gksrk
gSA
(3) HIV
,d vko`r okbjl gS] ftlds Hkhrj ,dy jTtqd okys RNA ds nks leku v.kq rFkk mRØe VªkalfØIVst ds nks v.kq gksrs
gSA
(4) HIV
,d vuko`r i'p okbjl gSA
Ans.
3
95.
Among the following edible fishes, which one is a marine fish having rich source of omega-3 fatty
acids?
(1) Mackerel
95.
(2) Myst.us
(3) Mangur
(4) Mrigala
fuEufyf[kr [kk| eNfy;ksa esa ls og dkSu&lh leqnzh eNyh gS] tks vksesxk&3 olk vEyksa dk mÙke lzksr gS\
(1)
esdsjy
(2)
feLVe
(3)
ekaxqj
(4)
fezxy
Ans.
1
96.
Match Column-I with Column-ll and select the correct option using the codes given below :
Column-l
Column-II
a.
Citric acid
(i)
Trichoderma
b.
Cyclosporin A
(ii)
Clostridium
c.
Statins
(iii)
Aspergillus
d.
Butyric acid
(iv)
Monascus
Codes:
a
b
c
d
(1)
(iii)
(iv)
(i)
(ii)
(2)
(iii)
(i)
(ii)
(iv)
(3)
(iii)
(i)
(iv)
(ii)
(4)
(i)
(iv)
(ii)
(iii)
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96.
dkWye-I vkSj dkWye-ll ds chp feyku dhft;s rFkk uhps fn;s x;s dwV dk iz;ksx dj lgh fodYi dks pqfu;s%
dkWye-l
dkWye-II
a.
flfVªd vEy
(i)
VªkbdksMekZ
b.
lkbDyksLiksfju A
(ii)
DykWfLMªfM;e
c.
LVsfVu
(iii)
,sLijftye
d.
C;wfVfjd vEy
(iv)
eksuSLdl
dwV:
(1)
Ans.
(2)
(3)
(4)
3
a
(iii)
b
(iv)
c
(i)
d
(ii)
(iii)
(iii)
(i)
(i)
(i)
(iv)
(ii)
(iv)
(ii)
(iv)
(ii)
(iii)
97.
Biochemical Oxygen Demand (BOD) may not be a good index for pollution for water bodies
receiving effluents from
(1) sugar industry
(2) domestic sewage
(3) dairy industry
(4) petroleum industry
97.
fuEyfyf[kr esa ls fdlds cfg%lzkoksZ ds dkj.k iznwf"kr gksus okys ty&fudk;ksa esa tSo jklk;fud vkWDlhtu ek¡x (BOD) iznw"k.k
ds fy, ,d vPNk lwpd ugha gS\
(1) 'kdZjk m|ksx
(2) ?kjsyw okfgr ey
(3) nqX/k m|ksx
(4) isVªksfy;e m|ksx
Ans.
4
98.
The principle of competitive exclusion was stated by
(1) Verhulst and Pearl
(3) G. F. Gause
(2) C. Danvin
(4) MacArthur
98.
Li/khZ vuU;rk ds fu;e dk izfriknu fdlus fd;k Fkk
(1) cjgqYLV vkSj iyZ
(2) lh- MkfoZu
(3) th- ,Q- xkWls
(4) eSd~vkFkZj
Ans.
3
99.
Which of the following National Parks is home to the famous musk deer or hangul?
(1) Dachigam National Park, Jammu & Kashmir
(2) Keibul Lamjao National Park, Manipur
(3) Bandhavgarh National Park, Madhya Pradesh
(4) Eaglenest Wildlife Sanctuary, Arunachal Pradesh
99.
fo[;kr dLrqjh e`x vFkok gaxqy fuEufyf[kr jk"Vªh; m|kuksa esa ls dgk¡ ik;k tkrk gS\
(1) Mkphxke jk"Vªh; m|ku] tEew vkSj d'ehj
(2) dhcqy yketkvks jk"Vªh; m|ku] ef.kiqj
(3) cka/kox<+ jk"Vªh; m|ku] e/; izns'k
(4) bZxysLV oU;tho 'kj.k&LFky] v:.kkpy izns'k
Ans.
1
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NEET - 2 Examination (2016) (Code - _)
(Page # 27)
100.
A lake which is rich in organic waste may result in
(1) mortality of fish due to lack of oxygen
(2) increased popu lation of aquatic organisms due to minerals
(3) drying of the lake due to algal bloom
(4) increased population of fish due to lots of nutrients
100.
dkcZfud vif'k"V ls Hkjh fdlh >hy esa D;k gks ldrk gS\
(1) vkWDlhtu dh deh ds dkj.k eNfy;ksa dk ej tkuk
(2) [kfutksa ds dkj.k tyh; thoksa dh lef"V esa o`f)
(3) 'kSoky&LQqVu ds dkj.k >hy dk lw[k tkuk
(4) vf/kd iks"kd inkFkksZ ds dkj.k eNfy;sa dh lef"V esa o`f)
Ans.
1
101.
The highest DDT concentration in aquatic food chain shall occur in
(1) eel
(2) phytoplankton
(3) seagull
(4) crab
101.
tyh; [kk|&J`a[kyk esa vf/kdre DDT dh lkanzrk fdlesa gksxh\
(1) bZy
(2) ikniIyod
(3) leqnzh xy
Ans.
3
102.
Which of the following sets of diseases is caused by bacteria?
(1) Herpes and influenza
(2) Cholera and tetanus
(3) Typhoid and smallpox
(4) Tetanus and mumps
102.
jksxksa dk fuEufyf[kr esa ls dkSu&lk lewg thok.kqvksa }kjk laØfer gksrk gS\
(1) gihZt vkSj bU¶yq,atk
(2) gStk vkSj fVVusl
(3) VkbQkWbM vkSj pspd ¼LekWyikWDl½
(4) fVVsul vkSj xylqvk
Ans.
2
103.
Match Column-I with Column-II for h ousefly classification and select the correct option using the
codes given below :
Column-I
Column-II
a.
Family
(i)
Diptera
b.
Order
(ii)
Arthropoda
c
Class
(iii)
Muscidae
d.
Phylum
(iv)
Insecta
Codes :
a
b
c
d
(1)
(iv)
(ii)
(i)
(iii)
(2)
(iii)
(i)
(iv)
(ii)
(3)
(iii)
(ii)
(iv)
(i)
(4)
(iv)
(iii)
(ii)
(i)
103.
?kjsyw eD[kh ds oxhZdj.k ds fy, dkWye-I vkSj dkWye-II esa feyku dhft;s rFkk uhps fn;s x;s dwV dk iz;ksx dj lgh fodYi
dks pqfu;s%
dkWye-I
dkWye-II
a.
dqy ¼QSfeyh½
(i)
fMIVsjk
b.
x.k ¼vkWMZj½
(ii)
vkFkzksZiksMk
c
oxZ ¼Dykl½
(iii)
efLlMh
d.
la/k ¼Qkbye½
(iv)
balsDVk
(4)
dsdM+k
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NEET - 2 Examination (2016) (Code - _)
(Page # 28)
Codes :
a
b
c
d
(1)
(2)
(3)
(4)
(ii)
(i)
(ii)
(iii)
(i)
(iv)
(iv)
(ii)
(iii)
(ii)
(i)
(i)
(iv)
(iii)
(iii)
(iv)
Ans.
2
104.
Choose the correct statement.
(1) All Pisces have gills covered by an operculum.
(2) All mammals are viviparous.
(3) All cyclostomes do not possess jaws and paired fins.
(4) All reptiles have a three-chambered heart.
104.
lgh dFku pqfu;saA
(1) lHkh eNfy;sa esa Dykse izPNn ls <¡ds gq, gksrs gSaA
(2) lHkh Lru/kkjh lthoiztd gSA
(3) lHkhlkbDyksLVkseksa esa tcM+s rFkk ;qfXer ia[k ugha gksrks gSaA
(4) lHkh ljhl`iksa esa rhu&d{kh; gn; gksrk gSA
Ans.
3
105.
Study the four statements (A-D) given below and select the two correct ones out of them :
A. Definition of biological species was given by Ernst Mayr.
B. Photoperiod does not affect reproduction in plants.
C. Binomial nomenclature system was given by R. H. Whittaker.
D. In unicellular organisms, reproduction is synonymous with growth.
The two correct statements are
(1) A and B
(2) Band C
(3) C and D
(4) A and D
105.
uhps fn;s x;s pkj dFkuksa (A-D) dk v/;;u dhft;s vkSj muesa ls nks lgh dFkuksa dks pqfu;s %
A. tSo Lih'khtksa dh ifjHkk"kk vULVZ ek;j us nh FkhA
B. izdk'kdky dk ikS/kksa ds tuu ij izHkko ugha iM+rk gSA
C. f}uke ukei)fr ra= vkj- ,p- fgVsdj us fn;k Fkk
D. ,ddksf'kdk thoksa esa tuu vkSj o`f) lekukFkZd gksrs gSaA
nks lgh dFku gSa
(1) A vkSj B
(2) B vkSj C
(3) C vkSj D
(4) A vkSj D
Ans.
4
106.
In male cockroaches, sperms are stored in which part of the reproductive system?
(1) Vas deferens
(2) Seminal vesicles
(3) Mushroom glands
(4) Testes
106.
uj frypês esa 'kqØk.kq tuura= ds fdl Hkkx esa HkaMkfjr jgrs gSa\
(1) 'kqØokgd
(2) 'kqØk'k;
(3) e'k:e xzafFk;k¡
(4) o`"k.k
Ans.
2
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NEET - 2 Examination (2016) (Code - _)
107.
107.
(Page # 29)
Smooth muscles are
(1) voluntary, spindle-shaped, uninucleate
(3) voluntary, multinucleate, cylindrical
(2) involuntary, fusiform, non-striated
(4) involuntary, cylindrical, striated
fpduh is'kh gksrh gSa
(1) ,sfPNd] rdqZ:ih] ,ddsUnzdh;
(3) ,sfPNd] cgqdsUnzdh;] csyukdkj
vuSfPNd] rdqZ:i] vjsf[kr
(4) vuSfPNd] csyukdj] jsf[kr
(2)
Ans.
2
108.
Oxidative phosphorylation is
(1) formation of ATP by energy released from electrons removed during substrate oxidation
(2) formation of ATP by transfer of phosphate group from a sUbstrate to ADP
(3) oxidation of phosphate group in ATP
(4) addition of phosphate group to ATP
108.
vkWDlhMsfVo QkWLQksfjys'ku gksrk gS
(1) fØ;k/kkj ds vkWDlhdj.k ds nkSjku bysDVªkWu ds vyx fd;s tkus ls mRiUu ÅtkZ }kjk ATP dk fuekZ.k
(2) ,d fØ;k/kkj ls ADP rd QkWLQsV oxZ ds LFkkukarj.k }kjk ATP dk fuekZ.k
(3) ATP esa QkWLQsV oxZ dk vkWDlhdj.k
(4) QkWLQsV oxZ dk ATP esa tqM+ tkuk
Ans.
1
109.
Which of the following is the least likely to be involved in stabilizing the three-dimensional
folding of most proteins?
(1) Ester bonds
(2) Hydrogen bonds
(3) Electrostatic interaction
(4) Hydrophobic interaction
109.
vf/kdka'k izksVhuksa ds f=foeh; oyu dks fLFkj j[kus esa fuEufyf[kr esa ls fdldh cgqr de laHkkouk gS\a
(1) ,LVj vkca/k
(2) gkbMªkstu vkca/k
(3) fLFkjoS/kqr ikjLifjd vfHkfØ;k
(4) tyHkh: ikjLifjd vfHkfØ;k
Ans.
1
110.
Which of the following describes graph correctly?
Potential energy
B
A
Substrate
Product
Reaction
(1)
(2)
(3)
(4)
Exothermic reaction with energy A in absence of enzyme and B in presence of enzyme
Endothermic reaction with energy A in presence of enzyme and B in absence of enzyme
Exothermic reaction with energy A in presence of enzyme and B in absence of enzyme
Endothermic reaction with energy A in absence of enzyme and B in presence of enzyme
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NEET - 2 Examination (2016) (Code - _)
(Page # 30)
110.
fuEufyf[kr esa ls dkSu&lk fn;s x;s xzkQ dk lgh o.kZu djrk gS\
foHko ÅtkZ
B
A
fØ;k/kkj
mRikn
vfHkfØ;k
(1)
ÅtkZ A ds lkFk ,atkbe dh vuqifLFkfr esa vkSj B ds lkFk ,atkbe dh mifLFkfr esa ckáÅ"eh vfHkfØ;k
(2)
Åtk A ds lkFk ,atkbe dh mifLFkfr esa vkSj B ,atkbe dh vuqifLFkfr esa vkarjÅ"eh vfHkfØ;k
(3)
ÅtkZ A ds lkFk ,atkbe dh mifLFkfr esa vkSj B ds lkFk ,tkabe dh vuqifLFkfr esa ckáÅ"eh vfHkfØ;k
(4)
ÅtkZ A ds lkFk ,atkbe dh vuqifLFkfrs esa vkSj B ds lkFk ,tkabe dh mifLFkfr esa vkarjÅ"eh vfHkfØ;k
Ans.
3
111.
When cell has stalled DNA replication fork, which checkpoint should be predominantly activated?
111.
(1) Both G2 / M and M
(2) G1/S
(3) G2/M
(4) M
tc dksf'kdk esa DNA izfrd`fr;u f}'kk[k :d tkrk gS] rc fdl tk¡p&fcUnq dks izHkkoh :i ls lfØf;r djuk pkfg;s\
(1) G2 / M
vkSj M nksuksa
(2) G1/S
(3) G2/M
(4) M
Ans.
2
112.
Match the stages of meiosis in Column-I to their characteristic featu res in Column-ll and select
the correct option using the codes given below :
Column-I
Column-II
a.
Pachytene
(i)
Pairing of homologous chromosomes
b.
Metaphase I
(ii)
Terminalization of chiasmata
c.
Diakinesis
(iii)
Crossing-over takes place
d.
Zygotene
(iv)
Chromosomes align at equatorial plate
Codes:
a
b
c
d
(1)
(iv)
(iii)
(ii)
(i)
(2)
(iii)
(iv)
(ii)
(i)
(3)
(i)
(iv)
(ii)
(iii)
(4)
(ii)
(iv)
(iii)
(i)
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NEET - 2 Examination (2016) (Code - _)
(Page # 31)
dkWye-I esa nh x;h v/kZlw=h foHkktu dh fofHkUu voLFkkvksa dk dkWye-II esa fn;s x;s muds fof'k"V y{k.kksa ds lkFk feyku dhft;s
rFkk uhps fn;s x;s dwV dk iz;ksx dj lgh fodYi dks pqfu;s %
dkW ye -I
dkWy e -II
a.
iSdhVhu
(i)
letkr xq.klw=ksa dk ;qXeu
b.
esVkQst I (e/;koLFkk I)
(ii)
dkb,sTesvk dk lekiu
c.
Mk;kdkbusfll
(iii)
thu fofue; gksrk gS
d.
tkbxksVhu
(iv)
xq.klw= e/;orhZ iêh ij O;ofLFkr gks tkrs gS
dwV :
(1)
(2)
(3)
(4)
a
(iv)
(iii)
(i)
(ii)
b
(iii)
(iv)
(iv)
(iv)
c
(ii)
(ii)
(ii)
(iii)
d
(i)
(i)
(iii)
(i)
Ans.
2
113.
Which hormones do stimulate the production of pancreatic juice and bicarbonate?
(1) Insulin and glucagon
(2) Angiotensin and epinephrine
(3) Gastrin and insulin
(4) Cholecystokinin and secretin
dkSuls gkWeksZu vXU;k'k; jl vkSj ckbdkcksZusV ds mRiknu dks míhfir djrs gS\
(1) balqfyu vkSj XywdSxkWu
(2) ,aft;ksVsaflu vkSj ,fiusfÝu
(3) xSfLVªu vkSj balqfyu
(4) dksysflLVksdkbfuu vkSj lsØsfVu
Ans.
4
114.
The partial pressure of oxygen in the alveoli of the lungs is
(1) less than that of carbon dioxide
(3) more than that in the blood
(2) equal to that in the blood
(4) less than that in the blood
QsQM+ksa dh dwfidkvksa esa vkWDlhtu dh vkaf'kd nkc gksrh gS
(1) dkcZu MkbvkWDlkbM dh vkaf'kd nkc ls de
(2) #f/kj esa vkWDlhtu dh vkaf'kd nkc ds cjkcj
(3) #f/kj esa vkWDlhtu dh vkaf'kd nkc ls vf/kd
(4) #f/kj esa vkWDlhtu dh vkaf'kd nkc ls de
Ans.
3
115.
Choose the correct statement.
(1) Receptors do not produce graded potentials.
(2) Nociceptors respond to changes in pressure.
(3) Meissner's corpuscles are thermoreceptors.
(4) Photoreceptors in the human eye are depolarized during darkness and become hyperpolarized in response to the light stimulus.
lgh dFku pqfu;sA
(1) xzkgh Øfed foHko mRiUu ugha djrs gSA
(2) uksfllsIVj nkc esa ifjorZuksa ds izfr vuqfØ;k djrs gSaA
(3) ehtuj df.kdk;sa rkixzkgh gksrh gSA
(4) ekuo us= esa izdk'kxzkgh vU/ksjs esa fo/kqzfor gks tkrs gSa vkSj izdk'k ds míhiu dh vuqfØ;k esa vfr/kqzfor gks tkrs gSA
Ans.
4
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NEET - 2 Examination (2016) (Code - _)
(Page # 32)
116.
Graves' disease is caused due to
(1) hypersecretion of adrenal gland
(2) hyposecretion of thyroid gland
(3) hypersecretion of thyroid gland
(4) hyposecretion of adrenal gland
xzsVl jksx dk dkj.k gksrk gS
(1) ,sMªhuy xzafFk dk vfrL=o.k
(3) FkkbjkWbN xzafFk dk vfrL=o.k
(2)
FkkbjkWbM xzafFk dk vYiL=o.k
(4) ,sMªhuy xzafFk dk vYiL=o.k
Ans.
3
117.
Name the ion responsible for unmasking of active sites for myosin for cross-bridge activity during
muscle contraction.
(1) Potassium
(2) Calcium
(3) Magnesium
(4) Sodium
is'kh ladqpu ds nkSjku ØkWl&fczt fØ;k ds fy, ek;ksflu ds lfØ; LFkyksa dks mtkxj djus ds fy, mRrkjnk;h vk;u dk uke
crkb;sA
(1) iksVSf'k;e
(2) dSfYl;e
(3) eSXuhf'k;e
(4) lksfM;e
Ans.
2
118.
Name the blood cells, whose reduction in number can cause clotting disorder, leading to excessive loss of blood from the body.
(1) Thrombocytes
(2) Erythrocytes
(3) Leucocytes
(4) Neutrophils
mu #f/kj dksf'kdkvksa ds uke crkb;s] ftudh la[;k esa deh gksus ij #f/kj&FkDdu izfØ;k esa xM+cM+ gks ldrk gS vkSj ftlds
dkj.k 'kjhj ls dkQh #f/kj cg ldrk gSA
(1) fcack.kq ¼FkzksEckslkbV½ (2) jäk.kq ¼,fjFkzkslkbV½ (3) 'osrk.kq ¼Y;wdkslkbV½ (4) mHk;jath ¼U;wVªksfQy½
Ans.
1
119.
Name a peptide hormone which acts mainly on hepatocytes, adipocytes and enhances cellular
glucose uptake and utilization.
(1) Gastrin
(2) Insulin
(3) Glucagon
(4) Secretin
ml isIVkbM gkWeksZu dk uke crkb;s] tks iz/kkur% ;d`rk.kqvksa ¼gsikVkslkbVksa½ vkSj olk.kqvksa ¼,fMikslkbVksa½ ij izHkko Mkyrk gS rFkk
dksf'kdk }kjk Xywdksl ds vo'kks"k.k rFkk muds mi;ksx dks c<+kok nsrk gSA
(1) xSfLVªu
(2) balqfyu
(3) XywdSxkWu
(4) lsØsfVu
Ans.
2
120.
Osteoporosis, an age-related disease of skeletal system, may occur due to
(1) accumulation of uric acid leading to inflammation of joints
(2) immune disorder affecting neuromuscular junction leading to fatigue
(3) high concentration of Ca++ and Na+
(4) decreased level of estrogen
vfLFkeqf"kjrk] tks dadky dk ,d vk;q&lEcU/kh jksx gs] fdlds dkj.k gks ldrk gS\
(1) ;wfjd vEy dk ,d=hdj.k] ftlds dkj.k tksM+ lwt tkrs gS
(2) izfrj{kk&fodklj] tks raf=is'kh; taD'ku ij izHkko Mkyrk gS] ftlds dkj.k Fkdku gksrh gS
(3) Ca++ vkSj Na+ dh mPp lkanzrk
(4) ,LVªkstu ds Lrj esa deh
Ans.
4
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NEET - 2 Examination (2016) (Code - _)
121.
Serum differs from blood in
(1) lacking antibodies
(3) lacking albumins
lhje #f/kj ls fHkUu gksrk gS] D;ksafd
(1) mlesa izfrfi.M ugha gksrs
(3) mlesa ,sYC;wfeu ugha gksrs
(Page # 33)
(2) lacking globulins
(4) lacking clotting factors
mlesa XyksC;wfyu ugha gksrs
(4) mlesa Ldanu dkjd ugha gksrs
(2)
Ans.
4
122.
Lungs do not collapse between breaths and some air always remains in the lungs which can
never be expelled because
(1) pressure in the lungs is higher than the atmospheric pressure
(2) there is a negative pressure in the lungs
(3) there is a negative intrapleural pressure pulling at the lung walls
(4) there is a positive intrapleural pressure
lk¡l ysus ds chp QsQM+s fpid ugha tkrs vkSj FkksM+h&cgqr gok QsQM+ksa esa lnk gh cuh jgrh gS ftls ckgj fudkyk ugha tk ldrk]
D;kasfd
(1) QsQM+ksa ds Hkhrj dh nkc] ok;qeaMy dh nkc ls vf/kd gksrh gS
(2) QsQM+ksa ds chp _.kkRed nkc gkrh gS
(3) _.kkRed var%QqIQqlh nkc gksrh gs] tks QsQM+ksa dh fHkfÙk;ksa dks ,d&nwljs ls nwj [khaprh jgrh gS
(4) /kukRed var%QqIQqlh nkc gksrh gS
Ans.
3
123.
The posterior pituitary gland is not a 'true' endocrine gland because
(1) it secretes enzymes
(2) it is provided with a duct
(3) it only stores and releases hormones
(4) it is under the regulation of hypothalamus
i'p fiV;wVjh xzafFk ^okLrfod^ var%L=koh xzafFk ugha gksrh gS] D;ksafd
(1) ;g ,atkbeksa dk L=ko djrh gS
(2) bldh ,d okfguh gksrh gS
(3) ;g gkWeksZuksa dks dsoy HkaMkfjr djrh gS vkSj fu"dkflr djrh gS
(4) ;g gkbiksFkSysel ds fu;eu ds v/khu gksrh gS
Ans.
3
124.
The part of nephron involved in active reabsorption of sodium is
(1) descending limb of Henle's loop
(2) distal convolu ted tubule
(3) proximal convoluted tubule
(4) Bowman's capsule
usÝkWu dk og Hkkx] tks lksfM;e ds lfØ; iqu%vo'kks"k.k dk dk;Z djrk gS] gS
(1) gsUys ik'kdqaMyh dk vojksgh ikn
(2) nwjLFk laofyr ufydk
(3) fudVLFk laofyr ufydk
(4) ckseu laiqV
Ans.
3
125.
Which of the following is hormone-releasing IUD?
(1) Cu7
(2) LNG-20
(3) Multiload 375
(4) Lippes loop
fuEufyf[kr esa ls dkSulk gkWeksZu&fueksapd IUD gksrk gS?
(1) Cu7
Ans.
(2) LNG-20
(3)
eYVhMkst 375
(4)
fyIil ik'kdqaMyh
2
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(Page # 34)
126.
NEET - 2 Examination (2016) (Code - _)
Which of the following is incorrect regarding vasectomy?
(1) Irreversible sterility
(2) No sperm occurs in seminal fluid
(3) No sperm occurs in epididymis
(4) Vasa deferentia is cut and tied
'kqØokgd&mPNsnu ds ckjs esa fuEufyf[kr esa ls dkSulk xyr gS\
(1) vuqRØe.kh; ca/;rk
(2) oh;Z esa 'kqØk.kq ugha gksrs
(3) ,fifMfMfel esa 'kqØk.kq ugha gksrs
(4) 'kqØokgd dks dkVdj ck¡/k fn;k tkrk gS
Ans.
3
127.
Embryo with more than 16 blastomeres formed due to in uitro fertilization is transferred into
(1) cervix
(2) uterus
(3) fallopian tube
(4) fimbriae
ik=s fu"kspu }kjk fufeZu 16 ls vf/kd dksjd[kaMksa ¼CykLVksfe;jksa½ okys Hkwz.k dks LFkkukarfjr dj fn;k tkrk gSA
(1) xzhok esa
(2) xHkkZ'k; esa
(3) QSyksih uyh esa
(4) >kyj esa
Ans.
2
128.
Which of the following depicts the correct pathway of transport of sperms?
(1) Efferent ductules  Rete testis  Vas deferens  Epididymis
(2) Rete testis  Efferent ductules  Epididymis  Vas deferens
(3) Rete testis  Epididymis  Efferent ductules  Vas deferens
(4) Rete testis  Vas deferens  Efferent ductules  Epididymis
fuEufyf[kr esa ls dkSulk 'kqØk.kqvksa ds ifjogu ds iFk dks lgh :i ls crkrk gS\
(1) viokgh okfgfudk;sa  o`"k.k tkfydk  'kqØokgd  ,fifMfMfel
(2) o`"k.k tkfydk  viokgh okfgfudk;sa  ,fifMfMfel  'kqØokgd
(3) o`"k.k tkfydk  ,fifMfMfel  viokgh okfgfudk;sa  'kqØokgd
(4) o`"k.k tkfydk  'kqØokgd  viokgh okfgfudk;sa  ,fifMfMfel
Ans.
2
129.
Match Column-I with Column-II and select the correct option using the codes given below:
Column-I
Column-II
a.
Mons pubis
(i)
Embryo formation
b.
Antrum
(ii)
Sperm
c.
Trophectoderm
(iii)
Female external genitalia
d.
Nebenkern
(iv)
Graafian follicle
Codes :
a
b
c
d
(1)
(i)
(iv)
(iii)
(ii)
(2)
(iii)
(iv)
(ii)
(i)
(3)
(iii)
(iv)
(i)
(ii)
(4)
(iii)
(i)
(iv)
(ii)
dkWye-I vkSj dkWye-II ds chp feyku dhft;s rFkk uhps fn;s x;s dwV dk iz;ksx dj lgh fodYi dks pqfu;s %
dkW ye -I
dkWy e -II
a.
ekSal I;wfcl
(i)
Hkwz.k cuuk
b.
xàj
(ii)
'kqØk.kq
c.
VªksQsDVksMeZ
(iii)
eknk cká tuusafnz;
d.
uscsUduZ
(iv)
xzkQh iqVd
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NEET - 2 Examination (2016) (Code - _)
(Page # 35)
dwV :
Ans.
130.
(1)
(2)
(3)
(4)
3
a
(i)
(iii)
(iii)
(iii)
b
(iv)
(iv)
(iv)
(i)
c
(iii)
(ii)
(i)
(iv)
d
(ii)
(i)
(ii)
(ii)
Several hormones like hCG, hPL, estrogen, progesterone are produced by
(1) pituitary
(2) ovary
(3) placenta
(4) fallopian tube
dqN gkWeksZu] tSls hCG, hPL, ,LVªkstu, izkstsLVsjkWu dgk¡ mRiUu gksrs gS\
(1) fiV;wVjh xzafFk
(2) vaMk'k;
(3) vijk
(4)
QSyksih uyh
Ans.
3
131.
If a colour-blind man man ies a woman who is homozygous for normal colour vision , the probability of their son being colour-blind is
(1) 1
(2) 0
(3) 0.5
(4) 0.75
,d o.kkZ/k iq#"k ,d ,slh L=h ls fookg djrk gS tks lkekU; jax n`f"V ds fy, le;qXeth gSA muds iq= ds o.kkZa/k gksus dh laHkkouk
D;k gksxh\
Ans.
132.
(1) 1
2
(2) 0
Genetic drift operates in
(1) slow reproductive population
(3) large isolated population
vkuqoaf'kd fopyu ¼viokg½ dgk¡ gksrk gS\
(1) ean :i ls tuuh; lef"V
(3) cM+h foyfxr lef"V
(3) 0.5
(4) 0.75
(2) small isolated population
(4) non-reproductive population
NksVh foyfxr lef"V
(4) vtuuh; lef"V
(2)
Ans.
2
133.
In Hardy-Weinberg equation, the frequency of heterozygous individual is represented by
(1) q2
(2) p2
(3) 2pq
(4) pq
gkMhZ&okbucxZ lehdj.k esa fo"ke;qXeth O;f"V dh izkf;drk dk fu:i.k fdlls gksrk gS\
Ans.
134.
(1) q2
3
(2) p2
(3) 2pq
(4) pq
The chronological order of human evolution from early to the recent is
(1) Australopithecus  Homo habilis  Ramapithecus  Homo erectus
(2) Austraiopithecus  Ramapithecus  Homo habilis  Homo erectus
(3) Ramapithecus  Austraiopithecus  Homo habilis  Homo erectus
(4) Ramapithecus  Homo habilis  Australopithecus  Homo erectus
vkfnekuo ls vfHkuo ekuo rd ekuo fodkl dk dkykuqØfed Øe gS
(1) vkWLVªsyksfiFksdl  gkseks gSfcfyl  jkekfiFksdl  gkseks bjsDVl
(2) vkWLVªsyksfiFksdl  jkekfiFksdl  gkseks gSfcfyl  gkseks bjsDVl
(3) jkekfiFksdl  vkWLVªsyksfiFksdl  gkseks gSfcfyl  gkseks bjsDVl
(4) jkekfiFksdl  gkseks gSfcfyl  vkWLVªsyksfiFksdl  gkseks bjsDVl
Ans.
3
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NEET - 2 Examination (2016) (Code - _)
(Page # 36)
135.
Which
I.
II.
III.
IV.
(1) II,
of the following is the correct sequence of events in the origin of life?
Formation of protobionts
Synthesis of organic monomers
Synthesis of organic polymers
Fonnation of DNA-based genetic systems
III, IV, I
(2) I, II, III, IV
(3) I, III, II, IV
(4) II, III, I, IV
fuEufyf[kr esa ls dkSulk thou dh mRifÙk esa ?kVukvksa dk lgh vuqØe gs\
I.
vkfnthoksa dk fuekZ.k
II.
dkcZfud eksuksejksa dk la'ys"k.k
III.
dkcZfud ikWyhejksa dk la'ys"k.k
IV.
DNA ij vk/kkfjr vkuqoaf'kd ra=ksa dk fuekZ.k
Ans.
(1) II, III, IV, I
4
(2) I, II, III, IV
(3) I, III, II, IV
(4) II, III, I, IV
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NEET - 2 Examination (2016) (Code - _)
[PHYSICS]
136.
(Page # 37)
A person can see clearly objects only when they lie between 50 cm and 400 cm from his eyes. In
order to increase the maximum distance of distinct vision to infinity, the type and power of the
correcting lens, the person has to use , will be
(1) convex, + 0.15 diopter
(2) convex, + 2.25 diopter
(3) concave, – 0.25 diopter
(3) concave, – 0.2 diopter
,d O
; fDr vi uhvka
[ k l sdsoy 50 cm r Fkk 400 cm nw
j hdschp fLFkr oLr q
v ksadksl q
Li "V ns[k l dr k gS
al q
Li "V n' kZ
u dh
vf/kdr e nw
j hdksvuUr r d dj usdsfy , ml O
; fDr dksfdl i zd kj dsvkS
j fdr uh' kfDr dsl a
' kks/kd y S
a
l dhvko ; dr k
gksxh\
(1) mR
r y , + 0.15 Mk; ksIVj
(2) mR
r y , + 2.25 Mk; ksIVj
(3) vor y , – 0.25 Mk; ks
IVj
(3) vor y , – 0.2 Mk; ks
IVj
Sol.
137.
3
For object at
u=–
v = –400
1
1
1
–
=
v
u
f
1
–0=
 400
 P = –0.25
 if image formed at 400 cm, the reason can see
1
f
D concave
A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length
60 cm. The apperature is illuminated normally by a parallel beam of wavelength 5 × 10–5 cm.
The distance of the first dark band of the diffraction pattern from the centre of the screen is
0.02 cm pkS
M+
kbZds, d j s[kh; } kj d dks60 cm Qksd l nw
j hdsfdl hy S
a
l l sl fUudV l keusj [ kk x; k gS
A } kj d dks5 ×
10–5 cm. r j a
xnS
/; Zdsi zd k' k dhl ekUr j fdj .ki q
a
t } kj k y Ecor ~i zd kf' kr fd; k x; k gS
A i zkIr foor Z
u iS
VuZdsi zFke vnhIr
cS
. M dhi nsZdsdsUnzl snw
j hgksxh&
Sol.
(1) 0.15
(3) 0.25
1
d sin  = 
(2) 0.10
(4) 0.20
dy

D
D
d
On solving distance = 0.15
y
138.
Electron of mass m with de - Broglie wavelength  fall on the target in an X -ray tube. The cutoff
wavelength (0) of the emitted X - ray is
fdl hX-fdj .k ufydk dsy{; i j  nS
&czkW
Xyhr j a
xnS
/; Zr Fkk m nzO
; eku dsbysDVªkW
u Vdj kr sgS
A mRl ft Z
r X-fdj .k dsl a
Lr C
/k
(va
r d) r j a
xnS
/; Z(0) dk eku gksxk &
(1) 0   0
(2)  0 
2mc 2
h
2h
mc
(4) 0 
2m2 c2 2
h2
(3) 0 
Sol.
2
Cut off wavelength 0 of x rays
hc
 0 = KE
....(1)
For de-broglle wavelength
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NEET - 2 Examination (2016) (Code - _)
(Page # 38)
h
=
2m(KE)
 KE =
h2
2m2
h2
put value in equation (1)
hc
 0 = 2m2
2m2c
0 =
h
139.
Sol.
140.
Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. the maximum
energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no
photoelectrons will reach the anode A, if the stopping potential of A relative to C is
fdl h i zd k' kfo| q
r l sy dsdS
FkksM (_ .kkxz) C i j 5eV Åt kZdsQksVkW
u vki fr r gksrsga
S
A mRl ft Z
r i zd kf' kd by sDVªkW
uksadh
vf/kdr e Åt kZ2 eV gS
A 6 eV Åt kZdsQksVkW
uksadsC i j vki fr r gksusi j dksbZHkhi zd kf' kd by sDVªkW
u , suksM (/kukxz) A r d
ughai gq
a
psxk] ; fn C dsl ki s{k A dk fuj ks/kh foHko gks&
(1) – 3 v
(2) + 3 V
(3) + 4 v
(4) – 1 V
1
(I) Ei = w +KE
5 = w +2 ev
w = 3ev
(II) 6 = 3 + KE
KE = 3ev
=–3v
If an electron in a hydrogen atoms jumps from the 3 rd orbit to the 2nd orbit, it emits a photo of
wavelength . when it jumps from he 4th orbit to the 3rd orbit, the corresponding wavelength of
the photon will be
fdl hgkbMªkst u i j ek.kqesat c , d by sDVªkW
u r r̀ h; d{kkl sf} r h; d{kkesal a
Øe.kdj r kgS
] r ks r j a
xnS
/; ZdkQksVkW
u mRl ft Z
r
gksrk gS
A ; fn by sDVªkW
u pr q
FkZd{kk l sr r̀ h; d{kk esal a
Øe.k dj s] r ksQksVkW
u dk l a
xr r j a
xnS
/; Zgksxk &
(1)
Sol.
20

13
(2)
16

25
(3)
9

16
(4)
20

7
4
1
1
1
= R  2  2 

 n1 n2 
1 1
1
= R 4  9



 5 
1

=R 

 36 
 
1
1

=R  
9
16
'


 7 
1
= R  16  9 
'


'
5
16  9
=
×

36
7
=
20

7
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NEET - 2 Examination (2016) (Code - _)
141.
Sol.
(Page # 39)
The half - life of a radioactive substance is 30 minutes. The time ( in minutes) taken between
40 % decay and 85% decay of the same radioactive substance is
fdl hj sfM; ksl fØ; i nkFkZdhv/kZ
&vk; q30 feuV gS
A bl hj sfM; ksl fØ; i nkFkZds40% {k; l s85% r d {K; gksusesay xk
l e; (feuVksesa
) gks
xk &
(1) 60
(2) 15
(3) 30
(4) 45
1
t1/2 = 30 min.
N = N0 2–t/T
 0.6 N0 = N0 2 t1 / 30
0.15 N0 = N0 2 t2 / 30
=
2 t1 / 30
2 t2 / 30
t / 30
log 2 1
log 0.6
=
 t / 30
log 0.15
log 2 2
30
 t1
0.6

× t =
30
0.15
2
t1
2
 t =
5
2

142.
Sol.
For CE transistor amplifier. The audio signal voltage across the collector resistance of 2k is 4v..
if the current amplification factor of the transistor is 100 and the base resistance is 1 k, then
the input signal voltage is
fdl hCE Vªka
ft LVj i zo/kZ
d esal a
xzkgd i zfr j ks/k 2k gS
A bl dsfl j ksadschp JO
; la
d sr (vkW
fM; ksfl xzy ) oksYVRk 4v gS
A ; fn
Vªka
ft LVj dk /kkj k i zo/kZ
d xq
. kka
d 100 r Fkk vk/kkj i zfr j ks/k 1 kgS
] r ksfuos'k l a
d sr oksYVRk dk eku gksxk &
(1) 15 mV
(2) 10 mV
(3) 20 mV
(4) 30 mV
3
IC
IB = 100
I0 =
2  10 3
= 100
IB
= 2 × 10–3
4
2000
IB = 2 × 10–5
VB
–5
R B = 2 × 10
VB
= 2 × 10–5
1000
VB = 2 × 10–2
= 20 mA
143.
The gi v en c i rcui t has tw o i d eal d i ode s connec ted as shown i n the fi gure bel ow.
The current flowing through the resistance R1 will be
nksvkn' kZMk; ksMksdksi fj i Fk esauhpsfn; svkj s[k esan' kkZ
; sxk; svuq
l kj t ksM+
k x; kgS
A R1 i zfr j ks/kl si zokfgr /kkj k dkeku gksxk
2
R1
D1
D2
3
2
10V
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NEET - 2 Examination (2016) (Code - _)
(Page # 40)
Sol.
144.
(1) 3.13 A
(3) 10.0 A
2
D1  R.B.
D2  F.B.
10
I=
= 2.5 A
4
(2) 2.5 A
(4) 1.43 A
What is the output Y in the following circuit when all the three inputs A, B,C are first 0 and then 1 ?
uhpsfn; sx; si fj i Fk esafuxZ
e Y D; k gksxk] t cfd r huksafuos'k A, B,C i zkj a
Hk esa0 (' kw
U; ) r Fkk fQj 1 (, d ) gS\
A
P
B
Q
Y
C
Sol.
(1) 1 , 1
(3) 0, 0
4
(2) 0, 1
(4) 1, 0
A
B
P
AB
Q
C
y
C
y = (ABC)
= ABC
at first y = 1
then y = 0
145.
Planck’s constant (h), speed of light in vacuum (c) and Newton’s gravitational constant (g) are
three fundamental constant, which of the following combinations of these has the dimension of
length?
; fn Iy ka
d fLFkj ka
d (h), fuokZ
r esai zd k' k dk osx (c) r Fkk U; w
Vu dk xq
: Roh; fLFkj ka
d (G) r hu ekS
ft y d fLFkj ka
d gks] r ks
fuEufy f[ kr esal sfdl dh foek oghgksxht ksy EckbZdh gksrhgS\
(1)
(3)
Sol.
Ge
h3 / 2
hG
c5 / 2
(2)
hG
c3 / 2
(4)
hc
G
2
M0L0T0 = hxCyGz
x=z
M1L2 T 1M1L3 T 2
L3 / 2 T 3 / 2
=
L5 / 2 T 3 / 2
L3 / 2 T 3 / 2
=L
146.
Two cars P and Q start from a point at the same time a straight line and their positions are
represented by xp(t) = at + bt2 and xQ(t) = ft – t2. At what time do the cars have the same v e locity ?
nksdj ksaP vkS
j Q , d ghl e; i j fdl hfcUnql sl j y j s[kk esapy uk i zkj a
Hk dj r hgSvkS
j mudhfLFkfr ; ksadksØe' k%xp(t)
= at + bt2 r Fkk xQ(t) = ft – t2 l sfu: fi r fd; k t kr k gS
A fdl l e; i j bu nksuksadkj ksadk osx l eku gksxk \
f a
(1) 2 1  b


(2)
af
1b
a f
(3) 2 b  1


af
(4) 2 1  b


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NEET - 2 Examination (2016) (Code - _)
Sol.
(Page # 41)
1
VP = a + 2bt
Va = f – 2t
a + 2bt = f – 2t
fa
=t
2b  2
In the given figure, a = 15 m/s2 represents the total acceleration of a particle moving in the
clockwise direction in a circle of radius R = 2.5 m at a given instant of time. The speed of the
particle is
n' kkZ
; sx; svkj s[k esaR = 2.5 m f=kT; k dsoR̀r kdkj i Fk i j nf{k.kkor Zxfr dj r sgq
, fdl hd.k dsdq
y Roj .k dksfdl h{k.k
2
a = 15 m/s l sfu: fi r fd; k t kr k gS
A bl d.k dh pky gksxh&
30°
147.
R
O
Sol.
(1) 6.2 m/s
(3) 5.0 m/s
4
a
(2) 4.5 m/s
(4) 5.7 m/s
ar2  a2t  225
tan 30 
at
ar
at
ar
ar  at 3
ar2 
a
ar2
 225
3
4 2
ar  225
3
ar2  225 
ar 
3
4
15
3
2
v2
15 3

2.5
2
v2 
148.
15  2.5  3
2
= 5.7
A rigid ball of mass m strikes a rigid all at 60° and gets reflected without loss of speed as shown
in the figure below. The value of impulse imparted by the wall on the ball will be
m nz
O
; eku dh, d l [ r xsa
n (ckW
y ) fdl hn`<+nhokj l suhpsvkj s[kesan' kkZ
; sx; svuq
l kj 60 ° i j Vdj kdj i j kofr Z
r gkst kr h
gS
A ; fn bl i zfØ; k esaxsa
n dhpky esadksbZgkfu ughgksrh gS
] r ksnhokj } kj k xsa
n i j y xsvkosx dk eku gksxk &
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NEET - 2 Examination (2016) (Code - _)
(Page # 42)
m
V
60°
60°
V
(1)
mV
3
(2) mV
(3) 2 mV
Sol.
(4)
mV
2
2
mvcos60°
60°
mvsin60°
60°
60°
mvcos60°
mvsin60°
dp = 2 mv cos 60°
= mv
149.
A bullet of mass 10 g moving horizontally with a velocity of 400 ms–1 strikes a wooden block of
mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre
of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after
it emerges out horizontally from the block will be
400 ms–1 ds{kS
fr t osx l spt yr hgq
bZ10 g nzO
; eku dh, d xksy h] 2 kg nzO
; eku dsydMhds, d xq
Vdsl sVdj kr hgS
A; g xq
Vdk
, d 5m yEchgYdhvfor kU; Mksjhl syVdkgS
A ; fn xksy hdsVdj kusdsi fj .kkeLo: i xq
Vdsdkxq
: Ro dsUnz10 cm Å/okZ
/kj Åi j
mB t kr k gS
] r ksxq
Vdsl s{kS
fr t fn' kkesackgj fudyusi j xksy hdhpky gksxh&
Sol.
(1) 160 ms–1
(3) 80 ms–1
4
10  400
10
=2V+
× V’
1000
1000
4 = 2V + 0.01 V’
1
2 V2 = 2 gh
2
2  9.8  0.1
V2 =
100
V = 1.4
4 = 2 × 1.4 + 0.01 V’
1.2 = 0.01 V’
(2) 100 ms–1
(4) 120 ms–1
1.2
0.01
= 120 m/sec
V’ =
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NEET - 2 Examination (2016) (Code - _)
150.
Sol.
(Page # 43)
Two identical balls A and B having velocities of 0.5 m/s and – 0.3 m/s respectively collide
elastically in one dimensiion. The velocities of B and A after the collision respectively will be
(1) 0.3 m/s and 0.5 m/s
(2) – 0.5 m/s and 0.3 m/s
(3) 0.5 m/s and – 0.3 m/s
(4) – 0.3 m/s and 0.5 m/s
nksl oZ
l e xsa
nksA vkS
j B dsosx Øe' k%0.5 m/s r Fkk –0.3 m/s gS
A ; s, d j s[kk dsvuq
fn' k py r sgq
, Vdj kr hgS
A ; fn ; g
VDdj i zR; kLFk gS
] r ksbl VDdj dsi ' pkr ~B r Fkk A dsosx gksa
xs] Øe' k%
(1) 0.3 m/s r Fkk 0.5 m/s
(2) – 0.5 m/s r Fkk 0.3 m/s
(3) 0.5 m/s r Fkk – 0.3 m/s
(4) – 0.3 m/s r Fkk 0.5 m/s
3
As the collision is elastic in nature, the velocities are interchanged.
0.5 m/s
0.3 m/s
A  0.3 m/s

B  0.5 m/s
151.






ˆ when a force of 4iˆ  3j
ˆ to 4j
ˆ  3k
ˆ N is applied. How
A particle moves from a point 2iˆ  5j
much work has been done by the force ?






fdl hd.k i j 4iˆ  3jˆ N cy y xkusi j og fcUnq 2iˆ  5jˆ l sfcUnq 4jˆ  3kˆ r d foLFkkfi r gkst kr k gS
A bl i zfØ; k
esacy } kj k fd; k x; k dk; Zfdr uk gksxk \
Sol.
(1) 2 J
(3) 11 J
4

r1 =  2 î  5 ĵ

r2 = 4 ĵ  3k̂

F = 4 î  3 ĵ



 r = r2  r1
(2) 8 J
(4) 5 J
= 4 ĵ  3k̂ – (2 î  5 ĵ)
= 2 î  ĵ  3k̂

w = F. r
= (4 î  3 ĵ) . (2 î  ĵ  3k̂)
=8–3
=5J
152.
Two rotating bodies A and B of masses m and 2 m with moments of inertial IA and IB(IB > IA) have
equal kinetic energy of rotation. If LA and LB be be their angular momenta respectively, then
?kw
. kZ
u dj r sgq
, nksfi a
MksA r FkkB dsnzO
; eku Øe' k%m r Fkk 2m vkS
j t M+
Ro vk?kw
. kZØe' k%IA r FkkIB(IB > IA) gS
A bu nksuksa
dh?kw
. kZ
u xfr t Åt kZ
; svki l esacj kcj gS
A ; fn budsdks.kh; l a
osx Øe' k%LA r Fkk LB gks] r ks&
(1) LA > LB
Sol.
4
A
m
IB > IA
K A = KB
(2) L A 
LB
2
(3) LA = 2LB
(4) LB > LA
B
2m
1
1
IA wA2c =
Iw 2
2
2 B B
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NEET - 2 Examination (2016) (Code - _)
(Page # 44)
2
2
 LA 
L 
 = I  B 
IA 
B
 IA 
 IB 
LA
=
LB
L
153.
IA
IB
I
A solid sphere of mass m and radius R is rotating about its diameter. A solid cylinder of the same
mass and same redius is also rotating about its geometrical axis with an angullar speed twice
that of the sphere . the ratio of their kinetic energies of rotation Esphere / Ecylinder  will be
nzO
; eku m r Fkkf=kT; kR dk, d Bksl xksy kvi usO
; kl dsi fj r %?kw
. kZ
u dj j gkgS
A ml hnzO
; eku r Fkkml hnzO
; eku r Fkkml h
f=kT; k dk , d Bksl csy u ¼
fl fy a
Mj ½Hkhvi usT; kfer h; v{k dsi fj r %?kw
. kZ
u dj j gk gS
A csy u ds?kw
. kZ
u dhdks.kh; pky xksy s
l snksxq
uk gS
A bu nksuksdh?kw
. kZ
u xfr t Åt kZ
v ksadk vuq
i kr Esphere / Ecylinder  gksxkA
Sol.
(1) 3 : 1
(3) 1 : 5
3
(2) 2 : 3
(4) 1 : 4
mr
2
mr 2
5
mr 2
2
(wC = 2ws)
12
mr2w2s
KS
2
5

KC
1 mr2
.4w2s
2 2
KS
1
KC = 5
154.
A light rod of length l ha s two masses m1 and m2 attached to its two ends. The moment of inertia
of the system about an axis perpendicular of the rod and passing through the centre of mass is
, d gYdhNM+dhy EckbZl gS
A bl dsnksfl j ksl sØe' k%m1 r Fkk m2 nzO
; eku dsfi a
Ml a
y xzgS
A bl NM+dsy Ecor ~r Fkk
bl dsl a
gfr dsUnzl sxq
t j r sgq
, v{k dsi fj r %bl fudk; dk t M+
Ro vk?kw
. kZgksxkA
(1)
Sol.
m1m2 l2
m1m2 2
(2) m  m l
1
2
(3)
m1  m2 2
l
m1m2
(4) m1  m2  l2
2
m2
m1
r1
r2
I = m1r12 + m2r22
m2r
m1r
r1 = m  m
r2 = m  m
1
2
1
2
m1m2
I = m  m r2
1
2
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NEET - 2 Examination (2016) (Code - _)
155.
(Page # 45)
Starting from the centre of the each having radius R, the variation of g( acceleration due to
gravity) is shown by
; fn i zFohdhf=kT; k R gS
] r ksi F̀ohdsdsUnzl si zkj a
Hk dj xq
: Roh; Roj .k g dsi fj or Z
u dksfuEuka
fdr esal sdkS
u & l k vkj s[k
¼
xzkQ½l ghn' kkZ
r k gS\
g
g
(1)
(2)
O
O
r
R
g
r
R
r
g
(3)
(4)
O
Sol.
R
r
R
O
3
gr
R
Gm
gs =
R2
gin =
go =
156.
gM
r2
r
mg0R2
(2) 2 R  h


mg0R2

(3) 2 R  h


(4)
2mg0R 2
R h
3
TE = KE + PE
=
1
GMm
mv
v02 –
2
R h
=
1
GM
GMm
m
–
2
R h
R h
=–
157.
R
A satellite of mass m is orbiting the earth ( or radius R ) at a height h from its surface. The total
energy of the satellite in term of g0. the value of acceleration due to gravity at the earth’s surface,
is
, d mi xzg] ft l dk nzO
; eku m gS
] i F̀ohdsi `"B l sh Å¡pkbZi j i F̀ohdhi fj Øek dj j gk gS
A ; fn i F̀ohdhf=kT; k R gSr Fkk
bl dsi `"B i j xq
: Roh; Roj .k dk eku g0 gS] r ksmi xzg dhdq
y Åt kZgksxhA
2mg0R 2
(1) 
R h
Sol.
g
GMm
gR 2m
=–
2(R  h)
2(R  h)
A rectangular film of liquid is extended from ( 4cm × 2 cm) to ( 5cm × 4 cm). if the work done
is 3×10–4 J, the value of the surface tension of the liquid is
fdl hnzo dhvk; r kdkj >YYh¼
fQYe½dk foLr kj ( 4cm × 2 cm) l sc<+
kdj ( 5cm × 4 cm). dj fn; k t kr k gS
A ; fn
bl i zfØ; k esafd; k x; k dk; Z3×10–4 J gks] r ksnzo dsi `"B r uko dk eku gksxk
(1) 8.0 N m–1
(2) 0.250 N m–1
(3) 0.125 N m–1
(4) 0.2 N m–1
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NEET - 2 Examination (2016) (Code - _)
(Page # 46)
Sol.
3
S = 2TA
3 × 10–4 = 2 × T (12 × 10–4)
T=
158.
3
1
=
= 0.125
24
8
Three liquids of densities 1, 2 and 3 (with 1 > 2 >3), having the same value of surface tension
T, rise to the same height in three identical capillaries. The angles of contance 1, 2 and 3 obey
r hu nzoksads?kuRo Øe' k%1, 2 r Fkk 3 (1 > 2 >3) gS
A r huksanzoksadk i `"B r uko Tl eku gS
A r hu l oZ
l e dsf' kdkvksaesa
r huksanzo l eku Å¡pkbZr d p<+
r sgS
A; fn bu nzoksadsfy , Li ' kZ& dks.kØe' k%1, 2 r Fkk3 gS
] r ksfuEufy f[ kr esal sdkS
u&l k
l EcU/k Bhd gksxk \
(1)   1  2  3 

2
(2)

 1  2  3  0
2
(3) 0  1  2  3 

2
(4)

 1  2  3  
2
Sol.
3
2T cos  = hrdg
T  same
r  same
h  same
cos   
159.
Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100°C, while the other one is at 0°C. if the two bodies are brought into
contact, then, assuming no heat loss, the final common temperature is
(1) 0° C
(2) 50° C
(3) more than 50° C
(4) less than 50° C but greater than 0°C
nksl oZ
l e fi a
M , d , sl si nkFkZdscusgSft udhÅ"ek /kkfj r k r ki dsl kFk c<+t kr hgS
A buea
sl s, d fi a
M dk r ku 100°C,
r Fkknw
l j sdk0°C. gS
A; fn bu nksuksadksl Ei dZesaj [ kkt k, vkS
j bl i zfØ; kesaÅ"ekdk{k; u gks]r ksnksuksafi a
MksdkmH; fu"B
r ki gksxk
Sol.
(1) 0° C
(3) 50° C l svf/kd
3
dQ = msdT
dQ  same
dT 
(2) 50° C
(4) 50° C
l sde i j Ur q0°C l svf/kd
1
ms
1
s
s for 100°C is more so change in temp.  , so final will more than 50°
dT 
160.
A body cools from a temperature 3t to 2T in 10 minutes. The room temperature is T. Assume that
Newton’s law of cooling is applicable. the temperature of the body at the end of next 10 minutes
will be
fdl hoLr qdk r ku 3T l s 2T r d fxj usea
s10 feUkV dk l e; y xr k gS
A dej sdk r ku T gS
A ; fn bl esaU; w
Vu ds' khr y u
fu; e dk vuq
i ky u gksrk gS
] r ksvxy s10 feuV dsvUr esaoLr qdk r ki gksxk
(1) T
Sol.
(2)
7
T
4
(3)
3
T
2
(4)
4
T
3
3
dT
= K (Tav.. – T)
dt
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NEET - 2 Examination (2016) (Code - _)
(Page # 47)
T
= K (2.5 T – T)
10
T
= K 1.5 T
10
K=
1
15
2 T  T'
1  2T  T'  T 
=
2
10
15 

6T – 3T’ = T’
6T
T’ =
4
=
161.
3
T
2
One mole of an ideal monatomic gas undergoes a process described by the equation PV3 =
constant. The heat capacity of the gas during this process is
fdl hi zØe esa, d i j ek.kq
d vkn' kZxS
l ds, d eksy dsi fj or Z
u dksl ehdj .k PV3 = fLFkj ka
d } kj k O
; Dr fd; k t kr k gS
A bl
i zØe dhvof/k esaxS
l dhÅ"ek /kkfj r k gksxh
(1) R
Sol.
3
R
2
(3)
5
R
2
(4) 2R
1
PV3 = cos 4T
C  Cv 

162.
(2)
R
1x
3
R
R
2
13
CV 

3
R
2
3
R
R R
2
2
The temperature inside a refrigerator is t2°C and the room temperature is t1°C. the amount of
heat delivered to the room for each joule of electrical energy consumed ideally will be
fdl h ' khr y d¼
j sfQzt j sVj ½dsHkhr j dk r ku t2°C gSvkS
j dej sdk r ku t1°C gS
A vkn' kZvoLFkk esai zfr t w
y fo| q
r Åt k
ZdsO
; ; gksusi j dej sdksLFkkukUr fj r Å"ek dk eku gksxkA
t1  t2
(1) t  273
1
Sol.
t1
(2) t  t
1
2
(3)
t1  273
t1  t2
(4)
t2  273
t1  t2
3
Room
Electro
energy w
t1
Q2
t2
Regulator
w  Q2  Q1
Q1 = Heat delivered to room
w = Energy electro
Q1
Q1
T1


w
Q1  Q2
T1  T2

t1  273
t1  t2
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NEET - 2 Examination (2016) (Code - _)
(Page # 48)
163.
A given sample of an ideal gas occupies a volume V at a pressure P and absolute temperature T.
the mass of each molecule of the gas is m. Which of the following gives the density of the gas ?
fdl hvkn' kZxS
l dsfun' kZdk nkc P r Fkk i j e r ki T gksusi j vk; r u V gS
A bl xS
l dsi zR; sd v.kqdk nzO
; eku m gS
A xS
l
dk ?kuRo gksxkA
Sol.
(1) mKT
(3) Pm / ( kT)
3
RT
P
M = molecular mass
M
M = NA m
RT
R

k
NAm
NA
P
164.
Sol.
kT
m

(2) P/ ( kT)
(4) P / ( kTV)
Pm
kT
A body of mass m is attached to the lower end of a spring whose upper end is fixed. the spring
has negligible mass. When the ma ss m is slightly pulled down and released, it oscillates with a
time period of 3s. when the mass m is increased by 1kg. the time period of oscillations becomes
5 s. The value of m in kg is
fdl hdekuhdk Åi j hfl j k fLFkj gSr Fkk fupy sfl j sl sm nzO
; eku dk , d fi a
M y Vdk gS
A dekuhdk vi uk nzO
; eku ux.;
gS
A dekuhdsfupy sfl j sdksFkksM+
k&l k [ khpdj NksM+nsusi j nzO
; eku m dk fi a
Mnksy u dj usy xr kgSvkS
j bl dsnksy uksadk
vkor Z
d ky 3s gS
A m dseku esa1kg c<+
kusi j nksy uksadk vkor Z
d ky 5s. gkst kr k gSm dk kg esaeku gS
(1)
9
16
(2)
3
4
(3)
4
3
(4)
16
9
1
m
=3
R
m1
2
=5
R
T = 2
m 1
5
=
m
3
1
25
=
n
9
1
16
=
m
9
1+
m=
165.
m
9
16
The second overtone of an open organ pipe has the same frequency as the first overtone of a
closed pipe L metre long. the length of the open pipe will be
fdl h[ kq
y svkxZ
u i kbi dsf} r h; vf/kLoj d dhvkof̀Ùk L ehVj y EcscUn i kbi dsi zFke vf/kLoj d dhvkof̀Ùk dscj kcj gS
A
[ kq
y si kbi dh y EckbZgksxhA
(1) 4L
Sol.
(2) L
(3) 2L
(4)
L
2
3
3V
3V
2l0 = 4lC
l0 = 2lC = 2l
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NEET - 2 Examination (2016) (Code - _)
166.
(Page # 49)
Three sound waves of equal amplitudes have frequencies (n – 1), n, ( n + 1). They superimpose
to give beats. The number of beats produced per second will be
l eku vk; ke dhr hu /ofu r j a
xksdhvkof̀Ùk; k¡ Øe' k%(n – 1), n, ( n + 1). gS
A budsv/; kj ksi.k l sfoLi Un mRiUu gksrs
gS
A i zfr l sd .M mRiUu foLi Unksadhl a
[ ; k gksxhA
Sol.
167.
Sol.
(1) 2
(2) 1
1
max. different n = 2
(4) 3
An electric dipole is placed at an angle of 30° with an electric field intensity 2× 105 N/C. it
experiences a torque equal to 4 Nm. the charge on the dipole, if the dipole length is 2 cm, is
, d fo| q
r f} /kzq
o dks2×105 N/C r hozrk dsfo| q
r {ks=k l s30° dks.k i j j [ kusl sml i j 4 Nm dk cy &vk?kw
. kZy xr k gS
A
; fn f} /kw
zo dh y EckbZ2 cm gks] r ksml i j vkos'k gksxkA
(1) 4C
(2) 5 mC
(3) 2 mC
(4) 5 mC
3
  PE sin 

4
P

 4  105
1
E sin 
5
2  10 
2
P = qd
q
168.
(3) 4
P 4  105

 2  103
d 2  102
= 2 mC
A parallel - plate capacitor of area A, plate separation d and capacitance C is file with four
dielectric materials having dielectric constants k1, k2, k3 and k4 as shown in the figure below. if a
single dielectric material is to be used to have the same capacitance C in the is capacitor. then its
dielectric constant k is given by
, d l ekUr j &i fêdkl a
/kkfj =kdk{ks=kQy A r Fkkbl dh/kkfj r kC gS
Abl dhnksIy sVksadschp dki F̀kdu ¼
nw
j h½d gS
Abl esaØe' k%
k1, k2, k3 r Fkk k4 i j koS
|q
r ka
d dspkj i j koS
|q
r i nkFkZ
] uhpsfn; svkj s[kesan' kkZ
; sx; svuq
l kj ] Hkj sx; sgS
a
A; fn bu pkj ksai j koS
|q
r
i nkFkksZ
adsLFkku i j bl l a
/kkfj =k esak i j koS
|q
r ka
d dk dsoy , d i nkFkZHkj kt k; ] r kfd bl dh/kkfj r kC ghgksr ks]r ksk dkeku
gksxk &
1
1
1
1
3
(1) k  k  k  k  2k
1
2
3
4
A/3
A/3
A/3
k1
k2
k3
d/2
(2) k = k1 + k2 + k3 + 3k4
(3) k 
d
2
k1  k 2  k 3   2k 4
3
k4
(4)
Sol.
2
3
1


k k1  k 2  k 3 k 4
A
4
1
1
1
c eq. - C1  C2  C3 + C 4
1
=
C eq.
1
A
A
A
K20
K 3 0
3 
3 
3
d
d
d
2
2
2
K 1 0
+
1
K 4 0A
d
2
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NEET - 2 Examination (2016) (Code - _)
(Page # 50)
d
2
d
d
2
K eq. 0 A = 0 AK1  K 2  K 3  + K  A
4 0
3
1
3
1
k eq. = 2K1  K 2  K 3  + 2K 4
2
3
1
K eq. = K1  K 2  K 3 + K 4
169.
The potential difference (VA – VB) between the points A and B in the given figure is
n' kkZ
; sx; svkj s[k esafcUnq
v ksaA r Fkk B dschp foHkokUr j (VA – VB) gksxkA
VA
Sol.
170.
Sol.
+
3V
(1) +9 V
(3) + 3 V
1
4+3+2
VA – VB = 4 + 3 + 2 = 9
–
VB
(2) – 3 V
(4) + 6 V
A filament bulb ( 500W, 100V) is to be used in a 230 V main supply. When a resistance R is
connected in series, it works perfectly and the bulb consumes 500 W. the value of R is
, d fQy kesa
V¼
r Ur q
½cYc (500W, 100V) dks230 V dhesu l Iy kbZesai z;q
Dr fd; k t kuk gS
A bl dsJs.khØe esaR i zfr j ks/
k t ksM+
usi j ; g cYc i w
. kZ
r %Bhd dk; Zdj r k gSr Fkk 500 W ' kfDr y srk gS
A R dk eku gS%
(1) 13 
(2) 230 
(3) 46 
(4) 26 
4
Rb
3
230V
P = 500 w
V = 100 v
Rb 
V2
P
R
V2 100  100

 20
P
500
i
P 500

5
V 100
i
230
5
Rb  R
230 = 5Rb + 5Rb
5R = 230 – 5Rb
= 230 – 100
R 
171.
130
 26
5
A long wire carrying a steady current is bent into a circular loop of one turn. the magnetic field at
the centre of the loop is B. it is then bent into a circular coil of n turns. the magnetic field at the
centre of this coil of n turns will be
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NEET - 2 Examination (2016) (Code - _)
(Page # 51)
fdl hy Ecsr kj l svi fj or hZfo| q
r &/kkj ki zokfgr gksj ghgS
A bl r kj dks, d QsjsdsoÙ̀kkdkj i k' k¼
yw
i ½esaeksM+
usi j bl dsdsUnz
i j pq
Ecdh; {ks=k dk eku B gS
A vxj bl hr kj dksn QsjksadhoÙ̀kkdkj dq
a
My hesaeksM+fn; k t kr k gS
] r ksbl n Qsa
j ksadhdq
a
My h
dsdsUnzi j pq
Ecdh; {ks=k gksxk %
Sol.
(1) 2n2 B
(3) n2 B
3
(2) nB
(4) 2nB
0i
2r = l
2r
Now same wire in n turns
B
n 2r2     2r
0ni 0ni
 i

 n2  0 
r 
2r2
 2r 
2
n
= n2B
B1 
172.
A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in
equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep
the magnet in this new position is
fdl h, dl eku {kS
fr t pq
Ecdh; {ks=kea
s, d i r y sl w
r h/kkxsl sy Vdk; kx; k, d na
Mpq
Ecd l kE; koLFkkesagS
Abl s60° l s?kq
ekus
d fy , vko' ; d Åt kZW gS
A vc bl pq
Ecd dksbl hu; hfLFkfr esacuk; sj [ kusdsfy , vko' ; d cy &vk?kw
. kZdk eku gksxk&
(1)
(3)
Sol.
2W
3
Sol.
3
3W
2
(4)
3W
3
1 = 0°
2 = 60°
w = PE [cos  – cos 60°]
1

PE
= PE 1   =
2


2
T = PE sin 60° = 2w 
173.
W
(2)
3
=
2
3 w
An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57
×10–2 T. if the value of e/m is 1.76 × 1011C/kg, the frequency of revolution of the electron is
3.57 ×10–2 T r hoz
r kdsvuq
i zLFkpq
Ecdh; {ks=kdsi zHkko esa, d by sDVªkW
u oÙ̀kh; d{kkesa?kw
. kZ
u dj j gkgS
A ; fn e/m dkeku
1.76 × 1011C/kg gks
] r ksby sDVªkW
u dsi fj Øe.k dhvkof̀Ùk gksxh&
(1) 6.28 MHz
(2) 1 GHz
(3) 100 MHz
(4) 62.8 MHz
2
2r 2mv
T

x
x
x
x
x
v
qBv
x
x
x
x
x
x
x
x
x
x
qB
x
x
x
x
x

f 
x
x
x
x
x
2m
qB
 1.76  1011
m
×3.57×10–2
6.28

 109
6.28
= 1 GHz
  2f 
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
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NEET - 2 Examination (2016) (Code - _)
(Page # 52)
174.
Sol.
175.
Which of the following combinations should be selected for better tunning of an L – C – R circuit
used for comminution ?
la
pkj gsrqfdl hL – C – R i fj i Fk dscsgr j l eLoj .k ¼
V~
;w
fua
x½dsfy , fuEufy f[ kr esal sdkS
u&l k l a
; kst u mi ; q
Dr gksxk\
(1) R = 25 , L = 1.5 H, C = 45 F
(2) R = 20 , L = 1.5 H, C = 35 F
(3) R = 25 , L = 2.5 H, C = 45 F
(4) R = 15 , L = 3.5 H, C = 30 F
4
Quality factor is greatest when resistance is minimum.
A uniform magnetic field is restricted within a region of radius r. the magnetic field changes with

dB
time at a rate
. Loop 1 of radius R > r encloses the region r and loop 2 of radius R is outside
dt
the region of magnetic field as shown in the figure below. then the e.m.f generated is
r
R
R
1
2

dB 2
r in loop 1 and zero in loop 2 (2) zero in loop 1 and zero in loop 2
(1) 
dt



dB 2
dB 2
dB 2
r in loop 1 and 
r in loop 2(4) 
r in loop 1 and zero in loop 2
(3) 
dt
dt
dt

dB
dksbZ, dl eku pq
Ecdh; {ks=kr dsfy , {ks=kesal hfer gS
A ; g pq
Ecdh; {ks=kl e; dsl kFk
dhnj l si fj ofr Z
r gksrkgS
A uhps
dt
fn; svkj s[kesan' kkZ
; sx; svuq
l kj f=kT; kR > r dki k' k ¼
yw
i ½1, r {ks=kdksi fj c) dj r k gSr FkkR f=kT; kdk i k' k2, pq
Ecdh;
{ks=k dhl hek dsckgj gS
A mRiUu fo| q
r okgd cy dk eku gksxkA
r
R
R
1

dB 2
r
dt

dB 2

r
(3) i k' k 1 es
a
dt
1
(1) i k' k 1 es
a
Sol.
r Fkk i k' k 2 esa' kw
U;
2
(2)

dB 2
r (4)
r Fkk i k' k 2 esa
dt
Induced emf  e = –
i k' k1 esa' kw
U; r Fkk i k' k 2 esa' kw
U;
i k' k 1 esa

dB 2
r
dt
r Fkk i k' k 2 esa' kw
U;
AdB
dt
for loop 2 emf is zero
dB
e = – r
dt
176.
The potential differences across the resistance, capacitance and inductance are 80V, 40 V and
100V respectively in an L - C - r circuit . The power factore of this circuit is
fdl hL–C–R i fj i Fk esai zfr j ks/kd] /kkfj r k r Fkk i zsjdRo dsfl j ksadschp foHkokur j Øe' k%80V, 40 V r Fkk 100V gS
A bl
i fj i Fk dk ' kfDr xq
. kka
d gksxk %
Sol.
(1) 1.0
(3) 0.5
4
VR = 80V
VC = 40 V
(2) 0.4
(4) 0.8
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NEET - 2 Examination (2016) (Code - _)
(Page # 53)
VL = 100 V
VR2  (VL  VC )2
Vnet =
VL-VC
(80)2  (60)2
=
= 100 volt
cos  =
=
177.
Sol.
VR
R
iR
=
=
Vnet
Z
iZ
VR
80
= 0.8
100
A 100  resistance and a capacitor of 100 reactance are connected in series across a 220 V
source. when the capacitor is 50% charged, the peak value of the displacement current is
100  dk , d i z
fr j ks/k r Fkk 100  i zfr ?kkr dk , d l a
/kkfj =k] fdl h220 V dsL=kksr l sJs.khØe esat q
M+
sgS
Al a
/kkfj =k ds
50% vkos
f' kr gksusi j foLFkki u /kkj k dk f' k[ kj eku gksxk %
(1) 11 2 A
(3) 11 A
2
R = 100
XC = 100
Z=
(2) 2.2 A
(4) 4.4 A
R 2  X2C
= 100 2
Irms =
220
100 2
=
22
10 2
22
2 × 10 2
= 2.2 amp.
Ipeak =
178.
Sol.
Two identical glass (g = 3/2) equiconvex lenses of focal length f each are kept in contact. The
space between the two lenses is filled with water (w = 4/3). The focal length of the combination
is
dk¡p (g = 3/2) dsnksl oZ
l e l eksÙky y sa
l ksaesai zR; sd dhQksd l nw
j hf gS
A budksl Ei dZesaj [ kdj budschp dsfj Dr LFkku
dkst y (w = 4/3) l sHkj fn; k t kr k gS
A bl i zd kj cusl a
; kst u dh Qksd l nw
j hgksxh%
(1) 3f/ 4
(2) f/3
(3) f
(4) 4f /3
1
Let radius of curvature of equiconvex lens be R
1
3
2
1
1
1
1
= f + f + f
f
1
2
3
1
3
 1 1
1
= f =   1   
2
f

 R R 
1
For water part
Given;
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NEET - 2 Examination (2016) (Code - _)
(Page # 54)
1
1
4
  1
 
  1 
f2 =  3

R
R
 

1
1   2 
f2 = 3  R 
1
1
1
f3 = f = R
1
1
=
f
R
f=
179.
Sol.
2
1
+
3R
R
3R
3f
=
4
4
An air bubble in a glass slab with refractive index 1.5 ( near normal incidence) is 5 cm deep when
viewed from one surface and 3 cm deep when viewed from the opposite face. the thickness (in
cm) of the slab is
dka
pkdhfdl hi fêdk] ft l dkvi or Z
uka
d 1.5 gS
] dsHkhr j ok; qdk, d cq
y cq
y kcUn gS
A i fêdkds, d i `"B l sy xHkx y Ecor ~
ns[kusi j bl cq
y cq
y sdhxgj kbZ5 cm i zrhr gksrhgS
A bl i fêdk dheksVkbZ(cm esa
) gks
xh %
(1) 16
(2) 8
(3) 10
(4) 12
4
d1’ = 5 cm
d2’ = 3cm
d1
n1
d2 = n2
1.5
1.5
×5
d2 =
×3
1
1
d1 = 7.5 cm
d2 = 4.5
dnet = d1 + d2 = 7.5 + 4.5 = 12 cm
d1 =
180.
The interference pattern is obtained with two coherent light sources of intensity ratio n. In the
interference pattern, the ratio
Imax  Imin
will be
Imax  Imin
i zd k' k dsnksdy kl Ec) L=kksrksadk r hozrk vuq
i kr n gS
A budsv/; kj ksi.k l si zkIr O
; fr dj .k i S
VuZesavuq
i kr
Imax  Imin
Imax  Imin
dk eku gksxk %
2 n
(1)
(3)
Sol.
(2)
2
n  1
n
n 1
n
2 n
n 1
(4)
2
n  1
3
 I1  I2 
Imax .


=
 I  I 
Imin .
2 
 1
2
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NEET - 2 Examination (2016) (Code - _)


 1  I2 

I1 


= 
I
1 2 

I1 

(Page # 55)
2
2
1  n 

= 

1  n 
Imax  Imin
(1  n )2  (1  n)2
=
Imax  Imin
(1  n)2  (1  n)2
=
4 n
2 n
=
2(1  n)
1n
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