Neet 2016 Paper With Solution Code C R Y

June 22, 2018 | Author: xanshah | Category: Meiosis, Stoma, Science, Botany, Biology
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NEET - 201601-05-2016 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 : [email protected] | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 1 (2) Both (a) and (b) are correct.ac. P.324005 Reg.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 2 . (4) (a) is true but (b) is false. Resonance Eduventures Ltd. 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.resonance. 6635555 | To Know more : sms RESO at 56677 Website : www.wjsfVd d kjd Ans. Relaxin & inhibin not antagonistic 4.No. | 01-05-2016 | Code-C. +2 Relaxin  causes pelvic musculature relaxation inhibin  inhibits FSH So. No. : +91-744-3012222. CORPORATE OFFICE : CG Tower.qXeksaesalsd kSu&lk .ac. : +91-022-39167222 Ph. Jawahar Nagar.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 . P. Which one of the following options is correct ? (1) Both (a) and (b) are false. Near City Mall.in | E-mail : contact@resonance.)-324005 | Ph. Jhalawar Road.y usfVª. decreses B.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) . Main Road. Kota (Raj. Office : J-2.| NEET-2016 3. A-46 & 52.fVª.) .R.: +91-744-3192222 | FAX No. IPIA. Kota (Raj. 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. (1) Sol. (3) (b) is true but (a) is false.sYMksLVsjkWu . 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. 6.Zd h lewg esaj[ksaxs\ (1) ukbVª kst u Ans.d ikni izd k'k 'olu lsgksusoky h gkfu lscprk gS] mld h t y mi. Sol. shows high rates of photosynthesis at high temperatures and has improved efficiency of nitrogen utilisation. Ans.Y ekbVksd kWf. 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.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 3 . (2) Lora =k izfrd `fr.in | E-mail : [email protected].) . lajpuk (4) LFkkukUrj.Mª.ksx d h n{krk mUur gSA vki bl ikni d ksfuEufy f[kr esalsfd l . Jhalawar Road. Sol. A-46 & 52.ac.R. (4) Mitochondria and chloroplast have their own ribosomes wwith help of which they can synthesize protein.d Iy kfTeM d k vfHky {k. fLFkfjd kjd (2) C3 (3) C4 (4) CAM (3) C4 plants have high rate of photosynthesis at higher temperature.k d h mPp nj d ksn'kkZrk gSvkSj mld h ukbVªkst u mi.ks X. Kota (Raj. : +91-744-3012222.k ughagS\ (1) .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.d y – jTt q d h. CORPORATE OFFICE : CG Tower. 6635555 | To Know more : sms RESO at 56677 Website : www. IPIA.ac.)-324005 | Ph. No. Near City Mall. Main Road.: +91-744-3192222 | FAX No.k . Jawahar Nagar.No. Sol.ksx d h n{krk mUur gS] og mPp rki ij izd k'k la 'y s"k. Kota (Raj.d d kf. Office : J-2. (1) Plasmide are double stranded DNA.324005 Reg.u (3) o` Ùkh. Resonance Eduventures Ltd.| NEET-2016 | 01-05-2016 | Code-C. : +91-022-39167222 Ph. A plant in your garden avoids photorespiratory losses. has improved water use efficiency. 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 . vkSj pØ h.ac.k izHkko vkSj y ky cwan ¼jsM Mªki½ fd ld h [kkst esaizeq[k .ac. 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. 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. Office : J-2. (2) when N/K is exactly one.No.d lkFk d k.) . (4) when N/K equals zero.Zd juk (4) iz d k'kQ kLQ ksfjy s'ku Ans. Main Road.| NEET-2016 7. by sDVªkWu vfHkxeu (3) nksiz d k'k rU=kksad k .R.)-324005 | Ph. 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. No. (2) 9.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 4 .in | E-mail : [email protected] Reg. : +91-022-39167222 Ph.: +91-744-3192222 | FAX No. (3) when N nears the carrying capacity of the habitat. | 01-05-2016 | Code-C. Sol. Kota (Raj. Jhalawar Road. A-46 & 52.resonance. Resonance Eduventures Ltd. vkLrj (2) fpd uh is'kh vka=k fHkfÙk (3) . IPIA. 6635555 | To Know more : sms RESO at 56677 Website : www. Jawahar Nagar. : +91-744-3012222. Kota (Raj.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.sfjvksy h Å rd d aM jk (4) ifjorhZmid y k ukfld kxz Ans. 8.a=k jgsgSa\ (1) vkW DlhMsfVo Q kLQ ksfjy s'ku (2) iz d k'kQ kLQ ksfjy s'ku vkSj vpØ h. CORPORATE OFFICE : CG Tower. Near City Mall. No.kks ad h cká Liksjksiksy sfuu d h cuh gksrh gS (4) cgq r lh t kfr.resonance. Ans. (3) Golden algae are also called desmids. Kota (Raj.kqHkh d gk t krk gSA (4) Eubacteria are called true bacteria.kq v ksa½ d ksvlR. (2) Dehiscence of anther occur due to stomium cells of endothecium 11. ughagS\ (1) nz for ukbVªkst u esaHk. 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. Main Road.: +91-744-3192222 | FAX No. Jawahar Nagar. (4) Eubacteria are also called false bacteria. Which one of the following statements is wrong ? (1) Phycomycetes are also called algal fungi.in | E-mail : [email protected] d jrsgq.ac. CORPORATE OFFICE : CG Tower.) .st k ld rsgSa (2) ijkxd ks "k d sLQ qVu esaVsihVe lgk.k xEHkhj izR.Y y kWft fLVd ekWM y d k vuql j.k ¼lqt hok.kZ okl (4) t c N/K 'kw U. Resonance Eduventures Ltd. t hok.k d ksuhy (3) Lof.wt Zrk iS nk d jrsgSa Ans.ksad sijkxd . : +91-022-39167222 Ph. Sol.d gks (3) t c N i. (2) Sol. A-46 & 52. 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. Ans.kZ e gfjr 'kSoky Hkh d grsgSaA 'kSoky ksad ksMs fLeM Hkh d grsgSaA (4) .k x.ac.rk d jrk gS (3) ijkxd . No. Jhalawar Road.uks cSDVhfj. Kota (Raj.q cSDVhfj.324005 Reg.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 5 .k gS% dN/dt = rN(1-N/K) : (1) t c t Uenj d h vis{kk e`R. (2) Cyanobacteria are also called blue-green algae. Near City Mall.qDr fd .k lsn'kkZ. Office : J-2. 6635555 | To Know more : sms RESO at 56677 Website : www.kst ukvksaesaiz.Mkfjr ijkxd . Sol. dN  N  rN  1   dt k  d h /kkfjrk {kerk d slehi gksA d scjkcj gksA dN  rN 1  1  0 dt 10.| NEET-2016 | 01-05-2016 | Code-C. IPIA. d scjkcj d c gksxh \ y kWft fLVd ekWM y d ks fuEufy f[kr lehd j.R.qnj vf/kd gksA (2) t c N/K Bhd .k] Q ly izt uu .)-324005 | Ph. : +91-744-3012222. fd lh lef"V d h o`f) nj 'kwU. Main Road.ac. Jhalawar Road.Y The Avena curvature is used for bioassay of: (1) Ethylene (2) ABA (3) GA3 (4) IAA . 13.fFky hu Ans. IPIA.| NEET-2016 12.resonance. Near City Mall. iz. Jawahar Nagar. : +91-744-3012222.ac. : +91-022-39167222 Ph. Kota (Raj.: +91-744-3192222 | FAX No. 6635555 | To Know more : sms RESO at 56677 Website : www.R.) . Office : J-2.)-324005 | Ph. Sol. 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. 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.qDr gksrh gS\ (1) .324005 Reg. Sol.in | E-mail : contact@resonance. CORPORATE OFFICE : CG Tower. No. Resonance Eduventures Ltd. (2) ABA (3) GA3 (4) IAA (4) Avena curvature bioassay is done to test function of IAA. Kota (Raj. | 01-05-2016 | Code-C.ohuk oØ rk fd ld st S o vkekiu d sfy .in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 6 . A-46 & 52.No. (4) Higher blood pressure in pulmonary vein than pulmonary artery is an abnormal condition leading to pulmonary hypertension & pulmonary oedema. (1) 14. ij gksA (4) v. : +91-022-39167222 Ph.k O.) . Identify the correct statement on 'inhibin' : (1) Is produced by nurse cells in testes and inhibits the secretion of LH.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. CORPORATE OFFICE : CG Tower. IPIA.kqv ksad k LFkkukUrj. 6635555 | To Know more : sms RESO at 56677 Website : www. esav.k xHkkZ'k. (3) the ovum and sperms are transported simultaneously to ampullary – isthmic junction of the fallopian tube.3) 16. (4) the ovum and sperms are transported simultaneously to ampullary – isthmic junction of the cervix. | 01-05-2016 | Code-C. In meiosis crossing over is initiated at : (1) Diplotene (2) Pachytene (3) Leptotene (4) Zygotene v) Zl w=kh foHkkt u esat hu fofue. (4) Is produced by granulose cells in ovary and inhibits the secretion of LH.u ufy d k ughaesaNksM +st kusd sBhd ckn gksA (3) v. (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 Q SYkksfi.aiq y jh&bLFkfed laxe ij .d gh le. Kota (Raj.| NEET-2016 15.324005 Reg.kqvkS j 'kqØ k.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 7 .kqd sfueqqZDr gksusd s48 ?kaVsd sHkhrj gksrk gksA (2) 'kq Ø k. Tonus Spasm  fatigue Tetanus 19.Mk. (2. A-46 & 52.d gh le.ac. Jawahar Nagar.kqd sQ SYkksfi. Office : J-2.kqvkS j 'kqØ k. Sol. (2) the sperms are transported into vagina just after the release of ovum in fallopian tube.kqv ksad k LFkkukUrj. 18. FSH and Prolactin.kqv ksad k LFkkukUrj. Near City Mall. (2) Inhibits the secretion of LH. Euglenoids. Chrysophytes. (4) Sol.kqv ksad k .u ufy d k d s.)-324005 | Ph.Mk.k gksusoky h nh?kZd kfy d is'kh lad qp u d gy krk gSaS% (1) Vks ul (2) . ij gksrk gksA Ans.k xzhok d s.: +91-744-3192222 | FAX No.k v.aiqy jh bLFkfed laxe ij .     low level activity of muscles at rest to maintain posture.ac.R. Main Road. fd l voLFkk esavkjEHk gksrk gS\ (1) f}iê (2) LFkw y iê (3) ruq iê (4) . Dinoflagellates and slime moulds are included in the kingdom: (1) Animalia (2) Monera (3) Protista (4) Fungi Ø kblksQ kbV] . Jhalawar Road.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.ksfu d sHkhrj LFkkukUrj.kogkfjd r% rHkh laHko gksxh t c% (1) xz hok uky d sHkhrj 'kqØ k.in | E-mail : contact@resonance. (3) Is produced by granulose cells in ovary and inhibits the secretion of FSH. (2) In pachytene recombination nodule is formend after which crossing over occur 17.a Bu ¼LikT+e½ s (3) Fkd ku (4) fVVs ul Ans.resonance. ekuoksaesafu"ksp u izfØ .Mk.No. Sudden involuntary muscle contraction decline in muscle activity Sustained muscle contraction in response to successive stimuli Resonance Eduventures Ltd. Kota (Raj.q Xeiê Ans. : +91-744-3012222. No.Mk. k iz/kkur% mRiUu gksusoky snh?kZd ky h 'olu&fod kj d k uke crkb. 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.: +91-744-3192222 | FAX No.klw=kh fod kj Ans.kd h. (3) 20. gS (4) Resonance Eduventures Ltd. % o`"k.FSH vkS j izksy SfDVu ò o. d h d f.k d kslanfer d jrk gSA Ans.k¡t y ok. 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. Kota (Raj.) .resonance.324005 Reg.R. : +91-022-39167222 Ph.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 .g v.rk (2) okrLQ hfr (3) vLFkek (4) 'olu vkEy jDrrk Ans. (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.k d kslanfer d jrk gSA (4) . Office : J-2.k d kslanfer d jrk gSA (2) LH. 6635555 | To Know more : sms RESO at 56677 Website : www. Kota (Raj.k lclsy Ecso`{kksaesals. Sol. Near City Mall. CORPORATE OFFICE : CG Tower. Jhalawar Road.No.Y igpkfu.kqd vkSj fo"kecht k.slHkh vuko`rcht h gSa (4) fld ks b. vuqd wfy r ughagksrh gSa A (2) vuko` rcht h] lecht k.g | 01-05-2016 | Code-C.ò o.ac.kqd ] nks uksaizd kj d sgksrsgSa (3) lkfYofu. Jawahar Nagar. (2) 21. Main Road. % (1) 'olu {kkje. IPIA.k d kslanfer d jrk gSA (3) .| NEET-2016 'ba fgfcu' d sckjsesalgh d Fku (1) .)-324005 | Ph.d Ans.kksad h /kk=kh ¼ulZ½ d ks f'kd kvksa}kjk mRiUu gksrk gSvkSj LH .k d k lclsvf/kd mi. d h d f.in | E-mail : [email protected] v. : +91-744-3012222.qDr o. d ks f'kd kvksa}kjk mRiUu gksrk gSvkSj LH ò o. Ginkgo and Pinus all are gymnosperms (4) Sequoia is one of the tallest trees lgh d Fku pqfu.ac. No. A-46 & 52.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 8 .Mk'k. d ksf'kd kvksa}kjk mRiUu gksrk gSA vkSj FSH ò o. % (1) vuko` rcht h ikniksad h ifÙk.kd h. 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 .Mk'k.qd h pjerk d sfy .k] ft a xksvkSj ikbul] . When the F1 plants were selfed the resulting genotypes were in the ratio of : . Kota (Raj.qXet h : ckSus (3) 1 : 2 : 1 : : y ECksfo"ke. vko'.resonance. CORPORATE OFFICE : CG Tower. (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.qXet h: ckSus (4) 3 : 1 : : y Ecs: ckS us (2) Resonance Eduventures Ltd.ac.Zd jusd sfy .q Xet h : y Ecsfo"ke. No.kA t c F1 ikniks ad ksLoijkfxr fd . Pure tall TT Pure dwarf tt T t F1 generation = Tt F2 generation Genothypes = 1 TT Pure Tall 24. Jawahar Nagar.k rkst hu izk: i d k ifjek. : +91-022-39167222 Ph.324005 Reg. 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.d gksxk \ (1) y SDVkst vkSj xSy sDVks t (2) Xy w d kst (3) xS y sDVkst Ans. 6635555 | To Know more : sms RESO at 56677 Website : www.k x.k fd l vuqikr esaFkk \ Ans. Office : J-2.Y y Sd izp ky s d d h vfHkO. | 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. Near City Mall.k x.) . (4) y S DVkst (4) As lae operon becomes active after inducing lactose but glucose & galactose can't do so. : +91-744-3012222.R. A-46 & 52.fDr d sfy .23. Main Road. fuEufy f[kr esalsd kSu .in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 9 . Jhalawar Road.in | E-mail : [email protected] izsjd d s: i esad k. IPIA.No. Kota (Raj.: +91-744-3192222 | FAX No. Sol.q Xet h : y Ecsle.d ckSusrn~: i izt uu m|ku eVj ikni lslad fjr d jk.d y ECksrn~: i iz t uu m|ku eVj ikni d ks.)-324005 | Ph.ac. 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.klw=k xfr (1) Synapsis is pairing of homologous chromosomes which occurs during meiosis but it is absent in mitos. Sol. (3) Mutations destroy telomerase inhibitor. (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. Main Road. Near City Mall.in | E-mail : contact@resonance. Sol. 27.resonance. CORPORATE OFFICE : CG Tower. 28. d j nsrsgSaA (1) Cancer will be caused by increased telomerase activity making the cancerous cells immortal & not by inhibition of telomerase production.ac. IPIA. Resonance Eduventures Ltd.) . A-46 & 52.qXeu (2) rd q Z: ih rarq (3) d s fUnzd k d k foy ksiu (4) xq .: +91-744-3192222 | FAX No. | NEET-2016 | 01-05-2016 | Which part of the tobacco plant is infected by Meloidogyne incognita ? Code-C. : +91-022-39167222 Ph. Office : J-2.Y Ans.k t krk \ mRifjorZu d slaca/k esad S al j d ks f'kd kvksad sfy . : +91-744-3012222. (4) Mutations inactivate the cell control. Cotyledon of maize grain is called : vf/kd rj d od ksaesad ksf'kd k fHkfÙk d k . Jawahar Nagar.o d kSu lk gS\ (2) Chitin (4) Cellulose (2) d kbfVu (4) ls Y.k d ksfuf"Ø . Kota (Raj. (1) Root (1) t M+ (1) 26.324005 Reg. (1) Hemicelluloses (3) Peptidoglycan (1) gs ehlsY.ac.k d snkSjku fuEufy f[kr es alsd kSu&lk y {k. Kota (Raj.d izeq[k vo.a=k. Which of the following is not a characteristic feature during mitosis in somatic cells? Ans. (2) Mutations in proto-oncogenes acceleration the cell cycle. One of the major components of cell wall of most fungi is : Ans. Sol.k ughaik. 6635555 | To Know more : sms RESO at 56677 Website : www. (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. Jhalawar Road.25.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 10 . (1) Synapsis (2) Spindle fibres (3) Disappearance of nucleolus (4) Chromosome movement (1) lw =k.d d ksf'kd kvksaes alelw =k.ww y kst eDd k d snkusd scht i=k d ksD.wy ks t (3) is IVhMksXy kbd u (2) 29.)-324005 | Ph.No.k d gk t krk gS\ Ans. No. fuEufy f[kr d Fkuksaesalsd kSu&lk lgh ughagS\ Ans.R. (1) Mutations inhibit production of telomeres. a (3) lgk.ac.r jks xt ud (3) cuk. (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 & . : +91-022-39167222 Ph. Kota (Raj.a=k. x.in | E-mail : contact@resonance. Sol. Ans. 6635555 | To Know more : sms RESO at 56677 Website : www.k d gk t krk gS\ Ans. (2) Guanosine and Retinol (4) Opsin and Retinol (2) Xokuks flu vkSj jsfVukWy ls (4) vks fIlu vkSj jsfVukWy ls 34.k t krk gS] gksrsgSa% Ans. Sol. | NEET-2016 | 01-05-2016 | Which of the following would appear as the pioneer organisms on bare rocks ? Code-C. Jawahar Nagar. Changes in GnRH pulse frequency in females is controlled by circulating levels of : eknkvksaesaGnRH iYl ckjackjrk cny ko d k fu. 33. A-46 & 52. No. 31.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 .ac. Resonance Eduventures Ltd. Jhalawar Road.: +91-744-3192222 | FAX No.resonance.d uXu pV~Vku ij . Kota (Raj.30. Sol.R. d ksf'kd kvksad ksD. (1) Attenuated pathogens (2) Activated pathogens (3) Harvested antibodies (4) Gamma globulin (1) {kh. (1) Transducin and Retinene (3) Opsin and Retinal' (1) Vª kaLM~. iz frj{kh (4) xkek Xy ks C.sxk \ converted into soil.324005 Reg.kSfxd cuk gksrk gS% Ans.d vxzxkeh t ho d s: i esafuEufy f[kr esalsd kSu vk. Office : J-2. Main Road.wflu vkSj jsfVuhu ls (3) vks fIlu vkSj jsfVuy ls (3) Rhodopsin is made of opsin & retinal.d d ks f'kd k.Y Ans.k fd ld sifjlap j. x.) .in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 11 . Sol.¡ (4) vko/kZRod d ks f'kd k. CORPORATE OFFICE : CG Tower. Near City Mall. jks xt ud (2) lfØ f. (1) Lenticels (2) Complementary cells (3) Subsidiary cells (4) Bulliform cells (1) okrjU/kz (2) iw jd d ksf'kd k.)-324005 | Ph.k d j fn. 32. Specialised epidermal cells surrounding the guard cells are called : }kj d ksf'kd kvksad ks?ksjusoky h fof'k"Vhd `r ckgkzRoph.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. : +91-744-3012222.ksd h cw¡nksaesaft Ugsaeq¡ g }kjk fny k. IPIA.No.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 . (3) reduce the blood supply to the brain (4) decrease the affinity of hemoglobin with oxygen. | NEET-2016 | 01-05-2016 | Which of the following features is not present in the Phylum .ac. (1) 3 .No.k t krk \ Ans. 37. t krsgSa? (1) 3 . Near City Mall. CORPORATE OFFICE : CG Tower. Kota (Raj. (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. gksrk gSaA jsIVhfy . Sol.Y fuEufy f[kr y {k. Jawahar Nagar.resonance.k (1) release bicarbonate ions by the liver.k Q kby e&vkFkzkZ siksMk easughaik. 36.k d e gkst k.Arthropoda? Code-C.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 12 . 6635555 | To Know more : sms RESO at 56677 Website : www.: +91-744-3192222 | FAX No.k ges'kk gh t arqv ksad svuq: ih oxZesaik. (1) .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.&Lia nu d h nj d e gkst k.k d kW M sZVk (2) Resonance Eduventures Ltd.d {k oky k ân.)-324005 | Ph. ft lesaviw.kkasesalsd kSu&lk y {k. Jhalawar Road.) .d ` r }kjk ckbd kcksZusV d k fu"d klu gksusy xsxkA (2) ân. Office : J-2. A-46 & 52. Kota (Raj. No. Ans.kZr% caVk gqv k .d fuy . (2) reduce the rate of heart beat. : +91-022-39167222 Ph. (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.in | E-mail : [email protected] Reg.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.sxh (3) efLr"d d k : f/kj la Hkj. : +91-744-3012222. IPIA.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.ac.35. Main Road.k t kuk eseSfy . kqd k (3) .kqv ksad k (2) rhu Xy hljkW y v.d izk: ih olk d k v.| NEET-2016 | 01-05-2016 | Code-C.q Xeut h t ho esad soy .Y Sol.ac.srFkk lgh fod Yi pqfu. 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.d Xy hljkW y vkSj .k d jrsgSa (b) lgiz Hkkfork (ii) fo"ke.)-324005 | Ph. 6635555 | To Know more : sms RESO at 56677 Website : www.resonance.kqvkSj rhu olk vEy v.ad ksiwjh rjg vfHkO.324005 Reg. Chordates can be gnathostomata & agnatha (without jaws).d olk vEy v.sy hy Lo. Near City Mall.kZu lseSp d hft . 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.Mklu Resonance Eduventures Ltd. Main Road.a=k. Only cartilaginous fishes (chondrichthyes) show cartilaginous endoskeleton without exception 38. Kota (Raj. Office : J-2. Kota (Raj.: +91-744-3192222 | FAX No.Dr d jrk gS (c) cgq izHkkfork fo"ke.kqv ksad k (4) .) .d y t hu vus d y {k.kqfd ld k cuk gksrk gS\ Ans.R.d gh . In mammals. No.ad ksvfHkO.k ny (3) la . A-46 & 52. IPIA.k d k fu.kqd k (3) fat is a triglyceride which is made up of 3 molecules of fatty acids and one molecule of glycerol 40. mud so.d Xy hljkW y v. prototheria group shows oviparity while metatheria & eutheria show viviparity. Sol.kqv ksavkSj .: 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 .sy hy Lo.d y y {k.kst d (1) (2) Anther (4) Placenta (2) ijkxd ks "k (4) cht k.qXeut h t ho esanksuksagh . (1) Thalamus or petal (3) Connective (1) iq "iklu . : +91-022-39167222 Ph.Dr d jrsgSaA (iv) . x. Reptilia has an order crocodilia which shows 4 chambered heart. : +91-744-3012222.ac.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 13 . (3) 39. A typical fat molecule is made up of : .d olk vEy v. Jawahar Nagar. Jhalawar Road.in | E-mail : [email protected] 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. CORPORATE OFFICE : CG Tower. (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.No. 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. : +91-744-3012222.d MkblSd sjkbM gSA (3) ls Y.a. No.d y jTt qd d slkFk t qM +k gksrk gS] D.d ikWfy lSd sjkbM gSA (4) . Which one of the following is a characteristic feature of cropland ecosystem? fuEufy f[kr esalsd kS u . Jhalawar Road.q Dr vehuksvEy gSA (2) lq Ø ksl . Sol. (4) Uracil is a pyrimidine. and the other at night.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 . | 01-05-2016 | Code-C.R.d d `f"kHkwfe ikfjrU=k d k vfHky {k.w jSfly .k. 43. 42.k jkr esagksrh gS A (2) nks uksaizfØ .kkad fHkUu gSA (4) mi.324005 Reg.q ZDr izfØ .k.qZDr d Fkuksaesa¼d kj.wy ksl . Near City Mall. (2) Both processes cannot happen simultaneously.| NEET-2016 41.d lYQ j. 6635555 | To Know more : sms RESO at 56677 Website : www. (4) The above processes happen only during nighttime.k d gy krk gS: (1) Okazaki fragment (2) Polysome (3) Polymer (4) Polypeptide (1) vks d kt kd h [k.k d snkSjku mlh jU/kzlsd kcZu MkbvkWDlkbM ikni lsfolfjr gksrh gSA mi. (1) One process occurs during day time. (3) Both processes can happen together because the diffusion coefficient ofwater and CO2 is different.kksaij fopkj d j½ .in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 14 . Office : J-2. IPIA.in | E-mail : contact@resonance. Jawahar Nagar.d iz fØ . Kota (Raj.)-324005 | Ph.M (2) ikW y hlkse (3) ikW y hej ¼cgqy d ½ (4) ikW y hisIVkbM (2) 44. %: Ans. 45.d lkFk gksld rh gS aD.exploitation (3) Alien species invasion (4) Habitat loss and fragmentation Resonance Eduventures Ltd.k xq.resonance. (1) Ecological succession (3) Least genetic diversity (1) ikfjrfU=kd vuq Ø e. (1) Xy kbflu .ac.k gS\ Ans.k.: +91-744-3192222 | FAX No. A-46 & 52.d lad qy t ksRNA d s. Main Road. (1) .No.k (3) U.) . 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. (2) Sucrose is a disaccharide. Kota (Raj. CORPORATE OFFICE : CG Tower.ad soy jkr esagksld rh gSa A (3) Ans. (1) Glycine is a sulphur containing amino acid. (3) Cellulose is a polysaccharide.Y Which one of the following statements is wrong? fuEufy f[kr esalsd kS u&lk d Fku xy r gS\ Ans. dioxide diffuses into the plant during photosynthesis. Through the same stomatal opening carbon.d fod Yi pqfu.ks afd t y vkSj CO2 d k folj.sa.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` .ac. A complex of ribosomes attached to a single strand of RNA is known as : jkbcksl kse d k .d lkFk ughagksld rhA (3) nks uksaizfØ . : +91-022-39167222 Ph.k fnu esarFkk nwl jh izfØ . d d h vko'.ksxSl &mRiknu d sfy . : +91-022-39167222 Ph. The primitive prokaryotes responsible for the production of biogas from the dung of ruminant animals.kZxqgk ifjo/kZ u d snkSjku vfu/kkZfjr vkSj vjh.kqv kasd s (2) y o. A-46 & 52.sehu lsfufeZr cká d ad ky Resonance Eduventures Ltd.) .kjkfx.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.: +91-744-3192222 | FAX No.in | E-mail : [email protected] (4) iz frca/ku . d kj.k d gk¡ik. (1) Antennae compIex (3) Lumen of thylakoids (1) .k (3) (2) Polymerase chain reaction (4) Restriction enzymes (2) ikW y hejst J`a[ky k vfHkfØ .| NEET-2016 | 01-05-2016 | Code-C.at kbe 48.&Mh–. : +91-744-3012222.R.– la d j.u-.)-324005 | Ph. fony u (4) N– .k isfjIy SusVk vesfjd kuk esaugh ik.k (3) ft a d vaxqfy fo'y s"k. Near City Mall.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-. Kota (Raj. IPIA. Jhalawar Road. Jawahar Nagar.Y t Urqv ksavkSj ikniksad h foy qfIr d k fuEufy f[kr esalsd kSu&lk . Main Road.d rk ughagksrh \ Ans.– vaxqfy Nkiu d h fd lh Hkh rd uhd d sfy . mÙkjnk.resonance.h t krh gS\ Ans.u–.No. In a chloroplast the highest number of protons are found in : gjfr y od esaizksVkWu d h vf/kd re la[. (4) 46.ls fVy Xy wd ksl . No.ac. CORPORATE OFFICE : CG Tower. 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.kh] t ks jks eUFkh t arqv kasd sxkscj ls ck. Kota (Raj.d lclseq[. fuEufy f[kr esalsfd l . Office : J-2.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 15 .ac. nsg esanh.u-.h gksrsgS] fd ld s varxZr vkrsgS\ (1) lq t hok. 47.kad s Ans. include the : (1) Eubacteria (2) Halophiles (3) Thermoacidophiles (4) Methanogens osvkfne izkd ~d sUnzd h izk.ks ad s (3) rki&vEy (4) ehFkS ut ud ksad s jkfx. 6635555 | To Know more : sms RESO at 56677 Website : www.324005 Reg. (4) 49. (1) DNA -DNA hybridization (3) Zinc finger analysis (1) Mh–.k t krk gS\ (1) fo[ka M 'k% [kafMr (2) ns gxqgk d s: i (3) Hkw z.kh. : +91-022-39167222 Ph.) .| NEET-2016 | 01-05-2016 | Code-C. Q ly ksad ksQ y hnkj ikS/kksa¼y sX. CORPORATE OFFICE : CG Tower. Office : J-2.we½ . IPIA. A-46 & 52.in | E-mail : contact@resonance. Periplaneta Americana shows indeterminate and spiral cleavage 50. Jawahar Nagar.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 16 . 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 . (3) Sol.k ?kkl pkjxkg d slkFk cny d j y xkusd ksD. (2) 51. : +91-744-3012222.ac. Kota (Raj.No. 6635555 | To Know more : sms RESO at 56677 Website : www. No.ac.Y Ans.: +91-744-3192222 | FAX No. 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 Resonance Eduventures Ltd.324005 Reg.R. Kota (Raj. Main Road.k d gk t krk gS\ (1) LFkkukUrjh Ñ f"k (2) y s[ks rh (3) leks Ppjs[kh.resonance. Near City Mall. (4) iV~ Vhnkj [ksrh [ksr Ans. Jhalawar Road.)-324005 | Ph. : +91-022-39167222 Ph. (1) (a).Ñ r&f'kjk Ans. Kota (Raj. IPIA. which blood vessel would normally carry largest amount of urea? (1) Hepatic Portal Vein (2) Renal Vein (3) Dorsal Aorta (4) HepaticVein. ? (1) lw{ket ho mRikn v uqiz. (c) Phenylketonuria is an autosomal recessive gene disorder.e y kbist rsy d s/kCcksad ksgVkuk C.: +91-744-3192222 | FAX No.Y uhpsnh x. 6635555 | To Know more : sms RESO at 56677 Website : www.in | E-mail : contact@resonance. (b) Down's syndrome is due to aneuploidy. (4) (a). No. Lru/kkfj. : +91-744-3012222.ac. Resonance Eduventures Ltd.h rkfy d k esaxy r fey k. A-46 & 52.) .wVk. Main Road. Office : J-2. CORPORATE OFFICE : CG Tower.wjh.R. (c)and (d) are correct.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.No.w fj. Kota (Raj. In mammals.r% lclsvf/kd . Pick out the correct statements : (a) Haemophilia is a sex-linked recessive disease.resonance. Jhalawar Road.Ñ r fuokfgd k f'kjk (2) o` Dd &f'kjk (3) i` "B egk/keuh (4) .ksaesad kSu&lh : f/kj&okfgd k lkekU.)-324005 | Ph. Near City Mall.ksx Dy kWLVªhfM.k ogu d jrh gS\ (1) .| NEET-2016 | 01-05-2016 | Code-C.ac. (b) and (c)are correct.324005 Reg.h enksad kspqfu.h x. Urea/Ornithine cycle takes place in liver so the vein leaving liver possesses maximum urea which is hepatic vein 53.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 17 . Jawahar Nagar. (3) (b)and (d) are correct. (d) Sickle cell anaemia is an X-linked recessive gene disorder. (1) 52. (4) Sol.fy d e (2) VªkbZ d ksM ekZiksy hLiksje lkbDy ks Liksfju A izfrj{kk laned vkS"kf/k (3) eksuSLd l ijI. (2) (a) and (d) are correct. Kota (Raj.ksl kse (2) i{ekHkks a] d 'kkHkksavkSj Ans.)-324005 | Ph. (1) Sol. A-46 & 52. Centrioles and Cilia (4) Centrioles. Nucleosome and Centrioles (2) Cilia.R. (b) vkS j (c) lgh gSA (2) (a) vkS j (d) lgh gSA (3) (b) vkSj (d) lgh gSA (4) (a). : +91-744-3012222. Main Road.resonance.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 18 .Ñ nXU. Kota (Raj.kl d h okfguh d sxz g. Office : J-2. Jhalawar Road. Jawahar Nagar. (1) 55.wfj.) .ks a] U. 6635555 | To Know more : sms RESO at 56677 Website : www. lw{eufy d k. No. IPIA.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. (c) vkS j (d) lgh gSA Ans.¡la?kVd gksrh gS: (1) rkjd k. Flagella and Peroxisomes (3) Spindle fibres. % (a) gheks Q hfy .in | E-mail : [email protected] 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). : +91-022-39167222 Ph.| NEET-2016 | 01-05-2016 | Code-C. 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.No.k½ . CORPORATE OFFICE : CG Tower.324005 Reg.ac. Sickle cell anemia is an autosomal codominant disorder 54. Microtubules are the constituents of : d ikV vojksf/kuh (1) Centrosome.: +91-744-3192222 | FAX No. Spindle fibres and Chromatin. vkSj rkjd d sUnzksd s ijvkWDlhd k. Near City Mall.wfDy .Y lgh d Fku pqfu.k gksrk gSA (c) Q s fuy d hVksuesg ¼fQ ukby d hVksU.k fy ax&lgy Xu vizHkkoh jksx gSA (b) Mkmu fla M ªkse vlqxqf. fuEufy f[kr esalsd kS u&lh lajpuk .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) Resonance Eduventures Ltd.krk d sd kj.ac. : +91-744-3012222.ac.56. Kota (Raj. In tender coconut.k ughagS? (1) vks iaf'k.Q y fHkfÙk Ans. (2) 58. CORPORATE OFFICE : CG Tower.kiwohZ e/.R. : +91-022-39167222 Ph.kiks"k (2) vUr%Q y fHkfÙk (3) xw nsnkj (4) LorU=k d s Unzd h Hkzw. Jawahar Nagar.lh (2) fy fy .y ikuh D.324005 Reg.: +91-744-3192222 | FAX No. Jhalawar Road. Near City Mall. Main Road. 6635555 | To Know more : sms RESO at 56677 Website : www.)-324005 | Ph.in | E-mail : contact@resonance. No. Pitcher of nepenthes is modification of leaf Resonance Eduventures Ltd. Tricarpellary.Y d Ppsukfj.resonance. Office : J-2.k gS\ (1) LorU=k d s Unzd h Hkzw. Kota (Raj.No. (2) Sol.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 19 .qDrk.) . 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 . syncarpous gynoecium is found in flowers of: (1) Poaceae (2) Liliaceae (3) Solanaceae (4) Fabaceae f=kd ks"Bd h] . d h piVh lajpuk (2) us iUFkht d k ?kV (3) flVª l d sd kVs (4) [khjsd siz rku Ans. | 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.y eas] ukfj. edible part is liquid endosperm that represents free nuclear endosperm 57.lh (3) lks y Sus lh (4) Q S csl h Ans.d rusd k : ikUrj.Mih t k.k¡x fd ld siq"i esagksrk gS\ (1) iks .ac. IPIA. A-46 & 52. (1) Sol. vkSj og Lo&d ksf'kd kvks ad ksu"V d jusy xsrksbld sifj.ac. If self & non-self recognization power is lost than immune cells can attack our own body cells and cause auto immune disease Resonance Eduventures Ltd.kasesa] izfrj{kk ra=k Lo&d ksf'kd kvksavkSj xSj&d ksf'kd kvksaesaHksn d j ld rk gSA .: +91-744-3192222 | FAX No. (4) 61.) . izfrj{kk (2) . No. (4) Sol.Ut kbe fd llsizkIr fd .kkZ Hk o`Ur (4) i.R.No.rk d sd kj. Stems modified into flat green organs performing the functions of leaves are known as : (1) Scales (2) Cladodes (3) Phyllodes (4) Phylloclades ifÙk.kkZ Hk LRkEHk Ans.ks cSfly l I. In higher vertebrates. IPIA.ksad k d k. Near City Mall.wfVMk Q sjksDlhMsUl (2) FkeZ l .kkZ Hk ioZ (3) i.)-324005 | Ph.| NEET-2016 59. : +91-744-3012222. (2) 60. CORPORATE OFFICE : CG Tower. Jhalawar Road. | 01-05-2016 | Code-C.w Mkseksukl (3) fFk.in | E-mail : [email protected] .k d gk t krk gS\ (1) 'kYd (2) ik.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 20 .kkeLo: i D. Office : J-2. If this property is lost due to genetic abnormality and it attacks self-cells. Kota (Raj. A-46 & 52.g xq.s y t hZvuqfØ .k gksxk\ (1) lfØ . : +91-022-39167222 Ph.k (3) fujks i vLohd kj d j nsuk (4) Loiz frj{kk fod kj Ans. 6635555 | To Know more : sms RESO at 56677 Website : www.resonance.DosfVd l (4) cS fly l lcfVfy l Ans. the immune system can distinguish self-cells and non-self. Jawahar Nagar. Main Road. then it leads to : (1) Active immunity (2) Allergic response (3) Graft rejection (4) Auto-immune disease mPprj d 'ks: fd . Kota (Raj.k t krk gS\ (1) L.ac.324005 Reg.Zd jusoky s ] piVsgjsvax esa: ikUrfjr rusd ksD.k u"V gkst k.fn ra=k d k vkuqoaf'kd vilkekU.Y The taq polymerase enzyme is obtained from : (1) Pseudomonas putida (2) Thermus aquaticus (3) Thiobacillus ferroxidans (4) Bacillus subtilis VSd ikWfy ejst . (4) It can be used for detection of Down syndrome. fd ld h vko'.d d Fku uke i) fr d sfu.qXed d svfHkxeu d sfy .eksad sfo: ) gS\ (1) uke d kst c gkFk lsfy [krsgSrks mlsjs[kkafd r fd . (1) Sol. (2) It is usually done when a woman is between 14 -16 weeks pregnant. : +91-022-39167222 Ph.62. Near City Mall.: +91-744-3192222 | FAX No. Jawahar Nagar. (2) 63. A-46 & 52. 6635555 | To Know more : sms RESO at 56677 Website : www. Office : J-2. No. Main Road.No.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 21 .ac. Which one of the following is contrary to the rules of nomenclature. fu. transport of male gametes requires : (1) Water (2) Wind (3) Insects (4) Birds czk. (1) When written by hand.Y Nomenclature is governed by certain universal rules. Jhalawar Road. IPIA.)-324005 | Ph.ksQ kbV vkSj VsfjMksQ kbV esauj . Resonance Eduventures Ltd. Kota (Raj.resonance.d rk gksrh gS\ (1) t y (2) iou (3) d hV (4) i{kh Ans. (3) It is used for prenatal sex determination.in | E-mail : [email protected] 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. : +91-744-3012222.) .R. 64.eksa}kjk fu/kkZfjr gks rh gSA fuEufy f[kr easalsd kSu lk . Kota (Raj. | NEET-2016 | 01-05-2016 | Code-C. CORPORATE OFFICE : CG Tower.324005 Reg. 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. 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. In context of Amniocentesis.ac. In bryophytes and pteridophytes.which of the following statement is incorrect ? (1) It can be used for detection of Cleft palate. 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. : +91-744-3012222.k (4) blsMkmu d sfy . (3) Resonance Eduventures Ltd.) . Near City Mall. Jawahar Nagar.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. 66. Main Road.g iSy sV½ d k irk y xkusd sfy .qDr fd .No. A-46 & 52.k t krk gSA flaM ªkse d k irk y xkusd sfy .ac. iz. IPIA. gastric acid is secreted by the : (1) acidic cells (2) gastrin secreting cells (3) parietal cells (4) peptic cells vkek'k. In the stomach. gastric acid is HCl secreted by parietal or oxyntic cells.k t krk gSA Ans.)-324005 | Ph.| NEET-2016 | 01-05-2016 | Code-C.k t krk gSA vkerkSj lsrc fd .qDr fd . 6635555 | To Know more : sms RESO at 56677 Website : www. Jhalawar Road. Cleft palate is a structural defect and cannot be determined by amniocentesis.324005 Reg. CORPORATE OFFICE : CG Tower.in | E-mail : [email protected]. iz.klw=k d svaR. d ksf'kd kvksals (4) is fIVd d ksf'kd kvksals Ans.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) . iz.R. Office : J-2. : +91-022-39167222 Ph.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 22 . (1) Sol.qDr fd .: +91-744-3192222 | FAX No. 65.klw=k d slw=kd sUæ ij ij Ans. Kota (Raj.klw=k d sd kbusVksl kse ij (2) xq . (3) Sol.klw=k d sd kbuksVksd ksj (4) xq . 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 .ka'k ij (3) xq . No. Kota (Raj.resonance. Kota (Raj. (3) Sol.LVj (2) gkbMª ks tu cU/k cU/k (4) lgla .R.)-324005 | Ph. Ans. Jhalawar Road. Jawahar Nagar. IPIA.No. (3) Sol. | NEET-2016 | 01-05-2016 | Code-C. No. Kota (Raj. 6635555 | To Know more : sms RESO at 56677 Website : www.Sol.kh d kSu lk gS? (1) leq æh.324005 Reg. 68.resonance. 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. : +91-744-3012222. : +91-022-39167222 Ph. CORPORATE OFFICE : CG Tower.ac.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 23 . all three are bounded by double membrane 69.kd k (3) gfjry od (4) y . Which is the National Aquatic Animal of India? (1) Sea. Main Road.: +91-744-3192222 | FAX No.ud k.d y d y k lsf?kjk gksrk gS? (1) d s Uæd (2) lw =kd f.ac.Y kinetochore of chromosome facilitates the attachment of spindle fibre (chromosomal fibre) and pole 67. t y h.?kksM +k (2) xa xk d h 'kkd Z (3) unh d h MkW fYQ u (4) Cy wg~ osy Ans.horse (2) Gangetic shark (3) River dolphin (4) Blue whale Hkkjr d k jk"Vªh.kst h cU/k Resonance Eduventures Ltd. 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 . Near City Mall. izk. Office : J-2.) .ksft r gksrsgSa\ (1) MkblYQ kbM lsrq (3) Q kLQ ks M kb. Except Lysosome. A-46 & 52.in | E-mail : contact@resonance. Office : J-2. fufd y . 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 . CORPORATE OFFICE : CG Tower.sfo"kk. eSaxuht Ans.k (3) vfHklkjh fod kl (2) vilkjh fod kl ds ds (4) lk>k oa 'kijaijk Resonance Eduventures Ltd.s alaØ e. RNA is of low molecular weight 72. manganese fuEufy f[kr esalsd kS u lHkh rhu c`gÙkiks"kd gSa\ (1) ukbVª kst u.afd l d kj. (1) Sol.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 24 . Analogous structures are a result of : (1) Stabilizing selection (2) Divergent evolution (3) Convergent evolution (4) Shared ancestry leo`fÙk lajpuk. : +91-022-39167222 Ph.k mRiUu gksrh gS\ (1) fLFkjd kjh oj. (1) or bonus Sol. molybdepum (4) Molybdenum. Near City Mall. manganese.) .k d jrsgSaA Ans. Q kLQ ksjl (2) cks jkWu ft ad .od Hkkj oky k gksrk gSA vkoj.k d k vHkko gksrk gSA (3) . (1) 70. Kota (Raj.: +91-744-3192222 | FAX No. 6635555 | To Know more : sms RESO at 56677 Website : www. Main Road. Jhalawar Road. eSXuhf'k. eksy hCMsue (4) eks y hCMsue. IPIA. eSaxt hu (3) y kS g.| NEET-2016 | 01-05-2016 | Code-C. all three are macronutrients ? (1) Nitrogen. In viroid. zinc. esxy r gS? (1) mud k RNA mPp (2) mues aizksVhu vkf.Y Ans.e.M d sfo"k.ac. phosphorus (2) boron. : +91-744-3012222.No.R. Jawahar Nagar. In which of the following.kq v ksalsvis{kkÑ r NksVsgksrsgSA (4) . nickel.324005 Reg. rkez. A-46 & 52. No answer is correct 71.resonance.in | E-mail : contact@resonance. magnesium.)-324005 | Ph. No. (3) Iron.copper.ac. Kota (Raj. Jawahar Nagar. (3) Insects that consume pollen or nectar without bringing about pollination are called pollen/ nectar robbers. Office : J-2.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. 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. Near City Mall. (4) LH and FSH decrease gradually during the follicular phase xy r d Fku d kspqfu.324005 Reg. No.ac. (4) Sol.| NEET-2016 | 01-05-2016 | Code-C.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. Kota (Raj.)-324005 | Ph.kqt uu eslgk. LH and FSH both increase during follicular phase. 74. (3) 73. but only one pollen tube of the same species grows into the style. Main Road. (2) Pollen grains of many species can germinate on the stigma of a flower. 75.resonance.k if{k.Y Ans. : +91-022-39167222 Ph. Which of the following statements is not correct? (1) Some reptiles have also been reported as pollinators in some plant species.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. A-46 & 52. d jrk gSt ks'kqØ k.R. IPIA.) . (4) Sol.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 25 .ac. : +91-744-3012222.No.d y {k.ksavkSj Lru/kkfj. 6635555 | To Know more : sms RESO at 56677 Website : www. (4) Pollen germination and pollen tube growth are regulated by chemical components of pollen interacting with those of the pistil Resonance Eduventures Ltd. . (1) LH y hfMx d ksf'kd kvksals.in | E-mail : [email protected] t krk gS (1) fu. (1) Warm blooded nature (2) Ossified endoskeleton (3) Breathing using lungs (4) Viviparity fuEufy f[kr y {k.: +91-744-3192222 | FAX No. birds are oviparous while mammals are oviparous(prototherians) and viviparous(metatherians and eutherians).kksaesalsd kSu lk . CORPORATE OFFICE : CG Tower. Kota (Raj.ksanksuksaesaughaik. Jhalawar Road. Which one of the following characteristics is not shared by birds and mammals? . k fd . No.in | E-mail : [email protected] rFkk L=khd sl j d h ikjLifjd fØ .k rFkk ijkx ufy d k o`f) ] ijkxd . 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 .af=kr gksrh gSA Ans. 6635555 | To Know more : sms RESO at 56677 Website : www. : +91-744-3012222. 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.d iq"i d sorhZd kxzij vad qfjr gksld rsgaSijUrqmlh t kfr d sijkxd .R.| NEET-2016 | 01-05-2016 | Code-C. prevent ovulation Resonance Eduventures Ltd.ac. : +91-022-39167222 Ph.k ed jan d ksxzg. CORPORATE OFFICE : CG Tower.k .ksad sijkxd .fud ?kVd ksa}kjk fu.d lad j. Jhalawar Road.k va d qj.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 26 .d ijkx–ufy d k ofrZd k esavkxsc<+rh gSA (3) d hV t ksfcuk ijkx.k d jrsgSmUgsaijkx/ ed jan pksj d grsgSa (4) ijkxd .k d sQ y Lo: i mRiUu jklk. suppress sperm motility and fertilizing capacity of sperms (4) Hormonal Prevent/retard entry of Contraceptives sperms.k (1) Sol. crk. (2) Sol. Office : J-2.k d jrsgq. Kota (Raj.Y fuEufy f[kr esal sd kS u lk d Fku lR.resonance.ac.sijkx .ksaeasijkx.sx.k gSa\ Ans. Kota (Raj.kksad h d soy .) . 77.324005 Reg.kq d t uu (3) eq d qy u (4) d kf. Main Road.sgSA (2) cgq r lkjh t kfr. IPIA. Near City Mall. (1) vla xt uu (2) cht k. ughagS\ (1) d q N lfjl`i d qN ikni t kfr.)-324005 | Ph. 76.No. A-46 & 52.: +91-744-3192222 | FAX No. Jawahar Nagar. Near City Mall.)-324005 | Ph.h .kqv ksd h xfr'khy rk . IPIA. Resonance Eduventures Ltd. 6635555 | To Know more : sms RESO at 56677 Website : www. CORPORATE OFFICE : CG Tower. A-46 & 52.No.ac.k¡ fu"ksp u jks d rh gSaA (3) var% xHkkZ'k. The amino acid Tryptophan is the precursor for the synthesis of : (1) Cortisol and Cortisone (2) Melatonin and Serotonin (3) Thyroxine and Triiodothyronine .R.resonance.jkW fDlu (4) bZ LVªkst u vkSj VªkbZv k.kqv ksad h Hk{kd ksf'kd rk c<+k nsrh gS] 'kqØ k.qfDr.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.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.324005 Reg. Jawahar Nagar.) .ac.ksM ksFkk.| NEET-2016 | 01-05-2016 | Code-C. : +91-022-39167222 Ph.in | E-mail : contact@resonance. (4) Estrogen and Progesterone vehuks avEy fVªIVksQ Su fd ld sla'y s"k.k d sfy .: +91-744-3192222 | FAX No.k¡ 'kqØ k. iwoZxkeh gksrk gS\ (1) d ks fVZl ksy vkSj d ksfVZl ksu (2) es y kVksfuu vkSj lsjksVksfuu (3) Fkk.oafu"ksp u {kerk d k eanu d jrk gSA (4) gkWeksZuh xHkZfujks/kd 'kqØ k. Office : J-2. Kota (Raj.kqt uu ughagksusnss rs (2) jks/k ¼csfj.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 27 . : +91-744-3012222. Jhalawar Road. (2) Sol. Main Road. Kota (Raj. No.j½ fof/k.jksfuu vkSj izkst sLVsjkWu Ans. Jawahar Nagar.resonance. (2) More abundant species will exclude the less abundant species through competition.324005 Reg. 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 Resonance Eduventures Ltd.k gksxk \ (1) vkW Dlht u d h d eh d sd kj.d unh esat c d kcZfud vif'k"V lsHkjiw j ?kjsy wokfgr ey cgd j fxjrk gks] rksmld k ifj. (4) An increased production of fish due to biodegradable nutrients .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[. Kota (Raj. Office : J-2. iks"k. Gause's principle of competitive exclusion states that: (1) Larger organisms exclude smaller ones through competition. (3) Competition for the same resources excludes species having different food preferences. : +91-022-39167222 Ph.k d sd kj.| NEET-2016 79. (2) Drying of the river very soon due to algal bloom.kWej t k. (4) No two species can occupy the same niche indefinitely for the same limiting resources. Near City Mall.kkke D.d gh fud sr esvlhfer vof/k d sfy . No. Jhalawar Road.sxhA (4) ck. 81. IPIA. Main Road. 6635555 | To Know more : sms RESO at 56677 Website : www.: +91-744-3192222 | FAX No.ac. Li/khZviot Zu d k xkWl sfu.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. CORPORATE OFFICE : CG Tower.sxkA (1) Sol.No. ughajg ld rh D. A-46 & 52.ks fMxzsM scy Ans.) .k eNy h d k mRiknu c<+t k. t kusoky h Lih'kht d ksvioft Zr d j nsxhA (3) leku lal k/kuksad sfy .in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 28 .k eNfy .k unh t Ynh gh lw[k t k.)-324005 | Ph. t kusoky h Lih'kht Li/kkZ}kjk d e la[.sxh A (3) t y h.ac.k esaik. Kota (Raj. 80.k esaik. 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 .in | E-mail : contact@resonance. (4) Sol. Hkkst u d h lef"V esao`f) gkst k.kasfd lhekd kjh lal k/ku leku gh gksrsgSaA Ans.R. | 01-05-2016 | Code-C. (3) Increased population of aquatic food web organisms. : +91-744-3012222.saxhA (2) 'kS oky izLQ qVu d sd kj. fd l uke lst kuk t krk gS\ (1) d ks jksuk (2) d S fjuk (3) iS il (4) oS Dlhy e Ans. Near City Mall. 6635555 | To Know more : sms RESO at 56677 Website : www. : +91-022-39167222 Ph. Kota (Raj. A-46 & 52. 83. (4) Standard or vexillum Wing or alae Keel or carina Sol.k gksrk gS\ (1) Q s Q Mksad sHkhrj ikuh . CORPORATE OFFICE : CG Tower.324005 Reg. Kota (Raj.in | E-mail : contact@resonance. Jhalawar Road. Main Road.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 29 . Office : J-2.: +91-744-3192222 | FAX No. 82.yt hZvfHkfØ .) .ac.resonance. No.k (3) Q s Q M+ksesekLV d ksf'kd kvksad h . 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. (3 or 4) Sol.| NEET-2016 | 01-05-2016 | Code-C.k (4) 'okluy h d h 'kks Fk Ans. IPIA. Papilionaceous corolla Which of the following is a restriction endonuclease? (1) RNase (2) Hind II (3) Protease Resonance Eduventures Ltd.Y vLFkek d k d kj. Jawahar Nagar.kq}kjk laØ e.)-324005 | Ph.No.k D. : +91-744-3012222.d f=kr gkst kuk (2) Q s Q Mksd k t hok.ac.R. asthma is an allergic disease caused by allergens and characterized by inflammation of tracheobronchial tree. : +91-744-3192222 | FAX No.u. Jawahar Nagar. CORPORATE OFFICE : CG Tower. (4) Both (a) and (b) are correct. 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.aM ksU. igkM+h ij p<+uk vklku gksrk gSA D.wfDy . Jhalawar Road.d izfrca /k .u. (2) It is easier to carry a small body weight. A-46 & 52.resonance.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 30 . (3) Sol. (3) (b) is correct but (a) is false. (2) Sol. Of the above statements which one of the following options is correct ? (1) Both (a) and (b) are false.in | E-mail : [email protected] Reg. (b) The first autotrophic organisms were the chemoautotrophs that never released oxygen. (4) Small animals have a lower O2 requirement.ksafd (1) Nks Vst arqv ksad seqd kcy sescM+st arqv ksad h isf'k.Y (4) DNase I fuEufy f[kr esalsd kS u lk . It is much easier for a small animal to run uphill than for a large animal.ac. Resonance Eduventures Ltd. 84.ac. Near City Mall. 6635555 | To Know more : sms RESO at 56677 Website : www.| NEET-2016 | 01-05-2016 | Code-C. IPIA. Main Road.t (4) Mh. Kota (Raj.No. (2) (a) is correct but (b) is false. smaller animals have higher BMR related with sustained energy production and delayed muscle fatigue 85. Kota (Raj.R. (3) Smaller animals have a higher metabolic rate. Office : J-2. : +91-744-3012222.) . cMs+vkd kj d st arqv ksad eqd kcy sesNksVsvkd kj d st arqv ksad fy .t I Ans. No. : +91-022-39167222 Ph. because: (1) The efficiency of muscles in large animals is less than in the small animals.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.d rk vis{kkÑ r d e gksrh gSA Ans.ksad h d k.)-324005 | Ph.h nj vis{kkÑ r vf/kd gksrh gSA (4) Nks Vsvkd kj d st arqv ksad hO2 vko'.t (2) fgUn II (3) iz ksfV.t gSA (1) vkj. )-324005 | Ph.kke D.in | E-mail : contact@resonance.: +91-744-3192222 | FAX No. : +91-022-39167222 Ph. d ksf'kd kvksalsfHkUu gSA blesad ksf'kd k Iy sV ugah curh vkSj bl d kj. This would result in (1) Polyteny (2) Aneuploidy (3) Polyploidy (4) Somaclonal variation [ksr lsy k.| NEET-2016 | 01-05-2016 | Code-C. There is no formation of cell plate and thus the cell is containing more number of chromosomes as compared to other dividing cells.Y t hou d h mRifÙk d slanHkZesnksd Fku fn. Jawahar Nagar. foHkkt u oky h d ksf'kd kvksad h vis{kk vf/kd xq. gSa% (a) i` Foh ij (b) iz Fke izd V gksusoky svkjafHkd Ùke t ho gjsughaFksvkSj laHkor. Main Road.hA og viusf'k{kd lsd grk gSfd .d ikni d ksf'kd k esa. A-46 & 52. Even the first autotroph was dependent on chemicals so oxygen is not released 86. Office : J-2.324005 Reg.k vok. No. (3) Sol.) . Resonance Eduventures Ltd.klw=k gSA bld k ifj.oh FksA izd V gksusoky sLoiks"kh t ho jlksLoiks"kh Fksft UgksausvkWDlht u d k mRlt Zu ughafd .d fo|kFkhZ}kjk vaR. He tells his teacher that this cell is not like other cells at telophase stage.ac.k gksxk\ (1) cgq iV~Vrk (2) vlq xqf.k bl d ks f'kd k esa vU.No. First originated organism was prokaryote chemoheterotroph and oxygen was not available on earth at that time so it must be anaerobic too.R.koLFkk ij vU.g d ksf'kd k vUR. 6635555 | To Know more : sms RESO at 56677 Website : www.sx. . CORPORATE OFFICE : CG Tower. Near City Mall. (4) Sol.ac.krk (3) cgq xq f. A cell at telophase stage is observed by a student in a plant brought from the field. : +91-744-3012222.koLFkk ns[kh x.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. Kota (Raj. Kota (Raj.resonance.krk (4) d k.sx. IPIA. Jhalawar Road.Dy ks uh fofHkUurk Ans.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 31 . (3) Sol.G. Kota (Raj.: +91-744-3192222 | FAX No. CORPORATE OFFICE : CG Tower.No.resonance. Kota (Raj. Vkal y s vksM e Resonance Eduventures Ltd.Odum (3) A.in | E-mail : [email protected] esafd l xSl d h d eh gksusij Ropk d sd Sal j d svolj c<+t k.| NEET-2016 87. Tansley (4) E.k Fkk\ (1) bZ . 6635555 | To Know more : sms RESO at 56677 Website : www. IPIA. Which one of the following is the starter codon? fuEufy f[kr esalsd kS u lk .P. : +91-744-3012222. Okfea x (2) bZ . Jhalawar Road.axsa\ (1) ehFks u (2) ukbVª l vkWDlkbM (3)vks t ksu (4) veks fu.R.k Ans.qDr ou izcU/ku d h /kkj. (3) . Haeckel bd ksflLVe (ikfjrU=k) 'kCn lclsigy sfd luscuk. Office : J-2. ih.k Hkkjr esfd l nkSjku izLrkfor d h x..d izkjEHkd izd wV gS\ (1) UAG (2) AUG (3) UGA (4) UAA Ans. fgd y t h. (2) 90.324005 Reg.ac. : +91-022-39167222 Ph.) . Joint Forest Management Concept was introduced in India during: la.)-324005 | Ph. (4) bZ . The term ecosystem was coined by : (1) E. Near City Mall. Main Road. Jawahar Nagar.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.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 32 .h Fkh\ (1) 1990s (2) 1960s (3) 1970s (4) 1980s Ans. | 01-05-2016 | Code-C.ac. Warming (2) E. (4) 89. A-46 & 52. No. 88. in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 33 . fd og ik'k d ksiw.0 ms –2 gksrsgS? i`Foh d h f=kT. the minimum speed required at the lowest point = 5gR So ans is (1) 92. the magnetic flux linked with –3 each turn of the solenoid is 4 × 10 Wb.| NEET-2016 | 01-05-2016 | Code-C. Office : J-2.k 6400 km y hft .4 × 107 J kg–2 rFkk 6.4 × 10 J –2 –2 kg and 6. CORPORATE OFFICE : CG Tower.kZd j ld s? R (1) (2) 5gR gR (3) 2gR (4) 3gR Ans. Kota (Raj. 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. The self-inductance of the solenoid is fd lh y Ech ifjukfy d k esa Q sjksa d h la[.fn nkslfn'kksad s. Jawahar Nagar.k d sfd lh Å /okZ/kj ik'k (y wi) esam nzO.M+d ksfd l fuEure osx lsiz os'k d juk pkfg.ac. the angle between these vectors is : . ¶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 Resonance Eduventures Ltd.resonance. Near City Mall. Roj.k d scjkcj gS] rksbu lfn'kksad schp d ks.in | E-mail : [email protected] Q sjslslac) pqEcd h. To complete the vertical loop.) . Kota (Raj.eku d sfd lh fi. A long solenoid has 1000 turns. (1) 2000 km (2) Sol. Jhalawar Road.sx d k ifjek. No. : +91-022-39167222 Ph. (3) PART B – PHYSICS 91.k 1000 gSA t c bl ifjukfy d k ls 4A /kkjk izokfgr gksrh gS] rc bl ifjukfy d k d sizR.ac.k gSA Ans. IPIA. r = 9000 km. 6635555 | To Know more : sms RESO at 56677 Website : www. Main Road.4 × 10 r GM =6 r2 dividing both the equations.324005 Reg. i`Foh d si`"B lsfd ruh Å ¡p kbZij xq : Roh. (1) Sol.No. 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.Y Ans.k mu nkslfn'kksad svUrj d sifjek.   = 90° 7 At what height from the surface of earth the gravitational potential and the value of g are –5. A-46 & 52.  (2) 2600 km (3) 1600 km (4) 1400 km GM 7 = 5. : +91-744-3012222. so height from the surface = 9000 – 6400 = 2600 km  94. When a current of 4A flows through it.A Ans. If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors. foHko vkSj xq: Roh.: +91-744-3192222 | FAX No.)-324005 | Ph.R.0 ms respectively ? Take the radius of earth as 6400 km.k g d seku Ø e'k% – 5. circuit is fd lh L=kksr ft ld k emf. 40 are connected in series across a source of emf V = 10 sin 340t.8 R=400 Sol. No. As a result.kkeLo: i xksy s. Then v varies as a function of the distance x between the spheres.d nwl jsd h vksj osx vlsvkrsgS A rc xksy ksad schp d h nwjh] x d sQ y u d s : i esaosx fopj. izfrd "kZ.89 W (2) (2) 0.k vkjEHk gksrk gS] vkSj bld sifj.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 34 . 50 F d k la/kkfj=k rFkk 40d k izfrjks/kd la. IPIA.resonance.51 W (3) 0.: +91-744-3192222 | FAX No. V = 10 sin 340 t gS] lsJs. as : (1) v  x –1 1/2 (2) v x (3) v x –1/2 (4) v x fd lh mHk.8 So | z | (40)2  (58. CORPORATE OFFICE : CG Tower.d nwl jslsd(d<< ) nwjh ij gSA nksuksagh xksy ksals.korhZ/kkjk ifjiFk es a'kfDr {k.ekughu Mks fj. self = Li –3 (4 × 10 )(1000) = (L)(4) L = 1 Henry 95.8)2  65 i0  v0 10  A | z | 65  irms  i0 2  10 65 2 2  10  2 Ploss  rrms R   40  0. 1/c=58. Jhalawar Road.C.| NEET-2016 | 01-05-2016 | Code-C.k fd l : i esagksrk gS? Resonance Eduventures Ltd. Office : J-2. Kota (Raj. The power loss in A.ac. Jawahar Nagar.ksalsfuy afcr] nksloZl e vkos f'kr xksy s] vU. 6635555 | To Know more : sms RESO at 56677 Website : www. (1) 0.R.k d sd kj.ksft r gS A bl izR.k] vkjEHk esa.Y Ans.No. (1) Sol. Near City Mall. Kota (Raj.kh esa20 mH d k izsjd .) .ksU. A-46 & 52.fu"B fcUnqls ] y EckbZd h nksæO. : +91-744-3012222. : +91-022-39167222 Ph. Two identical charged spheres suspended from a common point by two mass less strings of lengths . Main Road. An inductor 20 mH.r nj lsvkos'k d k {kj. gSA Ans. the spheres approach each other with a velocity v.d fu.46watt  65 2  So the nearest answer will be (2) 96. are initially at a distance d(d << ) a part because of their mutual repulsion. a capacitor 50 F and a resistor.8 – 6.ac.324005 Reg.67W (4) 0.in | E-mail : [email protected] L = 6.)-324005 | Ph. The charges begin to leak from both the spheres at a constant rate. ac. : +91-022-39167222 Ph.ac. Kota (Raj. Kota (Raj. (1) v  x (4) | 01-05-2016 | 1/2 –1 (2) v x Tcos (3) v x Code-C. Sol.No. CORPORATE OFFICE : CG Tower.324005 Reg. Near City Mall. 1 2 V 2F 8F A capacitor of 2F is charged as shown in the diagram.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 35 . 6635555 | To Know more : sms RESO at 56677 Website : www. No. Jawahar Nagar.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 qx dq 3 1/ 2  dx   x   dt 2  dt  dx –1/2 x dt  97. Main Road. Office : J-2.resonance.in | E-mail : contact@resonance. Jhalawar Road. IPIA.) . the percentage of its stored energy dissipated is : (1) 80% (2) 0% (3) 20% (4) 75% Resonance Eduventures Ltd. : +91-744-3012222.| NEET-2016 Ans.R. A-46 & 52. When the switch S is turned to position 2.)-324005 | Ph.: +91-744-3192222 | FAX No. Where  is a constant.k t krk gS] rksblesalafpr Å t kZd k izfr'kr {k.  (4) Velocity is perpendicular to r and acceleration is directed towards the origin. 6635555 | To Know more : sms RESO at 56677 Website : www.k bl izd kj xeu d jrk gSfd mld k fLFkfr lfn'k r  cos txˆ  sin tyˆ }kjk fu: fir fd . A-46 & 52. (3) 20% 5 4 V2 V2 × 100 = 80%  A particle moves so that its position vector is given by r  cos txˆ  sin tyˆ .| NEET-2016 1 2 | 01-05-2016 | Code-C. Jawahar Nagar.k nksuksagh r d sy Ecor~gSaA (3) os x  (4) os x r Ans.d fu.k nksuksagh r d slekUrj gSA d sy Ecor~gSrFkk Roj. No.k ewy fcUnqlsnwj d h vksj funsZf'kr gSA  (2) os x vkSj Roj. Jhalawar Road.) . Kota (Raj.ac.R. Initial energy stored in the 2F 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.gk¡ . Main Road.Y V 8F 2F vkjs[k esan'kkZ.k ewy fcUnqd h vksj funsZf'kr gSA (4) Resonance Eduventures Ltd.resonance. : +91-022-39167222 Ph.  (2) Velocity and acceleration both are perpendicular to r .: +91-744-3192222 | FAX No. Near City Mall.in | E-mail : contact@resonance. Office : J-2.  vkSj Roj. Which of the following is true?  (1) Velocity is perpendicular to r and acceleration is directed away from the origin.rkad gS? fuEufy f[kr esalsd kS u lk d Fku lR.324005 Reg.svuql kj 2F /kkfjrk d sfd lh la/kkfj=k d k vkos'ku fd .in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 36 . CORPORATE OFFICE : CG Tower.k x.k gS] .  d ksbZd .  (3) Velocity is acceleration both are parallel to r .No. gS?  (1) os x r d sy Ecor~gSrFkk Roj. IPIA.k x. : +91-744-3012222. gks xkA Ans.ac.k gSA t c fLop S d ksfLFkfr 2 ij ?kqek.)-324005 | Ph. Kota (Raj. (1) 80% (1) (2) 0% Sol. 324005 Reg. 2 (1) 9MR /32 (3) (2) 15MR /32 2 (3) 13MR /32 2 (4) 11 MR /32 M.ac. R M/4 R/2 Sol. ft ld h f=kT.k vkSj vkSl r ?kuRo i`Foh d h rqy uk es anksxqusgS? (1) 1 : Ans. 6635555 | To Know more : sms RESO at 56677 Website : www.V = 0 so r r V   and a  – 2 r  so a will be always aiming towards the origin.k R d h fd lh fMLd lsR O.in | E-mail : [email protected]. No. A-46 & 52. : +91-744-3012222. : +91-022-39167222 Ph.Sol. Kota (Raj. I1 = MR 2 2 2  M  R  2  4  2  3MR 2  M  R  I2 =           2 32  4  2  Inet = I1 – I2 = 100.k gS? 2 Ans. Jawahar Nagar. MR 2 3MR2 13MR2 –  2 32 32 so answer is 3.: +91-744-3192222 | FAX No. | NEET-2016 | 01-05-2016 |  ˆ V  cos t i  sin tjˆ  dr = – sint ˆi + cost ˆj V dt   dV 2 2 = –  cost ˆi –  sint ˆj a dt    since r.) .Y From a disc of radius R and mass M.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. whose rim passes through the centre is cut.kZD.u osx (vp) esaD.No. Near City Mall. passing through the centre ? nzO.eku M rFkk f=kT.u os x (ve) rFkk ml xzg ij iy k. 2 (2) 1 : 2 (3) 1 : 2 2 (4) 1 : 4 (3) Resonance Eduventures Ltd. 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. Office : J-2.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 37 . 99.)-324005 | Ph. What is the moment of inertia of the remaining part of the disc about at perpendicular axis. Main Road. IPIA. a circular hole of diameter R. Kota (Raj. Code-C. CORPORATE OFFICE : CG Tower. Jhalawar Road.k vuqikr gksxk. 101.ksft r fd .d &nwl jsd h foijhr fn'kkvksaesala.k x.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 38 .R. 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 .d nwl jsd h lgj. Kota (Raj. A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Main Road.) Ans. Kota (Raj.k gSA nks lsy ksad ksJs. Sol.rk d jrsgq.Y 4 R3 3 R Ve  R  R  V1  1 1 V2 R2 2  V1 1  V2 2 2 so answer is 3.r foHkokUrj cuk. Ve = 2G    2GM  R  | 01-05-2016 | Code-C. Jawahar Nagar. Office : J-2.)-324005 | Ph. 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. A-46 & 52. j[kk x.in | E-mail : [email protected] gksxh? (ok. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. Jhalawar Road. (1) 885 Hz (4) (2) 765 Hz (3) 800 Hz (4) 838 Hz Resonance Eduventures Ltd.resonance.| NEET-2016 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.ju fd lh izs{kd ls. 6635555 | To Know more : sms RESO at 56677 Website : www. vkSj fQ j . No. Two cells are connected in series first to support one another and then in opposite direction. The ratio of emf's is : fd lh fOkHkoekih d srkj d h y EckbZ100 cm gSrFkk bld sfljksad schp d ksbZfu.qea s/ofu d h pky = 330 ms–1 y hft . : +91-744-3012222.No.) .324005 Reg.khØ e esaigy s. Near City Mall.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.: +91-744-3192222 | FAX No.ac.&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.k gSA bu nksuksaizd j.ac. Then. IPIA. CORPORATE OFFICE : CG Tower. : +91-022-39167222 Ph.kksaesa'kwU. C = 0 (4) A = 1. B = 0. ifjiFk esa] fuxZr 1 izkIr d jusd sfy . 6635555 | To Know more : sms RESO at 56677 Website : www.No.Y 15 m/sec.1=1 Sol.u gS% A B Y C Ans. No.: +91-744-3192222 | FAX No. 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.ac. B = 1. B = 0.| NEET-2016 | 01-05-2016 | Code-C. Kota (Raj.resonance. C = 0 (2) A = 0. Near City Mall. : +91-022-39167222 Ph. Jhalawar Road.in | E-mail : contact@resonance. Office : J-2.ac. x. C = 0 Output = 1. Main Road. B = 1. B = 0. B=0 C=1 So ans will be (1) Resonance Eduventures Ltd. C = 0 uhpsfn. Sol. (1) A = 1. fuos'k d k lgh p. To get output 1 for the following circuit. C = 0 (4) A = 1.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 39 . Kota (Raj.) .324005 Reg. Jawahar Nagar. 103. the correct choice for the input is : A B Y C (1) A = 1. CORPORATE OFFICE : CG Tower. C = 0 (1) A=1 Output = 0 +1=1 (2) A = 0. C = 1 (3) A = 1. B = 1.R. : +91-744-3012222. A-46 & 52. B = 0.)-324005 | Ph. IPIA. C = 1 (3) A = 1. B = 1. (1) Sol. No..k mRiUu foorZu iSVuZesa30° d sd ksa..ac.No.)-324005 | Ph.d y f>jh ij 5000 Å rjaxnS/.Z gS% (1) 3  Ans.Zd sfofd j.fn blh i`"B d ks V 4 rajxnS/.Zd k izd k'k vkiru d jrk gS] rksf>jh d sd kj. (1) Sol.324005 Reg. the stopping potential is V .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. A-46 & 52. Office : J-2.(ii)  4  2 Solving equation (i) and (ii) = hc hc  3  th  th = 3 Resonance Eduventures Ltd. KEmax.k ij igy k fufEU"B fn[kkbZnsrk gSA igy k f}rh.Z2d sfofd j.kksalsiznhIr fd . If the same surface is illuminated with radiation of wavelength 2.d mfPP"B ft l d ksa.ac. Kota (Raj. | NEET-2016 | 01-05-2016 | Code-C... = eVst = eV = hc –  (2) 4    (3) 5   (4) 5  2 hc –  . Jhalawar Road.: +91-744-3192222 | FAX No. : +91-744-3012222.104. CORPORATE OFFICE : CG Tower. 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. 6635555 | To Know more : sms RESO at 56677 Website : www. 3  2 3 2 –1  3  ' = sin   4 When a metallic surface is illuminated with radiation of wavelength ' the stopping potential is V.) . Near City Mall.k t krk gS] rksfujks/kh foHko V gSA .kksalsiznhIr fd . 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/.resonance.k t k. Kota (Raj.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.R.in | E-mail : contact@resonance. 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 . : +91-022-39167222 Ph. IPIA.(i)  V  hc e   –  . Main Road. Jawahar Nagar.] t ksfujks/kh foHko gkst krk gSA bl /kkfRod i`"B d h nsgy h rajxnS/.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 40 . Kota (Raj. C  b o d. B  a o d. Near City Mall. A-46 & 52. B  b o c. C  b and d. C  b and d. D  a o d (2) A  b o c. D  c and d (4) A  a and d. | 01-05-2016 | Code-C. [Where m is the magnification produced by the mirror] d kWy e–1 d h laxr izfof"V. : +91-022-39167222 Ph. D  a and d (3) A  a and c.k (c) okLrfod izfrfcEc (d) vkHkklh izfrfcEc (2) Resonance Eduventures Ltd. C  a o b. (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. Jawahar Nagar.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. Sol.eku 'm' rFkk osx ' v ' lsxfreku d ksbZ-d .in | E-mail : [email protected]¡m niZ. IPIA. B  b and 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.ksad k fey ku d kWy e–2 d h izfof"V.ksalsd hft . D  b and c (1) A  c o d. Kota (Raj. Main Road. C  a and b.324005 Reg.)-324005 | Ph.sA [. D  b o c 1 2 1 2 1 2 (b) Concave mirror (c) Real image (d) Virtual image (a) mÙky niZ. Jhalawar Road. 6635555 | To Know more : sms RESO at 56677 Website : www. C  b and c. B  b o d. B  b and c.| NEET-2016 106.k (b) vory niZ.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 41 . B  b and c. D  a and d (2) A  b and c.ac. B  b o c.resonance.No.ac. C  b o d.: +91-744-3192222 | FAX No.kksa}kjk mRiUu vko/kZu gSa] Ans.) . No. D  a o d (3) A  a o c. 1 m  m 1 r or r 1 m Match the corresponding entries of column–1 with column–2. C  b o c. D  c o d (4) A  a o d. Office : J-2. (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. : +91-744-3012222. CORPORATE OFFICE : CG Tower. B  a and d. 108.1 m/s 2 (3) 0. oksYVrk V(t) ls180° vxzgSA (2) /kkjk (t).4 ls-eh– y Ech f=kT. (1) 0.ac. oks YVrkV(t) ls90° i'p gSA (3) .in | E-mail : contact@resonance. leads voltage V(t) by 180° (2) Current (t).h gS: (1) Current (t).4 × 10–2) = 8 × 10–4   at = 0. lags voltage V(t) by 90° (3) Over a full cycle the capacitor C does not consume any energy from the voltage source. Sol.ac.d iw . Main Road.: +91-744-3192222 | FAX No. Kota (Raj. Kota (Raj.k lsxfr d jrk gSA . Jawahar Nagar.k d h xfrt Å t kZ8 × 10–4 J gks t krh gS] rks bl Roj.k D.18 m/s W all = KE  1 2 (mat)(s) = mv 2 (10 × 10–3)(at) (4 × 6. will be (2) 109. | NEET-2016 | 01-05-2016 | Code-C.15 m/s 2 2 (4) 0. image is erect so image will be virtual. No. since image is inverted.R.) .4 cm with a constant tangential acceleration. Image is virtual and diminished. (C) M = +2. and the 2 mirror will be concave.2 m/s (2) (2) 0. : +91-744-3012222. image is magnified so the mirror will be concave. oks YVrkV(t) d h d y k esagSA Ans.k 6.r Li'kZ& js[kh. IPIA. Image is erect so it will be virtual.Sol. since image is i inverted so it will be real. Roj.kZpØ esala/kkfj=k C oks YVrk L=kks r lsd ksbZÅ t kZmiHkqDr ughad jrkA (4) /kkjk (t). Ans. Sol. CORPORATE OFFICE : CG Tower. so image is magnified and inverted. 1 (B) M =  . will be (2) A particle of mass 10 g moves along a circle of radius 6. (4) Current (t) is in phase with voltage V(t) (1) /kkjk (t).eku d k d ksbZd . A-46 & 52.324005 Reg.Y (A) m = –2.)-324005 | Ph. (3) Capacitor does not consume energy effectively over full cycles Resonance Eduventures Ltd. Jhalawar Road. Which is possible only for concave mirror.resonance. so it will be real. so the mirror 2 should be convex.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 42 . 6635555 | To Know more : sms RESO at 56677 Website : www. 1 (D) m =  . A small signal voltage V(t) = V0 sint is applied across an ideal capacitor C : d ksbZy ?kqflXuy oksYVrk V(t) = V0 sint fd lh vkn'kZla/kkfj=k C d sfljksaij vuqiz.No.saiwjh d jusij d .fn xfr vkjEHk d jusd si'pkr~nksifjØ ek.k d k ifjek.k d so`Ùk d svuqfn'k fd lh fu. : +91-022-39167222 Ph.1 m/s m/s Ans. so image is inverted and diminished. Near City Mall.qDr d h x. 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. Office : J-2.k gS\ 2 Ans. | NEET-2016 | 01-05-2016 | Code-C.: +91-744-3192222 | FAX No. gh le.saleku ijUrqæO.ac. (1) 54.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.)-324005 | Ph. To view an object 200 cm away from the objective. : +91-022-39167222 Ph.] nksuksay sal ksad schp d h nwjh gksuh pkfg.) .in | E-mail : [email protected] kj d sxq.k¡Ø e'k%  1 vkSj 2 gSaA .kkad Ø e'k% 1 vkSj 2 gSaA ihry vkSj LVhy d h NM+ksad h y EckbZ. Jawahar Nagar. A-46 & 52. If (2 – 1) is maintained same at all temperatures. 6635555 | To Know more : sms RESO at 56677 Website : www. Lengths of brass and steel rods are  1 and 2 respectively. No. Jhalawar Road. 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. 2 = 2 (1 + 2()) (2) 12 = 2 1 (3) 122 = 2 12 (4) 122 = 22 1 1 = 1(1 + 1()) 2– 1 =(2 – 1) + (22 – 11) As the length difference is independent of temperature difference hence 12 – 11 = 0  112.110.No.k t k. gS\ (1) 11 = 2 2 Ans. leku cuk.0 cm (4) 50. Kota (Raj. (1) Sol.fn (2 – 1) d kslHkh rkiksad sfy .d vkSj usf=kd k d h Q ksd l nwfj.s] rc uhpsfn. CORPORATE OFFICE : CG Tower.0 cm (1) fLFkr fd lh fcEc d ksns[kusd sfy . which one of the following relations holds good ? ihry ¼czkl½ vkSj LVhy d h NM+ksad svuqnS/.324005 Reg. IPIA. : +91-744-3012222. x. Kota (Raj.ac. (4) nks uksa.ekuks aij (2) fMLd fuHkZj d jrk gS (3) xks yk Ans.k¡Ø e'k% 40 cm vkSj 4 cm gSaA vfHkn`'. : (2) 37. 22 = 11 A astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. Sol.0 cm Resonance Eduventures Ltd. Main Road.d ls 200 cm nw j Ans.Y A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length.d  k2 t  1  2  R  k2 R2 111.k. igq¡p saxs (3) Time does not depend on mass.3 cm (3) 46.R.resonance. Office : J-2. the lenses must be separated by a distance : fd lh [kxksy h.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 43 . laca/kksaesals d kSu&lk lR. 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 . Near City Mall. nwjchu d svfHkn`'.eku fHkUu gSa] leku mUurka'k vkSj y EckbZd snksvkur lery ksaij y q<+d rsgSaA bu nksuksafi. resonance. Jhalawar Road.k gksxk y xHkx: Ans.0 s d si'pkr~ms–2 esabld k usV Roj. Jawahar Nagar.2 Joules y hft . Main Road. Kota (Raj.324005 Reg.65 W (4) 236.r d ks.P.Zd jrk gSA iz'khru fd .k mRiUu d j nsrk gSA 2. Lora=k gSA bl fMLd ij d ksbZcy vk?kw. 600 d S y ksjh Å "ek d ksizfr lsd .)-324005 | Ph. %: (1 cal = 4. Kota (Raj.2) 277  Effort put 26  Effort put = 236. at = R = (0.ac. Sol.r j[kusd sfy .0 (4) 6. = (2) 2. | NEET-2016 Tube length = v0 + fe | 01-05-2016 | Code-C. vko'. it is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. Its net acceleration in ms at the end of 2.: +91-744-3192222 | FAX No. (T2 < T1) effort put T1 – T2 for 1 second analysis  (600 )( 4. CORPORATE OFFICE : CG Tower.d 'kfDr pkfg.0 (3) 7. Roj.5 watt Resonance Eduventures Ltd. (1) 3.kh.d leku o`Ùkkd kj fMLd viusry d sy Ecor~vkSj d sUæ lsxqt jusoky s v{k d sifjr% ?kweusd sfy .O.Sol.s ) Ans.ac. No. it is subjected to a torque which produces a constant angular –2 –2 acceleration of 2. 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.d gksrk gSA bld sfy . A-46 & 52.0 rad s . t kusoky sLFkku d k rki fu.No.k d h d ksbZ.Zd jrk gS] t ksblesa2. : +91-022-39167222 Ph.  ac = 8m/s . : +91-744-3012222.5 J  Power = 236. Office : J-2. Near City Mall.0 s is approximately : fojkekoLFkk esafLFkr 50 cm f=kT.5 W T2 Heat extracted  .365 W (3) 23.kZd k.M ckgj fud ky uk vko'.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. C.R.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 44 . 6635555 | To Know more : sms RESO at 56677 Website : www. The power required is : (Take 1 cal = 4.0 rad s–2 d k fu. IPIA.5)(2)  anet = 8 2  12 ~ 8 A refrigerator works between 4°C and 30°C.5 m V = 2m/s at t =2 114.0 (2) The angular speed of disc increases with time.0 (2) 8. (1) 2365 W (4) Sol. and hence centripetal acceleration also anet = ac = a 2t  a 2c 2 R  = tangential speed R = Radius = 0.2 Joules) d ksbZjs fÝ t jsVj 4°C vkSj 30°C d schp d k.) .in | E-mail : contact@resonance. : +91-744-3012222.: +91-744-3192222 | FAX No. A-46 & 52.ru rd laihfM+r fd .kqd rk ij fuHkZj d jsxkA (2) xS l d kslerkih.5 2  2    0 cos2   4    (2. izfØ .k gksxh? (1) 0 2 (2) 0 Ans.ksx esamfPp"B d h rhozrk 0 gSA nksuks af>fj. Kota (Raj. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half.k esavf/kd d k.R.d rk gksxh] .d rk gks xhA Ans.ksx esa mi. IPIA.ksad schp d h nwjh d = 5gS.No. where  is the wavelength of light used is the experiment. 6635555 | To Know more : sms RESO at 56677 Website : www. : i lsmld svk/ksvk. . 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. izfØ .in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 45 .Y A gas is compressed isothermally to half its initial volume.) .ax d sfd lh f} f>jh iz. CORPORATE OFFICE : CG Tower.k t krk gSA blh xSl d ksi`Fkd : i ls: n~/kks"e izfØ . izd k'k d h rjaxnS/.d rk gksxhA (4) xS l d kslerkih. Office : J-2.ru rd laihfM+r fd . : +91-022-39167222 Ph.324005 Reg.ksx fd .Zd jusd h vko'. Main Road. No.k }kjk laihfMr d jsavFkok : n~/kks"e izfØ . Sol.resonance.in | E-mail : contact@resonance. Distance between two slits is d = 5.k }kjk la ihfMr d jusesavf/kd d k. (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.  y    max cos2       (3)  0 4 (4) 3 0 4 D  10 d y for a position in front of a slit   5    2.ac. Jhalawar Road. v vi The intensity at the maximum in Young's double slit experiment is 0. izfØ .k t krk gSA rc : (1) pkgslerkih. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d ? .Zd jusd h vko'.5 )    0 cos2    10  0 2 Resonance Eduventures Ltd.)-324005 | Ph. | NEET-2016 | 01-05-2016 | Code-C.115.g xSl d h ijek.Z d jusd h vko'.k vFkok : n~/kks"e izfØ . (1) Sol.Zd jusd h vko'.k }kjk mld svk/ksvk. Kota (Raj. x.gk¡iz. Jawahar Nagar.k }kjk laihfMr d jsa] fd l izd j.ZgSA fd lh f>jh d slkeusnwjh D = 10d ij fLFkr insZij rhozrk D.d rk gksxhA (3) xS l d ks: n~/kks"e izfØ .k }kjk laihfMr d jusesavf/kd d k.k nksuksaesagh leku d k.ac. Near City Mall. Sol. Jawahar Nagar.: +91-744-3192222 | FAX No.No.d nwl jsesafefJr u gks usoky snksnzo] ft ud s?kuRo  rFkk n(n > 1) gSa] fd lh ik=k esaHkjsgSA izR.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. Office : J-2. A-46 & 52. IPIA. The density d is equal to . Near City Mall. The value of current flowing through AB is : laf/k Mk.k R d h ofØ r lM+d ij xfreku gSA .324005 Reg. No.L) g = [A (L – pL) + n ApL] g On solving d = (1+ (n – 1)p)  118. CORPORATE OFFICE : CG Tower. 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. Jhalawar Road.resonance. : +91-744-3012222.ac.g lM+d d ks.A AB lsizokfgr /kkjk d k eku gS% A +4V (2) 0 A –3 Ans.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 .ac. n wt of body = upthrust by the two liquids If A = Area of section then (d A. | 01-05-2016 | Code-C. : +91-022-39167222 Ph.  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) Resonance Eduventures Ltd.jksavkSj lM+d d schp ?k"kZ. The height of each liquid is h. The maximum safe velocity on this road is: d ksbZd kj f=kT.R.k ij >qd h gSAd kj d sVk. 6635555 | To Know more : sms RESO at 56677 Website : www. Kota (Raj. The cylinder floats with its axis vertical and length pL (p < 1) in the denser liquid. The coefficient of friction between the tyres of the care and the road is s.ksM d ksvkn'kZekud j fopkj d hft . Consider the junction diode as ideal.kkad s gSabl lM+d ij d kj d k vf/kd re lqj{kk osx gS% (1) Ans.Y Two non-mixing liquids of densities  and n(n > 1) are put in container. Main Road.| NEET-2016 117.k xq. (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.) . Kota (Raj. (1) 10 A (3) For diode as ideal i= 119. The road is banked at an angle .in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 46 . A solid cylinder of length L and density d is put in this container.in | E-mail : contact@resonance.)-324005 | Ph. in | E-mail : [email protected]×10 m –1 107 m –1 4 2 If the velocity of a particle is  = At + Bt . 2 from the axis of the wire is: f=kT.: +91-744-3192222 | FAX No.ac.kh d h vafUre y kbu d h rjax la[. : +91-022-39167222 Ph.d leku : i lsforfjr gSA rkj d sv{k lsf=kT.  0i  i  1 = 0   B : B = 1 : 1 2r 2R  2  7 –1 Given the value of Rydberg constant is 10 m .k x.k gksxh % 7 Ans.h /kkjk I çokfgr gks jgh gSA bl rkj d h vuqçLFk d kV ij /kkjk . Kota (Raj. (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.resonance.k d k osx  = At + Bt2 gS. The ratio of the magnetic fields B and B.)-324005 | Ph. Kota (Raj. then the distance travelled by it between 1s and 2s is . CORPORATE OFFICE : CG Tower.gk¡A rFkk B fLFkjkad gS] rksbl d .R. : +91-744-3012222. (1) 2. Jhalawar Road.fn fd lh d . Sol.ac. Jawahar Nagar. 6635555 | To Know more : sms RESO at 56677 Website : www.Sol. at a radial distances a and 2a respectively.k a d sfd lh y Ecslh/ks rkj ls d ks bZLFkk.ksa a 2 vkSj 2a ij Ø e'k% pqEcd h.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 47 .h nwjh gS% (1) Ans.5 ×10 m –1 7 (4) 0. A B  2 3 (2) 3 A B 2 (3) 3A + 7B (4) 3 7 A B 2 3 (4) Resonance Eduventures Ltd. the wave number of the last line of the Balmer series in hydrogen spectrum will be : fjMcxZfu. 122.No.k nwfj. No. A-46 & 52. IPIA.025 ×10 m  –1 wave number = 7 (3) 0. where A and B are constants.rkad d k eku 107 m–1 fn.5×10 m (4) –1 1 1   1  R 2 – 2     2 4 (2) 0. The current is uniformly distributed over its cross-section. Near City Mall.Y A long straight wire of radius a carries a steady current I.324005 Reg. Office : J-2.k gS] gkbMªkst u LisDVªe d h ckej Js. Sol.k }kjk 1s vkSj 2s d schp py h x. Code-C.) . | 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. Main Road. {ks=kksaB vkSj B d k vuqikr gS& (1) 4 Ans. –1 the r. Jawahar Nagar. velocity of velocity of its molecules in ms is .wure fopy u d ks.p. 2 (4) 45º .s.eku d h xSl d sv.: +91-744-3192222 | FAX No. Kota (Raj. 1 2 (3) 30º . æO.k d sfy .k fçTe lsU.s. Near City Mall.kqv ks ad k oxZek/.Y Distance 2 2 s  vdt  At  Bt 2   1  123.No. are fçTe d sfd lh viorZd i`"V ij fd lh çd k'k fd j. When the temperature and pressure of the gas are respectively. viru d ks.ac.)-324005 | Ph. ewy osx gS & ms (1) –1 100 3 Ans.R. 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. 2 Ans. Main Road.m. velocity of 200 ms at 27ºC and 1.k d k eku 45º gSA fçTe d ks. 6635555 | To Know more : sms RESO at 56677 Website : www. VRMS  V1  V2 200  V2 2 (2) 100 (3) 400 (4) 3 100 2 3 3RT Mo T1 T2 300 400  V2  400 3  Resonance Eduventures Ltd. : +91-022-39167222 Ph.in | E-mail : contact@resonance. (3) Sol.) . rki 27ºC rFkk 1. Office : J-2. the angle of minimum deviation and refractive index of the material of the prism respectively. x. Sol.kqv ksad k oxZek/. 5 –1 –2 The molecules of a given mass of a gas have r.0×10 Nm 5 2 pressure. Kota (Raj.m. IPIA.05×105 Nm2 gS] rksms–1 esabl xSl d sv. A-46 & 52.resonance.ac.) osx 200 gSA t c bl xSl d srki vkSj nkc Ø e'k% 127ºC vkSj 0.m. | 01-05-2016 | Code-C. The angle of prism is 60º.wure fopfy r gksrh gS] rksU. 127ºC and 0. Jhalawar Road.05×10 Nm .| NEET-2016 Sol. CORPORATE OFFICE : CG Tower.g fd j.fn . : +91-744-3012222.k rFkk fçTe d sinkFkZd k viorZukad Ø e'k% gS% (1) 30º. 1 (2) 45º .in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 48 .0×105 Nm–2 ij fd lh fn.k d k eku 60º gSA .324005 Reg. If the ray suffers minimum deviation through the prism. ewy (r. 1 3A 7B  2 3 The angle incidence for a ray of light at a refracting surface of a prism is 45º. No. inkFkksZd sfy .d leku gSA buesalac) nsczkXy h rjaxnS/. The magnetic susceptibility negative for (1) paramagnetic and ferromagnetic materials (3) paramagnetic material only (2) diamagnetic material only (4) ferromagnetic material only pqEcd h. d soy vuq p qEcd h. : +91-744-3012222.ac.) . : +91-022-39167222 Ph. Jhalawar Road. Near City Mall. The next larger length of the column resonating with the same tuning fork is: .7 cm Sol. Ans.ksZd k vuqikr gS% Ans.resonance. (3) vkSj y kSg&pqEcd h.in | E-mail : contact@resonance. (2) Sol. A-46 & 52. lqxzkfgrk _ .kkRed gksrh gS% (1) vuq p qEcd h.Y An air column. resonantes with a tuning fork when the smallest length of the column is 50 cm.eku m d sby sDVªkWu rFkk fd lh Q ksVkWu d h Å t kZ. (4) 150 cm  4 First harmonic at 3rd harmonic (3) 100 cm 3 4 1st length = 50 cm 3rd harmonic length 150 cm 126. 6635555 | To Know more : sms RESO at 56677 Website : www.125.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 49 . inkFkZd sfy . (2) d soy çfrpqEcd h. vuqukn d jrk gS t c bl ok. Main Road.R. IPIA. inkFkZd sfy .ac. Office : J-2.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. No. | NEET-2016 | 01-05-2016 | Code-C. (1) 200 cm (4) (2) 66. An electron of mass m and a photon have same energy E. r = 1 + x appropriate is diamagnetic 127.324005 Reg. Jawahar Nagar. 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) Resonance Eduventures Ltd. The ratio of de-Brogli wavelengths associated with them is: æO. inkFkZd sfy . closed at one end and open at the other. Kota (Raj.d fljsij cUn rFkk nwl jsfljsij [kqy k d ksbZok.)-324005 | Ph. (4) d s oy y kSg&pqEcd h. CORPORATE OFFICE : CG Tower.qLrEHk fd lh Lofj=k f}Hkqt d slkFk ml le. E .: +91-744-3192222 | FAX No.No. Kota (Raj. resonance. CORPORATE OFFICE : CG Tower.M fd lh d ky fJr cy F  (2t ˆi  3t 2 ˆj)N . : +91-744-3012222. t d slkFk fopj.R. 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 .in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 50 . Near City Mall. 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 Resonance Eduventures Ltd.eku d k d ks bZfi.| NEET-2016 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.No.)-324005 | Ph.gk¡ ˆi rFkk ˆj . (t2 i + t3 ˆj ) power = 2t3 + 3 t5 129. Sol. No. x vkSj y v{k d svuqfn'k ek=kd lfn'k gS] d sv/khu xfr vkjEHk d jrk gS] rksle. A-46 & 52.k Q = at – bt2 d s: i esagksrk gS] t gk¡a rFkk b /kukRed fu. Jhalawar Road. Main Road. when ˆi and ˆj are unit vectors along x and y axis.: +91-744-3192222 | FAX No.) . : +91-022-39167222 Ph.ac. electron = For h | 01-05-2016 | Code-C. The total heat produced in R is : fd lh çfrjks/k R lsçokfgr vkos'k d k le.V.324005 Reg. IPIA. Sol. where a and b are positive constants. = (2ti + 3t2 ˆj ). t ij bl cy }kjk fod flr 'kfDr D. What power will be developed by the force at the time t ?  1 kg æO. Kota (Raj.ac. 6635555 | To Know more : sms RESO at 56677 Website : www.in | E-mail : contact@resonance. Kota (Raj.k gksxh? 3 Ans. Office : J-2.Y …(1) 2ME photon E = h = hc photon …(2) from these two ratio obtained by dividing these (2)  1 : 2 = 128. . 2 The charge flowing through a resistance R varies with time t as Q = at – bt . Jawahar Nagar.rkad gSA R esamRié d qy Å "ek gS% (1) Ans. 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.Y 2 a3 R 6b A npn transistor is connected in common emittet configuration in a given amplifier. If the current amplification factor is 0. Sol. (1) 4. 4 800 192 power gain = [current gain] [resistance gain] [.69.No.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.) .8 V foHkoikr gSA . Sol. Near City Mall.4 × 105 J/Kg rFkk g = 10 N/kg gSa] rksÅ ¡p kbZh d k eku gS% Ans.96 and the input resistance of the circuit is 192 .ac.84 (2) Voltage gain = [current gain] [resistance gain] [. A-46 & 52. Office : J-2.84 131. Jhalawar Road. (1) 68 km (4) (2) 34 km (3) 544 km (4) 136 km Mg h = mL 4 h= 4L = 136 km g Resonance Eduventures Ltd. ço/kZd esad ksbZnpn Vªkaft LVj mHk. the voltage gain and the power gain of the amplifier will respectively be : fd lh fn.kl esala.fn cQ Zd h xqir Å "ek 3. a piece of ice falls from a height h so that it melts completely.ksft gSA 800  d ksd ks bZy ksM çfrjks/k laxzkgd ifjiFk esals.kZr% fi?ky t krk gSA mRiUu gksusoky h m"ek d k d soy .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. Jawahar Nagar. 6635555 | To Know more : sms RESO at 56677 Website : www.ac.)-324005 | Ph.| NEET-2016 | 01-05-2016 | a/2b it   (a – 2bt)2 Rdt =  a2 t  0 Put t  130. A load resistance of 800  is connected in the collector circuit and the voltage drop across it is 0.3.in | E-mail : contact@resonance. : +91-022-39167222 Ph. Kota (Raj.ksft r gSvkSj bld sfljksaij 0. x. a 2b  H= 2 3 4b t 4bat – 3 2 Code-C.324005 Reg. : +91-744-3012222.k t krk gS rFkk cQ Z d h leLr Å t kZ bld s fxjrs le. No. Å "ek esa : ikUrfjr gkst krh gSA .69 (2) 4.kkad 0.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 51 .fu"B mRlt Zd foU. Main Road. Kota (Raj.96] (3) 3. 3. CORPORATE OFFICE : CG Tower.: +91-744-3192222 | FAX No.resonance.fn /kkjk ço/kZd xq .8 V.R.3.d &pkSFkkbZ Hkkx gh cQ Z }kjk vo'kksf"kr fd .96] [4] = 3.84 (4) 4. IPIA. The value of h is : 5 [Latent heat of ice is 3. | NEET-2016 | 01-05-2016 | Code-C. The ratio 1/2 is : (1) m1  m2 m1 (2) m1 m2 (3) m1  m2 m2 (4) m2 m1 nzO.d leku jLlh fd lh n`< Vsd lsÅ /okZ/kj y Vd h gS A bl jLlh d seqDr fljsls nzO. Main Road. A-46 & 52. 6635555 | To Know more : sms RESO at 56677 Website : www. 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.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 52 .eku m1 rFkk y EckbZL d h d ksbZ.fn jLlh d s'kh"kZrd igq¡p usij bl LiUn d h rjaxnS /. A block of mass m2 is attached to the free end of the rope. The wavelength of the pulse when it reaches the top of the rope is 2. A transverse pulse of wavelength 1 is produced at the lower end of the rope. IPIA. iz okfgr gksjgh gS] fd lh y Ecslh/kspky d XY ft lls/kkjk I izokfgr gksjgh gSd sfud V .R. is placed near and coplanar with a long straight conductor XY carrying a current I. Near City Mall.No.132. m1  m2 m1 (2) m1 m2 (3) m1  m2 m2 (4) m2 m1 (3) Resonance Eduventures Ltd. : +91-022-39167222 Ph.in | E-mail : [email protected]. No.resonance.eku m2 d k d ksbZxqVd k t qM k gSA jLlh d seqDr fljsij rjaxnS/. the net force on the loop will be : d ksbZoxkZd kj ik'k (y wi) ABCD ft lls/kkjk i. Jhalawar Road. Office : J-2.k t krk gSA .Z2 gkst krh gSA rc vuikr 1/2 d k eku gS% (1) Ans. 0 iL 2 (2) 20 i 3 D A L/2 (1) C B L (3)  0 i 2 (4) 20 iL 3 (2) FBC C B FAB FCD D A FAD Sol. : +91-744-3012222. Kota (Raj.ac.Y A square loop ABCD carrying a current i.d gh ry esaj[kk gSaA bl ik'k ij y xusoky k y sV cy gks xk % Y L  i X Ans. FBC cancels FAD FNet = FAB – FCD = 133.Z1 d k d ksbZvuqiz LFk LiUn mRiUu fd .324005 Reg.) . CORPORATE OFFICE : CG Tower.: +91-744-3192222 | FAX No. Kota (Raj. Jawahar Nagar.)-324005 | Ph. kksad h Å t kZ] rjaxnS/.)-324005 | Ph.ac.k d kSu lk lac/k lgh gS\ Ans.k x.No. Wien's constant. rjax mRiUu d jusesafd . x.in | E-mail : contact@resonance. Office : J-2.k t k ld rk gS\ (1) d ks bZRofjr vkos'k (2) fu. CORPORATE OFFICE : CG Tower.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 53 .Z 500 nm ij U2 rFkk rja xnS/. No. 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.Y T m/ v 1  M2 2  M2  M1 Tension = M2g Tension = M2g T2 = (M1 + M2)g M1 T1 = M2g M2 2  1 134.kd k 5760 K rki ij gSaA bl fi.resonance.| NEET-2016 Sol. b = 2.88 × 10 nmK. vkos'k (4) vkos 'kghu d .z1000 nm ij U3 gSA ohu&fu. Main Road. : +91-744-3012222. (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.M }kjk mRlft Zr fofd j. fod Yiksaesalsfd ld k mi.Z250 nm ij U1 rja xnS/. Jawahar Nagar.324005 Reg.R.r os x lsxfreku d ksbZvkos'k (3) fLFkj Ans. Kota (Raj.ac.d lap fjr fo|qr pqEcd h.: +91-744-3192222 | FAX No.k (1) Resonance Eduventures Ltd. Near City Mall.88 × 106 nmK gSA uhpsfn. Which of the following is correct ? (1) U2 > U1 (2) U1 = 0 (3) U3 = 0 (4) U1 > U2 d ksbZd `f". 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. Jhalawar Road. | 01-05-2016 | Code-C. 6635555 | To Know more : sms RESO at 56677 Website : www. A-46 & 52. M1  M2 M2 A block body is at a temperature of 5760 K.ksx . : +91-022-39167222 Ph.) . IPIA. Kota (Raj.rkad ] b = 2. Office : J-2.kkRed y sfd u H .059 log 2 2 10 7    (In order to make log1 = 0) (log1 = 0 d jusd sfy .kkRed gks rsgSA (4) G . No.ac. H and S all are negative (4) G and H are negative but S is positive fuEUkfy f[kr y {k. G = H  TS Adsorption is a spontaneous process (where S < 0.oaS /kukRed gksrsgSA (3) G. H . CORPORATE OFFICE : CG Tower.kksaesalsd kSu lk vf/k'kks"k.u – (aq.k) PH2 0.) E = Eº – 0=0– vfHkfØ .oaH _ .| NEET-2016 | 01-05-2016 | Code-C. d jusd sfy .oaS lHkh _ .059 log 2 2  H(aq. Kota (Raj.ac. 6635555 | To Know more : sms RESO at 56677 Website : www. (3) Sol.)    PH2 0. G < 0 and H < 0) fxCl gsYegkWV~t lehd j.resonance. 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. IPIA. Main Road.k d svuql kj G = H  T S vf/k'kks"k. 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.324005 Reg.R.Y PART C – CHEMISTRY . A-46 & 52.kkRed y sfd u S /kukRed gksrk gSA Ans.No. 2H + + 2e  H2 (reduction reaction) (vip.: +91-744-3192222 | FAX No.in | E-mail : [email protected] . According to Gibbs Helmholtz equation. Jawahar Nagar. (1) 10 atm (2) 10–14 atm (3) 10–12 atm (4) 10–10 atm (2) Sol. G < 0 rFkk H < 0) 137.) . : +91-022-39167222 Ph.)-324005 | Ph.d H2 d k nkc gSA –4 Ans.) -7 2 PH2 = (10 ) = 10–14 atm Resonance Eduventures Ltd.svko'. Kota (Raj. Near City Mall. 136.kkRed y sfd u H /kukRed gksrk gSA (2) G _ . Jhalawar Road.d Lor% izØ e gksrk gS(t gk¡ S < 0.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 54 . : +91-744-3012222.k lslEcfU/kr gS\ (1) G rFkkS _ . k. Å "ek xfrd h.ac. IPIA.. G .k Å t kZd ksizHkkfor d jrk gS] vU. Jhalawar Road. Near City Mall. 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. (4) (a) is elimination.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 55 . fuEu vfHkfØ .resonance. (b) is substitution and (c) is addition reaction.)-324005 | Ph. (b) and (c) are substitution reaction.s aFkSYih Å t kZ] (1) Catalyst can affect only activation energy of the chemical reaction and cannot alter any thermodynamic parameters : (ie.k vkSj (c) .k gSA (4) (a) foy ks iu Ans. gS? (1) (a) iz frLFkkiu. (b) vkSj (c) izfrLFkkiu vfHkfØ .¡gSA (3) Resonance Eduventures Ltd.fud vfHkfØ . (b) and (c) are addition reactions.R. S ) 139.k. H .M d ksugha% (vFkkZ r~H .¡gSrFkk (c) .k gSA (3) (a) foy ks iu vfHkfØ .ac. | NEET-2016 | 01-05-2016 | Code-C. Main Road.fud vfHkfØ .k. (b) izfrLFkkiu vfHkfØ .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.in | E-mail : contact@resonance.) .ksXkt vfHkfØ .k.ks xt vfHkfØ .k esamRis zsjd d s.s UVªkWih (3) vka rfjd (4) .: +91-744-3192222 | FAX No. CORPORATE OFFICE : CG Tower. S ) mRizsjd d soy jklk. No. vfHkfØ .k Å t kZ (2) . (2) (a) and (b) are elimination reactions and (c) is addition reaction. G .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..138.ksxt vfHkfØ . Kota (Raj. Sol.No.ksx lsfuEUkfy f[kr esalsd kSu lh ek=kk cny rh gSA Ans. ekin.kvksad sfy .k. 6635555 | To Know more : sms RESO at 56677 Website : www. Kota (Raj. (3) (a) is elimination.¡gSA (2) (a) vkS j (b) foy ksiu vfHkfØ .324005 Reg. A-46 & 52.k d h lfØ . : +91-022-39167222 Ph. : +91-744-3012222. (1) lfØ . Jawahar Nagar. (b) vkSj (c) . Office : J-2. k lar`Ir . Main Road. Kota (Raj.k Br ijek.k d ks.k + KBr OH fd lh lewg d k nwl jslewg }kjk izfrLFkkiu gksusij mlsizfrLFkkiu vfHkfØ .) .k d grsgSA CH3 (b) H3C + KOH  Br iz frLFkkiu CH3 H3C vfHkfØ .ks xt vfHkfØ .kSfxd esa: ikUrj.k lewgksad sgVusij vfHkfØ .No.324005 Reg. IPIA.kSfxd d k vlar`Ir .k d grsgSA Br (c) + Br2  .)-324005 | Ph. | 01-05-2016 | Code-C. CORPORATE OFFICE : CG Tower. Jawahar Nagar.k d grsgSA  140.k d ksfoy ksiu vfHkfØ .k lscuk mRikn gSA: (1) .R.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 56 .fYMgkbM/d hVksu f'kQ {kkj Resonance Eduventures Ltd. Near City Mall.ac. : +91-744-3012222. 6635555 | To Know more : sms RESO at 56677 Website : www.oaizkFkfed .sfYMgkbM .k lewgksad st qM +usij vlar`Ir .| NEET-2016 Sol.ac.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. 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 .kSfxd d slar`Ir .resonance.kSfxd esa: ikUrj.: +91-744-3192222 | FAX No.ks xt vfHkfØ .sehu d h vfHkfØ . Office : J-2.s jksesfVd vEy (2) f'kQ ~{kkj (3) d hVks u Ans. No. Kota (Raj. : +91-022-39167222 Ph.in | E-mail : [email protected]. Sol.kq. (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 . (a) CH3CH2CH2Br + KOH CH3CH=CH2 + KBr + H2O foy ks iu -ca /k d k fuekZ. Jhalawar Road. A-46 & 52.k rFkk ijek. 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 Resonance Eduventures Ltd.R. Main Road.s jhy . 6635555 | To Know more : sms RESO at 56677 Website : www. Jawahar Nagar. IPIA.324005 Reg.qXe foLFkkfud `r ughagksrsgSA . Office : J-2.k .sehu lkekU.fjy .knk {kkjh.ehu) Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape.s jhy . (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. Jhalawar Road.ksfd ukbVªkst u d s. | NEET-2016 | 01-05-2016 | Code-C. gSD.) . CORPORATE OFFICE : CG Tower.sjksesfVd oy . gSA (2) NH2 Sol.sehu lsT.sehu lsd e {kkjh.ac.sfYd y . .sehu lsT.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 57 .fjy .sehu lsT. d s by s DVªkWu d slkFk foLFkkfud `r gksrsgSA (3) .qXe by sDVªksu . gSD.r% . Kota (Raj.141. RNH2  (Alkyl amine) NH2 .knk {kkjh. A-46 & 52.ksfd ukbVªkst u d s.fYd y 142.sjhy .d kd h-.resonance. 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.sfYd y .sehu lkekU.)-324005 | Ph.in | E-mail : contact@resonance. d slkFk foLFkkfud `r gksrsgSA .sfYd y . lewg d sd kj.sjhy . : +91-022-39167222 Ph.ksfd .ehu esaukbVªkst u d sby sDVªkWu .: +91-744-3192222 | FAX No.qXe by sDVªks u .fYd y .r% .Y The correct statement regarding the basicity of arylamines is : (1) Arylamines are generally more basic than alkylamines.d kd h . gksrh gSA D. (4) Arylamines are generally more basic than alkylamines because of aryl group .s fjy Ans.qXe csUt hu oy .sehu lkekU.No.sehu esaukbVªkst u ijek. : +91-744-3012222. Kota (Raj.kqsp-lad fjr gSA (2) .sehu lkekU. because the nitrogen atom in arylamines is sp-hybridized.ehu  RNH2   (.sfYd y .slgh d Fku gS: (1) .sjhy .ehu esaukbVªks t u d sby sDVªkWu . d s by sDVªksu d slkFk foLFkkfud `r ugh gksrsgSA (4) .r% .r% .sjksesfVd oy . (2) Arylamines are generally less basic than alkylamines because the nitrogen lone pair electrons are delocalized by interaction with the aromatic ring electrons system. No.s jhy . Near City Mall.sehu d s{kkjd rk d sfy .ac.knk {kkjh. u d j ld rsgSA gkbMªkst u d svk/ksiy k. (2) The staggered conformation of ethane is less stable than eclipsed conformation.d ik=k esaj[kk x.h gSt cfd xzLr la: i. : +91-744-3012222.k gSA t ksfd lw{e fNnzd s}kjk iy k.k] xzLr la: i.k esa.h gS D.k] xzLr la: i.Fksu d k xzLr la: i. (1) Resonance Eduventures Ltd.324005 Reg. Kota (Raj. Office : J-2.k esa.h gSD. Kota (Raj. because eclipsed conformation has no torsional strain.k d h rqy uk d sfy .resonance. because staggered conformation has torsional strain.ksafd xzLr la: i.u esay xsle.k lsvf/kd LFkk. A-46 & 52.sgSblfy . because staggered conformation has no torsional strain.) .| NEET-2016 | 01-05-2016 | Code-C.saBu ruko ¼ ejksM +h fod `rh½ gSA (3) .oavkWDlt hu xSl ksad sleku eksy ksd ks.k lsd e LFkk.u d jsxkA? Ans.h gSD. 6635555 | To Know more : sms RESO at 56677 Website : www.slgh d Fku gS& (1) .: +91-744-3192222 | FAX No.saBu ruko ¼ ejksM +h fod `rh½ gSA Ans. No.k] lkarfjr la: i.Fksu d k lkarfjr la: i.k ls vf/kd LFkk.u gq.Fksu d k lkarfjr la: i.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 58 . MH2 MO2 1 2 = 4 32 = = 1 4  nO2 x = 1 8 fraction of oxygen escaped = 1 .R. (4) The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain. (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.saBu ruko ¼ ejksM +h fod `rh½ ughagSA (4) . vkW Dlht u d k va'k = 1 8 The correct statement regarding the comparison of staggered and eclipsed conformations of ethane.sx. CORPORATE OFFICE : CG Tower. is: (1) The staggered conformation of ethane is more stable than eclipsed conformation.Fksu d k xzLr la: i.oaxz Lr la: i.k] lkarfjr la : i.ksafd lkarfjr la: i.No. Jawahar Nagar.ac.k esa.ac. Jhalawar Road.ksafd lkarfjr la: i. (3) The eclipsed conformation of ethane is more stable than staggered conformation. : +91-022-39167222 Ph. IPIA. esavkWDlht u d k fd ruk va'k iy k. .in | E-mail : contact@resonance. Sol. 8 iy k.)-324005 | Ph.k lsvf/kd LFkk.saBu ruko ¼ ejks M +h fod `rh½ ughagSA (2) . Main Road.k es a . vkaf'kd nkc leku gS(ekuk fd x gS) rO2 rH2 = nO2 / t x /t 2 nO2 / t x /t 2 143.Y gkbMªkst u . Near City Mall.Fksu d slkarfjr . Sol.saBu ruko) d sd kj. : +91-744-3012222.sxq. 333. CORPORATE OFFICE : CG Tower. Kota (Raj.E  B < C < O < N correct order of increasing electron gain Enthalpy I < Br < F < Cl (in magnitude) Values (in KJ/mol)  296. A-46 & 52. 349.UFkSYih) Ans.in | E-mail : contact@resonance. Kota (Raj.R. 6635555 | To Know more : sms RESO at 56677 Website : www. No. 349. Resonance Eduventures Ltd.)-324005 | Ph.k/keZd sifjorZu d svuql kj lger ughagS\ (1) Li < Na < K < Rb (c<+ rh gqbZ/kkfRod 3+ (2) Al < Mg 2+ + f=kT. 325.k d h rqy uk esad e LFkk.k H H H xzflr U.qXe – cfU/kr .wesu la: i.k) eku (KJ/mol esa)  296. vlR.fud . Jawahar Nagar.| NEET-2016 H H H H | 01-05-2016 | Code-C.fud – vkd kj) (3) B < C < N < O (c<+ rk gqv k izFke vk.h gksrk gSA 144. 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.UFkSYih d k lgh c<+rk Ø e I < Br < F < Cl (d soy ifjek.uu Å t kZd k lgh c<+rk Ø e  B < C < O < N by sDVª kWu y fC/k .resonance.k xzflr la: i.k(.No.ac.svk.: +91-744-3192222 | FAX No. 333.Y H Sol. : +91-022-39167222 Ph. 325. Main Road. H H H H H H H H H lkarfjr U.k cfU/kr . 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.324005 Reg.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 59 . Office : J-2.k) < Na < F (c<+ rsgq.sx.) .k] lkarfjr la: i.qXe izfrd "kZ. fod Yi 3 & 4 gS izFke vk. IPIA. (3 & 4) Incorrect option are 3 & 4 Correct order of increasing Ist I. Near City Mall.wesu la: i. Jhalawar Road.UFkSYih) (4) I < Br < Cl < F (c<+ rh gqbZby sDVªksu y fC/k .ac. : +91-022-39167222 Ph.d izFke d ksfV d h vfHkfØ . The half-life period of the reaction is : . NO rFkk NO2 (2) Cu + 4HNO3  Cu(NO3)2 + 2NO2 + 2H2O conc Brown gas Cu + conc HNO3 NO2 Cu + dil HNO3 NO 147.3(0. CORPORATE OFFICE : CG Tower.in | E-mail : contact@resonance. Jawahar Nagar.) n2 n2   10 k n4 / 3 2.03 mol  Ans.k d k osx vfHkfØ .6 – 0.1 s (2) (2) 24.1 s d ksfV vfHkfØ .k izkjEHk gksusd s10 sec ckn 0. Kota (Raj. –1 s –1 gSA bl vfHkfØ . Main Road. (1) Cu(NO3)2 rFkk N2O (2) Cu(NO3)2 rFkk NO2 (3) Cu(NO3)2 rFkk NO (4) Cu(NO3)2 . 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 . vkca/k d s}kjk (4) is IVkbZM vkca/k d s}kjk (4) Resonance Eduventures Ltd. Kota (Raj.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. 6635555 | To Know more : sms RESO at 56677 Website : www.04 mol  –1 s –1 Code-C.s ehuksvEy . NO and NO2 d kWij d kslkUnzHNO3 d slkFk xeZd jusij curk gS& Ans. When copper is heated with conc.3  0.Y at 10 seconds and 0. A-46 & 52.1 s (4) 44.resonance.4 sec.k d sfy .k d h v) Zvk.R.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. 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.3  10 2. No. | 01-05-2016 | The rate of a first-order reaction is 0.kqesafofHkUu . Jhalawar Road.477) = 24.No.04 mol –1 s–1 rFkk 20 sec ckn 0. : +91-744-3012222.qd ky gS& (1) 54.) .in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 60 . Near City Mall. 146.03 mol  –1 –1 s at 20 seconds after initiation of the reaction.324005 Reg.| NEET-2016 145. Sol. IPIA. Sol.03 10 3 t1/ 2  = (3) 34.ac.04 1 4 n  n 20 – 10 0.: +91-744-3192222 | FAX No.)-324005 | Ph. Office : J-2.ac. CORPORATE OFFICE : CG Tower.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 61 .Y + H2 N — CH — C — OH O O amino acid (. Kota (Raj.No.ac.k ek/. Jhalawar Road. Jawahar Nagar. Near City Mall. R1 R2 H2N — CH — C — NH — CH — COOH O Peptide Bond (isIVkbM cU/k) 148. A-46 & 52. 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) Resonance Eduventures Ltd.ac.resonance.: +91-744-3192222 | FAX No.e xSl gSA 149. 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. : +91-744-3012222.324005 Reg. Main Road. IPIA.) .ehuksvEy ½ Sol.| NEET-2016 R1 H2N — CH — C — OH | 01-05-2016 | R2 Code-C. 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 . Kota (Raj. No. 6635555 | To Know more : sms RESO at 56677 Website : www. Office : J-2.in | E-mail : [email protected] gS& (1) xS l Ans. : +91-022-39167222 Ph. (2) Sol.)-324005 | Ph. Which one given below is a non-reducing sugar ? (1) Sucrose (2) Maltose (3) Lactose (4) Glucose fuEu esalsd kSulh . Jhalawar Road.saA lgh lad sr i) fr gS& LrEHkI (a) (b) (c) (d) LrEHkII lkW. CORPORATE OFFICE : CG Tower. The correct statement regarding RNA and DNA.R.d 'kd Zjk gS\ (1) lq Ø ksl Ans. A-46 & 52.d vuvipk.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. : +91-744-3012222.d xSl vipk. 6635555 | To Know more : sms RESO at 56677 Website : www. Near City Mall.slgh d Fku Ø e'k% gS& Resonance Eduventures Ltd. respectively is : (1) The sugar component in RNA is 2'-deoxyribose and the sugar component in DNA is arabinose.ac.k (iv) Au d k fu"d "kZ . Sol.k eaM y ifj"d j. (1) (2) (3) (4) (2) Sol. (3) The sugar component in RNA is ribose and the sugar component in DNA is 2'-deoxyribose. Main Road. (4) The sugar component in RNA is arabinose and the sugar component in DNA is ribose. RNA rFkk DNA d sfy .)-324005 | Ph. bl izd kj d h 'kd Zjk vuvipk.uksesfjd d kcZu ijLij cfU/kr gSA blfy . Cyanide process Leaching process of Au Au + 2NaCN aq. (2) ekYVks l (3) y s DVksl (4) Xy q d ksl (1) Sucrose is Non Reducing sugar. Ge 150.| NEET-2016 | 01-05-2016 | Code-C. (2) The sugar component in RNA is arabinose and the sugar component in DNA is 2'-deoxyribose. Kota (Raj.Y LrEHk I d smYy s[k d ksLrEHkII d smYy s[k lsfey k. Jawahar Nagar.u (iii) Al d k fu"d "kZ . No. 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.d gksrh gSA) 151. (both the anomeric carbon are bonded to each other than such sugars are non reducing) lqØ ksl .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. IPIA. Kota (Raj.) .resonance.324005 Reg. Office : J-2.ac.No. : +91-022-39167222 Ph.: +91-744-3192222 | FAX No.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 62 .d 'kd Zjk gSA (nksuksa . | 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 rFkkS = 0 (3) H > 0 rFkkS < 0 (4) H < 0 rFkkS > 0 Ans. (4) Sol. G = H – TS 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) 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 # 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? 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 # 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–CC sp hybridisation Steric Number (1 + 1–vecharge) sp CH3–CC  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 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 # 65 qXe Ans. A-46 & 52. fuEufy f[kr esalsd kS ulk d Fku lR. Kota (Raj.) .d kd h . 6635555 | To Know more : sms RESO at 56677 Website : www.qXe > vkca/kh .d kd h . DokaVe la[.k t krk gS\ (1) gjkCr2(SO4)3 curk gS A (2) foy .u jaxghu gkst krk gSA (4) SO2 vipfr.qXe – .d kd h . Office : J-2.qXe > .q Xe – .qXe > .qXe – . DokaVe la[.qXe > .)-324005 | Ph.q Xe – vkca/kh .k (1) Resonance Eduventures Ltd. | NEET-2016 | 01-05-2016 | Code-C.u esalsikl fd .qXe > vkca/kh . K2Cr2O7 d sfoy .k (4) fnxa 'kh. gS] t c SO2 d ksvEy h. Jhalawar Road.d kd h . No.qXe – vkca/kh .d kd h .in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 66 . (2) The order of repulsion force according to VSEPR theory : VSEPR fl) kUr d svuql kj izfrd "kZ.d kd h . (1) Sol.qXe > . IPIA.k (3) pq Ecd h.qXe > .k gkst krk gSA 4 es (K2Cr2O7 + SO2 + H2SO4   Cr2(SO4)3 + H2O + K2SO4) 159.qXe – vkca/kh . (2) eq [.qXe – vkca/kh . Jawahar Nagar.d kd h . Near City Mall.qXe (4) vkca /kh .qXe (2) .u uhy k iM+t krk gSA (3) foy .d kd h .k Ans.q Xe – .in | E-mail : [email protected] Reg.qXe – vkca/kh . gks rk gSA Ans. Main Road. DokaVe la [.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.: +91-744-3192222 | FAX No.k t k ld rk gS\ (1) iz p Ø .k cy d k Ø e gS: lone pair – lone pair > lone pair – bond pair > bond pair – bond pair 160.d kd h . (2) The solution turns blue (3) The solution is decolourized. K2Cr2O7   Cr2(SO4)3 green solution obtain where as SO2 oxidise into sulphate SO2– 4 K2Cr2O7   Cr2(SO4)3 gjk foy .qXe > vkca/kh .No. 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 .u izkIr gks rk gSt gk¡SO2 d k SO2– avkWDlhd j.qXe – vkca /kh .d kd h .qXe (3) .158. Kota (Raj.qXe – vkca /kh .d kd h .d gh d {kd esagSA buesavUrj fd ld s}kjk fd .k DokaVe la[.qXe – vkca/kh .resonance. : +91-022-39167222 Ph.d kd h .ac. : +91-744-3012222. 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) . (4) SO2 is reduced. CORPORATE OFFICE : CG Tower. Sol.ac. 8 I2 151. CORPORATE OFFICE : CG Tower. (1) F2 > Cl2 > Br2 > I2 (3) CI2 > Br2 > F2 > I2 (3) Sol.Sol.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 . 6635555 | To Know more : sms RESO at 56677 Website : www.k gksrk gSA (calcium cyanamide and carbon mixture).ac.k Dok. The product obtained as a result of a reaction of nitrogen with CaC2 is : ukbVªkst u d h CaC2 d slkFk vfHkfØ .: +91-744-3192222 | FAX No.k) 161. Sol.kst u . (2) It is fact 163. : +91-744-3012222.1 (kJ/mole) Resonance Eduventures Ltd.8 (2) I2 > Br2 > CI2 > F2 (4) Br2 > I2 > F2 > CI2 > F2 158.UFkSYih) CI2 242.oaVªkal &foU. Jawahar Nagar. Jhalawar Road.) . IPIA.)-324005 | Ph.Ve la[. Main Road. : +91-022-39167222 Ph. Bond dissociation enthalpy (cU/k fo.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 67 .R. Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules? Ans.kl gSA (3) lHkh Vª kUl&foU.k 1100ºC ij gksuslsukbVªkWfy e d k fuekZ.slgh gS\ > Br2 192.resonance.UFkSYih d sfy .ks tu .No.k d sizkIr mRikn gS& Ans. A-46 & 52. 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.6 > fuEufy f[kr esalsd kS u Ø e gSy kst u v. (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).fer fll~& . nksfHkUu eku gksld rsgSA (+½ rFkk –½) (pØ .oaVªkal &foU.k d sfy . Near City Mall. Kota (Raj. Office : J-2.kl gSA Ans.in | E-mail : [email protected] gSA (2) lHkh fll~ & foU.324005 Reg. Kota (Raj. No. Sol.kl gSA (4) . CaC2 + N2  CaCN2 + C (No answer in matching) CaC2 rFkk ukbVª kWt u d h vfHkfØ . CaC2 + N2  CaCN2 + C (No answer in matching) 162. | NEET-2016 | 01-05-2016 | Code-C.ac.kqv ksad h vkca/k fo.d kUrj fll~ & . Yd ks gy la'y s"k. d= ZA NA  a3 3 530 kg/m (2) 154 pm for BCC = (3) 352 pm (4) 527 pm Z=2 2  6.eku 6. (1) 264 pm (3) Sol.) . Jawahar Nagar.d d d ksf"Bd k d sd ksj d h y EckbZgSA (NA = 6. : +91-022-39167222 Ph. –3 –1 Lithium has bcc structure.94 g mol–1 gSA fy fFk. Main Road. IPIA.kqnzO.Elu . Near City Mall.Elu (3) .02  1023  a3 a3 = 43. No. NaH O Me Me–I Na O d ksoxhZd `r fd .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 .in | E-mail : [email protected] d hbbc lajpuk gSA bld k ?kuRo 530 kg m–3 rFkk ijek. 6635555 | To Know more : sms RESO at 56677 Website : www. Resonance Eduventures Ltd. A-46 & 52.k bZFkj la'y s "k.)-324005 | Ph.k & OH Ans. : +91-744-3012222.Yd ks gy (4) fut Z y hd j. OH + NaH O acid-ase Reaction 2 SN Na CH3—I O–CH3 This williamson ether synthesis 165. Calculate the 23 –1 edge length of a unit cell of Lithium metal. Kota (Raj.324005 Reg.02 × 1023 mol–1) Ans. Office : J-2.ac.94  10 3 6.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 68 .164. (NA = 6.52 × 10–10 m = 352 pm. Kota (Raj.k (2) Sol.resonance.k t k ld rk gS& (1) fofy . Jhalawar Road.No.94 g mol .k (2) fofy .k vfHkfØ .50 × 10–30 a = 3. | NEET-2016 The reaction : OH | 01-05-2016 | NaH O Code-C. CORPORATE OFFICE : CG Tower.02 × 10 mol ) fy fFk.e /kkrqd s.R.: +91-744-3192222 | FAX No.k fojpu vfHkfØ .ac.k vfHkfØ .k vfHkfØ . Its density is 530 kg m and its atomic mass is 6. (1) 103º C (2) 101º C (2) At B. CORPORATE OFFICE : CG Tower. = 100 + TB = 101ºC The electronic configurations of Eu (Atomic No.324005 Reg. : +91-744-3012222. Near City Mall.52 × × 1000 = 1 100 P0  Ps WA / MA = Ps WB / MB 168. la[.81 × 10–10 m gSA AB esaizR.81 10 10 (3) 4 (4) 8 = 0. [Xe]4f7 5d16s2 vkS j [Xe]4f8 5d16s2 (3) [Xe]4f 5d 6s .u d h leUo.P.98 × 10–10 m .52.: +91-744-3192222 | FAX No. [Xe]4f 6s and [Xe]4f 5d 6s 6 1 2 7 1 2 9 1 2 (3) [Xe]4f 5d 6s .kst u la[.54 Octahedral range (v"VQ y d h. A-46 & 52.81 × 10 m.resonance.P. No.5 / M = 732 100 /18 =1 On solving M = 32.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 69 . Kota (Raj. izR. Kota (Raj. d k 100 g t y esafoy .R.u d k 100º C ij ok"Ik nkc 732 mm gSA . So B.u d k DoFkukad gksxk % Ans.kl gS% 2 vkSj [Xe]4f 6s vkSj [Xe]4f8 5d16s2 1 vkSj [Xe]4f9 5d16s2 (4) [Xe]4f6 5d16s2.sd vk.oa1.sd vk.5 g foy s.5 g of a solute in 100 g water is 732 mm. the boiling point of this solution will be : . lhek) 0. (4) 102º C 760  732 6.fn Kb = 0. Sol.ac. Jhalawar Road. P0 = 760 torr for elevation of B. 64) and Tb (Atomic No. [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 .u d sfy . Main Road. Office : J-2.u d h milgla.98 × 10 m and 1.¡Ø e'k%0.uksad h vk.k gSA Ans.98  10 10 1. Sol.ac. 63) Gd (Atomic No. DoFkukad fcUnqij DoFkukad mUu. IPIA. 65) are : 7 2 7 1 2 9 2 (1) [Xe]4f 6s . | NEET-2016 | 01-05-2016 | Code-C.166.fud f=kT.732 r Co-ordination no.k NaCl lajpuk d sleku N% gSA 167. [Xe]4f 5d 6s Ans.Y + – –10 –10 The ionic radii of A and B ions are 0.P. [Xe]4f 6s 6 1 2 7 2 9 foU.d 6.5   32  TB = I Kbm = 1 × 0. 2 (1) Resonance Eduventures Ltd.oaB– vk.)-324005 | Ph.414  r < 0.in | E-mail : [email protected]. At 100º C the vapour pressure of a solution of 6.k. of each ion is 6 like NaCl structure. [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 (2) [Xe]4f 6s . (1) 2 (2) rA  rB  (2) 6 = 0.52 rksbl foy . (3) 100º C P0 = 760 torr  6. [Xe]4f 5d 6s and [Xe]4f 6s 7 2 8 2 8 1 2 (2) [Xe]4f 6s . If Kb = 0. 6635555 | To Know more : sms RESO at 56677 Website : www.) . The coordination number of each ion in AB is + A . Jawahar Nagar. : +91-022-39167222 Ph. kksaes a/kuk. y = 3-gsDlkbu (3) X = 2-C.Zughad jrk gSA (2) gkbMª kst u d srhu leLFkkfud gSft lesalsVªkbfV. (2) Resonance Eduventures Ltd.e (1H1) lokZf/kd izp fy r gSA 170.u. | 01-05-2016 | Code-C.: +91-744-3192222 | FAX No. Kota (Raj. IPIA.w Vkbu .NH3 H–CCH   x  y (2) CH CH Br (2) CH CH Br 3 2 3 2 X vkS j Y gS: (1) X = 1-C. Jawahar Nagar.resonance.ogkj ughad jrk gSA + (4) gkbMª ksfu.ac. 2 = 2-gsDlkbu Ans.)-324005 | Ph. Sol. 1 rFkk 2 fod Yi vlR. gS\ (1) Mkbgkbª kst u vipk.w Vkbu .NH3 (1)NaNH2 / liq.fud y o. Jhalawar Road.NH (1)NaNH / liq. 2 = 2-Hexyne vfHkfØ . v lR.ogkj d jrk gSA eg 3H2 + N2  2NH3 Correct – gkbMª kst u d srhu leLFkkfud gksrsgSA ft lesaizksVh. Near City Mall. Kota (Raj.u esaeqDr : i esagksrk gSA Ans.NH 3 3 2 3 2 3 H–CCH   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 . Office : J-2. + (4) Hydronium ion.d d s: i esad k. Main Road. H3O exists freely in solution. y = 2- gs Dlkbu (2) X = 1-C.u d h rjg O. : +91-744-3012222.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 70 . CORPORATE OFFICE : CG Tower. A-46 & 52. (2) Hydrogen has three isotopes of which tritium is the most common. (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.w Vkbu . H3O d k vfLrRo foy . In the reaction (1)NaNH / liq.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. 6635555 | To Know more : sms RESO at 56677 Website : www. y = 3-gsDlkbu (4) X = 2-C.e vk.w Vkbu .No. No. : +91-022-39167222 Ph.| NEET-2016 Sol.R.in | E-mail : contact@resonance. fuEufy f[kr esalsd kS u lk d Fku gkbMªkst u d sfy .d d sleku O. (3) Hydrogen never acts as cation in ionic salts. gSA Correct – MkbgkbMª kst u vipk. y = 3-Hexyne (3) X = 2-Butyne .k esa (1)NaNH2 / liq. y = 3-Hexyne (4) X = 2-Butyne .e izp qjrk esagSA (3) gkbMª kst u vk.) . 7 63Eu [Xe] 4f 6s 64Gd 65Tb 169.324005 Reg.ac. (4) The vapour will contain equal amounts of benzene and toluene. Main Road. : +91-744-3012222. IPIA. Office : J-2. Consider the following liquid-vapour equilibrium.)-324005 | Ph.ac.: +91-744-3192222 | FAX No. 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.koLFkk]   nzo   ok"i esalsd kSu lk lacU/k lgh gS\ (1) dlnP Hv  dT RT 2 (2) Hv dlnG dT 2 Ans.Y HCC —Na+CH3–CH2–Br S N2 NaNH2 / NH3 (a) CH3–CH2–CC   CH3–CH2–CCH 1-Butyne SCH3CH2–Br CH3–CH2–CC–CH2–CH3 Hex-3-yne 171. 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.No. A-46 & 52. No.   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. Vapour Pressure Data at 25ºC.in | E-mail : contact@resonance. (1) Sol. Kota (Raj. : +91-022-39167222 Ph. Jawahar Nagar.sx. Jhalawar Road.ac. (2) The vapour will contain a higher percentage of benzene.| NEET-2016 Sol. Benzene = 12.snzo & ok"i lkE. HCCH NaNH2 / NH3    NH3  | 01-05-2016 | Code-C. (3) The vapour will contain a higher percentage of toluene. (Given. toluene = 3. CORPORATE OFFICE : CG Tower.) . 6635555 | To Know more : sms RESO at 56677 Website : www.resonance.8kPa. Resonance Eduventures Ltd.85kPa) (1) Not enough information is given to make a prediction.R.324005 Reg. Kota (Raj. Near City Mall.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 71 . As a result the vapour will contain a higher percentage of Benzene.oh esaHksn d jsxk \ (1) .kst u d sfy .85kPa) (1) vi. Jawahar Nagar. No.oaVkWy wbZu gksxhA Ans.oaVkWy wbZu d s1 : 1 vkn'kZeksy j feJ.sok"i nkc 25ºC csUt hu = 12.No.) . (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. A-46 & 52.324005 Reg.ac.k d sok"i la. Kota (Raj. OH CH3 CH3 O C O CH3 CH3 Resonance Eduventures Ltd.s Y. Jhalawar Road.oabld sVªkal leko. vkWFkksZizfrLFkkfi ckbZfQ ukbZy esanksuksfjax .s l hVksu (3) vks t ksu Ans. Which of the following reagents would distinguish cis-cyclopenta-1. (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. Sol.e vkblksizksiksDlkbM (2) .)-324005 | Ph. Sol.d ry esaugh gSvr% blesaizd kf'k. : +91-022-39167222 Ph. VkWy wbZu = 3.kZ Ir lwp ukvksad sd kj. IPIA. csat hu d smPp vkaf'kd ok"i nkc d sd kj.| NEET-2016 | 01-05-2016 | Code-C. leko.diol from the trans-isomer ? (1) Aluminium isopropoxide (2) Acetone (3) Ozone (4) MnO2 fuEu esalsd kSu lk vfHkd eZd fll~& lkbDy ksisUVk-1.sx. Near City Mall.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 72 .resonance.qfefu.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 .in | E-mail : [email protected] gSA 174.k bld k eksy fHkUu vf/kd gksrk gSA blfy . 6635555 | To Know more : sms RESO at 56677 Website : www.k d ksbZiwokZuqeku ughay xk. Which of the following biphenyls is optically active fuEu esalsd kSu lk ckbZfQ uk. gS\ O 2N CH3 (1) (2) CH3 I Br Br I (3) Ans.R. CORPORATE OFFICE : CG Tower.2.sfuEufy f[kr esalsd kSu lk d Fku lR.8kPa. Kota (Raj. ok"i esacs at hu d h izfr'krrk vf/kd gksxhA 173.2-MkbZv kWy . : +91-744-3012222. Office : J-2. (4) MnO2 (2) OH O=C Sol. Main Road.: +91-744-3192222 | FAX No. gS\ d Yiuk d jsafd rkieku 25ºC ij fLFkj gSA (fn.Y csUt hu .y izd kf'kd lfØ . k] NH3 esaH–N–H vkca/k&d ks.k] NH3 esaH–N–H vkca/k d ks.Y The correct statement regarding a carbonyl compound with a hydrogen atom on its alpha-carbon.koLFkk esagksrsgSavkSj .: +91-744-3192222 | FAX No.sd Fkuksaesalsd kSu lk v l R. NH3 and H2O. (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.k lsvf/kd gSA (2) CH4 es aH–C–H vkca/k d ks.g bud s vuq: i bZukWy esa vklkuh ls lkE. Jhalawar Road.kSfxd ft uesa d kcZu ij gkbMªkst u ijek. 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 .kSfxd ft uesa  d kcZu ij gkbMªkst u ijek. Consider the molecules CH4. Office : J-2. lgh d Fku gS% (1) d kcks Zfuy .k d gy krk gSA Ans.g bud s vuq: i bZukWy ls d Hkh Hkh lkE. Kota (Raj.d gSA 176.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 73 .) .sx. No.g izØ e d kcksZfuy hd j.kSfxd ft uesa d kcZu ij gkbMªkst u mifLFkr gS] d sfy .kq mifLFkr gS] .koLFkk esagksrsgSvkSj .in | E-mail : [email protected]. d hVksbZukWy leko.koLFkk es gksrsgSvkSj .koLFkk esaughagksrsgSA (3) d kcksZfuy . Main Road. IPIA.g bud svuq: i bZukWy esavklkuh lslkE. (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. Kota (Raj.k] NH3 esaH–N–H vkca/k d ks.kq mifLFkr gS] .resonance. (3) H2O esaH–O–H vkca/k d ks.ac.suhpsfn. : +91-022-39167222 Ph.g bud svuq: i bZukWy lsvklkuh lslkE. gS\ es aH–C–H vkca/k d ks.kqmifLFkr gS] .ork d sfy .kSfxd ft uesa-d kcZu gkbMªkst u ijek. CORPORATE OFFICE : CG Tower.kqmifLFkr gS] .ac.g izØ e fd Vks& bZukW y py ko.k lsvf/kd gSa (4) H2O esaH–O–H vkca/k d ks.k] CH4 esaH–C–H vkca/k d ks. Near City Mall.)-324005 | Ph.R.324005 Reg.k lsd e gSA (3) Resonance Eduventures Ltd. | NEET-2016 | 01-05-2016 | Code-C. 6635555 | To Know more : sms RESO at 56677 Website : www.koLFkk d gy krk gS A (4) d kcks Zfuy .d gkbMªkst u d h mifLFkfr vko'. d kcksZfuy sp3 d kcZu ij d e lsd e . (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.kqv ksad sfy . Jawahar Nagar.sfYMgkbM&d hVksu lkE.ork d gy krhgS A (2) d kcks Zfuy . d kcksZfuy . NH3 vkS j H2O v.k] lHkh esa90º ls (1) CH4 vf/kd gSA Ans. : +91-744-3012222.g izØ e . (2) a carbonyl compound with a hydrogen atom on its alpha-carbon never equilibrates with its corresponding enol. Sol. 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) 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.k rFkk H2O esaH–O–H vkca/k d ks.No. A-46 & 52.kS fxd d kcZu ij gkbMªkst u ijek. kSfxd ksad ksmud slad j. A-46 & 52.ac.sx.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 74 . Office : J-2.in | E-mail : contact@resonance.| NEET-2016 Sol. : +91-022-39167222 Ph. No.k . Kota (Raj.s gSad ksfey k.No.ac.) . (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) Resonance Eduventures Ltd. Jawahar Nagar.324005 Reg.resonance. Near City Mall. Kota (Raj. 6635555 | To Know more : sms RESO at 56677 Website : www. Jhalawar Road. Main Road.)-324005 | Ph. (b) XeO3 (ii) oxZlery h (c) XeOF4 (iii) fijkfeMh (d) XeF4 (iv) oxZfijkfeMh d ksM : Ans.s.R.oavkd kj t ksfd LReHk II esafn. IPIA. 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.: +91-744-3192222 | FAX No. 3 CH4 C sp H 177. CORPORATE OFFICE : CG Tower.srFkk lgh fod Yi d ksfpfUgr d hft .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.sx. : +91-744-3012222.A LrEHk-I LrEHk II (a) XeF6 (i) foÑ r v"VQ y d h. sp H2O sp H Code-C. F  F F Xe F F F 3 XeO3 sp pyramidal fijkfeMh.)-324005 | Ph.| NEET-2016 Sol. Kota (Raj.ac. Main Road.No.   F Consider the nitration of benzene using mixed conc. Resonance Eduventures Ltd. H2SO4 and HNO3 . No.) .324005 Reg. 2+ (2) Mg ions from a complex with ATP. Jawahar Nagar. Sol. CORPORATE OFFICE : CG Tower. 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.   Xe O O O XeF4 F 3 2 square pyramidal 3 2 square planar oxZlery h. F   Xe F F O XeF4 F sp d F   Xe F 178.fn bl feJ. (4) Ca2+ ions are not important in maintaining the regular beating of the heart. Which of the following statement is false ? 2+ (1) Mg ions are important in the green parts of plants. Jhalawar Road.k t krk gSrksNO2+ d h lkUnzrk ?kVrh gSft llsukbfVªd j.in | E-mail : [email protected] v"VQ y d h.: +91-744-3192222 | FAX No.k d k osx gks xk% (1) nq xquk (2) rs t Ans.k d h nj /kheh gks t krh gSA 179. Near City Mall. .in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 75 . Office : J-2.fn KHSO4 d ksvf/kd ek=kk esafey k.k laknzH2SO4 . 6635555 | To Know more : sms RESO at 56677 Website : www. : +91-022-39167222 Ph. Kota (Raj.k esaT. IPIA.resonance. A-46 & 52. (3) /khjs (4) vifjofrZ r (3) If large amount of KHSO4 is added. sp d oxZfijkfeMh.knk ek=kk esaKHSO4 Mky rsgSrksukbVªhd j.ac. 3 3 XeF6 sp d | 01-05-2016 | distorted octahedral fod ` r Code-C. Concentration of NO2+ will decrease and hence the rate of nitration will be slower. 2+ (3) Ca ions are important in blood clotting. : +91-744-3012222.oaHNO3 d h mifLFkfr esagksjgk gSA .R. L.| NEET-2016 | 01-05-2016 | Code-C. Sol. gS\ 2+ (1) Mg vk.u jDr d kst ekusd sfy . Office : J-2.R.u jDr d kst ekusd sfy .Y fuEufy f[kr esalsd kS u lk d Fku vlR. CORPORATE OFFICE : CG Tower. 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.)-324005 | Ph.O of M–C  B.segRoiw .u .128Å) fuEufy f[kr esalsfd ld h C–O vkca/k y EckbZvf/kd re gS\ (eqDr C–O vkca/k y EckbZCO esa1.kZgSA Which of the following has longest C–O bond length ? (Free C–O bond length in CO is 1.fer j[kusesHkh egRoiw. of C–O ) Higher is negative charge on metal.ac.k /kkrqd kcZu cU/k y EckbZc<+rh gS(B. Near City Mall.u Ans. Sol. Jawahar Nagar. Kota (Raj.No.-Vh-ih. IPIA. 2+ Ca 180. xfr d ksfu.resonance. Jhalawar Road. 6635555 | To Know more : sms RESO at 56677 Website : www.srFkk ân. (4) +2 Ca are important in blood clotting and are also important in maintaining the regular beating of the heart.in | CIN : U80302RJ2007PLC024029 This solution was download from Resonance NEET 2016 Solution portal PAGE # 76 . No.ac. A-46 & 52.L.128Å gSA) + Ans. of C–O ) /kkrqij ft ruk vf/kd _ .324005 Reg. xfr d ksfu. (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.fer j[kusesegRoiw.: +91-744-3192222 | FAX No.kkos'k gksxk mruk vf/kd i'p vkcU/ku gksxkA blfy . : +91-022-39167222 Ph.) . : +91-744-3012222.O of M–C  B. Kota (Raj.kZgSA 2+ vk.u ikS /kks ad sgfjr Hkkxksad sfy . ân.segRoiw.O of C– CB.kZgSA (2) Mg 2+ (4) Ca2+ vk.O of C–CB. cU/k y EckbZHkh vf/kd gks xhA Resonance Eduventures Ltd. Main Road.in | E-mail : [email protected] slkFk ld qy cukrsgSA (3) Ca vk.kZughagSA vk. PRE-MEDICAL DIVISION etcwr 'kq:vkr Academic Session: 2016-17 csgrj Hkfo". Parvinder Singh: 9352880505 . ^^lathouh** ^^lathouh** ZERO FEE PROGRAM (ZFP)* in Yearlong Classroom Contact Programs (YCCPs) for AIIMS/ AIPMT 100 100 % % fu%'kqdk. June 2016 XI AIPMT April & May 2016 SAMPOORN (MD) 3. 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Mayank Tiwari: 9352529244 w Mr.No. XIII To know more about ZFP & Admission Process Please call at : w Mr. Documents Similar To Neet 2016 Paper With Solution Code C R YSkip carouselcarousel previouscarousel nextCLS Aipmt 16 17 XI Che Study Package 3 SET 2 Chapter 10uploaded by kalloliImportant NEET Chapters for 2017uploaded by Selvaraju ThangarajuNeets MC Questionsuploaded by Joshua Hernandezpractice test for neetuploaded by Mohammed Aftab AhmedCLS Aipmt 16 17 XIII Bot Study Package 2 SET 1 Chapter 6uploaded by prism1702Practice Test Neet 2uploaded by Mohammed Aftab AhmedCLS Aipmt 15 16 XI Che Study Package 3 SET 1 Chapter 9uploaded by sairajPhysics for Iit Jee - Tmhuploaded by Sanu KarmakarNEET Phase 2 2016 Question Paper Set AA.pdfuploaded by amitiffcoBansal Classesuploaded by microdotcdmCLS_aipmt-15-16_XIII_bot_Study-Package-1_Set-1_Chapter-2.pdfuploaded by Aakash Badtameez SharmaMock Test for NEET-2017uploaded by Nimish JainNEET 2013 Question Paperuploaded by pavaniCLS Aipmt 16 17 XI Che Study Package 3 SET 2 Chapter 11uploaded by kalloliNEET Model Papers Free Practice Set 2 Physicsuploaded by makmgmObjective Chemistryuploaded by Gabriel WilsonNEETS Q&Auploaded by Van Joseph MarianoXii Neet Physics Mcqsuploaded by Megan Knightxii-neet-chemistry-mcqs.pdfuploaded by Marcus RashfordCLS Aipmt 15 16 XIII Phy Study Package 1 Set 1 Chapter 4uploaded by divyashishtomarCLS Aipmt 15 16 XI Che Study Package 2 SET 1 Chapter 5uploaded by sairajxi-neet-iseet-chemistry.pdfuploaded by FATHIMA Biology Today June 2016uploaded by B.AdamBiology Today - May 2016uploaded by Umberto EllKVS Study Material Biology Class 12uploaded by Akshatha NayakNovember 2015uploaded by ΘωμαςΣτεφανιδηςBiology Study Material Finaluploaded by dharmalemj6291Biologyuploaded by MITHUN CHATTERJEEAllen Study Materialuploaded by Jitendra VermaNEET 40 Days Chy-syllabusuploaded by vatsadbgMore From xanshahSkip carouselcarousel previouscarousel nextGujcetuploaded by xanshahNeet 2016 Paper With Solution Code C R Yuploaded by xanshahGujcetuploaded by xanshahGujcetuploaded by xanshahGujcetuploaded by xanshahGUJCET - Surface Chemistryuploaded by xanshahOutline of the Vedic Textsuploaded by xanshahGujcetuploaded by xanshahGujcetuploaded by xanshahJEE MAIN 2016 Question Paper Solutionuploaded by xanshahOptics & Wave Mechanicsuploaded by xanshahPeriodic Tableuploaded by xanshahElectricity Magnetismuploaded by xanshahGUJCET - Solid Stateuploaded by xanshahGUJCET - Liquid Solutionsuploaded by xanshahGujcet - Electrochemistryuploaded by xanshahGujcet 2015 Chemuploaded by xanshahStructure of Atomuploaded by xanshahFooter MenuBack To TopAboutAbout ScribdPressOur blogJoin our team!Contact UsJoin todayInvite FriendsGiftsSupportHelp / FAQAccessibilityPurchase helpAdChoicesPublishersLegalTermsPrivacyCopyrightSocial MediaCopyright © 2018 Scribd Inc. .Browse Books.Site Directory.Site Language: English中文EspañolالعربيةPortuguês日本語DeutschFrançaisTurkceРусский языкTiếng việtJęzyk polskiBahasa indonesiaMaster your semester with Scribd & The New York TimesSpecial offer for students: Only $4.99/month.Master your semester with Scribd & The New York TimesRead Free for 30 DaysCancel anytime.Read Free for 30 DaysYou're Reading a Free PreviewDownloadClose DialogAre you sure?This action might not be possible to undo. 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