eenadupratibha.net · 2019-03-16 · EAMCET GRAND TEST (Total Syllabus) Page.No : 2 BOTANY 1. Correct statement among the following is ªH÷Ok "„\÷–Õ ‰‹iƒ‡Ø

$Page 1: eenadupratibha.net · 2019-03-16 · EAMCET GRAND TEST (Total Syllabus) Page.No : 2 BOTANY 1. Correct statement among the following is ªH÷Ok "„\÷–Õ ‰‹iƒ‡Ø$
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EAMCET GRAND TEST (Total Syllabus)

Page.No : 2

BOTANY

1. Correct statement among the following is ãH÷Ok "�\÷�Õ �¬i�³Ø°# "�¼Y¼.1) Autogamy and geitonogamy are not possible in dioecious species

UH�eOQêã�×�Ç°ì`Ç°��Õ P`Çà�¬~�Q®�¬O�¬~¡øO =°i�Çò UH�=$H�Æ �¬~¡�¬~�Q® �¬O�¬~¡øO [~¡Q®_¨xH÷ J=Hê�×O �è^Î°.2) Geitonogamy is not possible in monoecious species

kÞeOQêã�×�Ç°ì Ç°��Õ UH�=$H�Æ�¬~�Q®�¬O�¬~¡øO [~¡°Q®°@ä�½ J=Hê�×O �è Î°.3) Only autogamy is found in bisexual flowers

kÞeOQ®H� �¬ô��æ��Õ P Çà�¬~�Q® �¬O�¬~¡øO =¶ã Ç"Í° [~¡°Q®° Ç°Ok.4) Herkogamous flowers are unisexual

Ì�ì~ËøQ®q¶ KÇ¶�²OKÍ "³òH�ø�° UH�eOQêã�×�Ç°OQê LO\ì~ò.2. Choose the correct combinations �¬i�³Ø°# �¬=ü�¬ð�#° Q®°iëOKÇO_�.

Plant Family Characterstic feature "³òH�ø ä�½@°O|O =òY¼�H�Æ}OA. Red sanders Fabaceae Marginal placentation

Zã~¡KÇO Î#O �¦��è�² L��O Ç JO_È<�¼�¬OB. Thorn apple Solanaceae Carpels are arranged obliquely

L"³°à`Çë ªÚ�<Í�² �¦¬� Î�ì� U@"��° J=°iH�C. Meadow saffron Liliaceae Epigynous flowers

"³°_Ëª�ã�¦¬<£ ee�Í°�² JO_ÈHË�Õ�¬iH� �¬ô��æ�°D. Arachis Fabaceae Diadelphous stamens

J~�d�¹ �¦��è�² kÞ|O �ÎH� öH�¬~��°1) A & B only A, B =¶ã`Ç"Í° 2) C & D only C, D =¶ã`Ç"Í°3) A, B & D only A, B, D =¶ã`Ç"Í° 4) A, B, C & D

3. Artocarpus belongs to the family P~Ë�Hê~¡��¹ D ä�½@°O�ìxH÷ K³Ok#k1) Bromeliaceae ã�çq°e�Í°�² 2) Annonaceae J<Ë<Í�²3) Anacardiaceae J#HêiÛ�Í°�² 4) Moraceae "³¶ö~�²

4. Assertion (A) : Vascular cambium produces secondary xylem and secondary phloem

^�Î$_�È"�¼Y¼ (A) : <�oHê q��ì[¼ H�}ì=o kÞf�Ç° ¥~¡°=ô =°i�Çò kÞf�Ç° �é+¬H� H�}ì�ìxß U~¡æ~¡°ª�ë~ò.

Reason (R) : Only secondary meristems are responsible for producing the secondary tissues

Hê~¡}=ò (R) : kÞf�Ç° H�}ì�ì� L Çæuë�Õ kÞf�Ç° q��ì[¼ H�}ì�ì�° =¶ã Ç"Í° ��ç¾O\ì~ò.1) Both A and R are true and R is the correct explanation of A

A =°i�Çò R �¬i�³Ø°#q. =°i�Çò R J#°#k A ä�½ �¬i�³Ø°# q=~¡}.2) Both A and R are true and R is not the correct explanation of A

A =°i�Çò R �¬i�³Ø°#q. =°i�Çò R J#°#k A ä�½ �¬i�³Ø°# q=~¡}Hê^Î°.3) A is true and R is false A �¬i�³Ø°#k, R �¬i�³Ø°#k Hê Î°.4) A is false and R is true A �¬i�³Ø°#k Hê Î°, R �¬i�³Ø°#k.

5. The taxonomic unit �Phylum� in the classification of animals is equivalent to which hierarchiallevel in classfication of plants

[O Ç°=ô� =s¾H�~¡}�Õx =s¾H�~¡} ã�¬=¶}"³°Ø# Ì�¦á�"£° "³òH�ø� =s¾H�~¡}�Õx U =s¾H�~¡} ª��~ò ãH�=¶xH÷ �¬=¶#O

1) Class `Ç~¡Q®u 2) Order ãH�=°O 3) Division q��ìQ®O 4) Family ä�½@°O|O

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6. Cladophylls in Asparagus are P�¬æ~�Q®�¹�Õx HêÁ_Ë�¦²��° J#°#q1) Modified branches of unlimited growth Jxtó ÇOQê Ì�~¡°Q®° ~¡¶��O Ç~¡O K³Ok# �§Y�°2) Modified compound leaves ~¡¶��O Ç~¡O K³Ok# �¬O�ÇòH�ë �¬ã`��°3) Modified branches of limited growth xsâ Ç Ì�~¡°Q®° Î� Q®� ~¡¶��O Ç~¡O K³Ok# �§Y�°4) Modified inflorescence axis ~¡¶��O Ç~¡O K³Ok# �¬ô+¬æq<�¼ª�H�ÆO

7. In an inflorescence where flowers are borne laterally in an acropetal succession, the position ofthe youngest floral bud shall be

XH� �¬ô+¬æq<�¼�¬O #O^Î° �¬ô��æ� Pq~�Ä�=O ��~¡ÅÞOQê#° =°i�Çò JãQêa�ª�~¡ãH�=°O�Õ U~¡æ_�# �³°_È�, "�\÷�ÕJ Ç¼O Ç �è Ç �¬ô+¬æO �³òH�ø ª��#O1) Proximal �¬g°�¬�¬Ö 2) Distal ^Î¶~¡�¬�3) Intercalary =°^�Î¼�¬� 4) Any where Zª��#=ò�Õ<³á<�

8. In a typical complete, bisexual and hypogynous flower, the floral whorls from the outermost tothe innermost is:

XH� �¬O�¬î~¡â, kÞeOQ®H� =°i�Çò JO_ÈHË�§ �Î�²Ö Ç �¬ô+¬æO #O Î° �¬ô+¬æ�¬ã Ç =��Ç¶�° "³�°�¬e #°O_� �Õ�¬e"³á�¬ô#ä�½1) Calyx, corolla, androecium and gynoecium ~¡H�ÆH��¬ã`�=o, PH�~¡Â} �¬ã`�=o, öH�¬~�=o =°i�Çò JO_ÈHË�×O2) Calyx, corolla, gynoecium and androecium ~¡H�ÆH��¬ã`�=o,PH�~¡Â} �¬ã`�=o, JO_ÈHË�×O =°i�Çò öH�¬~�=o3) Gynoecium, androecium, corolla and calyx JO_ÈHË�×O, öH�¬~�=o,PH�~¡Â} �¬ã`�=o =°i�Çò ~¡H�ÆH��¬ã`�=o4) Androecium, gynoecium, corolla and calyx öH�¬~�=o, JO_ÈHË�×O, PH�~¡Â} �¬ã`�=o =°i�Çò ~¡H�ÆH��¬ã`�=o

9. Arrange the following type of fruits in correct ascending order based on the number of carpelsof pistil from which they are formed

D ãH÷Ok ~¡Hê� �¦¬�ì�° Zxß �¦¬� Î�ì�`Ë ä��_�# JO_ÈHË�×O #°O_� U~¡æ~¡KÇ|_È ��³¶ "�\÷ �¬OY¼ P �¥~¡OQê �¬i�³Ø°#P~Ë�¬ì} ãH�=°O�Õ J=°~¡óO_�.A) Pepo Ì��é B) Cypsela �²Ì��ìC) Nut Ì�Oä�½Q®� �¦¬�O D) Legume kÞq �¥~¡H� �¦¬�O1) A,B,C,D 2) D,B,A,C 3) C,D,A,B 4) C,B,A,D

10. A bivalent of meiosis-I consists of H�Æ�Ç°H�~¡}-I #O Îe �ÿá"��O\� H�ey�ÇòO_È°#k1) Two chromatids and one centromere ï~O_È° ãHù=¶\÷_£�° =°i�Çò XH� Ì�Oã\Õq°�Ç°~�2) Two chromatids and two centromeres ï~O_È° ãHù=¶\÷_£�° =°i�Çò ï~O_È° Ì�Oã\Õq°�Ç°~��°3) Four chromatids and two centromeres <��°Q®° ãHù=¶\÷_£�° =°i�Çò ï~O_È° Ì�Oã\Õq°�Ç°~��°4) Four chromatids and four centromeres <��°Q®° ãHù=¶\÷_£�° =°i�Çò <��°Q®° Ì�Oã\Õq°�Ç°~��°

11. Study the following table. D ãH÷Ok �¬\÷�H�#° J �Î¼�Ç°O KÍ�Çò#°I) Fucus Oogamous plant Diplontic life cycle

�¦¬î¼H�� ¹ JO_È�¬O�³¶Q® "³òH�ø ÎÞ�Ç°�²ÖuH� rq ÇKÇãH�OII) Marchantia Dioecious plant Haplo-diplontic life cycle

=¶~�øO+²�Ç¶ UH�eOQêã�×~ò "³òH�ø UH�Ð ÎÞ�Ç°�²ÖuH� rq ÇKÇãH�OIII) Cycas Dioecious plant Diplo-haplontic life cycle

Ì�áH��¹ UH�eOQêã�×~ò "³òH�ø ÎÞ�Ç°.UH��²ÖuH� rq ÇKÇãH�OIV) Zea mays Monoecious plant Diplo-haplontic life cycle

l�Ç¶"Í°�¹ kÞeOQêã�×~ò "³òH�ø ÎÞ�Ç°.UH��²ÖuH� rq ÇKÇãH�OWhich of the above non-spermatophytic plants show correct combinations.

Ì�á"�\÷�Õ �¬i�³Ø°# �¬OH��<��#° KÇ¶�¬ô c[~¡�²ì Ç "³òH�ø1) I & II 2) III & IV 3) I & III 4) II & IV

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12. The roots, the stem and the leaves are modified for the same purpose respectively in

"Í~¡°Á, HêO_ÈO =°i�Çò �¬ã`��° XöH P=�×¼H� Ç Hù~¡ä�½ ~¡¶��O Ç~¡O K³OkOKÇ°Hùx L#ß "³òH�ø�° =~¡°�¬Qê1) Dolichos, Dioscorea, Dionea _¨�H��¹, _È�Ç¶ªéøi�Ç¶, _È�³¶x�Ç¶2) Taeniophyllum, Euphorbia, Acacia \ìx�³¶�¦²�ÁO, �Çò�¦éiÄ�Ç¶, JöH+²�Ç¶3) Asparagus, Nerium, Bryophyllum P�¬æ�~�Q®�¹, hi�Ç°O, ã|�³¶�¦²�ÁO4) Vanda, Citrus, Nepenthes "�O_¨, �²ã@�¹, <³Ì�¦Ok��¹

13. Assertion (A) : Pollination precedes fertilization in the life cycle of angiosperms.

Î$_�È"�¼Y¼ (A) : P=$`Çc[ "³òH�ø� rq`ÇKÇãH�O#O^Î° �¦¬�nH�~¡}O#ä�½ =òO^Î°Qê �¬~�Q®�¬O�¬~¡øO [~¡°Q®°#°Reason (R) : Angiosperms are all heterosporous plants and their sporophytes are invariably dioecious.

Hê~¡}O (R) : P=$ Ç c[ "³òH�ø�xß�Çò a�#ß�² Îíc[ Ç "³òH�ø�° =°i�Çò "�\÷ �² Î�c[^¥�° qk�Qê UH�eOQêã�×�Ç¶�°1) A and R are true and R is the correct explanation of A.

A =°i�Çò R �¬i�³Ø°#q. =°i�Çò R J#°#k A ä�½ �¬i�³Ø°# q=~¡}.2) A and R are true and R is not the correct explanation of A.

A =°i�Çò R �¬i�³Ø°#q. =°i�Çò R J#°#k A ä�½ �¬i�³Ø°# q=~¡}Hê^Î°.3) A is true, R is false. A �¬i�³Ø°#k, R �¬i�³Ø°#k Hê Î°.4) A is false, R is true. A �¬i�³Ø°#k Hê^Î°, R �¬i�³Ø°#k.

14. Pickout the true statement from following.

D ãH÷Ok "�x�Õ �¬i�³Ø°# "�¼Y¼#° Q®°iëOKÇ°=ò.1) The microsporogenesis and megasporogenesis involve meiosis and result in the forma tion of

first cells of sporophytic generation

�¬¶H�Æà =°i�Çò �¬¶Ö� �²^Î�c*Õ`Çæ`Ç°ë�° H�Æ�Ç°H�~¡} q��[##° H�ey LO_�, �²^Îíc[^Î ^Î�×ä�½ã��~¡O��H�}ì�<Í~¡æ~¡KÇ°#°.

2) Hybridization is an artificial crossing which may involve either geitenogamy (or) xenogamy

�¬OH�sH�~¡}=ò J#°#k w\÷<ËQ®q° (�èH�) l<ËQ®q°�`Ë ä��_�# H�$ãu=° �¬OH�~¡}=ò.3) The endospermic and perispermic seeds of angiosperms contain the nutritive tissues of both

parental and next generations

JOä�½~¡KÇó^Î �¬�²ì`Ç =°i�Çò �¬iKÇó�^Î �¬�²ì`Ç P=$`Çc[ q`Çë#=ò [#H�`Ç~�xH÷ =°i�Çò =°~¡°�¬\÷ `Ç~�xH÷K³Ok# �é+¬} H�}ì�ì�#° H�ey LO_È°#°.

4) The double fertilization involves only syngamy but not triple fusion in some angiosperms which produce non-endospermic seeds

JOä�½~¡KÇó�^Î ~¡�²ì`Ç q`Çë<��#° L`ÇæuëKÍ�Çò P=$`Çc[ "³òH�ø�° kÞ�¦¬�nH�~¡}O #O^Î° �¬O�³¶Q®c*ì� �¬O�³¶Q®O=¶ã Ç"Í° [~¡°Q®°#°. Hêx ãu�¬O�³¶Q®O [~¡°Q® Î°.

15. Which of the following vegetation shows variations among the plants.

D ãH÷Ok "�x�Õ Ç=°�Õ`�=ò Zä�½ø= "³áq �Î¼=ò�#° KÇ¶�¬ô "³òH�ø� �¬=ò^¥�Ç°=òI) Sugarcane plantation in the field �¬O@�Ú�O�Õx K³�ä�½ "³òH�ø�°II) Strobilanthes Kunthiana stretches on hill tracks

HùO_ÈKÇi�Ç°� g°^Î ãªé�a�ìO^�Î�¹ ä�½Ou�Ç¶<� "³òH�ø�°III) Well spread water hyacinth in a water body

h\÷Hù�#° #O Î° �ìQê q�¬ëiOz L#ß L#ß SMìiß�Ç¶IV) Bamboo plantation in the forest J_Èq#O^Îe (�ìO|°) "³^Î°~¡° "³òH�ø�°1) I & II 2) III & IV 3) I & III 4) II & IV

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16. List - I *ìa`�- I List - II *ìa`�- II

A) Glyoxysome I) Synthesis of phospholipids

ïQáÁPH©�ªé"£° �¦�ªéæ�e�²_£� �¬O�õÁ+¬}B) Peroxysome II) Synthesis of lipids

Ì�~�H÷�ªé"£° e�²_£� �¬O�õÁ+¬}C) Smooth ER III) Autolysis of the cell under starvation conditions

#°#°�¬ô ER P�¬ð~¡�¬ôHù~¡ Ç �¬i�²� Ç°��Õ H�}O �¬Þ�Ç°Oqzó�uëD) Lysosome IV) Converts stored lipids to carbohydrates

�ÿáªéªé"£° x�=KÍ�Ç°|_�# e�²_£�#° Hê~ËÄÌ�áìã_Í\��°Qê =¶~¡°�¬°ëOkThe correct match is Wk �¬ï~á#*Õ_�O�¬ô

A B C D A B C D1) IV I III II 2) I IV II III

3) I IV III II 4) IV I II III

17. List -I *ìa`�- I List - II *ìa`�- IIA) Stipulate, compound leaves I) Brassicaceae

�¬ôKÇó��¬�²ì Ç, �¬O�ÇòH�ë �¬ã`��° ã�ì�²öH�²B) Obliquely arranged carpels II) Liliaceae

U@"��°Qê J=°i# �¦¬� Î�ì�° ee�Í°�²C) Homochlamydeous perianth III) Fabaceae

�¬=°�¬i�¬ã Ç�Çò ÇO �¦��è�²D) Ovary becomes bilocular due to the IV) Solanaceae

formation of false septum ªÚ�<Í�²J#$ Çä�½_È¼O U~¡æ_È_ÈO =�ÁJO_¨�×�Ç°O kÞa��Çò Ç=°=ô Ç°Ok

The correct match is Wk �¬ï~á#*Õ_�O�¬ôA B C D A B C D

1) IV III I II 2) IV III II I3) III IV II I 4) III IV I II

18. The correct ascending order of the following based on the total number of floral leaves present ineach flower of the following plants isãH÷Ok "³òH�ø� ã�¬u �¬ô+¬æO�Õ LO_Í �¬ô+¬æ �¬ã`�� "³ò ÇëO �¬OY¼ P �¥~¡OQê "�x �³òH�ø �¬ï~á# P~Ë�¬ì} ãH�=°OA) Onion h~¡°eÁ B) Sunnhemp [#°=òC) Kamanchi Hê=°Oz D) Mustard P=1) B,C,D,A 2) A,D,C,B 3) A,D,B,C 4) D,A,B,C

19. List - I *ìa`�- I List - II *ìa`�- II

A) Dicliny UH�eOQ® ÇÞO I) Solanum ªÚ�ì#"£°B) Self incompatibility �¬Þ�Ç°O q~¡° Î�O II) Sunflower �¬¶~¡¼HêO ÇOC) Protogyny ã�Ôë ��ìQ®ã�¬ �Î"³¶ Çæuë III) Abutilon J|°\÷�ì<£D) Protandry �¬ôO��ìQ® ã�¬ �Î"³¶ Çæuë IV) Hibiscus Ì�áìa�¬ø�¹E) Herkogamy Ì�ì~ËøQ®q° V) Castor P=ò ÎOThe correct match is Wk �¬ï~á#*Õ_�O�¬ô

A B C D E A B C D E1) III V I II IV 2) V III I II IV3) III V II I IV 4) V III II IV I

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20. Based on the number of cohorts arrange the following series (Note :recognised by Bentham &

Hooker) in an ascending order

ãH÷Ok ã�õ}°�#° (Q®=°xH� �ÿO^�¥"£° �¬�H�~��° Q®°iëOz#q) "�x�Õx HË�¬ð~�� ¬OY¼ P^�¥~¡OQê P~Ë�¬ì} ãH�=°O�ÕJ=°~¡óO_�.A) Bicarpellatae �ÿáHêï~æ�èÁ>ÿ B) Thalamiflorae �Î�ìq°�¦éÁï~C) Calyciflorae Hêe�²�¦éÁï~ D) Heteromerae Ì�ì\÷~Ëg°ï~1) B, C, A, D 2) D, A, C, B 3) A, D, C, B 4) D, A, B, C

21. Which of the following will occur during opening of photoactive stomata ?

�¦é\ÕUH÷�"£ �¬ã Ç~¡Oã �¥�° ³~¡KÇ°Hù<Í �¬=°�Ç°O�Õ D ãH÷Ok"�x�Õ UKÇ~¡¼�° [~¡°Q®°#° ?A) Efflux of K+ K+ |�²ìã�¬�=} E) Influx of K+ K+ JO`Çã�¬�=}B) Efflux of H+ H+ |�²ìã�¬�=} F) Influx of H+ H+ JO`Çã�¬�=}C) Efflux of Cl� Cl� |�²ìã�¬�=} G) Influx of Cl� Cl� JO`Çã�¬�=}D) Turgidity of guard cells ~¡H�ÆH� H�}ì� �Ôæ`Ç�²ÖuH) Flaccidity of guard cells ~¡H�ÆH� H�}ì� �×Á^�Î��²Öu1) B, C, G, E 2) A, C, F, H 3) B, E, G, D 4) B, D, E, H

22. The ratio between the number of reactions catalysed by aldolase and the number of reactions

catalysed by transketolase that take place in the regeneration phase of Calvin cycle is

HêeÞ<£ =��Ç°O�Õx �¬ô#~¡° Çæuë Î�×�Õ, P�ÕÛ�è*� Z<£*ÿá"£°KÍ Lã Íæi ÇO KÍ�Ç°|_È° =°i�Çò ã\ì<£�H©\ç�è*� Z<£*ÿá"£°KÍLã Íæi ÇO KÍ�Ç°|_È° KÇ~¡¼� =° �Î¼ Q®� x+¬æuë ZO Ç ?1) 1 : 1 2) 2 : 1 3) 1 : 2 4) 4 : 3

23. Reaction centres of PS-I and PS-II respectively are PS-I =°i�Çò PS-II� KÇ~�¼öHOã^¥�° =~¡°�¬Qê

1) P680, P700 2) P650, P680 3) P760, P700 4) P700, P680

24. A compound which is a product of decarboxylation and substrate of oxidative decarboxylation

in Krebs cycle is named as A. Read the following B, C and D

A J<Í �¬^¥~¡í=ò ãïH��=��Ç°O�Õ _�Hê~�ÄH©��è+¬<£ KÇ~¡¼ �³òH�ø L`�æk`ÇO =°i�Çò PH©�H�~¡} _�Hê~�ÄH÷��è+¬<£ä�½J^Î�¬��¬^¥~¡�O.D ãH÷Ok B, C =°i�Çò D �#° KÇ Î°=ô=òB is an aminoacid produced from A B J<Í J"³°Ø<ËP=°ÁO A #°O_� U~¡æ_�#kC is an amide produced from B C J<Í J"³°Ø_£ B #°O_� U~¡æ_�#kD is a substance produced from A D J<Í �¬^¥~¡�=ò A #°O_� U~¡æ_�#kIdentify BCD respectively BCD �#° =~¡°�¬Qê Q®°iëO�¬ô=ò

B C D1) a-ketoglutaric acid Glutamine Glutamic acid

a-H©\ÕQ®°Á\ìiH± P=°ÁO Q®°Á\ìq°<£ Q®°Á\ìq°H± P=°ÁO2) Glutamine Succinyl CoA Succinic acid

Q®°Á\ìq°<£ �¬H©�<³á�� HË-A �¬H©�xH± P=°ÁO3) Glutamine Glutamic acid Succinly CoA

Q®°Á\ìq°<£ Q®°Á\ìq°H± J=°ÁO �¬H©�<³á�� HË-A4) Glutamic acid Glutamine Succinyl CoA

Q®°Á\ìq°H± J=°ÁO Q®°Á\ìq°<£ �¬H©�<³á�� HË-A

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25. Find the correct combinations �¬i�³Ø°# "Í°�×qO�¬ô�#° H�#°Qù#°=òI) Auxin Went Parthenocarpy

PH÷�<£ "³O\� JxÀ+H��¦¬�#OII) Gibberellin Kurasawa Delay of senescence of fruits

l|Äï~eÁ<£ ä�½~¡°ª�= �¦¬�ì� r~¡â Ç#° P��¬¼ �¬~¡KÇ°#°III) Cytokinin Skoog Root hair formation

Ì�á\ÕïHáx<£ �¬¶øQ· =ü�öH�§�#° L ÇæuëKÍ�Çò#°IV) Ethylene Miller Fruit ripening

Zk�b<£ q°�Á~� �¦¬�ì� �¬H�Þ Ç1) I & II only I & II =¶ã`Ç"Í° 2) I, II and III I, II =°i�Çò III

3) III and IV III =°i�Çò IV 4) IV only IV =¶ã`Ç"Í°

26. Artificial synthesis of a DNA sequence coding for RNA was carried out by

1) Emerson 2) F Conrat 3) Watson and Crick 4) Hargobind Khorana

RNA �¬O�õÁ+¬}�Õ Ë_Èæ_Í DNA =~¡°�¬ ãH�=¶xß H�$ãu=°OQê �¬O�õÁ+¬} KÍ�²# �§ã�¬ë"Í Çë1) Z=°~¡�<£ 2) ã�¦�OH÷�� HË<£~�\�3) "�@�<£ =°i�Çò ãH÷H± 4) �¬ì~¡QËqO £ MÕ~�<�

27. Assertion (A) : Secondary succession is faster than primary succession.

Reason (R) : Secondary succession begins in areas where some soil is present or sediment is present.

1) A and R are true and R is the correct explanation of A.

2) A and R are true and R is not the correct explanation of A.

3) A is true, R is false. 4) A is false, R is true

xtó`Ç "�¼Y¼ (A) : "³òH�ø� kÞf�Ç° J#°ãH�=°O, ã�� Îq°H� J#°ãH�=°O H�O>ÿ "ÍQ®=O ÇOQê [~¡°Q®°#°.

Hê~¡}O (R) : HùO`Ç =°$uëH� LO_È° ã�¬^Í�§��Õ "³òH�ø� kÞf�Ç° J#°ãH�=°O ã��~¡O��=°Q®°#°.(1) A =°i�Çò R �° ï~O_È¶ x["³°Ø#q. R J#°#k A ä�½ �¬i�³Ø°# q=~¡}.(2) A =°i�Çò R �° ï~O_È¶ x["³°Ø#q. R J#°#k A ä�½ �¬i�³Ø°# q=~¡} Hê^Î°.(3) A X�¬ôC, R `Ç�¬ôC (4) A `Ç�¬ôC, R X�¬ôC

28. Codeine is an example for1) Alkaloid-Primary metabolite 2) Toxin-Secondary metabolite

3) Drug-Secondary metabolite 4) Alkaloid-Secondary metabolite

'HË_�<£Ñ J#°#k nxH÷ L^¥�¬ì~¡}1) P�ø�ì~ò_£ Ð ã�� Îq°H� r=ãH÷�Ç¶ L Çæ#ßO 2) \ìH÷��¹ Ð kÞf�Ç° r=ãH÷�Ç¶ L Çæ#ßO3) B+¬ �ÎO Ð kÞf�Ç° r=ãH÷�Ç¶ L Çæ#ßO 4) P�ø�ì~ò_£ Ð kÞf�Ç° r=ãH÷�Ç¶ L Çæ#ßO

29. Disease resistance in some plants is induced by an essential nutrient which is

1) Macro, mineral element 2) Micro, mineral element

3) Macro, non-mineral element 4) Micro, non-mineral element

Hùxß "³òH�ø��Õ "�¼k� x~Ë �ÎH� Ç Ì�O�ÚOkOKÇ_¨xH÷ Ë_Èæ_Í =ü�H�O D ~¡HêxH÷ K³Ok#k1) �¬¶Ö�, Yx[ =ü�H�O 2) �¬¶H�ÆàÐYx[ =ü�H�O3) �¬¶Ö�, Yx[O Hêx =ü�H�O 4) �¬¶H�ÆàÐYx[O Hêx =ü�H�O

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30. Incorrect statement of the following with reference to photosynthesis is

1) Cyclic electron transport takes place in stroma lamellae due to absence of PSII as well as NADP

reductase enzyme

2) Biosynthetic phase of photosynthesis utilizes ATP and NADPH2

3) Regeneration of primary acceptor of CO2 in C4 plants, takes place in bundle sheath cells

4) In C4

plants, CO2 fixation and Calvin cycle are separated in space, while in CAM plants they are

separated in time.

H÷~¡}[#¼ �¬O�³¶Q® ãH÷�Ç°ä�½ �¬O|OkOz D ãH÷Ok "�x�Õ �¬iHêx "�¼Y¼1) KÇãH©�Ç° Z�ãHê�<£ ~¡"�}ì , PSII =°i�Çò NADP H�Æ�Ç°H�~¡} Z<£*ÿá"£°�° �èx P=iâHê �¬@eH�� ã��O`ÇO�Õ

[~¡°Q®°#°2) r= �¬O�õÁ+¬} Î�×�Õ ATP =°i�Çò NADPH

2 �° qx�³¶yOKÇ|_È �~ò

3) C4 "³òH�ø��Õ ã��^�Îq°H� CO

2 ãQ®�Ôì`Ç �¬ô#~¡°`Çæuë, �¬ôO[�¬ô `ù_È°Q®° H�}ì��Õ [~¡°Q®°#°

4) C4 "³òH�ø��Õ CO

2 ª�Ö�¬# =°i�Çò ïHeÞ<£ =��Ç°O ª�Öh�Ç°OQê "Í~¡°KÍ�Ç°|_� LO_ÈQê, CAM "³òH�ø��Õ

Jq Hê��¬~¡OQê "Í~¡°KÍ�Ç°|_� LO\ì~ò.31. Assertion (A) : Pure proteins are never used as respiratory substrates.

Reason (R) : R.Q value for proteins is less than one.

1) A and R are true and R is the correct explanation of A.

2) A and R are true and R is not the correct explanation of A.

3) A is true, R is false. 4) A is false, R is true

xtó Ç "�¼Y¼ (A) : �×Ã Î�"³°Ø# ã�é\©<£�° �§Þ�¬ãH÷�Ç¶ J �Î�¬Ö �¬ ¥~�Ö�°Qê L�¬�³¶yOKÇ|_È=ô.

Hê~¡}O (R) : ã�é\©<£� �§Þ�¬ãH÷�Ç¶ HË+¬O\� q�°= XH�\÷H�O>ÿ Çä�½ø=

(1) A =°i�Çò R �° ï~O_È¶ x["³°Ø#q. R J#°#k A ä�½ �¬i�³Ø°# q=~¡}.(2) A =°i�Çò R �° ï~O_È¶ x["³°Ø#q. R J#°#k A ä�½ �¬i�³Ø°# q=~¡} Hê Î°.(3) A X�¬ôC, R `Ç�¬ôC (4) A `Ç�¬ôC, R X�¬ôC

32. Match the following with reference to Mendel's dihybrid cross F2 generation

"³°O_È�� kÞ�¬OH�~¡} F2 Ç~¡O ä�½ �¬O|OkOz D ãH÷Ok "�xx [ Ç�¬~¡KÇ°=ò Column-I Column-II

A) Any dominant phenotype I) 1/4

|�²ì~¡¾`Ç ^Î$�×¼~¡¶�¬O U^³á<�B) Any recessive phenotype II) 1/2

JO`Ç~¡¾`Ç ^Î$�×¼~¡¶�¬O U^³á<�C) Any homozygous genotype III) 3/4

�¬=°�ÇòQ®à[ [#°¼~¡¶�¬O U ³á<�D) Any heterozygous genotype IV) 1/4

q+¬=°�ÇòQ®à[ [#°¼~¡¶�¬O U ³á<� V) 2/3

A B C D A B C D

1) III V I II 2) III IV I II

3) III I IV V 4) V IV I II

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33. According to the modern scheme of photosynthetic non-cyclic electron transport

H÷~¡}[#¼ �¬O�³¶Q®¢H÷�Ç°�Õx JKÇ¢H©�Ç° Z�¢Hê�<£ ~¡"�}ìä�½ K³Ok# J �Î°xH� ��ì=# ¢�¬Hê~¡O A) O2 is involved in photolysis of water along with Mn and Ca

h\÷ HêOu q�õÁ+¬}�Õ Mn,Ca �`Ë ��@° O2 ä��_¨ ��ç¾O@°OkB) Many electron carriers are involved in each �uphill" step

¢�¬u ZQ®°_È° =¶~¡¾O�Õ#¶ K��ì Z�¢Hê�<£ "��¬ìHê�° ��ç¾O\ì~òC) ADP and Pi are coupled during the e� transport between QB and PQ

QB =°i�Çü PQ � =° �Î¼ [iöQ Z�¢Hê�<£ ~¡"�}ì�Õ ADP =°i�Çü Pi �° H��¬|_È �~òD) Electrons may move from FdR to PQ

Z�¢Hê�#°Á FdR #°O_� PQ ä�½ KÇeOKÇ=KÇ°ó1) C and D correct C =°i�Çü D �¬ï~á#q 2) A and C correct A =°i�Çü C �¬ï~á#q3) A,C and D correct A,C =°i�Çü D �¬ï~á#q 4) B and C correct B =°i�Çü C �¬ï~á#q

34. No. of NADPH+H+ utilised in each turn of Calvin cycle and No .of NADH+ H+ formed in eachturn of Krebs cycle respectively is

HêeÞ<£ =��Ç°=ò XH�øª�i =°i�Çü ¢ïH�� =��Ç°=ü XH�øª�i [iy`Í qx�³¶Q®�¬_Í NADPH+H+ ��¬OY¼=°i�Çò U~¡æ_Í NADH+H+ �¬OY¼�° =~¡°�¬Qê

1) 12 and 10 2) 6 and 6 3) 2 and 5 4) 2 and 3

35. Chose the incorrect statement from the following

D ¢H÷Ok "�x�Õ �¬ï~á# "�¼Y¼ Hêx Í Ë Q®°iëOKÇO_�

1) Glycolysis is not involved in the oxidation of fatty acids

Hù=ôÞ P=¶Á� PH©�H�~¡}�Õ ïQáÁHê��²�¹ [~¡Q® Î°2) Krebs cycle is involved in the oxidation of all kinds of substrates

Jxß~¡Hê� �¬ ¥~�� PH©�H�~¡}�Õ ¢ïH�� =��Ç°O ��ç¾O@°Ok3) Glycolysis is involved only in the oxidation of carbohydrates (Glucose)

Hê~ËÄÌ�áì¢_Í@Á (Q®¶ÁHË*�) PH©�H�~¡}�Õ =¶¢`Ç"Í° ïQáÁHê��²�¹ ��ç¾O@°Ok4) EMP or TCA cycle are involved in the oxidation of glycerol

yÁ[~�� PH©�H�~¡}�Õ EMP �è ¥ TCA =��Ç°O ��ç¾O\ì~ò36. Identity the correct ascending order of the following

D ¢H÷Ok "�x �¬ï~á# P~Ë�¬ì} ¢H�=¶xß Q®°iëOKÇO_�.A. No.of hierarchial levels in the ICTV scheme

ICTV q^�¥#O�Õx =s¾~¡} ª�Ö~ò� �¬OY¼B. No.of polypeptide chains per capsomer of TMV

TMV �Õ ¢�¬u Hê�é�"³°~��Õx ��bÌ�Ì��å_£ Qù�°�¬°� �¬OY¼

C. No.of pins on the base plate of T4 phage T4 �¦�*��Õ P �¥~¡�¦¬�H�O ä�½ LO_Í �²<£� �¬OY¼

D. No.of steps involved in lytic cycle �ÿá\÷H± =��Ç°O�Õ Q®� Î�×� �¬OY¼

E. No.of genomic copies in a virion of HIV HIV qi�Ç°<£ �Õx r<Ëq°H± #H��×Á �¬OY¼

1) B,E,A,D,C 2) D,C,A,E,B 3) B,E,D,A,C 4) E,B,A,D,C

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37. Choose the incorrect match with regard to pBR 322

pBR 322 ä�½ �¬O|Ok�Oz �¬iHêx H��~òH�#° Q®°iëOKÇ°=òA. Bam HI ________ tetR gene Bam HI ________ tetR [#°¼=ôB. PuuII _________ 'ori' sequence PuuII _________ 'ori' ¢H�=°OC. Puu I _________ ampR gene Puu I _________ ampR [#°¼=ôD. Sal I _______ 'rop' gene Sal I _______ 'rop' [#°¼=ô1) A and B 2) B and C 3) B and D 4) A and C

38. Observe the following mutation with respect to sickle cell anemia

�²ïH��Ì�� J<Íq°�Ç¶ä�½ �¬O|Ok�Oz D ãH÷Ok L Çæi=~¡ë<�xß �¬ijeOKÇ°=ò

DNA

mRNA

1 15 ......GAG.......3

1 13 ......CAC.......5

1 15 ......GAG.......3 1 15 ......GUG.......3

Protein O OA B O OO O O O

��È±�íj

ä�Û]ª$±

RNA

1 13 ......CTC.......5

1 15 ......GTG.......3

A & B in circles, respectively are A =°i�Çò B �° =~¡°�¬Qê1) Pro ; Glu ã�Ú ; Q®°Á 2) Pro ; Val ã�Ú ; "�� 3) Glu ; Leu Q®°Á ; �°¼ 4) Glu ; Val Q®°Á ; "��

39. Consider the cross AaBb q AaBb. If trait 'A' exhibits complete dominance and trait 'B' exhibitsco-dominance what phenotypic ratio is expected in the progeny ?

AaBb q AaBb �¬OH�~¡}ìxß �¬ijeOKÇ°=ò 'A' �H�Æ}=ò �¬O�¬î~¡â |�²ì~¡¾ Ç �Þxß =°i�Çò 'B' �H�Æ}O �¬�¬ì|�²ì~¡¾ Ç �ÞxßKÇ¶�²Oz# �³°_È�, Ì�á# À�~ùø#ß �¬OH�~¡}O ¥Þ~� U~¡æ_�# �¬O Çu �³òH�ø Î$�×¼~¡¶�¬ x+¬æuë ZO Ç?1) 1 : 1 : 1 : 1 2) 9 : 3 : 3 : 13) 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1 4) 3 : 6 : 3 : 1 : 2 : 1

40. Messelson and Stahl grew E.coli bacteria in the presence of heavy nitrogen (15NH4Cl) for manygenerations, and then exposed the bacteria to an excess of normal or light nitrogen (14NH4Cl),resulting in the production of generations 1 and 2. Which of the following percentages of light,intermediate (Hybrid) and heavy DNA did Messelson and Stahl see in their experiment.

"³°�¬��¬<£ =°i�Çò ª��° Z.HË�ÿá#° 15NH4Cl (15N J<Ík <³áã\Õ[<£ �³òH�ø ��ì~¡ Sªé\Õ�¹) =¶ã`Ç"Í° #ã`Ç[x�é+¬H�OQê Q®� �Ç¶#H�O�Õ J<ÍH� Ç~��° =~¡Ö#O [i�² P Ç~¡°"� Ç PH�}ì�#° ª� �¥~¡} 14NH4Cl Q®� �Ç¶#H�O �ÕxH÷=¶ió =~¡�#O [~¡�¬¬Qê `Ç~¡O-1 =°i�Çò `Ç~¡O-2 ��Õ U~¡æ_�# `ÍeH�, �¬OH�~¡O =°i�Çò ��ì~¡ DNA � �§`��#°Q®°iëOKÇ°=ò Percentages of DNA DNA �§`��°

Generation Light Intermediate Heavy

Ç~¡O ÍeH� =° �Î¼~¡H�O (�¬OH�~¡O) ��ì~¡1) 1 0% 100% 0%

2 50% 50% 0%

2) 1 50% 0% 50%

2 75% 0% 25%

3) 1 0% 100% 0%

2 33% 33% 33%

4) 1 0% 0% 0%

2 50% 50% 0%

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ZOOLOGY41. Match the following

A) Chordates with open circulatory system 1) Cyclostomes

B) Chordates without heart 2) Gnathostomes

C) Vertebrates without paired appendages 3) Lancelets

D) Vertebrates with three ear ossicles 4) Mammals

5) TunicatesãH÷Ok "�xx [ Ç�¬~¡°KÇ°=òA) q=$`Ç ã�¬�¬~¡} =¼=�¬� H�ey# Hêö~Û@°Á 1) Ì�áHËÁªé�=° r=ô�°B) Q®°O_³ �Õ�²Oz# Hêö~Û@°Á 2) <Í ËªÚ�"Í°\ì r=ô�°C) ÎÞO ÎÞ L��OQê�° �Õ�²Oz# �¬H��õ~¡°Hê�° 3) �ì<£��ÿ\��°D) =ü_È° H�~�â�²�YO_¨�° H�ey# �¬H��õ~¡°Hê�° 4) H©Æ~¡ ¥�°

5) @¶¼xöH\��°A B C D A B C D

1) 5 1 3 2 2) 5 3 1 4

3) 3 5 1 4 4) 5 3 2 4

42. Branch that deals with the study of continuous genetic adaptations of organisms to the envi-ronment is

1) Genetics 2) Ecology 3) Evolution 4) Ethology

�¬i�¬~��ä�½ J#°Q®°}OQê x~¡O Ç~¡O r=ô��Õ H�eöQ [#°¼ J#°ä��<�� ¬~¡"³°Ø# =¶~¡°æ�#° Q®°iOz# J �Î¼�Ç°#O1) [#°¼�§ã�¬ëO 2) r"�=~¡} �§ã�¬ëO 3) �¬i}ì=° �§ã�¬ëO 4) W �¥�r

43. Choose the correct statementsI) In diploblastic animals, the cells in each tissue layer are functionally interdependent.II) In Parazoans, different types of cells are functionally isolatedIII) Cells of metazoans are capable of independent existence1) I, II 2) II, III 3) II only 4) I, III

�¬iJ~ò# "�¼Y¼�#° Q®°iëOKÇ°=òI) kÞ�¬ëi Ç r=ô��Õ ã�¬u�¬ë~¡=ò �Õx H�}ì� =° �Î¼ �¬~¡�¬æ~¡ ãH÷�Ç¶`ÇàH� �¬=°#Þ�Ç°O LO@°OkII) ��~�*Õ=#Á��Õ qq �Î ~¡Hê� H�}ì�° ãH÷�Ç¶ ÇàH�OQê q=H�ë Ç K³Ok LO\ì~òIII) |�¬�H�} r=ô� �³òH�ø H�}ì�° �¬Þ ÇOã ÇOQê =°#°Q®_È ª�yOKÇQ®�=ô.1) I, II 2) II, III 3) II =¶ã`Ç"Í° 4) I, III

44. The intercellular junctions which act as rivets binding the cells together into strong sheets, areanchored in the cytoplasm by

1) Microfilaments made of keratin 2) Microtubules made of tubulin

3) Intermediate filaments made of keratin 4) Microfilaments made of actin

Î$_�È"³°Ø# �¬�H��#° |Ok�OKÍ iq@Á =�ÿ H�} ÇÞK��#° |Ok�OKÍ H�}ìO Ç~¡ �¬O �Î°�#° r= �¬ ¥~¡�O�Õ JuH÷Oz LOKÍq1) ïH~¡\÷<£ Ë U~¡æ_�# �¬¶H�Æà ÇO Ç°=ô�° 2) @¶¼|°¼e<£ Ë U~¡æ_�# �¬¶H�Æà<�oH��°3) ïH~¡\÷<£`Ë U~¡æ_�# =¶^�Î¼q°H� ÇO`Ç°=ô�° 4) UH÷�<£`Ë U~¡æ_�# �¬¶H�Æà`ÇO`Ç°=ô�°

45. Identify the incorrect statement related to Periplaneta1) The number of abdominal longitudinal tracheal trunks is equal to the number of gonapophyses in

female cockroach

2) Number of segments in one maxillary palp is equal to the number of podomeres in one leg

3) Number of retinulae in one ommatidium is equal to the number of abdominal ganglia

4) Number of podomeres in one leg is equal to the number of tarsomeres in one leg

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ãH÷Ok"�x�Õ Ì�i��Á<³\ìä�½ �¬O|Ok�Oz �¬iHêx "�¼Y¼#° Q®°iëOKÇ°=ò1) �çkíOH� L Î~¡=ò�Õx P�Ç° Ç "��Çò<��ì� �¬OY¼,P_È�çkíOH��Õx Qù<��éÌ�¦á�²�¹� �¬OY¼ä�½ �¬=¶#O2) XH� [Oa�Hê �¬æ~�ÅOQ®O�Õx YO_� �� ¬OY¼ XH� Hê�°�Õx �� ÎYO_� �� ¬OY¼ä�½ �¬=¶#O3) XH� <Íã`�O�×O�Õx <Íã Ç�¬@� H�}ì��¬OY¼ L Î~¡ <�_��¬O �Î°� �¬OY¼ä�½ �¬=¶#O4) XH� Hê�°�Õx �� ÎYO_� �� ¬OY¼ XH� Hê�°�Õx \ì~Ë�q°�Ç°~¡Á �¬OY¼ä�½ �¬=¶#O

46. Arrange the following ganglia in cockroah in a correct sequence from anterior to posterior

A) Frontal ganglion B) Hypocerebral ganglion

C) Proventricular ganglion D) Ingluvial ganglion

E) Brain1) B A C D E 2) B A E C D 3) E A B D C 4) A E B D C

�çkíOH��Õ <�_��¬O �Î°�#° �¬î~¡Þ��ìQ®O#°Oz �¬~¡��ìQ®Oä�½ XH� ãH�=°�¬ Îíu�Õ J=°~¡°ó=òA) ��ì\÷Hê <�_��¬Ok� B) J^�Ë=°�²ë+¬ø <�_��¬Ok�C) �¬î~¡ÞãQ®O �Î°� [~�¡iHê <�_��¬Ok� D) WOQ®¶Áq�Ç°�� <�_��¬Ok�E) "³°^Î_È°1) B A C D E 2) B A E C D 3) E A B D C 4) A E B D C

47. If amount of energy ulilised by plants for their respiration is 150 kJ. Then find out the energy ina secondary cornivore(Note: Plants use 25% of GPP for their respiration)1) 4.5 KJ 2) 0.045 3) 45 KJ 4) 0.45KJ

"³òH�ø�° �§¼�¬ãH÷�Ç°�Õ 150 kJ �×H÷ëx qx�³¶yOKÇ°ä�½#ß@Á~ò`Í kÞf�Ç° =¶Oª��¬ð~¡°��Õ ZO`Ç �×H÷ë LO@°Ok(Q®=°xH� : "³òH�ø�° GPP�Õ 25% =¶ã`Ç"Í° �§Þ�¬ãH÷�Ç°ä�½ L�¬�³¶yOKÇ°ä�½<�ß~ò)1) 4.5 KJ 2) 0.045 3) 45 KJ 4) 0.45KJ

48. Identify the correct statements from the followingA) Number of premolars is equal to the number of incisors in adult humansB) Monophyodont teeth in humans are pre molars and last molarsC) Dentine is secreted by ameloblasts of mesodermal origin1) A & C 2) A & B 3) B & C 4) A, B & C

D ãH÷Ok "�x�Õ �¬iJ~ò#q Q®°iëOKÇO_�.A) ã�Ï_�È =¶#=ô_��Õ JãQ®KÇ~¡Þ}Hê� �¬OY¼ ä�½O ÇHê� �¬OY¼ä�½ �¬=¶#OB) =¶#=ô_��Õ JãQ®KÇ~¡Þ}Hê�° =°i�Çò z=i KÇ~¡Þ}Hê�° UH�"�~¡ ÎO`��°C) =° �Î¼�¬ëÞKÇ xià Ç Zq°�Õ�ìÁ�¬°� H�}ì� #°O_� _³O\÷<£ ã�¬qOKÇ|_È° Ç°Ok1) A & C 2) A & B 3) B & C 4) A, B & C

49. Which of the following are unpaired cartilages in the wall of larynx1) Thyroid, cricoid and arytenoid 2) Arytenoid, cricoid and corniculate3) Thyroid, cricoid and epiglottis 4) Arytenoid, cunieform and corniculate

�¬æ~¡À�\÷H��Õ Q®� J ÎÞO ÎÞ =°$ Î°�ì�¬°��°1) ^�³á~�~ò_£, ãH÷Hê~ò_£ =°i�Çò J~¡\÷<�~ò_£ 2) J~¡\÷<�~ò_£, ãH÷Hê~ò_£ =°i�Çò Hêißä�½¼�è\�3) ^�³á~�~ò_£, ãH÷Hê~ò_£ =°i�Çò L�¬l�¬ìÞ 4) J~¡\÷<�~ò_£, ä�½¼x�¦�"£° =°i�Çò Hêißä�½¼�è\�

50. Assertion (A) : Impulse transmission across an electrical synapse is always faster than chemicalsynpase

Reason (R) : In a myelinted axon, the voltage gated Na+ and K+ channels are concentrated atthe nodes of Ranvier

1) Both A and R are correct. R is the correct explanation of A.

2) Both A and R are correct. R is not the correct explanation of A.

3) A is true, but R is false 4) A is false, but R is true

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Assertion (A) : q Î°¼ � <�_�H�}�¬Ok� ¥Þ~� <�_�ã�¬KË Î#, ~¡ª��Ç°# <�_�H�}�¬Ok� H�O>è Zä�½ø= "ÍQ®OQê [~¡°Q®° Ç°Ok.Reason (R) : =°�Ç°b<£ �¬�²ì Ç ÇOãuHêHÆê��Õ "Ë�è�*� öQ>ÿ_£ Na+ =°i�Çò K+ K��#�×Ã¤ ~�xÞ�Ç°~� H�}°�¬ô� = Îí

öHOãnH�$ Ç"³°Ø LO\ì~ò.1) A =°i�Çò R �¬ï~á#q, R J#°#k A ä�½ �¬ï~á# q=~¡}2) A =°i�Çò R �¬ï~á#q, R J#°#k A ä�½ �¬ï~á# q=~¡} Hê Î°3) A X�¬C, Hêx R `Ç�¬C 4) A `Ç�¬C,Hêx R X�¬C

51. Match the followingSet - I Set - IIA) Mononuclear phagocytes 1) NeutrophilsB) Antigen presenting cell 2) Synovial cellsC) Polymorpho nuclear phagocytes 3) PlateletsD) Inflammatory mediators 4) Dendritic cell1)A - 2,B - 4, C - 1, D - 3 2) A - 1, B - 4, C - 3,D - 23) A - 2,B - 4, C - 3, D - 1 4) A - 4,B - 2, C - 1, D - 3ãH÷Ok "�xx [ Ç�¬~¡°KÇ°=ò�¬\÷�H�- I �¬\÷�H�- IIA) UH�öHOã^ÎH� �¦�QËÌ�á\��° 1) #¶¼ã\Õ�¦²��°B) ã�¬u[#H� �¬=°~¡æH� H�}ì�° 2) Ì�á<Ëq�Ç°�� H�}ì�°C) |�¬�~¡¶�¬ öHOã^ÎH� �¦�QËÌ�á\��° 3) ~¡H�ë�¦¬�H÷H��°D) L[Þ�# =° �Î¼=ië ÇÞ �¬ ¥~��° 4) _³Oã_³á\÷H± H�}O1)A - 2,B - 4, C - 1, D - 3 2) A - 1, B - 4, C - 3,D - 23) A - 2,B - 4, C - 3, D - 1 4) A - 4,B - 2, C - 1, D - 3

52. Match the following and choose the correctList�I List�II

A) Ingression I) Urinogenital organsB) Involution II) EndodermC) Delamination III) MesodermD) Mesomere IV) HypoblastE) Myotome V) Voluntary muscles

A B C D E A B C D E1) III II IV I V 2) II III IV I V3) II III I IV V 4) II I III IV VD ãH÷Ok "�xx [ Ç�¬~¡°KÇ°=ò�¬\÷�H �-I �¬\÷�H �-II

A) ã�¬"Í�×O I) =üã Ç[#<ÍOãk�Ç° J=�Ç°"��°B) JO Ç~¡Þ�#O II) JO Ç�¬ëÞKÇOC) _��ìq°<Í+¬<£ III) =° �Î¼�¬ëÞKÇOD) g°ªéq°�Ç°~� IV) Ì�áì�é�ìÁ�¹�E) =°�³¶\Õ"£° V) x�Ç°Oãu Ç H�O_È~��°

A B C D E A B C D E1) III II IV I V 2) II III IV I V3) II III I IV V 4) II I III IV V

53. In a population which is in Hardy�Weinberg equilibrium, Drosophila flies with ressive blackcoloured are 36. The allelic frequency of black colour is 0.3. Then find out dominant grey colouredflies in that population1) 400 2) 364 3) 800 4) 264

�¬ìsÛ - "³~ò<£|~�¾ �¬=°`��²Öu�Õ #°#ß XH� [<��ì�Õ JO Ç~¡¾ Ç #�°�¬ô Í�¬ì =~¡âO�Õ #°#ß ã_Ëªé�¦²�ì� �¬OY¼.36. #�°�¬ô =~¡âJO Ç~¡¾ Ç �ÇòQ®àqH��æ �Ï#��¬ô#¼=ò 0.3. J~ò Í |�²ì~¡¾ Ç |¶_� Î =~¡âO�Õ LO_Í ã_Ëªé�¦²�ì� �¬OY¼ ZO Ç ?1) 400 2) 364 3) 800 4) 264

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54. This clinical inferences from ECG indicate the myocardial infarction (MI)

1) Depressed S-T segment & Prolonged Q-T interval

2) Elevated S-T segment & Prolonged Q-T interval

3) Prolonged P-R interval & Shortened Q-T interval

4) Tall T�wave & Prolonged Q-T interval

ECG H÷ÁxH�� J#Þ�Ç¶��Õ =°�³¶HêiÛ�Ç°�� W<£�¦�~�H�Æ<£ �¬¶zOKÍk1) S-T YO_�`ÇO J}KÇ|_�#@Á~ò`Í Ê Q-T JO Ç~¡O Zä�½ø= À��¬ô L#ß@Á~ò Í2) S-T YO_� ÇO Ì�iy#@Á~ò Í Ê Q-T JO Ç~¡O Zä�½ø= À��¬ô L#ß@Á~ò Í3) P-R JO Ç~¡O Hê�ì=k� Ì�iy#@Á~ò Í Ê Q-T JO Ç~¡O Zä�½ø= L#ß@Á~ò Í4) Z`³á# T�`Ç~¡OQ®O & Q-T JO Ç~¡O Zä�½ø= L#ß@Á~ò Í

55. The non - feeding and non pathogenic stage of Entamoeba histolytica in lumen of gut, in whichchromatoid bars first appear=¶#=ôx�Õ x=�²OKÍ P�¬ð~¡ À�H�~¡} [~¡�¬x, "�¼k� H�eyOKÇx, ãHË=¶\ì~ò_£ Í�¬ð�° "³ò Î@Qê U~¡æ_Í ZO@g°�ì�²ìªé��ÿá\÷Hê Î�×1) trophozoite �é+¬H� Î�× 2) precystic stage �¬î~¡Þ HË�×�¬Ö Î�×3) tetranucleate cyst KÇ Ç°+¬ø öHOã ÎH� Î�× 4) metacystic stage "³°\ì�²�¹�H± Î�×

56. Observe the statements and select correct ones ãH÷Ok "�¼Y¼�#° �¬ijeOz �¬i�³Ø°# ^¥xx ZO�²H� KÍ�Çò=òI) Two equal copulatory spicules present in male Wuchereria

�¬ô~¡°+¬ LKÇö~i�Ç¶�Õ ï~O_È° �¬=¶# �¬O�¬~¡ø H�O@Hê�° LO\ì~òII) Cocaine interferes with transport of dopamine and causes euphoria

HùïHá<£, _Ë�¬"³°Ø<£ J<Í <�_� Ja�"��¬ìH�O ~¡"�}ì�Õ *ÕH�¼O KÍ�¬°Hùx L�ìÁ�¬�²Öux H�ey�¬°ëOkIII) The lips, fingers and nails may turn grey to bluish in severe cases of pneumonia

#¶¼"³¶x�Ç¶ "�¼k� ãQ®�¬°ë��Õ Ì� Î=ô�°, "Í�×Ã¤ =°i�Çò QË�×Ã¤ |¶_� Î ~¡OQ®° #°O_� he =~¡âO�ÕH÷ =¶~¡°`�~òIV) 2nd and 3rd moults of Ascaris occur in alveoli of lungs of man.

Pª�øi�¹ 2, 3 x~ËàKÇ<��° =¶#=ôx T�²iu Ç°ë� "��ÇòHË�§��Õ [~¡°Q®°#°1) I, III and IV 2) I, II and IV 3) II, III and IV 4) I and IV only

57. The character which is not related to chondrichthyes 'HêOã_�H±k��¹Ñ ä�½ �¬O|Ok�Oz �¬iHêxk1) presence of claspers �¬O�¬~¡ø ÎO_¨�° LO_È°@ 2) placoid scales ��ÁHê~ò_£ �Ú�°�¬°�°3) filamentous gills ÇO Ç°~¡¶�¬ "³ò�¬æ�° 4) heterocercal caudal fin q+¬=° ��e�¬ôKÇó�"�[O

58. Identify the correct combinations �¬i�³Ø°#*Õ_� Q®°iëOKÇ°=òA) Pelecypoda � Glochidium larva Ì�e�²�Ú_È � QËÁv_��Ç°O �ì~�ÞB) Monoplacophora � Neometra "³¶<Ë��ÁHË�é~¡ � x�³¶"³°ã\ìC) Hemichordata � Dorsal heart Ì�ìg°Hêö~Û@ � �¬$+¬� �¬ì$ Î�Ç°OD) Echinoidea � Aristotle's lantern ZH÷<�~ò_��Ç° � Jiª��\÷�� n�¬O (�ìO Ç~¡°)1) A, C and D 2) A, B and C 3) A and B 4) B, C and D

59. Identify mismatch from the following

ãH÷Ok "�x�Õ �¬iHêx �¬O|O �¥xß Q®°iëOKÇO_�1) Khasi and Jaintia hills - Meghalaya Mìt =°i�Çò *ÿá~òO\÷�Ç¶ HùO_È�°-"Í°�¦¬¶��Ç°2) Sarguja, Bastar - Madhya pradesh �¬~¡°¾[, |�¬ë~�-=° �Î¼ã�¬ Í��3) Siberian crane - Grus leucogeranus Ì�áci�Ç° HùOQ®-ãQ®�¹ �¶¼HËÿ~�#�¹4) Wild life protection act - 197 =#¼ã��}÷ �¬O~¡H�Æ} KÇ@�O-1972

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60. Choose correct combination �¬i�³Ø°# H��~òH�#° ZO�²H� KÍ�Çò=òTissue character exampleH�}ì�O �H�Æ}O L^¥�¬ì~¡}

I) Stratified squamous covers dry skin surface pharynx, oesophagus

keratinised epithelium

ïH~¡\÷<£ �¬�²ì Ç�¬ëi Ç �×�ø� �Ú_� KÇ~¡à L�¬i Ç�O ãQ®�¬x, P�¬ì~¡ "��²ìH�L�¬H��× H��²æ LOKÇ°@

II) White adipose tissue metabolically not active common in adults

³�°�¬ô Hù=ôÞ H�}ì�O r=# ãH÷�Ç°��Õ ãH÷�Ç¶j�OQê Ì� Îí "�i�Õ ª� �¥~¡}O LO_È Î°III) Fibrous cartilage perichondrium absent intervertebral discs

ÇO Ç°�Çò Ç =°$ Ç°�ì�²Ö �¬i=°$ Î°�ì�²Ö �Õ�²OKÇ°@ JO Ç~¡ H��õ~¡°H� KÇãH÷H��°1) I, II and III I, II =°i�Çò III 2) II and III II =°i�Çò III

3) I and II I =°i�Çò II 4) I and III I =°i�Çò III

61. Gustatory receptors of cockroach are located on �çkíOH��Õ ~¡°z ãQê�¬ìHê�° LO_Ík1) Antennae, labrum �¬æ~¡Å�×$OQê�°, F+¬�O2) Labial palps, maxillary palps and antennae J �Î~¡�¬æ~�ÅOQê�° , [Oa�Hê�¬æ~�ÅOQê�° =°i�Çò �¬æ~¡Å�×$OQê�°3) Labrum, maxillary palps, labial palps F+¬�O, [Oa�Hê�¬æ~�ÅOQê�° =°i�Çò J �Î~¡�¬æ~�ÅOQê�°4) Maxillary palps, labial palps, mandibles [Oa�Hê�¬æ~�ÅOQê�°, J �Î~¡�¬æ~�ÅOQê�°, �¬ì#°=ô�°

62. Which of the following are green house gases

ãH÷Ok "�x�Õ �¬ìi Ç Q®$�¬ì "��Çò=ô�°1) SO2 , Ozone SO2 , F*Õ<£ 2) Methane, O2 g° �Í<£, O2

3) CO2, Methane CO2, g° �Í<£ 4) Ozone, NO2 F*Õ<£, NO2

63. Identify the set of hormones that are not antagonistic in function

q~¡° Î� �¬Þ��ì=O H�ey�ÇòO_Èx �¬ð~Ëà<£�#° Q®°iëOKÇO_�1) Melanocyte stimulating hormone (MSH) - Melatonin

"³°�<ËÌ�á\� Ln��¬H� �¬ð~Ëà<£ (MSH) - "³°�\Õx<£2) Calcitonin - Parathormone Hêe�\Õx<£-��~� �¥~Ëà<£3) Adrenalin - Noradrenalin Jã_�#e<£-<�~�Zã_�#e<£4) Insulin - Glucagon W<£�e<£-Q®¶ÁH�Qê<£

64. The cells that have CD8 markers on the cell membrane are

H�} L�¬i Ç�OÌ�á CD8 =¶~¡ø~��#° H�ey�ÇòO_Íq1) B - Lymphocytes B -eO�¦éÌ�á@°�°2) Natural Killer cells �¬�¬ì[ �¬ìO ÇH� H�}ì�°3) T cytotoxic cells T Ì�á\Õ\ìH÷�H± H�}ì�° (T H�}ì�°)4) T Helper cells T �¬�¬ð�Ç° H�}ì�° (TH H�}ì�°)

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65. Match the following �¬iQê [ Ç�¬~¡°KÇ°=òList - I List - II�¬\÷�H� - I �¬\÷�H� - IIA) Founder effect I) Long necked giraffes

ª�Ö�¬H� r=ô� ã�¬��ì=O �Ú_È=ô "³°_Èl~��¦ÔB) Bottleneck effect II) Heterozygous for sickle cell anaemia

J"�O Ç~¡ ã�¬��ì=O Hù_È=e H�} ~¡H�ë �Ôì# Çä�½ q+¬=°�ÇòQ®à[OC) Genetic load III) O+ve Blood group in Red Indian population

[#°¼��ì~¡O ï~_£ WO_��Ç°<£ =¶#= [<��ì�Õ O+ve ~¡H�ë �¬=ò^¥�Ç°OD) Centrifugal selection IV) Polydactylic dwarf individuals in old order Amish population

J�¬öHOã Î =~¡}O L Çë~�k Q®[�Ô��

V) Biston betularia

a�¬�<£ �ÿ@°¼�èi�Ç¶A B C D A B C D

1) III IV II I 2) IV II V III3) II I III IV 4) III IV II V

66. Euryhaline organisms �ÇüsÌ�ì�ÿá<£ r=ô�°1) all fishes Jhß KÍ�¬�°2) migratory fishes ã�¬"��¬ KÍ�¬�°3) fresh water animals =°Ozh\÷ [O Ç°=ô�°4) marine fishes �¬=òã Î KÍ�¬�°

67. The blood group of a woman is 'O'. she has two brothers; one with 'A' group and the other with

'B' group. The genotypes of her parents are:

'O' ~¡H�ë =~¡¾O H�ey# XH� ¢�ÔëH÷ H�� W Îí~¡° ªé Î~¡°��Õ XH�~¡° 'A' =~¡¾=ò =°i�³òH�~¡° 'B' =~¡¾=ò H�ey �Çò#ß@Á~ò Í,P"³° ÇeÁ ÎOã_È°� ~¡H�ë =~¡¾ [#°¼ ~¡¶��°1) IO IO -IA IA 2) IA IO - IA IB 3) IA IO - IB IO 4) IA IB - IO IO

68. Nearly all essential nutrients and 70 - 80% of electrolytes and water are reabsorbed in

J=�¬~¡"³°Ø# �é+¬Hê�°, �¬°=¶~¡° 70-80% q Î°¼ � q�õÁ+¬¼Hê�° =°i�Çò h~¡° D ��ìQ®=ò�Õ �¬ô#��Õ+¬}O QêqOKÇ_È|`�~ò1) DCT Î¶~¡�¬Ö �¬O=o Ç <�oH� 2) PCT �¬g°�¬ �¬O=o Ç <�oH�3) Collecting duct �¬OãQ®�¬ì} <��×O 4) Henle's loop Ì�ìhÁtH�àO

69. Match the contraceptive methods given under column - I with the column - II

�¬\÷�H�- I =°i�Çò �¬\÷�H�- II ��Õ W=Þ|_�# Q®~¡Ä�x~Ë �ÎH� �¬ Îí Ç°�#° [ Ç�¬~¡°KÇ°=òColumn - I Column - II

�¬\÷�H�- I �¬\÷�H�- II

A) Saheli �¬À�ìe I) tubectomy & vasectomy @¶¼�ÿH��q° Ê "Í�¬H��q°B) Barrier J=~Ë �ÎO II) once a week non steroid medicine

"�~�xH÷ XH�ª�i <�<£�Ô�~�~ò_£ =°O Î°�°C) IUD's Q®~�Ä��×�Ç¶O Ç~¡ ª� �Î<��° III) prevent meeting of sperms with ovum

JO_ÈO`Ë �×ÃãH�H�}ì� H��~òH�#° x~Ëk�OKÇ°@D) Sterilisation =O �Î¼H�~¡}O IV) make uterus unsuitable for implantation

Q®~�Ä��×�Ç°O �²O_È ã�¬uª�Ö�¬#ä�½ J#°=ôQê �èä�½O_¨ KÍ�Çò@V) removal of ovary ã�Ôë c[ HË�§�° ù�yO�¬ô

1) A - II, B - IV, C - III, D - V 2) A - II, B - III, C - IV, D - I3) A - II, B - III, C - IV, D - V 4) A - I, B - II, C - III, D - IV

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70. Observe the following characters D ãH÷Ok �H�Æ}ì�#° �¬ijeOKÇO_�a) very small or poorly developed eyes JO ÇQê Ja�=$k� K³O Îx �èH� z#ß <Íã`��°b) cleft palate `��°=ô peH�c) brain & spinal cord abnormality "³° Î_È° =°i�Çò "³#°ß��=ò�Õ Jª� �¥~¡}Od) trisomy of autosomes ³á�²ìH� ãHË"³¶*Õ"£° ã>ÿØªéq°H±These characters are associated with the disorder

D Ì�á �H�Æ}ì�° U J�¬�²ÖuH÷ �¬O|Ok�Oz#q1) Cat - cry syndrome �²eÁ J~¡°�¬ô �²Oã_Ë"£° 2) Edwards syndrome Z_ÈÞ~�Û �²Oã_Ë"£°3) Down's syndrome _ñ<£ �²Oã_Ë"£° 4) Patau's syndrome �¬\º �²Oã_Ë"£°

71. Match the following D ãH÷Ok "�\÷x [ Ç�¬~¡KÇ°=òA) Hyline, poona pearls Ì�áì�ÿá<£, �¬î<�Ì�~¡�� I) layers �è�Ç°~��°B) Sardines, salmon ª�~�_³á<£�°, ª��à<£ II) reduce cholesterol Hù�ÿãª��#° Çy¾OKÇ°@C) ADA deficiency ADA �Õ�¬O III) SCID fã= �¬q°ào Ç "�¼k� x~Ë �ÎH� �Õ�¬OD) H5N1 Virus IV) Bird flu |~�Û �¦¬îÁ H5N1 "³á~¡�¹ V) broilers ã�ì~ò�~��°

A B C D1) I III IV V2) I II III IV3) V II IV I4) V III II I

72. Based on :

# ratio, arrange the following Drosophila flies in ascending order

ã_¨ªé�¦²�ì�Õ :

#x+¬æuë q�°= P �¥~¡OQê ãH÷Ok "�\÷x P~Ë�¬ì} ãH�=°O�Õ J=°~¡°ó=ò

A) Normal male ª�^�¥~¡} �¬ô~¡°+¬rq B) Meta male Jk��¬ô~¡°+¬ rqC) Meta female Jk�ã�Ôë rq D) Inter sex �¬=°eOQ®rqE) Normal female ª�^�¥~¡} ã�Ôë rq1) A lB lClDl E 2) BlAlDl ElC3) ElDlClBlA 4) AlDlBlCl E

73. Papillae present on the posterior surface or base of tongue are

<��°H� P �¥~¡��ìQ®O �è ¥ �¬~�O Ç L�¬i Ç�OÌ�á LO_Í �¬¶HÆêàOä�½~��°1) filiform ÇO Ç°~¡¶�¬ �¬¶HÆêàOä�½~��° 2) circum vallate �¬iø"£° "��è\� �¬¶HÆêàOä�½~��°3) fungiform �¦¬Ow �¦�"£° �¬¶HÆêàOä�½~��°4) foliate �¦ée�³¶\� �¬¶HÆêàOä�½~��°

74. Atria and ventricles are seperated by a deep transverse groove known as

H�iâH��#° [~�¡iH��° #°O_� "Í~¡°KÍÀ� �Õ ³á# J_È°ÛQê_�x U=°O Î°~¡°1) auricular appendix Piä�½¼�ì~� Z�²O_�H±� 2) coronary sulcus H�~Ë#i �¬�ø�¹3) interatrial septum H�iâHêO`Ç~¡ q��ì[H�O 4) foramen ovale �¦Ú~¡q°<£ F"³��

75. Reserpine, a drug extracted from Rauwolfia vomitoria, is used in the treatment of

1) Hypersensitivity 2) Hyperglycemia 3) Hypertension 4) Hyperacidity

~�=ôeæ��Ç° "Ëq°\Õi�Ç¶ "³òH�ø #°O_� �¬OãQ®�²ìOKÍ ï~�¬iæ<£ =°O Î°#° nx zH÷ Ç�ä�½ L�¬�³¶yª�ë~¡°.

1) Ju�¬°xß Ç ÇÞO 2) Ì�áì�¬~�ïQåÁ�²q°�Ç¶ 3) Ì�áì�¬~�>ÿ#Â<£ 4) Ì�áì�¬~� Z�²_�\÷

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76. Which of the following animal has a chitinous exoskeleton and respires through tracheae but

lacks antennae?

1) Periplaneta 2) Palamnaeus 3) Limulus 4) Aranea

ïHá\÷<£ xià Ç �ì�¬ì¼ J�²Ö�¬O[~¡O H�ey LO_�, "��Çò<��× �§Þ�¬ãH÷�Ç° [iÀ�, �¬æ~¡Å �×$OQê�° �Õ�²Oz# [O Ç°=ô

1) Ì�i��Á<³\ì 2) À��ìq°ß�Ç°�¹ 3) e=ò¼��¹ 4) Z~�x�Ç¶77. Which of the following ganglia is not related to the �cranial outflow� of the autonomic nervous

system?

1) Ciliary ganglion 2) Otic ganglion

3) Pterygopalatine ganglion 4) Coeliac ganglion

H÷Ok "�\÷�Õ �¬Þ�Ç°OKËk Ç <�_� =¼=�¬� �³òH�ø �H�� ì�¬ì¼ ã�¬"��¬ìO� ä�½ �¬O|O �ÎO �èxk1) �ÜáeHê <�_��¬Ok� 2) ã�×=} <�_��¬Ok�3) ã Ç�Ç°OyH� ��=¼ <�_��¬Ok� 4) L Î~¡ <�_� �¬Ok�

78. Perilymph occurs in

1) Scala vestibuli and scala media 2) Scala vestibuli and scala tympani

3) Scala tympani and scala media 4) Scala tympani and Eustachian tube

�¬i��²Hê ã Î=O g\÷�Õ LO@°Ok.1) ª�ø�ì "³�²�|°¼e =°i�Çò ª�ø�ì g°_��Ç¶ 2) ª�ø�ì "³�²�|°¼e =°i�Çò ª�ø�ì \÷O��x3) ª�ø�ì \÷O��x =°i�Çò ª�ø�ì g°_��Ç¶ 4) ª�ø�ì \÷O��x =°i�Çò �ÇüÀ��+²�Ç°<£ <��×O

79. Identify the correct chronological sequence of the periods in Palaeozoic Era.

1) Cambrian g Silurian g Ordovician g Devonian g Carboniferous g Permian

2) Cambrian g Ordovician g Devonian g Silurian g Carboniferous g Permian

3) Cambrian g Ordovician g Silurian g Devonian g Carboniferous g Permian

4) Cambrian g Ordovician g Silurian g Carboniferous g Devonian g Permian

Hê�ãH�=¶#°ª�~¡OQê À�e�³¶ì~òH± =°�¬ð�ÇòQ®O�Õx �ÇòQê� �¬i�³Ø°# =~¡°�¬ãH�=°O

1) öHOãa�Ç°<£ g Ì�á�¶i�Ç°<£ g P~ËÛg�²�Ç°<£ g _�"Ëx�Ç°<£ g Hê~ËÄxÌ�¦~¡�¹ g �¬ià�Ç°<£

2) öHOãa�Ç°<£ g P~Ëíq�²�Ç°<£ g _�"Ëx�Ç°<£ g Ì�á�¶i�Ç°<£ g Hê~ËÄxÌ�¦~¡�¹ g �¬ià�Ç°<£

3) öHOãa�Ç°<£ g P~Ëíq�²�Ç°<£ g Ì�á�¶i�Ç°<£ g _�"Ëx�Ç°<£ g Hê~ËÄxÌ�¦~¡�¹ g �¬ià�Ç°<£

4) öHOãa�Ç°<£ g P~Ëíq�²�Ç°<£ g Ì�á�¶i�Ç°<£ g Hê~ËÄxÌ�¦~¡�¹ g _�"Ëx�Ç°<£ g �¬ià�Ç°<£

80. Tumors are characterised by

1) Contact inhibition 2) Angiogenesis

3) Anchorage dependence 4) Apoptosis

H�}÷ Ç°�° ã�¬ ÎiÅOKÍ �H�Æ}O1) �¬æ~¡Å x~Ë^�ÎO 2) UOl�³¶*ÿx�²�¹3) �OQ®~¡° P^�¥~¡O 4) ã�¬}ìoHê|^Îí H�}=°~¡}O

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Page.No : 19

PHYSICS

81. The coefficient of viscosity � �#NQI $Z E �

$K � where 'x' is distance. The dimensional formula of �

#%

$is

�²ßQ®Ö � Q®°}H�O � �#NQI $Z E �

$K � WH�ø_È 'x'J#°#k Î¶~¡=ò J~ò#

�

#%

$q°u�¦�~¡°à�ì

1) � �/. 6

� � 2) �/6

� 3) �/.6

� 4) �/6

�

82. Velocity and acceleration of a particle at some instant of time are � � �8 �K L �M OU� � �G

and

� � �C K � L M OU

� � �G

, then the component of acceleration along the direction of velocity is ......

XHê<ùH � � ¬=°�Ç°O�Õ XH � H �}=ò �³òH �ø " ÍQ ®=ò =°i�Çò ` ÇÞ~ ¡}O�° =~ ¡ °� ¬Qê

� � �8 �K L �M OU

� � �G

=°i�Çò � � �C K � L M OU

� � �G

, J~ò# "ÍQ®k�×�Õx ÇÞ~¡} JO�×O.

1) Increasing at the rate of 2ms-1 2) Decreasing at the rate of 2ms-1

2ms-1Kù�¬ôæ# Ì�~¡°Q®°#° 2ms-1 Kù�¬ôæ# ÇQ®°¾#°3) Increasing at the rate of 4ms-1 4) Decreasing at the rate of 4 ms-1

4ms-1 Kù�¬ôæ# Ì�~¡°Q®°#° 4 ms-1Kù�¬ôæ# `ÇQ®°¾#°83. A body is projected vertically up from the top of the tower. While falling down velocities at a

height 'h' above the top of the tower and at a depth 'h' below the top of the tower are in the ratio

1:2 Then the velocity of projection is

XH� QË�¬ô~¡O tY~¡O #°O_� XH� =�¬°ë=ô#° x@�x�°=ôQê Ì�áH÷ q�²i<�~¡°. P =�¬°ë=ô ãH÷OkH÷ �¬_È°#�¬ô_È° QË�¬ô~¡ tY~¡O#°O_� 'h'Z Ç°ë�Õ =°i�Çò 'h'kQ®°=# "ÍQ®=ò� x+¬æuë 1:2 J~ò# P =�¬°ë=ô#° q�²i#"ÍQ®O

1) �IJ

�2)

IJ

�3)

��IJ

�4)

�IJ

�

84. A body is projected froma point, on the surface of a planet. The horizontal and vertical displace-

ment 'x' and 'y' (in metre) respectively vary with time't' (in second) as Z �� V and �[ ��V V � .

The maximum height attained by the body is

XH� ãQ®�¬ìO �³òH�ø L�¬i Ç�OÌ�á XH� aO Î°=ô #°O_� XH� =�¬°ë=ô ã�¬öHÆ�²OKÇ|_�#k. H÷Æu[�¬=¶O Ç~¡ =°i�Çò x@�x�°=ôª�Ö# ã��O�×=ò�° =~¡°�¬Qê 'x'=°i�Çò 'y' �° 't' �¬=°�Ç°O`Ë��@° Z �� V =°i�Çò �

[ ��V V � . Qê =¶~¡°#°.

J~ò# =�¬°ë=ô KÍ~¡° Q®i+¬� Z`Ç°ë ZO`Ç?1) 200m 2) 100m 3) 5m 4) 25m

85. A bowler throws a ball horizontally along east direction with a speed of 144km/hr. The batsmanhits the ball such that it deviates from the initial direction of motion by 740 north of east directionwithout changing its speed. If the mass of ball is 1/3kg and time of contact between bat and ball is0.02 seconds, average force applied by batsman on ball is

XH� �º�~� `Ç¶~¡°æk�×Qê 144 H÷.g°/Q®O@ "ÍQ®O`Ë q�²i# |Oux, �ì¼\��=°<£ Hù\÷��#�¬ô_È°, |Ou J^Í=_�`Ë `Ç¶~¡°æ#°O_� 740 HË}O`Ë L Çë~�xH÷ uiyOk. |Ou ã Î=¼~�t 1/3 öHl =°i�Çò Ja��¦¬¶ Ç Hê�=ò 0.02Ì�. J~ò Í |OuÌ�á�¬xKÍ�²# �¬~��¬i |�=ò1) 800N 530 east of North 2) 800N 530 North of east

3) 800N 530 North of west 4) 800N 530 West of North

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Page.No : 20

86. 'F' is the force required to just prevent a body sliding down an incline of inclination 450. If U

��P theminimum force required to move the body up the incline is

"��°HË}O 450, U

��P Q®� XH� "��°`Ç�OÌ�á L#ß =�¬°ë=ô ãH÷O^Îä�½ *ì~¡ä�½O_È°#@°Á KÍ�Ç°=�ÿ##ß J=�¬~¡=°Q®°H�h�¬|�O 'F'. J Í =�¬°ë=ô#° "��° Ç�OÌ�áH÷ �ìQ®°@ä�½ J=�¬~¡=°Q®° H�h�¬|�O1) 4F 2) 2F 3) F 4) 8F

87. A massless plat form is placed on a vertical light elastic spring A particle of mass 0.1kg is droppedon the plat form form a height of 0.24m above the platform and on striking the particle the springcompressess by 0.01m. Then the height from which another particle of mass 0.2kg is dropped tocause a compression of 0.04m is:

ÍeH�~ò# �²Öuª�Ö�¬H� x\ì~¡° ã�²æOQ®° Ì�á# �ÇòOz# ã Î=¼~�t ~¡�²ì Ç�¬�Ü¤O�ÕxH÷ 0.1öHl ã Î=¼~�tQ®� H�}O#°0.24g°Z Ç°ë#°O_� *ì~¡q_�z#, ã�²æOQ®° 0.01g° �¬OHËzOKÇ°#°. W Íã�²æOQ®° 0.04g° �¬OHËzOKÇ=��Çò##ß 0.2öHl ã Î=¼~�tQ®�H�}O#° nxÌ�áH÷ ZO Ç Z Ç°ë#°O_� *ì~¡q_È°==��Çò#°?1) 3.96m 2) 2.56m 3) 2.16m 4) 1.96m

88. A particle is dropped on to a hard floor from aheight of 4.9m. At every bounce its kinetic energyis reduced by 19%. The time for which the ball is in motion is

4.9m Z`Ç°ë #°O_� XH� H�}=ò <Í� Ì�áH÷ *ì~¡q_�z<�~¡°. ã�¬u Ja��¦¬¶`Ç=ò#ä�½ ¥x Q®u[�×H÷ë 19% ÇQ®°¾#°. J~ò#|Ou ZO ÇHê�O KÇ�#O�Õ LO@°Ok?1) 1s 2) 18s 3) 19s 4) 2s

89. Two bodies of masses 4kg and 2kg are moving towards each other due to mutual gravitationalforce. If their velocities at an instant are 10ms-1 and 5ms-1 , velocity of centre of mass when theymeet is ... ms-1

�¬~¡�¬æ~¡ Q®°~¡°`�ÞH�~¡Â}|�O =�# 4kg , 2kg ã^Î=¼~��×Ã� Q®� ï~O_È° =�¬°ë=ô�° XH�^¥xHùH�\÷ �¬g°�²�¬°ë<�ß~ò. XH��¬=°�Ç°O�Õ "�\÷"ÍQê�° 10ms-1 =°i�Çò 5ms-1 J~ò`Í, "�\÷ ã^Î=¼~�t öHOã^ÎO "ÍQ®O...g°/Ì�1) zero 2) 5 3) 8.2 4) 10

90. ( K � L �M�T �K � L ��M � � � �G GG G G

G and for what value of 'a' angular momentum is conserved?

( K � L �M�T �K � L ��M � � � �G GG G G

G J~ò Í 'a'�³òH�ø U q�°=ä�½ HË}©�Ç° ã Î=¼"ÍQ®O x Ç¼ ÇÞO J=ô Ç°Ok?

1) -1 2) 0 3) 1 4) 2

91. A particle undergoes simple hormoinic motion having time-period T. The time taken for 3/8thoscillation is

XH� H�}O T _Ë�<�=~¡ë#Hê�O Ë �¬~¡�×�¬ì~� ÇàH� KÇ�#O�Õ L#ßk. 3/8= =O Ç° _Ë�<�xß �¬îiëKÍ�Çò@ä�½ �¬@°�Hê�"³°O Ç?

1) �

6�

§ ·¨ ¸© ¹

2) �

6�

§ ·¨ ¸© ¹

3) �

6��

§ ·¨ ¸© ¹

4) �

6��

§ ·¨ ¸© ¹

92. The kinetic energy given to a body is K, so that it moves from the surface of the earth to infinity.If only 20% of this kinetic energy is given to the same body on the surface of the earth, it rises toa hieght nR, where R is radius of te earth, then n =

��¶q°#°O_� XH� =�¬°ë=ô q�×ÞO Ç~��×O�ÕH÷ "³�×Ã¤@ä�½ Hê=��²# Q®u[�×H÷ë KJ~ò Í P Q®u[�×H÷ë�Õ 20% �§ ÇO =¶ã Ç"Í°P =�¬°ë=ôä�½ Wzó#�¬ô_È° Jk KÍ~¡° Z`Ç°ë nRWH�ø_È RJ#°#k ��¶q° "�¼ª�~¡ÖO J~ò`Í n =

1) 1/4 2) 1/2 3) 3/4 4) 1/6

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Page.No : 21

93. A wire AB is suspended about the end A and marks C and D are present at a distance 40cm aparton the wire. When load is suspended from B, then the marks C and D are displaced by 0.3mm and0.5mm respectively. The original length of AC is (Assume mass of the wire is neglisible )

A Hù# #°O_� ã"Í�ì_Èn�Çò|_�# fQ® AB Ì�á 40cm ^Î¶~¡O�Õ C , D� =^Îí ï~O_È° Q®°~¡°ë�°#ßq. B Hù##°O_� XH��ì~�xß ã"Í�ì_Èn�Çò@KÍ C=°i�Çò D �° 0.3mm ,0.5mm ª�Ö#ã|O�×O<ùOk Í AC�³òH�ø ùe�Ú_È=ô......1) 40cm 2) 20cm 3) 60cm 4) 80cm

94. A metallic wire of density 'd' floats on water. The maximum diameter of the wire so that it may

not sink and maximum length that can float at that diameter are respectively.

'd'ª�Oã Î Ç Q®� XH� �Õ�¬ì�¬ô fQ® h\÷Ì�á Í�°KÇ°#ßk. Jk h\÷�Õ =ò#Q®ä�½O_¨ LO_È=�ÿ#O>è ¥xH÷ Hê=��²# Q®i+¬�"�¼�¬O =°i�Çò P Q®i+¬� "�¼�¬�¬O = Îí ¥xH÷ LO_È=��²# Q®i+¬��Ú_È=ô� q�°=

1) �6 �6

�FI FIS S

2) �6

FIS,any length U�Ú_È=ôïHá##°

3) any diamter U "�¼�¬OïHá##°�6

FIS4)

�6

FISany lengthU �Ú_È=ôïHá##°

95. List-I �¬\÷�H� List-II�¬\÷�H�

a) Equation of continuity ª�`ÇO`Ç¼ �¬g°H�~¡}O e) �

RT

�NX

SK

b) Bernoullis equation �è~Òße�¹ �¬g°H�~¡}O f) ( � XT SK

c) Poiseulle's equation ��~òÌ�eÁ �¬g°H�~¡}O g) � � � �

# 8 # 8

d) Stoke's formula ãªé�H± �¬¶ã ÇO h) ��2 X IJ

�� U � U constant(�²Ö~�OH�=ò)

1) a - g, b - h, c - f , d - e 2) a - g, b - h, c - e, d - f

3) a - h, b - g, c - e, d - f 4) a - h, b - g, c - f, d - e

96. Two rods A and B of same length and radius are joined together end to end. The coefficient of

thermal conductivity of A and B are 2K and K respectively. If temperature, differene between the

open ends of A and B is 36oC, the temperature difference across 'A' is

XöH�Ú_È=ô =°i�Çò XöH =ãH�`� "�¼ª�~¡Ö=ò Q®� ï~O_È° H�_�Û�° A =°i�ÇòB �#° XH� ^¥x`ËXH�\÷ "�\÷ JOKÇ°�°H �ey�Çò#ß@° ÁQê H �e��~ ¡° A =°i�ÇòB �³òH �ø L+¬ â"�� ¬ ìH �Q ® °}Hê�° =~¡°� ¬Qê 2K =°i�Çò K

A =°i�Çò B ³iz# z=~¡� =°^�Î¼L�éâãQ®`� �è^�Î=ò 36oCJ~ò# 'A'�³òH�ø JOKÇ°� =°^�Î¼ L�éâãQ®`� ��è^Î=ò1) 12oC 2) 18oC 3) 24oC 4) 9oC

97. A vessel contains air of mass 6gm at 127oC. It has a small hole through which air leakes out. At

some stage, the pressure becomes half of the original value and the temperature falls to 27oC.

The mass of air that has escaped is

127oC= Îí XH� ��ã Ç�Õ 6ãQê ã Î=¼~�t Q®� "��Çò=ô L#ßk. XH� z#ß ~¡Oã ÎO #°O_� "��Çò=ô "³�°�¬eH÷ �é=ôKÇ°#ßk.D �²Öu�Õ �Ô_È#O. ùe�Ô_È#O �¬QêxH÷ Çy¾, L�éâãQ® Ç 27oCä�½ Çy¾# "³�°�¬eH÷ �é~ò# "��Çò=ô ã Î=¼~�t1) 4gm 2) 6gm 3) 2gm 4) 8gm

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Page.No : 22

98. 5gm of water at to C is mixed with 2.5 gm of ice of at 0oC in a calorimeter of negligible thermal

capacity. The resultant mixture is found to contain 7.0gm of water at 0oC, the value of 't' is

�¬iQ®}÷OKÇ�èx `Çä�½ø= L+¬â^¥~¡} ª�=°~¡Ö¼=ò H�ey# ïH�Õi g°@~� #O^Î° to C=^Îí Q®� 5ãQê� h\÷x 0oC=^ÎíQ®�2.5ãQê� =°OKÇ°ä�½ H�e�²<�~¡°. �¦¬e`Ç=òQê ïH�Õi g°@~��Õ 0oC=^Îí Q®� 7ãQê. h~¡° L#ßKË 't' q�°=1) 100oC 2)80oC 3) 72oC 4) 32oC

99. An ideal gas occupies a volume of 0.5m3 at a pressure of 105Pa and at a temperature of 27oC.

When 12500J of heat is added to it the temperature increases to 87oC. If the gas undergoes

isobaric change then the change in internal energy is

�Ô_È#=ò 105Pa=°i�Çò 27oC L�éâãQ® Ç = Îí P Î~¡Å"��Çò=ô �¦¬°#�¬i=¶}=ò 0.5g°3 . "��Çò=ôä�½ 12500J L+¬â~�tx�¬~¡�¦¬~� KÍ�²#�¬ô_È° L�éâãQ®`Ç�Õ Ì�~¡°Q®°^Î� 87oC"��Çò=ô�Õx =¶~¡°æ (isobaric)�²Ö~¡�Ô_È# ã�¬ãH÷�Ç° J~ò# =�Çò=ô�³òH�ø JO Ç~¡¾ Ç�×H÷ë�Õx =¶~¡°æ1) -2500J 2) -10,000J 3) 2500J 4) 12500J

100. One kg of a diatomic gas is at a pressure of 8x104N/m2. The density of the gas is 4 kg/m3. What

is the energy of the gas due to its thermal motion?

1H÷.ãQê kÞ�¬~¡=¶}° "��Çò=ô �Ô_È#O 8x104N/m2=°i�Çò ª�Oã^Î`Ç 4 kg/m3 J~ò`Í P "��Çò=ô L+¬âQ®uH� KÇ�#�×H÷ëZO Ç?1) 3 x 104F 2) 5 x 104 J 3) 6 x 104J 4) 7 x 104J

101. A string in fundamental mode under a tension of 129.6N produces 8 beats/s when it is vibratedalong with a tuning fork. When the tension in the string is increased to 160N, it sounds in unisionwith the same tuning fork. The frequency of the tuning fork is

ã�� Îq°H� �Ï#:�¬ô#¼=ò�Õ H�O�¬#=ò K³O Î° Ç°#ß fQ®�Õ Ç#¼ Ç 129.6N L#ß�¬ôæ_È°, JkH�O�¬#=ò K³O Î° Ç°#ß XH��×$u^ÎO_È=ò�Õ 8 q�¬æO^Î<��° Ì�H�#°ä�½ L`Çæuë KÍªéëOk. `Ç#¼`Ç#° 160Nä�½ Ì�Oz#�¬ô_È°, Jk �×$u^ÎO_È=ò`ËJ#°<� Î=ò�Õ L#ßk. �×$u ÎO_È=ò Ç~¡KÇ° Î#=ò1) 80 Hz 2) 100 Hz 3) 140Hz 4) 200Hz

102. At an instant a source of sound is moving with velocity 20m/s along line �K � L� and observer ismoving along line �K � L� � with velocity 10m/s. If source produces sound of frequency 1000Hzthe frequency heard by listener at that instant is

XH� ^ÎÞx[#H�O �K � L� ö~Y"³O|_� 20m/s "ÍQ®O`Ë =°i�Çò �¬ij�ä�½_È° �K � L� � ö~Y"³O|_� 10m/s"ÍQ®O`Ë

ã�¬�Ç¶}÷�¬°ë<�ß~¡°. �ÎÞxx[ �Ï#:�¬ô#¼O 1000Hz J~ò Í, �¬ij�ä�½_È° q<Í Î$�×¼�Ï#:�¬ô#Ê¼O ZO Ç?1) zero 2) 1000 Hz 3) < 1000 Hz 4) > 1000 Hz

103. Two converging glass lenses, 'A' and 'B' have focal lengths in the ratio 1:2. The radius of curva-ture of second surface of lens 'A' is 4 times of its first surface where as the radius of curvature offirst surface of lens 'B' is twice that of its second surface. Then the ratio between the radii of thesecond surfaces of 'A' and 'B' is

ï~O_È° ä�½O��ìHê~¡ H�@H�=ò�° 'A'=°i�Çò 'B'"�\÷ <��ì¼O Ç~�� x+¬æuë 1:2 'A'H�@H�=ò �³òH�ø ï~O_È= Ç�O =ãH� �"�¼ª�~¡Ö=ò "³ò^Î\÷ `Ç�=ò �³òH�ø =ãH�`� "�¼ª�~¡Ö=ò 4 ï~@°Á =°i�Çò 'B'H�@H�=ò "³ò^Î\÷ `Ç�O �³òH�ø =ãH�`�"�¼ª�~¡Ö=ò ï~O_È= Ç�O "�¼ª�~�ÖxH÷ 2 ï~@°Á J~ò# 'A' =°i�Çò'B' H�@H�=ò�° ï~O_È= Ç�=ò� =ãH� � "�¼ª�~¡Ö=ò�x+¬æuë:1) 5:3 2) 3:5 3) 1:2 4) 5:12

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Page.No : 23

104. A point object emitting white light is placed at the principle focus of a convex lens made of glass.It is observed that the green coloured image is formed at infinity then

QêA`Ë KÍ�Ç°|_�# ä�½O��ìHê~¡ H�@H�O ã�¬ �¥# <�a�= Îí ³�Áx HêOu[#Hêxß LOz#�¬ô_È°, Pä�½�¬KÇó~¡OQ®° ã�¬uaO|OJ#O Ç Î¶~¡O�Õ U~¡æ_�Ok. J~ò Í1) violet & red images are also formed at infinity

T^¥ =°i�Çò Z~¡°�¬ô ~¡OQ®° ã�¬uaO�ì�° ä��_¨ J#O Ç Î¶~¡O�Õ U~¡æ_È �~ò2) violet image is real but red image is virtual

T^¥ ~¡OQ®° ã�¬uaO|O x[ ã�¬uaO|O, Hêx Z~¡°�¬ô~¡OQ®° ã�¬uaO|O q° �¥¼ã�¬uaO|O3) Red image is real but violet image is virtual

Z~¡°�¬ô ~¡OQ®° ã�¬uaO|O x[ ã�¬uaO|O, Hêx T^¥~¡OQ®° ã�¬uaO|O q° �¥¼ã�¬uaO|O4) Image of all colours are real but formed at different distances

Jxß~¡OQ®°� ã�¬uaO�ì�° x[ã�¬uaO�ì�° Hêx "Íö~Þ~¡° Î¶~��Õ U~¡æ_È �~ò105. If light is polarized by reflection, then the angle between reflected and refracted light is

�¬~�=~¡ë#O^¥Þ~� HêOu �Î$=}O [iy Í, �¬~�=~¡ë# =°i�Çò =ãH©��=# H÷~¡}ì� =° �Î¼HË}=ò

1) S 2) �

S3) �S 4)

�

S

106. The closest distance of approach of an D -particle travelling with a velocity V towards a station-ary nucleus is 'd'. If the velocity of particle is increased to � times the closest distance of ap-proach twoards a stationary nucleus of double the charge will beV"ÍQ®=ò`Ë ã�¬�Ç¶}÷OKÇ°KÇ°#ß XH� D -H�}O 'Q'P"Í�×OQ®� XH� öHOã^ÎHêxß �¬g°�²OKÇ°^Î¶~¡O 'd'W�¬ôæ_È° D -H�}O�³òH�ø "ÍQ®O � ï~@°Á J~ò# Jk '2Q'P"Í�×OQ®� öHOã ÎHêxß �¬g°�²OKÇ° Î¶~¡O1) d 2) 2d 3) d/2 4) d/3

107. A � (P capacitor is charged to 80V and another 6 (P capacitor is charged to 30V. When they areconnected with similar plates together, the change in the energy stored of � (P capacitor is� (P ïH��²@~¡° 80V ä�½ 6 (P ïH�²�²@~� 30Vä�½ P"Í�×�¬~¡KÇ|_�<�~ò. P ï~O_�O\÷ �¬*ìu �¬�H��° XH�\÷Qê H�e�² Í, � (P

ïH��²@~��Õx �×H÷ë�Õ =¶~¡°æ ZO Ç?1) 0.8mJ 2) 1.2mJ 3) 2.4 mJ 4) 4.8 mJ

108. With the resistance wires in the two gaps of a meter bridge, the balance point was found to be (1/3m) from the left end. When a �: coil is connected in series with the smaller of the two resis-tances, the point is shifted to 2/3m from the same end. Find the resistance of the two wires.ï~O_È° x~Ë^�ÎH��¬ô fQ®�#° g°@~¡° ãa_�û �³òH�ø ï~O_È° Mìm��Õ LOz<�~¡°. J�¬ô_È° �¬O`Ç°�# aO^Î°=ô Z_È=°"³á�¬ô#°O_� 1/3g° Î¶~¡O�Õ =zó#k. Çä�½ø= x~Ë^�ÎOQ®� fQ®ä�½�: x~Ë^ÎOQ®� fQ® KÇ°@�x ã�õ}÷�Õ H�e�²#�¬ô_È° �¬O`Ç°�aO Î°=ô 2/3 g° Î¶~¡O�Õ =zó#k. J~ò# P ï~O_È° fQ®� x~Ë ÎO�°1) � ��: : 2) � ��: : 3) � ��: : 4) � ��: :

109. Four batteries are connected in series as shown in the figure, the terminal voltage of each battery

V1,V2,V3,V4 are

I V1 20V V2 8V

I V3 6V V4 2V

1: 2:

1: 1:

�¬@O�Õ KÇ¶�²# q �ÎOQê =��Ç°O�Õ 4 �ì¼@s�#° ã�õ}÷�Õ H�e�²<�~¡°. "�x >ÿiß#�� Ú@xÂ�Ç°�� =~¡°�¬QêV1,V2,V3,V4

J~ò# P q�°=ô1) 16V, 0,10V,6V 2) 6V, 10V, 10V, 16V 3) 6V, 10V, 0,16V 4) 20V, 8V, 6V, 2V

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Page.No : 24

100. A proton and an alpha particle enter in a uniform magnetic field with the same velocity. Theperiod of rotation of alpha particle will be

ã�é\ì#°, P�ìæH�}=ò�° XöH"ÍQ®O�Õ UH�su J�Ç°ª�øO`Ç öHÆã`Ç=ò�ÕxH÷ ã�¬"ÍtOz#�¬ô_È° P�ìæ�H�}=ò �³òH�øã��=°}Hê�=ò1) four times that of proton ã�é\ì#° ã��=°}Hê�ìxH÷ 4 ï~@°Á2) two times that of proton ã�é\ì#° ã��=°}Hê�ìxH÷ 2 ï~@°Á3) three times that of proton ã�é\ì#° ã��=°}Hê�ìxH÷ 3 ï~@°Á4) same as that of proton ã�é\ì#° ã��=°}Hê�ìxH÷ �¬=¶#=ò

111. A galvanometer is shunted so that only 1% of the total current flows through it. If the resistance

of the galvanometer is ��: . This shunt resistance required is

Qê�Þ<�g°@~¡°ä�½ XH� +¬O@° x~Ë^¥xß H��°�¬ô@ =�# "³ò`ÇëO q^Î°¼`Ç°ë�Õ 1% q^Î°¼`� Qê�Þ<�g°@~¡° ^¥Þ~�ã�¬=�²ìOKÇ°KÇ°#ßk. Qê�Þ<�g°@~¡° x~Ë ÎO ��: .J~ò# +¬O@° x~Ë ÎO1) �: 2) �: 3) ��: 4) �:

112. A bar magnet produces a null point at P. If the magnet is turned through 180o, the magnetic

needle at P oscillates with a time period T. If the magnet is removed, the period of oscillation of

needle at P would be

XH� ÎO_È�Ç°ª�øO Ç=ò PJ#° aO Î°=ô = Îí Ç@�¬ÖaO Î°=ô#° U~¡æ~¡z#k. J�Ç¶ª�øO`�xß180o ãu�²æ#�¬ô_È° P=^Îí#°Oz# XH� J�Ç°ª�øO Ç �¬¶k _Ë�<�=~¡ë#Hê�=ò T J�Ç°ª�øO`�xß �¬îiëQê ù�yOz# P= Îí #°#ß J�Ç°ª�øO Ç�¬¶k_Ë�<�=~¡ë#Hê�=ò

1) �6 2) 6

�3) �6 4) � �6

113. The network shown in figure is part of a complete circuit. If at a certain instant the current (I) is

5A and is decreasing at a rate of 103 A/s, then VB-VA is

5mH

A

15VB1:

�¬O=$`Ç=��Ç°O�Õx XH� ��ìQ®=ò �¬@O�Õ KÇ¶�¬|_�Ok. XH� �¬=°�Ç°=ò�Õ q^Î°¼`� ã�¬"��¬ì=ò 5A=°i�Çò 103 A/

s=O`Ç°# `ÇQ®°¾KÇ°O>è VB-VA ZO Ç?1) 5V 2) 10V 3) 15V 4) 20V

114. The natural frequency of an LC-circuit is 1,25,000 cycles per second. Then the capacitor 'C' is

replaced by another capacitor with a dielectric medium of dielectric constant 'K' In this case, the

frequency decreases by 25kHz

The value of K is

XH� LC =��Ç°O �¬�¬ì[�Ï#:�¬ô#¼O 1,25,000/Ì�. D=��Ç°O�Õx ïH��²@~¡° ª�Ö#O�Õ 'K' ~Ë^�ÎH� �²Ö~�OH�=òQ®�~Ë^�ÎH�O L#ß =°~ùH� ïH��²@~¡°#° LOz#�¬ô_È°, �Ï#:�¬ô#¼O 25kHz Çy¾Ok. J~ò`Í 'K' q�°= ZO Ç?1) 3.0 2) 2.1 3) 1.56 4) 1.7

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Page.No : 25

115. When light with certain wavelength is incident on a metal plate with work function W, the deBroglie wavelength of electrons emitted with maximum K.E is equal to the wavelength of incidentlight. If the incident light wave length is halved, then de Broglie wavelength of electrons emittedwith maximum K.E also halved. The energy of initial inident light is�¬xã�¬"Í°�Ç°=ò W Q®� �Õ�¬ì�¬ô`Ç�OÌ�á HêOu �¬`Ç#=°~ò#�¬ô_È° Q®i+¬�Q®u[�×H÷ë`Ë "³�°=_�# Z�ãHê�#°� _�ã�ìwÁÇ~¡OQ® ³á~¡É ¼=ò �¬ Ç#HêOu Ç~¡OQ® ³á~¡É ¼=ò#ä�½ �¬=¶#=ò. �¬ Ç#HêOu Ç~¡OQ® ³á~�É ¼xß �¬QêxH÷ Çy¾Oz#�¬ô_È°, Z�ãHê�#°�

_�ã�ìwÁ Ç~¡OQ® ³á~¡É ¼=ò ä��_¨ �¬QêxH÷ Çy¾Ok. J~ò Í ù�° Ç �¬ Ç#=°~ò# HêOu �×H÷ë ZO Ç?

1) 9

�2) �9 3)

�9

�4)

�9

�

116. Compare momentum of 105 ev X-ray photon (px) with momentum of a 105eV electron (pe)

105 ev �×H÷ëQ®� X-H÷~¡}�¦é\ì<£ ã^Î=¼"ÍQ®=ò (px)=°i�Çò 105eV �×H÷ëQ®� Z�ãHê�#° ã^Î=¼"ÍQ®=ò (pe) � x+¬uë ZO Ç?

1) G

Z

R �

R � 2)

G

Z

R ��

R � 3)

G

Z

R �

R � 4)

G

Z

R �

R �

117. If K is the neutron multiplication factor in chain reaction,

StatementA: If K < 1, the number of neutrons in successive generations decreases and chainreaction can continue

StatementB: If K = 1, the chain reaction will proceed at steady rate

Statement C: If K > 1, the number of neutrons increases and the reaction is super called criticalstate

KJ#°#k #¶¼ã\ì<£ ã�¬ Ç°¼`�æ Î# Hê~¡H�O J~ò#ã�¬=KÇ#=ò A : K < 1J~ò Í L`�æ Î#�Õ q_È° Î�ÿá# #¶¼ã\ì#Á �¬OY¼ Çä�½ø= =°i�Çò �×$OH��KÇ~¡¼ [~¡°Q®°#°ã�¬=KÇ#=ò B:K = 1J~ò Í �²Ö~¡"³°Ø# ö~@°`Ë �×$OY� KÇ~¡¼ Jqzó#ßOQê [~¡°Q®°#°ã�¬=KÇ#=ò C:K > 1J~ò Í #¶¼ã\ì#Á �¬OY¼ Ì�~¡°Q®° Ç¶ LO@°Ok. D �²Öux �¬¶�¬~� ãH÷\÷H�� À��\� JO\ì~¡°1) All A,B and C are correct 2) Only A and B are correct

A,B =°i�Çò C Jxß �¬i�³Ø°#q A =°i�Çò B �° =¶ã Ç"Í° �¬ï~á#k3) Only A and C are correct 4) Only B and C are correct

A =°i�Çò C �° =¶ã Ç"Í° �¬ï~á#k B =°i�Çò C �° =¶ã Ç"Í° �¬ï~á#q118. An electron orbits a stationary nucleus of charge +ze, where z is consant and e is the magnitude

of electron charge. It requires 47.2ev to excite the electron from the second Bohr orbit to thirdBohr orbit. The value of 'z' is

+ze P"Í�×OQ®� �²Ö~¡öHOã ÎH�O KÇ°@¶� Z�ãHê�<£ �¬iã��q°�¬°ë#ßk. (z«�²Ö~¡�¬OY¼ e « Z�ãHê�#° P"Í�×O)Z�ãHê�#°#° ï~O_È=�Õ~�H�H�Æ#°O_� =ü_È=�Õ~� H�H�Æ¼�ÕxH÷ L`Íë[�¬~¡KÇ\ìxH÷ 47.2ev�×H÷ë J=�¬~¡=°~ò`Í 'z'q�°= ZO Ç?1) 5 2) 4 3) 3 4) 2

119. The variations of electric and magnetic fields are represented by 'JG

and $G at every point in space

Assertion(A): The electromagnetic wave travels in the direction of $ 'uG G

Reason(R): Electromagnetic waves propagate perpendicular to both $G and '

JG

q Î°¼ � =°i�Çò J�Ç°ª�øO ÇöHÆã`��Õ =¶~¡°æ�#° 'JG

=°i�Çò $G

�¬¶zH�OHêÎ$_È"�¼Y¼: q Î°¼ Î�Ç°ª�øO Ç Ç~¡OQ®=ò $ 'u

G G

k�×�Õ ã�¬�Ç¶}÷OKÇ°#°Hê~¡}=ò: q Î°¼ Î�Ç¶ª�øO Ç Ç~¡OQê�° $

G

=°i�Çò 'G

ï~O_�O\÷H© �O|OQê LO_È°#°1) A and R are true and R is the correct explnation of AA=°i�ÇòR �¬i�³Ø°#q. =°i�Çò RJ#°#k Aä�½ �¬i�³Ø°# q=~¡}2) A and R are true and R is not the correct explanation of AA=°i�ÇòR �¬i�³Ø°#q =°i�Çò RJ#°#k A ä�½ �¬i�³Ø°# q=~¡}Hê^Î°3) A is true R is false 4) A is false , R is true

A �¬i�³Ø°#k R �¬i�³Ø°#kHê Î° A �¬i�³Ø°#kHê Î°, R �¬i�³Ø°#k

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Page.No : 26

120. Consider the following statements A and B and identity the correct answer

ãH÷Ok"�Hê¼�#° �¬ijeOz �¬i�³Ø°# �¬=¶ �¥<�xß Q®°iëOKÇO_�A) Germanium is preferred over silicon in the construction of zener diod

r#~� _È�³¶_£ Ç�Ç¶~¡° KÍ�Çò@ä�½ �²eHê<£ H�<�ß [ö~àx�Ç°O L�¬�³¶Q®H�~¡=òB) Zener diode is connected in parallel to load resistance

r#~�_È�³¶_£ #° ��ì~¡ x~Ë �¥xH÷ �¬=¶O Ç~¡OQê H��°�¬ô`�~¡°.1) A is correct, B is wrong A �¬i�³Ø°#k B `Ç�¬ôæ2) A is wrong, B is correct A `Ç�¬ôæ B �¬i�³Ø°#k3) Both A and B are correct A =°i�Çò B �° ï~O_È° �¬i�³Ø°#q4) Both A and B are wrong A =°i�Çò B �° ï~O_È° �¬i�³Ø°#q Hê=ô

CHEMISTRY

121. An alkene on ozonolysis gives two compounds A, B. Both are functional isomers. 'A' can give

positive test with tollens reagent where 'B' does not then the original alkene is

XH� Pbø<£ F*ç<��²�¹ KÇ~¡¼�Õ A =°i�Çò B J#° ï~O_È° �¬"Í°à�×<��#° W�¬°ëOk. 'A' =°i�Çò 'B' �° ã�¬"Í°�Ç°�¬=ü�¬ì ª� Î$�×¼Hê�°. \ì�ÿ<£� Hê~¡H�O`Ë A KÇ~¡¼#° W�¬°ëOk Hêh B W=Þ Î°. J~ò# P Pbø<£1) 2 � methyl � pentene 2 � q°^�³á�� Ì�O\©<£2) 2 � methyl � 2 � pentene 2 � q°^�³á�� 2 � Ì�O\©<£3) 3 � methyl � 2 � pentene 3 � q°^�³á�� 2 � Ì�O\©<£4) 3 � methyl � 3 � pentene 3 � q°^�³á�� 3 � Ì�O\©<£

122. Methyl group of toluene is o, p � director. This is best explained by

\Õ�ÿ<£�Õx q°^�³á�� ¬=ü�¬ì=ò o, p � ª�Ö# xö~í�×H�=ò, nxx KÇH�øQê q=iOKÇ°#k.1) Mesomeric effect g°ªé"³°iH± ã�¬��ì==ò2) Hyper conjugation Ju�¬O�ÇòQ®à=ò3) Inductive effect ãÀ�ö~�¬H� ã�¬��ì==ò4) Electromeric effect Z�¢HË�"³°iH± ã�¬��ì==ò

123. The Hamailtonian is numerically equal to �¬ì"³°�Õ�x�Ç°<£ P�¬ö~@~� ä�½ �¬OMì¼ ÇàH�OQê �¬=¶#"³°Ø#k

1)

� � � �

� � � �

J8

� O Z [ \

ª º� w w w� � \�« »S w w w¬ ¼

2)

� � � �

� � � �

J8

� O Z [ \

ª ºw w w� � �« »S w w w¬ ¼

3)

� � � �

� � � �

J8

� Z [ \

ª º� w w w� � �« »S w w w¬ ¼

4)

� � � �

� � � �

J8

� O Z [ \

ª º� w w w� � �« »S w w w¬ ¼

124. Some orders are given below ãH÷O^Î Hùxß ãH�=°=ò�° W=Þ|_�#q.

a) E2s(H) > E2s (Li) > E2s(Na) > E2s(K)

b) For H atom 1s < 2s = 2p < 3s = 3p = 3d < H �¬~¡=¶}°=ôä�½c) For Na atom 1s < 2s < 2p < 3s < 3p Na �¬~¡=¶}°=ôä�½d) Number of nodal planes 1s > 2s > 3s > 4s <Ë_È�� Ç�ì� �¬OY¼1) All are correct Jxß �¬i�³Ø°#q2) Only a, b and c are correct a, b =°i�Çò c =¶ã`Ç"Í° �¬i�³Ø°#q3) Only a, c and d are correct a, c =°i�Çò d =¶ã`Ç"Í° �¬i�³Ø°#q4) Only b, c and d are correct b, c =°i�Çò d =¶ã`Ç"Í° �¬i�³Ø°#q

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Page.No : 27

125. Select the incorrect order �¬iHêx ãH�=°=ò#° Z#°ßHËO_�.1) Na < Al < Mg < Si � First ionisation ã�¬^�Î=° J�Ç°hH�~¡}O2) O > F > N > C � Second ionisation kÞu�Ç° J�Ç°hH�~¡}O3) C < N < O < F � Electronegativity |°°} q Î°¼^¥ ÇàH� Ç4) N2O5 > Cl2O7 > Al2O3 > MgO � Acidic nature of oxides PïHá�_£� P=°Á�¬Þ��ì==ò

126. Which of the following is viable particulate DãH÷Ok"�x�Õ rqOKÇQ®e¾# H�}*ì Ç°�°

1) Moulds tbOã �¥�° 2) Cigarette smoke �²Q®ï~\� �ÚQ®3) Pulverised coal �çQ®°¾ �Ú_� 4) Both 2 and 3 2 =°i�Çò 3

127. Hydrolysis of the following followed by polymerisation produces a straight chain silicone polymer

DãH÷Ok"�x�Õ [�q�õÁ+¬} Ç~�Þ Ç, �Úe"³°sH�~¡}=ò�Õ �§v�Ç° �×$OY� �²eHË<£ ��e=°~�#° U~¡æ~¡°KÇ°#k.

1) MeSiCl3 2) Me2SiCl2 3) Me3SiCl 4) SiCl4

128. Assertion (A) : Boric acid is a weak mono basic acid.

^�Î$_�È"�¼Y¼ (A) : �ÕiH± P=°ÁO |��Ôì# UH�HÆê~¡H� P=°ÁO.

Reason (R) : H3BO3 has three O�H groups attached to Boron.

Hê~¡}=ò (R) : H3BO3 �Õ �Õ~�<£ä�½ H��¬|_�# =ü_È° O�H �¬=ü�¬ð�° LO\ì~ò.1) A and R are true and R is the correct explanation of A.


A =°i�Çò R �¬i�³Ø°#q. =°i�Çò R J#°#k A ä�½ �¬i�³Ø°# q=~¡}Hê^Î°.3) A is true, R is false. A �¬i�³Ø°#k, R �¬i�³Ø°#k Hê^Î°.4) A is false, R is true. A �¬i�³Ø°#k Hê^Î°, R �¬i�³Ø°#k.

129. Hydrogen gas can be stored in a tank made up of

Ì�áìã_Ë[<£ "��Çò=ô#° ��ã Î�¬~¡°KÇ° \ìOH± nx`Ë Ç�Ç¶~¡°KÍ�Ç°|_È°#°.1) Mg � MgH

22) Pt � 1r 3) Alnico J��xHË 4) NaPtF

6

130. A beaker contain 200 ml of 0.01M CH3COOH [Ka of CH3COOH is 10�6]. If 100 ml 0.01M NaOH

is added to this beaker, the change in pH

XH� cH�~¡°�Õ 200 q°.b.� 0.01M CH3COOH L#ßk.[CH3COOH �³òH�ø Ka q�°= 10�6]. 100 q°.b.�0.01M NaOH #° D cH�~¡°ä�½ H��¬Qê pH �Õ =KÇ°ó =¶~¡°æ1) 6 2) 5 3) 4 4) 2

131. If a small quantities of ferric nitrate is added to the system,

� � � � � ��

��

CS CS CS( G 5%0 ( G 5%0

�

� � ª º� ¬ ¼U

W=Þ|_�# =¼=�¬Öä�½ HùkíQê Ì�¦ãiH± <³áã>è\�#° H��¬Qê1) Intensity of yellow colour increases �¬�¬°�¬ô~¡OQ®° fã=`Ç Ì�~¡°Q®°#°2) The solution becomes colour less ã^¥=}=ò q=~¡â=ò JQ®°#°3) Intensity of deep red colour increases =ò^Î°~¡° Z~¡°�¬ô ~¡OQ®° fã=`Ç Ì�~¡°Q®°#°4) Intensity of deep blue colour increases =ò Î°~¡° h�=ò ~¡OQ®° fã= Ç Ì�~¡°Q®°#°

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132. The IUPAC name of is �³òH�ø IUPAC <�=°=ò

1) 1�ethyl�3, 3�dimethyl cyclohexane 1�W^�³á��3, 3�_³áq°^�³á�� Ì�áHËÁÌ�ìöH�<£2) 3�ethyl�1, 1�dimethyl cyclopentane 3�W^�³á��1, 1�_³áq°^�³á�� Ì�áHËÁÌ�O>è<£3) 3�ethyl�1, 1�dimethyl cyclohexane 3�W^�³á��1, 1�_³áq°^�³á�� Ì�áHËÁÌ�ìöH�<£4) 1�ethyl�3, 3�dimethyl cycloheptane 1�W^�³á��3, 3�_³áq°^�³á�� Ì�áHËÁÌ�ìÀ��<£

133. Colourless chemical among the following is

DãH÷Ok"�x�Õ =~¡â~¡�²ì Ç ~¡ª��Ç°#=ò

1) (NH4)3PO4.12M0O3 2) CuSO4

3) Fe4[Fe(CN)6]3 4) [Fe(CN)5NOS]

134. � � � ��I ��* 1

* % ��* 1 * %� oA A*' = � 69.01 KJ/mole ........ (a)

� � � ��I ��* 1

* % ��* 1 * %� oA A*' = � 72.03 KJ/mole ........ (b)

� � � ��I ��* 1

* % ��* 1 * %� oA A*' = � 72.79 KJ/mole ........ (c)

Then select the correct statement. �¬i�³Ø°# "�¼Y¼#° Z#°ßHËO_�.1) enthalpy of dilution is a � b qb<Ë+¬â=òä�½ �¬O|Ok�Oz#k a � b

2) enthalpy of dilution is b � a qb<Ë+¬â=òä�½ �¬O|Ok�Oz#k b � a

3) to enthalpy of dilution is a � c qb<Ë+¬â=òä�½ �¬O|Ok�Oz#k a � c

4) c corresponds to enthalpy of dilution c qb<Ë+¬â=òä�½ �¬O|Ok�Oz#k135. Assertion (A) : Enthalpy of a compound is equal to standard enthalpy of formation of that com-

pound.

�Î$_�È"�¼Y¼ (A) : XH� �¬"Í°à�×#=ò �³òH�ø ZO �¥eæ ¥x ã��=¶}÷H� �¬O�õÁ+¬} ZO �¥bæH÷ �¬=¶#O.

Reason (R) : Enthalpies of elements in their standard state is arbitarly taken as zero.

Hê~¡}=ò (R) : =ü�Hê�ä�½ "�\÷ ã��=¶}÷H� �²Ö`Ç°��Õ ZO^�¥eæ �¬°<�ßQê f�¬°ä�½O\ì~¡°.1) A and R are true and R is the correct explanation of A.


A =°i�Çò R �¬i�³Ø°#q. =°i�Çò R J#°#k A ä�½ �¬i�³Ø°# q=~¡}Hê^Î°.3) A is true, R is false. A �¬i�³Ø°#k, R �¬i�³Ø°#k Hê^Î°.4) A is false, R is true. A �¬i�³Ø°#k Hê^Î°, R �¬i�³Ø°#k.

136. The best method to prepare BeF2 is

BeF2 #° Ç�Ç¶~¡°KÍ�Çò@ä�½ "�_È° J`Ç¼O`Ç =°Oz �¬^Îíu1) Decomposition of (NH

4)

2BeF

4(NH

4)

2BeF

4 �³òH�ø q�¦¬°@#=ò

2) Heating BeO with coke and F2

BeO #° HËH± =°i�Çò F2 `Ë "Í_�KÍ�Çò@

3) Heating BeCl2 with F

2BeCl

2 #° F

2 `Ë "Í_�KÍ�Çò@

4) Heating Be with F2 Be #° F2 `Ë "Í_�KÍ�Çò@137. Critical temperature is highest for J`Ç¼k�H� �¬OkQ®� L�éâãQ®`Ç H��k.

1) CO2 2) NH3 3) H2O 4) He

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Page.No : 29

138. Incorrect match among the following.

DãH÷Ok"�x�Õ �¬iHêx [ Ç1) Tribromooctoxidel +6, +4, +4 = Oxidation state of bromine

¢>ÿØã�Õ"³¶PHê�ïH�å_£l+6, +4, +4 = ã�Õq°<£ �³òH�ø PH©�H�~¡} �²Ö`Ç°�°2) Sodium tetra thionatel +5, 0, 0, +5 = Oxidation state of sulphur

ªé_��Ç°O >ÿã\ì^�Î�³¶<Í\�l +5, 0, 0, +5 = �¬�æ�~� �³òH�ø PH©�H�~¡} �²Ö`Ç°�°3) Potassium triodidel 0, 0, �1 = Oxidation state of iodine

�Ú\ì+²�Ç°O ¢>ÿØJ�³ò_³á_£l 0, 0, �1 = J�³ò_�<£ �³òH�ø PH©�H�~¡} �²Ö`Ç°�°4) Hydrazoic acidl �1/3, �1/3, �1/3 = Oxidation state of nitrogen

Ì�áìã_Ë*ì~òH± P=°ÁOl �1/3, �1/3, �1/3 = <³áã\Õ[<£ �³òH�ø PH©�H�~¡} �²Ö`Ç°�°139. In the titration involving Cr2O7

2�, the indicator used and the end point is

Cr2O72� Ë KÍ�Çò JO�×=¶�¬#=ò�Õ, �¬¶zH� =°i�Çò JOu=° ª�Ö#=ò

1) Diphenyl amine � appearance of intense blue colour

_³á�¦²<³á�� Z"³°Ø<£ Ð =ò^Î°~¡° h�=ò ~¡OQ®° U~¡æ_È°@2) Diphenyl amine � appearance of pink colour _³á�¦²<³á�� Z"³°Ø<£ Ð T^¥ ~¡OQ®° U~¡æ_È°@3) Starch � appearance of yellow colour ª��~�ó Ð �¬�¬°�¬ô ~¡OQ®° U~¡æ_È°@4) Phenolpthalein � appearance of pink colour �¦²<��¦¬ëb<£ Ð T^¥ ~¡OQ®° U~¡æ_È°@

140. If 'Z' is overlapping axis, then anti bonding molecular orbital formed from two pX orbital is

represented as

'Z' J#°#k Ju��`Ç JH�Æ=ò J#°Hù#ß@Á~ò`Í, ï~O_È° pX PiÄ\ì�� #°O_� U~¡æ_�# J�¬|O Î J}° PiÄ\ì��#°DãH÷Ok q �ÎOQê KÇ¶�²OKÇ=KÇ°ó.

1) +- 2) - + - 3)

++-

- 4) +

-+

-141. At 293K, the Henry constant KH is highest for [KH/K.Bar]

293K =^Îí Ì�ìãh �²Ö~�OH�=ò KH J`Ç¼k�H�OQê H��k [KH/K.Bar]

1) O2 2) N2 3) Ar 4) CO2

142. The air tanks used by Scuba divers contain He, N2 and O2 in the following percentages

Q®[ D`ÇQê�×Ã¤ "�_È° Qêe \ì¼OH±��Õ LO_È° He, N2 =°i�Çò O2 "��Çò=ô� �§`��° =~¡°�¬Qê1) 11.7, 38.3, 50.1 2) 11.7, 56.2, 32.1 3) 56.2, 12.7, 28.1 4) 11.7, 46.8, 42

143. Conversion A to B follows 2nd order kinetics. 10M of A is reduced to 5M in 20 min. What will be

the time (in min) required for the conversion of 10M of A to 2.5 M.

A #°Oz B =¶~¡° KÇ~¡¼ kÞf�Ç° ãH�=¶OH�KÇ~¡¼. 10M Qê_�È Ç Q®� A #° 20 xII��Õ 5M ä�½ Çy¾OKÇ_È"³°Ø#k. 10M Qê_�È ÇQ®� A #° 2.5 M ä�½ `Çy¾OKÇ°@ä�½ �¬@°�Hê�O (xII ��Õ)1) 40 2) 30 3) 60 4) 20

144. Phsophorus acid on thermal decomposition gives ortho phosphoric acid. In this reaction, the

equivalent weight of phosphorus acid is

�¦��¬æ�~¡�¹ P=°ÁO L+¬â q�¦¬°@#=ò K³Ok P~Ë��¦�ª�æ�iH± P=°ÁO#° U~¡æ~¡z#k. D KÇ~¡¼�Õ �¦��¬æ�~¡�¹ P=°ÁO �³òH�ø Ç°�¼��ì~¡=ò1) 17.8 2) 27.8 3) 49.2 4) 54.6

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Page.No : 30

145. The resistance of a solution A is ��: and that of B is ��: . When equal volumes of these two

are mixed what will be the resistance of the resulting solution [Z* remain same]

A, B J#° ï~O_È° ã^¥=}ì� �³òH�ø x~Ë �¥�° =~¡°�¬Qê ��: , ��: . �¬=¶# �¦¬°#�¬i=¶}ì��Õ D ï~O_È° ã^¥=}ì�#°H��°�¬Qê U~¡æ_�# �¦¬e`Ç ã^¥=}=ò �³òH�ø [Z* �Õ =¶~¡°æ�è^Î°]1) ��: 2) ��: 3) ��: 4) ��:

146. Roasting of FeS2 produces SO2 gas. In this reaction the equivalent weight of FeS2 is

[Atomic weights: Fe = 56, S = 32]

FeS2 #° ~Ë�²�OQ· KÍ�²#�¬ô_È° SO2 "��Çò=ô U~¡æ_È°`Ç°Ok. D KÇ~¡¼�Õ FeS2 �³òH�ø Ç°�¼��ì~¡=ò[Atomis weight Fe = 56, S = 32]

1) 10.9 2) 120 3) 60 4) 31.8

147. Which of the following corresponds to a graph drawn between log x/m (taken on y axis) verses

log P

DãH÷Ok"�x�Õ log x/m (yÐJH�ÆOÌ�á) =°i�Çò log P � =° �Î¼ wz# ãQê�¦¹#° ³e�Ç°*è�Çò#k.

1) 2) 3) 4)

148. Which of the following is used as a bleaching agent for paper pulp.

Hêy Ç�¬ô Q®°Aû#° q~¡O[#=ò KÍ�Çò@ä�½ "�_È° q~¡O[# Hê~¡H�=ò1) Cl2O 2) ClO2 3) SO2 4) ClF3

149. Which of the following resumble copper in conductivity and appearence

DãH÷Ok"�x�Õ q Î°¼ � "��¬ìH� Ç =°i�Çò ~¡¶�¬=ò�Õ Hê�¬~� �Õ�¬ìO`Ë �éeH� LO_È°#k.1) TiO2 2) TiO 3) ReO3 4) CrO2

150. Select the incorrect set among the following.

DãH÷Ok"�x�Õ �¬iHêx [ Ç#° Z#°ßHËO_�.

1) Cr < Mo < WlMelting points Cr < Mo < Wl ã^Îg��=# ª�Ö#O

2) Mn < Tc < RelMelting points Mn < Tc < Rel ã^Îg��=# ª�Ö#O

3) Ti < V < Crl M3+/M2+ potential Ti < V < Crl M3+/M2+ �×H�àO

4) Ti < V < CrlM2+/M potential Ti < V < CrlM2+/M �×H�àO

151. Observed colour change in the convension

� ��

� � CS

0 K* 1�

ª º¬ ¼ to � � � ��

� � CS

0 K* 1 GP�

ª º¬ ¼

D KÇ~¡¼�Õ ~¡OQ®°�Õx =¶~¡°æ1) Green to pale blue Pä�½�¬KÇó #°Oz �è`Ç h�=ò ~¡OQ®°2) Green to purple Pä�½�¬KÇó #°Oz T^¥ ~¡OQ®°3) Purple to green T^¥ ~¡OQ®° #°Oz Pä�½�¬KÇó ~¡OQ®°4) Green to violet Pä�½�¬KÇó ~¡OQ®° #°Oz =ò^Î°~¡° h�=ò

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Page.No : 31

152.

1^

% 0 *� � � linkage is not observed in

1^

% 0 *� � � J#° |O �Î=ò nx�Õ H�x�²OKÇ Î°.

1) Nylon 6, 6 and Nylon 6 <³á�ì<£ 6, 6 =°i�Çò <³á�ì<£ 6

2) Nylon 2 � Nylon - 6 <³á�ì<£ 2 � <³á�ì<£ - 6

3) Nylon 6, 6 and Nylon 2 � Nylon � 6 <³á�ì<£ 6, 6 =°i�Çò <³á�ì<£ 2 � <³á�ì<£ � 6

4) Dacron and Melamine _¨ãHê<£ =°i�Çò "³°�"³°Ø<£

153. In the extraction of copper the reaction � � � �

�%W 5 �1 �%W 1 �51� o � , is observed in

Hê�¬~� �Õ�¬ì �¬OãQ®�¬ì}�Õ� � � �

�%W 5 �1 �%W 1 �51� o � J#° KÇ~¡¼ nx�Õ Q®=°xOKÇ=KÇ°ó.

1) In both roasting and bessimerisation process ��~¡û#KÇ~¡¼ =°i�Çò a�¬�=°ï~á*è+¬<£ q^�¥#O�Õ

2) Only in roasting process öH=�O ��~¡û# KÇ~¡¼�Õ

3) Only in bessimerisation process öH=�O a�¬�=°ï~á*è+¬<£ q �¥#O�Õ

4) Only in smelting process to get matte =¶\� �ÚO Î°@ä�½ Ì�àe�OQ·�Õ =¶ã Ç"Í°

154. Which of the following traid corresponds to neutral essential amino acid traid

DãH÷Ok ã Ç�Ç¶��Õ Uk Ç@�¬Ö Zq°<Ë P=¶Á�¬ô P=�×¼H� ã Ç�Ç°O

1) Glycine, Arginine, Threonine ïQÁå�²<£, Jiû<³á<£, ãk��³¶<³á<£

2) Histidine, Proline, Serine �²ì�²�_³á<£, ã�Ú�ÿá<£, Ì�ï~á<£

3) Glutamine, Valine, Alanine Q®°Á\ì"³°<£, "��ÿá<£, J�<³á<£

4) Leucine, Isoleucine, Valine �¶¼�²<£, Sªé�¶¼�²<£, "��ÿá<£

155. The purpose of glycerol in shaving soaps is À+qOQ· �¬|°Ä��Õ LO_È° yÁ�¬~�� L�¬�³¶Q®O

1) To produce more leather Zä�½ø= #°~¡Q® ~�=_ÈO HË�¬O

2) To prevent rapid drying ÇÞ~¡Qê �Ú_��ì~¡ä�½O_¨ LOKÇ_ÈO HË�¬O

3) To soften the hair "³Oã@°H��#° "³° Çë|~¡°KÇ°@ä�½

4) To act as antiseptic �Ç¶O\÷Ì��²�H±Qê �¬xKÍ�¬°ëOk

156. Which of the following compound will have a high rate in SN1 mechanism

DãH÷Ok"�x �¬ ¥~¡Ö��Õ SN1 KÇ~�¼j� Ç J Ç¼k�H�OQê H��k.

1) (CH3)

2CHCl 2)

CH2CH2

Cl

3) (CH3)

3C

Cl 4) (CH

3)

2 CH � CH

2Cl

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Page.No : 32

157. The correct decreasing order of acidic nature is

P=°Á�¬Þ��ì=O ÇQ®°¾ �¬i�³Ø°# ãH�=°O

1)

OH

> H2O > C

2H

5OH > C

2H

2

2)

OH

> H2O > C2H2 > C2H5OH

3)

OH

> C2H

5OH > C

2H

2 > H

2O

4)

OH

> C2H2 > C2H5OH > H2O

158.

NH2

2 2 4

2

1) HC2) HNO / H SO

3) H / Ni······l

A

'X'. Here the product 'X' is D KÇ~¡¼�Õ L`Çæ#ßO 'X' J#°#k

1)

NH2

NO2

2)

NO2

3NH�3)

NH2

3NH�4)

NH2

NH2

159. Carboxylic acid is a not a product in the following sequence of reaction

DãH÷Ok KÇ~¡¼ ãH�=°=ò�Õ Hê~�ÄH÷�eH± P=°ÁO XH� L Çæ#ßO Hêxk.

1) 3H O3 2 2 5CH CO C H

�

¶¶¶l 2) 4MnO /OH� �

%¶¶¶¶¶l

3) 4MnO3 2 3

/HCH COCH CH

� �

¶¶¶¶¶l 4) 3H O3 2 2CH CH CONH

�

� � ¶¶¶l

160. FKN0C1*

��%* %*1 # $

'��o ��o .

The number of sp2 hybridised carbon atoms in the organic compound 'B' will be

'B' J#° H�~¡Ä# �¬"Í°à�×#=ò�Õx sp2 �¬OH�sH�~¡}=ò K³Ok# Hê~¡Ä<£ �¬~¡=¶}°=ô� �¬OY¼1) 4 2) 3 3) 2 4) Zero �¬°#ß

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Page.No : 33

KEY SHEET

BOTANY

1) 1 2) 1 3) 4 4) 3 5) 3 6) 3 7) 2 8) 1 9) 2 10) 3

11) 1 12) 2 13) 3 14) 3 15) 4 16) 4 17) 3 18) 2 19) 2 20) 2

21) 3 22) 1 23) 4 24) 4 25) 1 26) 4 27) 1 28) 4 29) 2 30) 3

31) 2 32) 2 33) 2 34) 4 35) 4 36) 1 37) 3 38) 4 39) 4 40) 1

ZOOLOGY

41) 2 42) 3 43) 1 44) 3 45) 3 46) 4 47) 4 48) 2 49) 3 50) 2

51) 1 52) 2 53) 2 54) 2 55) 2 56) 3 57) 3 58) 1 59) 2 60) 2

61) 3 62) 3 63) 3 64) 3 65) 1 66) 2 67) 3 68) 2 69) 2 70) 4

71) 2 72) 2 73) 2 74) 2 75) 3 76) 4 77) 4 78) 2 79) 3 80) 2

PHYSICS

81) 2 82) 1 83) 3 84) 4 85) 3 86) 1 87) 4 88) 3 89) 2 90) 3

91) 3 92) 1 93) 3 94) 4 95) 2 96) 1 97) 3 98) 4 99) 3 100) 2

101) 1 102) 2 103) 1 104) 2 105) 2 106) 4 107) 4 108) 1 109) 1 110) 2

111) 3 112) 1 113) 3 114) 3 115) 3 116) 2 117) 4 118) 1 119) 4 120) 2

CHEMISTRY

121) 2 122) 2 123) 4 124) 2 125) 4 126) 1 127) 2 128) 2 129) 1 130) 4

131) 3 132) 3 133) 2 134) 2 135) 1 136) 1 137) 3 138) 4 139) 1 140) 3

141) 2 142) 2 143) 3 144) 4 145) 2 146) 1 147) 2 148) 2 149) 3 150) 3

151) 1 152) 4 153) 1 154) 4 155) 2 156) 3 157) 1 158) 3 159) 2 160) 2

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Page.No : 34

HINTS AND SOLUTIONS

BOTANY

1. Option (1) is correct: Autogamy is possible only in bisexual flowers and geitonogamy is taking place

in flowers of monoecious plants and plants with bisexual flowers.

kÞeOQ® �¬ô��æ��Õ =¶ã Ç"Í° P Çà�¬~�Q®�¬O�¬~¡øO [~¡°Q®°#°. UH�eOQêã�×~ò "³òH�ø��Õ kÞeOQ® �¬ô��æ�° H�e¾# "³òH�ø��ÕUH� �¬ô+¬æ�¬~�Q®�¬O�¬~¡øO (ïQá@<ùQ®q°) [~¡°Q®°#°.

2. Option (1) is correct: Marginal placentation is a characteristic feature of Fabaceae members and obliquearrangment of carpels and clockwise rotation of gynoecium to 450 is seen in solanceae.

�¦��è�² ä�½@°O| "³òH�ø�ÕÁ L��O Ç JO_È<�¼�¬O, ªÚ�<Í�² ä�½@°O| "³òH�ø��Õ JO_ÈHË�×O�çx �¦¬� Î�ì� U@"��°J=°iH�, JO_ÈHË�×O 450 HË}O�Õ �¬=¼k�×�ç u~¡°Q®°@ H�xæOKÇ°#°.

3. Option (4 ) is correct: Artocarpus belongs to Moraceae.

P~Ë�Hê~¡æ�¹ "³¶ö~�² ä�½@°O�ìxH÷ K³Ok#k.

4. Option (3) is correct: (A) is correct.

(R) is wrong because intrafascicular cambium of dicot stems is a parimary meristem but it is useful forsecondary growth.

^�Î$_�È"�¼Y¼ (Z) : �¬i�³Ø°#k

Hê~¡}=ò (P~�) �¬i�³Ø°# q=~¡} Hê^Î°. kÞ^Î�×c[ HêO_¨��Õx �¬ôO*ìO`Ç~¡ q��ì[¼ H�}*ì�=ò, ã��^�Îq°H�q��ì[¼H�}*ì�"³°Ø#�¬æ\÷H÷, kÞf�Ç° =$k�H÷ �¬�¬ì�Ç°�¬_È°#°.

5. Option (3) is correct: Phylum is equal to Divison in plant systamatics.

[O`Ç° =s¾H�~¡}�Õx Ì�¦á�"£° P#°=s¾H�~¡} ã�¬=¶}=ò, =$H�Æ =s¾H�~¡}�Õx q��ìQ®=ò#ä�½ �¬=¶#=ò.6. Option (3) is correct: These are modified green branches of limitted growth ( axillary braches) ment for

photosynthesis.

Wq �¬iq°u =$k�KÇ¶�¬ô / xsâ`Ç =$k� KÇ¶�¬ô Pä�½�¬KÇóx �§Y�°. g\÷x �¬ã`��HêO_¨�°Qê Q®°iëª�ë~¡°. (Wq ãw=�§Y�°)

7. Option (2) is correct: Youngest floral bud is at the distal region i.e, near apical region of the peduncle.

�¬ô��æq<�¼ª�H�Æ ^Î¶~¡�¬� Hù#�"³á�¬ôä�½ �è`Ç� "³òQ®¾�° J=°iH� Ð JãQêa�ª�~�#°ãH�=° P=°iH� �H�Æ}=ò.

8. Option (1) is correct: Starting from calyx the peripheral whorl, it follows m corolla m androeciumm Gynoecium

�¬ô��æ�¬#O Ì�á �¬ô��æ�¬ã`��° �¬ik� "³á�¬ô#°O_� öHOã^Î=ò �"³á�¬ôä�½ =~¡°�¬Qê, ~¡H�ÆH��¬ã`�=om PH�~¡ÂH� �¬ã`�=omöH�¬~�=o m JO_ÈHË�×O, Qê J=°iLO_È°#°.

9. Option (2) is correct: Legune (monocarpillary) ----- Cypsela ( bicarpellary) ---- Pepo ( tricarpellary) ----- Nut ( multicarpellary)

JO_ÈHË�×O �Õx �¦¬�^Î�ì� �¬OY¼#° J#°�¬iOz P~Ë�¬ì} ãH�=°O�Õ Ð kÞq^¥~¡H� �¦¬�O (UH� �¦¬�^Î�×) Ð �²Ì��ì(kÞ�¦¬�^Î�×) Ð Ì��Ú (ãu�¦¬�^Î�×) Ð Ì�Oä�½Q®� �¦¬�=ò (|�¬��¦¬�^Î�×).

10. Option (3) correct: Bivalent contains two chromosome. Each chromosome of a bivalent possesses onecentromere and two chromatid. Therefore the bivalent has two chromosomes, two centromeres andfour chromatids.

�ÿá=�ÿO\��Õ ï~O_È° ãHË"³¶*Õ"£°�° LO\ì~ò. �ÿá=�ÿO\� �Õ ã�¬u ãHË"³¶*Õ"£°ä�½, XH� Ì�Oã\çq°�Ç°~�, ï~O_È° ãHù=¶\÷_£�LO_È°#°. H�#°H� �ÿá=�ÿO\��Õ <��°Q®° ãHù=¶\÷_£�, ï~O_È° Ì�Oã\çq°�Ç°~�� LO_È°#°.

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Page.No : 35

11. Option (1) correct: Fucus and marchantia are non - spermatophytic plants. Fucus shows diplontic lifecycle because, except gametes which are haploid, all stages in the life cycle shows diploid condition.

Marchantia shows halo diplontic life cycle because the diploid sporophyte dependent on gametophytewhich is longlasting and photo synthetic.

�¬î¼H��¹ =°i�Çò =¶~�øOxÂ�Ç¶ c[~¡�²ì`Ç "³òH�ø�°. �¬î¼H��¹ JO_È �¬O�³¶Q® "³òH�ø, nx�Õ �¬O�³¶Q®c*ì�° `Ç�¬æ,Jxß Î�×�° ÎÞ�Ç°�²ÖuH�O Ð ÎÞ�Ç°�²ÖuH�Ðrq ÇKÇãH�O. =¶~�øOxÂ�Ç¶ UH� eOQêã�×~ò "³òH�ø, nx�Õ ÎÞ�Ç°�²ÖuH� �² Î�c[ ÎÎ�× �¬O�³¶Q® c[ Î Î�×Ì�á P^¥~¡�¬_� LO_È°@KÍ Ð UH� Ð ÎÞ�Ç°�²Öu Ç rq ÇKÇãH�O

12. Option (2) is correct: Photosynthetic rootes in Taeniophyllum, stem is a phylloclade in Euphorbia,

petile of the leaf is modified into phyllode in Acacia.

\©x�³¶�¦²�ÁO�Õ "Í�×Ã¤, �Çü�¦éiÄ�Ç¶�Õ �¬ã`��ì HêO_¨�°, JöH�²�Ç¶�Õ ã�¬��ì�¬#=ò (�¬ã`Ç=$O`Ç ~¡¶��O`Ç~¡O)�¬ã`Ç�¬ìi`Ç �Çò`ÇO =°i�Çò H÷~¡}[#¼ �¬O�³¶Q®ãH÷�Ç°ä�½ L�¬�³¶Q®�¬_È°#°.

13. Option (3) is correct: (A ): Is correct for all Angiosperms,

(R): Is wrong because the sporophytes in different angiosperms are monoecious, dioecious and someare polygamous.

�Î$_�È"�¼Y¼ (Z) Ð �¬i�³Ø°#k Hêx Hê~¡}O (J~�) �¬i�³Ø°#kHê Î°, ZO Î°H�#Qê qq �Î P=$ Ç c["³òH�ø�ÕÁ �² ÎÖc[ Î"³òH�ø�°, UH�eQêã�×�Ç°, kÞeOQêã�×�Ç°,qq �ÎeOQêã�×~ò�°Qê LO_È°#°.

14. Option (3) is correct: Reminent of nucellus is the perisperm (2n) and it belongs to parental generation.

Endosperm (3n) belongs to next generation.

q Çë#O�Õ �¦¬�n�H�~¡}ì#O Ç~¡O q°ye�é~ò# JO_¨O Ç�H�}*ì�O (2Z<£) �¬iK³ó� Î=òQê =¶~¡°#°. Wk =¶ Ç$ Ç~�xH÷K³Ok#k. �¦¬�nH�~¡}ì#O Ç~¡O U~¡æ_�# JOä�½~¡KÇó� Î=ò (3Z<£) Ç~�Þ Ç Ç~�xH÷ K³Ok#k.

15. Option (4) is correct: Sugarcane is propagated vegetatively by setts and water hyacinth is mostly

propagated by offsets - hence these plants donot show variations, therfore II and IV are correct as they

are monocarpic and shows variation in their progeny.

K³�ä�½ "³òH�ø, =°i�Çò SHêiß�Ç¶ `Ç~¡KÇ°Qê �§v�Ç° ã�¬`Ç°¼`Çæuë [~¡°�¬ô@KÍ g\÷ �¬O`Çu�Õ "³áq^�¥¼�° K��ì`Çä�½ø=.ãª��a�ìO^�Î�¹ =°i�Çò "³^Î°~¡° "³òH�ø�ÕÁ �¬ô+²æOKÇ°@, �ÿáOyH� ã�¬`Ç°¼`Çæuë ¥Þ~� �¬O`�#O U~¡æ_È°@KÍ g\÷�Õ "³áq^�¥¼�°H�xæOKÇ°#°.

16. Option (4) is correct: As the function of these cell organells are correctly matched

*ìa`Ç I �Õ W=Þ|_�# H�}ìOQê�ä�½ �¬O��OkOz *ìa Ç II �Õ W=Þ|_�# q �Î°�° Ð P�¬Â<£ Ð 4 ^¥Þ~� �¬iQêQ®°iëOKÇ|_�#q.

17. Option 3 is correct: As the characters suggested in list -I correctly matched to the members of respectivefamilies.

P�¬Â<£ Ð 3 �¬i�³Ø°#k, nx ã�¬Hê~¡O *ìa`Ç I �Õx �H�Æ}ì�°, *ìa`Ç II �Õx ä�½@°O�ì��Õx "³òH�ø�`Ë �¬O|O �Î=òH�e¾#q.

18. Option (2) is correct: (A)Onion (15) ------(D) Mustard (16) ------- (C)Kamanchi (17) ------ (B)Sun

hemp(21).

P�¬Â<£ Ð 2 �¬i�³Ø°#k. nx ã�¬Hê~¡O Ð h~¡°eÁ (A) 15 Ð P= (D) 16 Ð Hê=°Oz (C) 17 Ð [#°=ò (B) Ð 21�¬ô+¬æã`�� ¬OY¼Ì�á P^¥~¡�¬_� P~Ë�¬ì} ãH�=°O�ç J=°~¡ó|_�#q. A, D, C, B

19. Option (2) is correct: As characters in list - I, correctly matched to the examples.

P�¬Â<£ Ð 2 �¬i�³Ø°#k. nx ã�¬Hê~¡O *ìa`Ç I �Õx �H�Æ}ì�° *ìa`Ç I I �Õx "³òH�ø�`Ë �¬i�ée�Çò#ßq.

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Page.No : 36

20. Option - 2 is correct. D - 3 cohorts, A - 4 cohorts, C - 5 cohorts, B - 6 cohorts. The arrangement as per

ascending order D, A, C, B.

P�¬Â<£ Ð 2 �¬i�³Ø°#k. ZO^Î°=�# J#Qê (_�) Ð 3 HË�¬ð~��, (Z) Ð 4 HË�¬ð~��, (�²) Ð 5 HË�¬ð~�� (a) Ð 6 HË�¬ð~��.nx ã�¬Hê~¡O D, A, C, B P~Ë�¬ì} ãH�=°O�ç J=°~¡ó|_�#q.

21. Option (3) is correct: Opening of photoactive stomata

(B) Efflux of H+ (active) -------- (E) Influx of K+------- (G) In flux of Cl------ (D) Turgidity of guardcells---Stoma opens.

�¦é\ÕUH÷�"£ �¬ã`Ç~¡Oã^�¥�° `³~¡KÇ°Hù<Í ãH�=°=ò P�¬Â<£ Ð 3 B Ð E Ð G Ð D. �¬ï~á#ãH�=°=ò.

22. Option (1) is correct: During regeneration phase, two reactions are catalysed by Aldolase and two

reactions are catalysed by Transketolase.

HêeÞ<£ =��Ç°O�Õx �¬ô#~¡° Çæuë Î�×�Õ, P�ÕÛ�è*� ï~O_È°KÇ~¡¼�#°, ã\ì<£�H©\Õ�è*� ï~O_È° KÇ~¡¼�#° L Íæ~¡}O KÍ�Çò@KÍ,g\÷ =°^�Î¼ x+¬æuë 1 1 (P�¬Â<£ Ð 1)

23. Option (4) is correct: In PSI 700P and in PSII 680P are reaction centres.

PS I �Õ 700P =°i�Çò PS II �Õ 680P J<Í KÇ~�¼öHOã Î=ò��Õ �¬ã`��¬ìi Ç Ð Z, J}°=ô�°Qê L#ßq. P�¬Â<£ Ð 4,Ð �¬i�³Ø°#k.

24. Option (4) is correct: (A) is D - ketoglutaric acid

(B) is glutamic acid produced from (A) and Glutamine (C) is an amide formed from glutamic acid.

(D) is succinyl - Co A produced from (A) by oxidative decarboxylation.

P�¬Â<£ Ð 4 �¬i�³Ø°#k. (Z) D - H©\ÕQ®°Á\ìiH± P=°ÁO (a) J#°#k (Z) #°O_� U~¡æ_�# Q®°Á\ìq°H± P=°ÁO (�²) Q®°Á\ìq°<£J#°#k (a) #°O_� U~¡æ_�#k. (_�) �¬H÷�<³á�� HËZ J#°#k PH÷�_Í\÷�"£_�Hê~ËÄH÷��è+¬<£ ãH÷�Ç°^¥Þ~� (Z) #°O_� U~¡æ_È°#°.

25. Option (1) is correct: Showing I and II are correct with respect to their functions and scientists.

P�¬Â<£ Ð 1 . WO Î°�Õ =°i�Çò =$H�Æ�¬ì~Ëà<£ ~¡Hê�°, �§ã�¬ë"Í ëÇ�°, "�\÷ ãH÷�Ç°�ä�½ �¬O|Ok�z# �¬i�³Ø°# "Í°�×qO�¬ô�#°KÇ¶�¬ôKÇ°#ßq.

26. Option (4) is correct: H.Khorana is correct.

P�¬Â<£ Ð 4 Ì�ìK�.MÕ~�<�.27. Option (1) correct: (A) is correct : (R) explains (A) because secondary succession is faster because

there is already available soil for promoting plant growth.

P�¬Â<£ Ð 1 �¬i�³Ø°#k. �Î$_È"�¼Y¼ (Z) �¬i�³Ø°#k. Hê~¡}=ò (P~�) �Î$_È"�¼Y¼ä�½ �¬ï~á# q=~¡}. ZO Î°H�#Qê =°$uëH�L Çæuë H�e¾# P=~¡} ã�¬ Í�×O�Õ kÞf�Ç° J#°ãH�=°=ò "ÍQ®=O ÇOQê [~¡°Q®°#°.

28. Option (4) is correct: As per text, codine is alkoloid and is secondary metabolite

P�¬Â<£ Ð 4 �¬i�³Ø°#k. HË_�<£ J#°#k kÞf�Ç° r=ãH÷�Ç° L Çæ#ßO =°i�Çò P�ø�ì~ò_£.

29. Option (2) is correct: Nickel is the essential micro mineral element which induce disease resistance insome plants.

P�¬Â<£ Ð 2: xïH�� J#° �¬¶H�Æà P=�×¼H� Yx[ �é+¬H�=ò. Wk Hùxß "³ò�H�ø�Õ "�¼k� x~Ë �ÎH� ÇÞ=ò#° H��°Q®KÍ�Çò#°.30. Option (3) is correct: The primary acceptor of 2CO is PEP, which is produced in mesophyll cells, but

not in bundle sheath cells. Therefore option (3) is incorrect

P�¬Â<£ Ð 3: �¬ã`�O Ç~¡ H�}ì��Õ �¦�ªéæ�D<�� Ì�á~¡¶À�\� 2CO #° ã��^�Îq°H� �ÔÞH�~¡ë¡ëQê �¬xKÍ�Çò#°. JO`ÍHêx�¬ôOA�¬ô`ù_È°Q®° H�}ì��Õ Hê^Î°.

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Page.No : 37

31. Option (2) is correct: Pure proteins and Pure Fats never used as respiratory substrates.

Proteins before participating, degraded into aminoacids and involve in respiration. Fats also brokendown

into fattyacid and glycerol and acts as respiratory substrates. Therfore (A) is correct.

For complete oxidation of proteins more amount of oxygen is required. So utized 2O number is more

than released 2CO . So RQ value is less than one. But R is not correct explanation to A.

P�¬Â<£ Ð 2: (Z) �×Ã^Î�"³°Ø# ã�é\÷<£�° �§Þ�¬ãH÷�Ç° J^Î�¬Ö�¬^¥~¡ÖOQê L�¬�³¶Q®�¬_È=ô. ã�é\÷<£�°, P"³°Ø<ËP=¶Á�°Qêq�õÁ+¬}O K³Ok �§Þ�¬ãH÷�Ç°�Õ L�¬�³¶Q®�¬_È°#°.Hê=ô# (Z) �¬i�³Ø°#k.

(J~�) ã�Ú\©<£�° �¬îiëQê PH©�H�~¡}O K³O^Î°@ä�½ Zä�½ø= PH÷�[<£ J=�¬~¡O. Hê|\÷� q_È°^Î� J�Í°¼ 2CO �H�<�ß qx�³¶yOK³ 2O � �¬OY¼ Zä�½ø=. JO Î°=�# q�°= XH�\÷ H�<�ß Çä�½ø=. Hê=ô# (P~�) �¬i�³Ø°#k J~ò#�¬æ\÷H÷(Z) ä�½ q=~¡} Hê Î°.

32. Option (2) is correct

A question related to �Monohybrid hybridization� experiments

P�¬Â<£ Ð 2: UH� �¬OH�~¡ �¬OH�~¡} ã�¬�³¶Qê�ä�½ �¬O|Ok�Oz#k.

33. Option (2) is correct: A and C are correct as per �Z� scheme, flowchart.

P�¬Â<£ Ð 2: Ñ[_£Ñ �Ôø�O ã�¬Hê~¡=ò Z =°i�Çò �² �¬i�³Ø°#k.

34. Option (4) is correct: One 2CO involve in each turn of Calvin cycle, therefore it requires 2NADPH+

H+. In each turn of kerb�s cycle 3NADH+H+are produced.

P�¬Â<£ Ð 4: XH� 2CO J}°=ô ïHeÞ<£ =��Ç°O�Õ ��ç¾#ß�³ò_È� 2NADPH+ H+. qx�³¶yO�¬¬_È°#°. ãïH��=��Ç°OXH�øª�i [iy#KË 3NADH+H+ L`Çæuë JQ®°#°.

35. Option 4 is incorrect.

P�¬Â<£ Ð 4: �¬i�³Ø°#k Hê^Î°.

36. Option (1) is correct:

P�¬Â<£ Ð 4: �¬i�³Ø°#k

(A)- Three hiearchial levels in ICTV. classification

(A) "³á~¡�¹ =s¾H�~¡} q �¥#O�Õ Ð3 ª�Ö~ò�°H��=ô

(B) - one, (C) - six, (D)- four, (E)- two

aÐXH�\÷ �² Ð P~¡° : _� Ð <��°Q®° W Ð ï~O_È°

aW Z_��² P~Ë�¬ì}ìãH�=°=ò

37. Option (3) is correct: B and D are Incorrect see the 322BRP

P�¬Â<£ (3) : WO^Î°�Õ a =°i�Çò _� �¬i�³Ø°#q Hê=ô. 322BRP �¬ijeOKÇO_�

38. Option (4) is correct: A is mutated to U. Glutamine is replaced by Valine, hence Sickle cell anaemia is

resulted.

P�¬Â<£ 2 : A P#°#k U Qê L Çæi=~¡ë# K³O Î°@KÍ Q®°Á\ìq°<£ä�½ | Î°�°, P ��eÌ�>ÿØæ_£�Õ "��ÿá<£ KÍ~¡°@KÍ Ð ÑHË_È=eH�} Ð ~¡H�ë�Ôì#`Ç "�Þk�H÷ Hê~¡}�"³°Ø#k.

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Page.No : 38

39. Option (4) is correct:

P�¬Â<£ Ð 4 : �¬i�³Ø°#k. 3 : 6 : 3 : 1 : 2 : 1

: : : : :

AABB AABb AAbb aaBB aaBb aabb

AaBB AABb Aabb aaBb

AABB AaBb Aabb

AaBb

AaBb

AaBb

40. Option (1)is correct: As pre the experiment, in the first generation all are intermediate types. In the

secound generation 50% light (N14) and 50% (N14 & N15) intermediat types are formed but no 15N(heavy) types

P�¬Â<£ Ð 1 : "³ò^Î\÷`Ç~¡O�Õ Jhß =°^�Î¼~¡HêxH÷ (N14 & N15) K³Ok#q. ï~O_Ë= Ç~¡O�Õ 50% ((N14 & N14)

`ÍeH�~¡HêxH÷, 50% (N14 & N15) =° �Î¼~¡HêxH÷ K³Ok#q. (N15 & N15) ��ì~¡~¡H�O U~¡æ_È�è^Î°.

ZOOLOGY

41. Only cephalochordates are without heart among chordates and agnatha members of vertebrates with-

out paired appendages.

Ì��¦¬�Õ Hêö~�\ì r=ô�° =¶ã Ç"Í° Q®°O_³ �èx Hêö~�\ì�°. J�¬ì#°�¬H��õ~¡°Hê��Õ ÎÞO ÎÞ L��OQê�° =ôO_È=ô.

42. Evolution is due to adaptation

�¬i�¬~��ä�½ J#°ä��<��° U~¡æ_È@O =�# �¬i}ì=°O [~¡°Q®° Ç°Ok.

43. Metazoan cells are not capable of independent existance

"³°\ì*Õ"�<£ H�}ì�° �¬ÞO ÇOã ÇOQê LO_È�è=ô.

44. Because of keratin they anchor cells closely.

ïH~�\÷<£ =�# ÎQ®¾~¡Qê |Ok�Oz LO\ì~ò.

45. abdominal ganglia are 6 and retinulae are 7 each ammatidium

ãH÷Ok L Î~¡ <�_� �¬O �Î°�° 6, <Íã Ç�¬@� Hê}ì�° 7 LO\ì~~ò.

46. A) Before brain B) After the brain

"³ò Î_È°ä�½ =òO Î° "³ò Î_È° Ç~�Þ Ç LO@°Ok

C) Over gizzard D) Over crop

JO Ç~¡ [~�¡~¡OÌ�á J<�ß�×�Ç°OÌ�á

47. If 25% of GPP is 150, then 150 100

100 60025

s

� � then 75% of GPP = 450 NSP of secondar carni-

vores 450

0.451000

� �

GPP �Õ 25% 150 kJ J~ò#@Á~ò`Í, 150 100600

25GPP

s

� � . JO Î°�Õ 75% 450kJ. ^¥x=�Á kÞf�Ç°

=¶Oª��¬ð~¡°��Õ 4000.45

1000NSP � kJ

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Page.No : 39

48. Milk set dental formula is 2102

2102, so last molars and pre molars formed only once.

�� ÎO`��Õ ÎO Ç q<�¼�¬=ò2102

2102. z=i KÇ~¡¼}Hê�°, �¬î~¡ÞKÇ~¡¼}Hê�° XH�ª�i =¶ã Ç"Í° U~¡æ_È �~ò.

49. There 3 are unpaired, others are paired

D 3 J ÎÞO ÎÞ =°$ Î°�ì�¬°Ö�° q°ye#q ÎÞO ÎÞ =°$ Î°�ì�¬°Ö�°50. There is no synaptic delay in electrical transmission.

q Î°¼ �<�_� �¬Ok��Õ P��¬¼O LO_È Î°.51. Dendritic cells, macrophages, B- lymphocytes act as antigen presenting cells.

_³Oã_³\©H± H�}ì�°, �¬¶Ö�|H�ÆHê�°, B- eO�¦éÌ�á\� H�}ì�° ã�¬u[#H� ã�¬ Î~¡ÅH� H�}ì�°Qê �¬xKÍª�ë~ò.52. Mesomere forms execretory and repraductive system parts.

g°ªéq°�Ç°~� H�}=ò q�¬~¡ûH�, ã�¬`Ç°¼`Çæuë =¼=�¬Ö ��ìQê�#° U~¡æ_È°�¬°ëOk.

53. Recessive genotypic frequency is 0.09 (=36). Total organisms 36

4000.09

� � .

400 - 36 = 3364 dominant.

JO Ç~¡¾ Ç [#°¼~¡¶�¬=ò �Ï#��¬ô#¼=ò 0.09 («36) P#Qê "³ò Çë=ò r=ô� �¬OY¼ 36

4000.09

� �

|�²ì~¡¾ Çr=ô� �¬OY¼ 364

54. Tall T walle represents hyperkalemia, PR interval prolonged can lead to bradycardia

Z`³á# T Ð Ç~¡OQ®O Jk�H� �é\ì+²�Ç°O#°, PR =¼=k� n~¡É=°~ò# ã�ì_� HêiÛ�Ç¶#° �¬¶zOKÇ°#°.

55. Encystment occurs after development of chromatoid bar .

ãHù=°\ì~ò_£ �ì~��° U~¡æ_�# Ç~¡°"�`Ç HËjH�~¡} [~¡°Q®°`Ç°Ok.

56. Wucheraria male has 2 unequal copulatory spicules.

L=°Q®Ì�¦á�è~¡�Ç°�� ¬ô~¡°Q®°�Õ 2 J�¬=¶# �¬O�¬~¡ø H�O@Hê�°O\ì~ò.

57. This character found in osteichthys.

D �H�Æ}=ò P�²ÖH÷Ö�¹#O Î° LO@°Ok.

58. Neometra belong to crinoidea

x�³¶"³°ã\ì ãH÷<�~ò_��Ç¶H÷ K³Ok#k.

59. Sarguja & bastar are found in chattisgarh

�¬~¡°¾[ |�¬ë~� Ð J#°#q KÇfë<£Q®_£ä�½ K³Ok#q.

60. Pharynx, oesophagus have non keratinised squamous epithelium.

ãQ®�¬x, P�¬ð~¡"��²ìH��° ïH~�\÷<£ ~¡�²ì`Ç �¬Öi`Ç �×�ø� L�¬H��× =ôO@°Ok.

61. Both olfactory and gustatory receptors found on palps. Antenna donot have gustatory

receptors.

~¡°z, ã�¦¬¶} ãQê�¬ìHê�° ï~O_È° �¬æ~��OQê�Ì�á LO\ì~ò. �¬æ~¡æ�×$OQê�Ì�á ~¡°zãQê�¬ìHê�°O_È=ô.

62. 2CO , methane, 2NO are green house gases

2CO , g°^�Í< £, 2NO , J#°#q �¬ìi ÇQ®$�¬ì Hê�°+¬¼Hê�°

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Page.No : 40

63. Adrenaline, Noradrenaline have similar effects.

Jã_�<�e<£, <�~�Jã_�<�e<£ XöH q^�Î"³°Ø# ã�¬��ì"��°O\ì~ò.

64. HT cells have 4CD molecules

HT H�}ì�ä�½ 4CD =¶~¡ø~��#° H�ey LO\ì~ò.

65. Biston betularia is an example for industrial melanism.

a�¬�<£ a@°¼�èi�Ç¶ J#°#k ��~¡ã�§q°H� �§¼=°�ìxH÷ L^¥�¬ì~¡}

66. Migratory fishes, brackish water fishes have to tolerate fluctuations in salinity so they are curyhaline.

=��¬ KÍ�¬�°, L�¬ôæ h\÷ H��Ç°¼��° x=iOKÍ KÍ�¬�° Zä�½ø= =¶~¡°æ�#° Ç@°�HËQ®ey =ôO_¨e.

67. 'O' group is pssible in homozygous so she has to get allele for 'O' from both parents.

'O' ~¡H�ë =~¡¾=ò JO`Ç~¡¾`Ç=òD H�#°H� J"³° `ÇeÁ^ÎOã_È°�° W^Îíi #°Ok P �H�Æ}ìxß �¬OãH�q°OKÇ°HË"�e.

68. maximum reabsoption of substances occur in PCT

Jk�H� ��ìQ®O �¬ ¥~�Ö�° �Õ+¬}K³O Ík �¬g°�¬ �¬O=o Ç <��×O�Õ.

69. Saheli was developed by CDRI Lucknow.

�¬À�ìe J#°#q CDRI �HùßKÍ Ja�=$kí KÍ�Ç°|_�#k.

70. It is a autosomal recessive disease due to trisomy of 13 chromosome.

13= ãHË"³¶ªé"£°�° ã>ÿØªéq°� =�Á U~¡æ_Í JO Ç~¡¾ Ç �Õ�¬O.

71. Sardines,salmons have omega 3 fattyacids.

ª�ï~åÛ<£�°, ª��ìà<£��Õ X"³°Qê 3 �¦�\÷ P=¶Á�° =ôO\ì~ò.72. (A) 0.5 (B) 0.33 (C) 1.5 (D) 0.67 (E) 1.00

73. Crcumvallate are seen near base.

�¬iø"£° "��è\� <��°H� P^�¥~¡ ��ìQ®O�Õ LO\ì~ò.

74. It is external seperation of atria and ventricles.

Jk �ì�¬ì¼OQê "Í~¡° KÍ�¬°ëOk.

75. It reduces blood pressure.

Jk�H� ~¡H�ë �Ô_È<�xß `Çy¾�¬°ëOk.

76. Antennae are absent in chelicerata.

ïHeÌ�s\ì�Õ �¬æ~¡Å �×$OQê�° LO_È=ô.

77. 1, 2, 3 - belong to sympathetic system (thoracolumbar outflow)

1, 2, 3 J#°#q �¬�¬ð#°��¼`Ç <�_� =¼=�¬�ä�½ (L~¡� Ð H�\÷ J=ô\� �¦éÁ)

78. Scala media has endolymph.

Ì�ø�ì g°_��Ç¶�° ZO_ËeO�¦¹ =ôO@°Ok.

79. These are periods of paleozoic in a sequence

Wq �¬ô~�r= =°�¬ì�ÇòQ®O�Õx �×Hê�° �¬ï~á# =~¡°�¬�Õ.

80. New blood vessel growth generation seen in tumors.

Hù Çë ~¡H�ë<��ì�#° ãÀ�ö~�²OKÍq @¶¼=°~��°.

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Page.No : 41

PHYSICS

81. � Q Q Q � �� #

$ . �% / . 6 � /. 6Z $

� � �

� � � �

� �

#% # % �� /. 6 /6

$ $$ .

� � �

�

u

82. Component of aG ( a

G �³òH�ø JO�×=ò)X�C � � �

�X � � �

� �

� �

GG

G

83.

A(V)

h

U

h

B(2V)

AO � �7 8 �IJ�

OB � ��8 7 �IJ� ��7 ��IJ�

��IJ7

�

84. ��Z 7EQU �V� [ 7UKP �V IV

� T T �

� � � ��

7UKP ��* ��O

�I � �

T

u

85.�

��K- JJG G

Q �

��EQU�� K ��UKP�� L- �

JJG G G

� ��

� �X X ��UKP�� L �� K � EQU��'- � � �

JJG G G

G G

�

Q Q ��

�� UKP�� EQU�� L �� K �UKP�

'- u � uJJG G G

� � �� L �� K �

� � �� u u � u u

G G

� ��

�� L �K��

'- �G G G

�� O �U

��'- u

� �( O� �� 0

�V �

��

'- u u

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Page.No : 42

86. � �OI UKP EQU (T �P T

Q �

Q Q

UKP �� EQU�� (

ZUKP �� EQU��

� u�

� u

�� (Z �(

�� Z� �

87. ��OIJ MZ

�

� � � ��

�� I �� M ��

u u � u

� � � �� I J �� M ��

�u u � u

� ��

J �� O� J ��

� � u �

J ��O

88. � ��

OX OW GW W G ��

u � u �

�J � G � �� 6 ��U

I � G ��

� u �§ · § · ¨ ¸ ¨ ¸� �© ¹ © ¹

89.� � � �

� �

O 7 O 78

O O

�

�

� �� 8

� �

u � u

�

8 �O � U

90. Angular momentum is constant when torgque is zero T (W uG

GG

so TG and (

G

are parallel

C �C �

� � �

91.�

��u oscillation =

��#

�u

�# ##

� � �

T/4

O

2

A A

6

T

6 6 �6V

� � ��

92.�� )/O

- OX� 4

§ · ¨ ¸

© ¹and

�� )/O )/O )/O

�� 4 4 4 J

� �§ ·u � ¨ ¸ �© ¹

� )/O )/O �4 44 J J

� 4 4 J � �

��

�

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Page.No : 43

93. Since �C� displaced by 0.3mm, elongation in AC=0.3mm Then �D� also displaces by 0.3mm Butsince displacement if �D� is 0.5mm, i.e. elongation in CD=0.2mm

A

C

D

B

#% #%

%& %&

N NN N

N N

'' D �

'

#%

#%

N��N ��EO

��

94. � �� 66 �N OI 6 �N T �N�F I T

FIu � u S �

S

Diameter = �6�T

FI

S

independent of length

95. Formulae

96.

A

1T 2T 3T

B

Q

� �� %T � T

Rate of flow of heat is same

� � � �� M�# M#

N N

T � T T � T

� � � �� T � T T � T

� �� T � T T � T �

Q

� �� %T � T

97.� � �

� � � �

2 O 6 2 � ��2X OT6

22 O 6 O ��

�

� u � u§ ·¨ ¸© ¹

2 4m gm�

Mass of gas escaped = 6-4=2gm

98. � � Q

�O 5 V O. � � V � � �� V �� %' � u u � u �

99. ��

�

8 ��8 6 8 ��O

�� D � �

� �FY 2�FX �� , u

F7 F3 FY �� , � �

100. Volume = �OCUU �

O �FGPUKV[ �

Energy = � 2X�

= �� ,u

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Page.No : 44

101. Let �n�be the frequency of tuning fork. For stretched wire, frequency P � ��

6P ��

�D �

P � �P ��*\

P ��

� �

102.U

8G

and �

8G

are perpendicular to each other so no doppler effect i.e. apparent frequency = true fre-

quency

103. � �## #

� � ��

4H 4

�

ª º« »

P � �« »« »« »¬ ¼

� �$ $ $

� � ��

H �4 4

ª º P � �« »

¬ ¼

#$

#

$

�

4H

H �

�4

§ ·¨ ¸© ¹� § ·¨ ¸© ¹

$

#

�4� �

� 4 �� u #

$

4 �

4 ��

104. Focal length H� �DO and XKQNGV 4GF

O � O

so X 4H H!

105. Reflected and refracted are perpendicular to each other

106.��

�

3 �3� �OX

� F �

S�

� ��

� �

�

�3 �3� �O ��X

� Z �

S�

Z � FZ

�F � ��

107.� ��

8 ��8� �

u � u

�and � � ��

' � ��

� ª º' u u �¬ ¼

��O,

108.

�

Z [� Z�[ �

��

�

�

�[�

�� [ ��[

��

�

� :

� and Z � :

109. Applying kirchoff�s second law, 8 �� and ��

K �#� � � �

� � �

� �8 �� 8�8 � � � �8 � u � u

� �8 � � � ��8�8 � � � �8 � u � u

110. R

R

6� O �O 36 � 6 �6

$3 6 O �3

D

D

S � u �

111.) ��

5P � ��

:� �

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Page.No : 45

112. At null point,*

$ $ � Then J

+6 �

O �$ S

u

�

*

+6 �

O $ S

u

�6 � 6�

113. � ��

# $8 � � �� 8 �� u � � u � �

$ #8 8 ��8�

114.� �� -% ��

H - �� % �� .%

� � S

115.

�

� �

�

J�

OX

�

-OO � �- -

O -§ ·¨ ¸© ¹

�

�

�' 9 O

� � -

�

�

��' 9 � O

� � u -

' Y � �Y'

�' Y � �

��

�

116. ��

G G GR �O ' � � �� u u u u u and

� ��

Z �

' �� 2

E � ��

�u u

u

G

Z

2 ��

2 ��

117. Principle of nuclear reactor

118. � � �

*' \ ��G8' u

�� \ �� \ �u �

119. % ' $ uG G G

120. Silicon is preferred to Germanium

CHEMISTRY

121. � �1 �* 1 � <P

� � � � � � �

�

%* % %* %* %* %* %1%* %* %* %*1^%* $ #

� � � ��o �

122. Hyper conjugation explains

123. Schrodinger's wave equation

124. a) As Z increases the energy of 2s electron decreases

b) In H�atom energy depends only on 'n'

c) For multi electron system energy depends both and n and A

d) For any s orbital number of nodal planes = 0

125.� � � � � �

/I1 # 1 0 1 % 1� � �A A should be the correct order or acidic nature

126. Moulds are viable particulates

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Page.No : 46

127.

/G

5K

/G

%N %Nhydrolysis¶¶¶¶l

/G

5K

/G

*1 1*polymerisation¶¶¶¶¶¶l

/G

5K

/G

1 1

128. Boric acid is an electron pair acceptor but not proton donor

129. Magnesium � magnesium hydride storage tank is used for hydrogen

130. pH of > @� C

�%* %11* R- QI%

� � A

= � ��

�ª º� �¬ ¼ = 4

After addition it will be a buffer solution

��R* � QI

�� A = 6

Change in pH = 6 � 4 = 2

131. The system moves in the forward direction. Therefore � � �

(G 5%0�

ª º¬ ¼ increases

Hence intensity of red colour increases

132. Least sum rule is to be followed

133. Androus CuSO4 is colourless, because d�d splitting does not takes place

134. � �� * 1

*% ��* 1 *%� ��oA A * ��' � KJ/mole ......... (b)

� ��* 1

*% *% ��* 1��o �A A * ��' KJ/mole ......... (� a)

(b � a) corresponds to enthalpy of dilution, � � � �� * 1 ��* 1*% *%��oA A *' �3.02 KJ/mol

135. � � � � � �ITCRJKVG � I � I% 1 1� ��o

�

H* *' ' of CO2; But

2 4* * *' �

�

H*' of CO2 = HP � HR =

� � �%1 ITCRJKVG 1 %1

* * * *ª º� � ¬ ¼

136. Decomposing (NH4)2 BeF4, we get BeF2.

137. More extent of intermolecular forces

138. N3H exist as

N

NN H with oxidation number of N as 0, 0 and �1.

139. Diphenylamine indicator and dark blue colour indication of end point

140. Lateral overlapping

141. Nitrogen has highest value of KH

142. 11% He and 32% Oxygen

143. # $o reaction proceeds in second order kinetics

�� V

Cv

10Ml 5M in 20 min and 5 Ml 2.5M in 40 minTherefore, 10Ml 2.5M in 60 min

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Page.No : 47

144. � � �

� � � � ��* 21 �* 21 2*

� � �

��o �

Number of electrons transferred = 6

Equivalent weight = �/ � ��

��

u

145. �

�

�- <

��

�

�- <

��

� � �� < < <

4 ��

Therefore, resistance 4 �� :

When the volume is doubled the concentration is reduced to one�half

The resultant resistance 4 �� :

146.� � � � �

�(G5 ��1 �(G 1 �51� o �

11q32g of O2l

4q120g of FeS2

8g of O2 =� � ��

��

u u

u

147. Straight line with a positive slope and positive intercept

148. Pulp is usually bleached using chlorine dioxide

149. Oxide ReO3 behaves like metallic copper

150. The correct order of reduction potential (3+ to 2+state) is Cr < Ti < V

151. NiCH2O)63+ is green. Ni(H2O)4(en)3+ is blue

152. Amide linkage is present in polyamide polymer. Dacron and Melamine are not amide polymers

153. Given chemical process occur both in roasting and bessimerisation

154. Leucine, isoleucine and valine are neutral essential amino acids.

They do not have an additional � NH2 or � COOH group. Hence they are neutral

155. Glycerol prevents rapid drying

156. 30 carbocation is mole stable. Hence it has highest SN

1 reactivity

157. Based on Ka values

158.

NH2

HCl Nitration

3NH� 3NH�

NO2

Reduction

3NH�

NH2

159. Side chain does not contain benzylic hydrogen as �CH� attached to benzene nucleus

160. FKN�0C1*

� � �

�

*^ � � � �

�%* %*1 %* % 1* %* %* %* %*1^%*^%*1

%¶¶¶¶l � � ¶¶l � � �

1, 2 and 3rd carbon atoms are sp2 hybridised

III

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