NEET - 2016
01-05-2016 Time: 3 Hrs.
l e;: 3 ?kaVs
CODE-C / R / Y
Max. Marks : 720
v f/kd re v ad
: 720
INSTRUCTIONS (funs 'k) Z Important Instructions:
egRoiw.kZ funsZ'k %
1.
1.
The Answer Sheet is inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars on Side-1 and Side-2 carefully with blue/black ball point pen only.
2.
The test is of 3 hours duration and Test Booklet contains 180 questions. Each question carries 4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response, one mark will be deducted from the total scores. The maximum marks are 720.
3.
Use Blue/Black Ball Point Pen only for writing particulars on this page/marking responses.
4.
Rough work is to be done on the space provided for this purpose in the Test Booklet only.
5.
On completion of the test, the candidate must havdover the Answer Sheet to the invigilator in the Room/Hall. The candidates are allowed to take away this Test Booklet with them.
6.
The CODE for this Booklet is C. Make sure that the CODE printed on Side-2 of the Answer Sheet is the same as that on this Booklet. In case of discrepancy, the candidate should immediately report the matter to the Invigilator for replacement of both the Test Booklets and the Answer Sheets.
7.
The Candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on the Answer Sheet. Do not write your roll no. anywhere else except in the specified space in the Test Booklet/Answer Sheet.
8.
Use of white fluid for correction is NOT permissible on the Answer Sheet.
mÙkj i=k bl ijh{kk iqfLrd k d svUnj j[kk gSA t c vkid ks ijh{kk iqfLrd k [kksy usd ksd gk t k,] rksmÙkj i=k fud ky d j i`"B-1 ,oai`"B-2 ij d soy uhy s@ d ky sckWy ikWbaV isu lsfooj.k HkjsaA 2. ijh{kk d h vof/k 3 ?ka VsgS,oaijh{kk iq fLrd k es a180 iz 'u gS A iz a R;s d iz 'u 4 va d d k gS A iz R;s d lgh mÙkj d sfy, ijh{kkFkhZd ks4 va d fn, t k,a xs A iz R;s d xyr mÙkj d sfy, dq y ;ks x es als,d va d ?kVk;k t k,xkA vf/kd re va d 720 gSaA 3. bl i` "B ij fooj.kva fd r d jus,a o mÙkj i=k ij fu'kku yxkus d sfy, d osy uhys @ d kysckyW ikbWVa is u d k iz ;kxs d js Aa jQ d k;Zbl ijh{kk iqfLrd k esafu/kkZfjr LFkku ij gh d jsaA 5. ijh{kk lEiUu gks usij] ijh{kkFkhZd {k@gkWy NksM uslsiwoZ mÙkj i=k d {k fujh{kd d ksv o'; lkSai nsaA ijh{kkFkhZv ius lkFk iz'u iqfLrd k d ksy st k ld rsgSaA 4.
bl iqfLrd k d k lad sr gSC A ;g lqfuf'pr d j ysafd bl iqfLrd k d k lad sr] mÙkj i=k d si`"B-2 in Nislad sr ls feyrk gSA vxj ;g fHkUu gks rks ijh{kkFkhZ nwl jh ijh{kk iqfLrd k vkSsj mÙkj i=k ysus d s fy, fujh{kd d ks rqjUr voxr d jk,aA 7. ijh{kkFkhZlq fuf'pr d jsafd bl mÙkj i=k d kseksM+k u t k, ,oaml ij d ksbZvU; fu'kku u yxk,aA ijh{kkFkhZviuk vuqØ ekad iz'u iqfLrd k@mÙkj i=k es fu/kkZfjr LFkku d s vfrfjDr vU;=k uk fy[ksaA 6.
8.
mÙkj i=k ij fd lh izd kj d sla'kks/ku gsrqOgkbV ¶+y wbM d s iz;ksx d h vuqefr ughagSA
In case of any ambiguity in translation of any question, English version shall be treated as final.
iz'uksad sv uqokn esafd l h v Li"Vrk d h fLFkfr esa] v axzst h l aLd j.k d ksgh v fUre ekuk t k;sxkA Name of the Candiate (in Capital letters) : ____________________________________________________________ Roll Number : in figures :
in words : _______________________________________________
Name of Examination Centre (in Capital letters) : ________________________________________ Candidate's Signature : ______________________________ Invigilator's Signature : ___________________________________
ADMISSION ANNOUNCEMENT Academic Session: 2016-17
Selections (from 20012-2015)
Selections @ 2015
66
35
(YCCP: 51 | DLP+eLP: 15)
(YCCP: 20 | DLP+eLP: 15)
AIPMT Selections (from 20012-2015)
Selections @ 2015
1559
447
(YCCP: 1128 | DLP+eLP: 431)
(YCCP: 337 | DLP+eLP: 110) rd
Selection of every 3 student from classroom coaching program in AIPMT 2015
For Classes: XI, XII & XII+ Target: AIIMS / AIPMT
Resonance Medical Optional Scholarship Test (ResoMOST)
th
08 May, 2016 To know more about ZFP & Admission Process Please call at :
w Mr. Mayank Tiwari: 9352529244 w Mr. Chanakya Dadhich : 9529006055 w Mr. Parvinder Singh: 9352880505
Toll Free: 1800 258 5555 | E-mail:
[email protected] | Website: www.medical.resonance.ac.in
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
PART A – BIOLOGY. 1.
In a testcross involving F1 dihybrid flies, more parental-type offspring were produced than the recombinant-type offspring. This indicates : (1) Both of the characters are controlled by more than one gene. (2) The two genes are located on two different chromosomes. (3) Chromosomes failed to separate during meiosis. (4) The two genes are linked and present on the same chromosome.
ijh{kkFkhZizl ad j.k esa] ft lesaF1 f}lad j efD[k;k¡'kkfey Fkha] iqu;ksZ xt izd kj d h larfr;ksad h rqy uk esat ud &izd kj d h larfr;k¡vf/kd mRiUu gq;hA blesalad sr fey rsgSafd % (1) nks uksagh y {k.kksad k fu;a=k.k ,d (2) nkst hu
lsvf/kd t huksa}kjk gksrk gSA
nksvy x xq.klw=kksaij fLFkr gSaA
(3) v/kZ l w=k.k d snkSjku
xq.klw=k i`Fkd ughagksik,A (4) nkst hu lgy Xu gS avkSj ,d gh xq.klw=k ij fo|eku gSaA
Ans. Sol.
(4) If a plant with genotype Aa Bb is crossed with aabb then Independent Assortment would result in production of 4 type of offsprings in equal proportion. Aa Bb – Gametes AB Ab aB ab aa bb – Gametes ab ab ab ab offspring according to independent assortment AaBb 1 : (parental)
Aabb aaBb 1 : 1 (Recombinants)
:
aabb 1 (Parental)
Since parental percentage is more then recombinants it is due to linkage between genes A and B. 2.
Water soluble pigments found in plant cell vacuoles are: (1) Anthocyanins (2) Xanthophylls (3) Chlorophylls (4) Carotenoids
ikni d ks f'kd k d h jl/kkuh esat y ?kqfy r o.kZd d kSu lsgksrsgSa\ (1) ,UFkks l k;fuu (2) t S UFkksfQ y (3) i.kZ gfjr (4) d S jksfVukbM Ans. Sol.
(1) Anthocyanin in stored in vacuole
Resonance Eduventures Ltd. CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Reg. Office : J-2, Jawahar Nagar, Main Road, Kota (Raj.)-324005 | Ph. No.: +91-744-3192222 | FAX No. : +91-022-39167222 Ph.No. : +91-744-3012222, 6635555 | To Know more : sms RESO at 56677 Website : www.resonance.ac.in | E-mail :
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PAGE # 1
| NEET-2016 3.
| 01-05-2016 |
Code-C,R,Y
Which of the following pairs of hormones are not antagonistic (having opposite effects) to each other ? (1)
Relaxin
Inhibin
(2) (3) (4)
Parathormone Insulin Aldosterone
Calcitonin Glucagon Atrial Natriuretic Factor
gkWeksZuksad sfuEufy f[kr ;qXeksaesalsd kSu&lk ;qXe ,d nwl jsd sfojks/kh ¼foijhr izHkko oky k½ ughagS\ (1)
fjy SfDlu
bfUgfcu
(2)
iSjkFkkseksZu
d SfYlVksfuu
(3)
bal qfy u
Xy qd SxkWu
(4)
,sYMksLVsjkWu
,fVª;y usfVª;wjsfVd d kjd
Ans.
(1)
Sol.
Parathormone Calcitonin
increases blood Ca level +2 decreases blood Ca level
insulin glucagon
decreases blood glucose level increases blood glucose level
Aldosterone ANF
increases B. P. decreses B. P.
+2
Relaxin causes pelvic musculature relaxation inhibin inhibits FSH So, Relaxin & inhibin not antagonistic 4.
Mitochondria and chloroplast are : (a) semi-autonomous organelles (b) formed by division of pre-existing organelles and they contain DNA but lack protein synthesizing machinery. Which one of the following options is correct ? (1) Both (a) and (b) are false. (2) Both (a) and (b) are correct. (3) (b) is true but (a) is false. (4) (a) is true but (b) is false.
Resonance Eduventures Ltd. CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Reg. Office : J-2, Jawahar Nagar, Main Road, Kota (Raj.)-324005 | Ph. No.: +91-744-3192222 | FAX No. : +91-022-39167222 Ph.No. : +91-744-3012222, 6635555 | To Know more : sms RESO at 56677 Website : www.resonance.ac.in | E-mail :
[email protected] | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal
PAGE # 2
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
ekbVksd kWf.Mª;k vkSj Dy ksjksIy kLV ¼gfjry od ½ gSa% (a) v/kZ LoklÙk vaxd
gSaA
(b) iw oZorhZvaxd ksad sfoHkkt u
lscursgSavkSj muesaDNA gksrk gS] y sfd u izksVhu&la'y s"kh iz.kky h d k vHkko gksrk gSA fuEufy f[kr fod Yiksaesalsd kS u&lk lgh gS\ (1) (a) vkS j (b) nksuksagh xy r
gaSA
(2) (a) vkS j (b) nksuksagh lgh gaSA (3) (b)
lgh gSy sfd u (a) xy r gS A
(4) (a) lgh gSy s fd u (b) xy r
gSA
Ans. Sol.
(4) Mitochondria and chloroplast have their own ribosomes wwith help of which they can synthesize protein.
5.
Which of the following is not a feature of the plasmids ? (1) Single – stranded (2) Independent replication (3) Circular structure (4) Transferable
fuEufy f[kr esalsd kS ulk ,d Iy kfTeM d k vfHky {k.k ughagS\ (1) ,d y – jTt q d h; (2) Lora =k izfrd `fr;u (3) o` Ùkh;
lajpuk
(4) LFkkukUrj.k ;ks X; Ans. Sol.
(1) Plasmide are double stranded DNA.
6.
A plant in your garden avoids photorespiratory losses, has improved water use efficiency, shows high rates of photosynthesis at high temperatures and has improved efficiency of nitrogen utilisation. In which of the following physiological groups would you assign this plant ? (1) Nitrogen fixer (2) C3 (3) C4 (4) CAM
vkid sm|ku esa,d ikni izd k'k 'olu lsgksusoky h gkfu lscprk gS] mld h t y mi;ksx d h n{krk mUur gS] og mPp rki ij izd k'k la 'y s"k.k d h mPp nj d ksn'kkZrk gSvkSj mld h ukbVªkst u mi;ksx d h n{krk mUur gSA vki bl ikni d ksfuEufy f[kr esalsfd l ,d d kf;Zd h lewg esaj[ksaxs\ (1) ukbVª kst u Ans. Sol.
fLFkfjd kjd
(2) C3
(3) C4 (4) CAM (3) C4 plants have high rate of photosynthesis at higher temperature.
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PAGE # 3
| NEET-2016 7.
| 01-05-2016 |
Code-C,R,Y
Emerson's enhancement effect and Red drop have been instrumental in the discovery of: (1) Oxidative phosphorylation (2) Photophosphorylation and non-cyclic electron transport (3) Two photosystems operating simultaneously (4) Photophosphorylation and cyclic electron transport
belZu nh?khZd j.k izHkko vkSj y ky cwan ¼jsM Mªki½ fd ld h [kkst esaizeq[k ;a=k jgsgSa\ (1) vkW DlhMsfVo
Q kLQ ksfjy s'ku
(2) iz d k'kQ kLQ ksfjy s'ku
vkSj vpØ h; by sDVªkWu vfHkxeu (3) nksiz d k'k rU=kksad k ,d lkFk d k;Zd juk
(4) iz d k'kQ kLQ ksfjy s'ku Ans. Sol. 8.
vkSj pØ h; by sDVªkWu vfHkxeu
(3) Red drop occur due decreased functioning of ps-II beyond 680 nm and when both ps I and ps II are functioning together their is enchancement in quantum yield. Which type of tissue correctly matches with its location ? Tissue
Location
(1)
Cuboidal epithelium
Lining of stomach
(2)
Smooth muscle
Wall of intestine
(3)
Areolar tissue
Tendons
(4)
Transitional epithelium
Tip of nose
d kSu&lk Å rd viuh fLFkfr lslgh&lgh lqesfy r gS\ Å rd
fLFkfr
(1)
?kukd kj mid y k
vkek'k; vkLrj
(2)
fpd uh is'kh
vka=k fHkfÙk
(3)
,sfjvksy h Å rd
d aM jk
(4)
ifjorhZmid y k
ukfld kxz
Ans.
(2)
9.
When does the growth rate of a population following the logistic model equal zero ? The logistic model is given as dN/dt = rN(1-N/K) : (1) when death rate is greater than birth rate. (2) when N/K is exactly one. (3) when N nears the carrying capacity of the habitat. (4) when N/K equals zero.
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PAGE # 4
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
y kWft fLVd ekWM y d k vuql j.k d jrsgq, fd lh lef"V d h o`f) nj 'kwU; d scjkcj d c gksxh \ y kWft fLVd ekWM y d ks fuEufy f[kr lehd j.k lsn'kkZ;k x;k gS% dN/dt = rN(1-N/K) : (1) t c
t Uenj d h vis{kk e`R;qnj vf/kd gksA (2) t c N/K Bhd ,d gks
(3) t c N i;kZ okl (4) t c N/K 'kw U; Ans.
(2)
Sol.
dN N rN 1 dt k
d h /kkfjrk {kerk d slehi gksA d scjkcj gksA
dN rN 1 1 0 dt 10.
Which one of the following statements is not true ? (1) Stored pollen in liquid nitrogen can be used in the crop breeding programmes (2) Tapetum helps in the dehiscence of anther (3) Exine of pollen grains is made up of sporopollenin (4) Pollen grains of many species cause severe allergies
fuEufy f[kr esalsd kS ulk d Fku lR; ughagS\ (1) nz for
ukbVªkst u esaHk.Mkfjr ijkxd .k] Q ly izt uu ;kst ukvksaesaiz;qDr fd ;st k ld rsgSa
(2) ijkxd ks "k d sLQ qVu
esaVsihVe lgk;rk d jrk gS (3) ijkxd .kks ad h cká Liksjksiksy sfuu d h cuh gksrh gS (4) cgq r
lh t kfr;ksad sijkxd .k xEHkhj izR;wt Zrk iS nk d jrsgSa
Ans. Sol.
(2) Dehiscence of anther occur due to stomium cells of endothecium
11.
Which one of the following statements is wrong ? (1) Phycomycetes are also called algal fungi. (2) Cyanobacteria are also called blue-green algae. (3) Golden algae are also called desmids. (4) Eubacteria are also called false bacteria.
fuEufy f[kr esalsd kSulk d Fku xy r gS\ (1) Q kbd ks ekbflVht
d ks'kSofy r d od Hkh d gk t krk gSA
(2) lk;uks cSDVhfj;k d ksuhy (3) Lof.kZ e
gfjr 'kSoky Hkh d grsgSaA
'kSoky ksad ksMs fLeM Hkh d grsgSaA
(4) ;q cSDVhfj;k ¼lqt hok.kq v ksa½ d ksvlR; Ans. Sol.
t hok.kqHkh d gk t krk gSA
(4) Eubacteria are called true bacteria.
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PAGE # 5
| NEET-2016 12.
| 01-05-2016 |
Code-C,R,Y
The Avena curvature is used for bioassay of: (1) Ethylene (2) ABA (3) GA3 (4) IAA
,ohuk oØ rk fd ld st S o vkekiu d sfy , iz;qDr gksrh gS\ (1) ,fFky hu
Ans. Sol. 13.
(2) ABA (3) GA3 (4) IAA (4) Avena curvature bioassay is done to test function of IAA. Which of the following structures is homologus to the wing of a bird ? (1) Flipper of Whale (2) Dorsal fin of the Shark (3) Wing of a Moth (4) Hind limb of Rabbit
fuEufy f[kr lajpukvksaesalsd kSulh lajpuk i{kh d sia[k d slet kr gS % (1) g~ osy
d k ¶y hij (2) 'kkd Zd h i` "B ia[k (3) 'ky Hk d k ia [k
(4) [kjxks 'k d k i'p
ikn
Ans.
(1)
14.
Blood pressure in the pulmonary artery is : (1) less than that in the venae cavae (2) same as that in the aorta (3) more than that in the carotid (4) more than that in the pulmonary vein
Q q¶Q ql /keuh d sHkhrj : f/kj nkc gksrk gS% (1) egkf'kjk d sHkhrj
ft ruk gksrk gSmllsd e gksrk gSA (2) mruk gh ft ruk egk/keuh d sHkhrj gks rk gSA (3) d S jksfVM
d sHkhrj ft ruk gksrk gSmllsvf/kd gksrk gSA (4) Q q ¶Q ql f'kjk d sHkhrj ft ruk gksrk gS] mllsvf/kd gksrk gS A Ans. Sol.
(4) Higher blood pressure in pulmonary vein than pulmonary artery is an abnormal condition leading to pulmonary hypertension & pulmonary oedema.
Resonance Eduventures Ltd. CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Reg. Office : J-2, Jawahar Nagar, Main Road, Kota (Raj.)-324005 | Ph. No.: +91-744-3192222 | FAX No. : +91-022-39167222 Ph.No. : +91-744-3012222, 6635555 | To Know more : sms RESO at 56677 Website : www.resonance.ac.in | E-mail :
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PAGE # 6
| NEET-2016 15.
| 01-05-2016 |
Code-C,R,Y
Fertilization in humans is practically feasible only if: (1) the sperms are transported into cervix within 48 hrs of release of ovum in uterus. (2) the sperms are transported into vagina just after the release of ovum in fallopian tube. (3) the ovum and sperms are transported simultaneously to ampullary – isthmic junction of the fallopian tube. (4) the ovum and sperms are transported simultaneously to ampullary – isthmic junction of the cervix.
ekuoksaesafu"ksp u izfØ ;k O;kogkfjd r% rHkh laHko gksxh t c% (1) xz hok uky d sHkhrj 'kqØ k.kqv ksad k LFkkukUrj.k xHkkZ'k; esav.Mk.kqd sfueqqZDr gksusd s48 ?kaVsd sHkhrj gksrk gksA (2) 'kq Ø k.kqv ksad k ;ksfu
d sHkhrj LFkkukUrj.k v.Mk.kqd sQ SYkksfi;u ufy d k ughaesaNksM +st kusd sBhd ckn gksA
(3) v.Mk.kqvkS j
'kqØ k.kqv ksad k LFkkukUrj.k Q SYkksfi;u ufy d k d s,aiq y jh&bLFkfed laxe ij ,d gh le; ij gksA
(4) v.Mk.kqvkS j
'kqØ k.kqv ksad k LFkkukUrj.k xzhok d s,aiqy jh bLFkfed laxe ij ,d gh le; ij gksrk gksA
Ans.
(2,3)
16.
In meiosis crossing over is initiated at : (1) Diplotene (2) Pachytene
(3) Leptotene
(4) Zygotene
v) Zl w=kh foHkkt u esat hu fofue; fd l voLFkk esavkjEHk gksrk gS\ (1) f}iê
(2) LFkw y iê
(3) ruq iê
(4) ;q Xeiê
Ans. Sol.
(2) In pachytene recombination nodule is formend after which crossing over occur
17.
Chrysophytes, Euglenoids, Dinoflagellates and slime moulds are included in the kingdom: (1) Animalia (2) Monera (3) Protista (4) Fungi
Ø kblksQ kbV] ;qXy hukW bM] Mkbuks¶y st sy sV vkSj voiad Q Q aqnh fd l t ho t xr~esalfEefy r gSa\ (1) t a rqt xr~ Ans. 18.
(2) eks usjk
(3) iz ksfVLVk
(4) d od
(3) Lack of relaxation between successive stimuli in sustained muscle contraction is known as : (1) Tonus (2) Spasm (3) Fatigue (4) Tetanus
mÙkjksÙkj mn~nhiuksad schp foJkafr d h d eh d sd kj.k gksusoky h nh?kZd kfy d is'kh lad qp u d gy krk gSaS% (1) Vks ul (2) ,a Bu ¼LikT+e½ s (3) Fkd ku (4) fVVs ul Ans.
(4)
Sol.
Tonus Spasm fatigue Tetanus
19.
Identify the correct statement on 'inhibin' : (1) Is produced by nurse cells in testes and inhibits the secretion of LH. (2) Inhibits the secretion of LH, FSH and Prolactin. (3) Is produced by granulose cells in ovary and inhibits the secretion of FSH. (4) Is produced by granulose cells in ovary and inhibits the secretion of LH.
low level activity of muscles at rest to maintain posture. Sudden involuntary muscle contraction decline in muscle activity Sustained muscle contraction in response to successive stimuli
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PAGE # 7
| NEET-2016 'ba fgfcu' d sckjsesalgh d Fku (1) ;g
| 01-05-2016 |
Code-C,R,Y
igpkfu, %
o`"k.kksad h /kk=kh ¼ulZ½ d ks f'kd kvksa}kjk mRiUu gksrk gSvkSj LH - ò o.k d kslanfer d jrk gSA
(2) LH,FSH vkS j
izksy SfDVu ò o.k d kslanfer d jrk gSA
(3) ;g
v.Mk'k; d h d f.kd h; d ksf'kd kvksa}kjk mRiUu gksrk gSA vkSj FSH ò o.k d kslanfer d jrk gSA (4) ;g v.Mk'k; d h d f.kd h; d ks f'kd kvksa}kjk mRiUu gksrk gSvkSj LH ò o.k d kslanfer d jrk gSA Ans.
(3)
20.
Name the chronic respiratory disorder caused mainly by cigarette smoking: (1) Respiratory alkalosis (2) Emphysema (3) Asthma (4) Respiratory acidosis
/kweziku d jusd sd kj.k iz/kkur% mRiUu gksusoky snh?kZd ky h 'olu&fod kj d k uke crkb, % (1) 'olu {kkje;rk (2) okrLQ hfr (3) vLFkek
(4) 'olu
vkEy jDrrk
Ans.
(2)
21.
Which of the following most appropriately describes haemophilia ? (1) Dominant gene disorder (2) Recessive gene disorder (3) X-linked recessive gene disorder (4) Chromosomal disorder
fuEufy f[kr esalsd kS ulk gheksQ hfy ;k d k lclsvf/kd mi;qDr o.kZu izLrqr d jrk gSA (1) iz Hkkoh t hu
d k fod kj (3) X-lgy Xu viz Hkkoh t hu d k fod kj
(2) viz Hkkoh t hu
d k fod kj (4) xq .klw=kh fod kj
Ans. Sol.
(3) Gene related with haemophilia is always present on X chromosome and it is present on X chromosome and it is recessive gene disorder as it express itself in females when comes an homonzygous condition
22.
Select the correct statement: (1) The leaves of gymnosperms are not well adapted to extremes of climate (2) Gymnosperms are both homosporous and heterosporous (3) Salvinia, Ginkgo and Pinus all are gymnosperms (4) Sequoia is one of the tallest trees
lgh d Fku pqfu, % (1) vuko` rcht h ikniksad h ifÙk;k¡t y ok;qd h pjerk d sfy ,
vuqd wfy r ughagksrh gSa A (2) vuko` rcht h] lecht k.kqd vkSj fo"kecht k.kqd ] nks uksaizd kj d sgksrsgSa (3) lkfYofu;k] ft a xksvkSj
ikbul] ;slHkh vuko`rcht h gSa
(4) fld ks b;k lclsy Ecso`{kksaesals,d Ans.
gS
(4)
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PAGE # 8
23.
| NEET-2016 | 01-05-2016 | Which of the following is required as inducer(s) for the expression of Lac operon? (1) lactose and galactose (2) glucose (3) galactose (4) lactose
Code-C,R,Y
y Sd izp ky s d d h vfHkO;fDr d sfy , fuEufy f[kr esalsd kSu ,d izsjd d s: i esad k;Zd jusd sfy , vko';d gksxk \ (1)
y SDVkst vkSj xSy sDVks t
(2) Xy w d kst
(3) xS y sDVkst Ans. Sol.
(4) y S DVkst
(4) As lae operon becomes active after inducing lactose but glucose & galactose can't do so.
Pure tall TT
Pure dwarf tt
T
t
F1 generation = Tt
F2 generation
Genothypes = 1 TT Pure Tall 24.
T TT Tt
T t
:
2 Tt
t Tt tt
:
heterogygous tall
1 tt pure dwarf
A tall true breeding garden pea plant is crossed with a dwarf true breeding garden pea plant. When the F1 plants were selfed the resulting genotypes were in the ratio of :
,d y ECksrn~: i iz t uu m|ku eVj ikni d ks,d ckSusrn~: i izt uu m|ku eVj ikni lslad fjr d jk;k x;kA t c F1 ikniks ad ksLoijkfxr fd ;k x;k rkst hu izk: i d k ifjek.k fd l vuqikr esaFkk \
Ans.
(1) 3 : 1 : : Dwarf : Tall (2) 1 : 2 : 1 : : Tall homozygous : Tall heterozygous : Dwarf (3) 1 : 2 : 1 : : Tall heterozygous : Tall homozygous : Dwarf (4) 3 : 1 : : Tall : Dwarf (1) 3 : 1 : : ckS us: y ECks (2) 1 : 2 : 1 : : y Ecsle;q Xet h : y Ecsfo"ke;qXet h : ckSus (3) 1 : 2 : 1 : : y ECksfo"ke;q Xet h : y Ecsle;qXet h: ckSus (4) 3 : 1 : : y Ecs: ckS us (2)
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25.
| NEET-2016 | 01-05-2016 | Which part of the tobacco plant is infected by Meloidogyne incognita ?
Code-C,R,Y
Ans.
(1) Root (1) t M+ (1)
26.
Which of the following is not a characteristic feature during mitosis in somatic cells?
Ans. Sol.
(1) Synapsis (2) Spindle fibres (3) Disappearance of nucleolus (4) Chromosome movement (1) lw =k;qXeu (2) rd q Z: ih rarq (3) d s fUnzd k d k foy ksiu (4) xq .klw=k xfr (1) Synapsis is pairing of homologous chromosomes which occurs during meiosis but it is absent in mitos.
27.
Which of the following statements is not true for cancer cells in relation to mutations?
rEckd wd sikS/ksd k d kS u&lk Hkkx fey ks bMks xkbu bUd ks fXuVk }kjk laØ fer gksrk gS\ (2) Flower (2) iq "i
(3) Leaf (3) iRrh
(4) Stem (4) ruk
d kf;d d ksf'kd kvksaes alelw =k.k d snkSjku fuEufy f[kr es alsd kSu&lk y {k.k ughaik;k t krk \
mRifjorZu d slaca/k esad S al j d ks f'kd kvksad sfy , fuEufy f[kr d Fkuksaesalsd kSu&lk lgh ughagS\
Ans. Sol.
(1) Mutations inhibit production of telomeres. (2) Mutations in proto-oncogenes acceleration the cell cycle. (3) Mutations destroy telomerase inhibitor. (4) Mutations inactivate the cell control. (1) mRifjorZ u Vhy ksejst d smRiknu d kslanfer d j nsrsgSaA (2) iz kd ~d Sal jt huksaesamRifjorZu d ksf'kd k&pØ d ksRofjr d j nsrsgSA (3) mRifjorZ u Vhy ksejst laned d ksu"V d j nsrsgSA (4) mRifjorZ u d ksf'kd k&fu;a=k.k d ksfuf"Ø ; d j nsrsgSaA (1) Cancer will be caused by increased telomerase activity making the cancerous cells immortal & not by inhibition of telomerase production.
28.
One of the major components of cell wall of most fungi is :
Ans.
(1) Hemicelluloses (3) Peptidoglycan (1) gs ehlsY;wy ks t (3) is IVhMksXy kbd u (2)
29.
Cotyledon of maize grain is called :
vf/kd rj d od ksaesad ksf'kd k fHkfÙk d k ,d izeq[k vo;o d kSu lk gS\ (2) Chitin (4) Cellulose (2) d kbfVu (4) ls Y;ww y kst
eDd k d snkusd scht i=k d ksD;k d gk t krk gS\
Ans. Sol.
(1) scutellum (2) Plumule (3) coleorhiza (4) coleoptile (1) Ld q Vsy e (2) iz kad q j (3) ew y kad qj&pksy (4) iz kad q j&pksy (1) In maize grains single large shield shaped cotyledon is called scutellum.
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30.
| NEET-2016 | 01-05-2016 | Which of the following would appear as the pioneer organisms on bare rocks ?
Code-C,R,Y
Ans. Sol.
(1) Green algae (2) Lichens (3) Liverworts (4) Mosses (1) gfjr 'kS oky (2) y kbd s u (3) fy ojoVZ (4) ekW l (2) Lichens are pioneer organisms on bare rocks as they corrode the rocks by secreting enzyme &
,d uXu pV~Vku ij ,d vxzxkeh t ho d s: i esafuEufy f[kr esalsd kSu vk;sxk \
converted into soil. 31.
Ans. Sol. 32.
Changes in GnRH pulse frequency in females is controlled by circulating levels of : eknkvksaesaGnRH iYl ckjackjrk cny ko d k fu;a=k.k fd ld sifjlap j.k&Lrjkas}kjk gksrk gS\ (1) progesterone and inhibin (2) estrogen and progesterone (3) estrogen and inhibin (4) progesterone only (1) iz kst sLVsjkWu vkSj bafgfcu (2) bZ LVªkst u vkSj izkst sLVsjkWu (3) bZ LVªkst u vkSj bafgfcu (4) d s oy izkst sLVsjkWu (2) GnRH pulse frequency in controlled by estrogen and progesterone both after puberty Antivenom injection contains preformed antibodies while polio drops that are administered into the body contain :
izfrvkfo"k Vhd ksaesaiwoZfufeZr izfrj{kh gksrsgSat cfd iksfy ;ksd h cw¡nksaesaft Ugsaeq¡ g }kjk fny k;k t krk gS] gksrsgSa%
Ans.
33.
(1) Attenuated pathogens (2) Activated pathogens (3) Harvested antibodies (4) Gamma globulin (1) {kh.k d j fn, x, jks xt ud (2) lfØ f;r jks xt ud (3) cuk, x, iz frj{kh (4) xkek Xy ks C;qfy u (1) OPV is of 2 types : (i) OPV sabin – Live attenuated vaccine (ii) OPV salk – Killed vaccine Photosensitive compound in human eye is made up of:
ekuo us=k esaizd k'klaosnh ;kSfxd cuk gksrk gS%
Ans. Sol.
(1) Transducin and Retinene (3) Opsin and Retinal' (1) Vª kaLM~;wflu vkSj jsfVuhu ls (3) vks fIlu vkSj jsfVuy ls (3) Rhodopsin is made of opsin & retinal.
(2) Guanosine and Retinol (4) Opsin and Retinol (2) Xokuks flu vkSj jsfVukWy ls (4) vks fIlu vkSj jsfVukWy ls
34.
Specialised epidermal cells surrounding the guard cells are called :
}kj d ksf'kd kvksad ks?ksjusoky h fof'k"Vhd `r ckgkzRoph; d ksf'kd kvksad ksD;k d gk t krk gS\
Ans. Sol.
(1) Lenticels (2) Complementary cells (3) Subsidiary cells (4) Bulliform cells (1) okrjU/kz (2) iw jd d ksf'kd k,a (3) lgk;d d ks f'kd k,¡ (4) vko/kZRod d ks f'kd k,a (3) As subsidiary cells & guard cells both are modification of epidermal cells in which guard cells have chloroplasts which is absent in surroundings subsidiary cells.
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PAGE # 11
35.
| NEET-2016 | 01-05-2016 | Which of the following features is not present in the Phylum - Arthropoda?
Code-C,R,Y
fuEufy f[kr y {k.kkasesalsd kSu&lk y {k.k Q kby e&vkFkzkZ siksMk easughaik;k t krk \
Ans. Sol. 36.
Ans. 37.
(1) Jointed appendages (3) Metameric segmentation (1) la f/kr mikax (3) fo[ka M h [kaM hHkou (4) Parapodia is a characteristic of Annelida.
(2) Chitinous exoskeleton (4) Parapodia (2) d kbfVuh ckgkz d ad ky (4) ik'oZ ikn
Reduction in pH of blood will : : f/kj d spH esagksusoky h d eh d sd kj.k (1) release bicarbonate ions by the liver, (2) reduce the rate of heart beat. (3) reduce the blood supply to the brain (4) decrease the affinity of hemoglobin with oxygen. (1) ;d ` r }kjk ckbd kcksZusV d k fu"d klu gksusy xsxkA (2) ân;&Lia nu d h nj d e gkst k;sxh (3) efLr"d d k : f/kj la Hkj.k d e gkst k;sxkA (4) vkW Dlht u d slkFk gheksXy ksfcu d h ca/kqrk ?kV t k;sxhA (4) Which of the following characteristic features always holds true for the corresponding group of animals? fuEufy f[kr esalsd ks u&lsfof'k"V y {k.k ges'kk gh t arqv ksad svuq: ih oxZesaik, t krsgSa? (1)
3 - chambered heart with one incompletely divided ventricle
Reptilia
(2)
Cartilaginous endoskeleton
Chondrichthyes
(3)
Viviparous
Mammalia
(4)
Possess a mouth with an
Chordata
upper and a lower jaw
Ans.
(1)
3 - d {k oky k ân;
ft lesaviw.kZr% caVk gqv k ,d fuy ; gksrk gSaA
jsIVhfy ;k
(2)
mikfLFky var%d ad ky
d kW fMªDFkht
(3) (4)
lt ho izt d Å ijh vkSj fupy st cM+soky k eq[k d k ik;k t kuk
eseSfy ;k d kW M sZVk
(2)
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PAGE # 12
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
Sol.
Reptilia has an order crocodilia which shows 4 chambered heart. In mammals, prototheria group shows oviparity while metatheria & eutheria show viviparity. Chordates can be gnathostomata & agnatha (without jaws). Only cartilaginous fishes (chondrichthyes) show cartilaginous endoskeleton without exception
38.
Match the terms in Column I with their description in Column IIand choose the correct option: d kWy e I d h 'kCnksad ksd kWy e II esafn, x, mud so.kZu lseSp d hft ;srFkk lgh fod Yi pqfu,: Column I Column II (a) Dominance (i) Many genes govern a single character (b) Codominance (ii) In a heterozygous organism only one allele expresses itself (c) Pleiotropy (iii) In a heterozygous organism both alleles express themselves fully (d) Polygenic inheritance (iv) A single gene influences many characters LrEHkI LrEHkII (a) iz Hkkfork (i) vus d t hu ,d y y {k.k d k fu;a=k.k d jrsgSa (b) lgiz Hkkfork (ii) fo"ke;q Xeut h t ho esad soy ,d gh ,sy hy Lo;ad ksvfHkO;Dr d jrk
gS (c) cgq izHkkfork
fo"ke;qXeut h t ho esanksuksagh ,sy hy Lo;ad ksiwjh rjg vfHkO;Dr d jrsgSaA (iv) ,d y t hu vus d y {k.kksad ksizHkkfor d jrk gSA
(iii)
(d) cgq t huh oa 'kkxfr Code: (1) (2) (3) (4)
(a) (iv) (ii) (ii) (iv)
(b) (ill) (i) (ill) (i)
(c) (i) (iv) (iv) (ii)
(d) (ii) (ill) (i) (ill)
Ans.
(3)
39.
A typical fat molecule is made up of :
,d izk: ih olk d k v.kqfd ld k cuk gksrk gS\
Ans. Sol.
(1) Three glycerol and three fatty acid molecules (2) Three glycerol molecules and one fatty acid molecule (3) One glycerol and three fatty acid molecules (4) One glycerol and one fatty acid molecule (1) rhu Xy hljkW y vkSj rhu olk vEy v.kqv ksad k (2) rhu Xy hljkW y v.kqv ksavkSj ,d olk vEy v.kqd k (3) ,d Xy hljkW y v.kqvkSj rhu olk vEy v.kqv ksad k (4) ,d Xy hljkW y vkSj ,d olk vEy v.kqd k (3) fat is a triglyceride which is made up of 3 molecules of fatty acids and one molecule of glycerol
40.
Proximal end of the filament of stamen is attached to the :
iqad s l j d srUrqd k fud VLFk fljk fd llst qM +k gksrk gS\
Ans.
(1) Thalamus or petal (3) Connective (1) iq "iklu ;k ny (3) la ;kst d (1)
(2) Anther (4) Placenta (2) ijkxd ks "k (4) cht k.Mklu
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PAGE # 13
| NEET-2016 41.
| 01-05-2016 |
Code-C,R,Y
Which one of the following statements is wrong?
fuEufy f[kr esalsd kS u&lk d Fku xy r gS\
Ans. Sol. 42.
(1) Glycine is a sulphur containing amino acid. (2) Sucrose is a disaccharide. (3) Cellulose is a polysaccharide. (4) Uracil is a pyrimidine. (1) Xy kbflu ,d lYQ j;q Dr vehuksvEy gSA (2) lq Ø ksl ,d MkblSd sjkbM gSA (3) ls Y;wy ksl ,d ikWfy lSd sjkbM gSA (4) ;w jSfly ,d fifjfeMhu gSA (1) Glycine is the simplest amino acid which is devoid of sulpher content Water vapour comes out from the plant leaf through the stomatal opening. Through the same stomatal opening carbon. dioxide diffuses into the plant during photosynthesis. Reason out the above statements using one of following options :
ikni iÙkh lst y ok"i jU/kzksad s}kjk ckgj vkrk gSizd k'k la'y s"k.k d snkSjku mlh jU/kzlsd kcZu MkbvkWDlkbM ikni lsfolfjr gksrh gSA mi;qZDr d Fkuksaesa¼d kj.kksaij fopkj d j½ ,d fod Yi pqfu, %:
Ans. 43.
(1) One process occurs during day time, and the other at night. (2) Both processes cannot happen simultaneously. (3) Both processes can happen together because the diffusion coefficient ofwater and CO2 is different. (4) The above processes happen only during nighttime. (1) ,d iz fØ ;k fnu esarFkk nwl jh izfØ ;k jkr esagksrh gS A (2) nks uksaizfØ ;k,sa,d lkFk ughagksld rhA (3) nks uksaizfØ ;k,a,d lkFk gksld rh gS aD;ks afd t y vkSj CO2 d k folj.k xq.kkad fHkUu gSA (4) mi;q ZDr izfØ ;k,ad soy jkr esagksld rh gSa A (3)
Ans.
A complex of ribosomes attached to a single strand of RNA is known as : jkbcksl kse d k ,d lad qy t ksRNA d s,d y jTt qd d slkFk t qM +k gksrk gS] D;k d gy krk gS: (1) Okazaki fragment (2) Polysome (3) Polymer (4) Polypeptide (1) vks d kt kd h [k.M (2) ikW y hlkse (3) ikW y hej ¼cgqy d ½ (4) ikW y hisIVkbM (2)
44.
Which one of the following is a characteristic feature of cropland ecosystem?
fuEufy f[kr esalsd kS u ,d d `f"kHkwfe ikfjrU=k d k vfHky {k.k gS\
Ans. 45.
(1) Ecological succession (3) Least genetic diversity (1) ikfjrfU=kd vuq Ø e.k (3) U;w ure vkuqoaf'kd fofo/krk (3)
(2) Absence of soil organisms (4) Absence of weeds (2) e` nk t hoksad h vuqifLFkfr (4) vir` .kksad h vuqifLFkfr
Which of the following is the most important cause of animals and plants being driven to extinction? (1) Co – extinctions (2) Over - exploitation (3) Alien species invasion (4) Habitat loss and fragmentation
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PAGE # 14
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
t Urqv ksavkSj ikniksad h foy qfIr d k fuEufy f[kr esalsd kSu&lk ,d lclseq[; d kj.k gS\ (1) lg&lekfIr (2) vfr nks gu (3) fons 'kh t kfr d h p<+kbZ (4) vkokl gkfu vkS j [kaM u Ans.
(4)
46.
In a chloroplast the highest number of protons are found in :
gjfr y od esaizksVkWu d h vf/kd re la[;k d gk¡ik;h t krh gS\
Ans. 47.
(1) Antennae compIex (3) Lumen of thylakoids (1) ,s UVsuk leqPp (3) Fkkby s d ksbM d h vod kf'kd k (3)
(2) Stroma (4) Inter membrane space (2) ihfBd k (4) vUrj d y k LFkku
Which of the following is not required for any of the techniques of DNA fingerprinting available at present?
Mh-,u-,– vaxqfy Nkiu d h fd lh Hkh rd uhd d sfy , fuEufy f[kr esalsfd l ,d d h vko';d rk ughagksrh \
Ans.
(1) DNA -DNA hybridization (3) Zinc finger analysis (1) Mh–,u–,&Mh–,u-,– la d j.k (3) ft a d vaxqfy fo'y s"k.k (3)
(2) Polymerase chain reaction (4) Restriction enzymes (2) ikW y hejst J`a[ky k vfHkfØ ;k (4) iz frca/ku ,at kbe
48.
The primitive prokaryotes responsible for the production of biogas from the dung of ruminant animals, include the : (1) Eubacteria
(2) Halophiles
(3) Thermoacidophiles
(4) Methanogens
osvkfne izkd ~d sUnzd h izk.kh] t ks jks eUFkh t arqv kasd sxkscj ls ck;ksxSl &mRiknu d sfy , mÙkjnk;h gksrsgS] fd ld s varxZr vkrsgS\ (1) lq t hok.kqv kasd s
(2) y o.kjkfx;ks ad s
(3) rki&vEy
(4) ehFkS ut ud ksad s
jkfx;kad s
Ans.
(4)
49.
Which of the following features is not present in Periplaneta americana? (1) Metamerically segmented body (2) Schizocoelom as body cavity (3) Indeterminate and radial cleavage during embryonic development (4) Exoskeleton composed of N-acetylglucosamine
fuEufy f[kr esalsd kS u&lk y {k.k isfjIy SusVk vesfjd kuk esaugh ik;k t krk gS\ (1) fo[ka M 'k% [kafMr (2) ns gxqgk d s: i (3) Hkw z.kh;
nsg esanh.kZxqgk
ifjo/kZ u d snkSjku vfu/kkZfjr vkSj vjh; fony u
(4) N– ,ls fVy Xy wd ksl ,sehu
lsfufeZr cká d ad ky
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PAGE # 15
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
Ans.
(3)
Sol.
Periplaneta Americana shows indeterminate and spiral cleavage
50.
A system of rotating crops with legume or gras pasture to improve soil structure and fertility is called: (1) Shifting agriculture (2) Ley farming (3) Contour farming (4) Strip farming
e`nk lajpuk vkSj moZjd rk es alq/kkj y kusd sfy , Q ly ksad ksQ y hnkj ikS/kksa¼y sX;we½ ;k ?kkl pkjxkg d slkFk cny d j y xkusd ksD;k d gk t krk gS\ (1) LFkkukUrjh Ñ f"k (2) y s[ks rh (3) leks Ppjs[kh; (4) iV~ Vhnkj
[ksrh
[ksr
Ans.
(2)
51.
Which of the following is wrongly matched in the given table?
Microbe
Product
Application
(1)
Clostridium butylicum
Lipase
removal of oil stains
(2)
Trichoderma polysporum
Cyclosporin A
immunosuppressive drug
(3)
Monascus purpureus
Statins
lowering of blood cholesterol
(4)
Streptococcus
Streptokinase
removal of clot from blood vessel
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| NEET-2016
| 01-05-2016 |
Code-C,R,Y
uhpsnh x;h rkfy d k esaxy r fey k;h x;h enksad kspqfu, ?
(1)
lw{ket ho
mRikn
v uqiz;ksx
Dy kWLVªhfM;e
y kbist
rsy d s/kCcksad ksgVkuk
C;wVk;fy d e (2)
VªkbZ d ksM ekZiksy hLiksje
lkbDy ks Liksfju A
izfrj{kk laned vkS"kf/k
(3)
eksuSLd l ijI;wjh;l
LVsfVal
: f/kj&d ksy sLVªkWy d ksd e d juk
(4)
LVsªIVksd kWd l
LVªsIVksd buSt
: f/kj&okfgd k lsFkDd sd ksgVkuk
Ans.
(1)
52.
In mammals, which blood vessel would normally carry largest amount of urea? (1) Hepatic Portal Vein (2) Renal Vein (3) Dorsal Aorta (4) HepaticVein.
Lru/kkfj;ksaesad kSu&lh : f/kj&okfgd k lkekU;r% lclsvf/kd ;w fj;k ogu d jrh gS\ (1) ;Ñ r
fuokfgd k f'kjk
(2) o` Dd &f'kjk (3) i` "B
egk/keuh
(4) ;Ñ r&f'kjk Ans.
(4)
Sol.
Urea/Ornithine cycle takes place in liver so the vein leaving liver possesses maximum urea which is hepatic vein
53.
Pick out the correct statements : (a) Haemophilia is a sex-linked recessive disease. (b) Down's syndrome is due to aneuploidy. (c) Phenylketonuria is an autosomal recessive gene disorder. (d) Sickle cell anaemia is an X-linked recessive gene disorder. (1) (a), (b) and (c)are correct. (2) (a) and (d) are correct. (3) (b)and (d) are correct. (4) (a), (c)and (d) are correct.
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| NEET-2016
| 01-05-2016 |
Code-C,R,Y
lgh d Fku pqfu, % (a) gheks Q hfy ;k fy ax&lgy Xu vizHkkoh jksx gSA (b) Mkmu
fla M ªkse vlqxqf.krk d sd kj.k gksrk gSA (c) Q s fuy d hVksuesg ¼fQ ukby d hVksU;wfj;k½ ,d vfy ax lw=kh vizHkkoh t hu fod kj gSA
(d) nk=k d ks f'kd k jDrkYirkX-lgy Xu
vizHkkoh t hu fod kj gSA
(1) (a), (b) vkS j (c) lgh gSA (2) (a) vkS j (d) lgh gSA (3) (b)
vkSj (d) lgh gSA
(4) (a), (c) vkS j (d) lgh gSA Ans.
(1)
Sol.
Sickle cell anemia is an autosomal codominant disorder
54.
Which of the following guards the opening of hepatopancreatic duct into the duodenum? (1) Sphincter of Oddi
(2) Semilunar valve
(3) Ileocaecal valve
(4) Pyloric sphincter.
fuEufy f[kr esalsd kS u&lh lajpuk ;Ñ nXU;kl d h okfguh d sxz g.kh esa[kqy usoky sja/kzd h ns[kHkky d jrh gS\ (1) vks M kbZd ksvojksf/kuh
(2) v/kZ p anzkd kj
(3) f=kd ka =k d ikV
(4) t BjfuxZ e
Ans.
(1)
55.
Microtubules are the constituents of :
d ikV
vojksf/kuh
(1) Centrosome, Nucleosome and Centrioles (2) Cilia, Flagella and Peroxisomes (3) Spindle fibres, Centrioles and Cilia (4) Centrioles, Spindle fibres and Chromatin.
lw{eufy d k,¡la?kVd gksrh gS: (1) rkjd k;ks a] U;wfDy ;ksl kse (2) i{ekHkks a] d 'kkHkksavkSj
Ans.
vkSj rkjd d sUnzksd s
ijvkWDlhd k;ksad s
(3) rd Z q: ih js'kksa] rkjd d sUnzksavkSj
i{ekHkksad s
(4) rkjd d s Unzks] rd Z: ih js'kksavkSj
Ø kseSfVu d s
(3)
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PAGE # 18
56.
| NEET-2016 | 01-05-2016 | The coconut water from tender coconut represents : (1) Free nuclear endosperm
(2) Endocarp
(3) Fleshy mesocarp
(4) Free nuclear proembryo
Code-C,R,Y
d Ppsukfj;y eas] ukfj;y ikuh D;k gS\ (1) LorU=k d s Unzd h Hkzw.kiks"k
(2) vUr%Q y fHkfÙk
(3) xw nsnkj
(4) LorU=k d s Unzd h Hkzw.kiwohZ
e/;Q y fHkfÙk
Ans.
(1)
Sol.
In tender coconut, edible part is liquid endosperm that represents free nuclear endosperm
57.
Tricarpellary, syncarpous gynoecium is found in flowers of: (1) Poaceae (2) Liliaceae (3) Solanaceae (4) Fabaceae
f=kd ks"Bd h] ;qDrk.Mih t k;k¡x fd ld siq"i esagksrk gS\ (1) iks ,lh (2) fy fy ,lh (3) lks y Sus lh (4) Q S csl h Ans.
(2)
58.
Which of the following is not a stem modification? (1) Flattened structures of Opuntia (2) Pitcher of Nepenthes (3) Thorns of citrus (4) Tendrils of cucumber
fuEufy f[kr esalsd kS u ,d rusd k : ikUrj.k ughagS? (1) vks iaf'k;
d h piVh lajpuk
(2) us iUFkht
d k ?kV
(3) flVª l
d sd kVs
(4) [khjsd siz rku Ans.
(2)
Sol.
Pitcher of nepenthes is modification of leaf
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PAGE # 19
| NEET-2016 59.
| 01-05-2016 |
Code-C,R,Y
The taq polymerase enzyme is obtained from : (1) Pseudomonas putida
(2) Thermus aquaticus
(3) Thiobacillus ferroxidans
(4) Bacillus subtilis
VSd ikWfy ejst ,Ut kbe fd llsizkIr fd ;k t krk gS\ (1) L;w Mkseksukl (3) fFk;ks cSfly l
I;wfVMk Q sjksDlhMsUl
(2) FkeZ l
,DosfVd l
(4) cS fly l
lcfVfy l
Ans.
(2)
60.
Stems modified into flat green organs performing the functions of leaves are known as : (1) Scales
(2) Cladodes
(3) Phyllodes
(4) Phylloclades
ifÙk;ksad k d k;Zd jusoky s ] piVsgjsvax esa: ikUrfjr rusd ksD;k d gk t krk gS\ (1) 'kYd
(2) ik.kkZ Hk ioZ
(3) i.kkZ Hk o`Ur
(4) i.kkZ Hk LRkEHk
Ans.
(4)
61.
In higher vertebrates, the immune system can distinguish self-cells and non-self, If this property is lost due to genetic abnormality and it attacks self-cells, then it leads to : (1) Active immunity (2) Allergic response (3) Graft rejection (4) Auto-immune disease
mPprj d 'ks: fd ;kasesa] izfrj{kk ra=k Lo&d ksf'kd kvksavkSj xSj&d ksf'kd kvksaesaHksn d j ld rk gSA ;fn ra=k d k vkuqoaf'kd vilkekU;rk d sd kj.k ;g xq.k u"V gkst k, vkSj og Lo&d ksf'kd kvks ad ksu"V d jusy xsrksbld sifj.kkeLo: i D;k gksxk\ (1) lfØ ;
izfrj{kk
(2) ,s y t hZvuqfØ ;k (3) fujks i
vLohd kj d j nsuk
(4) Loiz frj{kk fod kj Ans.
(4)
Sol.
If self & non-self recognization power is lost than immune cells can attack our own body cells and cause auto immune disease
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PAGE # 20
62.
| NEET-2016 | 01-05-2016 | Code-C,R,Y Nomenclature is governed by certain universal rules. Which one of the following is contrary to the rules of nomenclature. (1) When written by hand, the names are to be underlined (2) Biological names can be written in any language (3) The first word in a biological name represents the genus name, and the second is a specific epithet (4) The names are written in Latin and are italicised
uke&i) fr d qn fo'ks"k lkoZt fud ekU; fu;eksa}kjk fu/kkZfjr gks rh gSA fuEufy f[kr easalsd kSu lk ,d d Fku uke i) fr d sfu;eksad sfo: ) gS\ (1) uke
d kst c gkFk lsfy [krsgSrks mlsjs[kkafd r fd ;k t krk gS
(2) t S fod
uke d ksfd lh Hkh Hkk"kk easfy [kk t k ld rk gSA
(3) t S fod
uke esaigy k 'kCn oa'k uke vkSj nwl jk 'kCn t kfr lad sr in d ksiznf'kZr d jrk gSSA
(4) ukeks ad ksy SfVu
Hkk"kk esavkSj frjNsv{kjksaesafy [kk t krk gS
Ans.
(2)
63.
In bryophytes and pteridophytes, transport of male gametes requires : (1) Water (2) Wind (3) Insects (4) Birds
czk;ksQ kbV vkSj VsfjMksQ kbV esauj ;qXed d svfHkxeu d sfy , fd ld h vko';d rk gksrh gS\ (1) t y (2) iou (3) d hV (4) i{kh Ans.
(1)
Sol.
64.
In context of Amniocentesis,which of the following statement is incorrect ? (1) It can be used for detection of Cleft palate. (2) It is usually done when a woman is between 14 -16 weeks pregnant. (3) It is used for prenatal sex determination. (4) It can be used for detection of Down syndrome.
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PAGE # 21
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
mYcos/ku d slanHkZesa] fuEufy f[kr esalsd kSu lk d Fku xy r gS\ (1) bls[ka M rky q¼Dy s¶V (2) ;g
iSy sV½ d k irk y xkusd sfy , iz;qDr fd ;k t krk gSA
vkerkSj lsrc fd ;k t krk gSt c L=kh d ks14 -16 lIrkg d schp d k xHkZgks rk gS A
(3) blsiz l oiwoZfy ax&fu/kkZj.k (4) blsMkmu
d sfy , iz;qDr fd ;k t krk gSA
flaM ªkse d k irk y xkusd sfy , iz;qDr fd ;k t krk gSA
Ans.
(1)
Sol.
Cleft palate is a structural defect and cannot be determined by amniocentesis.
65.
In the stomach, gastric acid is secreted by the : (1) acidic cells (2) gastrin secreting cells (3) parietal cells (4) peptic cells
vkek'k; esat Bj jl d k L=kko gksrk gS% (1) vEy
d ksf'kd kvksals
(2) xS fLVªu
d k L=kko d jusoky h d ks f'kd kvksals
(3) fHkÙkh;
d ksf'kd kvksals
(4) is fIVd
d ksf'kd kvksals
Ans. (3) Sol.
66.
gastric acid is HCl secreted by parietal or oxyntic cells.
Spindle fibres attach on to: (1) Kinetosome of the chromosome (2) Telomere of the chromosome (3) Kinetochore of the chromosome (4) Centromere of the chromosome
rd Zq: ih rarqy xrsgSA (1) xq .klw=k d sd kbusVksl kse
ij
(2) xq .klw=k d svaR;ka'k ij (3) xq .klw=k d sd kbuksVksd ksj (4) xq .klw=k d slw=kd sUæ
ij
ij
Ans. (3)
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PAGE # 22
Sol.
| NEET-2016 | 01-05-2016 | Code-C,R,Y kinetochore of chromosome facilitates the attachment of spindle fibre (chromosomal fibre) and pole
67.
Which is the National Aquatic Animal of India? (1) Sea- horse (2) Gangetic shark (3) River dolphin (4) Blue whale
Hkkjr d k jk"Vªh; t y h; izk.kh d kSu lk gS? (1) leq æh- ?kksM +k (2) xa xk d h 'kkd Z (3) unh d h MkW fYQ u (4) Cy wg~ osy Ans.
(3)
Sol.
68.
Which one of the following cell organelles is enclosed by a single membrane? (1) Nuclei (2) Mitochondria (3) Chloroplasts (4) Lysosomes
fuEufy f[kr eslsd kSu lk d ksf'kd kax d soy ,d y d y k lsf?kjk gksrk gS? (1) d s Uæd (2) lw =kd f.kd k (3) gfjry od (4) y ;ud k; Ans.
(3)
Sol.
Except Lysosome, all three are bounded by double membrane
69.
The two polypeptides of human insulin are linked togetherby : (1) Disulphide bridges
(2) Hydrogen bonds
(3) Phosphodiester bond
(4) Covalentbond
ekuo bUlqfy u d snksikWy hisIVkbM vkil esfd ld s}kjk la;ksft r gksrsgSa\ (1) MkblYQ kbM
lsrq
(3) Q kLQ ks M kb,LVj
(2) gkbMª ks tu
cU/k
cU/k
(4) lgla ;kst h cU/k
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PAGE # 23
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
Ans. (1) 70.
In which of the following, all three are macronutrients ? (1) Nitrogen, nickel, phosphorus (2) boron, zinc, manganese. (3) Iron,copper, molybdepum (4) Molybdenum, magnesium, manganese
fuEufy f[kr esalsd kS u lHkh rhu c`gÙkiks"kd gSa\ (1) ukbVª kst u, fufd y , Q kLQ ksjl (2) cks jkWu
ft ad , eSaxt hu
(3) y kS g, rkez, eksy hCMsue (4) eks y hCMsue, eSXuhf'k;e, eSaxuht Ans.
(1) or bonus
Sol.
No answer is correct
71.
Which of the following statements is wrong for viroids? (1) Their RNA is of high molecular weight (2) They lack a protein coat (3) They are smaller than viruses (4) They cause infections
fuEufy f[kr esalsd kS u lk d Fku okbjkW ;M d sfo"k; esxy r gS? (1) mud k RNA mPp (2) mues aizksVhu
vkf.od Hkkj oky k gksrk gSA
vkoj.k d k vHkko gksrk gSA
(3) ;sfo"kk.kq v ksalsvis{kkÑ r
NksVsgksrsgSA
(4) ;s alaØ e.k d jrsgSaA Ans.
(1)
Sol.
In viroid, RNA is of low molecular weight
72.
Analogous structures are a result of : (1) Stabilizing selection
(2) Divergent evolution
(3) Convergent evolution
(4) Shared ancestry
leo`fÙk lajpuk,afd l d kj.k mRiUu gksrh gS\ (1) fLFkjd kjh oj.k (3) vfHklkjh fod kl
(2) vilkjh fod kl
ds
ds
(4) lk>k oa 'kijaijk
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PAGE # 24
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
Ans. (3) 73.
Select the incorrect statement: (1) LH triggers secretion of androgens from the Leydig cells (2) FSH stimulates the sertoli cells which help in spermiogenesis (3) LH triggers ovulation in ovary. (4) LH and FSH decrease gradually during the follicular phase
xy r d Fku d kspqfu, (1) LH y hfMx
d ksf'kd kvksals,aM ªkst u d sL=kko d ksiz sfjr d jrk gSA
(2) FSH lVks Zy h d ksf'kd kvksad ksmíhfir (3) LH va M k'k;
d jrk gSt ks'kqØ k.kqt uu eslgk;rk d jrk gSA
esvaM ksRlt Zu d ksizsfjr d jrk gSA
(4) LH vkS j FSH iqVd
voLFkk d snkSjku /khsjs-/khjs?kVrk t krk gSA
Ans.
(4)
Sol.
LH and FSH both increase during follicular phase.
74.
Which one of the following characteristics is not shared by birds and mammals? . (1) Warm blooded nature
(2) Ossified endoskeleton
(3) Breathing using lungs
(4) Viviparity
fuEufy f[kr y {k.kksaesalsd kSu lk ,d y {k.k if{k;ksavkSj Lru/kkfj;ksanksuksaesaughaik;k t krk gS (1) fu;rrkih iz Ñ fr
(2) vfLFkHkw r
var% d ad ky
(3) Q s Q M+ksa}kjk 'olu
(4) lt hoiz t d rk
Ans.
(4)
Sol.
birds are oviparous while mammals are oviparous(prototherians) and viviparous(metatherians
and
eutherians).
75.
Which of the following statements is not correct? (1) Some reptiles have also been reported as pollinators in some plant species. , (2) Pollen grains of many species can germinate on the stigma of a flower, but only one pollen tube of the same species grows into the style. (3) Insects that consume pollen or nectar without bringing about pollination are called pollen/ nectar robbers. (4) Pollen germination and pollen tube growth are regulated by chemical components of pollen interacting with those of the pistil
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| NEET-2016
| 01-05-2016 |
Code-C,R,Y
fuEufy f[kr esal sd kS u lk d Fku lR; ughagS\ (1) d q N
lfjl`i d qN ikni t kfr;ksaeasijkx.k d jrsgq, crk;sx;sgSA
(2) cgq r
lkjh t kfr;ksad sijkxd .k ,d iq"i d sorhZd kxzij vad qfjr gksld rsgaSijUrqmlh t kfr d sijkxd .kksad h
d soy ,d ijkx–ufy d k ofrZd k esavkxsc<+rh gSA (3) d hV
t ksfcuk ijkx.k fd ;sijkx ;k ed jan d ksxzg.k d jrsgSmUgsaijkx/ ed jan pksj d grsgSa
(4) ijkxd .k va d qj.k rFkk ijkx
ufy d k o`f) ] ijkxd .k rFkk L=khd sl j d h ikjLifjd fØ ;k d sQ y Lo: i mRiUu
jklk;fud ?kVd ksa}kjk fu;af=kr gksrh gSA Ans.
(2)
Sol.
76.
Seed formation without fertilization in flowering plants involves the process of : (1) Apomixis
(2) Sporulation
(3) Budding
(4) Somatic hybridization
iq"ih ikniksaesfcuk fu"ksp u d scht cuuk fuEufy f[kr esalsd kSu lh izfØ ;k gSa\
Ans.
(1) vla xt uu
(2) cht k.kq d t uu
(3) eq d qy u
(4) d kf;d
lad j.k
(1)
Sol.
77.
Which of the following approaches does not give the defined action of contraceptive? (1)
Vasectomy
prevents spermatogenesis
(2)
Barrier methods
prevent fertilization
(3)
Intra uterine devices
increase phagocytosis of sperms, suppress sperm motility and fertilizing capacity of sperms
(4)
Hormonal
Prevent/retard entry of
Contraceptives
sperms, prevent ovulation
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PAGE # 26
| NEET-2016
| 01-05-2016 |
Code-C,R,Y and fertilization
fuEufy f[kr mikxeksaesalsd kSu lk mikxe fd lh xHkZfujks/kd d ksifjHkkf"kr ughad jrk? (1)
'kqØ okgd mPNsnu
'kqØ k.kqt uu ughagksusnss rs
(2)
jks/k ¼csfj;j½ fof/k;k¡
fu"ksp u jks d rh gSaA
(3)
var% xHkkZ'k;h ;qfDr;k¡
'kqØ k.kqv ksad h Hk{kd ksf'kd rk c<+k nsrh gS] 'kqØ k.kqv ksd h xfr'khy rk ,oafu"ksp u {kerk d k eanu d jrk gSA
(4)
gkWeksZuh xHkZfujks/kd
'kqØ k.kqv ksad sizos'k d ks jksd rsgS/ mld h nj d ks /khek d j nsrsgS] vaM ksRlxZ vkSj fu"ksp u ughagksusnsrs
Ans
(1)
Sol.
vasectomy causes sterilization by preventing transfer of sperms into semen
78.
The amino acid Tryptophan is the precursor for the synthesis of : (1) Cortisol and Cortisone (2) Melatonin and Serotonin (3) Thyroxine and Triiodothyronine . (4) Estrogen and Progesterone
vehuks avEy fVªIVksQ Su fd ld sla'y s"k.k d sfy , iwoZxkeh gksrk gS\ (1) d ks fVZl ksy
vkSj d ksfVZl ksu
(2) es y kVksfuu
vkSj lsjksVksfuu
(3) Fkk;jkW fDlu (4) bZ LVªkst u
vkSj VªkbZv k;ksM ksFkk;jksfuu
vkSj izkst sLVsjkWu
Ans. (2) Sol.
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PAGE # 27
| NEET-2016 79.
| 01-05-2016 |
Code-C,R,Y
A river with an inflow of domestic sewage rich in organic waste may result in : (1) Death of fish due to lack of oxygen. (2) Drying of the river very soon due to algal bloom. (3) Increased population of aquatic food web organisms. (4) An increased production of fish due to biodegradable nutrients
,d unh esat c d kcZfud vif'k"V lsHkjiw j ?kjsy wokfgr ey cgd j fxjrk gks] rksmld k ifj.kkke D;k gksxk \ (1) vkW Dlht u
d h d eh d sd kj.k eNfy ;kWej t k;saxhA
(2) 'kS oky
izLQ qVu d sd kj.k unh t Ynh gh lw[k t k;sxh A
(3) t y h;
Hkkst u d h lef"V esao`f) gkst k;sxhA
(4) ck;ks fMxzsM scy Ans.
iks"k.k d sd kj.k eNy h d k mRiknu c<+t k;sxkA
(1)
Sol.
80.
Gause's principle of competitive exclusion states that: (1) Larger organisms exclude smaller ones through competition. (2) More abundant species will exclude the less abundant species through competition. (3) Competition for the same resources excludes species having different food preferences. (4) No two species can occupy the same niche indefinitely for the same limiting resources.
Li/khZviot Zu d k xkWl sfu;e d grk gSfd (1) vis {kkÑ r
cM+svkd kj d st ho Li/kkZ}kjk NksVst arqv ks ad ksckgj fud ky nsrsgSaA
(2) vf/kd
la[;k esaik, t kusoky h Lih'kht Li/kkZ}kjk d e la[;k esaik, t kusoky h Lih'kht d ksvioft Zr d j nsxhA
(3) leku
lal k/kuksad sfy , Li/kkZml Lih'kht d ksvioft Zr d j nsxh t ksfHkUu izd kj d sHkkst u ij Hkh t hfor jg
ld rh gSA (4) d ks bZHkh nksLih'kht
,d gh fud sr esvlhfer vof/k d sfy , ughajg ld rh D;kasfd lhekd kjh lal k/ku leku gh
gksrsgSaA Ans.
(4)
Sol. 81.
Asthma may be attributed to : (1) accumulation of fluid in the lungs (2) bacterial infection of the lungs (3) allergic reaction of the mast cells in the lungs (4) inflammation of the trachea
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PAGE # 28
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
vLFkek d k d kj.k D;k gksrk gS\ (1) Q s Q Mksad sHkhrj
ikuh ,d f=kr gkst kuk
(2) Q s Q Mksd k t hok.kq}kjk laØ e.k (3) Q s Q M+ksesekLV
d ksf'kd kvksad h ,yt hZvfHkfØ ;k
(4) 'okluy h d h 'kks Fk Ans.
(3 or 4)
Sol.
asthma is an allergic disease caused by allergens and characterized by inflammation of tracheobronchial tree.
82.
The standard petal of a papilionaceous corolla is also called (1) Corona (2) Carina (3) Pappus (4) Vexillum
iSfify vksusl h oky sny iqa t esekud ny d ksvU; fd l uke lst kuk t krk gS\ (1) d ks jksuk (2) d S fjuk (3) iS il (4) oS Dlhy e Ans.
(4)
Standard or vexillum
Wing or alae Keel or carina Sol.
83.
Papilionaceous corolla
Which of the following is a restriction endonuclease? (1) RNase (2) Hind II (3) Protease
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PAGE # 29
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
(4) DNase I
fuEufy f[kr esalsd kS u lk ,d izfrca /k ,aM ksU;wfDy ,t gSA (1) vkj,u,t (2) fgUn II (3) iz ksfV,t (4) Mh,u,t I Ans.
(2)
Sol.
84.
It is much easier for a small animal to run uphill than for a large animal, because: (1) The efficiency of muscles in large animals is less than in the small animals. (2) It is easier to carry a small body weight. (3) Smaller animals have a higher metabolic rate. (4) Small animals have a lower O2 requirement.
cMs+vkd kj d st arqv ksad eqd kcy sesNksVsvkd kj d st arqv ksad fy , igkM+h ij p<+uk vklku gksrk gSA D;ksafd (1) Nks Vst arqv ksad seqd kcy sescM+st arqv ksad h isf'k;ksad h d k;Z{kerk d e (2) Nks Vs'kjhj (3) Nks V
gksrh gSA
d sHkkj d ksÅ ij y st kuk vis{kkÑ r vklku gksrk gSA
vd kj oky si'kqv ksad h mikip;h nj vis{kkÑ r vf/kd gksrh gSA
(4) Nks Vsvkd kj
d st arqv ksad hO2 vko';d rk vis{kkÑ r d e gksrh gSA
Ans.
(3)
Sol.
smaller animals have higher BMR related with sustained energy production and delayed muscle fatigue
85.
Following are the two statements regarding the origin of life : (a) The earliest organisms that appeared on the earth were non-green and presumably anaerobes. (b) The first autotrophic organisms were the chemoautotrophs that never released oxygen. Of the above statements which one of the following options is correct ? (1) Both (a) and (b) are false. (2) (a) is correct but (b) is false. (3) (b) is correct but (a) is false. (4) Both (a) and (b) are correct.
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PAGE # 30
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
t hou d h mRifÙk d slanHkZesnksd Fku fn;sx, gSa% (a) i` Foh ij (b) iz Fke
izd V gksusoky svkjafHkd Ùke t ho gjsughaFksvkSj laHkor;k vok;oh FksA
izd V gksusoky sLoiks"kh t ho jlksLoiks"kh Fksft UgksausvkWDlht u d k mRlt Zu ughafd ;k A
mijksDr d Fkuksaesalsd kSu lk fuEufy f[kr d Fku lgh gS\ (1) (a) vkS j (b)
nksuksagh xy r gSaA
(2) (a) lgh gSy s fd u (b) xy r
gSA
(3) (b) lgh gSy s fd u (a) xy r
gSA
(4) (a) vkS j (b)
nksuksagh lgh gSaA
Ans.
(4)
Sol.
First originated organism was prokaryote chemoheterotroph and oxygen was not available on earth at that time so it must be anaerobic too. Even the first autotroph was dependent on chemicals so oxygen is not released
86.
A cell at telophase stage is observed by a student in a plant brought from the field. He tells his teacher that this cell is not like other cells at telophase stage. There is no formation of cell plate and thus the cell is containing more number of chromosomes as compared to other dividing cells. This would result in (1) Polyteny (2) Aneuploidy (3) Polyploidy (4) Somaclonal variation
[ksr lsy k;sx, ,d ikni d ksf'kd k esa,d fo|kFkhZ}kjk vaR;koLFkk ns[kh x;hA og viusf'k{kd lsd grk gSfd ;g d ksf'kd k vUR;koLFkk ij vU; d ksf'kd kvksalsfHkUu gSA blesad ksf'kd k Iy sV ugah curh vkSj bl d kj.k bl d ks f'kd k esa vU; foHkkt u oky h d ksf'kd kvksad h vis{kk vf/kd xq.klw=k gSA bld k ifj.kke D;k gksxk\ (1) cgq iV~Vrk (2) vlq xqf.krk (3) cgq xq f.krk (4) d k;Dy ks uh fofHkUurk Ans.
(3)
Sol.
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PAGE # 31
| NEET-2016 87.
| 01-05-2016 |
Code-C,R,Y
Depletion of which gas in the atmosphere can lead to an increased incidence of skin cancers: (1) Methane (2) Nitrous oxide (3) Ozone (4) Ammonia
okrkoj.k esafd l xSl d h d eh gksusij Ropk d sd Sal j d svolj c<+t k,axsa\ (1) ehFks u (2) ukbVª l
vkWDlkbM
(3)vks t ksu (4) veks fu;k Ans.
(3)
Sol.
88.
Joint Forest Management Concept was introduced in India during:
la;qDr ou izcU/ku d h /kkj.k Hkkjr esfd l nkSjku izLrkfor d h x;h Fkh\ (1) 1990s (2) 1960s (3) 1970s (4) 1980s Ans.
(4)
89.
Which one of the following is the starter codon?
fuEufy f[kr esalsd kS u lk ,d izkjEHkd izd wV gS\ (1) UAG
(2) AUG
(3) UGA
(4) UAA
Ans.
(2)
90.
The term ecosystem was coined by : (1) E. Warming
(2) E.P.Odum
(3) A.G. Tansley
(4) E. Haeckel
bd ksflLVe (ikfjrU=k) 'kCn lclsigy sfd luscuk;k Fkk\ (1) bZ . Okfea x
(2) bZ . ih.
(3) ,.
(4) bZ . fgd y
t h. Vkal y s
vksM e
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PAGE # 32
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
Ans. (3)
PART B – PHYSICS 91.
What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop ?
f=kT;k d sfd lh Å /okZ/kj ik'k (y wi) esam nzO;eku d sfd lh fi.M+d ksfd l fuEure osx lsiz os'k d juk pkfg, fd og ik'k d ksiw.kZd j ld s? R
(1)
(2)
5gR
gR
(3)
2gR
(4)
3gR
Ans.
(1)
Sol.
To complete the vertical loop, the minimum speed required at the lowest point =
5gR
So ans is (1) 92.
If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is :
;fn nkslfn'kksad s;sx d k ifjek.k mu nkslfn'kksad svUrj d sifjek.k d scjkcj gS] rksbu lfn'kksad schp d ks.k gSA Ans. Sol.
(1) 180° (3) A B A B
(2) 0°
(3) 90°
(4) 45°
(A)2 + (B)2 + 2(A)(B)cos = (A)2 + (B)2 – 2(A)(B)cos 2cos = 0 93.
= 90° 7
At what height from the surface of earth the gravitational potential and the value of g are –5.4 × 10 J –2 –2 kg and 6.0 ms respectively ? Take the radius of earth as 6400 km.
i`Foh d si`"B lsfd ruh Å ¡p kbZij xq : Roh; foHko vkSj xq: Roh; Roj.k g d seku Ø e'k% – 5.4 × 107 J kg–2 rFkk 6.0 ms
–2
gksrsgS? i`Foh d h f=kT;k 6400 km y hft ,A
Ans.
(1) 2000 km (2)
Sol.
(2) 2600 km
(3) 1600 km
(4) 1400 km
GM 7 = 5.4 × 10 r
GM =6 r2 dividing both the equations, r = 9000 km. so height from the surface = 9000 – 6400 = 2600 km
94.
A long solenoid has 1000 turns. When a current of 4A flows through it, the magnetic flux linked with –3 each turn of the solenoid is 4 × 10 Wb. The self-inductance of the solenoid is
fd lh y Ech ifjukfy d k esa Q sjksa d h la[;k 1000 gSA t c bl ifjukfy d k ls 4A /kkjk izokfgr gksrh gS] rc bl ifjukfy d k d sizR;sd Q sjslslac) pqEcd h; ¶y Dl 4 × 10–3 Wb gksrk gSA bl ifjukfy d k d k Lo&izsjd Ro gS: (1) 1H
(2) 4H
(3) 3 H
(4) 2H
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PAGE # 33
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
Ans.
(1)
Sol.
self = Li –3 (4 × 10 )(1000) = (L)(4) L = 1 Henry
95.
An inductor 20 mH, a capacitor 50 F and a resistor. 40 are connected in series across a source of emf V = 10 sin 340t. The power loss in A.C. circuit is
fd lh L=kksr ft ld k emf, V = 10 sin 340 t gS] lsJs.kh esa20 mH d k izsjd , 50 F d k la/kkfj=k rFkk 40d k izfrjks/kd la;ksft r gS A bl izR;korhZ/kkjk ifjiFk es a'kfDr {k; gSA Ans.
(1) 0.89 W (2)
(2) 0.51 W
(3) 0.67W
(4) 0.76W
L = 6.8
R=400
Sol.
1/c=58.8 So | z | (40)2 (58.8 – 6.8)2 65
i0
v0 10 A | z | 65
irms
i0 2
10 65 2
2
10 2 Ploss rrms R 40 0.46watt 65 2 So the nearest answer will be (2) 96.
Two identical charged spheres suspended from a common point by two mass less strings of lengths , are initially at a distance d(d << ) a part because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres, as : (1) v x
–1
1/2
(2) v x
(3) v x
–1/2
(4) v x
fd lh mHk;fu"B fcUnqls ] y EckbZd h nksæO;ekughu Mks fj;ksalsfuy afcr] nksloZl e vkos f'kr xksy s] vU;ksU; izfrd "kZ.k d sd kj.k] vkjEHk esa,d nwl jslsd(d<< ) nwjh ij gSA nksuksagh xksy ksals,d fu;r nj lsvkos'k d k {kj.k vkjEHk gksrk gS] vkSj bld sifj.kkeLo: i xksy s,d nwl jsd h vksj osx vlsvkrsgS A rc xksy ksad schp d h nwjh] x d sQ y u d s : i esaosx fopj.k fd l : i esagksrk gS?
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PAGE # 34
| NEET-2016 Ans. Sol.
(1) v x (4)
| 01-05-2016 | 1/2
–1
(2) v x
Tcos
(3) v x
Code-C,R,Y –1/2
(4) v x
T
kq2
Tsin
x2
x
mg
Tsin =
kq2 x2
Tcos = mg Dividing the equations tan =
kq2 mgx 2
here tan sin =
x 2
x kq2 = 2 2 x 3/2
qx dq 3 1/ 2 dx x dt 2 dt
dx –1/2 x dt
97.
1 2
V 2F
8F
A capacitor of 2F is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is : (1) 80% (2) 0% (3) 20% (4) 75%
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| NEET-2016 1 2
| 01-05-2016 |
Code-C,R,Y
V 8F
2F
vkjs[k esan'kkZ;svuql kj 2F /kkfjrk d sfd lh la/kkfj=k d k vkos'ku fd ;k x;k gSA t c fLop S d ksfLFkfr 2 ij ?kqek;k t krk gS] rksblesalafpr Å t kZd k izfr'kr {k; gks xkA Ans.
(1) 80% (1)
(2) 0%
Sol.
Initial energy stored in the 2F capacitor is = Energy less = Eloss =
(4) 75%
1 2 2 (2)V = V J 2
2 8 (V – 0)2 C1C2 2 (V1 – V2) = 2 C1 C2 2 2 8
5 2 V J 4
% loss =
98.
(3) 20%
5 4 V2 V2
× 100 = 80%
A particle moves so that its position vector is given by r cos txˆ sin tyˆ . Where is a constant. Which of the following is true? (1) Velocity is perpendicular to r and acceleration is directed away from the origin. (2) Velocity and acceleration both are perpendicular to r . (3) Velocity is acceleration both are parallel to r . (4) Velocity is perpendicular to r and acceleration is directed towards the origin. d ksbZd .k bl izd kj xeu d jrk gSfd mld k fLFkfr lfn'k r cos txˆ sin tyˆ }kjk fu: fir
fd ;k x;k gS] ;gk¡
,d
fu;rkad gS? fuEufy f[kr esalsd kS u lk d Fku lR; gS? (1) os x r
d sy Ecor~gSrFkk Roj.k ewy fcUnqlsnwj d h vksj funsZf'kr gSA (2) os x vkSj Roj.k nksuksagh r d sy Ecor~gSaA (3) os x
(4) os x r Ans.
vkSj Roj.k nksuksagh r d slekUrj gSA d sy Ecor~gSrFkk Roj.k ewy fcUnqd h vksj funsZf'kr gSA
(4)
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Sol.
99.
| NEET-2016 | 01-05-2016 | ˆ V cos t i sin tjˆ dr = – sint ˆi + cost ˆj V dt dV 2 2 = – cost ˆi – sint ˆj a dt since r.V = 0 so r r V and a – 2 r so a will be always aiming towards the origin.
Code-C,R,Y
From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about at perpendicular axis, passing through the centre ?
nzO;eku M rFkk f=kT;k R d h fd lh fMLd lsR O;kl d k d ksbZo`Ùkkd kj fNnzbl izkd j d kVk t krk gSfd mld h usfe fMLd d sd sUnzlsxqt jsA fMCcsd s'ks"k Hkkx d k] fMLd d sy Ecor~mld sd sUnzlsxqt jusoky sv{k d sifjr% t M+Ro vk?kw.kZD;k gS? 2
Ans.
2
(1) 9MR /32 (3)
(2) 15MR /32
2
(3) 13MR /32
2
(4) 11 MR /32
M, R M/4
R/2
Sol. I1 =
MR 2 2 2
M R 2 4 2 3MR 2 M R I2 = 2 32 4 2 Inet = I1 – I2 =
100.
MR 2 3MR2 13MR2 – 2 32 32
so answer is 3.
The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and mean density are twice as that of earth is :
i`Foh ij iy k;u os x (ve) rFkk ml xzg ij iy k;u osx (vp) esaD;k vuqikr gksxk, ft ld h f=kT;k vkSj vkSl r ?kuRo i`Foh d h rqy uk es anksxqusgS? (1) 1 : Ans.
2
(2) 1 : 2
(3) 1 : 2 2
(4) 1 : 4
(3)
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PAGE # 37
| NEET-2016
Sol.
Ve =
2G
2GM R
| 01-05-2016 |
Code-C,R,Y
4 R3 3
R
Ve R
R V1 1 1 V2 R2 2
V1 1 V2 2 2
so answer is 3. 101.
A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf's is :
fd lh fOkHkoekih d srkj d h y EckbZ100 cm gSrFkk bld sfljksad schp d ksbZfu;r foHkokUrj cuk, j[kk x;k gSA nks lsy ksad ksJs.khØ e esaigy s,d nwl jsd h lgj;rk d jrsgq, vkSj fQ j ,d &nwl jsd h foijhr fn'kkvksaesala;ksft r fd ;k x;k gSA bu nksuksaizd j.kksaesa'kwU;&fo{ksi fLFkfr rkj d s/kukRed fljsls50 cm vkSj 10 cm nwjh ij izkIr gksrh gSA nksuksalsy ksad h emf d k vuqikr gS: Ans. Sol.
(1) 3 : 2 (1) E1 + E2 = K(50) E1 – E2 = K(10)
(2) 5 : 1
(3) 5 : 4
(4) 3 : 4
E1 E2 5 E1 E2 1
102.
E1 3 E2 2
A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed –1 of 15 ms . Then, the frequency of sound that the observer hears in the echo reflected from the cliff is : –1 (Take velocity of sound in air = 330 ms ) 800 Hz
vko` fÙk d h /ofu mRiUu d jusoky k d ksbZlk;ju fd lh izs{kd ls,d pV~Vku d h vksj 15 ms–1 d h pky ls
xfreku gSA rc ml /ofu d h vko`fÙk] ft lspV~Vku lsijkofrZr izfr/ofu d s: i esaog izs{kd lwurk gS] D;k gksxh? (ok;qea s/ofu d h pky = 330 ms–1 y hft ,) Ans.
(1) 885 Hz (4)
(2) 765 Hz
(3) 800 Hz
(4) 838 Hz
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PAGE # 38
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
15 m/sec. Sol.
f0
Frequency at the wall will be
v v0 330 0 f ' f0 800 v v 330 15 s
330 f' = 800 = 838 Hz 315 Since the observer and the wall are stationary so frequency of echo observed by the observer will also be 838 Hz. 103.
To get output 1 for the following circuit, the correct choice for the input is : A B
Y
C (1) A = 1, B = 0, C = 1 (3) A = 1, B = 0, C = 0
(2) A = 0, B = 1, C = 0 (4) A = 1, B = 1, C = 0
uhpsfn, x, ifjiFk esa] fuxZr 1 izkIr d jusd sfy , fuos'k d k lgh p;u gS% A B
Y
C
Ans.
(1) A = 1, B = 0, C = 1 (3) A = 1, B = 0, C = 0 (1) A=1 Output = 0 +1=1
(2) A = 0, B = 1, C = 0 (4) A = 1, B = 1, C = 0
Output = 1.1=1
Sol.
B=0 C=1
So ans will be (1)
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PAGE # 39
104.
| NEET-2016 | 01-05-2016 | Code-C,R,Y In a diffraction pattern due to a single slit of width 'a' the first minimum is observed at an angle 30° when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of : –1 3 (1) sin 4
1 4
–1
(2) sin
(3) sin
2 3
–1
–1 2 (4) sin 3
t c pkS M +kbZ'a' d h fd lh ,d y f>jh ij 5000 Å rjaxnS/;Zd k izd k'k vkiru d jrk gS] rksf>jh d sd kj.k mRiUu foorZu iSVuZesa30° d sd ksa.k ij igy k fufEU"B fn[kkbZnsrk gSA igy k f}rh;d mfPP"B ft l d ksa.k ij fn[kkbZnsxk] og gS% 3 (1) sin–1 4
1 4
–1
(2) sin
(3) sin
2 3
–1
–1 2 (4) sin 3
Ans.
(1)
Sol.
Path difference between the extreme rays at first minima = a sin = a sin(30°) =
a = 2
Path difference between the extreme rays at first secondary maxima = a sin' = (2)sin' =
105.
3 2
3 2
–1 3 ' = sin 4
When a metallic surface is illuminated with radiation of wavelength ' the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2, the stopping potential is
V . The threshold 4
wavelength for the metallic surface is : (1) 3
(2) 4
(3) 5
(4)
5 2
t c fd lh /kkfRod i`"B d ksrajxnS/;Zd sfofd j.kksalsiznhIr fd ;k t krk gS] rksfujks/kh foHko V gSA ;fn blh i`"B d ks V 4
rajxnS/;Z2d sfofd j.kksalsiznhIr fd ;k t k,] t ksfujks/kh foHko
gkst krk gSA bl /kkfRod i`"B d h nsgy h rajxnS/;Z
gS% (1) 3 Ans.
(1)
Sol.
KEmax. = eVst = eV =
hc –
(2) 4
(3) 5
(4)
5 2
hc – ...(i)
V hc e – ...(ii) 4 2 Solving equation (i) and (ii)
=
hc hc 3 th
th = 3
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PAGE # 40
| NEET-2016 106.
| 01-05-2016 |
Code-C,R,Y
When an -particle of mass 'm' moving with velocity ' v ' bombards on a heavy nucleus of charge 'Ze' its distance of closet approach from the nucleus depends on m as : (1) m
(2)
1 m
1
(3)
m
(4)
1 m2
t c æO;eku 'm' rFkk osx ' v ' lsxfreku d ksbZ-d .k 'Ze' vkos'k d sfd lh Hkkjh ukfHkd ij ceckjh d jrk gS] rks mld h ukfHkd lsfud Vre mixeu d h nwjh m ij bl izd kj fuHkZj d jrh gS% (1) m Ans. Sol.
(2)
1
(3)
m
(4)
1 m2
(2) At closest approach KE gets converted to PE
1 k(2e)(ze) mV 2 2 r 107.
1 m
m
1 r
or
r
1 m
Match the corresponding entries of column–1 with column–2. [Where m is the magnification produced by the mirror]
d kWy e–1 d h laxr izfof"V;ksad k fey ku d kWy e–2 d h izfof"V;ksalsd hft ;sA [;gk¡m niZ.kksa}kjk mRiUu vko/kZu gSa]
Ans.
(A)
Column–1 m = –2
(B)
m=
Column–2 (a) Convex mirror
(C)
1 2 m = +2
(D)
m=
(A)
m = –2
(B)
m=
(C)
m = +2
(D)
m=
(1)
A c and d;
B b and d;
C b and c; D a and d
(2)
A b and c;
B b and c;
C b and d; D a and d
(3)
A a and c;
B a and d;
C a and b; D c and d
(4)
A a and d;
B b and c;
C b and d; D b and c
(1)
A c o d;
B b o d;
C b o c; D a o d
(2)
A b o c;
B b o c;
C b o d; D a o d
(3)
A a o c;
B a o d;
C a o b; D c o d
(4)
A a o d;
B b o c;
C b o d; D b o c
1 2 1 2
1 2
(b)
Concave mirror
(c)
Real image
(d)
Virtual image
(a)
mÙky niZ.k
(b)
vory niZ.k
(c)
okLrfod izfrfcEc
(d)
vkHkklh izfrfcEc
(2)
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PAGE # 41
Sol.
108.
| NEET-2016 | 01-05-2016 | Code-C,R,Y (A) m = –2, so image is magnified and inverted. Which is possible only for concave mirror. since image is i inverted so it will be real. 1 (B) M = , so image is inverted and diminished. since image is inverted, so it will be real, and the 2 mirror will be concave. (C) M = +2, image is magnified so the mirror will be concave. Image is erect so it will be virtual. 1 (D) m = , image is erect so image will be virtual. Image is virtual and diminished, so the mirror 2 should be convex. Ans. will be (2) A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to –4 8 × 10 J by the end of the second revolution after the beginning of the motion ? 10 g
æO;eku d k d ksbZd .k 6.4 ls-eh– y Ech f=kT;k d so`Ùk d svuqfn'k fd lh fu;r Li'kZ& js[kh; Roj.k lsxfr d jrk
gSA ;fn xfr vkjEHk d jusd si'pkr~nksifjØ ek,saiwjh d jusij d .k d h xfrt Å t kZ8 × 10–4 J gks t krh gS] rks bl Roj.k d k ifjek.k D;k gS\ 2
Ans. Sol.
(1) 0.2 m/s (2)
(2) 0.1 m/s
2
(3) 0.15 m/s
2
2
(4) 0.18 m/s
W all = KE 1 2 (mat)(s) = mv 2 (10 × 10–3)(at) (4 × 6.4 × 10–2) = 8 × 10–4
at = 0.1 m/s m/s Ans. will be (2)
109.
A small signal voltage V(t) = V0 sint is applied across an ideal capacitor C :
d ksbZy ?kqflXuy oksYVrk V(t) = V0 sint fd lh vkn'kZla/kkfj=k C d sfljksaij vuqiz;qDr d h x;h gS: (1) Current (t), leads voltage V(t) by 180° (2) Current (t), lags voltage V(t) by 90° (3) Over a full cycle the capacitor C does not consume any energy from the voltage source. (4) Current (t) is in phase with voltage V(t) (1) /kkjk (t),
oksYVrk V(t) ls180° vxzgSA
(2) /kkjk (t), oks YVrkV(t) ls90° i'p
gSA
(3) ,d
iw .kZpØ esala/kkfj=k C oks YVrk L=kks r lsd ksbZÅ t kZmiHkqDr ughad jrkA (4) /kkjk (t), oks YVrkV(t) d h d y k esagSA Ans. Sol.
(3) Capacitor does not consume energy effectively over full cycles
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PAGE # 42
110.
| NEET-2016 | 01-05-2016 | Code-C,R,Y A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first ?
d sbZfMLd vkSj d ksbZxksy k] ft ud h f=kT;k,saleku ijUrqæO;eku fHkUu gSa] leku mUurka'k vkSj y EckbZd snksvkur lery ksaij y q<+d rsgSaA bu nksuksafi.Mksaesalsry h rd igy sd kSu igq¡p sxk? (1) Depends on their masses (3) Sphere
(2) Disk (4) both reach at the same time
(1) bud sæO;ekuks aij
(2) fMLd
fuHkZj d jrk gS
(3) xks yk Ans. Sol.
(4) nks uksa,d
k2 t 1 2 R k2 R2 111.
gh le; igq¡p saxs
(3) Time does not depend on mass, else
is least for sphere and hence least time is taken by sphere
Coefficient of linear expansion of brass and steel rods are 1 and 2 . Lengths of brass and steel rods are 1 and 2 respectively. If (2 – 1) is maintained same at all temperatures, which one of the following relations holds good ?
ihry ¼czkl½ vkSj LVhy d h NM+ksad svuqnS/;Zizl kj d sxq.kkad Ø e'k% 1 vkSj 2 gSaA ihry vkSj LVhy d h NM+ksad h y EckbZ;k¡Ø e'k% 1 vkSj 2 gSaA ;fn (2 – 1) d kslHkh rkiksad sfy , leku cuk;k t k;s] rc uhpsfn, x, laca/kksaesals d kSu&lk lR; gS\ (1) 11 = 2 2 Ans.
(1)
Sol.
2 = 2 (1 + 2())
(2) 12 = 2 1
(3) 122 = 2 12
(4) 122 = 22 1
1 = 1(1 + 1()) 2– 1 =(2 – 1) + (22 – 11) As the length difference is independent of temperature difference hence 12 – 11 = 0 112.
22 = 11
A astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance :
fd lh [kxksy h; nwjchu d svfHkn`';d vkSj usf=kd k d h Q ksd l nwfj;k¡Ø e'k% 40 cm vkSj 4 cm gSaA vfHkn`';d ls 200 cm nw j Ans.
(1) 54.0 cm (1)
fLFkr fd lh fcEc d ksns[kusd sfy ,] nksuksay sal ksad schp d h nwjh gksuh pkfg, : (2) 37.3 cm
(3) 46.0 cm
(4) 50.0 cm
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PAGE # 43
Sol.
| NEET-2016 Tube length = v0 + fe
| 01-05-2016 |
Code-C,R,Y
1 1 1 – V0 u 0 f
for objective
put u0 = –200 and f = 40 cm we get v0 = 50 cm L = 54 cm
113.
A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. it is subjected to a torque which produces a constant angular –2 –2 acceleration of 2.0 rad s . Its net acceleration in ms at the end of 2.0 s is approximately :
fojkekoLFkk esafLFkr 50 cm f=kT;k d h d ksbZ,d leku o`Ùkkd kj fMLd viusry d sy Ecor~vkSj d sUæ lsxqt jusoky s v{k d sifjr% ?kweusd sfy , Lora=k gSA bl fMLd ij d ksbZcy vk?kw.kZd k;Zd jrk gS] t ksblesa2.0 rad s–2 d k fu;r d ks.kh; Roj.k mRiUu d j nsrk gSA 2.0 s d si'pkr~ms–2 esabld k usV Roj.k gksxk y xHkx: Ans. Sol.
(1) 3.0 (2) 8.0 (3) 7.0 (4) 6.0 (2) The angular speed of disc increases with time, and hence centripetal acceleration also anet = ac =
a 2t a 2c
2 R
= tangential speed R = Radius = 0.5 m V = 2m/s at t =2
114.
ac = 8m/s ; at = R = (0.5)(2)
anet =
8 2 12 ~ 8
A refrigerator works between 4°C and 30°C. it is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is : (Take 1 cal = 4.2 Joules)
d ksbZjs fÝ t jsVj 4°C vkSj 30°C d schp d k;Zd jrk gSA iz'khru fd , t kusoky sLFkku d k rki fu;r j[kusd sfy , 600 d S y ksjh Å "ek d ksizfr
lsd .M ckgj fud ky uk vko';d gksrk gSA bld sfy , vko';d 'kfDr pkfg, %:
(1 cal = 4.2 Joules y hft ;s ) Ans.
(1) 2365 W (4)
Sol.
C.O.P. =
(2) 2.365 W
(3) 23.65 W
(4) 236.5 W
T2 Heat extracted ; (T2 < T1) effort put T1 – T2
for 1 second analysis
(600 )( 4.2) 277 Effort put 26
Effort put = 236.5 J
Power = 236.5 watt
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PAGE # 44
115.
| NEET-2016 | 01-05-2016 | Code-C,R,Y A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then : (1) Which of the case (whether compression through isothermal or through adiabatic process) requires more work will depend upon the atomicity of the gas (2) Compressing the gas isothermally will require more work to be done (3) Compressing the gas through adiabatic process will require more work to be done (4) Compressing the gas isothermally or adiabatically will require the same amount of work
fd lh xSl d kslerkih; : i lsmld svk/ksvk;ru rd laihfM+r fd ;k t krk gSA blh xSl d ksi`Fkd : i ls: n~/kks"e izfØ ;k }kjk mld svk/ksvk;ru rd laihfM+r fd ;k t krk gSA rc : (1)
pkgslerkih; izfØ ;k }kjk laihfMr d jsavFkok : n~/kks"e izfØ ;k }kjk laihfMr d jsa] fd l izd j.k esavf/kd d k;Z
d jusd h vko';d rk gksxh] ;g xSl d h ijek.kqd rk ij fuHkZj d jsxkA (2) xS l d kslerkih; izfØ ;k }kjk laihfMr d jusesavf/kd d k;Zd jusd h vko';d rk gksxhA (3) xS l
d ks: n~/kks"e izfØ ;k }kjk la ihfMr d jusesavf/kd d k;Zd jusd h vko';d rk gksxhA (4) xS l d kslerkih; izfØ ;k vFkok : n~/kks"e izfØ ;k nksuksaesagh leku d k;Zd jusd h vko';d rk gks xhA Ans. Sol.
(3) Directly from graph the magnitude of work done = Area under p-v plot is larger for adiabatic compression p Adiabatic Iso thermal
vf
116.
v
vi
The intensity at the maximum in Young's double slit experiment is 0. Distance between two slits is d = 5, where is the wavelength of light used is the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d ?
;ax d sfd lh f} f>jh iz;ksx esamfPp"B d h rhozrk 0 gSA nksuks af>fj;ksad schp d h nwjh d = 5gS, ;gk¡iz;ksx esa mi;ksx fd , x, izd k'k d h rjaxnS/;ZgSA fd lh f>jh d slkeusnwjh D = 10d ij fLFkr insZij rhozrk D;k gksxh? (1)
0 2
(2) 0
Ans.
(1)
Sol.
y max cos2
(3)
0 4
(4)
3 0 4
D 10 d
y for a position in front of a slit
5 2.5 2 2 0 cos2 4
(2.5 ) 0 cos2 10
0 2
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PAGE # 45
| NEET-2016 117.
| 01-05-2016 |
Code-C,R,Y
Two non-mixing liquids of densities and n(n > 1) are put in container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL (p < 1) in the denser liquid. The density d is equal to
,d nwl jsesafefJr u gks usoky snksnzo] ft ud s?kuRo rFkk n(n > 1) gSa] fd lh ik=k esaHkjsgSA izR;sd nzo d h Å ¡p kbZh gSA y EckbZL vkSj ?kuRo d d sfd lh csy u d ksbl ik=k esaj[kk t krk gSA ;g csy u ik=k esabl izd kj jSrjk gS] fd bld k v{k Å /okZ/kj jgrk gSrFkk bld h y EckbZpL (p < 1) l?ku nzo esagksrh gSA ?kuRo d d k eku gSA Ans.
(1) {1 + (n – 1)p} (1)
(2) {1 + (n + 1)p}
(3) {2+(n + 1)p}
(4) {2 + (n – 1)p}
L–pL pL
Sol.
n wt of body = upthrust by the two liquids If A = Area of section then (d A.L) g = [A (L – pL) + n ApL] g On solving d = (1+ (n – 1)p)
118.
Consider the junction diode as ideal. The value of current flowing through AB is :
laf/k Mk;ksM d ksvkn'kZekud j fopkj d hft ,A AB lsizokfgr /kkjk d k eku gS% A +4V (2) 0 A
–3
Ans. Sol.
(1) 10 A (3) For diode as ideal i=
119.
1k
B –6V
(3) 10
–2
A
–1
(4) 10 A
V 4 – (–6 ) –2 = 10 A 3 R 10
A car is negotiating a curved road of radius R. The road is banked at an angle . The coefficient of friction between the tyres of the care and the road is s. The maximum safe velocity on this road is:
d ksbZd kj f=kT;k R d h ofØ r lM+d ij xfreku gSA ;g lM+d d ks.k ij >qd h gSAd kj d sVk;jksavkSj lM+d d schp ?k"kZ.k xq.kkad s gSabl lM+d ij d kj d k vf/kd re lqj{kk osx gS% (1) Ans.
s tan R 1 – s tan g
2
(2)
gR 2
s tan (3) 1 – s tan
gR
s tan 1 – s tan
(4)
g s tan R 1 – s tan
(3)
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PAGE # 46
Sol.
| NEET-2016 | 01-05-2016 | For maximum speed the tendency of body is to slip up the incline hence
2 Vmax tan = Rg 1 – tan
tan Rg 1 – tan
or Vmax =
120.
Code-C,R,Y
A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B, at a radial distances
a and 2a respectively, 2
from the axis of the wire is:
f=kT;k a d sfd lh y Ecslh/ks rkj ls d ks bZLFkk;h /kkjk I çokfgr gks jgh gSA bl rkj d h vuqçLFk d kV ij /kkjk ,d leku : i lsforfjr gSA rkj d sv{k lsf=kT;k nwfj;ksa
a 2
vkSj 2a ij Ø e'k% pqEcd h; {ks=kksaB vkSj B d k vuqikr
gS& (1) 4 Ans. Sol.
(2)
1 4
(3)
1 2
(4) 1
(4) If r = radial separation
0i R i 1 i B = Binside = 0 2 r = = 0 2 2 2 R 2 ( 2 R ) 2 R B = Boutside =
121.
0i i 1 = 0 B : B = 1 : 1 2r 2R 2 7
–1
Given the value of Rydberg constant is 10 m , the wave number of the last line of the Balmer series in hydrogen spectrum will be :
fjMcxZfu;rkad d k eku 107 m–1 fn;k x;k gS] gkbMªkst u LisDVªe d h ckej Js.kh d h vafUre y kbu d h rjax la[;k gksxh % 7
Ans. Sol.
122.
(1) 2.5×10 m (4)
–1
1 1 1 R 2 – 2 2
4
(2) 0.025 ×10 m
–1
wave number =
7
(3) 0.5 ×10 m
–1
7
(4) 0.25×10 m
–1
107 m –1 4
2
If the velocity of a particle is = At + Bt , where A and B are constants, then the distance travelled by it between 1s and 2s is
;fn fd lh d .k d k osx = At + Bt2 gS;gk¡A rFkk B fLFkjkad gS] rksbl d .k }kjk 1s vkSj 2s d schp py h x;h nwjh gS% (1) Ans.
A B 2 3
(2)
3 A B 2
(3) 3A + 7B
(4)
3 7 A B 2 3
(4)
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PAGE # 47
| NEET-2016 Sol.
| 01-05-2016 |
Code-C,R,Y
Distance 2
2
s vdt At Bt 2
1
123.
1
3A 7B 2 3
The angle incidence for a ray of light at a refracting surface of a prism is 45º. The angle of prism is 60º. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are
fçTe d sfd lh viorZd i`"V ij fd lh çd k'k fd j.k d sfy , viru d ks.k d k eku 45º gSA fçTe d ks.k d k eku 60º gSA ;fn ;g fd j.k fçTe lsU;wure fopfy r gksrh gS] rksU;wure fopy u d ks.k rFkk fçTe d sinkFkZd k viorZukad Ø e'k% gS% (1) 30º;
1
(2) 45º ;
2 Ans. Sol.
1 2
(3) 30º ;
2
(4) 45º ;
2
(3) Give A = 60 and i = e = 60 min = i + e – A = 45 + 45 – 60 = 30
A sin m 2 2 A sin 2 124.
5
–1
–2
The molecules of a given mass of a gas have r.m.s. velocity of 200 ms at 27ºC and 1.0×10 Nm 5 2 pressure. When the temperature and pressure of the gas are respectively, 127ºC and 0.05×10 Nm , –1 the r.m.s. velocity of velocity of its molecules in ms is ;
rki 27ºC rFkk 1.0×105 Nm–2 ij fd lh fn, x, æO;eku d h xSl d sv.kqv ks ad k oxZek/; ewy (r.m.p.) osx 200 gSA t c bl xSl d srki vkSj nkc Ø e'k% 127ºC vkSj 0.05×105 Nm2 gS] rksms–1 esabl xSl d sv.kqv ksad k oxZek/; ewy osx gS & ms
(1)
–1
100 3
Ans.
(3)
Sol.
VRMS V1 V2
200 V2
2
(2) 100
(3)
400
(4)
3
100 2 3
3RT Mo T1 T2
300 400
V2
400 3
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PAGE # 48
125.
| NEET-2016 | 01-05-2016 | Code-C,R,Y An air column, closed at one end and open at the other, resonantes with a tuning fork when the smallest length of the column is 50 cm. The next larger length of the column resonating with the same tuning fork is:
,d fljsij cUn rFkk nwl jsfljsij [kqy k d ksbZok;qLrEHk fd lh Lofj=k f}Hkqt d slkFk ml le; vuqukn d jrk gS t c bl ok;qLrEHk d h d e lsd e yEckbZ50 lseh gks rh gSA blh Lofj=k f}Hkqt d slkFk vuqukn d jusoky h LrEHk d h vxy h cM+h y EckbZgS& Ans.
(1) 200 cm (4)
(2) 66.7 cm
Sol.
(4) 150 cm
4
First harmonic at
3rd harmonic
(3) 100 cm
3 4
1st length = 50 cm 3rd harmonic length 150 cm 126.
The magnetic susceptibility negative for (1) paramagnetic and ferromagnetic materials (3) paramagnetic material only
(2) diamagnetic material only (4) ferromagnetic material only
pqEcd h; lqxzkfgrk _ .kkRed gksrh gS% (1) vuq p qEcd h; (3)
vkSj y kSg&pqEcd h; inkFkksZd sfy ,
d soy vuq p qEcd h; inkFkZd sfy ,
(2)
d soy çfrpqEcd h; inkFkZd sfy ,
(4) d s oy
y kSg&pqEcd h; inkFkZd sfy ,
Ans.
(2)
Sol.
r = 1 + x appropriate is diamagnetic
127.
An electron of mass m and a photon have same energy E. The ratio of de-Brogli wavelengths associated with them is:
æO;eku m d sby sDVªkWu rFkk fd lh Q ksVkWu d h Å t kZ, E ,d leku gSA buesalac) nsczkXy h rjaxnS/;ksZd k vuqikr gS%
Ans.
1
1
1
1 2m 2 (1) c E
1 E 2 (2) c 2m
E 2 (3) 2m
(4) c(2mE )1 / 2
(2)
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PAGE # 49
| NEET-2016 Sol.
electron = For
h
| 01-05-2016 |
Code-C,R,Y
…(1)
2ME
photon
E = h =
hc photon
…(2)
from these two ratio obtained by dividing these (2) 1 : 2 =
128.
1 E c 2M
1/ 2
A body of mass 1 kg begins to move under the action of a time dependent force F (2t ˆi 3 t 2 ˆj )N , when ˆi and ˆj are unit vectors along x and y axis. What power will be developed by the force at the time t ? 1 kg æO;eku d k d ks bZfi.M fd lh d ky fJr cy F (2t ˆi 3t 2 ˆj)N , ;gk¡ ˆi rFkk ˆj , x vkSj y v{k d svuqfn'k ek=kd
lfn'k gS] d sv/khu xfr vkjEHk d jrk gS] rksle; t ij bl cy }kjk fod flr 'kfDr D;k gksxh? 3
Ans. Sol.
5
(1) (2t + 3t )W (1) M = 1 kg
2
3
2
(2) (2t + 3t )W
a=
F 2t ˆ 3 t 2 ˆ i j M (1) 1
V=
adt = 2t dt 1 + 3t dt
4
(3) (2t + 4t )W
3
4
(4) (2t + 3t )W
2
2 3 V = t ˆi + t ˆj
Power = F.V. = (2ti + 3t2 ˆj ). (t2 i + t3 ˆj ) power = 2t3 + 3 t5 129.
2
The charge flowing through a resistance R varies with time t as Q = at – bt , where a and b are positive constants. The total heat produced in R is :
fd lh çfrjks/k R lsçokfgr vkos'k d k le; t d slkFk fopj.k Q = at – bt2 d s: i esagksrk gS] t gk¡a rFkk b /kukRed fu;rkad gSA R esamRié d qy Å "ek gS% (1) Ans. Sol.
a 3R b
(2)
a 3R 6b
(3)
a 3R 3b
(4)
a 3R 2b
(2) 2 Q = at – bt i=
dQ = a – 2bt dt t
H=
0
a 3R i Rdt = 6b 2
i=0t=
a 2b
t 2
Rdt 0
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PAGE # 50
| NEET-2016
| 01-05-2016 |
a/2b
it
(a – 2bt)2 Rdt = a2 t
0
Put t
130.
a 2b
H=
2 3
4b t 4bat – 3 2
Code-C,R,Y
2
a3 R 6b
A npn transistor is connected in common emittet configuration in a given amplifier. A load resistance of 800 is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuit is 192 , the voltage gain and the power gain of the amplifier will respectively be :
fd lh fn, x, ço/kZd esad ksbZnpn Vªkaft LVj mHk;fu"B mRlt Zd foU;kl esala;ksft gSA 800 d ksd ks bZy ksM çfrjks/k laxzkgd ifjiFk esals;ksft r gSvkSj bld sfljksaij 0.8 V foHkoikr gSA ;fn /kkjk ço/kZd xq .kkad 0.96 gSA rFkk ifjiFk d k fuos'k çfrjks/k 192 gSA rksbl ço/kZd d h oksYVrk y fC/k rFkk 'kfDr y fC/k Ø e'k% gksxh% Ans. Sol.
(1) 4,3.69 (2) 4,3.84 (2) Voltage gain = [current gain] [resistance gain] [.96]
(3) 3.69, 3.84
(4) 4, 4
800 192
power gain = [current gain] [resistance gain] [.96] [4] = 3.84 131.
a piece of ice falls from a height h so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted in to heat during its fall. The value of h is : 5 [Latent heat of ice is 3.4 × 10 J/Kg and g = 10 N/kg] (1) 68 km (2) 34 km (3) 544 km (4) 136 km
cQ Zd k d ksbZVqd Mk Å ¡p kbZ h lsbl izd kj fxjrk gSfd og iw.kZr% fi?ky t krk gSA mRiUu gksusoky h m"ek d k d soy ,d &pkSFkkbZ Hkkx gh cQ Z }kjk vo'kksf"kr fd ;k t krk gS rFkk cQ Z d h leLr Å t kZ bld s fxjrs le; Å "ek esa : ikUrfjr gkst krh gSA ;fn cQ Zd h xqir Å "ek 3.4 × 105 J/Kg rFkk g = 10 N/kg gSa] rksÅ ¡p kbZh d k eku gS% Ans. Sol.
(1) 68 km (4)
(2) 34 km
(3) 544 km
(4) 136 km
Mg h = mL 4 h=
4L = 136 km g
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PAGE # 51
132.
| NEET-2016 | 01-05-2016 | Code-C,R,Y A square loop ABCD carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be :
d ksbZoxkZd kj ik'k (y wi) ABCD ft lls/kkjk i, iz okfgr gksjgh gS] fd lh y Ecslh/kspky d XY ft lls/kkjk I izokfgr gksjgh gSd sfud V ,d gh ry esaj[kk gSaA bl ik'k ij y xusoky k y sV cy gks xk % Y
L
i
X
Ans.
0 iL 2
(2)
20 i 3
D
A L/2
(1)
C
B
L
(3)
0 i 2
(4)
20 iL 3
(2) FBC C
B FAB
FCD
D
A FAD
Sol.
FBC cancels FAD FNet = FAB – FCD =
133.
2 0 Ji 0 iL iL 2 0 i – 0 = = 3 3 3L L 2 2 2 2
A uniform rope of length L and mass m1 hangs vertically from a rigid support. A block of mass m2 is attached to the free end of the rope. A transverse pulse of wavelength 1 is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is 2. The ratio 1/2 is : (1)
m1 m2 m1
(2)
m1 m2
(3)
m1 m2 m2
(4)
m2 m1
nzO;eku m1 rFkk y EckbZL d h d ksbZ,d leku jLlh fd lh n`< Vsd lsÅ /okZ/kj y Vd h gS A bl jLlh d seqDr fljsls nzO;eku m2 d k d ksbZxqVd k t qM k gSA jLlh d seqDr fljsij rjaxnS/;Z1 d k d ksbZvuqiz LFk LiUn mRiUu fd ;k t krk gSA ;fn jLlh d s'kh"kZrd igq¡p usij bl LiUn d h rjaxnS /;Z2 gkst krh gSA rc vuikr 1/2 d k eku gS% (1) Ans.
m1 m2 m1
(2)
m1 m2
(3)
m1 m2 m2
(4)
m2 m1
(3)
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PAGE # 52
| NEET-2016 Sol.
| 01-05-2016 |
Code-C,R,Y
T m/
v 1
M2
2
M2 M1 Tension = M2g
Tension = M2g
T2 = (M1 + M2)g M1
T1 = M2g M2 2 1 134.
M1 M2 M2
A block body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 6 250 nm is U1 at wavelength 500 nm is U2 and that at 1000 nm is U3. Wien's constant, b = 2.88 × 10 nmK. Which of the following is correct ? (1) U2 > U1 (2) U1 = 0 (3) U3 = 0 (4) U1 > U2
d ksbZd `f".kd k 5760 K rki ij gSaA bl fi.M }kjk mRlft Zr fofd j.kksad h Å t kZ] rjaxnS/;Z250 nm ij U1 rja xnS/;Z 500 nm ij U2 rFkk rja xnS/;z1000 nm ij U3 gSA
ohu&fu;rkad ] b = 2.88 × 106 nmK gSA uhpsfn;k x;k d kSu lk
lac/k lgh gS\ Ans.
(1) U2 > U1 (1)
Sol.
min T = b
(2) U1 = 0
(3) U3 = 0
(4) U1 > U2
1 T 4
u (T)
1 ( )4
so u1 > u2 135.
Out of the following options which one can be used to produce a propagating electromagnetic wave ? (1) An accelerating charge (2) A charge moving at constant velocity (3) A stationary charge (4) A chargeless particle
uhpsfn, x, fod Yiksaesalsfd ld k mi;ksx ,d lap fjr fo|qr pqEcd h; rjax mRiUu d jusesafd ;k t k ld rk gS\ (1) d ks bZRofjr vkos'k (2) fu;r os x lsxfreku d ksbZvkos'k (3) fLFkj Ans.
vkos'k
(4) vkos 'kghu
d .k
(1)
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PAGE # 53
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
PART C – CHEMISTRY . 136.
Which one of the following characteristics is associated with adsorption ? (1) G and S are negative but H is positive (2) G is negative but H and S are positive (3) G, H and S all are negative (4) G and H are negative but S is positive
fuEUkfy f[kr y {k.kksaesalsd kSu lk vf/k'kks"k.k lslEcfU/kr gS\ (1) G rFkkS _
.kkRed y sfd u H /kukRed gksrk gSA
(2) G _
.kkRed y sfd u H ,oaS /kukRed gksrsgSA (3) G, H ,oaS lHkh _ .kkRed gks rsgSA (4) G ,oaH _
.kkRed y sfd u S /kukRed gksrk gSA
Ans.
(3)
Sol.
According to Gibbs Helmholtz equation, G = H TS Adsorption is a spontaneous process (where S < 0, G < 0 and H < 0)
fxCl gsYegkWV~t lehd j.k d svuql kj G = H T S vf/k'kks"k.k ,d Lor% izØ e gksrk gS(t gk¡ S < 0, G < 0 rFkk H < 0) 137.
The pressure of H2 required to make the potential of H2-electrode zero in pure water at 298 K is : –4 (1) 10 atm –14 (2) 10 atm –12 (3) 10 atm –10 (4) 10 atm 298 K ij
'kq ) t y esaH2 by sDVªksM d k foHko 'kwU; d jusd sfy ;svko';d H2 d k nkc gSA
–4
Ans.
(1) 10 atm (2) 10–14 atm (3) 10–12 atm (4) 10–10 atm (2)
Sol.
2H
+
+ 2e H2 (reduction reaction) (vip;u –
(aq.)
E = Eº –
0=0–
vfHkfØ ;k)
PH2 0.059 log 2 2 H(aq.)
PH2 0.059 log 2 2 10 7
(In order to make log1 = 0) (log1 = 0 d jusd sfy ,) -7 2
PH2 = (10 )
= 10–14 atm
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PAGE # 54
138.
| NEET-2016 | 01-05-2016 | Code-C,R,Y The addition of a catalyst during a chemical reaction alters which of the following quantities ? (1) Activation energy (2) Entropy (3) Internal energy (4) Enthalpy
fd lh jklk;fud vfHkfØ ;k esamRis zsjd d s;ksx lsfuEUkfy f[kr esalsd kSu lh ek=kk cny rh gSA
Ans. Sol.
(1) lfØ ;.k Å t kZ
(2) ,s UVªkWih
(3) vka rfjd
(4) ,s aFkSYih
Å t kZ]
(1) Catalyst can affect only activation energy of the chemical reaction and cannot alter any thermodynamic parameters : (ie. H , G , S )
mRizsjd d soy jklk;fud vfHkfØ ;k d h lfØ ;.k Å t kZd ksizHkkfor d jrk gS] vU; Å "ek xfrd h; ekin.M d ksugha% (vFkkZ r~H , G , S ) 139.
For the following reaction : (a) CH3CH2CH2Br + KOH CH3CH=CH2 + KBr + H2O
CH3
(b) H3C
Br (c)
H3C
CH3 + KBr
+ KOH
OH
Br + Br2
Br Which of the following statements is correct ? (1) (a) is substitution, (b) and (c) are addition reactions. (2) (a) and (b) are elimination reactions and (c) is addition reaction. (3) (a) is elimination, (b) is substitution and (c) is addition reaction. (4) (a) is elimination, (b) and (c) are substitution reaction.
fuEu vfHkfØ ;kvksad sfy ;s% (a) CH3CH2CH2Br + KOH CH3CH=CH2 + KBr + H2O
CH3
(b) H3C
Br (c)
H3C + KOH
CH3 + KBr OH
Br + Br2
Br
fuEu esalsd kSu lk d Fku lR; gS? (1) (a) iz frLFkkiu, (b) vkSj (c) ;ksXkt vfHkfØ ;k,¡gSA (2) (a) vkS j (b) foy ksiu
vfHkfØ ;k,¡gSrFkk (c) ;ks xt vfHkfØ ;k gSA (3) (a) foy ks iu vfHkfØ ;k, (b) izfrLFkkiu vfHkfØ ;k vkSj (c) ;ksxt vfHkfØ ;k gSA (4) (a) foy ks iu Ans.
vfHkfØ ;k, (b) vkSj (c) izfrLFkkiu vfHkfØ ;k,¡gSA
(3)
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PAGE # 55
| NEET-2016 Sol.
| 01-05-2016 |
Code-C,R,Y
(a) CH3CH2CH2Br + KOH CH3CH=CH2 + KBr + H2O Elimination reaction Formation of -bond and conversion of saturated compound into unsaturated compound by the removal of groups or atoms is known as Elimination reaction
CH3
(b) H3C
Br
CH3
H3C
Substitution Reaction
+ KBr
+ KOH
OH
Replacement of one group by other group known as Substitution Reaction Br (c) addition reaction + Br2 Br Conversion of unsaturated compound into saturated compound by the addition of groups or atoms is called as addition reaction. (a) CH3CH2CH2Br + KOH CH3CH=CH2 + KBr + H2O foy ks iu -ca /k d k fuekZ.k rFkk ijek.kq;k lewgksad sgVusij
vfHkfØ ;k
lar`Ir ;kSfxd d k vlar`Ir ;kSfxd esa: ikUrj.k d ksfoy ksiu
vfHkfØ ;k d grsgSA CH3
(b) H3C
+ KOH
Br
iz frLFkkiu
CH3
H3C
vfHkfØ ;k
+ KBr OH
fd lh lewg d k nwl jslewg }kjk izfrLFkkiu gksusij mlsizfrLFkkiu vfHkfØ ;k d grsgSA Br
(c)
+ Br2
;ks xt
vfHkfØ ;k
Br
ijek.kq;k lewgksad st qM +usij vlar`Ir ;kSfxd d slar`Ir ;kSfxd esa: ikUrj.k d ks;ks xt vfHkfØ ;k d grsgSA 140.
The product formed by the reaction of an aldehyde with a primary amine is : (1) Aromatic acid (2) Schiff base (3) Ketone (4) Carboxylic acid
,sfYMgkbM ,oaizkFkfed ,sehu d h vfHkfØ ;k lscuk mRikn gSA: (1) ,s jksesfVd vEy (2) f'kQ ~{kkj (3) d hVks u Ans. Sol.
(4) d kcks ZfDlfy d
vEy
(2)
C=O + H2N–R
C=N–R
Aldehyde/ketone
Schiff's Base
C=O + H2N–R
C=N–R
,fYMgkbM/d hVksu
f'kQ {kkj
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PAGE # 56
141.
| NEET-2016 | 01-05-2016 | Code-C,R,Y The correct statement regarding the basicity of arylamines is : (1) Arylamines are generally more basic than alkylamines, because the nitrogen atom in arylamines is sp-hybridized. (2) Arylamines are generally less basic than alkylamines because the nitrogen lone pair electrons are delocalized by interaction with the aromatic ring electrons system. (3) Arylamines are generally more basic than alkylamines because the nitrogen lone pair electrons are not delocalized by interaction with the aromatic ring electron system. (4) Arylamines are generally more basic than alkylamines because of aryl group
,sjhy ,sehu d s{kkjd rk d sfy ;slgh d Fku gS: (1) ,s jhy ,sehu
lkekU;r% ,sfYd y ,sehu lsT;knk {kkjh; gSD;ksfd ,sjhy ,sehu esaukbVªkst u ijek.kqsp-lad fjr gSA
(2) ,s jhy ,sehu
lkekU;r% ,sfYd y ,sehu lsd e {kkjh; gSD;ksfd ukbVªkst u d s,d kd h ;qXe by sDVªks u ,sjksesfVd oy ; d s by s DVªkWu d slkFk foLFkkfud `r gksrsgSA
(3) ,s jhy ,sehu
lkekU;r% ,sfYd y ,sehu lsT;knk {kkjh; gksrh gSA D;ksfd ukbVªkst u d s,d kd h-;qXe by sDVªksu ,sjksesfVd
oy ; d s by sDVªksu d slkFk foLFkkfud `r ugh gksrsgSA (4) ,s fjy Ans.
lewg d sd kj.k ,sjhy ,sehu lkekU;r% ,sfYd y ,sehu lsT;knk {kkjh; gSA
(2)
NH2
Sol.
Delocalised lone pair of nitrogen atom with Benzene ring in aryl amine
aryl amine
lone pair of electrons of nitrogen atom are not delocalized in alkyl amine. RNH2 (Alkyl amine)
NH2
,fjy ,ehu esaukbVªks t u d sby sDVªkWu ;qXe csUt hu oy ; d slkFk foLFkkfud `r gksrsgSA
,fjy ,ehu
RNH2 (,fYd y
142.
,fYd y ,ehu esaukbVªkst u d sby sDVªkWu ;qXe foLFkkfud `r ughagksrsgSA
,ehu)
Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape ? (1) 1/2 (2) 1/8 (3) 1/4 (4) 3/8
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PAGE # 57
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
gkbMªkst u ,oavkWDlt hu xSl ksad sleku eksy ksd ks,d ik=k esaj[kk x;k gSA t ksfd lw{e fNnzd s}kjk iy k;u d j ld rsgSA gkbMªkst u d svk/ksiy k;u esay xsle; esavkWDlht u d k fd ruk va'k iy k;u d jsxkA?
Ans. Sol.
(1) ½ (2) 1/8 (3) ¼ (4) 3/8 (2) Equal moles are given so partial pressure is equal (let = x)
leku eksy fn;sx;sgSblfy , vkaf'kd nkc leku gS(ekuk fd x gS) rO2 rH2
=
nO2 / t x /t 2 nO2 / t x /t 2
143.
MH2 MO2 1 2 = 4 32
=
=
1 4
nO2 x
=
1 8
fraction of oxygen escaped =
1 . 8
iy k;u gq, vkW Dlht u d k va'k =
1 8
The correct statement regarding the comparison of staggered and eclipsed conformations of ethane, is: (1) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain. (2) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain. (3) The eclipsed conformation of ethane is more stable than staggered conformation, because eclipsed conformation has no torsional strain. (4) The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain.
,Fksu d slkarfjr ,oaxz Lr la: i.k d h rqy uk d sfy ;slgh d Fku gS& (1)
,Fksu d k lkarfjr la: i.k] xzLr la: i.k ls vf/kd LFkk;h gS D;ksafd lkarfjr la: i.k es a ,saBu ruko ¼ ejks M +h
fod `rh½ ughagSA (2)
,Fksu d k lkarfjr la: i.k] xzLr la: i.k lsd e LFkk;h gSD;ksafd lkarfjr la: i.k esa,saBu ruko ¼ ejksM +h fod `rh½
gSA (3)
,Fksu d k xzLr la: i.k] lkarfjr la: i.k lsvf/kd LFkk;h gSD;ksafd xzLr la: i.k esa,saBu ruko ¼ ejksM +h fod `rh½
ughagSA (4)
,Fksu d k xzLr la: i.k] lkarfjr la : i.k lsvf/kd LFkk;h gSt cfd xzLr la: i.k esa,saBu ruko ¼ ejksM +h fod `rh½
gSA Ans.
(1)
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PAGE # 58
| NEET-2016 H
H
H
H
| 01-05-2016 |
Code-C,R,Y
H
Sol. H
H H
H
H Staggered Newmann conformation
H H
Eclipsed Newmann conformation
due to bond pair – bond pair repulsion (Torsional strain) Eclipsed conformation is less stable than staggered conformation. H
H
H
H
H
H H
H H lkarfjr U;wesu la: i.k
H
H H
xzflr U;wesu la: i.k
cfU/kr ;qXe – cfU/kr ;qXe izfrd "kZ.k(,saBu ruko) d sd kj.k xzflr la: i.k] lkarfjr la: i.k d h rqy uk esad e LFkk;h gksrk gSA 144.
In which of the following options the order of arrangement does not agree with the variation of property indicated against it ? (1) Li < Na < K < Rb (increasing metallic radius) 3+ 2+ + – (2) Al < Mg < Na < F (increasing ionic size) (3) B < C < N < O (increasing first ionization enthalpy) (4) I < Br < Cl < F (increasing electron gain enthalpy)
fuEufy f[kr esalsd kS u lk Ø e fn;sx;sxq.k/keZd sifjorZu d svuql kj lger ughagS\ (1) Li < Na < K < Rb (c<+ rh gqbZ/kkfRod 3+
(2) Al
< Mg
2+
+
f=kT;k)
< Na < F (c<+ rsgq;svk;fud –
vkd kj)
(3) B < C < N < O (c<+ rk gqv k izFke
vk;fud ,UFkSYih) (4) I < Br < Cl < F (c<+ rh gqbZby sDVªksu y fC/k ,UFkSYih) Ans. Sol.
(3 & 4) Incorrect option are 3 & 4 Correct order of increasing Ist I.E B < C < O < N correct order of increasing electron gain Enthalpy I < Br < F < Cl (in magnitude) Values (in KJ/mol) 296, 325, 333, 349.
vlR; fod Yi 3 & 4 gS izFke vk;uu Å t kZd k lgh c<+rk Ø e B < C < O < N by sDVª kWu y fC/k ,UFkSYih d k lgh c<+rk Ø e I < Br < F < Cl (d soy ifjek.k) eku (KJ/mol esa) 296, 325, 333, 349.
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PAGE # 59
| NEET-2016 145.
| 01-05-2016 |
The rate of a first-order reaction is 0.04 mol
–1
s
–1
Code-C,R,Y at 10 seconds and 0.03 mol
–1
–1
s
at 20 seconds
after initiation of the reaction. The half-life period of the reaction is :
,d izFke d ksfV d h vfHkfØ ;k d k osx vfHkfØ ;k izkjEHk gksusd s10 sec ckn 0.04 mol –1 s–1 rFkk 20 sec ckn 0.03 mol Ans. Sol.
–1
s
–1
gSA bl vfHkfØ ;k d h v) Zvk;qd ky gS&
(1) 54.1 s (2)
(2) 24.1 s
r2 C2 ( for first order reaction) (iz Fke r1 C1 k=
C r 1 1 n 2 n 2 t 2 – t1 C1 t 2 – t1 r1
k=
1 0.04 1 4 n n 20 – 10 0.03 10 3
t1/ 2 =
(3) 34.1 s
(4) 44.1 s
d ksfV vfHkfØ ;k d sfy ,)
n2 n2 10 k n4 / 3
2.3 0.3 10 2.3(0.6 – 0.477)
= 24.4 sec. 146.
When copper is heated with conc. HNO3 it produces : (1) Cu(NO3)2 and N2O (2) Cu(NO3)2 and NO2 (3) Cu(NO3)2 and NO (4) Cu(NO3)2 , NO and NO2
d kWij d kslkUnzHNO3 d slkFk xeZd jusij curk gS&
Ans. Sol.
(1) Cu(NO3)2 rFkk N2O
(2) Cu(NO3)2 rFkk NO2
(3) Cu(NO3)2 rFkk NO
(4) Cu(NO3)2 , NO rFkk NO2
(2) Cu + 4HNO3 Cu(NO3)2 + 2NO2 + 2H2O conc
Brown gas
Cu + conc HNO3 NO2 Cu + dil HNO3 NO
147.
In a protein molecule various amino acids are linked together by : (1) dative bond
(2) -glycosidic bond
(3) -glycosidic bond
(4) peptide bond
izksVhu v.kqesafofHkUu ,s ehuksvEy ,d nwl jslst qM +sjgrsgS& (1) nkrk vkca /k d s}kjk (2) -Xy kbZ d ksflfMd vkca/k d s}kjk (3) -Xy kbZ d ksflfMd Ans.
vkca/k d s}kjk
(4) is IVkbZM
vkca/k d s}kjk
(4)
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| NEET-2016
R1 H2N — CH — C — OH
| 01-05-2016 | R2
Code-C,R,Y
+ H2 N — CH — C — OH O
O amino acid (,ehuksvEy ½ Sol.
R1
R2
H2N — CH — C — NH — CH — COOH O Peptide Bond (isIVkbM cU/k) 148.
Fog is a Colloidal solution of : (1) Gas in gas (2) Liquid in gas
(3) Gas in liquid
(4) Solid in gas
(3) nz o
(4) xS l
/kqa/k d ksy kWbMh foy ;u gS& (1) xS l Ans.
(2)
Sol.
Fog
esaxSl d k
(2) xS l
esanzo d k
esaxSl d k
esaBksl d k
Dispersed phase is liquid Dispersion medium is gas
/kqa/k
ifjf{kIr izkoLFkk nzo gSA ifj{ksi.k ek/;e xSl gSA
149.
Match items of Column I with the items of Column II and assign the correct code : Column I
(a) (b) (c) (d)
Column II
Cyanide process Froth floatation process Electrolytic reduction Zone refining
(i) (ii) (iii) (iv) (v)
Ultrapure Ge Dressing of ZnS Extraction of Al Extraction of Au Purification of Ni
Code d ks M +: (1) (2) (3) (4)
(a) (iii) (iv) (ii) (i)
(b) (iv) (ii) (iii) (ii)
(c) (v) (iii) (i) (iii)
(d) (i) (i) (v) (iv)
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PAGE # 61
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
LrEHk I d smYy s[k d ksLrEHkII d smYy s[k lsfey k;saA lgh lad sr i) fr gS& LrEHkI (a) (b) (c) (d)
LrEHkII
lkW;ukbM izØ e
(i)
vfr'kq) Ge
Q su Iy ou fof/k
(ii)
ZnS d k iz l k/ku
fo|qr vi?kVuh vip;u
(iii)
Al d k fu"d "kZ .k
eaM y ifj"d j.k
(iv)
Au d k fu"d "kZ .k
(v)
Ni d k 'kks /ku
d ksM +: (a) (iii) (iv) (ii) (i)
(b) (iv) (ii) (iii) (ii)
(c) (v) (iii) (i) (iii)
(d) (i) (i) (v) (iv)
Ans.
(1) (2) (3) (4) (2)
Sol.
Cyanide process Leaching process of Au Au + 2NaCN aq.
O2
–
Au(CN)2 + Na
+
Froth floatation process Pressing of ZnS (It is applicable for concentration of sulphide are) Electrolytic reduction Extraction of Al Zone refining Purification of Si, Ge 150.
Which one given below is a non-reducing sugar ? (1) Sucrose (2) Maltose (3) Lactose
(4) Glucose
fuEu esalsd kSulh ,d xSl vipk;d 'kd Zjk gS\ (1) lq Ø ksl Ans. Sol.
(2) ekYVks l
(3) y s DVksl
(4) Xy q d ksl
(1) Sucrose is Non Reducing sugar. (both the anomeric carbon are bonded to each other than such sugars are non reducing)
lqØ ksl ,d vuvipk;d 'kd Zjk gSA (nksuksa ,uksesfjd d kcZu ijLij cfU/kr gSA blfy , bl izd kj d h 'kd Zjk vuvipk;d gksrh gSA) 151.
The correct statement regarding RNA and DNA, respectively is : (1) The sugar component in RNA is 2'-deoxyribose and the sugar component in DNA is arabinose. (2) The sugar component in RNA is arabinose and the sugar component in DNA is 2'-deoxyribose. (3) The sugar component in RNA is ribose and the sugar component in DNA is 2'-deoxyribose. (4) The sugar component in RNA is arabinose and the sugar component in DNA is ribose.
RNA rFkk DNA d sfy ;slgh d Fku
Ø e'k% gS&
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PAGE # 62
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
(1) RNA es a'kd Zjk ?kVd 2'-fMvkWDlhjkbcksl
vkSj DNA es a'kd Zjk ?kVd vjSfcuksl gSA (2) RNA es a'kd Zjk ?kVd vjSfcuksl gSvkSj DNA esa'kd Zjk ?kVd 2'-fMvkWDlhjkbcksl gSA (3) RNA es a'kd Zjk ?kVd
jkbcksl gSvkSj DNA esa'kd Zjk ?kVd 2'-fMvkWDlhjkbcksl gSA (4) RNA es a'kd Zjk ?kVd vjSfcuksl gSvkSj DNA esa'kd Zjk ?kVd jkbcksl gSA Ans. Sol.
152.
(3) DNA
RNA
De-oxy Ribose sugar
Ribose Sugar
The correct thermodynamic conditions for the spontaneous reaction at all temperatures is : (1) H < 0 and S < 0
(2) H < 0 and S = 0
(3) H > 0 and S < 0
(4) H < 0 and S > 0
lHkh rki ij Lor% vfHkfØ ;k d sfy , lgh Å "ekxfrd h; 'krsZgS& (1) H < 0 rFkk S < 0
(2) H < 0 rFkkS = 0
(3) H > 0 rFkkS < 0
(4) H < 0 rFkkS > 0
Ans.
(4)
Sol.
G = H – TS For spontaneous process (G = -Ve) at all temperature, H < 0 & S > 0.
lHkh rki ij Lor% izØ e d sfy, (G = -Ve), H < 0 & S > 0. 153.
Which is the correct statement for the given acids? (1) Phosphinic acid is a diprotic acid while phosphonic acid is a monoprotic acid. (2) Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid. (3) Both are diprotic acids. (4) Both are triprotic acids
fuEufy f[kr esalsd kS ulk d Fku fn;sx;svEy ks ad sfy ;slgh gSa\ (1) Q kW fLQ fud
vEy f}izksVh vEy gSt cfd Q kWLQ ksfud vEy ,d izksVh vEy gSA
(2) Q kW fLQ fud
vEy ,d izksVh vEy gSt cfd Q kWLQ ksfud vEy f}izksVh vEy gSA (3) nks ukasf}izksv h vEy gSA (4) nks uksaf=kizksVh vEy
gSA
Ans.
(2)
Sol.
Phosphoric acid (Phosphonic acid) H3PO3 (dibasic)
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PAGE # 63
| NEET-2016
| 01-05-2016 | O
Code-C,R,Y
P H OH OH Hypophosphorous acid (Phosphinic acid) H3PO2 (monobasic) O P H 154.
H
OH –13
MY and NY3, two nearly insoluble salts, have the same Ksp values of 6.2 x 10 at room temperature, which statements would be true in regard to MY and NY3? (1) The addition of the salt of KY to solution of MY and NY3 will have no effect on their solubilities. (2) The molar solubilities of MY and NY3 in water are identical. (3) The molar solubility of MY in water is less than that of NY3. (4) The salts MY and NY3 are more soluble in 0.5 M KY than in pure water. MY
,oaNY3, nksy xHkx vfoy s; y o.kksad k d ejsd srki ij Ksp d k eku 6.2 x 10–13 ,d leku gSA fuEu esals
d kSulk d Fku MY ,oaNY3 d slanHkZesalR; gS\ (1) KY y o.k d ksMY ,oaNY3 d sfoy ;u es aMky usij bud h foy s;rk ij d ksbZiz Hkko ughaiM+rk gSA (2) MY ,oaNY3 d h t y (3) MY d h t y
esaeksy j foy s;rk leku gSA
esaeksy j foy s;rk NY3 lsd e gSA
(4) MY ,oa NY3 d sy o.k 'kq ) Ans.
(3)
Sol.
MY KSP = S12 = 6.2 × 10
–13
t y d h rqy uk esa0.5 M KY esaT;knk foy s; gSA
= 62 × 10
–14
S1 = 7.9 × 10–7 mole/lt = Solubility in pure water MY3 KSP = 27 S24 = 6.2 × 10–13 = 62 × 10–14 ~ 10–3.5 mole/lt = Solubility in pure water S2 _ rd
Solubility of NY3 > solubility of MY so 3 statement is true Addition of KY will decrease the solubility due to common ion effect. Sol.
MY KSP = S12 = 6.2 × 10
–13
= 62 × 10
S1 = 7.9 × 10–7 eks y /y hVj = 'kq) MY3 KSP = 27
S24
= 6.2 × 10
~ 10–3.5 eks S2 _ y /y hVj = 'kq)
–13
–14
t y esafoy s;rk
t y es afoy s;rk
NY3 d h foy s ;rk > MY d h foy s;rk blfy , KY fey kusij
155.
–14
= 62 × 10
rhljk d Fku lR; gSA
levk;u izHkko d sd kj.k foy s ;rk ?kVrh gSA
Which of the following in an analgesic?
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PAGE # 64
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
(1) Chloromycetin (2) Novalgin (3) Penicillin (4) Streptomycin
fuEu esalsd kSulh nok ,d ihMkgkjh gS\ (1) Dy ks jksekblhfVu (2) uks oy ft u (3) is fuflfy u (4) LVª sIVksekbflu Ans. Sol.
(2) Novalgin is an analgesic it is a fact.
156.
The pair of electron in the given carbanion, CH3C C , is present in which of the following orbitals?
fn;sx;sd kcZ& _ .kk;u CH3C C , d s;qXe by sDVªkWu fuEu esalsfd l d {kd esamifLFkr gS\
Ans.
(1) sp (2) 2p 3 (3) sp 2 (4) sp (1)
Sol.
CH3–CC sp hybridisation
Steric Number (1 + 1–vecharge)
sp
CH3–CC
sp lad j.k
f=kfoe la[;k (1 + 1_ .kkos'k) 157.
sp
Among the following, the correct order of acidity is :
fuEu esalsvEy rk d k lgh Ø e gS&
Ans. Sol.
(1) HCIO4 < HCIO2 < HCIO < HCIO3 (2) HCIO3 < HCIO4 < HCIO2 < HCIO (3) HCIO < HCIO2 < HCIO3 < HCIO4 (4) HCIO2 < HCIO < HCIO3 < HCIO4 (3) As oxidation number of central atom increases, acidic nature increases.
d sUnzh; ijek.kqd h vkWDlhd j.k la[;k c<+usij vEy h; izd `fr c<+rh gSA HClO < HClO2 < HClO3 < HClO4
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PAGE # 65
158.
| NEET-2016 | 01-05-2016 | Code-C,R,Y Which one of the following statements is correct when SO2 is passed through acidified K2Cr2O7 solution? (1) Green Cr2(SO4)3 is formed. (2) The solution turns blue (3) The solution is decolourized. (4) SO2 is reduced.
fuEufy f[kr esalsd kS ulk d Fku lR; gS] t c SO2 d ksvEy h; K2Cr2O7 d sfoy ;u esalsikl fd ;k t krk gS\ (1) gjkCr2(SO4)3 curk gS A (2) foy ;u
uhy k iM+t krk gSA
(3) foy ;u
jaxghu gkst krk gSA (4) SO2 vipfr; gks rk gSA
Ans.
(1)
Sol.
K2Cr2O7 Cr2(SO4)3 green solution obtain where as SO2 oxidise into sulphate SO2– 4 K2Cr2O7 Cr2(SO4)3 gjk foy ;u
izkIr gks rk gSt gk¡SO2 d k SO2– avkWDlhd j.k gkst krk gSA 4 es
(K2Cr2O7 + SO2 + H2SO4 Cr2(SO4)3 + H2O + K2SO4) 159.
Predict the correct order among the following : (1) Ione pair – bond pair > bond pair – bond pair > lone pair – lone pair (2) lone pair – lone pair > lone pair – bond pair > bond pair – bond pair (3) lone pair – lone pair > bond pair – bond pair > lone pair – bond pair (4) bond pair – bond pair > lone pair – bond pair > lone pair – lone pair
fuEu esalslgh Ø e gksxk & (1) ,d kd h ;q Xe – vkca/kh ;qXe > vkca/kh ;qXe – vkca/kh ;qXe > ,d kd h ;qXe – ,d kd h ;qXe (2) ,d kd h ;q Xe – ,d kd h ;qXe > ,d kd h ;qXe – vkca/kh ;qXe >
vkca/kh ;qXe – vkca /kh ;qXe (3) ,d kd h ;q Xe – ,d kd h ;qXe > vkca/kh ;qXe – vkca/kh ;qXe > ,d kd h ;qXe – vkca /kh ;qXe (4) vkca /kh ;qXe – vkca/kh ;qXe > ,d kd h ;qXe – vkca/kh ;qXe > ,d kd h ;qXe – ,d kd h ;qXe Ans. Sol.
(2) The order of repulsion force according to VSEPR theory : VSEPR fl) kUr
d svuql kj izfrd "kZ.k cy d k Ø e gS:
lone pair – lone pair > lone pair – bond pair > bond pair – bond pair 160.
Two electrons occupying the same orbital are distinguished by : (1) Spin quantum number (2) Principal quantum number (3) Magnetic quantum number (4) Azimuthal quantum number
nksby sDVªkWu t ksfd ,d gh d {kd esagSA buesavUrj fd ld s}kjk fd ;k t k ld rk gS\ (1) iz p Ø .k DokaVe
la[;k (3) pq Ecd h; DokaVe la [;k Ans.
(2) eq [;
DokaVe la[;k (4) fnxa 'kh; DokaVe la[;k
(1)
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PAGE # 66
Sol.
| NEET-2016 | 01-05-2016 | Code-C,R,Y – Same orbital can have two different values of spin of e of +½ and –½ (spin quantum number)
leku d {kd esaby sDVªkWu d spØ .k d sfy , nksfHkUu eku gksld rsgSA (+½ rFkk –½) (pØ .k Dok.Ve la[;k) 161.
The product obtained as a result of a reaction of nitrogen with CaC2 is :
ukbVªkst u d h CaC2 d slkFk vfHkfØ ;k d sizkIr mRikn gS&
Ans. Sol.
(1) Ca2CN (2) Ca(CN)2 (3) CaCN (4) CaCN3 (Bonus) Reaction of CaC2 and nitrogen at 1100ºC form nitrolim (calcium cyanamide and carbon mixture). CaC2 + N2 CaCN2 + C (No answer in matching) CaC2 rFkk ukbVª kWt u
d h vfHkfØ ;k 1100ºC ij gksuslsukbVªkWfy e d k fuekZ.k gksrk gSA (calcium cyanamide and
carbon mixture). CaC2 + N2 CaCN2 + C (No answer in matching) 162.
Natural rubber has : (1) Random cis – and trans–configuration (2) All cis–configuration (3) All trans–configuration (4) Alternate cis– and trans–configuration
izkd `frd jcj esa& (1) vfu;fer
fll~& ,oaVªkal &foU;kl gSA (2) lHkh fll~ & foU;kl gSA
(3) lHkh Vª kUl&foU;kl
gSA (4) ,d kUrj fll~ & ,oaVªkal &foU;kl gSA
Ans. Sol.
(2) It is fact
163.
Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules?
Ans.
(1) F2 > Cl2 > Br2 > I2 (3) CI2 > Br2 > F2 > I2 (3)
Sol.
Bond dissociation enthalpy (cU/k fo;ks tu
,UFkSYih)
CI2 242.6
>
fuEufy f[kr esalsd kS u Ø e gSy kst u v.kqv ksad h vkca/k fo;kst u ,UFkSYih d sfy ;slgh gS\
>
Br2 192.8
(2) I2 > Br2 > CI2 > F2 (4) Br2 > I2 > F2 > CI2
>
F2 158.8
I2 151.1
(kJ/mole)
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PAGE # 67
164.
| NEET-2016 The reaction : OH
| 01-05-2016 |
NaH
O
Code-C,R,Y Me
Me–I
Na
can be classified as : (1) Williamson alcohol synthesis reaction (3) Alcohol formation reaction
O
(2) Williamson ether synthesis reaction (4) Dehydration reaction
vfHkfØ ;k & OH
Ans.
NaH
O
Me
Me–I
Na
O
d ksoxhZd `r fd ;k t k ld rk gS& (1) fofy ;Elu ,Yd ks gy la'y s"k.k vfHkfØ ;k
(2) fofy ;Elu
(3) ,Yd ks gy
(4) fut Z y hd j.k vfHkfØ ;k
fojpu vfHkfØ ;k
bZFkj la'y s "k.k vfHkfØ ;k
(2)
Sol.
OH + NaH
O
acid-ase Reaction 2
SN
Na
CH3—I
O–CH3 This williamson ether synthesis 165.
–3
–1
Lithium has bcc structure. Its density is 530 kg m and its atomic mass is 6.94 g mol . Calculate the 23 –1 edge length of a unit cell of Lithium metal. (NA = 6.02 × 10 mol )
fy fFk;e d hbbc lajpuk gSA bld k ?kuRo 530 kg m–3 rFkk ijek.kqnzO;eku 6.94 g mol–1 gSA fy fFk;e /kkrqd s,d d d ksf"Bd k d sd ksj d h y EckbZgSA (NA = 6.02 × 1023 mol–1) Ans.
(1) 264 pm (3)
Sol.
d=
ZA NA a3 3
530 kg/m
(2) 154 pm
for BCC
=
(3) 352 pm
(4) 527 pm
Z=2
2 6.94 10 3 6.02 1023 a3
a3 = 43.50 × 10–30 a = 3.52 × 10–10 m = 352 pm.
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PAGE # 68
166.
| NEET-2016 | 01-05-2016 | Code-C,R,Y + – –10 –10 The ionic radii of A and B ions are 0.98 × 10 m and 1.81 × 10 m. The coordination number of each ion in AB is +
A
,oaB– vk;uksad h vk;fud f=kT;k,¡Ø e'k%0.98 × 10–10 m ,oa1.81 × 10–10 m gSA AB esaizR;sd vk;u d h
milgla;kst u la[;k gSA Ans. Sol.
(1) 2 (2)
rA rB
(2) 6
=
0.98 10 10 1.81 10 10
(3) 4
(4) 8
= 0.54
Octahedral range (v"VQ y d h;
lhek) 0.414
r < 0.732 r
Co-ordination no. of each ion is 6 like NaCl structure.
izR;sd vk;u d h leUo; la[;k NaCl lajpuk d sleku N% gSA 167.
At 100º C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be :
,d 6.5 g foy s; d k 100 g t y esafoy ;u d k 100º C ij ok"Ik nkc 732 mm gSA ;fn Kb = 0.52 rksbl foy ;u d k DoFkukad gksxk % Ans. Sol.
(1) 103º C (2) 101º C (2) At B.P. P0 = 760 torr
for elevation of B.P.
DoFkukad fcUnqij
DoFkukad mUu;u d sfy ,
(3) 100º C
P0 = 760 torr
6.5 32 TB = I Kbm = 1 × 0.52 × × 1000 = 1 100
P0 Ps WA / MA = Ps WB / MB
168.
(4) 102º C
760 732 6.5 / M = 732 100 /18
=1
On solving M = 32.
So B.P. = 100 + TB = 101ºC
The electronic configurations of Eu (Atomic No. 63) Gd (Atomic No. 64) and Tb (Atomic No. 65) are : 7 2 7 1 2 9 2 (1) [Xe]4f 6s , [Xe]4f 5d 6s and [Xe]4f 6s 7 2 8 2 8 1 2 (2) [Xe]4f 6s , [Xe]4f 6s and [Xe]4f 5d 6s 6 1 2 7 1 2 9 1 2 (3) [Xe]4f 5d 6s , [Xe]4f 5d 6s and [Xe]4f 5d 6s 6 1 2 7 1 2 8 1 2 (4) [Xe]4f 5d 6s , [Xe]4f 5d 6s and [Xe]4f 5d 6s Eu (i-l-63) Gd (i-l-64) vkS j Tb (i-l-65) d sby sDVªksfud 7
2
7
1
7
2
8
2
(1) [Xe]4f 6s , [Xe]4f 5d 6s (2) [Xe]4f 6s , [Xe]4f 6s 6
1
2
7
2
9
foU;kl gS%
2
vkSj [Xe]4f 6s
vkSj [Xe]4f8 5d16s2 1
vkSj [Xe]4f9 5d16s2 (4) [Xe]4f6 5d16s2, [Xe]4f7 5d16s2 vkS j [Xe]4f8 5d16s2
(3) [Xe]4f 5d 6s , [Xe]4f 5d 6s Ans.
2
(1)
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PAGE # 69
| NEET-2016 Sol.
7
63Eu [Xe] 4f 6s 64Gd 65Tb
169.
| 01-05-2016 |
Code-C,R,Y
2
7
1
2
9
0
2
[Xe] 4f 5d 6s [Xe] 4f 5d 6s
Which of the following statements about hydrogen is incorrect ? (1) Dihydrogen does not act as a reducing agent. (2) Hydrogen has three isotopes of which tritium is the most common. (3) Hydrogen never acts as cation in ionic salts. + (4) Hydronium ion, H3O exists freely in solution.
fuEufy f[kr esalsd kS u lk d Fku gkbMªkst u d sfy , v lR; gS\ (1) Mkbgkbª kst u
vipk;d d s: i esad k;Zughad jrk gSA (2) gkbMª kst u d srhu leLFkkfud gSft lesalsVªkbfV;e izp qjrk esagSA
(3) gkbMª kst u
vk;fud y o.kksaes a/kuk;u d h rjg O;ogkj ughad jrk gSA + (4) gkbMª ksfu;e vk;u, H3O d k vfLrRo foy ;u esaeqDr : i esagksrk gSA Ans. Sol.
(1 & 2) 1 and 2 option are incorrect Correct – Dihydrogen act as reducing agent for eg 3H2 + N2 2NH3 1
Correct – Hydrogen has three isotopes of which protium (1H ) is the most common. 1 rFkk 2 fod Yi
vlR; gSA
Correct – MkbgkbMª kst u
vipk;d d sleku O;ogkj d jrk gSA eg 3H2 + N2 2NH3 Correct – gkbMª kst u d srhu leLFkkfud gksrsgSA ft lesaizksVh;e (1H1) lokZf/kd izp fy r gSA 170.
In the reaction (1)NaNH / liq.NH
(1)NaNH / liq.NH
3
3
2 3 2 3 H–CCH x y (2) CH CH Br (2) CH CH Br 2
2
X and Y are : (1) X = 1-Butyne ; y = 2-Hexyne (2) X = 1-Butyne ; y = 3-Hexyne (3) X = 2-Butyne ; y = 3-Hexyne (4) X = 2-Butyne ; 2 = 2-Hexyne
vfHkfØ ;k esa (1)NaNH2 / liq.NH3 (1)NaNH2 / liq.NH3 H–CCH x y (2) CH CH Br (2) CH CH Br 3
2
3
2
X vkS j Y gS: (1) X = 1-C;w Vkbu ; y = 2-
gs Dlkbu (2) X = 1-C;w Vkbu ; y = 3-gsDlkbu
(3) X = 2-C;w Vkbu ; y = 3-gsDlkbu (4) X = 2-C;w Vkbu ; 2 = 2-gsDlkbu Ans.
(2)
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PAGE # 70
| NEET-2016 Sol.
HCCH
NaNH2 / NH3 NH3
| 01-05-2016 |
Code-C,R,Y
HCC —Na+CH3–CH2–Br
S N2 NaNH2 / NH3 (a) CH3–CH2–CC CH3–CH2–CCH 1-Butyne
SCH3CH2–Br
CH3–CH2–CC–CH2–CH3 Hex-3-yne 171.
Consider the following liquid-vapour equilibrium.
Liquid Vapour Which of the following relations is correct ? (1)
dlnP Hv dT RT 2
(2)
Hv
dlnG dT 2
RT 2
RT 2
(3)
dlnP Hv dT RT
(4)
(3)
dlnP Hv dT RT
(4)
dlnP dT 2
Hv T2
uhpsfn;sx;snzo & ok"i lkE;koLFkk] nzo ok"i esalsd kSu lk lacU/k lgh gS\ (1)
dlnP Hv dT RT 2
(2)
Hv
dlnG dT 2
Ans.
(1)
Sol.
The variation of vapour pressure and temperature is nP = on differentiate
dT 2
Hv T2
Hº + constant RT
d(nP) Hº =+ +0 dT RT 2
ok"inkc d k rki d slkFk ifjorZ u n P = on differentiate
dlnP
Hº + constant RT
d(nP) Hº =+ +0 dT RT 2
d( nP) Hº = dT RT 2 172.
Which of the following statements about the composition of the vapour over an ideal 1 : 1 molar mixture of benzene and toluene is correct ? Assume that the temperature is constant at 25ºC. (Given, Vapour Pressure Data at 25ºC, Benzene = 12.8kPa, toluene = 3.85kPa) (1) Not enough information is given to make a prediction. (2) The vapour will contain a higher percentage of benzene. (3) The vapour will contain a higher percentage of toluene. (4) The vapour will contain equal amounts of benzene and toluene.
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PAGE # 71
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
csUt hu ,oaVkWy wbZu d s1 : 1 vkn'kZeksy j feJ.k d sok"i la;kst u d sfy ;sfuEufy f[kr esalsd kSu lk d Fku lR; gS\ d Yiuk d jsafd rkieku 25ºC ij fLFkj gSA (fn;sx;sok"i nkc 25ºC csUt hu = 12.8kPa, VkWy wbZu = 3.85kPa) (1) vi;kZ Ir
lwp ukvksad sd kj.k d ksbZiwokZuqeku ughay xk;k t k ld rk gSA (2) ok"i es acsUt hu d h vf/kd izfr'krrk gksxh (3) ok"i
esaVkWy wbZu d h vf/kd izfr'krrk gksxh (4) ok"i es aleku ek=kk esacsUt hu ,oaVkWy wbZu gksxhA
Ans. Sol.
(2) Due to high partial vapour pressure of Benzene as compare to that of toluene so the mole fraction of Benzene will be higher than that of toluene. As a result the vapour will contain a higher percentage of Benzene.
csat hu d smPp vkaf'kd ok"i nkc d sd kj.k bld k eksy fHkUu vf/kd gksrk gSA blfy , ok"i esacs at hu d h izfr'krrk vf/kd gksxhA 173.
Which of the following biphenyls is optically active
fuEu esalsd kSu lk ckbZfQ uk;y izd kf'kd lfØ ; gS\ O 2N
CH3 (1)
(2)
CH3
I
Br Br
I
(3)
Ans. Sol.
(4)
I I I (3) O-substituted biphenyls are optically active as both the rings are not in one plane hence their mirror mages are non-super imposable.
vkWFkksZizfrLFkkfi ckbZfQ ukbZy esanksuksfjax ,d ry esaugh gSvr% blesaizd kf'k; leko;ork gSA 174.
Which of the following reagents would distinguish cis-cyclopenta-1,2- diol from the trans-isomer ? (1) Aluminium isopropoxide (2) Acetone (3) Ozone (4) MnO2
fuEu esalsd kSu lk vfHkd eZd fll~& lkbDy ksisUVk-1,2-MkbZv kWy ,oabld sVªkal leko;oh esaHksn d jsxk \ (1) ,s Y;qfefu;e
vkblksizksiksDlkbM
(2) ,s l hVksu
(3) vks t ksu Ans.
(4) MnO2
(2) OH O=C
Sol.
OH
CH3 CH3
O C O
CH3 CH3
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PAGE # 72
175.
| NEET-2016 | 01-05-2016 | Code-C,R,Y The correct statement regarding a carbonyl compound with a hydrogen atom on its alpha-carbon, is : (1) a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this processes is known as keto-enol tautomerism . (2) a carbonyl compound with a hydrogen atom on its alpha-carbon never equilibrates with its corresponding enol. (3) a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as aldehyde-ketone equilibration. (4) a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known a carbonylation.
d kcksZfuy ;kSfxd ft uesa d kcZu ij gkbMªkst u mifLFkr gS] d sfy , lgh d Fku gS% (1) d kcks Zfuy ;kSfxd ft uesa d kcZu ij gkbMªkst u ijek.kq mifLFkr gS] ;g bud s vuq: i bZukWy esa vklkuh ls lkE;koLFkk esagksrsgSavkSj ;g izØ e fd Vks& bZukW y py ko;ork d gy krhgS A (2) d kcks Zfuy ;kSfxd ft uesa d kcZu ij gkbMªkst u ijek.kq mifLFkr gS] ;g bud s vuq: i bZukWy ls d Hkh Hkh lkE;koLFkk esaughagksrsgSA (3)
d kcksZfuy ;kS fxd d kcZu ij gkbMªkst u ijek.kqmifLFkr gS] ;g bud svuq: i bZukWy esavklkuh lslkE;koLFkk es
gksrsgSvkSj ;g izØ e ,sfYMgkbM&d hVksu lkE;koLFkk d gy krk gS A (4) d kcks Zfuy ;kSfxd ft uesa-d kcZu gkbMªkst u ijek.kqmifLFkr gS] ;g bud svuq: i bZukWy lsvklkuh lslkE;koLFkk esagksrsgSvkSj ;g izØ e d kcksZfuy hd j.k d gy krk gSA Ans. Sol.
(1) It is known that basic need for the existance of Keto-enol tautomers is the presence of at least one 3 hydrogen atom at adjacent sp carbon of carbonyl carbon.
d hVksbZukWy leko;ork d sfy , d kcksZfuy sp3 d kcZu ij d e lsd e ,d gkbMªkst u d h mifLFkfr vko';d gSA 176.
Consider the molecules CH4, NH3 and H2O. Which of the given statement is false ? (1) The H–C–H bond angle is CH4 is larger than the H–N–H bond angle is NH3 (2) The H–C–H bond angle is CH4 the H–N–H bond angle in NH3 and the H–O–H bond angle in H2O are all greater than 90º. (3) Then H–O–H bond angle in H2O is larger than the H–C–H bond angle in CH4 (4) The H–O–H bond angle in H2O is smaller than the H–N–H bond angle in NH3 CH4, NH3 vkS j H2O v.kqv ksad sfy ;suhpsfn;sx;sd Fkuksaesalsd kSu
lk v l R; gS\
es aH–C–H vkca/k d ks.k] NH3 esaH–N–H vkca/k d ks.k lsvf/kd gSA (2) CH4 es aH–C–H vkca/k d ks.k] NH3 esaH–N–H vkca/k d ks.k rFkk H2O esaH–O–H vkca/k d ks.k] lHkh esa90º ls
(1) CH4
vf/kd gSA
Ans.
(3) H2O
esaH–O–H vkca/k d ks.k] CH4 esaH–C–H vkca/k d ks.k lsvf/kd gSa
(4) H2O
esaH–O–H vkca/k d ks.k] NH3 esaH–N–H vkca/k&d ks.k lsd e gSA
(3)
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PAGE # 73
| NEET-2016
Sol.
3
CH4
C
sp
H
177.
sp
H2O
sp
H
Code-C,R,Y
Bond angle = 109º 281 H
N
3
NH3
| 01-05-2016 | H
H
H
H
O
3
H
H
Bond angle = 107º Bond angle = 104º 5
Match the compound given in column I with the hybridization and shape given in column II and mark the correct option. Column-I
Column-II
(a)
XeF6
(i)
distorted octahedral
(b)
XeO3
(ii)
square planar
(c)
XeOF4
(iii)
pyramidal
(d)
XeF4
(iv)
square pyramidal
Code : (a)
(b)
(c)
(d)
(1)
(iv)
(i)
(ii)
(iii)
(2)
(i)
(iii)
(iv)
(ii)
(3)
(i)
(ii)
(iv)
(iii)
(4)
(iv)
(iii)
(i)
(ii)
LrEHk esaI esafn;sx;s;kSfxd ksad ksmud slad j.k ,oavkd kj t ksfd LReHk II esafn;sx;s gSad ksfey k;srFkk lgh fod Yi d ksfpfUgr d hft ,A LrEHk-I LrEHk II (a)
XeF6
(i)
foÑ r v"VQ y d h;
(b)
XeO3
(ii)
oxZlery h
(c)
XeOF4
(iii)
fijkfeMh
(d)
XeF4
(iv)
oxZfijkfeMh
d ksM :
Ans.
(a)
(b)
(c)
(d)
(1)
(iv)
(i)
(ii)
(iii)
(2)
(i)
(iii)
(iv)
(ii)
(3)
(i)
(ii)
(iv)
(iii)
(4) (2)
(iv)
(iii)
(i)
(ii)
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PAGE # 74
| NEET-2016 Sol.
3 3
XeF6
sp d
| 01-05-2016 | distorted octahedral fod ` r
Code-C,R,Y
v"VQ y d h;
F
F
F
Xe F
F F
3
XeO3
sp
pyramidal fijkfeMh;
Xe O
O O
XeF4 F
3 2
square pyramidal
3 2
square planar oxZlery h;
sp d
oxZfijkfeMh;
F
Xe
F
F O XeF4 F
sp d
F
Xe
F 178.
F
Consider the nitration of benzene using mixed conc. H2SO4 and HNO3 . If a larger amount of KHSO4 is added to the mixture the rate of nitration will be : (1) doubled (2) faster (3) slower (4) unchanged
csUt hu d k ukbVªhd j.k laknzH2SO4 ,oaHNO3 d h mifLFkfr esagksjgk gSA ;fn bl feJ.k esaT;knk ek=kk esaKHSO4 Mky rsgSrksukbVªhd j.k d k osx gks xk% (1) nq xquk (2) rs t Ans. Sol.
(3) /khjs
(4) vifjofrZ r
(3) If large amount of KHSO4 is added, Concentration of NO2+ will decrease and hence the rate of nitration will be slower.
;fn KHSO4 d ksvf/kd ek=kk esafey k;k t krk gSrksNO2+ d h lkUnzrk ?kVrh gSft llsukbfVªd j.k d h nj /kheh gks t krh gSA 179.
Which of the following statement is false ? 2+ (1) Mg ions are important in the green parts of plants. 2+ (2) Mg ions from a complex with ATP. 2+ (3) Ca ions are important in blood clotting. (4) Ca2+ ions are not important in maintaining the regular beating of the heart.
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PAGE # 75
| NEET-2016
| 01-05-2016 |
Code-C,R,Y
fuEufy f[kr esalsd kS u lk d Fku vlR; gS\ 2+ (1) Mg vk;u ikS /kks ad sgfjr Hkkxksad sfy ;segRoiw.kZgSA 2+
vk;u ,-Vh-ih- d slkFk ld qy cukrsgSA (3) Ca vk;u jDr d kst ekusd sfy ;segRoiw .kZgSA
(2) Mg
2+
(4) Ca2+ vk;u Ans. Sol.
(4) +2 Ca are important in blood clotting and are also important in maintaining the regular beating of the heart. 2+
Ca 180.
ân; xfr d ksfu;fer j[kusesegRoiw.kZughagSA
vk;u jDr d kst ekusd sfy ;srFkk ân; xfr d ksfu;fer j[kusesHkh egRoiw.kZgSA
Which of the following has longest C–O bond length ? (Free C–O bond length in CO is 1.128Å)
fuEufy f[kr esalsfd ld h C–O vkca/k y EckbZvf/kd re gS\ (eqDr C–O vkca/k y EckbZCO esa1.128Å gSA) +
Ans. Sol.
(1) [Mn(CO)6] (4) Fe C O
(2) Ni(CO)4
(3) [Co(CO)4]
(4) [Fe(CO)4]
2–
Due to back bonding between metal-carbon bond length of C–O increase (B.O of M–C B.O of C– CB.L. of C–O ) Higher is negative charge on metal, higher is back bonding (synergic effect) so bond 2– length is higher so answer is [Fe(CO)4] Fe
C
O
i'p vkcU/ku d sd kj.k /kkrqd kcZu cU/k y EckbZc<+rh gS(B.O of M–C B.O of C–CB.L. of C–O ) /kkrqij ft ruk vf/kd _ .kkos'k gksxk mruk vf/kd i'p vkcU/ku gksxkA blfy , cU/k y EckbZHkh vf/kd gks xhA
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PAGE # 76
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