NEET 2020 Question Paper (Hindi)

¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê ‚¥∑§Ã No. :
Test Booklet Code KANHA ß‚ ¬ÈÁSÃ∑§Ê ◊¥ 44 ¬Îc∆ „Ò¥–
This Booklet contains 44 pages.

E1
Hindi+English

ß‚ ¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê ∑§Ê Ã’ Ã∑§ Ÿ πÊ‹¥ ¡’ Ã∑§ ∑§„Ê Ÿ ¡Ê∞–
Do not open this Test Booklet until you are asked to do so.
ß‚ ¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê ∑§ Á¬¿‹ •Êfl⁄áÊ ¬⁄ ÁŒ∞ ÁŸŒ¸‡ÊÊ¥ ∑§Ê äÿÊŸ ‚ ¬…∏¥–
Read carefully the Instructions on the Back Cover of this Test Booklet.
◊„àfl¬Íáʸ ÁŸŒ¸‡Ê — Important Instructions :
1. The Answer Sheet is inside this Test Booklet. When
1. ©ûÊ⁄ ¬òÊ ß‚ ¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê ∑§ •ãŒ⁄ ⁄πÊ „Ò– ¡’ •Ê¬∑§Ê ¬⁄ˡÊÊ
you are directed to open the Test Booklet, take out the
¬ÈÁSÃ∑§Ê πÊ‹Ÿ ∑§Ê ∑§„Ê ¡Ê∞, ÃÊ ©ûÊ⁄ ¬òÊ ÁŸ∑§Ê‹ ∑§⁄U äÿÊŸ¬Ífl¸∑§ Answer Sheet and fill in the particulars on
¬Îc∆-1 ∞fl¥ ¬Îc∆-2 ¬⁄ ∑§fl‹ ŸË‹ / ∑§Ê‹ ’ÊÚ‹ ¬ÊÚߥ≈ ¬Ÿ ‚ side-1 and side-2 carefully with blue/black ball point
Áflfl⁄áÊ ÷⁄¥– pen only.
2. The test is of 3 hours duration and Test Booklet contains
2. ¬⁄ˡÊÊ ∑§Ë •flÁœ 3 ÉÊ¥≈ „Ò ∞fl¥ ¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê ◊¥ 180 ¬˝‡Ÿ „Ò¥– 180 questions. Each question carries 4 marks. For each
¬˝àÿ∑§ ¬˝‡Ÿ 4 •¥∑§ ∑§Ê „Ò– ¬˝àÿ∑§ ‚„Ë ©ûÊ⁄ ∑§ Á‹∞ ¬⁄UˡÊÊÕ˸ ∑§Ê correct response, the candidate will get 4 marks. For
4 •¥∑§ ÁŒ∞ ¡Ê∞¥ª– ¬˝àÿ∑§ ª‹Ã ©ûÊ⁄ ∑§ Á‹∞ ∑ȧ‹ ÿÊª ◊¥ ‚ each incorrect response, one mark will be deducted
from the total scores. The maximum marks are 720.
∞∑§ •¥∑§ ÉÊ≈ÊÿÊ ¡Ê∞ªÊ– •Áœ∑§Ã◊ •¥∑§ 720 „Ò¥–
3. Use Blue/Black Ball Point Pen only for writing
3. ß‚ ¬Îc∆ ¬⁄ Áflfl⁄áÊ •¥Á∑§Ã ∑§⁄Ÿ ∞fl¥ ©ûÊ⁄ ¬òÊ ¬⁄ ÁŸ‡ÊÊŸ ‹ªÊŸ ∑§ particulars on this page/marking responses.
Á‹∞ ∑§fl‹ ŸË‹ / ∑§Ê‹ ’ÊÚ‹ ¬ÊÚß≈¥ ¬Ÿ ∑§Ê ¬˝ÿÊª ∑§⁄–¥ 4. Rough work is to be done on the space provided for
4. ⁄»§ ∑§Êÿ¸ ß‚ ¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê ◊¥ ÁŸœÊ¸Á⁄à SÕÊŸ ¬⁄ „Ë ∑§⁄¥– this purpose in the Test Booklet only.
5. On completion of the test, the candidate must hand
5. ¬⁄ˡÊÊ ‚ê¬ÛÊ „ÊŸ ¬⁄, ¬⁄ˡÊÊÕ˸ ∑§ˇÊ / „ÊÚ‹ ¿Ê«∏Ÿ ‚ ¬Ífl¸ ©ûÊ⁄
over the Answer Sheet to the invigilator before leaving
¬òÊ ∑§ˇÊ ÁŸ⁄ˡÊ∑§ ∑§Ê •fl‡ÿ ‚ÊÒ¥¬ Œ¥– ¬⁄ˡÊÊÕ˸ •¬Ÿ ‚ÊÕ the Room/Hall. The candidates are allowed to take
¬˝‡Ÿ ¬ÈÁSÃ∑§Ê ∑§Ê ‹ ¡Ê ‚∑§Ã „Ò¥– away this Test Booklet with them.
6. ß‚ ¬ÈÁSÃ∑§Ê ∑§Ê ‚¥∑§Ã „Ò E1– ÿ„ ‚ÈÁŸÁ‡øà ∑§⁄ ‹¥ Á∑§ ß‚ 6. The CODE for this Booklet is E1. Make sure that the
CODE printed on Side-2 of the Answer Sheet is the
¬ÈÁSÃ∑§Ê ∑§Ê ‚¥∑§Ã, ©ûÊ⁄ ¬òÊ ∑§ ¬Îc∆-2 ¬⁄ ¿¬ ‚¥∑§Ã ‚ Á◊‹ÃÊ same as that on this Test Booklet. In case of discrepancy,
„Ò– •ª⁄ ÿ„ Á÷ÛÊ „Ê ÃÊ ¬⁄ˡÊÊÕ˸ ŒÍ‚⁄Ë ¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê •ÊÒ⁄ ©ûÊ⁄ the candidate should immediately report the matter to
¬òÊ ‹Ÿ ∑§ Á‹∞ ÁŸ⁄ˡÊ∑§ ∑§Ê ÃÈ⁄ãà •flªÃ ∑§⁄Ê∞¥– the Invigilator for replacement of both the Test Booklet
and the Answer Sheet.
7. ¬⁄ˡÊÊÕ˸ ‚ÈÁŸÁ‡øà ∑§⁄¥ Á∑§ ß‚ ©ûÊ⁄ ¬òÊ ∑§Ê ◊Ê«∏Ê Ÿ ¡Ê∞ ∞fl¥ ©‚
7. The candidates should ensure that the Answer Sheet is
¬⁄ ∑§Ê߸ •ãÿ ÁŸ‡ÊÊŸ Ÿ ‹ªÊ∞¥– ¬⁄ˡÊÊÕ˸ •¬ŸÊ •ŸÈ∑˝§◊Ê¥∑§ ¬˝‡Ÿ not folded. Do not make any stray marks on the Answer
¬ÈÁSÃ∑§Ê / ©ûÊ⁄ ¬òÊ ◊¥ ÁŸœÊ¸Á⁄à SÕÊŸ ∑§ •ÁÃÁ⁄Äà •ãÿòÊ ŸÊ Sheet. Do not write your Roll No. anywhere else except
Á‹π¥– in the specified space in the Test Booklet/Answer
Sheet.
8. ©ûÊ⁄ ¬òÊ ¬⁄ Á∑§‚Ë ¬˝∑§Ê⁄ ∑§ ‚¥‡ÊÊœŸ „ÃÈ √„Êß≈ $ç‹Íß« ∑§ ¬˝ÿÊª 8. Use of white fluid for correction is NOT permissible on
∑§Ë •ŸÈ◊Áà Ÿ„Ë¥ „Ò– the Answer Sheet.

¬˝‡ŸÊ¥ ∑§ •ŸÈflÊŒ ◊¥ Á∑§‚Ë •S¬c≈UÃÊ ∑§Ë ÁSÕÁà ◊¥, •¥ª˝¡Ë ‚¥S∑§⁄UáÊ ∑§Ê „Ë •¥ÁÃ◊ ◊ÊŸÊ ¡ÊÿªÊ–
In case of any ambiguity in translation of any question, English version shall be treated as final.
¬⁄ˡÊÊÕ˸ ∑§Ê ŸÊ◊ (’«∏ •ˇÊ⁄Ê¥ ◊¥) —
Name of the Candidate (in Capitals) :
•ŸÈ∑§˝ ◊Ê¥∑§ — •¥∑§Ê¥ ◊¥
Roll Number : in figures
— ‡ÊéŒÊ¥ ◊¥
: in words
¬⁄ˡÊÊ ∑§ãŒ˝ (’«∏ •ˇÊ⁄Ê¥ ◊¥) —
Centre of Examination (in Capitals) :
¬⁄ˡÊÊÕ˸ ∑§ „SÃÊˇÊ⁄ — ÁŸ⁄ˡÊ∑§ ∑§ „SÃÊˇÊ⁄ —
Candidate’s Signature : Invigilator’s Signature :
Facsimile signature stamp of
Centre Superintendent :

E1 2 Hindi+English

1. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ∞∑§ ¡Ëfl ‚¥ÅÿÊ ∑§Ê ∞∑§ ªÈáÊ Ÿ„Ë¥ „Ò? 1. Which of the following is not an attribute of a
population ?
(1) Á‹¥ª •ŸÈ¬ÊÃ
(1) Sex ratio
(2) ¡ã◊ Œ⁄U (2) Natality
(3) ◊ÎàÿÈ Œ⁄U (3) Mortality
(4) ¡ÊÁà ¬⁄US¬⁄U Á∑˝§ÿÊ (4) Species interaction

2. flÎÁh ∑§Ë ¬˝Á∑˝§ÿÊ •Áœ∑§Ã◊ Á∑§‚ ŒÊÒ⁄UÊŸ „ÊÃË „Ò? 2. The process of growth is maximum during :
(1) ‹ÊÚª ¬˝ÊflSÕÊ (1) Log phase
(2) ¬‡øÃÊ ¬˝ÊflSÕÊ (2) Lag phase
(3) Senescence
(3) ¡ËáʸÃÊ
(4) Dormancy
(4) ¬˝‚ÈÁåÃ
3. The roots that originate from the base of the stem
3. ß ∑§ •ÊœÊ⁄U ‚ ©à¬ÛÊ „ÊŸ flÊ‹Ë ¡«∏Ê¥ ∑§Ê ÄÿÊ ∑§„Ê ¡ÊÃÊ „Ò? are :
(1) ¤Ê∑§«∏Ê ¡«∏ (1) Fibrous roots
(2) ¬˝ÊÕÁ◊∑§ ¡«∏ (2) Primary roots
(3) •flSÃ¥÷ ¡«∏ (3) Prop roots
(4) Lateral roots
(4) ¬Ê‡fl¸ ¡«∏
4. Match the following diseases with the causative
4. ÁŸêŸ ⁄UÊªÊ¥ ∑§Ê ©Ÿ∑§ ¬ÒŒÊ ∑§⁄UŸ flÊ‹ ¡ËflÊ¥ ∑§ ‚ÊÕ Á◊‹ÊŸ ∑§⁄U organism and select the correct option.
‚„Ë Áfl∑§À¬ ∑§Ê øÿŸ ∑§⁄UÊ–
Column - I Column - II
SÃ¥÷ - I SÃ¥÷ - II
(a) Typhoid (i) Wuchereria
(a) ≈UÊß»§ÊÚß«U (i) flÈø⁄UÁ⁄UÿÊ
(b) Pneumonia (ii) Plasmodium
(b) ãÿÍ◊ÊÁŸÿÊ (ii) å‹Òí◊ÊÁ«Uÿ◊
(c) Filariasis (iii) Salmonella
(c) »§Êß‹Á⁄U∞Á‚‚ (iii) ‚ÊÀ◊ÊŸ‹Ê
(d) ◊‹Á⁄UÿÊ (iv) „Ë◊ÊÁ»§‹‚ (d) Malaria (iv) Haemophilus
(a) (b) (c) (d) (a) (b) (c) (d)
(1) (i) (iii) (ii) (iv) (1) (i) (iii) (ii) (iv)
(2) (iii) (iv) (i) (ii) (2) (iii) (iv) (i) (ii)
(3) (ii) (i) (iii) (iv) (3) (ii) (i) (iii) (iv)
(4) (iv) (i) (ii) (iii) (4) (iv) (i) (ii) (iii)
5. ÁŸêŸ ◊¥ Á∑§‚ Ã∑§ŸË∑§ ∑§Ë ‚„ÊÿÃÊ ‚ ∞‚Ë ÁSòÊÿÊ° ¡Ê ª÷¸œÊ⁄UáÊ 5. In which of the following techniques, the embryos
Ÿ„Ë¥ ∑§⁄U ‚∑§ÃË, ◊¥ ÷˝ÍáÊ ∑§Ê SÕÊŸÊ¥ÃÁ⁄Uà Á∑§ÿÊ ¡ÊÃÊ „Ò? are transferred to assist those females who cannot
(1) ZIFT ∞fl¥ IUT conceive ?
(1) ZIFT and IUT
(2) GIFT ∞fl¥ ZIFT
(2) GIFT and ZIFT
(3) ICSI ∞fl¥ ZIFT
(3) ICSI and ZIFT
(4) GIFT ∞fl¥ ICSI (4) GIFT and ICSI
6. ¡ËŸ ‘I’ ¡Ê ABO ⁄UÄà flª¸ ∑§Ê ÁŸÿ¥òÊáÊ ∑§⁄UÃÊ „Ò ©‚∑§ ‚¥Œ÷¸ 6. Identify the wrong statement with reference to
◊¥ ª‹Ã ∑§ÕŸ ∑§Ê ¬„øÊÁŸ∞– the gene ‘I’ that controls ABO blood groups.
(1) ¡ËŸ (I) ∑§ ÃËŸ ∞‹Ë‹ „ÊÃ „Ò¥– (1) The gene (I) has three alleles.
(2) ∞∑§ √ÿÁÄà ◊¥ ÃËŸ ◊¥ ‚ ∑§fl‹ ŒÊ ∞‹Ë‹ „Ê¥ª– (2) A person will have only two of the three
alleles.
(3) ¡’ IA ∞fl¥ IB ŒÊŸÊ¥ ß∑§_ „ÊÃ „Ò¥, ÿ ∞∑§ ¬˝∑§Ê⁄U ∑§Ë (3) When IA and IB are present together, they
‡Ê∑¸§⁄UÊ •Á÷√ÿÄà ∑§⁄UÃ „Ò¥– express same type of sugar.
(4) ‘i’ ∞‹Ë‹ ∑§Ê߸ ÷Ë ‡Ê∑¸§⁄UÊ ©à¬ÛÊ Ÿ„Ë¥ ∑§⁄UÃÊ– (4) Allele ‘i’ does not produce any sugar.

Hindi+English
3 E1
7. ÁŸêŸÁ‹Áπà ◊¥ ‚ ‚„Ë ÿÈÇ◊ ∑§Ê øÈÁŸ∞ — 7. Choose the correct pair from the following :

(1) ‹Êߪ¡ - ŒÊ «UË.∞Ÿ.∞. ∑§ (1) Ligases - Join the two DNA
•áÊÈ•Ê¥ ∑§Ê ¡Ê«∏ÃÊ „Ò molecules

(2) ¬ÊÚÁ‹◊⁄U¡ - «UË.∞Ÿ.∞. ∑§Ê πá«UÊ¥ (2) Polymerases - Break the DNA into
◊¥ ÃÊ«∏ÃÊ „Ò fragments
(3) ãÿÍÁÄ‹ÿ¡ - «UË.∞Ÿ.∞. ∑§ ŒÊ (3) Nucleases - Separate the two strands
⁄UîÊÈ∑§Ê¥ ∑§Ê ¬ÎÕ∑§ of DNA
∑§⁄UÃÊ „Ò
(4) Exonucleases - Make cuts at specific
(4) ∞Ä‚ÊãÿÍÁÄ‹ÿ¡ - «Ë.∞Ÿ.∞. ◊¥ ÁflÁ‡Êc≈U
positions within DNA
SÕÊŸÊ¥ ¬⁄U ∑§Ê≈U
‹ªÊÃÊ „Ò
8. Select the correct match.

8. ‚„Ë Á◊‹ÊŸ ∑§Ê øÿŸ ∑§⁄UÊ– (1) Haemophilia - Y linked

(1) „Ë◊Ê»§ËÁ‹ÿÊ - Y ‚¥‹ÇŸ (2) Phenylketonuria - Autosomal
dominant trait
(2) $»§ÁŸ‹∑§Ë≈UÊãÿÍÁ⁄UÿÊ - •Á‹¥ª ∑˝§Ê◊Ê‚Ê◊
(3) Sickle cell anaemia - Autosomal
¬˝÷ÊflË ‹ˇÊáÊ recessive trait,
(3) ŒÊòÊ ∑§ÊÁ‡Ê∑§Ê •⁄UÄÃÃÊ - •Á‹¥ª ∑˝§Ê◊Ê‚Ê◊ chromosome-11
•¬˝÷ÊflË ‹ˇÊáÊ, (4) Thalassemia - X linked
∑˝§Ê◊Ê‚Ê◊-11
9. Match the following columns and select the
(4) ÕÒ‹‚ËÁ◊ÿÊ - X ‚¥‹ÇŸ
correct option.

9. ÁŸêŸ SÃ¥÷Ê¥ ∑§Ê Á◊‹ÊŸ ∑§⁄U ‚„Ë Áfl∑§À¬ ∑§Ê øÿŸ ∑§⁄UÊ– Column - I Column - II

SÃ¥÷ - I SÃ¥÷ - II (a) Gregarious, polyphagous (i) Asterias
pest
(a) ÿÍÕ, ’„È„Ê⁄UË ¬Ë«U∑§ (i) ∞S≈UÁ⁄Uÿ‚
(b) Adult with radial (ii) Scorpion
(b) √ÿS∑§Ê¥ ◊¥ •⁄UËÿ ‚◊Á◊Áà ∞fl¥ (ii) Á’ë¿ÈU symmetry and larva
‹Êflʸ ◊¥ Ám¬Ê‡fl¸ ‚◊Á◊Áà with bilateral symmetry

(c) ¬ÈSà »È§å»È§‚ (iii) ≈UËŸÊå‹ÊŸÊ (c) Book lungs (iii) Ctenoplana

(d) ¡Ëfl‚¥ŒËÁåà (iv) ‹Ê∑§S≈UÊ (d) Bioluminescence (iv) Locusta

(a) (b) (c) (d) (a) (b) (c) (d)
(1) (i) (iii) (ii) (iv) (1) (i) (iii) (ii) (iv)
(2) (iv) (i) (ii) (iii) (2) (iv) (i) (ii) (iii)
(3) (iii) (ii) (i) (iv)
(3) (iii) (ii) (i) (iv)
(4) (ii) (i) (iii) (iv)
(4) (ii) (i) (iii) (iv)
10. å‹Òí◊ÊÁ«Uÿ◊ ∑§Ë ‚¥∑˝§◊∑§ •flSÕÊ ¡Ê ◊ÊŸfl ‡Ê⁄UË⁄U ◊¥ ¬˝fl‡Ê
∑§⁄UÃË „Ò, „Ò — 10. The infectious stage of Plasmodium that enters
the human body is :
(1) ¬Ê·ÊáÊÈ
(1) Trophozoites
(2) ¡ËflÊáÊÈ¡ (2) Sporozoites
(3) ◊ÊŒÊ ÿÈÇ◊∑§¡Ÿ∑§ (3) Female gametocytes
(4) Ÿ⁄U ÿÈÇ◊∑§¡Ÿ∑§ (4) Male gametocytes

E1 4 Hindi+English

11. ©Ÿ ¬ŒÊÕÊZ ∑§Ê ¬„øÊÁŸ∞, Á¡Ÿ∑§Ë ‚¥⁄UøŸÊ•Ê¥ ◊¥ ∑˝§◊‡Ê— 11. Identify the substances having glycosidic bond and
peptide bond, respectively in their structure :
Ç‹Êß∑§Ê‚ÊßÁ«U∑§ ’¥œ •ÊÒ⁄U ¬å≈UÊß«U ’¥œ ¬Êÿ ¡ÊÃ „Ò¥ —
(1) Chitin, cholesterol
(1) ∑§ÊßÁ≈UŸ, ∑§Ê‹S≈U⁄UÊÚ‹
(2) Glycerol, trypsin
(2) ÁÇ‹‚⁄UÊÚ‹, Á≈˛UÁå‚Ÿ
(3) Cellulose, lecithin
(3) ‚‹È‹Ê¡, ‹Á‚ÁÕŸ
(4) Inulin, insulin
(4) ߟÈÁ‹Ÿ, ߥ‚ÈÁ‹Ÿ
12. The plant parts which consist of two generations -
12. ¬ÊŒ¬ ∑§Ê fl„ ÷ʪ ∑§ÊÒŸ-‚Ê „Ò Á¡‚◊¥ ŒÊ ¬Ë…∏Ë - ∞∑§ ¬Ë…∏Ë one within the other :
ŒÍ‚⁄U ∑§ •ãŒ⁄U „ÊÃË „Ò? (a) Pollen grains inside the anther
(a) ¬⁄Uʪ∑§Ê‡Ê ∑§ •ãŒ⁄U ¬⁄Uʪ∑§áÊ
(b) Germinated pollen grain with two male
(b) ŒÊ Ÿ⁄U ÿÈÇ◊∑§Ê¥ flÊ‹Ë •¥∑ȧÁ⁄Uà ¬⁄Uʪ∑§áÊ gametes
(c) »§‹ ∑§ •ãŒ⁄U ’Ë¡ (c) Seed inside the fruit
(d) ’Ë¡Êá«U ∑§ •ãŒ⁄U ÷Í˝áÊ-∑§Ê·
(d) Embryo sac inside the ovule
(1) ∑§fl‹ (a)
(1) (a) only
(2) (a), (b) •ÊÒ⁄U (c)
(2) (a), (b) and (c)
(3) (c) •ÊÒ⁄U (d) (3) (c) and (d)
(4) (a) •ÊÒ⁄U (d) (4) (a) and (d)

13. »§‹ËŒÊ⁄U »§‹Ê¥ flÊ‹ ¬ÊŒ¬Ê¥ ∑§Ë ¡«∏ ª˝ÁãÕ∑§Ê•Ê¥ ◊¥ ŸÊß≈˛UÊÁ¡Ÿ¡ 13. The product(s) of reaction catalyzed by nitrogenase
mÊ⁄UÊ ©à¬˝Á⁄Uà •Á÷Á∑˝§ÿÊ ∑§Ê/∑§ ©à¬ÊŒ ∑§ÊÒŸ ‚Ê/‚ „Ò/„Ò¥? in root nodules of leguminous plants is/are :
(1) ∑§fl‹ •◊ÊÁŸÿÊ (1) Ammonia alone

(2) ∑§fl‹ ŸÊß≈˛U≈U (2) Nitrate alone

(3) •◊ÊÁŸÿÊ •ÊÒ⁄U •ÊÚÄ‚Ë¡Ÿ (3) Ammonia and oxygen
(4) Ammonia and hydrogen
(4) •◊ÊÁŸÿÊ •ÊÒ⁄U „Êß«˛UÊ¡Ÿ

14. •¥Ã⁄UÊflSÕÊ ∑§Ë G1 ¬˝ÊflSÕÊ (ªÒ¬ 1) ∑§ ’Ê⁄U ◊¥ ‚„Ë ∑§ÕŸ ∑§Ê 14. Identify the correct statement with regard to
G1 phase (Gap 1) of interphase.
øÿŸ ∑§⁄UÊ–
(1) DNA synthesis or replication takes place.
(1) «UË.∞Ÿ.∞. ‚¥‡‹·áÊ ÿÊ ¬˝ÁÃ∑ΧÁÃ∑§⁄UáÊ „ÊÃÊ „Ò–
(2) Reorganisation of all cell components takes
(2) ‚÷Ë ∑§ÊÁ‡Ê∑§Ê •flÿflÊ¥ ∑§Ê ¬ÈŸª¸∆UŸ „ÊÃÊ „Ò– place.
(3) ∑§ÊÁ‡Ê∑§Ê ©¬Ê¬øÿË ‚Á∑˝§ÿ „ÊÃË „Ò, flÎÁh ∑§⁄UÃË „Ò (3) Cell is metabolically active, grows but does
‹Á∑§Ÿ DNA ∑§Ë ¬˝ÁÃ∑ΧÁà Ÿ„Ë¥ ∑§⁄UÃË– not replicate its DNA.
(4) Nuclear Division takes place.
(4) ∑§ãŒ˝∑§ Áfl÷Ê¡Ÿ „ÊÃÊ „Ò–
15. Cuboidal epithelium with brush border of microvilli
15. ‚͡◊Ê¥∑ȧ⁄UÊ¥ ∑§ ’˝È‡Ê ’Ê«¸U⁄U flÊ‹Ë ÉÊŸÊ∑§Ê⁄U ©¬∑§‹Ê ¬ÊÿË ¡ÊÃË „Ò —
is found in :
(1) •Ê¥òÊ ∑§ •ÊSÃ⁄U ◊¥ (1) lining of intestine
(2) ‹Ê⁄U ª˝¥ÁÕ ∑§Ë flÊÁ„∑§Ê ◊¥ (2) ducts of salivary glands
(3) flÎÄ∑§ÊáÊÈ ∑§Ë ‚◊ˬSÕ ‚¥flÁ‹Ã ŸÁ‹∑§Ê ◊¥ (3) proximal convoluted tubule of nephron
(4) ÿÍS≈U∑§ËÿŸ ŸÁ‹∑§Ê ◊¥ (4) eustachian tube

Hindi+English
5 E1
16. •¥ÃÁfl¸c≈U ∑§ÊÿÊ¥ ∑§ Áfl·ÿ ◊¥ ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ 16. Which of the following statements about inclusion
bodies is incorrect ?
ª‹Ã „Ò?
(1) They are not bound by any membrane.
(1) ÿ Á∑§‚Ë Á¤ÊÀ‹Ë ‚ ÁÉÊ⁄U Ÿ„Ë¥ „ÊÃ–
(2) These are involved in ingestion of food
(2) ÿ πÊl ∑§áÊÊ¥ ∑§ •¥Ãª˝¸„áÊ ◊¥ ‡ÊÊÁ◊‹ „ÊÃ „Ò¥– particles.
(3) ÿ ∑§ÊÁ‡Ê∑§ÊŒ˝√ÿ ◊¥ SflÃ¥òÊ M§¬ ◊¥ „ÊÃ „Ò¥– (3) They lie free in the cytoplasm.
(4) ÿ ∑§ÊÁ‡Ê∑§ÊŒ˝√ÿ ◊¥ ÁŸÁøà ¬ŒÊÕ¸ ∑§Ê √ÿÄà ∑§⁄UÃ „Ò¥– (4) These represent reserve material in
cytoplasm.
17. ‚È∑§ãŒ˝∑§Ë ∑§ÊÁ‡Ê∑§Ê•Ê¥ ◊¥ Ç‹Êß∑§Ê¬˝Ê≈UËŸ •ÊÒ⁄U Ç‹Êß∑§ÊÁ‹Á¬«U ∑§
ÁŸ◊ʸáÊ ∑§Ê ◊ÈÅÿ SÕ‹ ∑§ÊÒŸ ‚Ê „Ò? 17. Which is the important site of formation of
glycoproteins and glycolipids in eukaryotic cells ?
(1) •¥ÃŒ˝¸√ÿË ¡ÊÁ‹∑§Ê (1) Endoplasmic reticulum
(2) ¬⁄UÊÄ‚Ë‚Ê◊ (2) Peroxisomes
(3) ªÊÀ¡Ë ∑§Êÿ (3) Golgi bodies
(4) ¬Ê‹Ë‚Ê◊ (4) Polysomes

18. ¡‹ ß‹Ä≈˛UÊ»§Ê⁄UÁ‚‚ ◊¥, ¬ÎÕ∑§ „È∞ «UË.∞Ÿ.∞. ∑§ πá«UÊ¥ ∑§Ê 18. In gel electrophoresis, separated DNA fragments
Á∑§‚∑§Ë ‚„ÊÿÃÊ ‚ ŒπÊ ¡Ê ‚∑§ÃÊ „Ò? can be visualized with the help of :
(1) ø◊∑§Ë‹ ŸË‹ ¬˝∑§Ê‡Ê ◊¥ ∞‚Ë≈UÊ∑§ÊÁ◊¸Ÿ ‚ (1) Acetocarmine in bright blue light
(2) Ethidium bromide in UV radiation
(2) UV ÁflÁ∑§⁄UáÊ ◊¥ ∞ÁÕÁ«Uÿ◊ ’˝Ê◊Êß«U ‚
(3) Acetocarmine in UV radiation
(3) UV ÁflÁ∑§⁄UáÊ ◊¥ ∞‚Ë≈UÊ∑§ÊÁ◊¸Ÿ ‚
(4) Ethidium bromide in infrared radiation
(4) •fl⁄UÄà ÁflÁ∑§⁄UáÊ ◊¥ ∞ÁÕÁ«Uÿ◊ ’˝Ê◊Êß«U ‚
19. Identify the wrong statement with reference to
19. •ÊÚÄ‚Ë¡Ÿ ∑§ ¬Á⁄Ufl„Ÿ ∑§ ‚¥Œ÷¸ ◊¥ ª‹Ã ∑§ÕŸ ∑§Ê ¬„øÊŸÊ– transport of oxygen.
(1) •ÊÚÄ‚Ë¡Ÿ ∑§Ë „Ë◊ÊÇ‹ÊÁ’Ÿ ‚ ’¥œÃÊ ◊ÈÅÿ× O2 ∑§ (1) Binding of oxygen with haemoglobin is
•Ê¥Á‡Ê∑§ ŒÊ’ ‚ ‚¥’¥ÁœÃ „Ò– mainly related to partial pressure of O2.
(2) CO2 ∑§Ê •Ê¥Á‡Ê∑§ ŒÊ’ „Ë◊ÊÇ‹ÊÁ’Ÿ ‚ ’¥œŸ flÊ‹Ë (2) Partial pressure of CO2 can interfere with
O2 binding with haemoglobin.
O2 ◊¥ ’ÊœÊ «UÊ‹ ‚∑§ÃÊ „Ò–
(3) Higher H+ conc. in alveoli favours the
(3) flÊÿÈ ∑ͧÁ¬∑§Ê ◊¥ H+ ∑§Ë ©ìÊ ‚Ê¥ŒÃ˝ Ê •ÊÚÄ‚Ë„Ë◊ÊÇ‹ÊÁ’Ÿ formation of oxyhaemoglobin.
’ŸŸ ◊¥ ‚„Êÿ∑§ „ÊÃË „Ò– (4) Low pCO2 in alveoli favours the formation
(4) flÊÿÈ ∑ͧÁ¬∑§Ê ◊¥ ∑§◊ pCO2 •ÊÚÄ‚Ë„Ë◊ÊÇ‹ÊÁ’Ÿ ’ŸŸ of oxyhaemoglobin.
◊¥ ‚„Êÿ∑§ „ÊÃË „Ò– 20. Ray florets have :
20. •⁄U-¬Èc¬∑§ ◊¥ ÄÿÊ „ÊÃÊ „Ò? (1) Inferior ovary
(1) •œÊflÃ˸ •¥«UʇÊÿ (2) Superior ovary
(3) Hypogynous ovary
(2) ™§äfl¸flÃ˸ •¥«UʇÊÿ
(4) Half inferior ovary
(3) ¡ÊÿÊ¥ªÊœ⁄U •¥«UʇÊÿ
(4) •h¸ •œÊflÃ˸ •¥«UʇÊÿ 21. The specific palindromic sequence which is
recognized by EcoRI is :
21. ߸∑§Ê •Ê⁄U I mÊ⁄UÊ ¬„øÊŸ ¡ÊŸ flÊ‹Ê ¬ÒÁ‹ã«˛UÊÁ◊∑§ ∑˝§◊ „Ò — (1) 5' - GAATTC - 3'
(1) 5' - GAATTC - 3' 3' - CTTAAG - 5'
3' - CTTAAG - 5'
(2) 5' - GGAACC - 3'
(2) 5' - GGAACC - 3'
3' - CCTTGG - 5' 3' - CCTTGG - 5'
(3) 5' - CTTAAG - 3' (3) 5' - CTTAAG - 3'
3' - GAATTC - 5' 3' - GAATTC - 5'
(4) 5' - GGATCC - 3' (4) 5' - GGATCC - 3'
3' - CCTAGG - 5' 3' - CCTAGG - 5'

E1 6 Hindi+English

22. ¬˝ÁÃ’¥œŸ ∞¥¡Êß◊Ê¥ ∑§ Áfl·ÿ ◊¥ ª‹Ã ∑§ÕŸ ∑§Ê ¬„øÊÁŸ∞– 22. Identify the wrong statement with regard to
Restriction Enzymes.
(1) ¬˝àÿ∑§ ¬˝ÁÃ’¥œŸ ∞¥¡Êß◊ «UË.∞Ÿ.∞. ∑˝§◊ ∑§Ë ‹ê’Ê߸ ∑§Ê
(1) Each restriction enzyme functions by
ÁŸ⁄UˡÊáÊ ∑§⁄U∑§ ∑§Êÿ¸ ∑§⁄UÃ „Ò¥– inspecting the length of a DNA sequence.
(2) ÿ «UË.∞Ÿ.∞. ∑§Ë ‹«∏Ë ∑§Ê ¬ÒÁ‹ã«˛UÊÁ◊∑§ SÕ‹Ê¥ ¬⁄U (2) They cut the strand of DNA at palindromic
∑§Ê≈UÃ „Ò¥– sites.

(3) ÿ •ÊŸÈfl¥Á‡Ê∑§ ߥ¡ËÁŸÿÁ⁄¥Uª ◊¥ ©¬ÿÊªË „Ò¥– (3) They are useful in genetic engineering.

(4) Sticky ends can be joined by using DNA
(4) Áø¬Áø¬ Á‚⁄U «UË.∞Ÿ.∞. ‹Êߪ¡ mÊ⁄UÊ ¡Ê«∏ ¡Ê ‚∑§Ã ligases.
„Ò¥–
23. Which of the following is put into Anaerobic sludge
23. ÁŸêŸ ◊¥ ∑§ÊÒŸ flÊÁ„Ã◊‹ ©¬øÊ⁄U ∑§ Á‹∞ •flÊÿflËÿ •Ê¬¥∑§ digester for further sewage treatment ?
‚¥¬ÊÁøòÊ ◊¥ «UÊ‹Ê ¡ÊÃÊ „Ò? (1) Primary sludge
(1) ¬˝ÊÕÁ◊∑§ •Ê¬¥∑§ (2) Floating debris
(2) ÃÒ⁄UÃ „È∞ ∑ͧ«∏-∑§⁄U∑§≈U (3) Effluents of primary treatment

(3) ¬˝ÊÕÁ◊∑§ ©¬øÊ⁄U ∑§ ’Á„—dÊfl (4) Activated sludge

(4) ‚¥Á∑˝§ÿËà •Ê¬¥∑§
24. Select the correct events that occur during
inspiration.
24. •¥Ã—‡fl‚Ÿ ∑§ ŒÊÒ⁄UÊŸ „ÊŸ flÊ‹Ë ‚„Ë ÉÊ≈UŸÊ•Ê¥ ∑§Ê øÿŸ ∑§⁄UÊ–
(a) Contraction of diaphragm
(a) «UÊÿÊ»˝§Ê◊ ∑§Ê ‚¥∑ȧøŸ
(b) Contraction of external inter-costal muscles
(b) ’Ês •¥Ã⁄U¬‡Êȸ∑§ ¬Á‡ÊÿÊ¥ ∑§Ê ‚¥∑ȧøŸ
(c) Pulmonary volume decreases
(c) »È§å»È§‚ ∑§Ê •Êÿß ∑§◊ „ÊŸÊ
(d) Intra pulmonary pressure increases
(d) •¥Ã⁄UÊ »È§å»È§‚Ë ŒÊ’ ∑§Ê ’…∏ŸÊ
(1) (a) and (b)
(1) (a) ∞fl¥ (b)
(2) (c) and (d)
(2) (c) ∞fl¥ (d)
(3) (a), (b) and (d)
(3) (a), (b) ∞fl¥ (d)
(4) only (d)
(4) ∑§fl‹ (d)
25. If the head of cockroach is removed, it may live for
25. ÿÁŒ ÁËø^ ∑§Ê Á‚⁄U „≈UÊ ÁŒÿÊ ¡Ê∞ ÃÊ ÿ„ ∑ȧ¿U ÁŒŸÊ¥ Ã∑§ few days because :
¡ËÁflà ⁄U„ ‚∑§ÃÊ „Ò ÄÿÊ¥Á∑§ — (1) the supra-oesophageal ganglia of the
cockroach are situated in ventral part of
(1) ÁËø^ ∑§ •Áœª˝Á‚∑§Ê ªÈÁë¿U∑§Ê ©Œ⁄U ∑§ •œ⁄U ÷ʪ abdomen.
◊¥ ÁSÕà „ÊÃ „Ò¥–
(2) the cockroach does not have nervous system.
(2) ÁËø^ ◊¥ Ã¥ÁòÊ∑§Ê Ã¥òÊ Ÿ„Ë¥ „ÊÃÊ– (3) the head holds a small proportion of a nervous
(3) Á‚⁄U ◊¥ Ã¥ÁòÊ∑§Ê Ã¥òÊ ∑§Ê ∑§fl‹ ¿UÊ≈UÊ ÷ʪ „ÊÃÊ „Ò ¡’Á∑§ system while the rest is situated along the
ventral part of its body.
‡Ê· ‡Ê⁄UË⁄U ∑§ •œ⁄U ÷ʪ ◊¥ ÁSÕà „ÊÃÊ „Ò–
(4) the head holds a 1/3rd of a nervous system
(4) Á‚⁄U ◊¥ Ã¥ÁòÊ∑§Ê Ã¥òÊ ∑§Ê 1/3 ÷ʪ „ÊÃÊ „Ò ¡’Á∑§ ‡Ê· while the rest is situated along the dorsal
‡Ê⁄UË⁄U ∑§ ¬Îc∆U ÷ʪ ◊¥ „ÊÃÊ „Ò– part of its body.

Hindi+English
7 E1
26. ‚¥ÉÊ ∑§ÊÚ«¸U≈UÊ ∑§ Á‹∞ ∑§ÊÒŸ ‚ ∑§ÕŸ ‚„Ë „¥Ò? 26. Which of the following statements are true for
the phylum-Chordata ?
(a) ÿÍ⁄UÊ∑§ÊÚ«¸U≈UÊ ◊¥ ¬Îc∆U⁄UîÊÈ Á‚⁄U ‚ ¬Í¥¿U Ã∑§ »Ò§‹Ë „ÊÃË „Ò (a) In Urochordata notochord extends from
•ÊÒ⁄U ÿ„ ¡ËflŸ ∑§ •¥Ã Ã∑§ ’ŸË ⁄U„ÃË „Ò– head to tail and it is present throughout
(b) fl≈U˸’˝≈UÊ ◊¥ ¬Îc∆U⁄UîÊÈ ∑§fl‹ ÷˝ÍáÊËÿ ∑§Ê‹ ◊¥ ©¬ÁSÕà their life.
„ÊÃË „Ò– (b) In Vertebrata notochord is present during
the embryonic period only.
(c) ∑§ãŒ˝Ëÿ Ã¥ÁòÊ∑§Ê Ã¥òÊ ¬Îc∆UËÿ ∞fl¥ πÊπ‹Ê „ÊÃÊ „Ò– (c) Central nervous system is dorsal and
(d) ∑§ÊÚ«¸U≈UÊ ∑§Ê ÃËŸ ©¬‚¥ÉÊÊ¥ ◊¥ Áfl÷ÊÁ¡Ã Á∑§ÿÊ „Ò — hollow.
„◊Ë∑§ÊÚ«¸U≈UÊ, ≈˜UÿÍÁŸ∑§≈UÊ ∞fl¥ ‚»Ò§‹Ê∑§ÊÚ«¸U≈UÊ– (d) Chordata is divided into 3 subphyla :
Hemichordata, Tunicata and
(1) (d) ∞fl¥ (c) Cephalochordata.
(2) (c) ∞fl¥ (a) (1) (d) and (c)
(3) (a) ∞fl¥ (b) (2) (c) and (a)
(3) (a) and (b)
(4) (b) ∞fl¥ (c)
(4) (b) and (c)
27. ¡Ëfl ∑§Ê ©Ÿ∑§ ¡Òfl¬˝ÊÒlÊÁª∑§Ë ◊¥ ©¬ÿÊª ∑§ Á‹∞ ‚È◊Á‹Ã 27. Match the organism with its use in biotechnology.
∑§ËÁ¡∞–
(a) Bacillus (i) Cloning vector
(a) ’ÒÁ‚‹‚ ÕÈÁ⁄¥UÁ¡ÁŸÁ‚‚ (i) Ä‹ÊÁŸ∑§ flÄ≈U⁄U thuringiensis
(b) Õ◊¸‚ ∞ÄflÁ≈U∑§‚ (ii) ¬˝Õ◊ rDNA •áÊÈ ∑§Ê (b) Thermus (ii) Construction of
ÁŸ◊ʸáÊ aquaticus first rDNA
(c) ∞ª˝Ê’ÒÄ≈UËÁ⁄Uÿ◊ (iii) «Ë.∞Ÿ.∞. ¬ÊÚÁ‹◊⁄U¡ molecule
≈˜UÿÈÁ◊»§Á‚∞¥‚ (c) Agrobacterium (iii) DNA polymerase
(d) ‚ÊÀ◊ÊŸ‹Ê (iv) Cry ¬˝Ê≈UËŸ tumefaciens
≈Êß»§ËêÿÈÁ⁄Uÿ◊ (d) Salmonella (iv) Cry proteins
ÁŸêŸÁ‹Áπà ◊¥ ‚ ‚„Ë Áfl∑§À¬ øÈÁŸ∞ — typhimurium
(a) (b) (c) (d) Select the correct option from the following :
(1) (ii) (iv) (iii) (i) (a) (b) (c) (d)
(2) (iv) (iii) (i) (ii) (1) (ii) (iv) (iii) (i)
(3) (iii) (ii) (iv) (i) (2) (iv) (iii) (i) (ii)
(4) (iii) (iv) (i) (ii) (3) (iii) (ii) (iv) (i)
28. •ÁŸflÊÿ¸ ÃàflÊ¥ •ÊÒ⁄U ¬ÊŒ¬Ê¥ ◊¥ ©Ÿ∑§ ∑§ÊÿÊZ ∑§ Áfl·ÿ ◊¥ ÁŸêŸÁ‹Áπà (4) (iii) (iv) (i) (ii)

∑§Ê ‚È◊Á‹Ã ∑§ËÁ¡∞ —
28. Match the following concerning essential elements
(a) ‹Ê„ (i) ¡‹ ∑§Ê ¬˝∑§Ê‡Ê •¬ÉÊ≈UŸ and their functions in plants :
(b) Á¡¥∑§ (ii) ¬⁄Uʪ ∑§Ê •¥∑ȧ⁄UáÊ (a) Iron (i) Photolysis of water
(c) ’Ê⁄UÊŸÚ (iii) Ä‹Ê⁄UÊÁ»§‹ ∑§ ¡Òfl ‚¥‡‹·áÊ (b) Zinc (ii) Pollen germination
∑§ Á‹∞ •Êfl‡ÿ∑§ (c) Boron (iii) Required for chlorophyll
biosynthesis
(d) ◊Ò¥ªŸË¡ (iv) •Ê߸.∞.∞. ¡Òfl ‚¥‡‹·áÊ
(d) Manganese (iv) IAA biosynthesis
‚„Ë Áfl∑§À¬ øÈÁŸ∞ — Select the correct option :
(a) (b) (c) (d) (a) (b) (c) (d)
(1) (ii) (i) (iv) (iii) (1) (ii) (i) (iv) (iii)
(2) (iv) (iii) (ii) (i) (2) (iv) (iii) (ii) (i)
(3) (iii) (iv) (ii) (i) (3) (iii) (iv) (ii) (i)
(4) (iv) (i) (ii) (iii) (4) (iv) (i) (ii) (iii)

E1 8 Hindi+English

29. ª‹Ã ∑§ÕŸ ∑§Ê øÈÁŸ∞– 29. Identify the incorrect statement.

(1) •¥Ã—∑§Êc∆U ¡‹ ∑§Ê øÊ‹Ÿ Ÿ„Ë¥ ∑§⁄UÃË, ¬⁄UãÃÈ ÿÊ¥ÁòÊ∑§ (1) Heart wood does not conduct water but gives
mechanical support.
‚„ÊÿÃÊ ¬˝ŒÊŸ ∑§⁄UÃË „Ò–
(2) Sapwood is involved in conduction of water
(2) ⁄U‚ŒÊM§ ¡«∏ ‚ ¬ûÊË Ã∑§ ¡‹ ∑§ øÊ‹Ÿ ◊¥ •ÊÒ⁄U πÁŸ¡Ê¥ and minerals from root to leaf.
∑§ øÊ‹Ÿ ◊¥ ‡ÊÊÁ◊‹ „ÊÃË „Ò– (3) Sapwood is the innermost secondary xylem
(3) ⁄U‚ŒÊM§ ‚’‚ ÷ËÃ⁄UË ÁmÃËÿ∑§ ŒÊM§ „ÊÃÊ „Ò •ÊÒ⁄U ÿ„ and is lighter in colour.
•¬ˇÊÊ∑Χà „À∑§ ⁄¥Uª ∑§Ë „ÊÃË „Ò– (4) Due to deposition of tannins, resins, oils etc.,
heart wood is dark in colour.
(4) ≈ÒUÁŸŸ, ⁄UÁ¡Ÿ, ÃÒ‹ •ÊÁŒ ∑§ ¡◊Ê „ÊŸ ∑§ ∑§Ê⁄UáÊ •¥Ã—∑§Êc∆U
ª„⁄U ⁄¥Uª ∑§Ë „ÊÃË „Ò– 30. Match the following :

30. ÁŸêŸÁ‹Áπà ∑§Ê ‚È◊Á‹Ã ∑§ËÁ¡∞ — (a) Inhibitor of catalytic (i) Ricin
(a) ©à¬˝⁄U∑§ Á∑˝§ÿÊ ∑§Ê ÁŸ⁄UÊœ∑§ (i) Á⁄UÁ‚Ÿ activity

(b) ¬å≈UÊß«U ’¥œ œÊ⁄U∑§ (ii) ◊Ò‹ÊŸ≈U (b) Possess peptide bonds (ii) Malonate
(c) Cell wall material in (iii) Chitin
(c) ∑§fl∑§Ê¥ ◊¥ ∑§ÊÁ‡Ê∑§Ê Á÷ÁûÊ (iii) ∑§ÊßÁ≈UŸ
fungi
¬ŒÊÕ¸
(d) Secondary metabolite (iv) Collagen
(d) ÁmÃËÿ∑§ ©¬Ê¬øÿ¡ (iv) ∑§Ê‹¡
Ò Ÿ
Choose the correct option from the following :
ÁŸêŸÁ‹Áπà ◊¥ ‚ ‚„Ë Áfl∑§À¬ øÈÁŸ∞ — (a) (b) (c) (d)
(a) (b) (c) (d)
(1) (ii) (iv) (iii) (i)
(1) (ii) (iv) (iii) (i)
(2) (iii) (i) (iv) (ii)
(2) (iii) (i) (iv) (ii)
(3) (iii) (iv) (i) (ii) (3) (iii) (iv) (i) (ii)
(4) (ii) (iii) (i) (iv) (4) (ii) (iii) (i) (iv)

31. ÁmÃËÿ∑§ •¥«U∑§ ∑§Ê •œ¸‚ÍòÊË Áfl÷Ê¡Ÿ ¬Íáʸ „ÊÃÊ „Ò — 31. Meiotic division of the secondary oocyte is
(1) •¥«UÊà‚ª¸ ‚ ¬„‹ completed :
(1) Prior to ovulation
(2) ‚¥÷Êª ∑§ ‚◊ÿ
(2) At the time of copulation
(3) ÿÈÇ◊Ÿ¡ ’ŸŸ ∑§ ’ÊŒ
(3) After zygote formation
(4) ‡ÊÈ∑˝§ÊáÊÈ ∞fl¥ •¥«UÊáÊÈ ∑§ ‚¥‹ÿŸ ∑§ ‚◊ÿ
(4) At the time of fusion of a sperm with an
32. ⁄UÊ’≈¸U ◊ ∑§ •ŸÈ‚Ê⁄U, Áfl‡fl ◊¥ ¡ÊÁà ÁflÁflœÃÊ ‹ª÷ª Á∑§ÃŸË ovum

„Ò?
(1) 1.5 Á◊Á‹ÿŸ 32. According to Robert May, the global species
diversity is about :
(2) 20 Á◊Á‹ÿŸ
(1) 1.5 million
(3) 50 Á◊Á‹ÿŸ
(2) 20 million
(4) 7 Á◊Á‹ÿŸ (3) 50 million

33. ≈˛UÊ¥‚‹‡ÊŸ (•ŸÈflÊŒŸ/SÕÊŸÊ¥Ã⁄UáÊ) ∑§Ë ¬˝Õ◊ •flSÕÊ ∑§ÊÒŸ ‚Ë (4) 7 million

„ÊÃË „Ò?
33. The first phase of translation is :
(1) ⁄UÊß’Ê‚Ê◊ ‚ mRNA ∑§Ê ’㜟 (1) Binding of mRNA to ribosome
(2) «UË.∞Ÿ.∞. •áÊÈ ∑§Ë ¬„øÊŸ (2) Recognition of DNA molecule
(3) tRNA ∑§Ê ∞◊ËŸÊ∞‚Ë‹‡ÊŸ (3) Aminoacylation of tRNA
(4) ∞∑§ ∞¥≈UË-∑§Ê«UÊÚŸ ∑§Ë ¬„øÊŸ (4) Recognition of an anti-codon

Hindi+English
9 E1
34. Áfl‡fl ∑§ ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê ˇÊòÊ •Áœ∑§Ã◊ ¡ÊÁà 34. Which of the following regions of the globe exhibits
highest species diversity ?
ÁflÁflœÃÊ Œ‡ÊʸÃÊ „Ò?
(1) Western Ghats of India
(1) ÷Ê⁄Uà ∑§Ê ¬Á‡ø◊Ë ÉÊÊ≈U (2) Madagascar
(2) ◊«UʪÊS∑§⁄U (3) Himalayas
(3) Á„◊Ê‹ÿ (4) Amazon forests
(4) ∞◊¡ÊÚŸ ∑§ ¡¥ª‹
35. Which of the following statements is not
35. ÁŸêŸ ◊¥ ∑§ÊÒŸ‚Ê ∑§ÕŸ ‚„Ë Ÿ„Ë¥ „Ò? correct ?
(1) In man insulin is synthesised as a
(1) ◊ŸÈcÿ ◊¥ ߥ‚ÈÁ‹Ÿ ¬˝Ê∑˜§-ߥ‚ÈÁ‹Ÿ ‚ ‚¥‡‹Á·Ã „ÊÃÊ „Ò– proinsulin.
(2) ¬˝Ê∑˜§-ߥ‚ÈÁ‹Ÿ ◊¥ ∞∑§ •ÁÃÁ⁄UÄà ¬å≈UÊß«U, Á¡‚ (2) The proinsulin has an extra peptide called
‚Ë-¬å≈UÊß«U ∑§„Ã „Ò¥, „ÊÃË „Ò– C-peptide.
(3) ∑§Êÿʸà◊∑§ ߥ‚ÈÁ‹Ÿ ◊¥ A ∞fl¥ B oÎ¥π‹Ê∞° „ÊÃË „Ò ¡Ê (3) The functional insulin has A and B chains
linked together by hydrogen bonds.
„Êß«˛UÊ¡Ÿ ’¥œ mÊ⁄UÊ ¡È«∏Ë „ÊÃË „Ò–
(4) Genetically engineered insulin is produced
(4) •ÊŸÈfl¥Á‡Ê∑§ ߥ¡ËÁŸÿ⁄UË ß¥‚ÈÁ‹Ÿ ߸-∑§Ê‹Ê߸ mÊ⁄UÊ ©à¬ÊÁŒÃ in E-Coli.
„ÊÃÊ „Ò–
36. The transverse section of a plant shows following
36. ∞∑§ ¬ÊŒ¬ ∑§Ë •ŸÈ¬˝SÕ ∑§Ê≈U ◊¥ ÁŸêŸÁ‹Áπà ‡ÊÊ⁄UËÁ⁄U∑§ ‹ˇÊáÊ anatomical features :
Œ‡Êʸÿ ªÿ — (a) Large number of scattered vascular bundles
(a) •Áœ∑§ ‚¥ÅÿÊ ◊¥ Á’π⁄U „È∞ ‚¥fl„Ÿ ’¥«U‹ ¡Ê ¬Í‹Êë¿UÊŒ surrounded by bundle sheath.
‚ ÁÉÊ⁄U „Ò¥– (b) Large conspicuous parenchymatous ground
tissue.
(b) S¬c≈U ’„Èà ◊ÎŒÍÃ∑§Ëÿ ÷⁄UáÊ ™§Ã∑§– (c) Vascular bundles conjoint and closed.
(c) ‚¥ÿÈÄà •ÊÒ⁄U •flœË¸ ‚¥fl„Ÿ ’¥«U‹– (d) Phloem parenchyma absent.
(d) ¬Ê·flÊ„ ◊ÎŒÍÃ∑§ ∑§Ê •÷Êfl– Identify the category of plant and its part :
(1) Monocotyledonous stem
ß‚ ¬ÊŒ¬ ∑§Ë üÊáÊË •ÊÒ⁄U ©‚∑§ ÷ʪ ∑§Ê ¬„øÊÁŸ∞ —
(2) Monocotyledonous root
(1) ∞∑§’Ë¡¬òÊË ÃŸÊ (3) Dicotyledonous stem
(2) ∞∑§’Ë¡¬òÊË ¡«∏ (4) Dicotyledonous root
(3) Ám’Ë¡¬òÊË ÃŸÊ
(4) Ám’Ë¡¬òÊË ¡«∏ 37. Match the following columns and select the
correct option.
37. ÁŸêŸ SÃ¥÷Ê¥ ∑§Ê Á◊‹ÊŸ ∑§⁄U ‚„Ë Áfl∑§À¬ ∑§Ê øÿŸ ∑§⁄UÊ– Column - I Column - II
SÃ¥÷ - I SÃ¥÷ - II (a) 6 - 15 pairs of (i) Trygon
(a) Ä‹Ê◊ Á¿UŒ˝Ê¥ ∑§ 6 - 15 (i) ≈˛UÊߪÊŸ gill slits
ÿÈÇ◊ (b) Heterocercal (ii) Cyclostomes
(b) „Ò≈U⁄UÊ‚∑¸§‹ ¬Èë¿U ¬π (ii) ‚ÊßÄ‹ÊS≈UÊê‚ caudal fin

(c) flÊÿÈ ∑§Ê· (iii) ∑§Ê¥«˛UËÄÕË¡ (c) Air Bladder (iii) Chondrichthyes
(d) Áfl· Œ¥‡Ê (iv) •ÊÁS≈UÄÕË¡ (d) Poison sting (iv) Osteichthyes
(a) (b) (c) (d) (a) (b) (c) (d)
(1) (ii) (iii) (iv) (i) (1) (ii) (iii) (iv) (i)
(2) (iii) (iv) (i) (ii) (2) (iii) (iv) (i) (ii)
(3) (iv) (ii) (iii) (i) (3) (iv) (ii) (iii) (i)
(4) (i) (iv) (iii) (ii) (4) (i) (iv) (iii) (ii)

E1 10 Hindi+English

38. ∞‚.∞‹. Á◊‹⁄U Ÿ •¬Ÿ ¬˝ÿÊª ◊¥ ∞∑§ ’¥Œ ç‹ÊS∑§ ◊¥ Á∑§‚∑§Ê 38. From his experiments, S.L. Miller produced amino
acids by mixing the following in a closed flask :
Á◊üÊáÊ ∑§⁄U ∞Á◊ŸÊ •ê‹ ©à¬ÛÊ Á∑§ÿ?
(1) CH4, H2, NH3 and water vapor at 8008C
(1) 8008C ¬⁄U CH4, H2, NH3 •ÊÒ⁄U ¡‹ flÊc¬
(2) CH3, H2, NH4 and water vapor at 8008C
(2) 8008C ¬⁄U CH3, H2, NH4 •ÊÒ⁄U ¡‹ flÊc¬
(3) CH4, H2, NH3 and water vapor at 6008C
(3) 6008C ¬⁄U CH4, H2, NH3 •ÊÒ⁄U ¡‹ flÊc¬
(4) CH3, H2, NH3 and water vapor at 6008C
(4) 6008C ¬⁄U CH3, H2, NH3 •ÊÒ⁄U ¡‹ flÊc¬

39. ∑˝§◊ʪà ©ÛÊÁà ∑§ Á‹∞ ÷Í˝áÊËÿ ¬˝◊ÊáÊ ∑§Ê Á∑§‚Ÿ •SflË∑§Ê⁄U 39. Embryological support for evolution was
disapproved by :
Á∑§ÿÊ ÕÊ?
(1) Karl Ernst von Baer
(1) ∑§Ê‹¸ •Ÿ¸S≈U flÊÚŸ ’ÿ⁄U
(2) Alfred Wallace
(2) •À»˝§«U flÊ‹‚
(3) Charles Darwin
(3) øÊÀ‚¸ «UÊÁfl¸Ÿ
(4) Oparin
(4) •Ê¬Á ⁄UŸ
40. The process responsible for facilitating loss of water
40. ⁄UÊÁòÊ ◊¥ ÿÊ ¬Íáʸ ¬˝Ê×∑§Ê‹ ◊¥ ÉÊÊ‚ ∑§Ë ¬ÁûÊÿÊ¥ ∑§ ‡ÊË·¸ ‚ ¡‹ ∑§ in liquid form from the tip of grass blades at night
Œ˝fl •flSÕÊ ◊¥ ÁŸ∑§‹Ÿ ∑§Ê ‚Ȫ◊ ’ŸÊŸ ◊¥ ∑§ÊÒŸ ‚Ë ¬˝Á∑˝§ÿÊ and in early morning is :
©ûÊ⁄UŒÊÿË „ÊÃË „Ò? (1) Transpiration
(1) flÊc¬Êà‚¡¸Ÿ (2) Root pressure
(2) ◊Í‹Ëÿ ŒÊ’ (3) Imbibition
(3) •¥Ã—‡ÊÊ·áÊ (4) Plasmolysis
(4) ¡ËflŒ˝√ÿ∑È¥§øŸ
41. Secondary metabolites such as nicotine, strychnine
41. ÁmÃËÿ∑§ ©¬Ê¬øÿ¡, ¡Ò‚ Á∑§ ÁŸ∑§Ê≈UËŸ, ÁS≈˛UÄŸËŸ •ÊÒ⁄U ∑Ò§»§ËŸ and caffeine are produced by plants for their :
∑§Ê ¬ÊÒœÊ¥ ∑§ mÊ⁄UÊ •¬Ÿ Á‹∞ ÄÿÊ¥ ©à¬ÊÁŒÃ Á∑§ÿÊ ¡ÊÃÊ „Ò? (1) Nutritive value
(1) ¬Ê·áÊ ◊¥ ©¬ÿÊª (2) Growth response
(2) flÎÁh ¬⁄U ¬˝÷Êfl (3) Defence action
(3) ⁄UˇÊÊ ¬⁄U •‚⁄U (4) Effect on reproduction
(4) ¬˝¡ŸŸ ¬⁄U ¬˝÷Êfl
42. The oxygenation activity of RuBisCo enzyme in
42. ¬˝∑§Ê‡Ê‡fl‚Ÿ ◊¥ RuBisCo ∞¥¡Êß◊ ∑§Ë •ÊÚÄ‚Ë¡ŸË∑§⁄UáÊ Á∑˝§ÿÊ photorespiration leads to the formation of :
‚ Á∑§‚∑§Ê ÁŸ◊ʸáÊ „ÊÃÊ „Ò? (1) 2 molecules of 3-C compound
(1) 3-C ÿÊÒÁª∑§ ∑§ 2 •áÊÈ (2) 1 molecule of 3-C compound
(2) 3-C ÿÊÒÁª∑§ ∑§Ê 1 •áÊÈ (3) 1 molecule of 6-C compound
(3) 6-C ÿÊÒÁª∑§ ∑§Ê 1 •áÊÈ (4) 1 molecule of 4-C compound and 1 molecule
of 2-C compound
(4) 4-C ÿÊÒÁª∑§ ∑§Ê 1 •áÊÈ •ÊÒ⁄U 2-C ÿÊÒÁª∑§ ∑§Ê 1 •áÊÈ

43. Bt ∑§¬Ê‚ ∑§Ë Á∑§S◊ ¡Ê ’ÒÁ‚‹‚ ÕÈÁ⁄¥UÁ¡ÁŸÁ‚‚ ∑§ Áfl· ¡ËŸ 43. Bt cotton variety that was developed by the
introduction of toxin gene of Bacillus thuringiensis
∑§Ê ‚◊ÊÁflc≈U ∑§⁄U∑§ ’ŸÊ߸ ªÿË „Ò, ¬˝ÁÃ⁄UÊœË „Ò — (Bt) is resistant to :
(1) ∑§Ë≈U ¬Ë«∏∑§Ê¥ ‚ (1) Insect pests
(2) ∑§fl∑§Ëÿ ⁄UÊªÊ¥ ‚ (2) Fungal diseases
(3) ¬ÊŒ¬ ‚ÍòÊ∑ΧÁ◊ ‚ (3) Plant nematodes
(4) ∑§Ë≈U ¬⁄U÷ˇÊË ‚ (4) Insect predators

Hindi+English
11 E1
44. ÁŸêŸ ◊¥ ∑§ÊÒŸ, ∞‚ ¡ËflÊ¥ ∑§ ‚„Ë ©ŒÊ„⁄UáÊÊ¥ ∑§Ê ‚¥ŒÁ÷¸Ã ∑§⁄UÃÊ „Ò 44. Which of the following refer to correct example(s)
of organisms which have evolved due to changes
¡Ê ◊ÊŸfl ∑§Ë Á∑˝§ÿÊ•Ê¥ mÊ⁄UÊ flÊÃÊfl⁄UáÊ ◊¥ ’Œ‹Êfl ∑§ ∑§Ê⁄UáÊ in environment brought about by anthropogenic
Áfl∑§Á‚à „È∞ „Ò? action ?
(a) ªÒ‹Ê¬ÒªÊ mˬ ◊¥ «UÊÁfl¸Ÿ ∑§Ë Á»¥§ø¥ (a) Darwin’s Finches of Galapagos islands.
(b) π⁄U¬ÃflÊ⁄UÊ¥ ◊¥ ‡ÊÊ∑§ŸÊ‡ÊË ∑§Ê ¬˝ÁÃ⁄UÊœ (b) Herbicide resistant weeds.
(c) ‚‚Ë◊∑§ãŒ˝∑§Ê¥ ◊¥ ŒflÊßÿÊ¥ ∑§Ê ¬˝ÁÃ⁄UÊœ (c) Drug resistant eukaryotes.
(d) ◊ŸÈcÿ mÊ⁄UÊ ’ŸÊÿË ¬Ê‹ÃÍ ¬‡ÊÈ ¡Ò‚ ∑ȧûÊÊ¥ ∑§Ë ŸS‹¥ (d) Man-created breeds of domesticated animals
like dogs.
(1) ∑§fl‹ (a)
(2) (a) ∞fl¥ (c) (1) only (a)

(3) (b), (c) ∞fl¥ (d) (2) (a) and (c)
(3) (b), (c) and (d)
(4) ∑§fl‹ (d)
(4) only (d)

45. ¬˝ÁÃ⁄UˇÊÊ ∑§ ‚¥Œ÷¸ ◊¥ ª‹Ã ∑§ÕŸ ∑§Ê ¬„øÊÁŸ∞–
45. Identify the wrong statement with reference to
(1) ¡’ ¬⁄U¬Ê·Ë ∑§Ê ‡Ê⁄UË⁄U (¡ËÁflà •ÕflÊ ◊ÎÃ) ¬˝Á០∑§ immunity.
‚¥¬∑¸§ ◊¥ •ÊÃÊ „Ò •ÊÒ⁄U ©‚∑§ ‡Ê⁄UË⁄U ◊¥ ¬˝ÁÃ⁄UˇÊË ©à¬ÛÊ (1) When exposed to antigen (living or dead)
„ÊÃ „Ò¥– ß‚ ““‚Á∑˝§ÿ ¬˝ÁÃ⁄UˇÊÊ”” ∑§„Ã „Ò¥– antibodies are produced in the host’s body.
It is called “Active immunity”.
(2) ¡’ ’Ÿ ’ŸÊ∞ ¬˝ÁÃ⁄UˇÊË ¬˝àÿˇÊ M§¬ ‚ ÁŒ∞ ¡ÊÃ „Ò¥, ß‚
(2) When ready-made antibodies are directly
““ÁŸÁc∑˝§ÿ ¬˝ÁÃ⁄UˇÊÊ”” ∑§„Ã „Ò¥– given, it is called “Passive immunity”.
(3) ‚Á∑˝§ÿ ¬˝ÁÃ⁄UˇÊÊ ¡ÀŒË „ÊÃË „Ò •ÊÒ⁄U ¬Íáʸ ¬˝ÁÃÁ∑˝§ÿÊ ŒÃË (3) Active immunity is quick and gives full
„Ò– response.

(4) ÷˝ÍáÊ ◊ÊÃÊ ‚ ∑ȧ¿U ¬˝ÁÃ⁄UˇÊË ¬˝Êåà ∑§⁄UÃÊ „Ò, ÿ„ ÁŸÁc∑˝§ÿ (4) Foetus receives some antibodies from
mother, it is an example for passive
¬˝ÁÃ⁄UˇÊÊ ∑§Ê ©ŒÊ„⁄UáÊ „Ò– immunity.

46. Á∑§‚ ÁflÁœ mÊ⁄UÊ ’Ë∑§ÊŸ⁄UË ∞flË¡ ∞fl¥ ◊Ò⁄UËŸÊ ⁄Uê‚ ‚ ÷«∏ ∑§Ë
46. By which method was a new breed ‘Hisardale’ of
Ÿß¸ ŸS‹ “Á„‚Ê⁄U«U‹” ÃÒÿÊ⁄U ∑§Ë ªÿË „Ò? sheep formed by using Bikaneri ewes and Marino
(1) ’Á„—¬˝¡ŸŸ rams ?
(1) Out crossing
(2) ©à¬Á⁄UfløŸ ¬˝¡ŸŸ
(2) Mutational breeding
(3) ‚¥∑§⁄UáÊ
(3) Cross breeding
(4) •¥Ã—¬˝¡ŸŸ (4) Inbreeding

47. ◊ÊŸfl ¬ÊøŸ Ã¥òÊ ‚ ‚¥ŒÁ÷¸Ã ‚„Ë ∑§ÕŸ ∑§Ê øÿŸ ∑§⁄UÊ–
47. Identify the correct statement with reference to
(1) ˇÊÈŒ˝Ê¥òÊ ¿UÊ≈UË •Ê¥Ã ◊¥ πÈ‹ÃÊ „Ò– human digestive system.
(2) Á‚⁄UÊ‚Ê •Ê„Ê⁄U ŸÊ‹ ∑§Ê ‚’‚ •ãŒ⁄U flÊ‹Ë ¬⁄Uà „ÊÃË (1) Ileum opens into small intestine.
„Ò– (2) Serosa is the innermost layer of the
alimentary canal.
(3) ˇÊÈŒ˝Ê¥òÊ •àÿÊÁœ∑§ ∑È¥§«UÁ‹Ã ÷ʪ „ÊÃÊ „Ò–
(3) Ileum is a highly coiled part.
(4) ∑ΧÁ◊M§¬ ¬Á⁄U‡ÊÁ·∑§Ê ª˝„áÊË ‚ ©à¬ÛÊ „ÊÃÊ „Ò–
(4) Vermiform appendix arises from duodenum.

E1 12 Hindi+English

48. ÁŸêŸ SÃ¥÷Ê¥ ∑§Ê Á◊‹ÊŸ ∑§⁄U ‚„Ë Áfl∑§À¬ ∑§Ê øÿŸ ∑§⁄UÊ– 48. Match the following columns and select the
correct option.
SÃ¥÷ - I SÃ¥÷ - II
Column - I Column - II
(a) Ä‹ÊS≈˛UËÁ«Uÿ◊ (i) ‚ÊßÄ‹ÊS¬ÊÁ⁄UŸ-∞
(a) Clostridium (i) Cyclosporin-A
éÿÍ≈UÊÿÁ‹∑§◊
butylicum
(b) ≈˛UÊß∑§Ê«U◊ʸ ¬ÊÚ‹ËS¬Ê⁄U◊ (ii) éÿÈÁ≈UÁ⁄U∑§ •ê‹
(b) Trichoderma (ii) Butyric Acid
(c) ◊ÊŸÊS∑§‚ ¬⁄UåÿÍ⁄UË•‚ (iii) Á‚Á≈˛U∑§ •ê‹ polysporum

(d) ∞S¬⁄UÁ¡‹‚ ŸÊߪ⁄U (iv) ⁄UÄÃ-∑§Ê‹S≈U⁄UÊ‹ ∑§◊ (c) Monascus (iii) Citric Acid
∑§⁄UŸ flÊ‹Ê ∑§Ê⁄U∑§ purpureus

(a) (b) (c) (d) (d) Aspergillus niger (iv) Blood cholesterol
(1) (iii) (iv) (ii) (i) lowering agent

(2) (ii) (i) (iv) (iii) (a) (b) (c) (d)

(3) (i) (ii) (iv) (iii) (1) (iii) (iv) (ii) (i)

(4) (iv) (iii) (ii) (i) (2) (ii) (i) (iv) (iii)

(3) (i) (ii) (iv) (iii)
49. ÁŸêŸ ◊¥ ◊ÍòÊ ∑§Ë ∑§ÊÒŸ‚Ë •flSÕÊ «UÊÿÊÁ’≈UË¡ ◊Á‹≈U‚ ∑§Ë •Ê⁄U (4) (iv) (iii) (ii) (i)
‚¥∑§Ã ∑§⁄UÃË „Ò?
(1) ÿÍ⁄UÁ◊ÿÊ ∞fl¥ ∑§Ë≈UÊŸÈÁ⁄UÿÊ 49. Presence of which of the following conditions in
urine are indicative of Diabetes Mellitus ?
(2) ÿÍ⁄UÁ◊ÿÊ ∞fl¥ ⁄UËŸ‹ ∑Ò§À∑ȧ‹Ë
(1) Uremia and Ketonuria
(3) ∑§Ë≈UÊŸÈÁ⁄UÿÊ ∞fl¥ Ç‹Êß∑§Ê‚ÍÁ⁄UÿÊ (2) Uremia and Renal Calculi
(4) ⁄UËŸ‹ ∑Ò§À∑ȧ‹Ë ∞fl¥ „Ê߬⁄UÇ‹ÊßÁ‚Á◊ÿÊ (3) Ketonuria and Glycosuria

(4) Renal calculi and Hyperglycaemia
50. ç‹Ê⁄UËÁ«UÿŸ ◊Ê°«U ∑§Ë ‚¥⁄UøŸÊ Á∑§‚∑§ ‚◊ÊŸ „ÊÃË „Ò?
(1) ◊Ê°«U •ÊÒ⁄U ‚‹È‹Ê¡ 50. Floridean starch has structure similar to :

(1) Starch and cellulose
(2) ∞◊Êß‹Ê¬Ä≈UËŸ •ÊÒ⁄U Ç‹Êß∑§Ê¡Ÿ
(2) Amylopectin and glycogen
(3) ◊ÒŸË≈UÊÚ‹ •ÊÒ⁄U ∞ÁÀ¡Ÿ
(3) Mannitol and algin
(4) ‹ÒÁ◊ŸÁ⁄UŸ •ÊÒ⁄U ‚‹È‹Ê¡
(4) Laminarin and cellulose

51. ÿÊÒŸ ‚¥øÁ⁄Uà ⁄UÊªÊ¥ ∑§ ‚„Ë Áfl∑§À¬ ∑§Ê øÿŸ ∑§⁄UÊ– 51. Select the option including all sexually transmitted
diseases.
(1) ‚È¡Ê∑§, Á‚Á»§Á‹‚, ¡ŸÁŸ∑§ ¬Á⁄U‚¬¸
(1) Gonorrhoea, Syphilis, Genital herpes
(2) ‚È¡Ê∑§, ◊‹Á⁄UÿÊ, ¡ŸÁŸ∑§ ¬Á⁄U‚¬¸
(2) Gonorrhoea, Malaria, Genital herpes
(3) AIDS, ◊‹Á⁄UÿÊ, »§Êß‹Á⁄UÿÊ
(3) AIDS, Malaria, Filaria
(4) ∑Ò¥§‚⁄U, AIDS, Á‚Á»§Á‹‚ (4) Cancer, AIDS, Syphilis

Hindi+English
13 E1
52. •h¸‚ÍòÊË Áfl÷Ê¡Ÿ ∑§ ‚¥Œ÷¸ ◊¥ ÁŸêŸÁ‹Áπà ∑§Ê ‚È◊Á‹Ã 52. Match the following with respect to meiosis :
∑§ËÁ¡∞ — (a) Zygotene (i) Terminalization
(a) ÿÈÇ◊¬≈˜U≈U •flSÕÊ (i) ©¬ÊãÃË÷flŸ (b) Pachytene (ii) Chiasmata
(b) SÕÍ‹¬≈˜U≈U •flSÕÊ (ii) ∑§Êß∞ï◊≈UÊ (c) Diplotene (iii) Crossing over
(c) Ám¬≈˜U≈U •flSÕÊ (iii) ¡ËŸ ÁflÁŸ◊ÿ (d) Diakinesis (iv) Synapsis
(d) ¬Ê⁄UªÁÃ∑˝§◊ (iv) ‚ÍòÊÿÈÇ◊Ÿ Select the correct option from the following :
(«UÊÿÊ∑§ÊߟÁ‚‚) (a) (b) (c) (d)
ÁŸêŸÁ‹Áπà ◊¥ ‚ ‚„Ë Áfl∑§À¬ øÈÁŸ∞ — (1) (iii) (iv) (i) (ii)
(a) (b) (c) (d) (2) (iv) (iii) (ii) (i)
(1) (iii) (iv) (i) (ii) (3) (i) (ii) (iv) (iii)
(2) (iv) (iii) (ii) (i)
(4) (ii) (iv) (iii) (i)
(3) (i) (ii) (iv) (iii)
(4) (ii) (iv) (iii) (i) 53. Which of the following pairs is of unicellular
algae ?
53. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê ÿÈÇ◊ ∞∑§ ∑§ÊÁ‡Ê∑§Ëÿ ‡ÊÒflÊ‹Ê¥ ∑§Ê (1) Laminaria and Sargassum
„Ò? (2) Gelidium and Gracilaria
(1) ‹ÒÁ◊ŸÁ⁄UÿÊ •ÊÒ⁄U ‚Ê⁄UªÊ‚◊ (3) Anabaena and Volvox
(2) ¡Á‹Á«Uÿ◊ •ÊÒ⁄U ª˝ÊÁ‚‹Á⁄UÿÊ (4) Chlorella and Spirulina
(3) ∞ŸÊ’ËŸÊ •ÊÒ⁄U flÊÚÀflÊÚÄ‚
54. Which of the following hormone levels will cause
(4) Ä‹Ê⁄U‹Ê •ÊÒ⁄U S¬ÊßL§‹ËŸÊ release of ovum (ovulation) from the graffian
follicle ?
54. ÁŸêŸ ∑§ ∑§ÊÒŸ ª˝Ê»§Ë ¬È≈U∑§ ‚ •¥«UÊáÊÈ ∑§Ê ◊ÊøŸ (•¥«UÊà‚ª¸) (1) High concentration of Estrogen
∑§⁄UªÊ? (2) High concentration of Progesterone
(1) ∞S≈˛UÊ¡Ÿ ∑§Ë ©ìÊ ‚Ê¥Œ˝ÃÊ (3) Low concentration of LH
(2) ¬˝Ê¡S≈U⁄UÊŸ ∑§Ë ©ìÊ ‚Ê¥Œ˝ÃÊ (4) Low concentration of FSH
(3) LH ∑§Ë ÁŸêŸ ‚Ê¥Œ˝ÃÊ
55. Match the following columns and select the
(4) FSH ∑§Ë ÁŸêŸ ‚Ê¥Œ˝ÃÊ correct option.
Column - I Column - II
55. ÁŸêŸ SÃ¥÷Ê¥ ∑§Ê Á◊‹ÊŸ ∑§⁄U ‚„Ë Áfl∑§À¬ ∑§Ê øÿŸ ∑§⁄UÊ–
(a) Bt cotton (i) Gene therapy
SÃ¥÷ - I SÃ¥÷ - II
(b) Adenosine (ii) Cellular defence
(a) ’Ë≈UË ∑§¬Ê‚ (i) ¡ËŸ ÁøÁ∑§à‚Ê
deaminase
(b) ∞«UËŸÊ‚ËŸ Á«U∞◊ËŸ¡ (ii) ∑§ÊÁ‡Ê∑§Ëÿ ‚È⁄UˇÊÊ
deficiency
∑§Ë ∑§◊Ë
(c) RNAi (iii) Detection of HIV
(c) •Ê⁄U.∞Ÿ.∞.•Ê߸ (iii) HIV ‚¥∑˝§◊áÊ ∑§Ê ¬ÃÊ
infection
‹ªÊŸÊ
(d) PCR (iv) Bacillus
(d) ¬Ë.‚Ë.•Ê⁄U. (iv) ’ÒÁ‚‹‚
thuringiensis
ÕÈÁ⁄¥UÁ¡ÁŸÁ‚‚
(a) (b) (c) (d) (a) (b) (c) (d)
(1) (iv) (i) (ii) (iii) (1) (iv) (i) (ii) (iii)
(2) (iii) (ii) (i) (iv) (2) (iii) (ii) (i) (iv)
(3) (ii) (iii) (iv) (i) (3) (ii) (iii) (iv) (i)
(4) (i) (ii) (iii) (iv) (4) (i) (ii) (iii) (iv)

E1 14 Hindi+English

56. ‚Ÿ˜ 1987 ◊¥ ◊ÊÚÁã≈˛Uÿ‹ ¬˝Ê≈UÊ∑§ÊÚ‹ Á∑§‚ ¬⁄U ÁŸÿ¥òÊáÊ ∑§ Á‹∞ 56. Montreal protocol was signed in 1987 for control
of :
„SÃÊˇÊÁ⁄Uà Á∑§ÿÊ ªÿÊ ÕÊ?
(1) Transport of Genetically modified organisms
(1) ∞∑§ Œ‡Ê ‚ ŒÍ‚⁄U Œ‡Ê ◊¥ •ÊŸÈfl¥Á‡Ê∑§Ã— M§¬Ê¥ÃÁ⁄Uà ¡ËflÊ¥ from one country to another
∑§ ¬Á⁄Ufl„Ÿ ∑§ Á‹∞
(2) Emission of ozone depleting substances
(2) •Ê$¡ÊŸ ∑§Ê ˇÊÁà ¬„È°øÊŸ flÊ‹ ¬ŒÊÕÊZ ∑§Ê ©à‚¡¸Ÿ
(3) Release of Green House gases
(3) „Á⁄Uà ªÎ„ ªÒ‚Ê¥ ∑§Ê ¿UÊ«∏ŸÊ
(4) Disposal of e-wastes
(4) e-flS≈U (e-∑ͧ«∏Ê ∑§⁄U∑§≈U) ∑§Ê ÁŸ¬≈UÊŸ

57. flÊÿ⁄UÊß«UÊ¥ ∑§ Áfl·ÿ ◊¥, ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ‚„Ë 57. Which of the following is correct about viroids ?

„Ò? (1) They have RNA with protein coat.

(1) ©Ÿ◊¥ •Ê⁄U.∞Ÿ.∞. ∑§ ‚ÊÕ ¬˝Ê≈UËŸ •Êfl⁄UáÊ „ÊÃÊ „Ò– (2) They have free RNA without protein coat.

(2) ©Ÿ◊¥ ¬˝Ê≈UËŸ •Êfl⁄UáÊ ∑§ Á’ŸÊ SflÃ¥òÊ •Ê⁄U.∞Ÿ.∞. „ÊÃÊ (3) They have DNA with protein coat.
„Ò– (4) They have free DNA without protein coat.
(3) ©Ÿ◊¥ ¬˝Ê≈UËŸ •Êfl⁄UáÊ ∑§ ‚ÊÕ «UË.∞Ÿ.∞. „ÊÃÊ „Ò–
(4) ©Ÿ◊¥ ¬˝Ê≈UËŸ •Êfl⁄UáÊ ∑§ Á’ŸÊ SflÃ¥òÊ «UË.∞Ÿ.∞. „ÊÃÊ 58. The ovary is half inferior in :
„Ò– (1) Brinjal

58. •h¸ •œÊflÃ˸ •¥«UʇÊÿ Á∑§‚◊¥ ¬ÊÿÊ ¡ÊÃÊ „Ò? (2) Mustard

(1) ’Ò¥ªŸ (3) Sunflower

(2) ‚⁄U‚Ê¥ (4) Plum

(3) ‚Í⁄U¡◊ÈπË
(4) •Ê‹Í’π È Ê⁄UÊ 59. The enzyme enterokinase helps in conversion of :
(1) protein into polypeptides
59. ∞¥≈U⁄UÊ∑§Êߟ¡ Á∑§‚∑§Ê ’Œ‹Ÿ ◊¥ ‚„ÊÿÃÊ ∑§⁄UÃÊ „Ò? (2) trypsinogen into trypsin
(1) ¬˝Ê≈UËŸ ∑§Ê ¬Êڋˬå≈UÊß«U ◊¥ (3) caseinogen into casein
(2) Á≈˛UÁå‚ŸÊ¡Ÿ ∑§Ê Á≈˛UÁå‚Ÿ ◊¥ (4) pepsinogen into pepsin
(3) ∑Ò§‚ËŸÊ¡Ÿ ∑§Ê ∑Ò§‚ËŸ ◊¥
(4) ¬Áå‚ŸÊ¡Ÿ ∑§Ê ¬Áå‚Ÿ ◊¥ 60. Match the trophic levels with their correct species
examples in grassland ecosystem.
60. ÉÊÊ‚ ÷ÍÁ◊ ¬ÊÁ⁄UÃãòÊ ◊¥ ¬Ê·Ë SÃ⁄UÊ¥ ∑§ ‚ÊÕ ¡ÊÁÃÿÊ¥ ∑§ ‚„Ë (a) Fourth trophic level (i) Crow
©ŒÊ„⁄UáÊ ∑§Ê ‚È◊Á‹Ã ∑§ËÁ¡∞–
(b) Second trophic level (ii) Vulture
(a) øÃÈÕ¸ ¬Ê·Ë SÃ⁄U (i) ∑§ÊÒflÊ
(c) First trophic level (iii) Rabbit
(b) ÁmÃËÿ ¬Ê·Ë SÃ⁄U (ii) Áªh
(c) ¬˝Õ◊ ¬Ê·Ë SÃ⁄U (iii) π⁄UªÊ‡Ê (d) Third trophic level (iv) Grass
(d) ÃÎÃËÿ ¬Ê·Ë SÃ⁄U (iv) ÉÊÊ‚ Select the correct option :
‚„Ë Áfl∑§À¬ øÈÁŸ∞ — (a) (b) (c) (d)
(a) (b) (c) (d) (1) (ii) (iii) (iv) (i)
(1) (ii) (iii) (iv) (i)
(2) (iii) (ii) (i) (iv)
(2) (iii) (ii) (i) (iv)
(3) (iv) (iii) (ii) (i) (3) (iv) (iii) (ii) (i)
(4) (i) (ii) (iii) (iv) (4) (i) (ii) (iii) (iv)

Hindi+English
15 E1
61. ◊¥«U‹ Ÿ SflÃ¥òÊ M§¬ ‚ ¬˝¡ŸŸ ∑§⁄UŸ flÊ‹Ë ◊≈U⁄U ∑§ ¬ÊÒœ ∑§Ë 61. How many true breeding pea plant varieties did
Mendel select as pairs, which were similar except
Á∑§ÃŸË Á∑§S◊Ê¥ ∑§Ê ÿÈÇ◊Ê¥ ∑§ M§¬ ◊¥ øÈŸÊ ¡Ê Áfl¬⁄UËà Áfl‡Ê·∑§Ê¥ in one character with contrasting traits ?
flÊ‹ ∞∑§ ‹ˇÊáÊ ∑§ •‹ÊflÊ ∞∑§ ‚◊ÊŸ ÕË? (1) 4
(1) 4
(2) 2
(2) 2
(3) 14
(3) 14
(4) 8
(4) 8
62. Match the following columns and select the
62. ÁŸêŸ SÃ¥÷Ê¥ ∑§Ê Á◊‹ÊŸ ∑§⁄U ©Áøà Áfl∑§À¬ ∑§Ê øÿŸ ∑§⁄UÊ– correct option.
SÃ¥÷ - I SÃ¥÷ - II Column - I Column - II
(a) •Êª¸Ÿ •ÊÚ»§ ∑§Ê≈Uʸ߸ (i) ◊äÿ ∑§áʸ ∞fl¥ »§Á⁄¥UÄ‚ (a) Organ of Corti (i) Connects middle
∑§Ê ¡Ê«∏ÃË „Ò ear and pharynx
(b) ∑§ÊÁÄ‹ÿÊ (ii) ‹’Á⁄¥UÕ ∑§Ê ÉÊÈ◊ÊflŒÊ⁄U (b) Cochlea (ii) Coiled part of the
÷ʪ labyrinth
(c) ÿÍS≈U∑§ËÿŸ ŸÁ‹∑§Ê (iii) •¥«UÊ∑§Ê⁄U Áπ«∏∑§Ë ‚ (c) Eustachian tube (iii) Attached to the
¡È«∏Ë „ÊÃË „Ò oval window
(d) S≈U¬Ë¡ (iv) ’Á‚‹⁄U Á¤ÊÀ‹Ë ◊¥
(d) Stapes (iv) Located on the
ÁSÕÃ „ÊÃË „Ò
basilar
(a) (b) (c) (d)
membrane
(1) (ii) (iii) (i) (iv)
(a) (b) (c) (d)
(2) (iii) (i) (iv) (ii)
(3) (iv) (ii) (i) (iii) (1) (ii) (iii) (i) (iv)
(4) (i) (ii) (iv) (iii) (2) (iii) (i) (iv) (ii)
(3) (iv) (ii) (i) (iii)
63. ¡‹∑ȧê÷Ë •ÊÒ⁄U ¡‹Á‹‹Ë ◊¥ ¬⁄UʪáÊ Á∑§‚∑§ mÊ⁄UÊ „ÊÃÊ „Ò? (4) (i) (ii) (iv) (iii)
(1) ∑§Ë≈U ÿÊ flÊÿÈ mÊ⁄UÊ
63. In water hyacinth and water lily, pollination takes
(2) ∑§fl‹ ¡‹ œÊ⁄UÊ•Ê¥ mÊ⁄UÊ place by :
(3) flÊÿÈ •ÊÒ⁄U ¡‹ mÊ⁄UÊ (1) insects or wind
(4) ∑§Ë≈U •ÊÒ⁄U ¡‹ mÊ⁄UÊ (2) water currents only
(3) wind and water
64. ©‚ flÎÁh ÁŸÿ¥òÊ∑§ ∑§Ê ŸÊ◊ ’ÃÊßÿ Á¡‚ ªÛÊ ∑§Ë »§‚‹ ¬⁄U (4) insects and water
Á¿U«∏∑§Ÿ ‚ ©‚∑§ ß ∑§Ë ‹ê’Ê߸ ◊¥ ’…∏ÊûÊ⁄UË „ÊÃË „Ò, ÃÕÊ ªÛÊ
64. Name the plant growth regulator which upon
∑§ »§‚‹ ∑§Ë ¬ÒŒÊflÊ⁄U ’…∏ÃË „Ò– spraying on sugarcane crop, increases the length
(1) ‚Êß≈UÊ∑§Êߟ˟ of stem, thus increasing the yield of sugarcane
crop.
(2) Á¡’⁄U‹ËŸ
(1) Cytokinin
(3) ∞ÁÕ‹ËŸ (2) Gibberellin
(4) ∞é‚ËÁ‚∑§ •ê‹ (3) Ethylene
(4) Abscisic acid
65. ¬˝∑§Ê‡Ê •Á÷Á∑˝§ÿÊ ◊¥, ß‹Ä≈˛UÊÚŸÊ¥ ∑§ SÕÊŸÊ¥Ã⁄UáÊ ∑§Ê å‹ÊS≈UÊÁÄflŸÊŸ
∑§„Ê° ‚ ‚Ȫ◊ ’ŸÊÃÊ „Ò? 65. In light reaction, plastoquinone facilitates the
transfer of electrons from :
(1) PS-II ‚ Cytb6f ‚Áê◊üÊ
(1) PS-II to Cytb6f complex
(2) Cytb6f ‚Áê◊üÊ ‚ PS-I
(2) Cytb6f complex to PS-I
(3) PS-I ‚ NADP+ (3) PS-I to NADP+
(4) PS-I ‚ ATP Á‚ãÕ¡ (4) PS-I to ATP synthase

E1 16 Hindi+English

66. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ∞∑§ ’Ë¡ ¬˝‚ÈÁåà ÁŸÿ¥ÁòÊà ∑§⁄UŸ flÊ‹Ê 66. Which of the following is not an inhibitory
substance governing seed dormancy ?
ÁŸ⁄UÊœ∑§ ¬ŒÊÕ¸ Ÿ„Ë¥ „Ò?
(1) Gibberellic acid
(1) Á¡’⁄UÁ‹∑§ •ê‹
(2) Abscisic acid
(2) ∞é‚ËÁ‚∑§ •ê‹
(3) Phenolic acid
(3) Á»§ŸÊÁ‹∑§ •ê‹
(4) Para-ascorbic acid
(4) ¬Ò⁄UÊ-∞S∑§ÊÚÁ’¸∑§ •ê‹

67. •ŸÈ‹πŸ ∑§ ‚◊ÿ «UË.∞Ÿ.∞. ∑§Ë ∑È¥§«U‹Ë ∑§Ê πÊ‹Ÿ ◊¥ ∑§ÊÒŸ‚Ê 67. Name the enzyme that facilitates opening of DNA
helix during transcription.
∞¥¡Êß◊ ◊ŒŒ ∑§⁄UÃÊ „Ò?
(1) DNA ligase
(1) «UË.∞Ÿ.∞. ‹Êߪ$¡
(2) DNA helicase
(2) «UË.∞Ÿ.∞. „Ò‹Ë∑§$¡
(3) DNA polymerase
(3) «UË.∞Ÿ.∞. ¬ÊÚ‹Ë◊⁄U$¡
(4) RNA polymerase
(4) •Ê⁄U.∞Ÿ.∞. ¬ÊÚÁ‹◊⁄U$¡

68. ÁŸêŸ ◊¥ ∑§ÊÒŸ ◊ÍòÊflÎÁh ∑§Ê ⁄UÊ∑§Ÿ ◊¥ ‚„ÊÿÃÊ ∑§⁄UªÊ? 68. Which of the following would help in prevention of
diuresis ?
(1) ADH ∑§ •À¬dfláÊ ‚ •Áœ∑§ ¡‹ ∑§Ê ¬ÈŸ⁄UÊfl‡ÊÊ·áÊ
(1) More water reabsorption due to
undersecretion of ADH
(2) ∞À«UÊS≈U⁄UÊŸ ∑§ ∑§Ê⁄UáÊ flÎÄ∑§ ŸÁ‹∑§Ê ‚ Na+ ∞fl¥ ¡‹
∑§Ê ¬ÈŸ⁄UÊfl‡ÊÊ·áÊ (2) Reabsorption of Na+ and water from renal
tubules due to aldosterone
(3) ∞Á≈˛Uÿ‹ ŸÁ≈˛UÿÈ⁄UÁ≈U∑§ ∑§Ê⁄U∑§ mÊ⁄UÊ flÊÁ„∑§Ê•Ê¥ ∑§Ê ‚¥∑§ËáʸŸ
„ÊŸÊ (3) Atrial natriuretic factor causes
vasoconstriction
(4) JG ∑§ÊÁ‡Ê∑§Ê•Ê¥ mÊ⁄UÊ ⁄UÁŸŸ ∑§Ê dÊfláÊ ∑§◊ „ÊŸÊ
(4) Decrease in secretion of renin by JG cells

69. ∞∑§ ¬ÊÁ⁄UÃãòÊ ◊¥ ‚∑§‹ ¬˝ÊÕÁ◊∑§ ©à¬ÊŒ∑§ÃÊ •ÊÒ⁄U Ÿ≈U ¬˝ÊÕÁ◊∑§
©à¬ÊŒ∑§ÃÊ ∑§ ‚¥’㜠◊¥, ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ‚„Ë 69. In relation to Gross primary productivity and Net
„Ò? primary productivity of an ecosystem, which one
of the following statements is correct ?
(1) ‚∑§‹ ¬˝ÊÕÁ◊∑§ ©à¬ÊŒ∑§ÃÊ ‚ŒÒfl Ÿ≈U ¬˝ÊÕÁ◊∑§ ©à¬ÊŒ∑§ÃÊ
(1) Gross primary productivity is always less
‚ ∑§◊ „ÊÃË „Ò– than net primary productivity.

(2) ‚∑§‹ ¬˝ÊÕÁ◊∑§ ©à¬ÊŒ∑§ÃÊ ‚ŒÒfl Ÿ≈U ¬˝ÊÕÁ◊∑§ ©à¬ÊŒ∑§ÃÊ (2) Gross primary productivity is always more
‚ •Áœ∑§ „ÊÃË „Ò– than net primary productivity.

(3) ‚∑§‹ ¬˝ÊÕÁ◊∑§ ©à¬ÊŒ∑§ÃÊ •ÊÒ⁄U Ÿ≈U ¬˝ÊÕÁ◊∑§ ©à¬ÊŒ∑§ÃÊ (3) Gross primary productivity and Net primary
∞∑§ „Ë „Ò •ÊÒ⁄U •Á÷ÛÊ „Ò– productivity are one and same.

(4) ‚∑§‹ ¬˝ÊÕÁ◊∑§ ©à¬ÊŒ∑§ÃÊ •ÊÒ⁄U Ÿ≈U ¬˝ÊÕÁ◊∑§ ©à¬ÊŒ∑§ÃÊ (4) There is no relationship between Gross
primary productivity and Net primary
∑§ ’Ëø ∑§Ê߸ ‚ê’㜠Ÿ„Ë¥ „Ò– productivity.

Hindi+English
17 E1
70. ÁŸêŸ SÃ¥÷Ê¥ ∑§Ê Á◊‹ÊŸ ∑§⁄U ‚„Ë Áfl∑§À¬ ∑§Ê øÿŸ ∑§⁄UÊ– 70. Match the following columns and select the
correct option.
SÃ¥÷ - I SÃ¥÷ - II
Column - I Column - II
(a) •¬⁄UÊ (i) ∞¥«U˛Ê¡
 Ÿ
(a) Placenta (i) Androgens
(b) $¡ÊŸÊ ¬ÀÿÈÁ‚«UÊ (ii) ◊ÊŸfl ¡⁄UÊÿÈ
ªÊŸÒ«UÊ≈˛UÊÁ¬Ÿ (b) Zona pellucida (ii) Human Chorionic
Gonadotropin
(c) ’À’Ê-ÿÍ⁄UÕ˝‹ ª˝¥ÁÕÿÊ° (iii) •¥«UÊáÊÈ ∑§Ë ¬⁄UÃ
(hCG)
(d) ‹ËÁ«Uª ∑§ÊÁ‡Ê∑§Ê∞° (iv) Á‡Ê‡Ÿ ∑§Ê SŸ„Ÿ
(c) Bulbo-urethral (iii) Layer of the ovum
(a) (b) (c) (d)
glands
(1) (iv) (iii) (i) (ii)
(2) (i) (iv) (ii) (iii) (d) Leydig cells (iv) Lubrication of the
(3) (iii) (ii) (iv) (i) Penis
(4) (ii) (iii) (iv) (i) (a) (b) (c) (d)
(1) (iv) (iii) (i) (ii)
71. S≈˛UÊÁ’‹Ê߸ ÿÊ ‡Ê¥∑ȧ Á∑§‚◊¥ ¬Êÿ ¡ÊÃ „Ò¥?
(2) (i) (iv) (ii) (iii)
(1) ‚ÊÁÀflÁŸÿÊ (3) (iii) (ii) (iv) (i)
(2) ≈UÁ⁄U‚ (4) (ii) (iii) (iv) (i)
(3) ◊Ê∑Z§Á‡ÊÿÊ
71. Strobili or cones are found in :
(4) ßÄflË‚Ë≈U◊ (1) Salvinia
72. ∑ȧ¿U Áfl÷ÊÁ¡Ã „Ê ⁄U„Ë ∑§ÊÁ‡Ê∑§Êÿ¥ ∑§ÊÁ‡Ê∑§Ê ø∑˝§áÊ ‚ ’Ê„⁄U (2) Pteris
ÁŸ∑§‹ ¡ÊÃË „Ò¥ •ÊÒ⁄U ∑§ÊÁÿ∑§ ÁŸÁc∑˝§ÿÃÊ •flSÕÊ ◊¥ ¬˝fl‡Ê ∑§⁄U (3) Marchantia
¡ÊÃË „Ò– ß‚ ‡Êʥà •flSÕÊ (G0) ∑§„Ê ¡ÊÃÊ „Ò– ÿ„ ¬˝Á∑˝§ÿÊ (4) Equisetum
Á∑§‚∑§ •ãà ◊¥ „ÊÃË „Ò? 72. Some dividing cells exit the cell cycle and enter
(1) M ¬˝ÊflSÕÊ vegetative inactive stage. This is called quiescent
stage (G0). This process occurs at the end of :
(2) G1 ¬˝ÊflSÕÊ
(1) M phase
(3) S ¬˝ÊflSÕÊ (2) G1 phase
(4) G2 ¬˝ÊflSÕÊ (3) S phase
(4) G2 phase
73. ¬¥ÁÇflŸ ∞fl¥ «UÊÚ‹Á»§Ÿ ∑§ ¬ˇÊ ©ŒÊ„⁄UáÊ „Ò —
(1) •ŸÈ∑ͧ‹Ë ÁflÁ∑§⁄UáÊ ∑§Ê 73. Flippers of Penguins and Dolphins are examples
of :
(2) •Á÷‚Ê⁄UË Áfl∑§Ê‚ ∑§Ê
(1) Adaptive radiation
(3) •ÊÒlÊÁª∑§ ◊Ò‹ÁŸí◊ ∑§Ê (2) Convergent evolution
(4) ¬˝Ê∑ΧÁÃ∑§ fl⁄UáÊ ∑§Ê (3) Industrial melanism
(4) Natural selection
74. ÿÁŒ ŒÊ ‹ªÊÃÊ⁄U ˇÊÊ⁄U ÿÈÇ◊Ê¥ ∑§ ’Ëø ∑§Ë ŒÍ⁄UË 0.34 nm „Ò •ÊÒ⁄U
∞∑§ S߬ÊÿË ∑§ÊÁ‡Ê∑§Ê ∑§Ë DNA ∑§Ë Ám∑È¥§«U‹Ë ◊¥ ˇÊÊ⁄U ÿÈÇ◊Ê¥ 74. If the distance between two consecutive base pairs
∑§Ë ∑ȧ‹ ‚¥ÅÿÊ 6.6×109 bp „Ò– Ã’ DNA ∑§Ë ‹ê’Ê߸ is 0.34 nm and the total number of base pairs of a
DNA double helix in a typical mammalian cell is
„ÊªË ‹ª÷ª — 6.6×10 9 bp, then the length of the DNA is
(1) 2.0 ◊Ë≈U⁄U approximately :
(2) 2.5 ◊Ë≈U⁄U (1) 2.0 meters
(2) 2.5 meters
(3) 2.2 ◊Ë≈U⁄U
(3) 2.2 meters
(4) 2.7 ◊Ë≈U⁄U
(4) 2.7 meters

E1 18 Hindi+English

75. ◊ÊŸ∑§ ߸.‚Ë.¡Ë. ∑§Ê ÄÿÍ.•Ê⁄U.∞‚. ‚Áê◊üÊ Œ‡ÊʸÃÊ „Ò — 75. The QRS complex in a standard ECG represents :

(1) •ÊÁ‹¥ŒÊ¥ ∑§Ê ¬ÈŸœ˝È¸fláÊ (1) Repolarisation of auricles
(2) Depolarisation of auricles
(2) •ÊÁ‹¥ŒÊ¥ ∑§Ê Áflœ˝ÈfláÊ
(3) Depolarisation of ventricles
(3) ÁŸ‹ÿÊ¥ ∑§Ê Áflœ˝ÈfláÊ (4) Repolarisation of ventricles
(4) ÁŸ‹ÿÊ¥ ∑§Ê ¬ÈŸœ˝È¸fláÊ
76. Match the following columns and select the
76. ÁŸêŸ SÃ¥÷Ê¥ ∑§Ê Á◊‹ÊŸ ∑§⁄U ©Áøà Áfl∑§À¬ ∑§Ê øÿŸ ∑§⁄UÊ– correct option.
SÃ¥÷ - I SÃ¥÷ - II Column - I Column - II
(a) ß•ÊÁ‚ŸÊÁ»§‹ (i) ¬˝ÁÃ⁄UˇÊÊ ¬˝ÁÃÁ∑˝§ÿÊ (a) Eosinophils (i) Immune response
(b) ’‚ÊÁ»§‹ (ii) ÷ˇÊáÊ ∑§⁄UŸÊ (b) Basophils (ii) Phagocytosis
(c) ãÿÍ≈˛UÊÁ»§‹ (iii) Á„S≈UÊÁ◊Ÿ$¡, (c) Neutrophils (iii) Release
ÁflŸÊ‡Ê∑§Ê⁄UË ∞¥¡Êß◊Ê¥ histaminase,
∑§Ê ◊ÊøŸ destructive
(d) Á‹¥»§Ê‚Êß≈U (iv) ∑§áÊ Á¡Ÿ◊¥ Á„S≈UÊÁ◊Ÿ enzymes
„ÊÃ „Ò¥ ∑§Ê ◊ÊøŸ ∑§⁄UŸÊ (d) Lymphocytes (iv) Release granules
(a) (b) (c) (d) containing
(1) (iii) (iv) (ii) (i) histamine
(2) (iv) (i) (ii) (iii)
(a) (b) (c) (d)
(3) (i) (ii) (iv) (iii)
(1) (iii) (iv) (ii) (i)
(4) (ii) (i) (iii) (iv)
(2) (iv) (i) (ii) (iii)
77. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ‚„Ë „Ò? (3) (i) (ii) (iv) (iii)
(1) ∞Á«UŸËŸ ŒÊ H-’¥œÊ¥ ∑§ mÊ⁄UÊ ÕÊÿ◊ËŸ ∑§ ‚ÊÕ ÿÈÇ◊ (4) (ii) (i) (iii) (iv)
’ŸÊÃÊ „Ò–
(2) ∞Á«UŸËŸ ∞∑§ H-’¥œ ∑§ mÊ⁄UÊ ÕÊÿ◊ËŸ ∑§ ‚ÊÕ ÿÈÇ◊ 77. Which of the following statements is correct ?

’ŸÊÃÊ „Ò– (1) Adenine pairs with thymine through two
H-bonds.
(3) ∞Á«UŸËŸ ÃËŸ H-’¥œÊ¥ ∑§ mÊ⁄UÊ ÕÊÿ◊ËŸ ∑§ ‚ÊÕ ÿÈÇ◊ (2) Adenine pairs with thymine through one
’ŸÊÃÊ „Ò– H-bond.
(4) ∞Á«UŸËŸ, ÕÊÿ◊ËŸ ∑§ ‚ÊÕ ÿÈÇ◊ Ÿ„Ë¥ ’ŸÊÃÊ– (3) Adenine pairs with thymine through three
H-bonds.
78. ∞∑§ flÄ≈U⁄U ◊¥ ‚„‹ÇŸË «UË.∞Ÿ.∞. ∑§Ë ¬˝Áà ∑§Ë ‚¥ÅÿÊ ∑§Ê (4) Adenine does not pair with thymine.
ÁŸÿ¥ÁòÊà ∑§⁄UŸ flÊ‹ •ŸÈ∑˝§◊ ∑§Ê ÄÿÊ ∑§„Ê ¡ÊÃÊ „Ò?
78. The sequence that controls the copy number of the
(1) øÿŸÿÈÄà ◊Ê∑¸§⁄U linked DNA in the vector, is termed :
(2) •Ê⁄UË ‚Êß≈U (1) Selectable marker
(3) ¬Ò‹Ë¥«˛UÊÁ◊∑§ •ŸÈ∑˝§◊ (2) Ori site
(4) Á⁄U∑§ÊÚǟˇʟ (¬„øÊŸ) ‚Êß≈U (3) Palindromic sequence
(4) Recognition site
79. ÁŸêŸ ◊¥ ˇÊÊ⁄UËÿ ∞◊ËŸÊ •ê‹ ∑§Ê ¬„øÊÁŸ∞–
79. Identify the basic amino acid from the following.
(1) ≈UÊÿ⁄UÊ‚ËŸ
(1) Tyrosine
(2) Ç‹È≈UÊÁ◊∑§ •ê‹
(2) Glutamic Acid
(3) ‹ÊßÁ‚Ÿ (3) Lysine
(4) flÒ‹ËŸ (4) Valine

Hindi+English
19 E1
80. ÁŸêŸ SÃ¥÷Ê¥ ∑§Ê Á◊‹ÊŸ ∑§⁄U ‚„Ë Áfl∑§À¬ ∑§Ê øÿŸ ∑§⁄UÊ– 80. Match the following columns and select the
correct option.
SÃ¥÷ - I SÃ¥÷ - II
Column - I Column - II
(a) ¬ËÿÍ· ª˝¥ÁÕ (i) ª˝fl‚ ⁄UÊª
(a) Pituitary gland (i) Grave’s disease
(b) ÕÊÿ⁄UÊÚß«U ª˝¥ÁÕ (ii) «UÊÿÊÁ’≈UË¡ ◊Á‹≈U‚
(b) Thyroid gland (ii) Diabetes mellitus
(c) •ÁœflÎÄ∑§ ª˝¥ÁÕ (iii) «UÊÿÊÁ’≈UË¡
ßã‚ËÁ¬«U‚ (c) Adrenal gland (iii) Diabetes insipidus
(d) •ÇãÿʇÊÿ (iv) ∞«UË‚Ÿ ⁄UÊª (d) Pancreas (iv) Addison’s disease
(a) (b) (c) (d)
(a) (b) (c) (d)
(1) (iv) (iii) (i) (ii)
(1) (iv) (iii) (i) (ii)
(2) (iii) (ii) (i) (iv)
(2) (iii) (ii) (i) (iv)
(3) (iii) (i) (iv) (ii)
(4) (ii) (i) (iv) (iii) (3) (iii) (i) (iv) (ii)

(4) (ii) (i) (iv) (iii)
81. ‚„Ë ∑§ÕŸ ∑§Ê øÿŸ ∑§⁄UÊ–
(1) Ç‹Í∑§Ê∑§ÊÚÁ≈¸U∑§ÊÚß«U Ç‹Í∑§ÊÁŸÿÊÁ¡ŸÁ‚‚ ∑§Ê ¬˝Á⁄Uà ∑§⁄UÃ 81. Select the correct statement.
„Ò¥– (1) Glucocorticoids stimulate gluconeogenesis.
(2) Ç‹Í∑§ªÊÚŸ „Ê߬ÊÇ‹Êß‚ËÁ◊ÿÊ ‚ ‚¥’¥ÁœÃ „Ò– (2) Glucagon is associated with hypoglycemia.
(3) ߥ‚ÈÁ‹Ÿ •ÇãÿʇÊÿË ∑§ÊÁ‡Ê∑§Ê•Ê¥ ∞fl¥ ∞«UˬÊ‚Êß≈UÊ¥ ¬⁄U (3) Insulin acts on pancreatic cells and
Á∑˝§ÿÊ ∑§⁄UÃÊ „Ò– adipocytes.

(4) ߥ‚ÈÁ‹Ÿ „Ê߬⁄UÇ‹Êß‚ËÁ◊ÿÊ ‚ ‚¥’¥ÁœÃ „Ò– (4) Insulin is associated with hyperglycemia.

82. ÁŸêŸ ◊¥ ∑§ÊÒŸ‚Ë ¬˝Ê≈UËŸ ¡ãÃÈ•Ê¥ ◊¥ ’„ÈÃÊÿà ‚ „ÊÃË „Ò? 82. Which one of the following is the most abundant
protein in the animals ?
(1) „Ë◊ÊÇ‹ÊÁ’Ÿ
(1) Haemoglobin
(2) ∑§Ê‹¡
 Ÿ
(2) Collagen
(3) ‹ÒÁÄ≈UŸ
(3) Lectin
(4) ߥ‚ÈÁ‹Ÿ
(4) Insulin

83. fl¥‡ÊʪÁà ∑§ ªÈáÊ‚ÍòÊ Á‚hÊãà ∑§Ê ¬˝ÊÿÊÁª∑§ ¬˝◊ÊáÊŸ Á∑§‚Ÿ
83. Experimental verification of the chromosomal
Á∑§ÿÊ ÕÊ? theory of inheritance was done by :
(1) ◊¥«U‹ (1) Mendel
(2) ‚≈UŸ (2) Sutton

(3) ’Êfl⁄UË (3) Boveri

(4) ◊ÊÚª¸Ÿ (4) Morgan

E1 20 Hindi+English

84. ÁŸêŸ SÃ¥÷Ê¥ ∑§Ê Á◊‹ÊŸ ∑§⁄U ‚„Ë Áfl∑§À¬ ∑§Ê øÿŸ ∑§⁄UÊ– 84. Match the following columns and select the
correct option.
SÃ¥÷ - I SÃ¥÷ - II
Column - I Column - II
(a) å‹ÊflË ¬‚Á‹ÿÊ° (i) ŒÍ‚⁄UË ∞fl¥ ‚ÊÃflË¥
(a) Floating Ribs (i) Located between
¬‚‹Ë ∑§ ’Ëø ÁSÕÃ
second and
„ÊÃË „Ò¥
seventh ribs
(b) ∞∑˝§ÊÁ◊ÿŸ (ii) sÍ◊⁄U‚ ∑§Ê ‡ÊË·¸
(b) Acromion (ii) Head of the
(c) S∑Ò§¬È‹Ê (iii) Ä‹Áfl∑§‹
Humerus
(d) Ç‹ËŸÊÚÿ«U ªÈ„Ê (iv) ©⁄UÊÁSÕ ‚ Ÿ„Ë¥ ¡È«∏ÃË
(c) Scapula (iii) Clavicle
(a) (b) (c) (d)
(d) Glenoid cavity (iv) Do not connect
(1) (ii) (iv) (i) (iii)
with the sternum
(2) (i) (iii) (ii) (iv)
(a) (b) (c) (d)
(3) (iii) (ii) (iv) (i)
(1) (ii) (iv) (i) (iii)
(4) (iv) (iii) (i) (ii)
(2) (i) (iii) (ii) (iv)
(3) (iii) (ii) (iv) (i)
85. Á‚Á≈˛U∑§ •ê‹ ø∑˝§ ∑§ ∞∑§ ÉÊÈ◊Êfl ◊¥ ∑§Êÿ¸Œ˝fl SÃ⁄U »§ÊS»§ÊÁ⁄U‹‡ÊŸÊ¥
(4) (iv) (iii) (i) (ii)
∑§Ë ‚¥ÅÿÊ ÄÿÊ „ÊÃË „Ò?
(1) ‡ÊÍãÿ 85. The number of substrate level phosphorylations
(2) ∞∑§ in one turn of citric acid cycle is :

(3) ŒÊ (1) Zero
(2) One
(4) ÃËŸ
(3) Two
(4) Three
86. Á‚Ÿå≈UÊŸË◊‹ ‚Áê◊üÊ ∑§Ê ÁflÉÊ≈UŸ „ÊÃÊ „Ò —
(1) SÕÍ‹¬^ ∑§ ŒÊÒ⁄UÊŸ 86. Dissolution of the synaptonemal complex occurs
during :
(2) ÿÈÇ◊¬^ ∑§ ŒÊÒ⁄UÊŸ
(1) Pachytene
(3) Ám¬^ ∑§ ŒÊÒ⁄UÊŸ
(2) Zygotene
(4) ßȬ^ ∑§ ŒÊÒ⁄UÊŸ (3) Diplotene
(4) Leptotene
87. Ám¬Ê‡fl¸ ‚◊Á◊Áà ∞fl¥ •ªÈ„Ëÿ ¡ãÃÈ•Ê¥ ∑§ ©ŒÊ„⁄UáÊ Á∑§‚ ‚¥ÉÊ ◊¥
„Ò¥? 87. Bilaterally symmetrical and acoelomate animals
(1) ≈UËŸÊ»§Ê⁄UÊ are exemplified by :
(1) Ctenophora
(2) å‹≈UË„ÒÁÀ◊¥ÕË¡
(2) Platyhelminthes
(3) ∞S∑§„ÒÁÀ◊¥ÕË¡ (3) Aschelminthes
(4) ∞ŸÁ‹«UÊ (4) Annelida

88. ’Ë¡Êá«U ∑§Ê Á¬¥«U, ’Ë¡Êá«U flΥà ‚ ∑§„Ê° ¬⁄U ‚¥‹Áÿà „ÊÃÊ „Ò? 88. The body of the ovule is fused within the funicle
at :
(1) ŸÊÁ÷∑§Ê
(1) Hilum
(2) ’Ë¡Êá«UmÊ⁄U (2) Micropyle
(3) ’Ë¡Êá«U∑§Êÿ (3) Nucellus
(4) ÁŸ÷ʪ (4) Chalaza

Hindi+English
21 E1
89. •Ê„Ê⁄U ŸÊ‹ ∑§Ë ªÊé‹≈U ∑§ÊÁ‡Ê∑§Ê∞° M§¬Ê¥ÃÁ⁄Uà „ÊÃË „Ò¥ — 89. Goblet cells of alimentary canal are modified
from :
(1) ‡ÊÀ∑§Ë ©¬∑§‹Ê ∑§ÊÁ‡Ê∑§Ê•Ê¥ ‚
(1) Squamous epithelial cells
(2) SÃ¥÷Ê∑§Ê⁄U ©¬∑§‹Ê ∑§ÊÁ‡Ê∑§Ê•Ê¥ ‚ (2) Columnar epithelial cells
(3) ©¬ÊÁSÕ ∑§ÊÁ‡Ê∑§Ê•Ê¥ ‚ (3) Chondrocytes
(4) ‚¥ÿÈÄà ©¬∑§‹Ê ∑§ÊÁ‡Ê∑§Ê•Ê¥ ‚ (4) Compound epithelial cells

90. •¥≈UÊ∑¸˜§Á≈U∑§ ˇÊòÊ ◊¥ Á„◊-•¥œÃÊ Á∑§‚ ∑§Ê⁄UáÊ „ÊÃË „Ò? 90. Snow-blindness in Antarctic region is due to :
(1) ÁŸêŸ Ãʬ mÊ⁄UÊ •Ê°π ◊¥ Œ˝fl ∑§ ¡◊Ÿ ∑§ ∑§Ê⁄UáÊ (1) Freezing of fluids in the eye by low
temperature
(2) UV-B ÁflÁ∑§⁄UáÊ ∑§Ë ©ìÊ ◊ÊòÊÊ ∑§ ∑§Ê⁄UáÊ ∑§ÊÚÁŸ¸ÿÊ ∑§Ê
(2) Inflammation of cornea due to high dose of
‡ÊÊÕ UV-B radiation
(3) Á„◊ ‚ ¬˝∑§Ê‡Ê ∑§Ê ©ìÊ ¬⁄UÊfløŸ (3) High reflection of light from snow
(4) •fl⁄UÄà Á∑§⁄UáÊÊ¥ mÊ⁄UÊ ⁄U≈UËŸÊ ◊¥ ˇÊÁà (4) Damage to retina caused by infra-red rays

91. ¬„øÊÁŸ∞ Á∑§ ∑§ÊÒŸ-‚ •áÊÈ ∑§Ê •ÁSÃàfl Ÿ„Ë¥ „Ò–
91. Identify a molecule which does not exist.
(1) He2
(1) He2
(2) Li2
(3) C2 (2) Li2

(4) O2 (3) C2
(4) O2
92. Ni(OH)2 ∑§Ë 0.1 M NaOH ◊¥ Áfl‹ÿÃÊ ôÊÊà ∑§ËÁ¡∞–
ÁŒÿÊ „Ò Á∑§ Ni(OH)2 ∑§Ê •ÊÿŸË ªÈáÊŸ»§‹ 2×10−15 „Ò– 92. Find out the solubility of Ni(OH)2 in 0.1 M NaOH.
(1) 2×10−13 M Given that the ionic product of Ni(OH)2 is
2×10−15.
(2) 2×10−8 M
(1) 2×10−13 M
(3) 1×10−13 M
(2) 2×10−8 M
(4) 1×108 M
(3) 1×10−13 M
93. ÁŸêŸÁ‹Áπà ◊¥ ‚ ©Áøà ∑§ÕŸ ¬„øÊÁŸ∞ — (4) 1×108 M
(a) CO2(g) ∑§Ê •Êß‚∑˝§Ë◊ •ÊÒ⁄U Á„◊‡ÊËÁÃà πÊl ∑§ Á‹∞
¬˝‡ÊËÃ∑§ ∑§ M§¬ ◊¥ ©¬ÿÊª Á∑§ÿÊ ¡ÊÃÊ „Ò– 93. Identify the correct statements from the
following :
(b) C60 ∑§Ë ‚¥⁄UøŸÊ ◊¥, ’Ê⁄U„ ¿U— ∑§Ê’¸Ÿ fl‹ÿ •ÊÒ⁄U ’Ë‚
(a) CO2(g) is used as refrigerant for ice-cream
¬Ê°ø ∑§Ê’¸Ÿ fl‹ÿ „ÊÃ „Ò¥– and frozen food.
(c) ZSM-5, ∞∑§ ¬˝∑§Ê⁄U ∑§Ê Á¡•Ê‹Êß≈U „Ò ¡Ê ∞À∑§Ê„ÊÚ‹ (b) The structure of C60 contains twelve six
∑§Ê ªÒ‚Ê‹ËŸ ◊¥ M§¬Ê¥ÃÁ⁄Uà ∑§⁄UŸ ◊¥ ©¬ÿÊª Á∑§ÿÊ ¡ÊÃÊ carbon rings and twenty five carbon rings.
„Ò– (c) ZSM-5, a type of zeolite, is used to convert
alcohols into gasoline.
(d) CO ⁄¥Uª„ËŸ •ÊÒ⁄U ª¥œ„ËŸ ªÒ‚ „Ò–
(d) CO is colorless and odourless gas.
(1) ∑§fl‹ (a), (b) •ÊÒ⁄U (c)
(1) (a), (b) and (c) only
(2) ∑§fl‹ (a) •ÊÒ⁄U (c) (2) (a) and (c) only
(3) ∑§fl‹ (b) •ÊÒ⁄U (c) (3) (b) and (c) only
(4) ∑§fl‹ (c) •ÊÒ⁄U (d) (4) (c) and (d) only

E1 22 Hindi+English

94. ‚È∑˝§Ê‚ ∑§Ê ¡‹-•¬ÉÊ≈UŸ ÁŸêŸÁ‹Áπà •Á÷Á∑˝§ÿÊ mÊ⁄UÊ ÁŒÿÊ 94. Hydrolysis of sucrose is given by the following
¡ÊÃÊ „Ò — reaction.

‚È∑˝§Ê‚+H2O ⇌ Ç‹Í∑§Ê‚+»˝§Ä≈UÊ‚
 Sucrose+H2O ⇌ Glucose+Fructose

ÿÁŒ 300 K ¬⁄U ‚Êêÿ ÁSÕ⁄UÊ¥∑§ (Kc) 2×1013 „Ê, ÃÊ ©‚Ë If the equilibrium constant (Kc) is 2×1013 at
300 K, the value of ∆rGs at the same temperature
Ãʬ ¬⁄U ∆rGs ∑§Ê ◊ÊŸ „ÊªÊ — will be :
(1) −8.314 J mol−1K−1×300 K×ln(2×1013) (1) −8.314 J mol−1K−1×300 K×ln(2×1013)
(2) 8.314 J mol−1K−1×300 K×ln(2×1013) (2) 8.314 J mol−1K−1×300 K×ln(2×1013)
(3) 8.314 J mol−1K−1×300 K×ln(3×1013) (3) 8.314 J mol−1K−1×300 K×ln(3×1013)
(4) −8.314 J mol−1K−1×300 K×ln(4×1013) (4) −8.314 J mol−1K−1×300 K×ln(4×1013)

95. •Á÷Á∑˝ § ÿÊ•Ê ¥ ∑ § ÁŸêŸÁ‹Áπà ∑˝ § ◊ ◊ ¥ X ÿÊÒ Á ª∑§ ∑§Ê 95. Identify compound X in the following sequence of
¬„øÊÁŸ∞ — reactions :

(1) (1)

(2) (2)

(3) (3)

(4) (4)

Hindi+English
23 E1
96. •ŸÈÁøà ‚È◊‹ ∑§Ê ¬„øÊÁŸ∞– 96. Identify the incorrect match.

ŸÊ◊ •Ê߸.ÿÍ.¬Ë.∞.‚Ë. •Áœ∑Χà ŸÊ◊ Name IUPAC Official Name

(a) •ŸÁŸ‹©ÁŸÿ◊ (i) ◊Ò¥«U‹ËÁflÿ◊ (a) Unnilunium (i) Mendelevium
(b) •ŸÁŸ‹≈˛UÊßÿ◊ (ii) ‹Ê⁄¥UÁ‚ÿ◊ (b) Unniltrium (ii) Lawrencium
(c) •ŸÁŸ‹„ÁÄ‚ÿ◊ (iii) ‚Ë’ÊÁª¸ÿ◊ (c) Unnilhexium (iii) Seaborgium
(d) •Ÿ•ŸÿÈÁŸÿ◊ (iv) «U◊¸S≈UU«U˜Á≈Uÿ◊ (d) Unununnium (iv) Darmstadtium
(1) (a), (i) (1) (a), (i)
(2) (b), (ii) (2) (b), (ii)
(3) (c), (iii) (3) (c), (iii)
(4) (d), (iv) (4) (d), (iv)

97. ∞∑§ Ãàfl ∑§Ë 288 pm ‚‹ ∑§Ê⁄U flÊ‹Ë ∑§Êÿ ∑§ÁãŒ˝Ã ÉÊŸËÿ
97. An element has a body centered cubic (bcc)
‚¥⁄UøŸÊ „Ò, ¬⁄U◊ÊáÊÈ ÁòÊíÿÊ „Ò —
structure with a cell edge of 288 pm. The atomic
3 radius is :
(1) × 288 pm
4
3
2 (1) × 288 pm
(2) × 288 pm 4
4
2
4 (2) × 288 pm
(3) × 288 pm 4
3
4
4 (3) × 288 pm
(4) × 288 pm 3
2
4
(4) × 288 pm
98. ÁŸêŸÁ‹Áπà ◊¥ ‚ •áÊÈ•Ê¥ ∑§ Á∑§‚ ‚◊ÈìÊÿ ∑§Ê ‡ÊÍãÿ Ámœ˝Èfl 2
•ÊÉÊÍáʸ „ÊÃÊ „Ò?
98. Which of the following set of molecules will have
(1) •◊Ê Á ŸÿÊ, ’ Á ⁄UÁ‹ÿ◊ «UÊßç‹È • Ê ⁄ UÊß«U, ¡‹, zero dipole moment ?
1,4-«UÊßÄ‹Ê⁄UÊ’ã$¡ËŸ
(1) Ammonia, beryllium difluoride, water,
(2) ’Ê⁄UÊÚŸ ≈˛UÊßç‹È•Ê⁄UÊß«U, „Êß«˛UÊ¡Ÿ ç‹È•Ê⁄UÊß«U, ∑§Ê’¸Ÿ 1,4-dichlorobenzene
«UÊß•ÊÚÄ‚Êß«U, 1,3-«UÊßÄ‹Ê⁄UÊ’ã$¡ËŸ (2) Boron trifluoride, hydrogen fluoride, carbon
dioxide, 1,3-dichlorobenzene
(3) ŸÊß≈˛UÊ¡Ÿ ≈˛UÊßç‹È•Ê⁄UÊß«U, ’Á⁄UÁ‹ÿ◊ «UÊßç‹È•Ê⁄UÊUß«U,
¡‹, 1,3-«UÊßÄ‹Ê⁄UÊ’ã$¡ËŸ (3) Nitrogen trifluoride, beryllium difluoride,
water, 1,3-dichlorobenzene
(4) ’Ê⁄UÊŸÚ ≈˛UÊßç‹È•Ê⁄UÊß«U, ’Á⁄UÁ‹ÿ◊ «UÊßç‹È•Ê⁄UÊß«U, ∑§Ê’¸Ÿ
(4) Boron trifluoride, beryllium difluoride,
«UÊß•ÊÚÄ‚Êß«U, 1,4-«UÊßÄ‹Ê⁄UÊ’ã$¡ËŸ carbon dioxide, 1,4-dichlorobenzene

99. å‹ÒÁ≈UŸ◊ (Pt) ß‹Ä≈˛UÊ«U ∑§Ê ©¬ÿÊª ∑§⁄UÃ „È∞ ÃŸÈ ‚ÀçÿÍÁ⁄U∑§ 99. On electrolysis of dil.sulphuric acid using
•ê‹ ∑§ flÒlÈà •¬ÉÊ≈UŸ ¬⁄U, ∞ŸÊ«U ¬⁄U ¬˝Êåà ©à¬ÊŒ „ÊªÊ — Platinum (Pt) electrode, the product obtained at
anode will be :
(1) „Êß«˛UÊ¡Ÿ ªÒ‚
(1) Hydrogen gas
(2) •ÊÚÄ‚Ë¡Ÿ ªÒ‚
(2) Oxygen gas
(3) H2S ªÒ‚
(3) H2S gas
(4) SO2 ªÒ‚ (4) SO2 gas

E1 24 Hindi+English

100. ∞‚Ë≈UÊŸ •ÊÒ⁄U ◊ÁÕ‹◊ÒÇŸËÁ‡Êÿ◊ Ä‹Ê⁄UÊß«U ∑§Ë •Á÷Á∑˝§ÿÊ •ÊÒ⁄U 100. Reaction between acetone and methylmagnesium
chloride followed by hydrolysis will give :
ÃଇøÊØ ¡‹-•¬ÉÊ≈UŸ ‚ ¬˝Êåà „ÊªÊ —
(1) Isopropyl alcohol
(1) •Êß‚Ê¬˝ÊÁ¬‹ ∞À∑§Ê„ÊÚ‹
(2) Sec. butyl alcohol
(2) ÁmÃËÿ∑§ éÿÍÁ≈U‹ ∞À∑§Ê„ÊÚ‹
(3) Tert. butyl alcohol
(3) ÃÎÃËÿ∑§ éÿÍÁ≈U‹ ∞À∑§Ê„ÊÚ‹
(4) Isobutyl alcohol
(4) •Êß‚ÊéÿÍÁ≈U‹ ∞À∑§Ê„ÊÚ‹
101. Which of the following oxoacid of sulphur has
101. ÁŸêŸÁ‹Áπà ◊¥ ‚ ‚À»§⁄U ∑§ Á∑§‚ •ÊÚÄ‚Ê•ê‹ ◊¥ −O−O− −O−O− linkage ?
’¥œŸ „Ò? (1) H2SO3, sulphurous acid
(1) H2SO3, ‚À$çÿÍ⁄U‚ •ê‹ (2) H2SO4, sulphuric acid
(2) H2SO4, ‚À$$çÿÍÁ⁄U∑§ •ê‹ (3) H2S2O8, peroxodisulphuric acid

(3) H2S2O8, ¬⁄U•ÊÚÄ‚Ê«UÊß‚À$çÿÍÁ⁄U∑§ •ê‹ (4) H2S2O7, pyrosulphuric acid

(4) H2S2O7, ¬Êß⁄UÊ‚À$çÿÍÁ⁄U∑§ •ê‹
102. Which of the following amine will give the
102. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ-‚Ë ∞◊ËŸ ∑§ÊÁ’¸‹∞◊ËŸ ¬⁄UˡÊáÊ ŒªË? carbylamine test ?

(1) (1)

(2) (2)

(3) (3)

(4) (4)

Hindi+English
25 E1
103. Cr2+ ∑§ Á‹∞, ∑§fl‹ ¬˝ø∑˝§áÊ øÈ¥’∑§Ëÿ •ÊÉÊÍáʸ ∑§Ê ¬Á⁄U∑§Á‹Ã 103. The calculated spin only magnetic moment of Cr2+
ion is :
◊ÊŸ „Ò —
(1) 3.87 BM (1) 3.87 BM
(2) 4.90 BM (2) 4.90 BM
(3) 5.92 BM (3) 5.92 BM
(4) 2.84 BM (4) 2.84 BM

104. Á∑§‚Ë •ÊŒ‡Ê¸ ªÒ‚ ∑§ L§hÊc◊ ¬Á⁄UÁSÕÁà ◊¥ ◊ÈÄà ¬˝‚⁄UáÊ ∑§ Á‹∞ 104. The correct option for free expansion of an ideal
©Áøà Áfl∑§À¬ „Ò — gas under adiabatic condition is :
(1) q=0, ∆T=0 •ÊÒ⁄U w=0 (1) q=0, ∆T=0 and w=0
(2) q=0, ∆T < 0 •ÊÒ⁄U w > 0 (2) q=0, ∆T < 0 and w > 0
(3) q < 0, ∆T=0 •ÊÒ⁄U w=0 (3) q < 0, ∆T=0 and w=0

(4) q > 0, ∆T > 0 •ÊÒ⁄U w > 0 (4) q > 0, ∆T > 0 and w > 0

105. ’ã$¡ËŸ ∑§Ê Á„◊Ê¥∑§ •flŸ◊Ÿ ÁSÕ⁄UÊ¥∑§ (Kf) 5.12 K kg mol−1 105. The freezing point depression constant (Kf) of
benzene is 5.12 K kg mol−1. The freezing point
„Ò– ’ã$¡ËŸ ◊¥ ∞∑§ ÁfllÈØ-•Ÿ¬ÉÊ≈˜Uÿ Áfl‹ÿ flÊ‹ 0.078 m depression for the solution of molality 0.078 m
◊Ê‹‹ÃÊ flÊ‹ Áfl‹ÿŸ ∑§Ê Á„◊Ê¥∑§ •flŸ◊Ÿ (ŒÊ Œ‡Ê◊‹fl containing a non-electrolyte solute in benzene is
SÕÊŸÊ¥ Ã∑§ ÁŸ∑§Á≈UÃ), „Ò — (rounded off upto two decimal places) :
(1) 0.20 K (1) 0.20 K
(2) 0.80 K (2) 0.80 K
(3) 0.40 K (3) 0.40 K
(4) 0.60 K
(4) 0.60 K
106. ªÁ‹Ã CaCl2 ‚ 20 g ∑Ò§ÁÀ‡Êÿ◊ ¬˝Êåà ∑§⁄UŸ ∑§ Á‹∞ •Êfl‡ÿ∑§
106. The number of Faradays(F) required to produce
»Ò§⁄UÊ«U(F) ∑§Ë ‚¥ÅÿÊ „Ò, 20 g of calcium from molten CaCl2 (Atomic mass
(Ca ∑§Ê ¬⁄U◊ÊáÊÈ Œ˝√ÿ◊ÊŸ=40 ª˝Ê◊/◊Ê‹) of Ca=40 g mol−1) is :
(1) 1 (1) 1
(2) 2
(2) 2
(3) 3
(3) 3
(4) 4
(4) 4
107. ’ã$¡ÒÁÀ«U„Êß«U •ÊÒ⁄U ∞‚Ë≈UÊ$»§ËŸÊŸ ∑§Ë ÃŸÈ NaOH ∑§Ë ©¬ÁSÕÁÃ
◊¥ •Á÷Á∑˝§ÿÊ ß‚ ¬˝∑§Ê⁄U ¡ÊŸË ¡ÊÃË „Ò — 107. Reaction between benzaldehyde and acetophenone
in presence of dilute NaOH is known as :
(1) ∞‹˜«UÊÚ‹ ‚¥ÉÊŸŸ
(1) Aldol condensation
(2) ∑Ò§ÁŸ$¡Ê⁄UÊ •Á÷Á∑˝§ÿÊ
(2) Cannizzaro’s reaction
(3) ∑˝§ÊÚ‚ ∑Ò§ÁŸ$¡Ê⁄UÊ •Á÷Á∑˝§ÿÊ (3) Cross Cannizzaro’s reaction
(4) ∑˝§ÊÚ‚ ∞‹˜«UÊÚ‹ ‚¥ÉÊŸŸ (4) Cross Aldol condensation

108. ∑§Êª$¡ fláʸ‹Áπ∑§Ë, ©ŒÊ„⁄UáÊ „Ò — 108. Paper chromatography is an example of :
(1) •Áœ‡ÊÊ·áÊ fláʸ‹Áπ∑§Ë ∑§Ê (1) Adsorption chromatography
(2) Áfl¬Ê≈UŸ fláʸ‹Áπ∑§Ë ∑§Ê (2) Partition chromatography
(3) ¬Ã‹Ë ¬⁄Uà fláʸ‹Áπ∑§Ë ∑§Ê (3) Thin layer chromatography
(4) SÃ¥÷ fláʸ‹Áπ∑§Ë ∑§Ê (4) Column chromatography

E1 26 Hindi+English

109. Á∑§‚Ë •Á÷Á∑˝§ÿÊ ∑§ •Á÷∑§Ê⁄U∑§Ê¥ ∑§Ë ‚Ê¥Œ˝ÃÊ ◊¥ flÎÁh ‚ ¬Á⁄UfløŸ 109. An increase in the concentration of the reactants
of a reaction leads to change in :
„ÊªÊ —
(1) activation energy
(1) ‚Á∑˝§ÿáÊ ™§¡Ê¸ ◊¥
(2) heat of reaction
(2) •Á÷Á∑˝§ÿÊ ∑§Ë ™§c◊Ê ◊¥
(3) threshold energy
(3) Œ„‹Ë ™§¡Ê¸ ◊¥
(4) collision frequency
(4) ‚¥ÉÊ≈˜U≈U •ÊflÎÁûÊ ◊¥
110. A mixture of N2 and Ar gases in a cylinder contains
110. ∞∑§ Á‚Á‹¥«U⁄U ◊¥ N2 •ÊÒ⁄U Ar ªÒ‚Ê¥ ∑§ ∞∑§ Á◊üÊáÊ ◊¥ N2 ∑§ 7 g of N2 and 8 g of Ar. If the total pressure of the
7 g •ÊÒ⁄U Ar ∑§ 8 g „Ò¥– ÿÁŒ Á‚Á‹¥«U⁄U ◊¥ ªÒ‚Ê¥ ∑§ Á◊üÊáÊ ∑§Ê mixture of the gases in the cylinder is 27 bar, the
partial pressure of N2 is :
∑ȧ‹ ŒÊ’ 27 bar „Ê, ÃÊ N2 ∑§Ê •Ê¥Á‡Ê∑§ ŒÊ’ „Ò,
[Use atomic masses (in g mol−1) : N=14, Ar=40]
[¬⁄U◊ÊáÊÈ Œ˝√ÿ◊ÊŸÊ¥ (g mol−1 ◊¥) : N=14, Ar=40 ©¬ÿÊª
∑§ËÁ¡∞] (1) 9 bar
(1) 9 bar (2) 12 bar
(2) 12 bar (3) 15 bar
(3) 15 bar (4) 18 bar
(4) 18 bar
111. Identify the correct statement from the
111. ÁŸêŸÁ‹Áπà ◊¥ ‚ ‚„Ë ∑§ÕŸ ¬„øÊÁŸ∞ — following :

(1) Á¬≈UflÊ° ‹Ê„Ê 4% ∑§Ê’¸Ÿ flÊ‹Ê •‡ÊÈh ‹Ê„Ê „ÊÃÊ „Ò– (1) Wrought iron is impure iron with
4% carbon.
(2) »§»§Ê‹ŒÊ⁄U ÃÊ¥’Ê, CO2 ∑§ ÁŸ∑§Ê‚ ∑§ ∑§Ê⁄UáÊ »§»§Ê‹ŒÊ⁄U
(2) Blister copper has blistered appearance due
‹ªÃÊ „Ò– to evolution of CO2.
(3) ÁŸ∑Ò§‹ ∑§ Á‹∞ flÊc¬ ¬˝ÊflSÕÊ ‡ÊÊœŸ flÒŸ •Ê∑¸§‹ ÁflÁœ (3) Vapour phase refining is carried out for
mÊ⁄UÊ Á∑§ÿÊ ¡ÊÃÊ „Ò– Nickel by Van Arkel method.

(4) ∑§ìÊ ‹Ê„ ∑§Ê ÁflÁ÷ÛÊ •Ê∑§Ê⁄UÊ¥ ◊¥ …UÊ‹Ê ¡Ê ‚∑§ÃÊ „Ò– (4) Pig iron can be moulded into a variety of
shapes.

112. ÁŸêŸÁ‹Áπà ◊¥ ‚ Á∑§‚∑§ ∑§Ê⁄UáÊ ∞∑§ ÃÎÃËÿ∑§ éÿÍÁ≈U‹
∑§Ê’Ê¸œŸÊÿŸ ∞∑§ ÁmÃËÿ∑§ éÿÍÁ≈U‹ ∑§Ê’Ê¸œŸÊÿŸ ‚ •Áœ∑§ 112. A tertiary butyl carbocation is more stable than a
SÕÊÿË „ÊÃÊ „Ò? secondary butyl carbocation because of which of
the following ?
(1) −CH3 ‚◊Í„Ê¥ ∑§ −I ¬˝÷ÊflU ∑§ ∑§Ê⁄UáÊ
(1) −I effect of −CH3 groups
(2) −CH3 ‚◊Í„Ê¥ ∑§ +R ¬˝÷ÊflU ∑§ ∑§Ê⁄UáÊ
(2) +R effect of −CH3 groups
(3) −CH3 ‚◊Í„Ê¥ ∑§ −R ¬˝÷Êfl ∑§ ∑§Ê⁄UáÊ
(3) −R effect of −CH3 groups
(4) •ÁÂ¥ÿÇÈ ◊Ÿ (4) Hyperconjugation

113. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ-‚Ê œŸÊÿŸË •¬◊Ê¡¸∑§ „Ò? 113. Which of the following is a cationic detergent ?
(1) ‚ÊÁ«Uÿ◊ ‹ÊÚ⁄UÊß‹ ‚À»§≈U (1) Sodium lauryl sulphate
(2) ‚ÊÁ«Uÿ◊ ÁS≈U∞⁄U≈U (2) Sodium stearate
(3) ‚Á≈U‹≈˛UÊß◊ÁÕ‹ •◊ÊÁŸÿ◊ ’˝Ê◊Êß«U (3) Cetyltrimethyl ammonium bromide

(4) ‚ÊÁ«Uÿ◊ «UÊ«UÁ‚‹’ã$¡ËŸ ‚À»§ÊŸ≈U (4) Sodium dodecylbenzene sulphonate

Hindi+English
27 E1
114. 2-’˝Ê◊Ê-¬ã≈UŸ ‚ ¬ã≈U-2-߸Ÿ ’ŸŸ ∑§Ë Áfl‹Ê¬Ÿ •Á÷Á∑˝§ÿÊ — 114. Elimination reaction of 2-Bromo-pentane to form
pent-2-ene is :
(a) β-Áfl‹Ê¬Ÿ •Á÷Á∑˝§ÿÊ „Ò
(a) β-Elimination reaction
(b) ¡≈U‚Ò»§ ÁŸÿ◊ ∑§Ê ¬Ê‹Ÿ ∑§⁄UÃË „Ò (b) Follows Zaitsev rule
(c) Áfl„Êß«˛UÊ„Ò‹Ê¡ŸË∑§⁄UáÊ •Á÷Á∑˝§ÿÊ „Ò (c) Dehydrohalogenation reaction
(d) ÁŸ¡¸‹Ë∑§⁄UáÊ •Á÷Á∑˝§ÿÊ „Ò (d) Dehydration reaction
(1) (a), (b), (c) (1) (a), (b), (c)
(2) (a), (c), (d)
(2) (a), (c), (d)
(3) (b), (c), (d)
(3) (b), (c), (d)
(4) (a), (b), (d)
(4) (a), (b), (d)
115. fl„ Á◊üÊáÊ ¡Ê ⁄UÊ©À≈U ÁŸÿ◊ ‚ œŸÊà◊∑§ Áflø‹Ÿ ¬˝ŒÁ‡Ê¸Ã
∑§⁄UÃÊ „Ò, „Ò — 115. The mixture which shows positive deviation from
Raoult’s law is :
(1) ∞ÕÊŸÊÚ‹+∞‚Ë≈UÊŸ (1) Ethanol+Acetone
(2) ’ã$¡ËŸ+≈UÊ‹Í߸Ÿ (2) Benzene+Toluene
(3) ∞‚Ë≈UÊŸ+Ä‹Ê⁄UÊ$»§ÊÚ◊¸ (3) Acetone+Chloroform
(4) Ä‹Ê⁄UÊ∞ ÕŸ+’˝Ê◊ Ê∞ÕŸ (4) Chloroethane+Bromoethane

116. ÁŸêŸÁ‹Áπà ◊¥ ‚, ‚◊ãflÿ ÿÊÒÁª∑§Ê¥ ∑§Ê ’ŸÊŸ ∑§ Á‹∞ ‚¥ÀÊÁÇãÊÿÊ¥ 116. Which of the following is the correct order of
∑§Ë ’…∏ÃË ˇÊòÊ ¬˝’‹ÃÊ ∑§Ê ∑§ÊÒŸ-‚Ê ‚„Ë ∑˝§◊ „Ò? increasing field strength of ligands to form
(1) SCN− < F− < C2O42− < CN− coordination compounds ?
(2) SCN− < F− < CN− < C2O42− (1) SCN− < F− < C2O42− < CN−
(3) F− < SCN− < C2O42−< CN− (2) SCN− < F− < CN− < C2O42−
(4) CN− < C2O42−< SCN− < F− (3) F− < SCN− < C2O42−< CN−
117. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ-‚Ê ∞∑§ ˇÊÊ⁄UËÿ ∞◊ËŸÊ •ê‹ „Ò? (4) CN− < C2O42−< SCN− < F−
(1) ‚⁄UËŸ
117. Which of the following is a basic amino acid ?
(2) ∞‹ÊÁŸŸ
(1) Serine
(3) ≈UÊß⁄UÊ‚ËŸ
(2) Alanine
(4) ‹Êß‚ËŸ (3) Tyrosine
118. HCl ∑§Ê CaCl2, MgCl2 •ÊÒ⁄U NaCl ∑§ Áfl‹ÿŸ ‚ ªÈ$¡Ê⁄UÊ (4) Lysine
ªÿÊ– ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ-‚Ê/∑§ÊÒŸ-‚ ÿÊÒÁª∑§ Á∑˝§S≈UÁ‹Ã 118. HCl was passed through a solution of CaCl2, MgCl2
„È•Ê/„È∞? and NaCl. Which of the following compound(s)
(1) MgCl2 •ÊÒ⁄U CaCl2 ŒÊŸÊ¥ crystallise(s) ?

(2) ∑§fl‹ NaCl (1) Both MgCl2 and CaCl2
(2) Only NaCl
(3) ∑§fl‹ MgCl2
(3) Only MgCl2
(4) NaCl, MgCl2 •ÊÒ⁄U CaCl2
(4) NaCl, MgCl2 and CaCl2
119. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ-‚Ê ∞∑§ ¬˝Ê∑ΧÁÃ∑§ ’„È‹∑§ „Ò?
119. Which of the following is a natural polymer ?
(1) Á‚‚-1,4-¬ÊÚÁ‹•Êß‚Ê¬˝ËŸ
(1) cis-1,4-polyisoprene
(2) ¬ÊÚÁ‹ (éÿÍ≈UÊ«UÊ߸Ÿ-S≈UÊß⁄UËŸ) (2) poly (Butadiene-styrene)
(3) ¬ÊÚÁ‹éÿÍ≈UÊ«UÊ߸Ÿ (3) polybutadiene
(4) ¬ÊÚÁ‹ (éÿÍ≈UÊ«UÊ߸Ÿ-∞Á∑˝§‹ÊŸÊß≈˛UÊß‹) (4) poly (Butadiene-acrylonitrile)

E1 28 Hindi+English

120. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ-‚Ê ∑§Ê’¸Ÿ ◊ÊŸÊÄ‚Êß«U ∑§ Á‹∞ ‚„Ë 120. Which of the following is not correct about carbon
monoxide ?
Ÿ„Ë¥ „Ò?
(1) It forms carboxyhaemoglobin.
(1) ÿ„ ∑§Ê’Ê¸Ä‚Ë„Ë◊ÊÇ‹ÊÁ’Ÿ ’ŸÊÃË „Ò–
(2) It reduces oxygen carrying ability of blood.
(2) ÿ„ ⁄UÄà ∑§Ë •ÊÚÄ‚Ë¡Ÿ fl„Ÿ ÿÊÇÿÃÊ ∑§Ê ÉÊ≈UÊ ŒÃË „Ò–
(3) The carboxyhaemoglobin (haemoglobin
(3) ∑§Ê’Ê¸Ä‚Ë„Ë◊ÊÇ‹ÊÁ’Ÿ (CO ‚ ’¥ÁœÃ „Ë◊ÊÇ‹ÊÁ’Ÿ), bound to CO) is less stable than
•ÊÚÄ‚Ë„Ë◊ÊÇ‹ÊÁ’Ÿ ‚ •SÕÊÿË „ÊÃÊ „Ò– oxyhaemoglobin.

(4) ÿ„ •¬Íáʸ Œ„Ÿ ∑§ ∑§Ê⁄UáÊ ©à¬ÛÊ „ÊÃË „Ò– (4) It is produced due to incomplete combustion.

121. ‚È∑˝§Ê‚ ¡‹-•¬ÉÊ≈UŸ ¬⁄U ŒÃÊ „Ò — 121. Sucrose on hydrolysis gives :
(1) β-D-Ç‹Í∑§Ê‚+α-D-»˝§Ä≈UÊ‚
 (1) β-D-Glucose+α-D-Fructose
(2) α-D-Ç‹Í∑§Ê‚+β-D-Ç‹Í∑§Ê‚ (2) α-D-Glucose+β-D-Glucose

(3) α-D-Ç‹Í∑§Ê‚+β-D-»˝§Ä≈UÊ‚
 (3) α-D-Glucose+β-D-Fructose

(4) α-D-»˝§Ä≈UÊ‚
 +β-D-»˝§Ä≈UÊ‚
 (4) α-D-Fructose+β-D-Fructose

122. ÁŸêŸÁ‹Áπà œÊÃÈ •ÊÿŸ •Ÿ∑§ ∞¥$¡Êß◊Ê¥ ∑§Ê ‚Á∑˝§Áÿà ∑§⁄UÃÊ „Ò, 122. The following metal ion activates many enzymes,
Ç‹Í∑§Ê‚ ∑§ •ÊÚÄ‚Ë∑§⁄UáÊ ‚ ATP ∑§ ©à¬ÊŒŸ ◊¥ •ÊÒ⁄U Na ∑§ participates in the oxidation of glucose to produce
‚ÊÕ Á‡Ê⁄UÊ ‚¥∑§ÃÊ¥ ∑§ ‚¥ø⁄UáÊ ∑§ Á‹∞ ©ûÊ⁄UŒÊÿË „Ò — ATP and with Na, is responsible for the
transmission of nerve signals.
(1) •Êÿ⁄UŸ
(1) Iron
(2) ÃÊ¥’Ê (∑§ÊÚ¬⁄U) (2) Copper
(3) ∑Ò§ÁÀ‡Êÿ◊ (3) Calcium
(4) ¬Ê≈ÒUÁ‡Êÿ◊ (4) Potassium

123. ÁŸêŸÁ‹Áπà ◊¥ ‚ Á∑§‚◊¥ ¬⁄U◊ÊáÊÈ•Ê¥ ∑§Ë ‚¥ÅÿÊ •Áœ∑§Ã◊
123. Which one of the followings has maximum number
„ÊªË? of atoms ?
(1) Ag(s) ∑§Ê 1 g [Ag ∑§Ê ¬⁄U◊ÊáÊÈ Œ˝√ÿ◊ÊŸ=108] (1) 1 g of Ag(s) [Atomic mass of Ag=108]
(2) Mg(s) ∑§Ê 1 g [Mg ∑§Ê ¬⁄U◊ÊáÊÈ Œ˝√ÿ◊ÊŸ=24] (2) 1 g of Mg(s) [Atomic mass of Mg=24]
(3) O2(g) ∑§Ê 1 g [O ∑§Ê ¬⁄U◊ÊáÊÈ Œ˝√ÿ◊ÊŸ=16] (3) 1 g of O2(g) [Atomic mass of O=16]
(4) Li(s) ∑§Ê 1 g [Li ∑§Ê ¬⁄U◊ÊáÊÈ Œ˝√ÿ◊ÊŸ=7] (4) 1 g of Li(s) [Atomic mass of Li=7]

175
124. 71 Lu ◊¥ ¬˝Ê≈UÊÚŸÊ¥, ãÿÍ≈˛UÊÚŸÊ¥ •ÊÒ⁄U ß‹Ä≈˛UÊÚŸÊ¥ ∑§Ë ‚¥ÅÿÊ∞°, ∑˝§◊‡Ê— 124. The number of protons, neutrons and electrons in
„Ò¥ — 175
71 Lu , respectively, are :
(1) 71, 104 •ÊÒ⁄U 71
(1) 71, 104 and 71
(2) 104, 71 •ÊÒ⁄U 71
(2) 104, 71 and 71
(3) 71, 71 •ÊÒ⁄U 104 (3) 71, 71 and 104
(4) 175, 104 •ÊÒ⁄U 71 (4) 175, 104 and 71

Hindi+English
29 E1
125. ÁŸêŸÁ‹Áπà •Á÷Á∑˝§ÿÊ ◊¥ ∑§Ê’¸Ÿ ∑§Ë •ÊÚÄ‚Ë∑§⁄UáÊ ‚¥ÅÿÊ ◊¥ 125. What is the change in oxidation number of carbon
in the following reaction ?
ÄÿÊ ¬Á⁄UfløŸ „ÊÃÊ „Ò?
CH4(g)+4Cl2(g) → CCl4(l)+4HCl(g)
CH4(g)+4Cl2(g) → CCl4(l)+4HCl(g)
(1) +4 to +4
(1) +4 ‚ +4
(2) 0 to +4
(2) 0 ‚ +4
(3) −4 to +4
(3) −4 ‚ +4
(4) 0 to −4
(4) 0 ‚ −4

126. Identify the incorrect statement.
126. ª‹Ã ∑§ÕŸ ∑§Ê ¬„øÊÁŸ∞–
(1) ¡‹ ◊¥, Cr2+(d4), Fe2+(d6) ‚ •Áœ∑§ ¬˝’‹ (1) Cr2+(d4) is a stronger reducing agent than
Fe2+(d6) in water.
•¬øÊÿ∑§ „Ò–
(2) The transition metals and their compounds
(2) ‚¥∑§˝ ◊áÊ œÊÃÈ∞° •ÊÒ⁄U ©Ÿ∑§ ÿÊÒÁª∑§ ©Ÿ∑§Ë ’„È •ÊÚÄ‚Ë∑§⁄UáÊ are known for their catalytic activity due to
•flSÕÊ•Ê¥ ∑§Ê ª˝„áÊ ∑§⁄UŸ ∑§Ë ˇÊ◊ÃÊ ∑§ ∑§Ê⁄UáÊ ©à¬˝⁄U∑§Ë their ability to adopt multiple oxidation
‚Á∑˝§ÿÃÊ •ÊÒ⁄U ‚¥∑ȧ‹ ÁŸ◊ʸáÊ ∑§ Á‹∞ ¡ÊŸ ¡ÊÃ „Ò¥– states and to form complexes.

(3) •¥Ã⁄UÊ∑§Ê‡ÊË ÿÊÒÁª∑§ fl „ÊÃ „Ò¥ ¡Ê œÊÃÈ•Ê¥ ∑§ Á∑˝§S≈U‹ (3) Interstitial compounds are those that are
formed when small atoms like H, C or N
¡Ê‹∑§Ê¥ ∑§ ÷ËÃ⁄U ¿UÊ≈U •Ê∑§Ê⁄U flÊ‹ ¬⁄U◊ÊáÊÈ•Ê¥ ¡Ò‚ H, are trapped inside the crystal lattices of
C ÿÊ N ∑§ »¥§‚Ÿ (≈˛ÒU¬) ¬⁄U ’ŸÃ „Ò¥– metals.

(4) ∑˝§ÊÁ◊ÿ◊ ∑§Ë, CrO24− •ÊÒ⁄U Cr2O72− ◊¥ ©¬øÿŸ (4) The oxidation states of chromium in CrO24−
•flSÕÊ∞° ‚◊ÊŸ Ÿ„Ë¥ „Ò¥–
and Cr2O72− are not the same.

127. •Á÷Á∑˝§ÿÊ, 2Cl(g) → Cl2(g), ∑§ Á‹∞ ©Áøà Áfl∑§À¬ „Ò —
127. For the reaction, 2Cl(g) → Cl2(g), the correct
(1) ∆rH > 0 •ÊÒ⁄U ∆rS > 0 option is :

(2) ∆rH > 0 •ÊÒ⁄U ∆rS < 0 (1) ∆rH > 0 and ∆rS > 0

(3) ∆rH < 0 •ÊÒ⁄U ∆rS > 0 (2) ∆rH > 0 and ∆rS < 0

(4) ∆rH < 0 •ÊÒ⁄U ∆rS < 0 (3) ∆rH < 0 and ∆rS > 0

(4) ∆rH < 0 and ∆rS < 0
128. $¡Ë≈UÊ Áfl÷fl ∑§Ê ◊ʬŸ ∑§Ê‹ÊÚß«UË Áfl‹ÿŸ ∑§ Á∑§‚ ªÈáÊœ◊¸ ∑§
ÁŸœÊ¸⁄UáÊ ◊¥ ©¬ÿÊªË „ÊÃÊ „Ò? 128. Measuring Zeta potential is useful in determining
which property of colloidal solution ?
(1) ‡ÿÊŸÃÊ
(1) Viscosity
(2) Áfl‹ÿÃÊ
(2) Solubility
(3) ∑§Ê‹ÊÚß«UË ∑§áÊÊ¥ ∑§Ë Áfl‹ÿÃÊ
(3) Stability of the colloidal particles
(4) ∑§Ê‹ÊÚß«UË ∑§áÊÊ¥ ∑§Ê •Ê◊ʬ
(4) Size of the colloidal particles

E1 30 Hindi+English

129. ÿÍÁ⁄UÿÊ ¡‹ ∑§ ‚ÊÕ •Á÷Á∑˝§ÿÊ mÊ⁄UÊ A ’ŸÊÃÊ „Ò ¡Ê ÁflÉÊÁ≈Uà 129. Urea reacts with water to form A which will
decompose to form B. B when passed through
„Ê∑§⁄U B ’ŸÃÊ „Ò– ¡’ B ∑§Ê Cu2+ (¡‹Ëÿ) ‚ ªÈ$¡Ê⁄UÊ Cu2+ (aq), deep blue colour solution C is formed.
¡ÊÃÊ „Ò, Ã’ C ∑§Ê ª„⁄U ŸË‹ ⁄¥Uª ∑§Ê Áfl‹ÿŸ ¬˝Êåà „ÊÃÊ „Ò– What is the formula of C from the following ?
ÁŸêŸÁ‹Áπà ◊¥ ‚ C ∑§Ê ‚ÍòÊ ÄÿÊ „Ò? (1) CuSO4
(1) CuSO4 (2) [Cu(NH3)4]2+
(2) [Cu(NH3)4]2+ (3) Cu(OH)2
(3) Cu(OH)2 (4) CuCO3⋅Cu(OH)2
(4) CuCO3⋅Cu(OH)2
130. Match the following and identify the correct
130. ÁŸêŸÁ‹Áπà ∑§Ê ‚È◊Á ‹Ã ∑§ËÁ¡∞ •ÊÒ⁄U ©Áøà Áfl∑§À¬ ¬„øÊÁŸ∞– option.
(a) CO(g)+H2(g) (i) Mg(HCO3)2+ (a) CO(g)+H2(g) (i) Mg(HCO3)2+
Ca(HCO3)2 Ca(HCO3)2

(b) ¡‹ ∑§Ë •SÕÊÿË (ii) ∞∑§ ß‹Ä≈˛UÊÚŸ ãÿÍŸ (b) Temporary (ii) An electron
hardness of deficient hydride
∑§∆UÊ⁄UÃÊ „Êß«˛UÊß«U water
(c) B2H6 (iii) ‚¥‡‹·áÊ ªÒ‚ (c) B2H6 (iii) Synthesis gas
(d) H2O2 (iv) •‚◊Ã‹Ë ‚¥⁄UøŸÊ (d) H2O2 (iv) Non-planar
structure
(a) (b) (c) (d)
(1) (iii) (i) (ii) (iv) (a) (b) (c) (d)

(2) (iii) (ii) (i) (iv) (1) (iii) (i) (ii) (iv)
(3) (iii) (iv) (ii) (i) (2) (iii) (ii) (i) (iv)
(4) (i) (iii) (ii) (iv) (3) (iii) (iv) (ii) (i)
(4) (i) (iii) (ii) (iv)
131. ÁŸêŸÁ‹Áπà ∑§Ê ‚È◊Á‹Ã ∑§ËÁ¡∞ —
131. Match the following :
•ÊÚÄ‚Êß«U ¬˝∑ΧÁÃ
Oxide Nature
(a) CO (i) ˇÊÊ⁄UËÿ
(a) CO (i) Basic
(b) BaO (ii) ©ŒÊ‚ËŸ
(b) BaO (ii) Neutral
(c) Al2O3 (iii) •ê‹Ëÿ
(c) Al2O3 (iii) Acidic
(d) Cl2O7 (iv) ©÷ÿœ◊˸
(d) Cl2O7 (iv) Amphoteric
ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ-‚Ê ‚„Ë Áfl∑§À¬ „Ò? Which of the following is correct option ?
(a) (b) (c) (d) (a) (b) (c) (d)
(1) (i) (ii) (iii) (iv) (1) (i) (ii) (iii) (iv)
(2) (ii) (i) (iv) (iii) (2) (ii) (i) (iv) (iii)
(3) (iii) (iv) (i) (ii) (3) (iii) (iv) (i) (ii)
(4) (iv) (iii) (ii) (i)
(4) (iv) (iii) (ii) (i)

132. ¬˝Õ◊ ∑§ÊÁ≈U ∑§Ë ∞∑§ •Á÷Á∑˝§ÿÊ ∑§ Á‹∞ flª ÁSÕ⁄UÊ¥∑§ 132. The rate constant for a first order reaction is
4.606×10−3 s−1 „Ò– •Á÷∑§Ê⁄U∑§ ∑§ 2.0 g ∑§Ê 0.2 g Ã∑§ 4.606×10−3 s−1. The time required to reduce
ÉÊ≈UŸ ◊¥ •Êfl‡ÿ∑§ ‚◊ÿ „Ò — 2.0 g of the reactant to 0.2 g is :
(1) 100 s (1) 100 s
(2) 200 s (2) 200 s
(3) 500 s (3) 500 s
(4) 1000 s (4) 1000 s

Hindi+English
31 E1
133. ∞∑§ ∞À∑§ËŸ •Ê$¡ÊŸÊÁ‹Á‚‚ mÊ⁄UÊ ∞∑§ ©à¬ÊŒ ∑§ M§¬ ◊¥ ◊ÕÒŸÒ‹ 133. An alkene on ozonolysis gives methanal as one of
ŒÃË „Ò– ß‚∑§Ë ‚¥⁄UøŸÊ „Ò — the product. Its structure is :

(1) (1)

(2) (2)

(3)
(3)

(4)
(4)

134. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ-‚Ë ∞À∑§Ÿ flÈ≈˜¸U$¡ •Á÷Á∑˝§ÿÊ mÊ⁄UÊ
134. Which of the following alkane cannot be made in
•ë¿UË ‹Áéœ ◊¥ Ÿ„Ë¥ ’ŸÊ߸ ¡Ê ‚∑§ÃË?
good yield by Wurtz reaction ?

(1) n-„ÒÄ‚Ÿ
(1) n-Hexane

(2) 2,3-«UÊß◊ÁÕ‹éÿÍ≈UŸ (2) 2,3-Dimethylbutane

(3) n-„å≈UŸ (3) n-Heptane

(4) n-éÿÍ≈UŸ (4) n-Butane

E1 32 Hindi+English

135. ∞ÁŸ‚ÊÚ‹ HI ∑§ ‚ÊÕ ÁflŒ‹Ÿ mÊ⁄UÊ ŒÃÊ „Ò — 135. Anisole on cleavage with HI gives :

(1) (1)

(2) (2)

(3) (3)

(4) (4)

136. ÁŸêŸÁ‹Áπà ◊¥ ‚ Á∑§‚∑§ Á‹∞ ’Ê⁄U ◊ÊÚ«U‹ flÒœ Ÿ„Ë¥ „Ò? 136. For which one of the following, Bohr model is not
valid ?
(1) „Êß«˛UÊ¡Ÿ ¬⁄U◊ÊáÊÈ
(1) Hydrogen atom
(2) ∞∑§œÊ •ÊÿÁŸÃ „ËÁ‹ÿ◊ ¬⁄U◊ÊáÊÈ (He+) (2) Singly ionised helium atom (He+)
(3) «˜UÿÍ≈U⁄UÊÚŸ ¬⁄U◊ÊáÊÈ (3) Deuteron atom
(4) ∞∑§œÊ •ÊÿÁŸÃ ÁŸÿÊÚŸ ¬⁄U◊ÊáÊÈ (Ne+) (4) Singly ionised neon atom (Ne+)

137. Á∑§‚Ë ÁfllÈà øÈê’∑§Ëÿ Ã⁄¥Uª ◊¥ øÈ¥’∑§Ëÿ ˇÊòÊ •ÊÒ⁄U ÁfllÈà ˇÊòÊ 137. The ratio of contributions made by the electric field
and magnetic field components to the intensity of
∑§ ÉÊ≈U∑§Ê¥ ∑§Ë ÃËfl˝ÃÊ•Ê¥ ∑§ ÿÊªŒÊŸÊ¥ ∑§Ê •ŸÈ¬Êà „ÊÃÊ „Ò — an electromagnetic wave is : (c=speed of
(c=ÁfllÈà øÈê’∑§Ëÿ Ã⁄¥UªÊ¥ ∑§Ê flª) electromagnetic waves)
(1) c:1 (1) c:1
(2) 1:1 (2) 1:1
(3) 1:c (3) 1:c
(4) 1 : c2 (4) 1 : c2
138. Á∑§‚Ë •ãÃ⁄UʬÎc∆U ∑§ Á‹∞ ’˝ÍS≈U⁄U ∑§ÊáÊ ib „ÊŸÊ øÊÁ„∞ — 138. The Brewsters angle ib for an interface should be :
(1) 08 < ib < 308 (1) 08 < ib < 308
(2) 308 < ib < 458 (2) 308 < ib < 458
(3) 458 < ib < 908 (3) 458 < ib < 908
(4) ib = 908 (4) ib = 908

Hindi+English
33 E1
139. Á∑§‚Ë Á‚Á‹á«U⁄U ◊¥ 249 kPa ŒÊ’ •ÊÒ⁄U 278C Ãʬ ¬⁄U „Êß«˛UÊ¡
 Ÿ 139. A cylinder contains hydrogen gas at pressure of
249 kPa and temperature 278C.
ªÒ‚ ÷⁄UË „Ò–
Its density is : (R=8.3 J mol−1 K−1)
ß‚∑§Ê ÉÊŸàfl „Ò — (R=8.3 J mol−1 K−1)
(1) 0.5 kg/m3
(1) 0.5 kg/m3
(2) 0.2 kg/m3
(2) 0.2 kg/m3
(3) 0.1 kg/m3
(3) 0.1 kg/m3
(4) 0.02 kg/m3 (4) 0.02 kg/m3

140. ∑§Ê߸ Á∑§⁄UáÊ ‹ÉÊÈ Á¬˝ï◊ ∑§ÊáÊ (Á¬˝ï◊ ∑§ÊáÊ A) ∑§ Á∑§‚Ë ∞∑§ 140. A ray is incident at an angle of incidence i on one
¬Îc∆U ¬⁄U •Ê¬ÃŸ ∑§ÊáÊ i ¬⁄U •Ê¬ÃŸ ∑§⁄U∑§ Á¬˝ï◊ ∑§ Áfl¬⁄UËà surface of a small angle prism (with angle of prism
»§‹∑§ ‚ •Á÷‹ê’flà ÁŸª¸Ã „ÊÃË „Ò– ÿÁŒ ß‚ Á¬˝ï◊ ∑§ A) and emerges normally from the opposite surface.
If the refractive index of the material of the prism
¬ŒÊÕ¸ ∑§Ê •¬fløŸÊ¥∑§ µ „Ò, ÃÊ •Ê¬ÃŸ ∑§ÊáÊ „Ò, ‹ª÷ª —
is µ, then the angle of incidence is nearly equal
A
(1) to :
2µ
2A A
(2) (1)
µ 2µ
(3) µA 2A
(2)
µ
µA
(4) (3) µA
2
µA
141. ‚◊ÊŸ œÊÁ⁄UÃÊ ∑§ ŒÊ Á‚Á‹á«U⁄U A •ÊÒ⁄U B ∞∑§ ŒÍ‚⁄U ‚ Á∑§‚Ë (4)
2
S≈UÊÚ¬ ∑§ÊÚ∑§ ‚ „ÊÃ „È∞ ¡È«∏ „Ò¥– A ◊¥ ◊ÊŸ∑§ Ãʬ •ÊÒ⁄U ŒÊ’ ¬⁄U
∑§Ê߸ •ÊŒ‡Ê¸ ªÒ‚ ÷⁄UË „Ò– B ¬Íáʸ× ÁŸflʸÁÃà „Ò– ‚◊Sà 141. Two cylinders A and B of equal capacity are
connected to each other via a stop cock. A contains
ÁŸ∑§Êÿ ™§c◊Ëÿ⁄UÊÁœÃ „Ò– S≈UÊÚ¬ ∑§ÊÚ∑§ ∑§Ê •øÊŸ∑§ πÊ‹ ÁŒÿÊ an ideal gas at standard temperature and pressure.
ªÿÊ „Ò– ÿ„ ¬˝Á∑˝§ÿÊ „Ò — B is completely evacuated. The entire system is
thermally insulated. The stop cock is suddenly
(1) ‚◊ÃÊ¬Ë opened. The process is :
(2) L§hÊc◊ (1) isothermal
(3) ‚◊•ÊÿÃŸË (2) adiabatic

(4) ‚◊ŒÊ’Ë (3) isochoric
(4) isobaric
142. Á∑§‚Ë ¬ŒÊÕ¸ ∑§ 0.5 g ∑§ ÃÈÀÿÊ¥∑§ ™§¡Ê¸ „Ò —
142. The energy equivalent of 0.5 g of a substance is :
(1) 4.5×1016 J
(1) 4.5×1016 J
(2) 4.5×1013 J
(2) 4.5×1013 J
(3) 1.5×1013 J
(4) 0.5×1013 J (3) 1.5×1013 J
(4) 0.5×1013 J
143. ¬ÎâflË ∑§ ¬Îc∆U ¬⁄U Á∑§‚Ë Á¬á«U ∑§Ê ÷Ê⁄U 72 N „Ò– ¬ÎâflË ∑§Ë
ÁòÊíÿÊ ∑§Ë •ÊœË ŒÍ⁄UË ∑§ ’⁄UÊ’⁄U ™°§øÊ߸ ¬⁄U ß‚ Á¬á«U ¬⁄U 143. A body weighs 72 N on the surface of the earth.
What is the gravitational force on it, at a height
ªÈL§àflÊ∑§·¸áÊ ’‹ Á∑§ÃŸÊ „ÊªÊ? equal to half the radius of the earth ?
(1) 48 N (1) 48 N
(2) 32 N (2) 32 N
(3) 30 N (3) 30 N
(4) 24 N
(4) 24 N

E1 34 Hindi+English

144. ¬˝ÁÃ⁄UÊœ ∑§ ´§áÊÊà◊∑§ Ãʬ ªÈáÊÊ¥∑§ flÊ‹ ∆UÊ‚ „ÊÃ „Ò¥ — 144. The solids which have the negative temperature
coefficient of resistance are :
(1) œÊÃÈ∞°
(1) metals
(2) ∑§fl‹ ⁄UÊœË (2) insulators only
(3) ∑§fl‹ •œ¸øÊ‹∑§ (3) semiconductors only
(4) ⁄UÊœË •ÊÒ⁄U •œ¸øÊ‹∑§ (4) insulators and semiconductors

145. ‚⁄U‹ •ÊflÃ˸ ªÁà ∑§⁄UÃ Á∑§‚Ë ∑§áÊ ∑§ ÁflSÕʬŸ •ÊÒ⁄U àfl⁄UáÊ ∑§ 145. The phase difference between displacement and
’Ëø ∑§‹ÊãÃ⁄U „ÊÃÊ „Ò — acceleration of a particle in a simple harmonic
(1) π rad motion is :

3π (1) π rad
(2) rad
2 3π
(2) rad
π 2
(3) rad
2 π
(3) rad
(4) ‡ÊÍãÿ 2
(4) zero
146. Á∑§‚Ë S∑˝Í§ ª$¡ ∑§Ê •À¬Ã◊Ê¥∑§ 0.01 mm „Ò ÃÕÊ ß‚∑§ flÎûÊËÿ
¬Ò◊ÊŸ ¬⁄U 50 ÷ʪ „Ò¥– 146. A screw gauge has least count of 0.01 mm and
there are 50 divisions in its circular scale.
ß‚ S∑˝Í§ ª$¡ ∑§Ê øÍ«∏Ë •ãÃ⁄UÊ‹ (Á¬ø) „Ò —
The pitch of the screw gauge is :
(1) 0.01 mm
(1) 0.01 mm
(2) 0.25 mm
(2) 0.25 mm
(3) 0.5 mm
(3) 0.5 mm
(4) 1.0 mm
(4) 1.0 mm
147. Á∑§‚Ë Áª≈UÊ⁄U ◊¥ ‚◊ÊŸ ¬ŒÊÕ¸ ∑§Ë ’ŸË ŒÊ «UÊÁ⁄UÿÊ¥ A •ÊÒ⁄U B ∑§
Sfl⁄U „À∑§ ‚ ◊‹ Ÿ„Ë¥ πÊ ⁄U„ „Ò¥ •ÊÒ⁄U 6 Hz •ÊflÎÁûÊ ∑§ 147. In a guitar, two strings A and B made of same
material are slightly out of tune and produce beats
ÁflS¬ãŒ ©à¬ÛÊ ∑§⁄U ⁄U„ „Ò¥– ¡’ B ◊¥ ßÊfl ∑§Ê ∑ȧ¿U ∑§◊ ∑§⁄U of frequency 6 Hz. When tension in B is slightly
ÁŒÿÊ ¡ÊÃÊ „Ò , ÃÊ ÁflS¬ãŒ •ÊflÎ Á ûÊ ’…∏ ∑ §⁄U decreased, the beat frequency increases to 7 Hz.
7 Hz „Ê ¡ÊÃË „Ò– ÿÁŒ A ∑§Ë •ÊflÎÁûÊ 530 Hz „Ò, ÃÊ B ∑§Ë If the frequency of A is 530 Hz, the original
frequency of B will be :
◊Í‹ •ÊflÎÁûÊ „Ò —
(1) 523 Hz
(1) 523 Hz
(2) 524 Hz
(2) 524 Hz
(3) 536 Hz
(3) 536 Hz
(4) 537 Hz
(4) 537 Hz

148. Two particles of mass 5 kg and 10 kg respectively
148. ©¬ˇÊáÊËÿ Œ˝√ÿ◊ÊŸ ∑§Ë 1 m ‹ê’Ë Á∑§‚Ë ŒÎ…∏ ¿U«∏ ∑§ ŒÊ Á‚⁄UÊ¥ are attached to the two ends of a rigid rod of length
‚ 5 kg •ÊÒ⁄U 10 kg Œ˝√ÿ◊ÊŸ ∑§ ŒÊ ∑§áÊ ¡È«∏ „Ò¥– 1 m with negligible mass.
5 kg ∑§ ∑§áÊ ‚ ß‚ ÁŸ∑§Êÿ ∑§ ‚¥„Áà ∑§ãŒ˝ ∑§Ë ŒÍ⁄UË (‹ª÷ª) The centre of mass of the system from the 5 kg
„Ò — particle is nearly at a distance of :
(1) 33 cm (1) 33 cm
(2) 50 cm (2) 50 cm
(3) 67 cm (3) 67 cm
(4) 80 cm (4) 80 cm

Hindi+English
35 E1
∧
149. Á∑§‚Ë ∑§áÊ, Á¡‚∑§Ê ÁSÕÁà ‚ÁŒ‡Ê 2 k m „Ò, ¬⁄U ¡’ ◊Í‹ 149. Find the torque about the origin when a force of
∧ ∧
Á’¥ŒÈ ∑§ ¬Á⁄U× 3 j N ∑§Ê ∑§Ê߸ ’‹ ∑§Êÿ¸ ∑§⁄UÃÊ „Ò, ÃÊ 3 j N acts on a particle whose position vector is
∧
’‹•ÊÉÊÍáʸ ôÊÊà ∑§ËÁ¡∞– 2k m .
∧ ∧
(1) 6i N m (1) 6i N m
∧ ∧
(2) 6j Nm (2) 6j Nm
∧ ∧
(3) −6 i N m (3) −6 i N m
∧ ∧
(4) 6k N m (4) 6k N m

150. 20 cm2 ˇÊòÊ»§‹ ∑§ Á∑§‚Ë •¬⁄UÊflÃ˸ ¬Îc∆U ¬⁄U 20 W/cm2 150. Light with an average flux of 20 W/cm2 falls on a
•ÊÒ‚Ã ç‹Ä‚ ∑§ ‚ÊÕ ¬˝∑§Ê‡Ê •Á÷‹ê’flà •Ê¬ÃŸ ∑§⁄UÃÊ „Ò– non-reflecting surface at normal incidence having
surface area 20 cm2. The energy received by the
1 Á◊Ÿ≈U ∑§Ë ‚◊ÿÊflÁœ ◊¥ ß‚ ¬Îc∆U ¬⁄U ¬˝Êåà ∑§Ë ªÿË ™§¡Ê¸ „Ò —
surface during time span of 1 minute is :
(1) 10×103 J
(1) 10×103 J
(2) 12×103 J
(2) 12×103 J
(3) 24×103 J
(3) 24×103 J
(4) 48×103 J
(4) 48×103 J

151. 10 cm ÁòÊíÿÊ ∑§ Á∑§‚Ë ªÊ‹Ëÿ øÊ‹∑§ ¬⁄U 3.2×10−7 C
151. A spherical conductor of radius 10 cm has a charge
•Êfl‡Ê ∞∑§‚◊ÊŸ M§¬ ‚ ÁflÃÁ⁄Uà „Ò– ß‚ ªÊ‹ ∑§ ∑§ãŒ˝ ‚
of 3.2×10−7 C distributed uniformly. What is
15 cm ŒÍ⁄UË ¬⁄U ÁfllÈà ˇÊòÊ ∑§Ê ¬Á⁄U◊ÊáÊ ÄÿÊ „Ò?
the magnitude of electric field at a point 15 cm
 1  from the centre of the sphere ?
 = 9× 109 N m2 /C2 
 4 π0 
 1 
(1) 1.28×104 N/C  = 9× 109 N m2 /C2 
 4 π0 
(2) 1.28×105 N/C
(1) 1.28×104 N/C
(3) 1.28×106 N/C
(4) 1.28×107 N/C (2) 1.28×105 N/C
(3) 1.28×106 N/C
152. •¥ÃÁ⁄UˇÊ ∑§ 0.2 m3 •Êÿß ∑§ Á∑§‚Ë ÁŸÁ‡øà ˇÊòÊ ◊¥ „⁄U (4) 1.28×107 N/C
SÕÊŸ ¬⁄U ÁfllÈà Áfl÷fl 5 V ¬ÊÿÊ ªÿÊ „Ò– ß‚ ˇÊòÊ ◊¥ ÁfllÈÃ
ˇÊòÊ ∑§Ê ¬Á⁄U◊ÊáÊ „Ò — 152. In a certain region of space with volume 0.2 m3,
the electric potential is found to be 5 V throughout.
(1) ‡ÊÍãÿ The magnitude of electric field in this region is :
(2) 0.5 N/C (1) zero
(3) 1 N/C (2) 0.5 N/C
(4) 5 N/C (3) 1 N/C
(4) 5 N/C
153. Á∑§‚Ë p-n ‚¥Áœ «UÊÿÊ«U ◊¥ •flˇÊÿ-ˇÊòÊ ∑§Ë øÊÒ«∏Ê߸ ◊¥ flÎÁh
∑§Ê ∑§Ê⁄UáÊ „Ò — 153. The increase in the width of the depletion region
(1) ∑§fl‹ •ª˝ÁŒÁ‡Ê∑§ ’Êÿ‚ in a p-n junction diode is due to :
(1) forward bias only
(2) ∑§fl‹ ¬‡øÁŒÁ‡Ê∑§ ’Êÿ‚
(2) reverse bias only
(3) •ª˝ÁŒÁ‡Ê∑§ •ÊÒ⁄U ¬‡øÁŒÁ‡Ê∑§ ’Êÿ‚ ŒÊŸÊ¥ (3) both forward bias and reverse bias
(4) •ª˝ÁŒÁ‡Ê∑§ œÊ⁄UÊ (current) ◊¥ flÎÁh (4) increase in forward current

E1 36 Hindi+English

154. 40 µF ∑§ Á∑§‚Ë ‚¥œÊÁ⁄UòÊ ∑§Ê 200 V, 50 Hz ∑§Ë ac •Ê¬ÍÁø 154. A 40 µF capacitor is connected to a 200 V, 50 Hz
ac supply. The rms value of the current in the
‚ ‚¥ÿÊÁ¡Ã Á∑§ÿÊ ªÿÊ „Ò– ß‚ ¬Á⁄U¬Õ ◊¥ œÊ⁄UÊ ∑§Ê flª¸ ◊Êäÿ circuit is, nearly :
◊Í‹ (rms) ◊ÊŸ „Ò, ‹ª÷ª — (1) 1.7 A
(1) 1.7 A
(2) 2.05 A
(2) 2.05 A
(3) 2.5 A
(3) 2.5 A
(4) 25.1 A (4) 25.1 A

155. Á∑§‚Ë ªÒ‚ ∑§ Á‹∞, Á¡‚∑§Ê •ÊÁáfl∑§ √ÿÊ‚ d ÃÕÊ ‚¥ÅÿÊ 155. The mean free path for a gas, with molecular
ÉÊŸàfl n „Ò, ◊Êäÿ ◊ÈÄà ¬Õ ∑§Ê ß‚ ¬˝∑§Ê⁄U √ÿÄà Á∑§ÿÊ ¡Ê diameter d and number density n can be expressed
‚∑§ÃÊ „Ò — as :
1 1
(1) (1)
2 n πd 2 n πd
1 1
(2) (2)
2 n πd2 2 n πd2
1 1
(3) (3)
2 n2 πd2 2 n2 πd2
1 1
(4)
2 n2 π2 d2 (4)
2 n2 π2 d2
156. ≈˛UÊ¥Á¡S≈U⁄U Á∑˝§ÿÊ ∑§ Á‹∞ ŸËø ÁŒÿÊ ªÿÊ ∑§ÊÒŸ‚Ê ∑§ÕŸ ‚„Ë „Ò? 156. For transistor action, which of the following
(1) •ÊœÊ⁄U, ©à‚¡¸∑§ •ÊÒ⁄U ‚¥ª˝Ê„∑§ ˇÊòÊÊ¥ ∑§Ë «UÊ¬Ÿ ‚Ê¥Œ˝ÃÊ∞° statements is correct ?
‚◊ÊŸ „ÊŸË øÊÁ„∞– (1) Base, emitter and collector regions should
have same doping concentrations.
(2) •ÊœÊ⁄U, ©à‚¡¸∑§ •ÊÒ⁄U ‚¥ª˝Ê„∑§ ˇÊòÊÊ¥ ∑§ ‚Êß$¡ ‚◊ÊŸ
(2) Base, emitter and collector regions should
„ÊŸ øÊÁ„∞– have same size.
(3) ©à‚¡¸∑§ ‚¥Áœ •ÊÒ⁄U ‚¥ª˝Ê„∑§ ‚¥Áœ ŒÊŸÊ¥ „Ë •ª˝ÁŒÁ‡Ê∑§ (3) Both emitter junction as well as the collector
’ÊÿÁ‚Ã „ÊÃË „Ò¥– junction are forward biased.
(4) •ÊœÊ⁄U ˇÊòÊ ’„Èà ¬Ã‹Ê •ÊÒ⁄U „À∑§Ê «UÊÁ¬Ã „ÊŸÊ øÊÁ„∞– (4) The base region must be very thin and lightly
doped.
157. Œ„‹Ë •ÊflÎÁûÊ ∑§Ë 1.5 ªÈŸË •ÊflÎÁûÊ ∑§Ê ¬˝∑§Ê‡Ê, ¬˝∑§Ê‡Ê ‚Ȫ˝Ê„Ë
¬ŒÊÕ¸ ¬⁄U •Ê¬ÃŸ ∑§⁄UÃÊ „Ò– ÿÁŒ ¬˝∑§Ê‡Ê ∑§Ë •ÊflÎÁûÊ •ÊœË ÃÕÊ 157. Light of frequency 1.5 times the threshold
frequency is incident on a photosensitive material.
©‚∑§Ë ÃËfl˝ÃÊ ŒÊ ªÈŸË ∑§⁄U ŒË ¡Ê∞, ÃÊ ¬˝∑§Ê‡Ê ÁfllÈà œÊ⁄UÊ What will be the photoelectric current if the
Á∑§ÃŸË „ÊªË? frequency is halved and intensity is doubled ?
(1) ŒÊ ªÈŸË (1) doubled
(2) øÊ⁄U ªÈŸË (2) four times
(3) one-fourth
(3) ∞∑§-øÊÒÕÊ߸
(4) zero
(4) ‡ÊÍãÿ
158. When a uranium isotope 235
92 U is bombarded with
158. ¡’ ÿÍ⁄UÁŸÿ◊ ∑§ Á∑§‚Ë ‚◊SÕÊÁŸ∑§ 235 92 U ¬⁄U ãÿÍ≈˛UÊÚŸ ’◊’Ê⁄UË
a neutron, it generates 89
36 Kr , three neutrons
∑§⁄UÃÊ „Ò, ÃÊ 36 Kr •ÊÒ⁄U ÃËŸ ãÿÍ≈˛UÊÚŸÊ¥ ∑§ ‚ÊÕ ©à¬ÛÊ „ÊŸ flÊ‹Ê
89
and :
ŸÊÁ÷∑§ „Ò —
144
(1) 144 (1) 56 Ba
56 Ba
91 91
(2) (2) 40 Zr
40 Zr
101 101
(3) 36 Kr
(3) 36 Kr
103 103
(4) 36 Kr (4) 36 Kr

Hindi+English
37 E1
159. DNA ◊¥ ∞∑§ ’¥œ ∑§Ê πÁá«Uà ∑§⁄UŸ ∑§ Á‹∞ •Êfl‡ÿ∑§ ™§¡Ê¸ 159. The energy required to break one bond in DNA is
10−20 J. This value in eV is nearly :
10−20 J „Ò– eV ◊¥ ÿ„ ◊ÊŸ „Ò, ‹ª÷ª —
(1) 6
(1) 6
(2) 0.6
(2) 0.6
(3) 0.06
(3) 0.06
(4) 0.006
(4) 0.006
160. Two bodies of mass 4 kg and 6 kg are tied to the
160. 4 kg •ÊÒ⁄U 6 kg Œ˝√ÿ◊ÊŸ ∑§ ŒÊ Á¬á«UÊ¥ ∑§ Á‚⁄UÊ¥ ∑§Ê Á∑§‚Ë ends of a massless string. The string passes over
Œ˝√ÿ◊ÊŸ⁄UÁ„à «UÊ⁄UË ‚ ’Ê¥œÊ ªÿÊ „Ò– ÿ„ «UÊ⁄UË Á∑§‚Ë ÉÊ·¸áÊ⁄UÁ„à a pulley which is frictionless (see figure). The
ÁÉÊ⁄UŸË ‚ ªÈ¡⁄UÃË „Ò (•Ê⁄Uπ ŒÁπ∞)– ªÈL§àflËÿ àfl⁄UáÊ (g) ∑§ acceleration of the system in terms of acceleration
¬ŒÊ¥ ◊¥ ß‚ ÁŸ∑§Êÿ ∑§Ê àfl⁄UáÊ „Ò — due to gravity (g) is :

(1) g (1) g
(2) g/2 (2) g/2
(3) g/5 (3) g/5
(4) g/10 (4) g/10

161. •ŸÈ¬˝SÕ ∑§Ê≈U ˇÊòÊ»§‹ A ÃÕÊ ‹ê’Ê߸ L ∑§Ê ∑§Ê߸ ÃÊ⁄U Á∑§‚Ë 161. A wire of length L, area of cross section A is hanging
SÕÊÿË ≈U∑§ ‚ ‹≈U∑§Ê „Ò– ß‚ ÃÊ⁄U ∑§ ◊ÈÄà Á‚⁄U ‚ Á∑§‚Ë from a fixed support. The length of the wire
Œ˝√ÿ◊ÊŸ M ∑§Ê ÁŸ‹¥Á’à ∑§⁄UŸ ¬⁄U ß‚∑§Ë ‹ê’Ê߸ L1 „Ê ¡ÊÃË changes to L1 when mass M is suspended from its
„Ò– ÿ¥ª-ªÈáÊÊ¥∑§ ∑§ Á‹∞ √ÿ¥¡∑§ „Ò — free end. The expression for Young’s modulus is :
MgL1 MgL1
(1) (1)
AL AL
Mg(L1 − L) Mg(L1 − L)
(2) (2)
AL AL
MgL MgL
(3) (3) AL1
AL1
MgL MgL
(4) (4) A(L1 − L)
A(L1 − L)

162. The average thermal energy for a mono-atomic gas
162. Á∑§‚Ë ∞∑§¬⁄U◊ÊáÊÈ∑§ ªÒ‚ ∑§Ë •ÊÒ‚Ã ÃʬËÿ ™§¡Ê¸ „ÊÃË „Ò —
is : (kB is Boltzmann constant and T, absolute
(’ÊÀ≈˜U‚◊ÊŸ ÁŸÿÃÊ¥∑§=kB ÃÕÊ ÁŸ⁄U¬ˇÊ Ãʬ=T)
temperature)
1
(1) kBT 1
2 (1) kBT
2
3
(2) kBT 3
2 (2) kBT
2
5
(3) kBT 5
2 (3) kBT
2
7 7
(4) kBT kBT
2 (4)
2

E1 38 Hindi+English

163. ŸËø ÁŒÿÊ ªÿÊ ∑§ÊÒŸ‚Ê ª˝Ê»§ ∑§ÊÚ¬⁄U ∑§ Á‹∞, Ãʬ (T) ∑§ ‚ÊÕ 163. Which of the following graph represents the
¬˝ÁÃ⁄UÊœ∑§ÃÊ (ρ) ∑§ Áflø⁄UáÊ ∑§Ê ÁŸM§Á¬Ã ∑§⁄UÃÊ „Ò? variation of resistivity (ρ) with temperature (T) for
copper ?

(1)
(1)

(2)
(2)

(3)
(3)

(4)
(4)

164. ŸËø Á∑§‚Ë ¬˝ÁÃ⁄UÊœ ∑§Ê fláʸ ∑§Ê«U ÁŒÿÊ ªÿÊ „Ò —
164. The color code of a resistance is given below :

ß‚∑§ ¬˝ÁÃ⁄UÊœ •ÊÒ⁄U ‚sÃÊ ∑§ ◊ÊŸ ∑˝§◊‡Ê— „Ò¥ — The values of resistance and tolerance, respectively,
(1) 470 kΩ, 5% are :
(2) 47 kΩ, 10% (1) 470 kΩ, 5%
(3) 4.7 kΩ, 5% (2) 47 kΩ, 10%
(4) 470 Ω, 5% (3) 4.7 kΩ, 5%
(4) 470 Ω, 5%
165. ÿ¥ª ∑§ ÁmÁ¤Ê⁄UË ¬˝ÿÊª ◊¥, ÿÁŒ ∑§‹Ê‚¥’h dÊÃÊ¥ ∑§ ’Ëø ∑§Ê
¬ÎÕ∑§Ÿ •ÊœÊ ÃÕÊ ¬Œ¸ ‚ ∑§‹Ê‚¥’h dÊÃÊ¥ ∑§Ë ŒÍ⁄UË ∑§Ê ŒÊ ªÈŸÊ 165. In Young’s double slit experiment, if the separation
∑§⁄U ÁŒÿÊ ¡Ê∞, ÃÊ Á»˝¥§¡ øÊÒ«∏Ê߸ „Ê ¡Ê∞ªË — between coherent sources is halved and the
distance of the screen from the coherent sources is
(1) ŒÊ ªÈŸË doubled, then the fringe width becomes :

(2) •ÊœË (1) double
(2) half
(3) øÊ⁄U ªÈŸË
(3) four times
(4) ∞∑§-øÊÒÕÊ߸ (4) one-fourth

Hindi+English
39 E1
166. Á∑§‚Ë ‚◊ÊãÃ⁄U ¬Á^∑§Ê ‚¥œÊÁ⁄UòÊ, Á¡‚◊¥ ◊Êäÿ◊ ∑§ M§¬ ◊¥ flÊÿÈ 166. The capacitance of a parallel plate capacitor with
air as medium is 6 µF. With the introduction of a
÷⁄UË „Ò, ∑§Ë œÊÁ⁄UÃÊ 6 µF „Ò– ∑§Ê߸ ¬⁄UÊflÒlÈà ◊Êäÿ◊ ÷⁄UŸ ¬⁄U dielectric medium, the capacitance becomes 30 µF.
ß‚∑§Ë œÊÁ⁄UÃÊ 30 µF „Ê ¡ÊÃË „Ò– ß‚ ◊Êäÿ◊ ∑§Ê ¬⁄UÊflÒlÈÃÊ¥∑§ The permittivity of the medium is :
„Ò — (e0=8.85×10−12 C2 N−1 m−2)
(e0=8.85×10−12 C2 N−1 m−2) (1) 0.44×10−13 C2 N−1 m−2
(1) 0.44×10−13 C2 N−1 m−2 (2) 1.77×10−12 C2 N−1 m−2
(2) 1.77×10−12 C2 N−1 m−2 (3) 0.44×10−10 C2 N−1 m−2
(3) 0.44×10−10 C2 N−1 m−2 (4) 5.00 C2 N−1 m−2
(4) 5.00 C2 N−1 m−2
167. Dimensions of stress are :
167. ¬˝ÁÃ’‹ ∑§Ë Áfl◊Ê∞° „Ò¥ — (1) [ M L T−2 ]
(1) [ M L T−2 ] (2) [ M L2 T−2 ]
(2) [ M L2 T−2 ] (3) [ M L0T−2 ]
(3) [ M L0T−2 ] (4) [ M L−1 T−2 ]
(4) [ M L−1 T−2 ]
168. Assume that light of wavelength 600 nm is coming
168. ÿ„ ◊ÊÁŸ∞ Á∑§ Á∑§‚Ë ÃÊ⁄U ‚ 600 nm Ã⁄¥UªŒÒäÿ¸ ∑§Ê ¬˝∑§Ê‡Ê •Ê from a star. The limit of resolution of telescope
whose objective has a diameter of 2 m is :
⁄U„Ê „Ò– ©‚ ŒÍ⁄UŒ‡Ê¸∑§ Á¡‚∑§ •Á÷ŒÎ‡ÿ∑§ ∑§Ê √ÿÊ‚ 2 m „Ò,
(1) 3.66×10−7 rad
∑§ Áfl÷ŒŸ ∑§Ë ‚Ë◊Ê „Ò —
(2) 1.83×10−7 rad
(1) 3.66×10−7 rad
(3) 7.32×10−7 rad
(2) 1.83×10−7 rad
(4) 6.00×10−7 rad
(3) 7.32×10−7 rad
(4) 6.00×10−7 rad
169. A series LCR circuit is connected to an ac voltage
169. ∑§Ê߸ üÊáÊË LCR ¬Á⁄U¬Õ Á∑§‚Ë ∞.‚Ë. flÊÀ≈UÃÊ dÊà ‚ ‚¥ÿÊÁ¡Ã source. When L is removed from the circuit, the
phase difference between current and voltage
„Ò– ¡’ L ∑§Ê „≈UÊ Á‹ÿÊ ¡ÊÃÊ „Ò, ÃÊ œÊ⁄UÊ •ÊÒ⁄U flÊÀ≈UÃÊ ∑§ π
π is . If instead C is removed from the circuit,
’Ëø 3 ∑§Ê ∑§‹ÊãÃ⁄U „ÊÃÊ „Ò– ÿÁŒ ß‚∑§ SÕÊŸ ¬⁄U ¬Á⁄U¬Õ ‚ 3
π
π the phase difference is again between current
C ∑§Ê „≈UÊÃ „Ò¥, ÃÊ ÷Ë œÊ⁄UÊ •ÊÒ⁄U flÊÀ≈UÃÊ ∑§ ’Ëø ∑§‹ÊãÃ⁄U 3
3
and voltage. The power factor of the circuit is :
„Ë ⁄U„ÃÊ „Ò– ß‚ ¬Á⁄U¬Õ ∑§Ê ‡ÊÁÄà ªÈáÊ∑§ „Ò —
(1) zero
(1) ‡ÊÍãÿ
(2) 0.5
(2) 0.5
(3) 1.0
(3) 1.0
(4) −1.0
(4) −1.0
170. A short electric dipole has a dipole moment of
170. Á∑§‚Ë ‹ÉÊÈ ÁfllÈà Ámœ˝Èfl ∑§Ê Ámœ˝Èfl •ÊÉÊÍáʸ 16×10−9 C m
16×10−9 C m. The electric potential due to the
„Ò– ß‚ Ámœ˝Èfl ∑§ ∑§Ê⁄UáÊ, ß‚ Ámœ˝Èfl ∑§ •ˇÊ ‚ 608 ∑§Ê ∑§ÊáÊ
dipole at a point at a distance of 0.6 m from the
’ŸÊŸ flÊ‹Ë Á∑§‚Ë ⁄UπÊ ¬⁄U ÁSÕà 0.6 m ŒÍ⁄UË ∑§ Á∑§‚Ë Á’ãŒÈ
centre of the dipole, situated on a line making an
¬⁄U, ÁfllÈà Áfl÷fl „ÊªÊ —
angle of 608 with the dipole axis is :
 1 
 = 9× 109 N m2 /C2   1 
 4 π 0   = 9× 109 N m2 /C2 
 4 π0 
(1) 50 V
(1) 50 V
(2) 200 V
(2) 200 V
(3) 400 V
(3) 400 V
(4) ‡ÊÍãÿ
(4) zero

E1 40 Hindi+English

171. 599 œÊ⁄UáʇÊË‹ÃÊ ∑§Ë Á∑§‚Ë ‹Ê„ ∑§Ë ¿U«∏ ¬⁄U 1200 A m−1 171. An iron rod of susceptibility 599 is subjected to a
magnetising field of 1200 A m −1 . The
ÃËfl˝ÃÊ ∑§Ê øÈê’∑§Ëÿ ˇÊòÊ ‹ªÊÿÊ ªÿÊ „Ò– ß‚ ¿U«∏ ∑§ ¬ŒÊÕ¸ permeability of the material of the rod is :
∑§Ë ¬Ê⁄UªêÿÃÊ „Ò — (µ0=4π×10−7 T m A−1)
(µ0=4π×10−7 T m A−1) (1) 2.4π×10−4 T m A−1
(1) 2.4π×10−4 T m A−1 (2) 8.0×10−5 T m A−1
(2) 8.0×10−5 T m A−1 (3) 2.4π×10−5 T m A−1
(3) 2.4π×10−5 T m A−1 (4) 2.4π×10−7 T m A−1
(4) 2.4π×10−7 T m A−1
172. A long solenoid of 50 cm length having 100 turns
172. 50 cm ‹ê’Ë Á∑§‚Ë ¬Á⁄UŸÊÁ‹∑§Ê, Á¡‚◊¥ 100 »§⁄U „Ò¥, ‚ carries a current of 2.5 A. The magnetic field at
2.5 A œÊ⁄UÊ ¬˝flÊÁ„à „Ê ⁄U„Ë „Ò– ß‚ ¬Á⁄UŸÊÁ‹∑§Ê ∑§ ∑§ãŒ˝ ¬⁄U the centre of the solenoid is :
øÈê’∑§Ëÿ ˇÊòÊ „Ò — (µ0=4π×10−7 T m A−1)
(µ0=4π×10−7 T m A−1) (1) 6.28×10−4 T
(2) 3.14×10−4 T
(1) 6.28×10−4 T
(2) 3.14×10−4 T (3) 6.28×10−5 T
(3) 6.28×10−5 T (4) 3.14×10−5 T
(4) 3.14×10−5 T 173. A charged particle having drift velocity of
7.5×10 −4 m s −1 in an electric field of
173. Á∑§‚Ë •ÊflÁ‡Êà ∑§áÊ, Á¡‚∑§Ê ÃËfl˝ÃÊ ∑§
3×10−10 Vm−1
3×10−10 Vm−1, has a mobility in m2 V−1 s−1
ÁfllÈà ˇÊòÊ ◊¥ •¬flÊ„ flª 7.5×10 m s „Ò, ∑§Ë
−4 −1
of :
m2 V−1 s−1 ◊¥ ªÁÇÊË‹ÃÊ „Ò — (1) 2.25×1015
(1) 2.25×1015 (2) 2.5×106
(2) 2.5×106 (3) 2.5×10−6
(3) 2.5×10−6 (4) 2.25×10−15
(4) 2.25×10−15
174. The quantities of heat required to raise the
174. r1 •ÊÒ⁄U r2 ÁòÊíÿÊ•Ê¥ (r1=1.5 r2) ∑§ ŒÊ ∑§ÊÚ¬⁄U ∑§ ∆UÊ‚ ªÊ‹Ê¥ temperature of two solid copper spheres of radii
∑§ Ãʬ ◊¥ 1 K ∑§Ë flÎÁh ∑§⁄UŸ ∑§ Á‹∞ •Êfl‡ÿ∑§ ™§c◊Ê•Ê¥ ∑§Ë r1 and r2 (r1=1.5 r2) through 1 K are in the
◊ÊòÊÊ•Ê¥ ∑§Ê •ŸÈ¬Êà „Ò — ratio :
27 27
(1) (1)
8 8
9 9
(2) (2)
4 4
3 3
(3) (3)
2 2
5 5
(4) (4)
3 3
175. Áfl⁄UÊ◊ÊflSÕÊ ∑§ Á∑§‚Ë ß‹Ä≈˛UÊÚŸ ∑§Ê V flÊÀ≈U ∑§ Áfl÷flÊãÃ⁄U ‚ 175. An electron is accelerated from rest through a
àflÁ⁄Uà Á∑§ÿÊ ªÿÊ „Ò– ÿÁŒ ß‚ ß‹Ä≈˛UÊÚŸ ∑§Ë Œ ’˝ÊÚÇ‹Ë Ã⁄¥UªŒÒäÿ¸ potential difference of V volt. If the de Broglie
1.227×10−2 nm „Ò, ÃÊ Áfl÷flÊãÃ⁄U „Ò — wavelength of the electron is 1.227×10−2 nm, the
potential difference is :
(1) 10 V
(2) 102 V (1) 10 V
(3) 103 V (2) 102 V
(4) 104 V (3) 103 V
(4) 104 V
176. ‚ÊÕ¸∑§ •¥∑§Ê¥ ∑§Ê ◊„àfl ŒÃ „È∞ 9.99 m−0.0099 m ∑§Ê ◊ÊŸ
176. Taking into account of the significant figures, what
ÄÿÊ „Ò? is the value of 9.99 m−0.0099 m ?
(1) 9.9801 m (1) 9.9801 m
(2) 9.98 m (2) 9.98 m
(3) 9.980 m (3) 9.980 m
(4) 9.9 m
(4) 9.9 m

Hindi+English
41 E1
177. Á∑§‚Ë ◊ËŸÊ⁄U ∑§ Á‡Êπ⁄U ‚ Á∑§‚Ë ª¥Œ ∑§Ê 20 m/s ∑§ flª ‚ 177. A ball is thrown vertically downward with a
velocity of 20 m/s from the top of a tower. It hits
™§äflʸœ⁄U •œÊ◊ÈπË »¥§∑§Ê ªÿÊ „Ò– ∑ȧ¿U ‚◊ÿ ¬‡øÊà ÿ„ ª¥Œ the ground after some time with a velocity of
œ⁄UÃË ‚ 80 m/s ∑§ flª ‚ ≈U∑§⁄UÊÃË „Ò– ß‚ ◊ËŸÊ⁄U ∑§Ë ™°§øÊ߸ 80 m/s. The height of the tower is : (g=10 m/s2)
„Ò — (g=10 m/s2) (1) 360 m
(1) 360 m (2) 340 m
(2) 340 m (3) 320 m
(3) 320 m (4) 300 m
(4) 300 m 178. A capillary tube of radius r is immersed in water
178. ÁòÊíÿÊ r ∑§Ë ∑§Ê߸ ∑§Á‡Ê∑§Ê Ÿ‹Ë ¡‹ ◊¥ «ÍU’Ë „Ò •ÊÒ⁄U ß‚◊¥ ¡‹ and water rises in it to a height h. The mass of
the water in the capillary is 5 g. Another capillary
™°§øÊ߸ h Ã∑§ ø…∏ ªÿÊ „Ò– ∑§Á‡Ê∑§Ê Ÿ‹Ë ◊¥ ÷⁄U ¡‹ ∑§Ê tube of radius 2r is immersed in water. The mass
Œ˝√ÿ◊ÊŸ 5 g „Ò– ÁòÊíÿÊ 2r ∑§Ë ∑§Ê߸ •ãÿ ∑§Á‡Ê∑§Ê Ÿ‹Ë ¡‹ of water that will rise in this tube is :
◊¥ «ÍU’Ë „Ò– ß‚ Ÿ‹Ë ◊¥ ™§¬⁄U ø…∏ ¡‹ ∑§Ê Œ˝√ÿ◊ÊŸ „Ò — (1) 2.5 g
(1) 2.5 g (2) 5.0 g
(2) 5.0 g (3) 10.0 g
(3) 10.0 g (4) 20.0 g
(4) 20.0 g 179. A resistance wire connected in the left gap of a
179. Á∑§‚Ë ◊Ë≈U⁄U ‚ÃÈ ∑§ ’Ê∞° •ãÃ⁄UÊ‹ ◊¥ ‚¥ÿÊÁ¡Ã ∑§Ê߸ ¬˝ÁÃ⁄UÊœ ÃÊ⁄U metre bridge balances a 10 Ω resistance in the
right gap at a point which divides the bridge wire
ß‚∑§ ŒÊ∞° •ãÃ⁄UÊ‹ ∑§ 10 Ω ¬˝ÁÃ⁄UÊœ ∑§Ê ©‚ Á’ãŒÈ ¬⁄U ‚¥ÃÈÁ‹Ã in the ratio 3 : 2. If the length of the resistance
∑§⁄UÃÊ „Ò ¡Ê ‚ÃÈ ∑§ ÃÊ⁄U ∑§Ê 3 : 2 ∑§ •ŸÈ¬Êà ◊¥ Áfl÷ÊÁ¡Ã ∑§⁄UÃÊ wire is 1.5 m, then the length of 1 Ω of the
„Ò– ÿÁŒ ¬˝ÁÃ⁄UÊœ ÃÊ⁄U ∑§Ë ‹ê’Ê߸ 1.5 m „Ò, ÃÊ ß‚ ¬˝ÁÃ⁄UÊœ resistance wire is :
(1) 1.0×10−2 m
ÃÊ⁄U ∑§Ë fl„ ‹ê’Ê߸ Á¡‚∑§Ê ¬˝ÁÃ⁄UÊœ 1 Ω „ÊªÊ, „Ò — (2) 1.0×10−1 m
(1) 1.0×10−2 m (3) 1.5×10−1 m
(2) 1.0×10−1 m (4) 1.5×10−2 m
(3) 1.5×10−1 m
(4) 1.5×10−2 m 180. For the logic circuit shown, the truth table is :
180. Œ‡Êʸ∞ ª∞ Ã∑¸§ ¬Á⁄U¬Õ ∑§ Á‹∞, ‚àÿ◊ÊŸ ‚Ê⁄UáÊË „Ò —

(1) A B Y
0 0 0
(1) A B Y
0 1 0
0 0 0
1 0 0
0 1 0
1 1 1
1 0 0
(2) A B Y
1 1 1
0 0 0
(2) A B Y
0 1 1
0 0 0
1 0 1
0 1 1
1 1 1
1 0 1
(3) A B Y
1 1 1
0 0 1
(3) A B Y
0 1 1
0 0 1
1 0 1
0 1 1
1 1 0
1 0 1
(4) A B Y
1 1 0
0 0 1
(4) A B Y
0 1 0
0 0 1
1 0 0
0 1 0
1 1 0
1 0 0
1 1 0
-o0o-
-o0o-