NEET 2021 Question Paper in Hindi

¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê ‚¥∑§Ã
Test Booklet Code
AGAJHA No. :

ß‚ ¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê ∑§Ê Ã’ Ã∑§ Ÿ πÊ‹¥ ¡’ Ã∑§ ∑§„Ê Ÿ ¡Ê∞–
Hindi+English

M4 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.
ß‚ ¬ÈÁSÃ∑§Ê ◊¥ 48 ¬Îc∆ „Ò¥–
This Booklet contains 48 pages.

◊„àfl¬Íáʸ ÁŸŒ¸‡Ê — Important Instructions :
1. ©ûÊ⁄ ¬òÊ ß‚ ¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê ∑§ •ãŒ⁄ ⁄πÊ „Ò– ¡’ •Ê¬∑§Ê 1. The Answer Sheet is inside this Test Booklet. When
you are directed to open the Test Booklet, take out the
¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê πÊ‹Ÿ ∑§Ê ∑§„Ê ¡Ê∞, ÃÊ ©ûÊ⁄ ¬òÊ ÁŸ∑§Ê‹ ∑§⁄U Answer Sheet and fill in the particulars on OFFICE Copy
äÿÊŸ¬Ífl¸∑§ ∑§Êÿʸ‹ÿ ¬˝ÁÃÁ‹Á¬ ¬⁄ ∑§fl‹ ŸË‹/∑§Ê‹ ’ÊÚ‹ ¬ÊÚߥ≈ carefully with blue/black ball point pen only.
¬Ÿ ‚ Áflfl⁄áÊ ÷⁄¥– 2. The test is of 3 hours duration and the Test Booklet
contains 200 multiple-choice questions (four options
2. ¬⁄UˡÊÊ ∑§Ë •flÁœ 3 ÉÊ¥≈Ê „Ò ∞fl¥ ¬⁄UˡÊÊ ¬ÈÁSÃ∑§Ê ◊¥ ÷ÊÒÁÃ∑§Ë,
with a single correct answer) from Physics, Chemistry
⁄U‚ÊÿŸ‡ÊÊSòÊ ∞fl¥ ¡ËflÁflôÊÊŸ (flŸS¬ÁÃÁflôÊÊŸ ∞fl¥ ¬˝ÊÁáÊÁflôÊÊŸ) and Biology (Botany and Zoology). 50 questions in
Áfl·ÿÊ¥ ‚ 200 ’„ÈÁfl∑§À¬Ëÿ ¬˝‡Ÿ „Ò¥ (4 Áfl∑§À¬Ê¥ ◊¥ ‚ ∞∑§ ‚„Ë each subject are divided into two Sections (A and B)
as per details given below :
©ûÊ⁄U „Ò)– ¬˝àÿ∑§ Áfl·ÿ ◊¥ 50 ¬˝‡Ÿ „Ò Á¡Ÿ∑§Ê ÁŸêŸ fláÊʸŸÈ‚Ê⁄U
(a) Section A shall consist of 35 (Thirty-five) Questions
ŒÊ •ŸÈ÷ʪÊ¥ (A ÃÕÊ B) ◊¥ Áfl÷ÊÁ¡Ã Á∑§ÿÊ ªÿÊ „Ò — in each subject (Question Nos – 1 to 35,
(a) •ŸÈ÷ʪ A ∑§ ¬˝àÿ∑§ Áfl·ÿ ◊¥ 35 (¬Ò¥ÃË‚) (¬˝‡Ÿ ‚¥ÅÿÊ 1 51 to 85, 101 to 135 and 151 to 185). All questions
‚ 35, 51 ‚ 85, 101 ‚ 135 ∞fl¥ 151 ‚ 185) are compulsory.
(b) Section B shall consist of 15 (Fifteen) questions in
¬˝‡Ÿ „Ò– ‚÷Ë ¬˝‡Ÿ •ÁŸflÊÿ¸ „Ò¥– each subject (Question Nos – 36 to 50, 86 to 100, 136
(b) •ŸÈ÷ʪ B ∑§ ¬˝àÿ∑§ Áfl·ÿ ◊¥ 15 (¬¥Œ˝„) (¬˝‡Ÿ ‚¥ÅÿÊ 36 to 150 and 186 to 200). In Section B, a candidate
‚ 50, 86 ‚ 100, 136 ‚ 150 ∞fl¥ 186 ‚ 200) ¬˝‡Ÿ „Ò– needs to attempt any 10 (Ten) questions out of 15
(Fifteen) in each subject.
•ŸÈ÷ʪ B ‚ ¬⁄UˡÊÊÁÕ¸ÿÊ¥ ∑§Ê ¬˝àÿ∑§ Áfl·ÿ ‚ 15 (¬¥Œ˝„) ◊¥ Candidates are advised to read all 15 questions
‚ ∑§Ê߸ 10 (Œ‚) ¬˝‡Ÿ ∑§⁄UŸ „Ê¥ª– in each subject of Section B before they start
¬⁄UˡÊÊÁÕ¸ÿÊ¥ ∑§Ê ‚ȤÊÊfl „Ò Á∑§ ¬˝‡ŸÊ¥ ∑§ ©ûÊ⁄U ŒŸ ∑§ ¬Ífl¸ •ŸÈ÷ʪ attempting the question paper. In the event of a
candidate attempting more than ten questions, the
B ◊¥ ¬˝àÿ∑§ Áfl·ÿ ∑§ ‚÷Ë 15 ¬˝‡ŸÊ¥ ∑§Ê ¬…∏¥– ÿÁŒ ∑§Ê߸
first ten questions answered by the candidate
¬⁄UˡÊÊÕ˸ 10 ¬˝‡Ÿ ‚ •Áœ∑§ ¬˝‡ŸÊ¥ ∑§Ê ©ûÊ⁄U ŒÃÊ „Ò ÃÊ ©‚∑§ mÊ⁄UÊ shall be evaluated.
©ûÊÁ⁄Uà ¬˝Õ◊ 10 ¬˝‡ŸÊ¥ ∑§Ê „Ë ◊ÍÀÿÊ¥∑§Ÿ Á∑§ÿÊ ¡Ê∞ªÊ– 3. Each question carries 4 marks. For each correct response,
the candidate will get 4 marks. For each incorrect
3. ¬˝àÿ∑§ ¬˝‡Ÿ 4 •¥∑§ ∑§Ê „Ò– ¬˝àÿ∑§ ‚„Ë ©ûÊ⁄ ∑§ Á‹∞ ¬⁄UˡÊÊÕ˸
response, one mark will be deducted from the total
∑§Ê 4 •¥∑§ ÁŒ∞ ¡Ê∞¥ª– ¬˝àÿ∑§ ª‹Ã ©ûÊ⁄ ∑§ Á‹∞ ∑ȧ‹ ÿÊª ◊¥ scores. The maximum marks are 720.
‚ ∞∑§ •¥∑§ ÉÊ≈ÊÿÊ ¡Ê∞ªÊ– •Áœ∑§Ã◊ •¥∑§ 720 „Ò¥– 4. Use Blue/Black Ball Point Pen only for writing
4. ß‚ ¬Îc∆ ¬⁄ Áflfl⁄áÊ •¥Á∑§Ã ∑§⁄Ÿ ∞fl¥ ©ûÊ⁄ ¬òÊ ¬⁄ ÁŸ‡ÊÊŸ ‹ªÊŸ particulars on this page/marking responses on Answer
Sheet.
∑§ Á‹∞ ∑§fl‹ ŸË‹/∑§Ê‹ ’ÊÚ‹ ¬ÊÚߥ≈ ¬Ÿ ∑§Ê ¬˝ÿÊª ∑§⁄¥– 5. Rough work is to be done in the space provided for this
5. ⁄»§ ∑§Êÿ¸ ß‚ ¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê ◊¥ ÁŸœÊ¸Á⁄à SÕÊŸ ¬⁄ „Ë ∑§⁄¥– purpose in the Test Booklet only.

¬˝‡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 :

M4 Hindi+English
2
•ŸÈ÷ʪ - A (÷ÊÒÁÃ∑§Ë) Section - A (Physics)

1. ÁŸêŸÁ‹Áπà ∑§ÕŸÊ¥ (A) ÃÕÊ (B) ¬⁄U ÁfløÊ⁄U ∑§ËÁ¡∞ ÃÕÊ 1. Consider the following statements (A) and (B)
and identify the correct answer.
‚„Ë ©ûÊ⁄U ∑§Ê ÁøÁã„à ∑§ËÁ¡∞–
(A) A zener diode is connected in reverse bias,
(A) ∞∑§ ¡Ÿ⁄U «UÊÿÊ« ©à∑˝§◊ •Á÷ŸÁà ◊¥ ¡È«∏Ê „Ò, ¡’ when used as a voltage regulator.
Áfl÷fl ÁŸÿãòÊ∑§ ∑§Ë Ã⁄U„ ¬˝ÿÈÄà „ÊÃÊ „Ò– (B) The potential barrier of p-n junction lies
(B) p-n ‚Á㜠∑§Ê Áfl÷fl ¬˝ Ê øË⁄U 0.1 flÊ À ≈U ÃÕÊ between 0.1 V to 0.3 V.
0.3 flÊÀ≈U ∑§ ’Ëø „ÊÃÊ „Ò– (1) (A) and (B) both are correct.

(1) ŒÊŸÊ¥ (A) ÃÕÊ (B) ‚àÿ „Ò¥– (2) (A) and (B) both are incorrect.
(3) (A) is correct and (B) is incorrect.
(2) ŒÊŸÊ¥ (A) ÃÕÊ (B) ª‹Ã „Ò¥–
(4) (A) is incorrect but (B) is correct.
(3) (A) ‚àÿ „Ò ¬⁄UãÃÈ (B) ª‹Ã „Ò–
(4) (A) ª‹Ã „Ò ¬⁄UãÃÈ (B) ‚àÿ „Ò– 2. A capacitor of capacitance ‘C’, is connected across
an ac source of voltage V, given by
2. ÁŒ∞ ªÿ ¬˝àÿÊflÃ˸ flÊÀ≈UÃÊ dÊà V=V0 sinωt ‚ ‘C’ œÊÁ⁄UÃÊ V=V0 sinωt
∑§Ê ∞∑§ œÊÁ⁄UòÊ ¡È«∏Ê „Ò– The displacement current between the plates of
the capacitor, would then be given by :
œÊÁ⁄UòÊ ∑§ å‹≈UÊ¥ ∑§ ’Ëø ÁflSÕʬŸ œÊ⁄UÊ „ÊªË —
(1) Id=V0 ωCcosωt
(1) Id=V0 ωCcosωt
V
(2) Id= 0 cosωt
V ωC
(2) Id= 0 cosωt
ωC
V
(3) Id= 0 sin ωt
V ωC
(3) Id= 0 sin ωt
ωC (4) Id=V0 ωCsinωt
(4) Id=V0 ωCsinωt
3. A body is executing simple harmonic motion with
frequency ‘n’, the frequency of its potential energy
3. ∞∑§ flSÃÈ ‘n’ •ÊflÎÁûÊ ‚ ‚⁄U‹ •Êflø ªÁà ∑§⁄UÃË „Ò– ß‚∑§Ë is :
ÁSÕÁá ™§¡Ê¸ ∑§Ë •ÊflÎÁûÊ „Ò — (1) n
(1) n (2) 2n
(2) 2n (3) 3n
(3) 3n (4) 4n
(4) 4n
4. The equivalent capacitance of the combination
4. ÁŒ∞ ªÿ ‚¥ÿÊ¡Ÿ ◊¥ ÃÈÀÿ œÊÁ⁄UÃÊ „Ò — shown in the figure is :

(1) 3C (1) 3C
(2) 2C (2) 2C
(3) C/2 (3) C/2
(4) 3C/2 (4) 3C/2

Hindi+English M4
3
5. Á¬˝í◊ ‚ ÁŸª¸Ã ∑§ÊáÊ ∑§ ◊ÊŸ ∑§Ê ôÊÊà ∑§ËÁ¡∞– ∑§Ê°ø ∑§Ê 5. Find the value of the angle of emergence from the
prism. Refractive index of the glass is 3 .
•¬fløŸÊ¥∑§ 3 „Ò —

(1) 608
(1) 608 (2) 308
(2) 308 (3) 458
(3) 458 (4) 908
(4) 908
6. A dipole is placed in an electric field as shown. In
which direction will it move ?
6. ÁøòÊÊŸÈ‚Ê⁄ ∞∑§ Ámœ˝Èfl ÁfllÈà ˇÊòÊ ◊¥ ⁄UπÊ ¡ÊÃÊ „Ò– ÿ„ Á∑§‚
ÁŒ‡ÊÊ ◊¥ ªÁà ∑§⁄UªÊ?

(1) towards the left as its potential energy will
increase.
(1) ’Ê°ÿË¥ Ã⁄U»§ ÄÿÊ¥Á∑§ ß‚∑§Ë ÁSÕÁá ™§¡Ê¸ ’…∏ªË– (2) towards the right as its potential energy will
(2) ŒÊÿË¥ Ã⁄U»§ ÄÿÊ¥Á∑§ ß‚∑§Ë ÁSÕÁá ™§¡Ê¸ ÉÊ≈UªË– decrease.
(3) towards the left as its potential energy will
(3) ’Ê°ÿË¥ Ã⁄U»§ ÄÿÊ¥Á∑§ ß‚∑§Ë ÁSÕÁá ™§¡Ê¸ ÉÊ≈UªË–
decrease.
(4) ŒÊÿË¥ Ã⁄U»§ ÄÿÊ¥Á∑§ ß‚∑§Ë ÁSÕÁá ™§¡Ê¸ ’…∏ªË– (4) towards the right as its potential energy will
increase.
7. SÃê÷ - I ∑§Ê SÃê÷ - II ‚ ‚È◊Á‹Ã ∑§ËÁ¡∞ ÃÕÊ ŸËø ÁŒ∞
7. Match Column - I and Column - II and choose
ªÿ Áfl∑§À¬Ê¥ ‚ ‚„Ë ‚È◊Á‹Ã ∑§Ê ¿UÊ°Á≈U∞ — the correct match from the given choices.
SÃê÷ - I SÃê÷ - II Column - I Column - II
1 1
(A) ªÒ‚ ∑§ •áÊÈ•Ê¥ ∑§Ê (P) nm v 2 (A) Root mean square (P) nm v 2
3 3
flª¸ ◊Êäÿ ◊Í‹ flª speed of gas molecules
3 RT
3 RT (B) Pressure exerted (Q)
(B) •ÊŒ‡Ê¸ ªÒ‚ mÊ⁄UÊ •Ê⁄UÊÁ¬Ã ŒÊ’ (Q) M
M by ideal gas
5
5 (C) Average kinetic energy (R) RT
(C) •áÊÈ ∑§Ë •ÊÒ‚Ã ªÁá ™§¡Ê¸ (R) RT 2
2 of a molecule
3 3
(D) 1 ◊Ê‹ Ám¬⁄U◊ÊáÊÈ∑§ ªÒ‚ ∑§Ë (S) kBT (D) Total internal energy (S) kBT
2 2
∑ȧ‹ •ÊãÃÁ⁄U∑§ ™§¡Ê¸ of 1 mole of a
diatomic gas
(1) (A) - (R), (B) - (P), (C) - (S), (D) - (Q) (1) (A) - (R), (B) - (P), (C) - (S), (D) - (Q)
(2) (A) - (Q), (B) - (R), (C) - (S), (D) - (P) (2) (A) - (Q), (B) - (R), (C) - (S), (D) - (P)
(3) (A) - (Q), (B) - (P), (C) - (S), (D) - (R) (3) (A) - (Q), (B) - (P), (C) - (S), (D) - (R)
(4) (A) - (R), (B) - (Q), (C) - (P), (D) - (S) (4) (A) - (R), (B) - (Q), (C) - (P), (D) - (S)

M4 Hindi+English
4
8. ‚◊ÊŸ •ˇÊ ∑§ •ŸÈÁŒ‡Ê ‘d’ ŒÍ⁄UË ¬⁄U ∞∑§ 20 ‚.◊Ë. »§Ê∑§‚ ŒÍ⁄UË 8. A convex lens ‘A’ of focal length 20 cm and a concave
lens ‘B’ of focal length 5 cm are kept along the
∑§Ê ©ûÊ‹ ‹ã‚ ‘A’ ÃÕÊ 5 ‚.◊Ë. »§Ê∑§‚ ŒÍ⁄UË ∑§Ê •flË same axis with a distance ‘d’ between them. If a
‹ã‚ ‘B’ ⁄Uπ „Ò¥– ÿÁŒ ‘A’ ¬⁄U •Ê¬ÁÃà ‚◊ÊãÃ⁄U ¬˝∑§Ê‡Ê ¬Èã¡, parallel beam of light falling on ‘A’ leaves ‘B’ as a
‘B’ ‚ ÁŸ∑§‹Ÿ ¬⁄U ÷Ë ‚◊ÊãÃ⁄U ¬Èã¡ ⁄U„ÃË „Ò¥, ÃÊ ŒÍ⁄UË ‘d’ parallel beam, then the distance ‘d’ in cm will be :
‚.◊Ë. ◊¥ „ÊªË — (1) 25
(1) 25 (2) 15
(2) 15 (3) 50
(3) 50 (4) 30
(4) 30
9. If force [ F ], acceleration [ A ] and time [ T ] are
9. ÿÁŒ ’‹ [ F ], àfl⁄UáÊ [ A ] ÃÕÊ ‚◊ÿ [ T ] ∑§Ê ◊ÈÅÿ ÷ÊÒÁÃ∑§ chosen as the fundamental physical quantities.
Find the dimensions of energy.
⁄UÊÁ‡ÊÿÊ° ◊ÊŸ Á‹ÿÊ ¡Ê∞, ÃÊ ™§¡Ê¸ ∑§Ë Áfl◊Ê ôÊÊà ∑§ËÁ¡∞–
(1) [F ] [A] [ T]
(1) [F] [A] [T]
(2) [ F ] [ A ] [ T2 ] (2) [ F ] [ A ] [ T2 ]

(3) [ F ] [ A ] [ T−1 ] (3) [ F ] [ A ] [ T−1 ]
(4) [ F ] [ A−1 ] [ T ] (4) [ F ] [ A−1 ] [ T ]

10. ∞∑§ ∑§¬ ∑§Ê»§Ë ‘t’ Á◊Ÿ≈U ◊¥ 908C ‚ 808C Ã∑§ ∆Uã«UË „ÊÃË „Ò, 10. A cup of coffee cools from 908C to 808C in t minutes,
¡’ ∑§◊⁄U ∑§Ê Ãʬ 208C „Ò– ©‚Ë ∑§◊⁄U ∑§ Ãʬ ¬⁄U ‚◊ÊŸ Ã⁄U„ when the room temperature is 208C. The time
taken by a similar cup of coffee to cool from 808C
∑§ ∑§¬ ◊¥ ∑§Ê»§Ë ∑§Ê 808C ‚ 608C Ã∑§ ∆Uã«UÊ ∑§⁄UŸ ◊¥ ‚◊ÿ to 608C at a room temperature same at 208C is :
‹ªÊ „ÊªÊ —
13
13 (1) t
(1) t 10
10
13
13 (2) t
(2) t 5
5
10
10 (3) t
(3) t 13
13
5
5 (4) t
(4) t 13
13

11. A particle is released from height S from the
11. ∞∑§ ∑§áÊ ¬ÎâflË ‚Ä ‚ S ™§°øÊ߸ ‚ Áª⁄UÊÿÊ ¡ÊÃÊ „Ò– ∑ȧ¿U surface of the Earth. At a certain height its kinetic
ÁŸÁ‡øà ™§°øÊ߸ ¬⁄U ß‚∑§Ë ªÁá ™§¡Ê¸ ß‚∑§Ë ÁSÕÁá ™§¡Ê¸ ∑§Ë energy is three times its potential energy. The
ÃËŸ ªÈŸÊ „ÊÃË „Ò– ß‚ ˇÊáÊ ∑§áÊ ∑§Ë ¬ÎâflË ‚Ä ‚ ™§°øÊ߸ ÃÕÊ height from the surface of earth and the speed of
∑§áÊ ∑§Ë øÊ‹ „ÊÃË „Ò — the particle at that instant are respectively :

S 3gS S 3gS
(1) (1) ,
4
,
2 4 2

S 3gS S 3gS
(2) (2) ,
4 2
,
4 2

S 3gS S 3gS
(3) , (3) ,
2 2 2 2

S 3gS S 3gS
(4) , (4) ,
4 2 4 2

Hindi+English M4
5
12. M Œ˝√ÿ◊ÊŸ ÃÕÊ d ÉÊŸàfl ∑§Ë ¿UÊ≈UË ª¥Œ ∑§Ê flª ÁÇ‹‚⁄UËŸ ‚ 12. The velocity of a small ball of mass M and density
÷⁄U ’øŸ ◊¥ «UÊ‹Ÿ ¬⁄U ∑ȧ¿U ‚◊ÿ ’ÊŒ •ø⁄U „Ê ¡ÊÃÊ „Ò– ÿÁŒ d, when dropped in a container filled with glycerine
becomes constant after some time. If the density
d
ÁÇ‹‚⁄UËŸ ∑§Ê ÉÊŸàfl „Ê, ÃÊ ª¥Œ ¬⁄U ‹ªŸ flÊ‹Ê ‡ÿÊŸ ’‹ of glycerine is d , then the viscous force acting on
2
2
„ÊªÊ — the ball will be :
Mg Mg
(1) (1)
2 2
(2) Mg (2) Mg
3 3
(3) Mg (3) Mg
2 2
(4) 2Mg
(4) 2 Mg
13. x-ÁŒ‡ÊÊ ◊¥ ‚¥øÁ⁄Uà ∞∑§ ‚◊Ë ÁfllÈà øÈê’∑§Ëÿ Ã⁄¥Uª ∑§ Á‹∞
13. For a plane electromagnetic wave propagating in
ÁŸêŸÁ‹Áπà ‚¥ÿÊ¡ŸÊ¥ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑˝§◊‡Ê— ÁfllÈà ˇÊòÊ (E)
x-direction, which one of the following combination
ÃÕÊ øÈê’∑§Ëÿ ˇÊòÊ (B) ∑§Ë ‚„Ë ‚ê÷fl ÁŒ‡ÊÊ•Ê¥ ∑§Ê ¬˝ŒÁ‡Ê¸Ã gives the correct possible directions for electric
∑§⁄UÃÊ „Ò? field (E) and magnetic field (B) respectively ?
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
(1) (1) j+k, j +k
j+k, j+k
∧ ∧ ∧ ∧
∧ ∧ ∧ ∧ (2) − j+k, − j−k
(2) − j+k, − j−k ∧ ∧ ∧ ∧
(3) j +k, − j−k
∧ ∧ ∧ ∧
(3) j +k, − j−k ∧ ∧ ∧ ∧
(4) − j+k, − j+k
∧ ∧ ∧ ∧
(4) − j+k, − j+k 14. Column - I gives certain physical terms associated
with flow of current through a metallic conductor.
14. SÃê÷ - I œÊàflËÿ øÊ‹∑§ ‚ ¬˝flÊÁ„à œÊ⁄UÊ ‚ ‚ê’ÁãœÃ ∑ȧ¿U Column - II gives some mathematical relations
÷ÊÒÁÃ∑§ Ãâÿ √ÿÄà ∑§⁄UÃÊ „Ò– SÃê÷ - II ‚◊ÊŸ ªÁáÊÃËÿ ‚ê’㜠involving electrical quantities. Match
Column - I and Column - II with appropriate
Á¡Ÿ◊¥ ÁfllÈà ⁄UÊÁ‡ÊÿÊ° ‚Áê◊Á‹Ã „ÊÃË „Ò, ∑§Ê √ÿÄà ∑§⁄UÃÊ „Ò– relations.
SÃê÷ - I ∑§Ê SÃê÷ - II ‚ ‚„Ë ‚ê’㜠mÊ⁄UÊ ‚È◊Á ‹Ã ∑§ËÁ¡∞ — Column - I Column - II
SÃê÷ - I SÃê÷ - II
m
m (A) Drift Velocity (P)
(A) •ŸÈª◊Ÿ flª (P) 2
ne2 ρ
ne ρ
(B) ÁfllÈÃËÿ ¬˝ÁÃ⁄UÊœ∑§ÃÊ (Q) nevd (B) Electrical Resistivity (Q) nevd
eE eE
(C) üÊÊ°Ã ∑§Ê‹ (R) τ (C) Relaxation Period (R) τ
m m
E
(D) œÊ⁄UÊ ÉÊŸàfl (S)
J
E
(D) Current Density (S)
(1) (A)-(R), (B)-(S), (C)-(P), (D)-(Q) J
(2) (A)-(R), (B)-(S), (C)-(Q), (D)-(P) (1) (A)-(R), (B)-(S), (C)-(P), (D)-(Q)
(3) (A)-(R), (B)-(P), (C)-(S), (D)-(Q) (2) (A)-(R), (B)-(S), (C)-(Q), (D)-(P)
(4) (A)-(R), (B)-(Q), (C)-(S), (D)-(P)
(3) (A)-(R), (B)-(P), (C)-(S), (D)-(Q)
15. ∞∑§ ÁS¬˝¥ª 10 ãÿÍ≈UŸ ∑§ ’‹ ‚ 5 ‚.◊Ë. Áπ¥øË „ÊÃË „Ò– ¡’ (4) (A)-(R), (B)-(Q), (C)-(S), (D)-(P)
2 Á∑§.ª˝Ê. Œ˝√ÿ◊ÊŸ ∑§Ê ß‚‚ ‹≈U∑§ÊÿÊ ¡ÊÃÊ „Ò, ÃÊ ŒÊ‹Ÿ ∑§Ê 15. A spring is stretched by 5 cm by a force 10 N. The
•Êflø∑§Ê‹ „ÊÃÊ „Ò — time period of the oscillations when a mass of 2 kg
(1) 0.0628 ‚∑§á«U is suspended by it is :
(1) 0.0628 s
(2) 6.28 ‚∑§á«U
(2) 6.28 s
(3) 3.14 ‚∑§á«U (3) 3.14 s
(4) 0.628 ‚∑§á«U (4) 0.628 s

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16. ∞∑§ ‚◊ÊãÃ⁄U å‹≈U œÊÁ⁄UòÊ ∑§ å‹≈UÊ¥ ∑§ ’Ëø ∞∑§‚◊ÊŸ 16. A parallel plate capacitor has a uniform electric
→ →
ÁfllÈà ˇÊòÊ ‘ E ’ „Ò– ÿÁŒ å‹≈UÊ¥ ∑§ ’Ëø ∑§Ë ŒÍ⁄UË ‘d’ ÃÕÊ ¬˝àÿ∑§ field ‘ E ’ in the space between the plates. If the
distance between the plates is ‘d’ and the area of
å‹≈U ∑§Ê ˇÊòÊ»§‹ ‘A’ „Ê ÃÊ œÊÁ⁄UòÊ ◊¥ ∞∑§ÁòÊà ™§¡Ê¸ each plate is ‘A’, the energy stored in the capacitor
„Ò — (ε0=ÁŸflʸà ∑§Ë ÁfllÈÇÊË‹ÃÊ) is : (ε0=permittivity of free space)
1
(1) ε E2 (1)
1
ε E2
2 0 2 0
(2) ε0EAd
1 (2) ε0EAd
(3) ε E2 Ad
2 0 1
(3) ε E2 Ad
2
E Ad 2 0
(4) ε0
E2 Ad
(4) ε0
17. ÁŒπÊ∞ ªÿ •Ÿ¥Ã ‹ê’Ê߸ ∑§ øÊ‹∑§ ◊¥ 5 ∞Áê¬ÿ⁄U ∑§Ë œÊ⁄UÊ
¬˝ fl ÊÁ„à „Ê Ã Ë „Ò – øÊ‹∑§ ∑ § ‚◊ÊãÃ⁄U ∞∑§ ß‹ Ä ≈˛ U Ê Ú Ÿ 17. An infinitely long straight conductor carries a
105 ◊Ë./‚. ∑§ øÊ‹ ‚ ªÁà ∑§⁄UÃÊ „Ò– ∞∑§ ˇÊáÊ ¬⁄U ß‹Ä≈˛UÊÚŸ current of 5 A as shown. An electron is moving
ÃÕÊ øÊ‹∑§ ∑§ ’Ëø ‹ê’flØ ŒÍ⁄UË 20 ‚.◊Ë. „Ò– ©‚ ˇÊáÊ ¬⁄U with a speed of 105 m/s parallel to the conductor.
The perpendicular distance between the electron
ß‹Ä≈˛UÊÚŸ mÊ⁄UÊ •ŸÈ÷fl Á∑§ÿ ¡ÊŸ flÊ‹ ’‹ ∑§ ¬Á⁄U◊ÊáÊ ∑§Ë and the conductor is 20 cm at an instant.
ªáÊŸÊ ∑§ËÁ¡∞– Calculate the magnitude of the force experienced
by the electron at that instant.

(1) 4×10−20 ãÿÍ≈UŸ
(2) 8π×10−20 ãÿÍ≈UŸ (1) 4×10−20 N
(3) 4π×10−20 ãÿÍ≈UŸ (2) 8π×10−20 N
(3) 4π×10−20 N
(4) 8×10−20 ãÿÍ≈UŸ
(4) 8×10−20 N
18. ∞∑§ fl΄à »§Ê∑§‚ ŒÍ⁄UË ÃÕÊ fl΄à mÊ⁄∑§ ∑§Ê ‹ã‚ ŒÍ⁄UŒ‡Ê˸ ∑§ 18. A lens of large focal length and large aperture is
•Á÷ŒÎ‡ÿ∑§ ∑§ Á‹∞ •àÿÁœ∑§ ©¬ÿÊªË „ÊÃÊ „Ò, ÄÿÊ¥Á∑§ — best suited as an objective of an astronomical
(1) ∞∑§ fl΄à mÊ⁄U∑§ ªÈáÊÃÊ ÃÕÊ ŒÎ‡ÿÃÊ ∑§ Á‹∞ ÿÊªŒÊŸ telescope since :
∑§⁄UÃÊ „Ò– (1) a large aperture contributes to the quality
and visibility of the images.
(2) ∞∑§ fl΄à ˇÊòÊ»§‹ ∑§Ê •Á÷ŒÎ‡ÿ∑§ ©¬ÿÈÄà ¬˝∑§Ê‡Ê ‚¥ª„˝ áÊ (2) a large area of the objective ensures better
ˇÊ◊ÃÊ ∑§Ê ∑§Ê⁄U∑§ „ÊÃÊ „Ò– light gathering power.
(3) ∞∑§ fl΄à mÊ⁄U∑§ ©ûÊ◊ Áfl÷ŒŸ ¬˝ŒÊŸ ∑§⁄UÃÊ „Ò– (3) a large aperture provides a better resolution.
(4) ©¬ÿȸÄà ◊¥ ‚÷Ë– (4) all of the above.

19. A radioactive nucleus AZ X undergoes spontaneous
19. ∞∑§ ⁄UÁ«UÿÊ‚Á∑˝§ÿ ŸÊÁ÷∑§ AZ X Sfl× ÁflÉÊÁ≈Uà „ÊÃÊ „Ò
decay in the sequence
A
Z
X → Z −1B → Z −3C → Z − 2 D ∑˝§◊ ◊¥, ¡„Ê° Z Ãàfl A
X → Z −1B → Z −3C → Z − 2 D , where Z is the
Z
X ∑§Ê ¬⁄U◊ÊáÊÈ ‚¥ÅÿÊ „Ò– ∑˝§◊ ◊¥ ÁflÉÊÁ≈Uà ‚ê÷fl ∑§áÊ „Ò¥, atomic number of element X. The possible decay
particles in the sequence are :
∑˝§◊‡Ê— —
(1) α, β−, β+
(1) α, β−, β+
(2) α, β+, β− (2) α, β+, β−
(3) β+, α, β− (3) β+, α, β−
(4) β−, α, β+ (4) β−, α, β+

Hindi+English M4
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20. ÁŒπÊÿ ªÿ ÁøòÊ ∑§ •ŸÈ‚Ê⁄U ∞∑§ L ¬˝⁄U∑§àfl ∑§Ê ¬˝⁄U∑§, ∞∑§ 20. An inductor of inductance L, a capacitor of
C œÊÁ⁄UÃÊ ∑§Ê œÊÁ⁄UòÊ ÃÕÊ ∞∑§ ‘R’ ¬˝ÁÃ⁄UÊœ ∑§Ê ¬˝ÁÃ⁄UÊœ∑§ capacitance C and a resistor of resistance ‘R’ are
connected in series to an ac source of potential
‘V’ flÊÀ≈U Áfl÷flÊãÃ⁄U ∑§ ¬˝àÿÊflÃ˸ dÊà ‚ üÊáÊË ∑˝§◊ ◊¥ ¡È«∏U „Ò¥– difference ‘V’ volts as shown in figure.
L, C, R ¬⁄U Áfl÷flÊãÃ⁄U ∑˝§◊‡Ê— 40 flÊÀ≈U, 10 flÊÀ≈U ÃÕÊU Potential difference across L, C and R is 40 V,
40 flÊ À ≈U „Ò ¥ – LCR üÊ á ÊË ¬Á⁄U ¬ Õ ◊ ¥ ¬˝ fl ÊÁ„à œÊ⁄U Ê 10 V and 40 V, respectively. The amplitude of
current flowing through LCR series circuit is
10 2 ∞Áê¬ÿ⁄U „Ò– ¬Á⁄U¬Õ ∑§Ê ¬˝ÁÃ’ÊœÊ „Ò —
10 2 A. The impedance of the circuit is :

(1) 4 2 •Ê◊
(1) 4 2 Ω
(2) 5 2 •Ê◊
(3) 4 •Ê◊ (2) 5 2 Ω
(4) 5 •Ê◊ (3) 4Ω
(4) 5Ω
21. ¬ÎâflË ‚Ä ‚ ¬‹ÊÿŸ flª v „Ò– ‚◊ÊŸ Œ˝√ÿ◊ÊŸ ÉÊŸàfl ÃÕÊ
¬ÎâflË ∑§ ÁòÊíÿÊ ∑§ øÊ⁄U ªÈŸÊ ÁòÊíÿÊ flÊ‹ ŒÍ‚⁄U ª˝„ ∑§ ‚Ä ‚ 21. The escape velocity from the Earth’s surface is v.
¬‹ÊÿŸ flª „ÊÃÊ „Ò — The escape velocity from the surface of another
planet having a radius, four times that of Earth
(1) v
and same mass density is :
(2) 2v
(3) 3v (1) v
(4) 4v (2) 2v
(3) 3v
22. ∞∑§ S∑˝Í§ª¡ ¡’ ∞∑§ ÃÊ⁄U ∑§ √ÿÊ‚ ∑§Ê ◊ʬŸ ∑§ Á‹∞ ¬˝ÿÈÄÃ
(4) 4v
Á∑§ÿÊ ¡ÊÃÊ „Ò, ÃÊ ÁŸêŸÁ‹Áπà ¬Ê∆KÊ¥∑§ ŒÃÊ „Ò —
◊ÈÅÿ ¬Ò◊ÊŸ ∑§Ê ¬Ê∆KÊ¥∑§ = 0 Á◊‹Ë◊Ë≈U⁄U 22. A screw gauge gives the following readings when
flÎûÊËÿ ¬Ò◊ÊŸ ∑§Ê ¬Ê∆KÊ¥∑§ = 52 πÊŸ used to measure the diameter of a wire
ÁŒÿÊ ªÿÊ „Ò Á∑§ ◊ÈÅÿ ¬Ò◊ÊŸÊ ¬⁄U 1 Á◊‹Ë◊Ë≈U⁄U, flÎûÊËÿ ¬Ò◊ÊŸÊ Main scale reading : 0 mm
∑§ 100 πÊŸÊ¥ ∑§ ‚¥ªÃ „ÊÃÊ „Ò– ©¬ÿȸÄà ÁŒ∞ ªÿ ¬˝ˇÊáÊÊ¥ ‚ Circular scale reading : 52 divisions
ÃÊ⁄U ∑§Ê √ÿÊ‚ „Ò — Given that 1 mm on main scale corresponds to
100 divisions on the circular scale. The diameter
(1) 0.52 ‚.◊Ë.
of the wire from the above data is :
(2) 0.026 ‚.◊Ë.
(1) 0.52 cm
(3) 0.26 ‚.◊Ë. (2) 0.026 cm
(4) 0.052 ‚.◊Ë. (3) 0.26 cm
23. ∞∑§ ≈U⁄U’Êߟ ∑§Ê ø‹ÊŸ ∑§ Á‹∞ 15 Á∑§ª˝Ê/‚. ∑§Ë Œ⁄U ‚ (4) 0.052 cm
60 ◊Ë. ™§°øÊ߸ ‚ ¬ÊŸË Áª⁄UÃÊ „Ò– ÉÊ·¸áÊ ∑§ ∑§Ê⁄UáÊ ¬˝Ê⁄UÁê÷∑§
23. Water falls from a height of 60 m at the rate of
ÁŸfl‡ÊË ™§¡Ê¸ ∑§ 10 ¬˝ÁÇÊà ∑§Ë „ÊÁŸ „ÊÃË „Ò– ≈U⁄U’Êߟ ∑§ mÊ⁄UÊ 15 kg/s to operate a turbine. The losses due to
Á∑§ÃŸË ‡ÊÁÄà ©à¬ãŸ ∑§Ë ¡ÊÃË „Ò? frictional force are 10% of the input energy. How
much power is generated by the turbine ?
(g=10 ◊Ë./‚.2)
(g=10 m/s2)
(1) 10.2 Á∑§‹ÊflÊ≈U
(1) 10.2 kW
(2) 8.1 Á∑§‹ÊflÊ≈U (2) 8.1 kW
(3) 12.3 Á∑§‹ÊflÊ≈U (3) 12.3 kW
(4) 7.0 Á∑§‹ÊflÊ≈U (4) 7.0 kW

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24. Áfl⁄UÊ◊ÊflSÕÊ (t=0) ‚ ∞∑§ ¿UÊ≈ UÊ é‹ÊÚ∑§ Áø∑§Ÿ ŸÃ‚◊Ë ‚ ŸËø 24. A small block slides down on a smooth inclined
∑§Ë •Ê⁄ Áπ‚∑§ÃÊ „Ò– ÿÁŒ •ãÃ⁄UÊ‹ t=n−1 ‚ t=n ∑§ ’Ëø plane, starting from rest at time t=0. Let Sn be
the distance travelled by the block in the interval
Sn Sn
é‹ÊÚ∑§ mÊ⁄UÊ ø‹Ë ªÿË ŒÍ⁄UË Sn „Ê, ÃÊ S ∑§Ê •ŸÈ¬Êà „ÊÃÊ „Ò — t=n−1 to t=n. Then, the ratio is :
n+1 Sn+1
2n−1 2n−1
(1) (1)
2n 2n
2n−1 2n−1
(2) (2) 2n+1
2n+1
2n+1 2n+1
(3) (3) 2n−1
2n−1
2n 2n
(4) (4) 2n−1
2n−1
25. Polar molecules are the molecules :
25. œ˝ÈflËÿ •áÊÈ ∞‚ •áÊÈ „ÊÃ „Ò¥ — (1) having zero dipole moment.
(1) Á¡Ÿ∑§Ê Ámœ˝Èfl •ÊÉÊÍáʸ ‡ÊÍãÿ „ÊÃÊ „Ò– (2) acquire a dipole moment only in the presence
(2) ¡Ê ÁfllÈà ˇÊòÊ ∑§ ©¬ÁSÕà ◊¥ „Ë ÁmœÈ˝fl •ÊÉÊÍáʸ ¬˝Êåà of electric field due to displacement of
∑§⁄UÃ „Ò¥, •Êfl‡ÊÊ¥ ∑§ ÁflSÕʬŸ ∑§ ∑§Ê⁄UáÊ– charges.
(3) ¡Ê Ámœ˝flÈ •ÊÉÊÍáʸ ∑§fl‹ Ã÷Ë ¬˝Êåà ∑§⁄UÃ „Ò, ¡’ øÈê’∑§Ëÿ (3) acquire a dipole moment only when magnetic
ˇÊòÊ •ŸÈ¬ÁSÕà „ÊÃÊ „Ò– field is absent.
(4) having a permanent electric dipole moment.
(4) Á¡Ÿ◊¥ SÕÊÿË ÁfllÈà Ámœ˝Èfl •ÊÉÊÍáʸ „ÊÃÊ „Ò–
26. A thick current carrying cable of radius ‘R’ carries
26. ∞∑§ ‘R’ ÁòÊíÿÊ ∑§Ë ◊Ê≈UË œÊ⁄UÊflÊ„Ë ∑§Á’‹ ◊¥ œÊ⁄UÊ ‘I’ ß‚∑§ current ‘I’ uniformly distributed across its
•ŸÈ¬˝SÕ ∑§Ê≈U ¬⁄U ‚◊ÊŸ M§¬ ‚ ÁflÃÁ⁄Uà „Ò– ∑§Á’‹ ∑§ ∑§Ê⁄UáÊ cross-section. The variation of magnetic field B(r)
øÈê’∑§Ëÿ ˇÊòÊ B(r) ∑§Ê ¬Á⁄UfløŸ ∑§Á’‹ •ˇÊ ‚ ‘r’ ŒÍ⁄UË ∑§ due to the cable with the distance ‘r’ from the axis
‚ʬˇÊ ¬˝ŒÁ‡Ê¸Ã Á∑§ÿÊ ¡ÊÃÊ „Ò — of the cable is represented by :

(1) ‚ (1)

(2) ‚ (2)

(3) ‚ (3)

(4) ‚ (4)

Hindi+English M4
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27. ∞∑§ Áfl÷fl◊Ê¬Ë ¬Á⁄U¬Õ ◊¥ 1.5 flÊÀ≈U Áfl.flÊ.’. ∑§Ë ∞∑§ ‚‹ 27. In a potentiometer circuit a cell of EMF 1.5 V gives
balance point at 36 cm length of wire. If another
36 ‚.◊Ë. ÃÊ⁄U ∑§ ‹ê’Ê߸ ¬⁄U ‚¥ÃÈÁ‹Ã Á’ãŒÈ ŒÃË „Ò– ÿÁŒ
cell of EMF 2.5 V replaces the first cell, then at
2.5 flÊÀ≈U Áfl.flÊ.’. flÊ‹Ë ŒÍ‚⁄UË ‚‹ ¬„‹Ë ‚‹ ∑§Ê ¬˝ÁÃSÕÊÁ¬Ã what length of the wire, the balance point occurs ?
∑§⁄UÃË „Ò, ÃÊ ÃÊ⁄U ∑§ Á∑§SÊ ‹ê’Ê߸ ¬⁄U ‚¥ÃÈÁ‹Ã Á’ãŒÈ ¬˝Êåà (1) 60 cm
„ÊªÊ? (2) 21.6 cm
(1) 60 ‚.◊Ë. (3) 64 cm
(2) 21.6 ‚.◊Ë. (4) 62 cm
(3) 64 ‚.◊Ë.
28. Two charged spherical conductors of radius R1 and
(4) 62 ‚.◊Ë.
R2 are connected by a wire. Then the ratio of
28. R1 ÃÕÊ R2 ÁòÊíÿÊ ∑§ ŒÊ •ÊflÁ‡Êà ªÊ‹Ëÿ øÊ‹∑§ ∞∑§ ÃÊ⁄U ‚ surface charge densities of the spheres (σ1/σ2) is :
¡Ê«∏U ÁŒ∞ ¡ÊÃ „Ò¥– ªÊ‹Ê¥ ∑§ ¬Îc∆U •Êfl‡Ê ÉÊŸàflÊ¥ (σ1/σ2) ∑§Ê R1
•ŸÈ¬Êà „ÊÃÊ „Ò — (1) R2
R1 R2
(1) R2 (2) R1
R2
(2) R1  R1 
(3)  
 R2 
 R1 
(3)  
 R2  R12
(4)
R12 R22
(4)
R22
29. An electromagnetic wave of wavelength ‘λ’ is
incident on a photosensitive surface of negligible
29. Ÿªáÿ ∑§Êÿ¸ »§‹Ÿ ∑§ ¬˝∑§Ê‡Ê ‚Ȫ˝Ê„Ë ‚Ä ¬⁄U ‘λ’ Ã⁄¥UªŒÒäÿ¸ ∑§Ë work function. If ‘m’ mass is of photoelectron
∞∑§ ÁfllÈÃøÈê’∑§Ëÿ Ã⁄¥Uª •Ê¬ÁÃà „ÊÃË „Ò– ÿÁŒ ‚Ä ‚ emitted from the surface has de-Broglie wavelength
©à‚Á¡¸Ã ‘m’ Œ˝√ÿ◊ÊŸ ∑§ »§Ê≈UÊß‹Ä≈˛UÊÚŸ ∑§Ë «UË-’˝ÊÇ‹Ë Ã⁄¥UªŒÒäÿ¸ λd, then :
λd „Ê, ÃÊ —
 2m  2
(1) λ= λd
 2m  2  hc 
(1) λ= λd
 hc 
 2mc  2
 2mc  2 (2) λd= λ
(2) λd= λ  h 
 h 
 2mc  2
 2mc  2 (3) λ= λd
(3) λ= λd  h 
 h 
 2h  2  2h  2
λ= (4) λ= λd
(4) λd  mc 
 mc 

30. n-≈UÊ߬ •œ¸øÊ‹∑§ ◊¥ ß‹Ä≈˛UÊÚŸ ∑§Ë ‚Ê¥Œ˝ÃÊ ©ÃŸÊ „Ë „Ò Á¡ÃŸÊ
30. The electron concentration in an n-type
p-≈UÊ߬ •œ¸øÊ‹∑§ ◊¥ ∑§Ê≈U⁄U ∑§Ë ‚Ê¥Œ˝ÃÊ „Ò– ŒÊŸÊ¥ ¬⁄U ’Ês semiconductor is the same as hole concentration
ÁfllÈà ˇÊòÊ ‹ªÊÿÊ ¡ÊÃÊ „Ò– ŒÊŸÊ¥ ◊¥ œÊ⁄UÊ•Ê¥ ∑§ •ŸÈ¬Êà ∑§Ë in a p-type semiconductor. An external field
ÃÈ‹ŸÊ ∑§ËÁ¡∞– (electric) is applied across each of them. Compare
the currents in them.
(1) n-≈UÊ߬ ◊¥ œÊ⁄UÊ=p-≈UÊ߬ ◊¥ œÊ⁄UÊ
(1) current in n-type=current in p-type.
(2) p-≈UÊ߬ ◊¥ œÊ⁄UÊ > n-≈UÊ߬ ◊¥ œÊ⁄UÊ ‚
(2) current in p-type > current in n-type.
(3) n-≈UÊ߬ ◊¥ œÊ⁄UÊ > p-≈UÊ߬ ◊¥ œÊ⁄UÊ ‚
(3) current in n-type > current in p-type.
(4) p-≈UÊ߬ ◊¥ ∑§Ê߸ œÊ⁄UÊ ¬˝flÊÁ„à Ÿ„Ë¥ „ÊªË, ∑§fl‹
(4) No current will flow in p-type, current will
n-≈UÊ߬ ◊¥ œÊ⁄UÊ ¬˝flÊÁ„à „ÊªË only flow in n-type.

M4 Hindi+English
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31. ∞∑§fláÊËZ 600 ŸÒŸÊ◊Ë≈U⁄U Ã⁄¥UªŒÒäÿ¸ flÊ‹ ¬˝∑§Ê‡Ê ‚ •Ê҂ß ¬˝Áà 31. The number of photons per second on an average
emitted by the source of monochromatic light of
‚∑§á«U ©à‚Á¡¸Ã »§Ê≈UÊŸÊ¥ ∑§Ë ‚¥ÅÿÊ „ÊªË ¡’ fl„ 3.3×10−3 wavelength 600 nm, when it delivers the power of
flÊ≈U ‡ÊÁÄà ©à‚Á¡¸Ã ∑§⁄UÃÊ „Ò — 3.3×10−3 watt will be : (h=6.6×10−34 Js)
(h=6.6×10−34 ¡Í‹×‚.) (1) 1018
(1) 1018 (2) 1017
(2) 1017
(3) 1016
(3) 1016
(4) 1015 (4) 1015

32. ÿÁŒ E ÃÕÊ G ∑˝§◊‡Ê— ™§¡Ê¸ ÃÕÊ ªÈL§àflÊ∑§·¸áÊ ÁŸÿÃÊ¥∑§ ∑§Ê 32. If E and G respectively denote energy and
E
gravitational constant, then has the dimensions
¬˝ŒÁ‡Ê¸Ã ∑§⁄UÃ „Ò¥, ÃÊ E ∑§Ë Áfl◊Ê „ÊÃË „Ò — of :
G
G
(1) [ M2 ] [ L−1 ] [ T0 ] (1) [ M2 ] [ L−1 ] [ T0 ]
(2) [ M ] [ L−1 ] [ T−1 ] (2) [ M ] [ L−1 ] [ T−1 ]
(3) [ M ] [ L0 ] [ T0 ] (3) [ M ] [ L0 ] [ T0 ]
(4) [ M2 ] [ L−2 ] [ T−1 ]
(4) [ M2 ] [ L−2 ] [ T−1 ]
33. ‚◊ÊŸ ‹ê’Ê߸, ‚◊ÊŸ •ŸÈ¬˝SÕ ∑§Ê≈U ∑§ ˇÊòÊ»§‹ ÃÕÊ ‚◊ÊŸ
33. The effective resistance of a parallel connection that
¬ŒÊÕ¸ ∑§ ‚◊ÊãÃ⁄U ∑˝§◊ ◊¥ ¡È«∏ øÊ⁄U ÃÊ⁄UÊ¥ ∑§Ê ÃÈÀÿ ¬˝ÁÃ⁄UÊœ consists of four wires of equal length, equal area of
0.25 •Ê◊ „Ò– ÿÁŒ ©Ÿ∑§Ê üÊáÊË ∑˝§◊ ◊¥ ¡Ê«∏U ÁŒÿÊ ¡Ê∞, ÃÊ cross-section and same material is 0.25 Ω. What
¬˝÷ÊflË ¬˝ÁÃ⁄UÊœ ÄÿÊ „ÊªÊ? will be the effective resistance if they are connected
in series ?
(1) 0.25 •Ê◊
(1) 0.25 Ω
(2) 0.5 •Ê◊
(2) 0.5 Ω
(3) 1 •Ê◊
(3) 1Ω
(4) 4 •Ê◊ (4) 4Ω
34. ∞∑§ ⁄UUÁ«UÿÊ‚Á∑˝§ÿ ãÿÍÄ‹Êß«U ∑§Ë •œ¸•ÊÿÈ 100 ÉÊ¥≈ „Ò¥– 34. The half-life of a radioactive nuclide is 100 hours.
150 ÉÊã≈U ∑§ ’ÊŒ ¬˝Ê⁄UÁê÷∑§ ‚Á∑˝§ÿÃÊ ∑§Ê ’øÊ „È•Ê Á÷ÛÊÊà◊∑§ The fraction of original activity that will remain
÷ʪ „ÊªÊ — after 150 hours would be :
(1) 1/2 (1) 1/2
1 1
(2) (2)
2 2 2 2
2
(3)
3 2
(3)
2 3
(4)
3 2 2
(4)
3 2
35. ∞∑§ 240 Œ˝√ÿ◊ÊŸ ‚¥ÅÿÊ ∑§Ê ŸÊÁ÷∑§, ¬˝àÿ∑§ Œ˝√ÿ◊ÊŸ ‚¥ÅÿÊ
120 ∑§ ŒÊ πá«UÊ¥ ◊¥ ≈ÍU≈UÃÊ „Ò– •πÁã«Uà ÃÕÊ πÁã«Uà 35. A nucleus with mass number 240 breaks into two
ŸÊÁ÷∑§Ê ¥ ∑§Ë ’㜟 ™§¡Ê¸ ¬˝ Á à ãÿÍ Á Ä‹ÿÊÚ Ÿ ∑˝ § ◊‡Ê— fragments each of mass number 120, the binding
7.6 ◊ªÊß‹Ä≈˛UÊÚŸ flÊÀ≈U (MeV) ÃÕÊ 8.5 ◊ªÊß‹Ä≈˛UÊÚŸ flÊÀ≈U energy per nucleon of unfragmented nuclei is
7.6 MeV while that of fragments is 8.5 MeV. The
(MeV) „Ò– ¬˝∑˝§◊ ◊¥ ∑ȧ‹ ¬˝Êåà ’㜟 ™§¡Ê¸ „ÊÃË „Ò —
total gain in the Binding Energy in the process is :
(1) 0.9 ◊ªÊß‹Ä≈˛UÊÚŸ flÊÀ≈U (1) 0.9 MeV
(2) 9.4 ◊ªÊß‹Ä≈˛UÊÚŸ flÊÀ≈U (2) 9.4 MeV
(3) 804 ◊ªÊß‹Ä≈˛UÊÚŸ flÊÀ≈U (3) 804 MeV
(4) 216 ◊ªÊß‹Ä≈˛UÊÚŸ flÊÀ≈U (4) 216 MeV

Hindi+English M4
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•ŸÈ÷ʪ - B (÷ÊÒÁÃ∑§Ë) Section - B (Physics)

36. R1 ÃÕÊ R2 ÁòÊíÿÊ•Ê¥ ∑§Ë ŒÊ øÊ‹∑§Ëÿ flÎûÊËÿ ‹Í¬ ∞∑§ Ë ◊¥ 36. Two conducting circular loops of radii R1 and R2
‚◊∑§ÁãŒ˝Ã ⁄UπË „Ò– ÿÁŒ R1> > R2 ÃÊ ©Ÿ∑§ ◊äÿ ¬Ê⁄US¬Á⁄U∑§ are placed in the same plane with their centres
coinciding. If R1> > R2 , the mutual inductance M
¬˝⁄U∑§àfl ‘M’ ‚◊ʟȬÊÃË „ÊÃÊ „Ò — between them will be directly proportional to :

R1
R1
(1) R2 ∑§ (1) R2

R2
R2
(2) R1 ∑§ (2) R1

2
R1 2
(3)
R2
∑§ R1
(3)
R2

2
R2 2
(4)
R1
∑§ R2
(4)
R1

37. ªÈáÊ∑§»§‹
37. In the product

→
(
F =q v × B
→ →
) →
(
F =q v × B
→ →
)
→
(
=q v × B i +B j+B0 k
∧ ∧ ∧
)
→
( ∧
=q v × B i +B j +B0 k
∧ ∧
)
→ ∧ ∧ ∧
→ ∧ ∧ ∧ For q=1 and v =2 i +4 j+6 k and
◊¥, q=1 ÃÕÊ v =2 i +4 j+6 k •ÊÒ⁄U
→ ∧ ∧ ∧
→ ∧ ∧ ∧
F=4 i −20 j+12 k ∑§ Á‹∞
F=4 i −20 j +12 k

→
→
B ∑§Ê ‚ê¬Íáʸ √ÿ¥¡∑§ ÄÿÊ „ÊªÊ? What will be the complete expression for B ?

∧ ∧ ∧ ∧ ∧ ∧
(1) (1) −8 i −8 j −6 k
−8 i −8 j −6 k

∧ ∧ ∧ ∧ ∧ ∧
(2) −6 i −6 j −8 k (2) −6 i −6 j −8 k

∧ ∧ ∧ ∧ ∧ ∧
(3) 8 i +8 j −6 k (3) 8 i +8 j−6 k

∧ ∧ ∧ ∧ ∧ ∧
(4) 6 i +6 j −8 k (4) 6 i +6 j−8 k

M4 Hindi+English
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38. ÁŒÿ ªÿ ¬Á⁄U¬Õ ◊¥, ÁŸfl‡ÊË Á«U¡Ë≈U‹ Á‚ªŸ‹ Á‚⁄UÊ¥ A, B ÃÕÊ 38. For the given circuit, the input digital signals are
C ¬⁄U •ŸÈ¬˝ÿÈÄà Á∑§ÿ ¡ÊÃ „Ò¥– Á‚⁄U y ¬⁄U ÁŸª¸Ã Á‚ªŸ‹ ÄÿÊ applied at the terminals A, B and C. What would
be the output at the terminal y ?
„ÊªÊ ?

(1) (1)

(2) (2)

(3)
(3)

(4)
(4)

39. ∞∑§ üÊáÊË LCR ¬Á⁄U¬Õ ◊¥ 5.0 „ãÊ⁄UË ∑§Ê ¬˝⁄U∑§, 80 ◊Êß∑˝§Ê
»Ò§⁄UÊ«U ∑§Ê œÊÁ⁄UòÊ ÃÕÊ 40 •Ê◊ ∑§Ê ¬˝ÁÃ⁄UÊœ∑§ 230 flÊÀ≈U ∑§ 39. A series LCR circuit containing 5.0 H inductor,
80 µF capacitor and 40 Ω resistor is connected to
¬Á⁄UfløŸËÿ •ÊflÎÁûÊ ∑§ ¬˝àÿÊflÃ˸ dÊà ‚ ¡È«∏Ê „Ò– •ŸÈŸÊŒ 230 V variable frequency ac source. The angular
∑§ÊáÊËÿ •ÊflÎÁûÊ ¬⁄U ‡ÊÁÄà ∑§Ë •ÊœË ‡ÊÁÄà SÕÊŸÊãÃÁ⁄Uà ∑§⁄UŸ frequencies of the source at which power
flÊ‹ dÊà ∑§Ë ∑§ÊáÊËÿ •ÊflÎÁûÊÿÊ° „ÊªË — transferred to the circuit is half the power at the
resonant angular frequency are likely to be :
(1) 25 ⁄UÁ«UÿŸ / ‚., 75 ⁄UÁ«UÿŸ / ‚.
(1) 25 rad/s and 75 rad/s
(2) 50 ⁄UÁ«UÿŸ / ‚., 25 ⁄UÁ«UÿŸ / ‚.
(2) 50 rad/s and 25 rad/s

(3) 46 ⁄UÁ«UÿŸ / ‚., 54 ⁄UÁ«UÿŸ / ‚. (3) 46 rad/s and 54 rad/s

(4) 42 ⁄UÁ«UÿŸ / ‚., 58 ⁄UÁ«UÿŸ / ‚. (4) 42 rad/s and 58 rad/s

Hindi+English M4
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40. ∞∑§ 200 ‚.◊Ë. ‹ê’Ê߸ ÃÕÊ 500 ª˝Ê◊ Œ˝√ÿ◊ÊŸ ∑§Ë ‚◊ÊŸ ¿U«∏ 40. A uniform rod of length 200 cm and mass 500 g is
∞∑§ fl¡ ∑§ 40 ‚.◊Ë. ÁŸ‡ÊÊŸ ¬⁄U ‚¥ÃÈÁ‹Ã „ÊÃË „Ò– ∞∑§ balanced on a wedge placed at 40 cm mark. A
mass of 2 kg is suspended from the rod at 20 cm
2 Á∑§.ª˝Ê. ∑§Ê Œ˝√ÿ◊ÊŸ ¿U«∏ ‚ 20 ‚.◊Ë. ¬⁄U ÁŸ‹Áê’à Á∑§ÿÊ and another unknown mass ‘m’ is suspended from
¡ÊÃÊ „Ò ÃÕÊ ŒÍ‚⁄UÊ •ôÊÊà Œ˝√ÿ◊ÊŸ ‘m’ ¿U«∏ ‚ 160 ‚.◊Ë. the rod at 160 cm mark as shown in the figure.
ÁŸ‡ÊÊŸ ‚ ÁŸ‹Áê’à Á∑§ÿÊ ¡ÊÃÊ „Ò– ôÊÊà ∑§ËÁ¡∞ ‘m’ ∑§Ê ◊ÊŸ Find the value of ‘m’ such that the rod is in
Á¡‚‚ ¿U«∏ ‚¥ÃÈ‹Ÿ •flSÕÊ ◊¥ ⁄U„– (g=10 ◊Ë./‚.2) equilibrium. (g=10 m/s2)

1 1
(1) kg
(1) Á∑§.ª˝Ê. 2
2
1 1
(2) kg
(2)
3
Á∑§.ª˝Ê. 3

1 1
(3) kg
(3)
6
Á∑§.ª˝Ê. 6

1 1
(4) Á∑§.ª˝Ê. (4) kg
12 12

41. 30 ‚.◊Ë. »§Ê∑§‚ ŒÍ⁄UË ∑§ ©ûÊ‹ ‹ã‚ ‚ 60 ‚.◊Ë. ŒÍ⁄UË ¬⁄U 41. A point object is placed at a distance of 60 cm from
a convex lens of focal length 30 cm. If a plane
∞∑§ Á’ãŒÈ flSÃÈ ©¬ÁSÕà „Ò– ÿÁŒ ∞∑§ ‚◊Ë Œ¬¸áÊ, ◊ÈÅÿ mirror were put perpendicular to the principal axis
•ˇÊ ∑§ ‹ê’flØ ÃÕÊ ß‚‚ 40 ‚.◊Ë. ŒÍ⁄UË ¬⁄U ⁄UπÊ ¡ÊÃÊ „Ò, ÃÊ of the lens and at a distance of 40 cm from it, the
•ÁãÃ◊ ¬˝ÁÃÁ’ê’ ¬ÊÿÊ ¡ÊÿªÊ ŒÍ⁄UË — final image would be formed at a distance of :

(1) 20 ‚.◊Ë. ‹ã‚ ‚, ÿ„ flÊSÃÁfl∑§ ¬˝ÁÃÁ’ê’ „ÊªÊ– (1) 20 cm from the lens, it would be a real
image.
(2) 30 ‚.◊Ë. ‹ã‚ ‚, ÿ„ flÊSÃÁfl∑§ ¬˝ÁÃÁ’ê’ „ÊªÊ–
(2) 30 cm from the lens, it would be a real
(3) 30 ‚.◊Ë. ‚◊Ë Œ¬¸áÊ ‚, ÿ„ •Ê÷Ê‚Ë ¬˝ÁÃÁ’ê’ image.
„ÊªÊ– (3) 30 cm from the plane mirror, it would be a
(4) 20 ‚.◊Ë. ‚◊Ë Œ¬¸áÊ ‚, ÿ„ •Ê÷Ê‚Ë ¬˝ÁÃÁ’ê’ virtual image.
„ÊªÊ– (4) 20 cm from the plane mirror, it would be a
virtual image.
42. ∞∑§ ∑§Ê⁄U Áfl⁄UÊ◊ÊflSÕÊ ‚ ¬˝Ê⁄Uê÷ ∑§⁄UÃË „Ò ÃÕÊ 5 ◊Ë./‚.2 ‚ 42. A car starts from rest and accelerates at 5 m/s2.
àflÁ⁄Uà „ÊÃË „Ò– t=4 ‚∑§á«UU ¬⁄U ∑§Ê⁄U ◊¥ ’Ò∆U √ÿÁÄà mÊ⁄UÊ ∞∑§ At t=4 s, a ball is dropped out of a window by a
ª¥Œ Áπ«∏∑§Ë ∑§ ’Ê„⁄U Áª⁄UÊÿË ¡ÊÃË „Ò– t=6 ‚∑§á«UU ¬⁄U ª¥Œ person sitting in the car. What is the velocity and
∑§Ê flª ÃÕÊ àfl⁄UáÊ ÄÿÊ „ÊÃÊ „Ò? (ÁŒÿÊ „Ò — g=10 ◊Ë./‚.2) acceleration of the ball at t=6 s ?
(Take g=10 m/s2)
(1) 20 ◊Ë./‚., 5 ◊Ë./‚.2
(1) 20 m/s, 5 m/s2
(2) 20 ◊Ë./‚., 0 (2) 20 m/s, 0
(3) 20 2 ◊Ë./‚., 0 (3) 20 2 m/s, 0
(4) 20 2 ◊Ë./‚., 10 ◊Ë./‚.2 (4) 20 2 m/s, 10 m/s2

M4 Hindi+English
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43. ‚◊ÊŸ ‚Êß¡ ∑§Ë 27 ’Í¥Œ¥ ¬˝àÿ∑§ 220 flÊÀ≈U ¬⁄U •ÊflÁ‡Êà „ÊÃË 43. Twenty seven drops of same size are charged at
„Ò– fl Á◊‹∑§⁄U ∞∑§ ’«∏Ë ’Í¥Œ ’ŸÊÃË „Ò– ’«∏Ë ’Í¥Œ ∑§ Áfl÷fl 220 V each. They combine to form a bigger drop.
∑§Ë ªáÊŸÊ ∑§ËÁ¡∞– Calculate the potential of the bigger drop.
(1) 660 flÊÀ≈U (1) 660 V
(2) 1320 flÊÀ≈U (2) 1320 V
(3) 1520 flÊÀ≈U (3) 1520 V
(4) 1980 V
(4) 1980 flÊÀ≈U
44. A uniform conducting wire of length 12a and
44. ∞∑§ 12a ‹ê’Ê߸ ÃÕÊ ¬˝ÁÃ⁄UÊœ ‘R’ ∑§Ê ‚◊ÊŸ øÊ‹∑§Ëÿ ÃÊ⁄U, resistance ‘R’ is wound up as a current carrying
(i) ‘a’ ÷È¡Ê ∑§ ‚◊’Ê„È ÁòÊ÷È¡ ÃÕÊ
coil in the shape of,
(ii) ‘a’ ÷È¡Ê ∑§ flª¸ ∑§ •Ê∑§Ê⁄U ∑§Ë œÊ⁄UÊflÊ„Ë ∑ȧá«U‹Ë ◊¥ (i) an equilateral triangle of side ‘a’.
◊Ê«∏Ê ¡ÊÃÊ „Ò– (ii) a square of side ‘a’.
¬˝àÿ∑§ ∑ȧá«U‹Ë ∑§Ê øÈê’∑§Ëÿ ÁmœÈ˝fl •ÊÉÊÍáʸ ∑˝§◊‡Ê— „Ò — The magnetic dipole moments of the coil in each
(1) 3 Ia2 ÃÕÊ 3 Ia2 case respectively are :
(2) 3 Ia2 ÃÕÊ Ia2 (1) 3 Ia2 and 3 Ia2
(3) 3 Ia2 ÃÕÊ 4 Ia2 (2) 3 Ia2 and Ia2
(4) 4 Ia2 ÃÕÊ 3 Ia2 (3) 3 Ia2 and 4 Ia2
(4) 4 Ia2 and 3 Ia2
45. ∞∑§ 0.15 Á∑§.ª˝Ê. ∑§Ë ª¥Œ 10 ◊Ë. ™§°øÊ߸ ‚ Áª⁄UÊÿË ¡ÊÃË „Ò 45. A ball of mass 0.15 kg is dropped from a height
ÃÕÊ ¡◊ËŸ ‚ ≈U∑§⁄UÊ∑§⁄U ‚◊ÊŸ ™§°øÊ߸ Ã∑§ ©¿U‹ÃË „Ò– ª¥Œ ¬⁄U 10 m, strikes the ground and rebounds to the same
‹ªÊÿ ªÿ •Êfl ª ∑§Ê ¬Á⁄U◊ÊáÊ „Ê Ã Ê „Ò , ‹ª÷ª — height. The magnitude of impulse imparted to
(g=10 ◊Ë./‚.2) the ball is (g=10 m/s2) nearly :
(1) 0 Á∑§.ª˝Ê.×◊Ë. / ‚. (1) 0 kg m/s
(2) 4.2 Á∑§.ª˝Ê.×◊Ë. / ‚. (2) 4.2 kg m/s
(3) 2.1 kg m/s
(3) 2.1 Á∑§.ª˝Ê.×◊Ë. / ‚. (4) 1.4 kg m/s
(4) 1.4 Á∑§.ª˝Ê.×◊Ë. / ‚. 46. A step down transformer connected to an ac mains
46. ∞∑§ •¬øÊÿË ≈˛UÊã‚»§Ê◊¸⁄U ¡Ê 220 flÊÀ≈U ◊ÈÅÿ ¬˝àÿÊflÃ˸ ¬ÍÁø ‚ supply of 220 V is made to operate at 11 V, 44 W
¡È«∏Ê „Ò, 11 flÊÀ≈U, 44 flÊ≈U ‹Òê¬ ¬⁄U ∑§Êÿ¸ ∑§⁄UÃÊ „Ò– ≈˛UÊã‚»§Ê◊¸⁄U lamp. Ignoring power losses in the transformer,
◊¥ ‡ÊÁÄà „ÊÁŸ ∑§Ê Ÿªáÿ ◊ÊŸÃ „È∞, ¬˝Ê⁄UÁê÷∑§ ¬Á⁄U¬Õ ◊¥ œÊ⁄UÊ what is the current in the primary circuit ?
ÄÿÊ „ÊÃË „Ò? (1) 0.2 A
(1) 0.2 ∞Áê¬ÿ⁄U (2) 0.4 A
(3) 2A
(2) 0.4 ∞Áê¬ÿ⁄U (4) 4A
(3) 2 ∞Áê¬ÿ⁄U 47. A particle moving in a circle of radius R with a
(4) 4 ∞Áê¬ÿ⁄U uniform speed takes a time T to complete one
47. ∞∑§ ∑§áÊ R ÁòÊíÿÊ ∑§ flÎûÊ ◊¥ ‚◊ÊŸ øÊ‹ ‚ ªÁà ∑§⁄UÃ „È∞ ∞∑§ revolution.
If this particle were projected with the same speed
øÄ∑§⁄U ¬Í⁄UÊ ∑§⁄UŸ ◊¥ T ‚◊ÿ ‹ÃÊ „Ò–
at an angle ‘θ’ to the horizontal, the maximum
ÿÁŒ ÿ„Ë ∑§áÊ ˇÊÒÁá ‚ ©‚Ë øÊ‹ ‚ ∑§ÊáÊ ‘θ’ ¬⁄U ¬˝ˇÊÁ¬Ã height attained by it equals 4R. The angle of
Á∑§ÿÊ ¡Ê∞, ÃÊ 4R ∑§ ’⁄UÊ’⁄U •Áœ∑§Ã◊ ™§°øÊ߸ ¬˝Êåà ∑§⁄UÃÊ „Ò– projection, θ, is then given by :
¬˝ˇÊ¬áÊ ∑§ÊáÊ ‘θ’ ÁŒÿÊ ¡ÊÃÊ „Ò —
1
1  gT2  2
 gT2  2 θ=cos −1
 
θ=cos−1   (1)  π2 R 
(1)  π2 R   
 
1 1
 π2 R  2 −1
 π2 R  2
θ=cos−1   (2) θ=cos  
(2)  gT2   gT2 
   
1 1
 π2 R  2  π2 R  2
(3) θ=sin−1   (3) θ=sin−1  
 gT2   gT2 
   
1 1
 2gT2  2  2gT2  2
(4) θ=sin−1   (4) θ=sin −1
 
 π2 R   π2 R 
   

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48. ∞∑§ ‘m’ Œ˝ √ ÿ◊ÊŸ ∑§Ê ∑§áÊ ¬Î â flË ‚Ä ‚ ‚◊ÊŸ fl ª 48. A particle of mass ‘m’ is projected with a velocity
v=kVe(k < 1) ‚ ¬˝ˇÊÁ¬Ã Á∑§ÿÊ ¡ÊÃÊ „Ò– v=kVe(k < 1) from the surface of the earth.
(Ve=¬‹ÊÿŸ flª) (Ve=escape velocity)
∑§áÊ ∑§ mÊ⁄UÊ ‚Ä ∑§ ™§¬⁄U ¬˝Êåà •Áœ∑§Ã◊ ™§°øÊ߸ „Ò — The maximum height above the surface reached
by the particle is :
2
 k  2
(1) R   k 
 1−k  (1) R 
 1−k 
2
 k  2
(2) R   k 
 1+k  (2) R 
 1+k 
R2 k
(3) R2 k
1+k (3)
1+k
Rk2 Rk2
(4) 2 (4)
1−k 1−k2
49. ÁŒ∞ ªÿ ¬Á⁄U¬Õ ∑§ •ŸÈ‚Ê⁄U r1, r2 ÃÕÊ r3 ¬˝ÁÃ⁄UÊœÊ¥ flÊ‹ ÃËŸ 49. Three resistors having resistances r1, r2 and r3
¬˝ÁÃ⁄UÊœ∑§ ¡Ê«∏ ªÿ „Ò¥– ¬Á⁄U¬Õ ◊¥ ¬˝ÿÈÄà ¬˝ÁÃ⁄UÊœÊ¥ ∑§ ¬ŒÊ¥ ◊¥ are connected as shown in the given circuit. The
i
i3 ratio 3 of currents in terms of resistances used
œÊ⁄UÊ•Ê¥ i ∑§Ê •ŸÈ¬Êà „Ò — i1
1 in the circuit is :

r1
(1) r1
r2 +r3 (1) r2 +r3
r2 r2
(2) r2 +r3 (2) r2 +r3
r1 r1
(3) r1+r2 (3) r1 +r2
r2 r2
(4) r1+r3 (4) r1 +r3
50. ‘M’ Œ˝√ÿ◊ÊŸ ÃÕÊ ‘R’ ÁòÊíÿÊ ∑§ ∞∑§ flÎûÊËÿ ¿UÀ‹ ‚ 908 50. From a circular ring of mass ‘M’ and radius ‘R’ an
‚Ä≈U⁄U ∑§ ‚¥ªÃ ∞∑§ øʬ (•Ê∑¸§) „≈UÊ ÁŒÿÊ ¡ÊÃÊ „Ò– ’ø „È∞ arc corresponding to a 908 sector is removed. The
¿UÀ‹ ∑§ ÷ʪ ∑§Ê ¡«∏àfl •ÊÉÊÍáʸ ¿UÀ‹ ∑§ ∑§ãŒ˝ ‚ ªÈ¡⁄UŸ flÊ‹Ë moment of inertia of the remaining part of the ring
about an axis passing through the centre of the
ÃÕÊ ¿UÀ‹ ∑§ Ë ∑§ ‹ê’flØ •ˇÊ ∑§ ‚ʬˇÊ ‘MR2’ ∑§Ê ‘K’ ring and perpendicular to the plane of the ring is
ªÈŸÊ „Ò– ‘K’ ∑§Ê ◊ÊŸ „Ò — ‘K’ times ‘MR2’. Then the value of ‘K’ is :
3 3
(1) (1)
4 4
7 7
(2) (2)
8 8
1 1
(3) (3)
4 4
1 1
(4) (4)
8 8

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•ŸÈ÷ʪ - A (⁄U‚ÊÿŸ‡ÊÊSòÊ) Section - A (Chemistry)
51. ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê Áfl∑§À¬ ∞∑§ ◊Ê‹ •ÊŒ‡Ê¸ ªÒ‚ ∑§ Á‹∞ 51. Which one among the following is the correct option
CP ∞fl¥ CV ∑§ ‚„Ë ‚¥’¥œ ∑§Ê √ÿÄà ∑§⁄UÃÊ „Ò? for right relationship between CP and CV for one
(1) CP+CV=R mole of ideal gas ?
(2) CP−CV=R (1) CP+CV=R
(3) CP=RCV (2) CP−CV=R
(4) CV=RCP (3) CP=RCV
(4) CV=RCP
52. ∑ȧ‹ 14 ¬˝∑§Ê⁄U ∑§ ’˝fl  ¡Ê‹∑§Ê¥ ∑§Ë •¥Ã—∑§ÁãŒ˝Ã ∞∑§∑§ ∑§ÊÁc∆U∑§Ê•Ê¥
∑§Ë ‚¥ÅÿÊ ∑§ Á‹∞ ‚„Ë Áfl∑§À¬ „Ò — 52. The correct option for the number of body centred
(1) 7 unit cells in all 14 types of Bravais lattice unit
(2) 5 cells is :
(3) 2 (1) 7
(4) 3 (2) 5
(3) 2
53. ©à∑Χc≈U ªÒ‚Ê¥ ∑§Ê ŸÊ◊ ©Ÿ∑§Ë ⁄UÊ‚ÊÿÁŸ∑§ •Á÷Á∑˝§ÿÊ ∑§ ¬˝Áà (4) 3
•Á∑˝§ÿÃÊ ∑§ ∑§Ê⁄UáÊ ¬«∏Ê „Ò– ©Ÿ‚ ‚ê’ÁãœÃ •‚àÿ ∑§ÕŸ ∑§Ê
¬„øÊŸ¥– 53. Noble gases are named because of their inertness
towards reactivity. Identify an incorrect
(1) ©à∑Χc≈U ªÒ‚¥ ¡‹ ◊¥ •À¬ Áfl‹ÿ „Ò¥– statement about them.
(2) ©à∑Χc≈U ªÒ‚Ê¥ ∑§ ª‹ŸÊ¥∑§ ∞fl¥ ÄflÕŸÊ¥∑§ •Áà ©ìÊ „ÊÃ (1) Noble gases are sparingly soluble in water.
„Ò¥– (2) Noble gases have very high melting and
(3) ©à∑Χc≈U ªÒ‚Ê¥ ◊¥ ŒÈ’¸‹ ¬Á⁄UˇÊ¬áÊ ’‹ „ÊÃ „Ò¥– boiling points.
(4) ©à∑Χc≈U ªÒ‚Ê¥ ∑§ ß‹Ä≈˛UÊÚŸ ‹Áéœ ∞ãÕÒÀ¬Ë ∑§Ê ◊ÊŸ ©ìÊ (3) Noble gases have weak dispersion forces.
œŸÊà◊∑§ „ÊÃÊ „Ò– (4) Noble gases have large positive values of
electron gain enthalpy.
54. 2-’˝Ê◊Ê ¬ã≈UŸ ∑§ Áfl„Êß«˛UÊ„Ò‹Ê¡ŸË∑§⁄UáÊ •Á÷Á∑˝§ÿÊ ∑§Ê ◊ÈÅÿ
©à¬ÊŒ ¬ã≈U-2-߸Ÿ „Ò– ©Äà ©à¬ÊŒ ∑§Ê ÁŸ◊ʸáÊ •ÊœÊÁ⁄Uà „ÊÃÊ 54. The major product formed in dehydrohalogenation
„Ò — reaction of 2-Bromo pentane is Pent-2-ene. This
(1) ‚≈U$¡»§ ÁŸÿ◊ ¬⁄U product formation is based on ?
(1) Saytzeff’s Rule
(2) „Èá«U ÁŸÿ◊ ¬⁄U
(2) Hund’s Rule
(3) „ÊÚ»§◊ÒŸ ÁŸÿ◊ ¬⁄U
(3) Hofmann Rule
(4) „∑§‹ ÁŸÿ◊ ¬⁄U (4) Huckel’s Rule
55. ◊äÿÊflÿflÃÊ ¬˝ŒÁ‡Ê¸Ã ∑§⁄UŸ flÊ‹Ê ÿÊÒÁª∑§ „Ò — 55. The compound which shows metamerism is :
(1) C5H12
(1) C5H12
(2) C3H8O (2) C3H8O
(3) C3H6O (3) C3H6O
(4) C4H10O (4) C4H10O
56. BF3 ∞∑§ ‚◊ËËÿ ∞fl¥ ß‹Ä≈˛UÊÚŸ ãÿÍŸ ÿÊÒÁª∑§ „Ò– ∑§ãŒ˝Ëÿ 56. BF3 is planar and electron deficient compound.
¬⁄U◊ÊáÊÈ ∑§Ê ‚¥∑§⁄UáÊ ∞fl¥ ©‚∑§ øÊ⁄UÊ¥ •Ê⁄U ß‹Ä≈˛UÊÚŸÊ¥ ∑§Ë ‚¥ÅÿÊ Hybridization and number of electrons around the
„Ò, ∑˝§◊‡Ê— — central atom, respectively are :
(1) sp3 ∞fl¥ 4 (1) sp3 and 4
(2) sp3 ∞fl¥ 6 (2) sp3 and 6
(3) sp2 ∞fl¥ 6 (3) sp2 and 6
(4) sp2 ∞fl¥ 8 (4) sp2 and 8

57. ““Á≈Uã«U‹ ¬˝÷Êfl ÁŸêŸ ∑§ mÊ⁄UÊ ¬˝ŒÁ‡Ê¸Ã Á∑§ÿÊ ¡ÊÃÊ „Ò–”” ‚„Ë 57. The right option for the statement “Tyndall effect
is exhibited by”, is :
Áfl∑§À¬ øÈŸ¥–
(1) NaCl solution
(1) NaCl Áfl‹ÿŸ
(2) Glucose solution
(2) Ç‹Í∑§Ê‚ Áfl‹ÿŸ (3) Starch solution
(3) S≈UÊø¸ Áfl‹ÿŸ (4) Urea solution
(4) ÿÍÁ⁄UÿÊ Áfl‹ÿŸ

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58. ÁŸêŸ ˇÊÊ⁄UËÿ ◊ÎŒÊ œÊÃÈ „Ò‹Êß«UÊ¥ ◊¥ ‚ ∑§ÊÒŸ ‚„‚¥ÿÊ¡Ë ∞fl¥ 58. Among the following alkaline earth metal halides,
∑§Ê’¸ÁŸ∑§ Áfl‹Êÿ∑§Ê¥ ◊¥ ÉÊÈ‹Ÿ‡ÊË‹ „Ò? one which is covalent and soluble in organic
solvents is :
(1) ∑Ò§ÁÀ‡Êÿ◊ Ä‹Ê⁄UÊß«U
(1) Calcium chloride
(2) S≈˛UÊÚÁã‡Êÿ◊ Ä‹Ê⁄UÊß«U
(2) Strontium chloride
(3) ◊ÒÇŸËÁ‡Êÿ◊ Ä‹Ê⁄UÊß«U (3) Magnesium chloride
(4) ’⁄UËÁ‹ÿ◊ Ä‹Ê⁄UÊß«U (4) Beryllium chloride
59. ÁŸêŸ ◊¥ ‚ Á∑§‚ ÁflÁœ ∑§Ê ©¬ÿÊª ∑§⁄U •àÿÁœ∑§ ‡ÊÈh œÊÃÈ ∑§Ê 59. Which one of the following methods can be used to
¬˝Êåà ∑§⁄U ‚∑§Ã „Ò¥ ¡Ê ∑§◊⁄U ∑§ Ãʬ ¬⁄U Œ˝fl „Ò? obtain highly pure metal which is liquid at room
(1) ÁfllÈà •¬ÉÊ≈UŸ temperature ?
(2) fláʸ‹Áπ∑§Ë (1) Electrolysis
(3) •Ê‚flŸ (2) Chromatography
(4) ◊¥«U‹ ¬Á⁄Uc∑§⁄UáÊ (3) Distillation
(4) Zone refining
60. Á∑§‚Ë •Á÷Á∑˝§ÿÊ A→B ∑§ Á‹∞ •Á÷Á∑˝§ÿÊ ∑§Ë ∞ãÕÒÀ¬Ë
−4.2 kJ mol−1 ∞fl¥ ‚Á∑˝§ÿáÊ ∑§Ë ∞ãÕÒÀ¬Ë 9.6 kJ mol−1 60. For a reaction A→B, enthalpy of reaction is
„Ò– •Á÷Á∑˝§ÿÊ ∑§ Á‹∞ ‚„Ë ÁSÕÁá ™§¡Ê¸ •Ê⁄Uπ ÁŸêŸ −4.2 kJ mol−1 and enthalpy of activation is
Áfl∑§À¬ ◊¥ ¬˝ŒÁ‡Ê¸Ã Á∑§ÿÊ ªÿÊ „Ò — 9.6 kJ mol−1. The correct potential energy profile
for the reaction is shown in option.

(1)
(1)

(2)
(2)

(3)
(3)

(4)

(4)

61. ÁŸêŸ ’„È‹∑§Ê¥ ◊¥ ‚ ∑§ÊÒŸ ÿÊªÊà◊∑§ ’„È‹∑§Ÿ ∑§ mÊ⁄UÊ ÁŸÁ◊¸Ã
61. Which one of the following polymers is prepared
Á∑§ÿÊ ¡ÊÃÊ „Ò? by addition polymerisation ?
(1) ≈Uç‹ÊÚŸ (1) Teflon
(2) ŸÊß‹ÊÚŸ-66 (2) Nylon-66
(3) ŸÊflÊ‹∑§ (3) Novolac
(4) «U∑˝§ÊÚŸ (4) Dacron

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62. ÁŸêŸ ⁄UÊ‚ÊÿÁŸ∑§ •Á÷Á∑˝§ÿÊ ◊¥ ◊ÈÅÿ ©à¬ÊŒ „Ò — 62. The major product of the following chemical
reaction is :

(1)
(1)

(2)
(2)

(3)
(3)

(4)
(4)

63. ’ÊÚÿ‹ ∑§ ÁŸÿ◊ ∑§Ê ‚„Ë ª˝Ê»§Ëÿ ÁŸM§¬áÊ øÈŸ¥ ¡Ê ÁflÁ÷ÛÊ ÃʬÊ¥ 63. Choose the correct option for graphical
¬⁄U ªÒ‚ ∑§Ê ŒÊ’ vs. •Êÿß ∑§Ê ¬˝ŒÁ‡Ê¸Ã ∑§⁄U ⁄U„Ê „Ê — representation of Boyle’s law, which shows a graph
of pressure vs. volume of a gas at different
temperatures :

(1)
(1)

(2)
(2)

(3)
(3)

(4)
(4)

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64. ∞ÕŸ ∑§ ÁŸêŸÃ◊ SÕÊÿË ‚¥M§¬áÊ ◊¥ ÁmË ∑§ÊáÊ „Ò — 64. Dihedral angle of least stable conformer of ethane
is :
(1) 1208
(1) 1208
(2) 1808
(2) 1808
(3) 608
(3) 608
(4) 08
(4) 08
65. ∞∑§ ∑§Ê’¸ÁŸ∑§ ÿÊÒÁª∑§ ◊¥ 78% (÷Ê⁄U mÊ⁄UÊ) ∑§Ê’¸Ÿ ∞fl¥ ‡Ê· 65. An organic compound contains 78% (by wt.) carbon
¬˝ÁÇÊà „Êß«˛UÊ¡Ÿ ∑§Ë ◊ÊòÊÊ „Ò– ß‚ ÿÊÒÁª∑§ ∑§ ◊͋ʟȬÊÃË ‚ÍòÊ and remaining percentage of hydrogen. The right
∑§Ê ‚„Ë Áfl∑§À¬ „Ò — [¬⁄U◊ÊáÊÈ ÷Ê⁄U — C=12, H=1] option for the empirical formula of this compound
(1) CH is : [Atomic wt. of C is 12, H is 1]
(2) CH2 (1) CH

(3) CH3 (2) CH2
(3) CH3
(4) CH4
(4) CH4
66. flÊàÿÊ ÷^Ë ◊¥ ¬˝Êåà Á∑§ÿÊ ¡Ê ‚∑§Ÿ flÊ‹Ê •Áœ∑§Ã◊ Ãʬ◊ÊŸ
„Ò — 66. The maximum temperature that can be achieved
in blast furnace is :
(1) 1200 K Ã∑§
(1) upto 1200 K
(2) 2200 K Ã∑§
(2) upto 2200 K
(3) 1900 K Ã∑§ (3) upto 1900 K
(4) 5000 K Ã∑§ (4) upto 5000 K

67. ÁŸêŸ Áfl‹ÿŸÊ¥ ∑§Ê ’ŸÊÿÊ ªÿÊ — 67. The following solutions were prepared by dissolving
250 ml ¡‹ ◊¥ 10 g Ç‹Í∑§Ê‚ (C6H12O6) ∑§Ê ÉÊÊ‹∑§⁄U (P1), 10 g of glucose (C6H12O6) in 250 ml of water (P1),
250 ml ¡‹ ◊¥ 10 g ÿÍÁ⁄UÿÊ (CH4N2O) ∑§Ê ÉÊÊ‹∑§⁄U (P2) 10 g of urea (CH4N2O) in 250 ml of water (P2) and
10 g of sucrose (C 12 H 22 O 11 ) in 250 ml of
∞fl¥ 250 ml ¡‹ ◊¥ 10 g ‚È∑˝§Ê‚ (C12H22O11) ∑§Ê ÉÊÊ‹∑§⁄U water (P3). The right option for the decreasing
(P3)– ߟ Áfl‹ÿŸÊ¥ ∑§ ¬⁄UÊ‚⁄UáÊ ŒÊ’Ê¥ ∑§ ÉÊ≈UÃ ∑˝§◊ ∑§Ê ‚„Ë order of osmotic pressure of these solutions is :
Áfl∑§À¬ „Ò — (1) P2 > P1 > P3
(1) P2 > P1 > P3 (2) P1 > P2 > P3
(2) P1 > P2 > P3 (3) P2 > P3 > P1
(3) P2 > P3 > P1 (4) P3 > P1 > P2
(4) P3 > P1 > P2
68. Zr (Z=40) and Hf (Z=72) have similar atomic and
68. Zr (Z=40) ∞fl¥ Hf (Z=72) ∑§ ¬⁄U◊ÊÁáfl∑§ ∞fl¥ •ÊÿÁŸ∑§ ionic radii because of :
ÁòÊíÿÊ∞° ‚◊ÊŸ „Ò¥– ß‚∑§Ê ∑§Ê⁄UáÊ „Ò — (1) belonging to same group
(1) ŒÊŸÊ¥ ‚◊ÊŸ ‚◊Í„ ∑§ ‚ŒSÿ „Ò¥ (2) diagonal relationship
(2) Áfl∑§áʸ ‚ê’㜠(3) lanthanoid contraction
(3) ‹ÒãÕŸÊÚÿ«U •Ê∑È¥§øŸ (4) having similar chemical properties

(4) ŒÊŸÊ¥ ∑§ ⁄UÊ‚ÊÿÁŸ∑§ ªÈáÊœ◊¸ ‚◊ÊŸ „Ò¥ 69. The molar conductance of NaCl, HCl and
CH3COONa at infinite dilution are 126.45, 426.16
69. NaCl, HCl ∞fl¥ CH3COONa ∑§Ë •Ÿ¥Ã ßÈÃÊ ¬⁄U ◊Ê‹⁄U and 91.0 S cm2 mol−1 respectively. The molar
øÊ‹∑§ÃÊ ∑˝§◊‡Ê— 126.45, 426.16 ∞fl¥ 91.0 S cm2 mol−1 conductance of CH3COOH at infinite dilution is.
„Ò– •Ÿ¥Ã ßÈÃÊ ¬⁄U CH3COOH ∑§Ë ◊Ê‹⁄U øÊ‹∑§ÃÊ „Ò– Choose the right option for your answer.
(1) 201.28 S cm2 mol−1 (1) 201.28 S cm2 mol−1
(2) 390.71 S cm2 mol−1 (2) 390.71 S cm2 mol−1
(3) 698.28 S cm2 mol−1 (3) 698.28 S cm2 mol−1
(4) 540.48 S cm2 mol−1
(4) 540.48 S cm2 mol−1

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70. Á∑§‚Ë ·≈˜U∑§ÊáÊËÿ •Êl („ĂʪÊŸ‹ Á¬˝Á◊Á≈Ufl) ∞∑§∑§ ∑§ÊÁc∆U∑§Ê 70. Right option for the number of tetrahedral and
◊¥ øÃÈc»§‹∑§Ëÿ ∞fl¥ •c≈U»§‹∑§Ëÿ Á⁄UÁÄÃÿÊ¥ ∑§Ë ‚¥ÅÿÊ „ÃÈ octahedral voids in hexagonal primitive unit cell
are :
‚„Ë Áfl∑§À¬ „Ò —
(1) 8, 4
(1) 8, 4
(2) 6, 12
(2) 6, 12
(3) 2, 1
(3) 2, 1
(4) 12, 6
(4) 12, 6
71. A particular station of All India Radio, New Delhi,
71. •ÊÚ‹ ߥÁ«UÿÊ ⁄UÁ«UÿÊ, Ÿß¸ ÁŒÀ‹Ë ∑§Ê ∞∑§ S≈U‡ÊŸ 1,368 kHz broadcasts on a frequency of 1,368 kHz (kilohertz).
(Á∑§‹Ê „≈˜U¸¡) ∑§Ë •ÊflÎÁûÊ ¬⁄U ¬˝‚Ê⁄UáÊ ∑§⁄UÃÊ „Ò– ‚¥øÊ⁄U∑§ The wavelength of the electromagnetic radiation
(≈˛UÊ¥‚◊Ë≈U⁄U) mÊ⁄UÊ ©à‚Á¡¸Ã ÁfllÈà øÈê’∑§Ëÿ ÁflÁ∑§⁄UáÊ ∑§Ê emitted by the transmitter is : [speed of light,
Ã⁄¥UªŒÒäÿ¸ „Ò — [¬˝∑§Ê‡Ê ∑§Ê flª, c=3.0×108 ms−1] c=3.0×108 ms−1]

(1) 219.3 m (1) 219.3 m
(2) 219.2 m
(2) 219.2 m
(3) 2192 m
(3) 2192 m
(4) 21.92 cm
(4) 21.92 cm
72. Ethylene diaminetetraacetate (EDTA) ion is :
72. ∞ÁÕ‹ËŸ «UÊß∞◊ËŸ≈U≈˛UÊ∞‚Ë≈U≈U (EDTA) •ÊÿŸ „Ò —
(1) Hexadentate ligand with four “O” and two
(1) øÊ⁄U “O” ∞fl¥ ŒÊ “N” ŒÊÃÊ ¬⁄U◊ÊáÊÈ•Ê¥ ∑§ ‚ÊÕ ·≈˜UŒ¥ÃÈ⁄U “N” donor atoms
Á‹ªã«U (2) Unidentate ligand
(2) ∞∑§Œ¥ÃÈ⁄U Á‹ªã«U (3) Bidentate ligand with two “N” donor atoms
(3) ŒÊ “N” ŒÊÃÊ ¬⁄U◊ÊáÊÈ•Ê¥ ∑§ ‚ÊÕ ÁmŒ¥ÃÈ⁄U Á‹ªã«U (4) Tridentate ligand with three “N” donor
(4) ÃËŸ “N” ŒÊÃÊ ¬⁄U◊ÊáÊÈ•Ê¥ ∑§ ‚ÊÕ ÁòÊŒ¥ÃÈ⁄U Á‹ªã«U atoms

73. ‘C–X’ ’¥œ ∑§Ë ’¥œ ∞ãÕÒÀ¬Ë ∑§Ê ‚„Ë ∑˝§◊ „Ò — 73. The correct sequence of bond enthalpy of ‘C–X’ bond
is :
(1) CH3−F < CH3−Cl < CH3−Br < CH3−I
(1) CH3−F < CH3−Cl < CH3−Br < CH3−I
(2) CH3−F > CH3−Cl > CH3−Br > CH3−I
(2) CH3−F > CH3−Cl > CH3−Br > CH3−I
(3) CH3−F < CH3−Cl > CH3−Br > CH3−I
(3) CH3−F < CH3−Cl > CH3−Br > CH3−I
(4) CH3−Cl > CH3−F > CH3−Br > CH3−I
(4) CH3−Cl > CH3−F > CH3−Br > CH3−I
74. T (K) ¬⁄U «UÊß◊ÁÕ‹∞◊ËŸ ∑§Ê pKb ∞fl¥ ∞‚ËÁ≈U∑§ •ê‹ ∑§Ê
74. The pKb of dimethylamine and pKa of acetic acid
pKa ◊ÊŸ ∑˝§◊‡Ê— 3.27 ∞fl¥ 4.77 „Ò– «UÊß◊ÁÕ‹•◊ÊÁŸÿ◊ are 3.27 and 4.77 respectively at T (K). The correct
∞‚Ë≈U≈U Áfl‹ÿŸ ∑§ pH ∑§Ê ‚„Ë Áfl∑§À¬ „Ò — option for the pH of dimethylammonium acetate
(1) 8.50 solution is :

(2) 5.50 (1) 8.50
(2) 5.50
(3) 7.75
(3) 7.75
(4) 6.25
(4) 6.25
75. ’⁄UËÁ‹ÿ◊ Ä‹Ê⁄UÊß«U ∑§Ë ∆UÊ‚ •flSÕÊ ∞fl¥ flÊc¬ •flSÕÊ ◊¥
‚¥⁄UøŸÊ∞° „Ò¥ — 75. The structures of beryllium chloride in solid state
and vapour phase, are :
(1) ∑˝§◊‡Ê— oÎ¥π‹Ê ∞fl¥ Ám‹∑§ (1) Chain and dimer, respectively
(2) ŒÊŸÊ¥ ◊¥ ⁄UπËÿ (2) Linear in both
(3) ∑˝§◊‡Ê— Ám‹∑§ ∞fl¥ ⁄UπËÿ (3) Dimer and Linear, respectively
(4) ŒÊŸÊ¥ ◊¥ oÎ¥π‹Ê (4) Chain in both

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76. ŸËø ŒÊ ∑§ÕŸ ÁŒ∞ ª∞ „Ò¥ —U 76. Given below are two statements :

∑§ÕŸ I : Statement I :
Aspirin and Paracetamol belong to the class of
∞ÁS¬Á⁄UŸ ∞fl¥ ¬Ò⁄UÊÁ‚≈UÊ◊ÊÚ‹ Sflʬ∑§ ¬Ë«∏Ê„Ê⁄UË (ŸÊ⁄U∑§ÊÁ≈U∑§ narcotic analgesics.
∞ŸÀ¡Á‚∑§) flª¸ ∑§ „Ò¥–
Statement II :
∑§ÕŸ II : Morphine and Heroin are non-narcotic analgesics.
◊ÊÚ»§Ë¸Ÿ ∞fl¥ „⁄UÊߟ •Sflʬ∑§ ¬Ë«∏Ê„Ê⁄UË (ŸÊÚŸ-ŸÊ⁄U∑§ÊÁ≈U∑§ In the light of the above statements, choose the
∞ŸÀ¡Á‚∑§) flª¸ ∑§ „Ò¥– correct answer from the options given below.

©¬ÿȸÄà ∑§ÕŸÊ¥ ∑§ ¬˝∑§Ê‡Ê ◊¥ ŸËø ÁŒ∞ „È∞ Áfl∑§À¬Ê¥ ◊¥ ‚ ‚„Ë (1) Both Statement I and Statement II are
true.
©ûÊ⁄U øÈŸ¥–
(2) Both Statement I and Statement II are
(1) ŒÊŸÊ¥ ∑§ÕŸ I ∞fl¥ ∑§ÕŸ II ‚„Ë „Ò¥– false.
(2) ŒÊŸÊ¥ ∑§ÕŸ I ∞fl¥ ∑§ÕŸ II ª‹Ã „Ò¥– (3) Statement I is correct but Statement II
is false.
(3) ∑§ÕŸ I ‚„Ë „Ò ‹Á∑§Ÿ ∑§ÕŸ II ª‹Ã „Ò–
(4) Statement I is incorrect but
(4) ∑§ÕŸ I ª‹Ã „Ò ‹Á∑§Ÿ ∑§ÕŸ II ‚„Ë „Ò– Statement II is true.

77. ‚ÍøË-I ∑§Ê Á◊‹ÊŸ ‚ÍøË-II ‚ ∑§⁄¥U– 77. Match List - I with List - II.
List - I List - II
‚ÍøË-I ‚ÍøË-II
(a) PCl5 (i) Square pyramidal
(a) PCl5 (i) flª¸ Á¬⁄UÊÁ◊«UË
(b) SF6 (ii) Trigonal planar
(b) SF6 (ii) ÁòÊ∑§ÊáÊËÿ ‚◊ËËÿ (c) BrF5 (iii) Octahedral
(c) BrF5 (iii) •c≈U»§‹∑§Ëÿ (d) BF3 (iv) Trigonal bipyramidal
(d) BF3 (iv) ÁòÊ∑§ÊáÊËÿ ÁmÁ¬⁄UÊÁ◊«UË Choose the correct answer from the options given
below.
ŸËø ÁŒ∞ ª∞ Áfl∑§À¬Ê¥ ◊¥ ‚ ‚„Ë ©ûÊ⁄U øÈŸ¥–
(1) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
(1) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii) (2) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
(2) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) (3) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
(3) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii) (4) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

(4) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i) 78. Statement I :
Acid strength increases in the order given as
78. ∑§ÕŸ I : HF << HCl << HBr << HI.
•ê‹Ëÿ ‚Ê◊âÿ¸ ÁŒ∞ ª∞ ∑˝§◊ÊŸÈ‚Ê⁄U ’…∏ÃÊ „Ò Statement II :
HF << HCl << HBr << HI. As the size of the elements F, Cl, Br, I increases
∑§ÕŸ II : down the group, the bond strength of HF, HCl,
HBr and HI decreases and so the acid strength
¡Ò‚-¡Ò‚ ‚◊Í„ ◊¥ ŸËø ¡ÊŸ ¬⁄U Ãàfl F, Cl, Br, I ∑§Ê •Ê∑§Ê⁄U increases.
’…∏ÃÊ ¡ÊÃÊ „Ò flÒ‚-flÒ‚ HF, HCl, HBr ∞fl¥ HI ∑§ ’¥œ ∑§Ë In the light of the above statements, choose the
¬˝’‹ÃÊ ÉÊ≈UÃË ¡ÊÃË „Ò– •Ã— •ê‹Ëÿ ‚Ê◊âÿ¸ ’…∏ÃÊ ¡ÊÃÊ „Ò– correct answer from the options given below.
©¬ÿȸÄà ∑§ÕŸÊ¥ ∑§ ¬˝∑§Ê‡Ê ◊¥ ŸËø ÁŒ∞ „È∞ Áfl∑§À¬Ê¥ ◊¥ ‚ ‚„Ë (1) Both Statement I and Statement II are
©ûÊ⁄U øÈŸ¥– true.
(2) Both Statement I and Statement II are
(1) ŒÊŸÊ¥ ∑§ÕŸ I ∞fl¥ ∑§ÕŸ II ‚„Ë „Ò¥– false.
(2) ŒÊŸÊ¥ ∑§ÕŸ I ∞fl¥ ∑§ÕŸ II ª‹Ã „Ò¥– (3) Statement I is correct but Statement II
is false.
(3) ∑§ÕŸ I ‚„Ë „Ò ‹Á∑§Ÿ ∑§ÕŸ II ª‹Ã „Ò–
(4) Statement I is incorrect but
(4) ∑§ÕŸ I ª‹Ã „Ò ‹Á∑§Ÿ ∑§ÕŸ II ‚„Ë „Ò– Statement II is true.

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79. ©‚ ÿÊÒÁª∑§ ∑§Ê ¬„øÊŸ¥ ¡Ê Á„ã‚’ª¸ •Á÷∑§◊¸∑§ ∑§ ‚ÊÕ Á∑˝§ÿÊ 79. Identify the compound that will react with Hinsberg’s
∑§⁄U∑§ ∆UÊ‚ ’ŸÊ∞ªÊ ¡Ê ˇÊÊ⁄U ◊¥ ÉÊÈ‹Ÿ‡ÊË‹ „Ò — reagent to give a solid which dissolves in alkali.

(1) (1)

(2) (2)

(3) (3)

(4) (4)

80. ÁŸêŸ ⁄UÊ‚ÊÿÁŸ∑§ •Á÷Á∑˝§ÿÊ ◊¥ ÁŸÁ◊¸Ã ∑§Ê’¸ÁŸ∑§ ÿÊÒÁª∑§ ∑§Ê 80. What is the IUPAC name of the organic compound
IUPAC ŸÊ◊ ÄÿÊ „Ò? formed in the following chemical reaction ?

(i) C H MgBr , ‡ÊÈc∑§ ߸Õ⁄ (i) C2H5MgBr , dry Ether Product
Acetone 
∞‚Ë≈UÊŸ 
2 5 → ©à¬ÊŒ →
(ii) H O, H+
2
(ii) H2O, H+
(1) 2-methyl propan-2-ol
(1) 2-◊ÁÕ‹ ¬˝Ê¬
 Ÿ -2-•ÊÚ‹
(2) pentan-2-ol
(2) ¬ã≈UŸ-2-•ÊÚ‹ (3) pentan-3-ol
(3) ¬ã≈UŸ-3-•ÊÚ‹ (4) 2-methyl butan-2-ol
(4) 2-◊ÁÕ‹ éÿÍ≈UŸ-2-•ÊÚ‹
81. Which of the following reactions is the metal
81. ÁŸêŸ •Á÷Á∑˝§ÿÊ•Ê¥ ◊¥ ‚ ∑§ÊÒŸ œÊÃÈ ÁflSÕʬŸ •Á÷Á∑˝§ÿÊ „Ò? displacement reaction ? Choose the right option.
‚„Ë Áfl∑§À¬ øÈŸ¥– (1)
∆
2KClO3 → 2KCl+3O2
∆
(1) 2KClO3 → 2KCl+3O2 (2)
∆
Cr2O3+2Al → Al2O3+2Cr
∆
(2) Cr2O3+2Al → Al2O3+2Cr (3) Fe+2HCl → FeCl2+H2↑
(3) Fe+2HCl → FeCl2+H2↑ (4) 2Pb(NO3)2 → 2PbO+4NO2+O2↑
(4) 2Pb(NO3)2 → 2PbO+4NO2+O2↑
82. The correct structure of 2,6-Dimethyl-dec-4-ene
82. 2,6-«UÊß◊ÁÕ‹-«U∑§-4-߸Ÿ ∑§Ë ‚„Ë ‚¥⁄UøŸÊ „Ò — is :

(1) (1)

(2) (2)

(3) (3)

(4) (4)

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83. ÁŸêŸ ◊¥ ‚ •‚àÿ ∑§ÕŸ „Ò — 83. The incorrect statement among the following
(1) ‹ÒãÕŸÊÚÿ«U •Ê∑È¥§øŸ ∑§Ë ÃÈ‹ŸÊ ◊¥ ∞∑§ Ãàfl ‚ ŒÍ‚⁄U is :
Ãàfl ∑§Ê ∞ÁÄ≈UŸÊÚÿ«U •Ê∑È¥§øŸ •Áœ∑§ „Ò– (1) Actinoid contraction is greater for element
to element than Lanthanoid contraction.
(2) •Áœ∑§Ê¥‡Ê ÁòÊ‚¥ÿÊ¡Ë ‹ÒãÕŸÊÚÿ«U •ÊÿŸ ∆UÊ‚ •flSÕÊ ◊¥ (2) Most of the trivalent Lanthanoid ions are
⁄¥Uª„ËŸ „ÊÃ „Ò¥– colorless in the solid state.
(3) ‹ÒãÕŸÊÚÿ«U ™§c◊Ê ∞fl¥ ÁfllÈà ∑§ •ë¿U øÊ‹∑§ „ÊÃ „Ò¥– (3) Lanthanoids are good conductors of heat and
(4) ∞ÁÄ≈UŸÊÚÿ«U •àÿÁœ∑§ •Á÷Á∑˝§ÿʇÊË‹ œÊÃÈ∞° „Ò¥, Áfl‡Ê· electricity.
M§¬ ‚ ¡’ fl ‚͡◊ Áfl÷ÊÁ¡Ã „Ò¥– (4) Actinoids are highly reactive metals,
especially when finely divided.
84. RBC ∑§Ë ∑§◊Ë, „ËŸÃÊ ¡ÁŸÃ ⁄UÊª „Ò —
84. The RBC deficiency is deficiency disease of :
(1) Áfl≈UÊÁ◊Ÿ B12 ∑§Ë
(1) Vitamin B12
(2) Áfl≈UÊÁ◊Ÿ B6 ∑§Ë (2) Vitamin B6
(3) Áfl≈UÊÁ◊Ÿ B1 ∑§Ë (3) Vitamin B1
(4) Áfl≈UÊÁ◊Ÿ B2 ∑§Ë (4) Vitamin B2
85. „Êß«˛UÊ¡Ÿ ∑§Ê ∞∑§ ⁄UÁ«UÿÊ∞ÁÄ≈Ufl ‚◊SÕÊÁŸ∑§, ≈˛UÊßÁ≈Uÿ◊, ÁŸêŸ 85. Tritium, a radioactive isotope of hydrogen, emits
◊¥ ‚ Á∑§‚ ∑§áÊ ∑§Ê ©à‚¡¸Ÿ ∑§⁄UÃÊ „Ò? which of the following particles ?
(1) ’Ë≈UÊ (β−) (1) Beta (β−)
(2) •À»§Ê (α) (2) Alpha (α)
(3) Gamma (γ)
(3) ªÊ◊Ê (γ)
(4) Neutron (n)
(4) ãÿÍ≈˛UÊÚŸ (n)
Section - B (Chemistry)
•ŸÈ÷ʪ - B (⁄U‚ÊÿŸ‡ÊÊSòÊ)
86. Match List - I with List - II.
86. ‚ÍøË-I ∑§Ê Á◊‹ÊŸ ‚ÍøË-II ‚ ∑§⁄¥U– List - I List - II
‚ÍøË-I ‚ÍøË-II (a) 2SO2(g)+O2(g) → (i) Acid rain
(a) 2SO2(g)+O2(g)→ (i) •ê‹ fl·Ê¸ 2SO3(g)
2SO3(g) hν
(b) HOCl(g)  → (ii) Smog
(b) HOCl(g) 
→
hν (ii) œÍ◊-∑§Ê„⁄UÊ i i
i i OH+Cl
OH+C l (c) CaCO3+H2SO4→ (iii) Ozone
(c) CaCO3+H2SO4→ (iii) •Ê¡ÊŸ ˇÊ⁄UáÊ CaSO4+H2O+CO2 depletion
CaSO4+H2O+CO2
(d) hν
NO2(g)  (iv) Tropospheric
→
(d) hν
NO2(g)  → (iv) ˇÊÊ÷◊¥«U‹Ëÿ
NO(g)+O(g) pollution
NO(g)+O(g) ¬˝Œ·Í áÊ Choose the correct answer from the options given
ŸËø ÁŒ∞ ª∞ Áfl∑§À¬Ê¥ ◊¥ ‚ ‚„Ë ©ûÊ⁄U øÈŸ¥– below.
(1) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv) (1) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
(2) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) (2) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
(3) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii) (3) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
(4) (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i) (4) (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)

NaOH, + ?
87.
NaOH, + ?
CH3CH2COO−Na+  → CH3CH3+ 87. CH3CH2COO−Na+  → CH3CH3+
ª◊¸ Heat
Na2CO3.
Na2CO3.
Consider the above reaction and identify the
©¬ÿÈĸ à •Á÷Á∑˝§ÿÊ ◊¥ •ŸÈ¬ÁSÕà •Á÷∑§◊¸∑§/⁄U‚ÊÿŸ ∑§Ê ¬„øÊŸ–¥ missing reagent/chemical.
(1) B2H6 (1) B2H6
(2) ‹Ê‹ »§ÊS»§Ê⁄U‚ (2) Red Phosphorus
(3) CaO (3) CaO
(4) DIBAL-H (4) DIBAL-H

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88. ÁŸêŸ ⁄UÊ‚ÊÿÁŸ∑§ •Á÷Á∑˝§ÿÊ ◊¥ ÁŸÁ◊¸Ã ©à¬ÊŒ „Ò — 88. The product formed in the following chemical
reaction is :

(1)
(1)

(2)
(2)

(3)
(3)

(4)
(4)

89. ‚ÍøË-I ∑§Ê Á◊‹ÊŸ ‚ÍøË-II ‚ ∑§⁄¥U–
‚ÍøË-I ‚ÍøË-II 89. Match List - I with List - II.
List - I List - II
(a) [Fe(CN)6]3− (i) 5.92 BM
(a) [Fe(CN)6]3− (i) 5.92 BM
(b) [Fe(H2O)6]3+ (ii) 0 BM
(b) [Fe(H2O)6]3+ (ii) 0 BM
(c) [Fe(CN)6]4− (iii) 4.90 BM
(c) [Fe(CN)6]4− (iii) 4.90 BM
(d) [Fe(H2O)6]2+ (iv) 1.73 BM
(d) [Fe(H2O)6]2+ (iv) 1.73 BM
ŸËø ÁŒ∞ ª∞ Áfl∑§À¬Ê¥ ◊¥ ‚ ‚„Ë ©ûÊ⁄U øÈŸ¥– Choose the correct answer from the options given
(1) (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii) below.
(2) (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i) (1) (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
(3) (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii) (2) (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
(3) (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
(4) (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
(4) (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
90. •ÊÿŸÊ¥ ∑§ ÁŸêŸÁ‹Áπà ÿÈÇ◊Ê¥ ◊¥ ‚ ∑§ÊÒŸ ∞∑§ ‚◊ß‹Ä≈˛UÊÚÁŸ∑§
ÿÈÇ◊ Ÿ„Ë¥ „Ò? 90. From the following pairs of ions which one is not
an iso-electronic pair ?
(1) O2−, F−
(1) O2−, F−
(2) Na+, Mg2+ (2) Na+, Mg2+
(3) Mn2+, Fe3+ (3) Mn2+, Fe3+
(4) Fe2+, Mn2+ (4) Fe2+, Mn2+

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91. ⁄UÊ‚ÊÿÁŸ∑§ •Á÷Á∑˝§ÿÊ ∑§ ÁŒ∞ ∑˝§◊ ◊¥ •Á÷∑§◊¸∑§ ‘R’ „Ò — 91. The reagent ‘R’ in the given sequence of chemical
reaction is :

(1) H2O
(1) H2O
(2) CH3CH2OH
(2) CH3CH2OH
(3) HI
(3) HI
(4) CuCN/KCN
(4) CuCN/KCN
92. 1 ‹Ë≈U⁄U •Êÿß ◊¥ 08C ¬⁄U ∞∑§ Á◊üÊáÊ Á¡‚◊¥ 4 g O2 ∞fl¥
92. Choose the correct option for the total pressure
2 g H2 ‹Ë ªß¸ „Ê, ©‚∑§Ê ∑ȧ‹ ŒÊ’ (atm ◊¥) ∑§ ‚„Ë (in atm.) in a mixture of 4 g O2 and 2 g H2 confined
Áfl∑§À¬ ∑§Ê øÈŸ¥– in a total volume of one litre at 08C is :
[ÁŒÿÊ ªÿÊ „Ò — R=0.082 L atm mol−1K−1, T=273 K] [Given R=0.082 L atm mol−1K−1, T=273 K]
(1) 2.518 (1) 2.518
(2) 2.602 (2) 2.602
(3) 25.18 (3) 25.18
(4) 26.02 (4) 26.02

93. ÁŒ∞ ª∞ •áÊÈ•Ê¥ ◊¥ ‚ ∑§ÊÒŸ •œ˝ÈflËÿ ¬˝∑ΧÁà ∑§Ê „Ò? 93. Which of the following molecules is non-polar in
nature ?
(1) POCl3
(1) POCl3
(2) CH2O
(2) CH2O
(3) SbCl5
(3) SbCl5
(4) NO2 (4) NO2

94. ÁŸêŸ ◊¥ ‚ Á∑§‚ √ÿflSÕÊ ◊¥, ©Ÿ∑§ ‚Ê◊Ÿ ’ÃÊ∞ ª∞ ªÈáÊœ◊¸ ∑§ 94. In which one of the following arrangements the
•ŸÈ‚Ê⁄U, ©Áøà ∑˝§◊ Ÿ„Ë¥ ÁŒÿÊ ªÿÊ „Ò? given sequence is not strictly according to the
(1) HF < HCl : •ê‹Ëÿ ‚Ê◊âÿ¸ ∑§ properties indicated against it ?
< HBr < HI ’…∏Ã ∑˝§◊ ◊¥ (1) HF < HCl : Increasing acidic
< HBr < HI strength
(2) H2O < H2S : pKa ◊ÊŸÊ¥ ∑§ ’…∏Ã ∑˝§◊
< H2Se < H2Te ◊¥ (2) H2O < H2S : Increasing pKa
< H2Se < H2Te values
(3) NH3 < PH3 : •ê‹Ëÿ ‹ˇÊáÊ ∑§ (3) NH3 < PH3 : Increasing
< AsH3 < SbH3 ’…∏Ã ∑˝§◊ ◊¥ < AsH3 < SbH3 acidic character
(4) CO2 < SiO2 : •ÊÚÄ‚Ë∑§⁄UáÊ ˇÊ◊ÃÊ ∑§ (4) CO2 < SiO2 : Increasing
< SnO2 < PbO2 ’…∏Ã ∑˝§◊ ◊¥ < SnO2 < PbO2 oxidizing power

95. 458C ¬⁄U ∞∑§ Áfl‹ÿŸ Á¡‚◊¥ ’ã¡ËŸ ∞fl¥ •ÊÚÄ≈UŸ ∑§Ê ◊Ê‹⁄U 95. The correct option for the value of vapour pressure
•ŸÈ¬Êà 3 : 2 „Ê, ©‚∑§ flÊc¬ ŒÊ’ ∑§ ◊ÊŸ ∑§Ê ‚„Ë Áfl∑§À¬ of a solution at 458C with benzene to octane in
„Ò — molar ratio 3 : 2 is :
[458C ¬⁄U ’ã¡ËŸ ∑§Ê flÊc¬ ŒÊ’ 280 mm Hg ÃÕÊ •ÊÚÄ≈UŸ [At 458C vapour pressure of benzene is
280 mm Hg and that of octane is 420 mm Hg.
∑§Ê flÊc¬ ŒÊ’ 420 mm Hg „Ò– •ÊŒ‡Ê¸ ªÒ‚ ◊ÊŸ¥] Assume Ideal gas]
(1) 160 mm Hg (1) 160 mm of Hg
(2) 168 mm Hg (2) 168 mm of Hg
(3) 336 mm Hg (3) 336 mm of Hg
(4) 350 mm Hg (4) 350 mm of Hg

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96. 0.007 M ∞‚ËÁ≈U∑§ •ê‹ ∑§Ë ◊Ê ‹ ⁄U øÊ‹∑§ÃÊ 96. The molar conductivity of 0.007 M acetic acid is
20 S cm2 mol−1 „Ò– ∞‚ËÁ≈U∑§ •ê‹ ∑§Ê ÁflÿÊ¡Ÿ ÁSÕ⁄UÊ¥∑§ 20 S cm 2 mol −1 . What is the dissociation
constant of acetic acid ? Choose the correct option.
ÄÿÊ „Ò? ‚„Ë Áfl∑§À¬ øÈŸ¥–
 Λ
= 350 S cm2 mol −1   Λ
= 350 S cm2 mol −1 
 H+   H+ 
Λ
2 −1  Λ
2 −1 

 CH3COO−
= 50 S cm mol 
  CH COO−= 50 S cm mol 
3

(1) 1.75×10−4 mol L−1 (1) 1.75×10−4 mol L−1
(2) 2.50×10−4 mol L−1 (2) 2.50×10−4 mol L−1
(3) 1.75×10−5 mol L−1 (3) 1.75×10−5 mol L−1

(4) 2.50×10−5 mol L−1 (4) 2.50×10−5 mol L−1

97. Á∑§‚Ë ¬˝Õ◊ ∑§ÊÁ≈U ∑§Ë •Á÷Á∑˝§ÿÊ ∑§ Á‹∞ •Ê⁄U¸ÁŸÿ‚ ‚◊Ë∑§⁄UáÊ  1
97. The slope of Arrhenius Plot  ln k v/s  of first
 1  T
 ln k v/s T  ∑§ …UÊ‹ ∑§Ê ◊ÊŸ −5×10 K „Ò– •Á÷Á∑˝§ÿÊ
3 order reaction is −5×103 K. The value of Ea of
  the reaction is. Choose the correct option for your
∑§ Á‹∞ Ea ∑§Ê ◊ÊŸ „Ò– ‚„Ë Áfl∑§À¬ øÈŸ¥– answer.
[ÁŒÿÊ ªÿÊ „Ò — R=8.314 JK−1mol−1] [Given R=8.314 JK−1mol−1]
(1) 41.5 kJ mol−1 (1) 41.5 kJ mol−1
(2) 83.0 kJ mol−1 (2) 83.0 kJ mol−1
(3) 166 kJ mol−1 (3) 166 kJ mol−1
(4) −83 kJ mol−1 (4) −83 kJ mol−1

98. ‚ÍøË-I ∑§Ê Á◊‹ÊŸ ‚ÍøË-II ‚ ∑§⁄¥U– 98. Match List - I with List - II.
‚ÍøË-I ‚ÍøË-II List - I List - II

(a) (i) Hell-Volhard-
(a) (i) „‹-»§Ê‹Ê«¸-
Zelinsky reaction
¡Á‹¥S∑§Ë •Á÷Á∑˝§ÿÊ

(b) (ii) ªÊ≈U⁄U◊ÊŸ-∑§Êπ (b) (ii) Gattermann-Koch
reaction
•Á÷Á∑˝§ÿÊ
(c) R−CH2−OH (iii) „Ò‹Ê»§Ê◊¸ •Á÷Á∑˝§ÿÊ (c) R−CH2−OH (iii) Haloform
+R'COOH +R'COOH reaction
‚ÊãŒ˝ H2SO4 Conc. H SO
 → →
2 4

(d) R−CH2COOH (iv) ∞S≈U⁄UË∑§⁄UáÊ (d) R−CH2COOH (iv) Esterification
(i) X 2 /‹Ê‹ P (i) X 2 /Red P
 →  →
(ii) H2O (ii) H2O

ŸËø ÁŒ∞ ª∞ Áfl∑§À¬Ê¥ ◊¥ ‚ ‚„Ë ©ûÊ⁄U øÈŸ¥– Choose the correct answer from the options given
below.
(1) (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
(1) (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
(2) (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv) (2) (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
(3) (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii) (3) (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)
(4) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) (4) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

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99. ÁŸêŸ ⁄UÊ‚ÊÿÁŸ∑§ •Á÷Á∑˝§ÿÊ ◊¥ ◊äÿflÃ˸ ©à¬ÊŒ ‘X’ „Ò — 99. The intermediate compound ‘X’ in the following
chemical reaction is :

(1)
(1)

(2)
(2)

(3)
(3)

(4)
(4)

100. ‚◊ÃʬËÿ ¬Á⁄UÁSÕÁÃÿÊ¥ ◊¥ Á∑§‚Ë •ÊŒ‡Ê¸ ªÒ‚ ∑§ •ŸÈà∑˝§◊áÊËÿ
¬˝‚⁄UáÊ ∑§ Á‹∞ ‚„Ë Áfl∑§À¬ „Ò — 100. For irreversible expansion of an ideal gas under
(1) ∆U=0, ∆Stotal=0 isothermal condition, the correct option is :
(2) ∆U ≠ 0, ∆Stotal ≠ 0 (1) ∆U=0, ∆Stotal=0
(3) ∆U=0, ∆Stotal ≠ 0 (2) ∆U ≠ 0, ∆Stotal ≠ 0
(4) ∆U ≠ 0, ∆Stotal=0 (3) ∆U=0, ∆Stotal ≠ 0
(4) ∆U ≠ 0, ∆Stotal=0
•ŸÈ÷ʪ - A (¡ËflÁflôÊÊŸ — flŸS¬ÁÃÁflôÊÊŸ)
Section - A (Biology : Botany)
101. ‚◊Ë∑§⁄UáÊ GPP−R=NPP ◊¥ R Á∑§‚ ÁŸM§Á¬Ã ∑§⁄UÃÊ „Ò?
(1) ÁflÁ∑§⁄UáÊ ™§¡Ê¸ 101. In the equation GPP−R=NPP
(2) ◊¥ŒŸ ∑§Ê⁄U∑§ R represents :
(3) ¬ÿʸfl⁄UáÊ ∑§Ê⁄U∑§ (1) Radiant energy
(2) Retardation factor
(4) ‡fl‚Ÿ „ÊÁŸ
(3) Environment factor
102. ¬˝∑ΧÁà ◊¥ •¥Ã⁄UÊ¡ÊÃËÿ ¬˝ÁÃS¬œÊ¸ ∑§ ’Êfl¡ÍŒ, ¬˝ÁÃS¬œÊ¸ ∑§⁄UŸ (4) Respiration losses
flÊ‹Ë ¡ÊÁÃÿÊ¥ Ÿ •¬ŸË ©ûÊ⁄U¡ËÁflÃÊ ∑§ Á‹∞ ∑§ÊÒŸ ‚Ë ÁflÁœ ∑§Ê
Áfl∑§Ê‚ Á∑§ÿÊ „ÊªÊ? 102. Inspite of interspecific competition in nature,
(1) ‚¥‚ÊœŸ Áfl÷Ê¡Ÿ which mechanism the competing species might
have evolved for their survival ?
(2) S¬œÊ¸ ÁŸ◊ȸÄÃ
(1) Resource partitioning
(3) ‚„Ê¬∑§ÊÁ⁄UÃÊ
(2) Competitive release
(4) ÷ˇÊáÊ
(3) Mutualism
103. •ãÃ⁄U¡ÊÃËÿ ¬⁄U¡ËÁflÃÊ ∑§Ê Á∑§‚ ¬˝∑§Ê⁄U ÁŸM§Á¬Ã Á∑§ÿÊ ¡Ê (4) Predation
‚∑§ÃÊ „Ò?
103. Amensalism can be represented as :
(1) ¡ÊÁÃ A (−) ; ¡ÊÁÃ B (0) (1) Species A (−) ; Species B (0)
(2) ¡ÊÁÃ A (+) ; ¡ÊÁÃ B (+) (2) Species A (+) ; Species B (+)
(3) ¡ÊÁÃ A (−) ; ¡ÊÁÃ B (−) (3) Species A (−) ; Species B (−)
(4) ¡ÊÁÃ A (+) ; ¡ÊÁÃ B (0) (4) Species A (+) ; Species B (0)

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104. ‚ÍøË - I ∑§Ê ‚ÍøË - II ∑§ ‚ÊÕ ‚È◊Á‹Ã ∑§ËÁ¡∞ — 104. Match List - I with List - II.

‚ÍøË - I ‚ÍøË - II List - I List - II
Cells with active cell Vascular
‚Á∑˝§ÿ ∑§ÊÁ‡Ê∑§Ê Áfl÷Ê¡Ÿ (a) (i)
(a) (i) ‚¥fl„Ÿ ™§Ã∑§ division capacity tissues
∑§Ë ˇÊ◊ÃÊ flÊ‹Ë ∑§ÊÁ‡Ê∑§Êÿ¥ Tissue having all cells
Meristematic
∞∑§ ™§Ã∑§ Á¡‚◊¥ ‚÷Ë (b) similar in structure (ii)
tissue
(b) ∑§ÊÁ‡Ê∑§Êÿ¥ ‚¥⁄UøŸÊ •ÊÒ⁄U (ii) Áfl÷íÿÊÃ∑§ and function
∑§Êÿ¸ ∑§Ë ŒÎÁc≈U ‚ ‚◊ÊŸ „Ò¥ Tissue having
(c) (iii) Sclereids
different types of cells
ÁflÁ÷㟠¬˝∑§Ê⁄U ∑§Ë
(c) (iii) ÁS∑§Á‹Á⁄U«U Dead cells with highly
∑§ÊÁ‡Ê∑§Ê•Ê¥ flÊ‹Ê ™§Ã∑§ (d) thickened walls and (iv) Simple tissue
•àÿÁœ∑§ ◊Ê≈UË Á÷ÁûÊ ∞fl¥ narrow lumen
(d) ‚¥∑§⁄UË ªÈÁ„∑§Ê flÊ‹Ë ◊Îà (iv) ‚⁄U‹ ™§Ã∑§ Select the correct answer from the options given
∑§ÊÁ‡Ê∑§Êÿ¥ below.
ŸËø ÁŒÿ ªÿ Áfl∑§À¬Ê¥ ◊¥ ‚ ‚„Ë ©ûÊ⁄U øÈÁŸ∞– (a) (b) (c) (d)
(a) (b) (c) (d) (1) (ii) (iv) (i) (iii)
(1) (ii) (iv) (i) (iii) (2) (iv) (iii) (ii) (i)
(2) (iv) (iii) (ii) (i) (3) (i) (ii) (iii) (iv)
(3) (i) (ii) (iii) (iv) (4) (iii) (ii) (iv) (i)
(4) (iii) (ii) (iv) (i)
105. The production of gametes by the parents,
105. ¡Ÿ∑§Ê¥ mÊ⁄UÊ ÿÈÇ◊∑§Ê¥ ∑§Ê ©à¬ÊŒŸ, ÿÈÇ◊¡ ∑§Ê ÁŸ◊ʸáÊ •ÊÒ⁄U formation of zygotes, the F1 and F2 plants, can be
F1 •ÊÒ⁄U F2 ¬ÊŒ¬Ê¥ ∑§Ê ÁŸêŸÁ‹Áπà ◊¥ ‚ Á∑§‚ ÁøòÊ mÊ⁄UÊ understood from a diagram called :
‚◊¤ÊÊ ¡Ê ‚∑§ÃÊ „Ò? (1) Bullet square
(1) ’È‹≈U flª¸ (2) Punch square
(2) ¬¥ø flª¸ (3) Punnett square
(3) ¬Ÿ≈U flª¸ (4) Net square
(4) Ÿ≈U flª¸
106. ¡‹ ¬⁄U ⁄Uπ, ∞ÁÕÁ«Uÿ◊ ’˝Ê◊Êß«U ‚ •Á÷⁄¥UÁ¡Ã «UË.∞Ÿ.∞. 106. DNA strands on a gel stained with ethidium
bromide when viewed under UV radiation, appear
⁄Uí¡È∑§Ê¥ ∑§Ê ¡’ ÿÈ.flË. ÁflÁ∑§⁄UáÊ ∑§ •ãê¸Ã ŒπÊ ¡ÊÃÊ „Ò Ã’
as :
fl ∑Ò§‚ ÁŒπÃ „Ò¥?
(1) ¬Ë‹Ë ¬Á^ÿÊ¥ (1) Yellow bands
(2) ø◊∑§Ë‹Ë ŸÊ⁄¥UªË ¬Á^ÿÊ¥ (2) Bright orange bands
(3) ª„⁄UË ‹Ê‹ ¬Á^ÿÊ¥ (3) Dark red bands
(4) ø◊∑§Ë‹Ë ŸË‹Ë ¬Á^ÿÊ¥ (4) Bright blue bands

107. ¡◊Ë Á∑§‚◊¥ ¬Êÿ ¡ÊÃ „Ò¥? 107. Gemmae are present in :
(1) ◊ÊÚ‚ ◊¥ (1) Mosses
(2) ≈UÁ⁄U«UÊ»§Êß≈U ◊¥ (2) Pteridophytes
(3) ∑ȧ¿U •ŸÊflÎÃ’Ë¡ËÿÊ¥ ◊¥ (3) Some Gymnosperms
(4) ∑ȧ¿U Á‹fl⁄Ufl≈¸U ◊¥ (4) Some Liverworts
108. ¬ÿʸfl⁄UáÊ ∑§ ¬˝àÿÈûÊ⁄U ◊¥ ¬ÊŒ¬ ÁflÁ÷㟠¬ÕÊ¥ ∑§Ê •ŸÈ‚⁄UáÊ ∑§⁄UÃ „Ò¥
108. Plants follow different pathways in response to
ÿÊ ÁflÁ÷㟠¬˝∑§Ê⁄U ∑§Ë ‚¥⁄UøŸÊ•Ê¥ ∑§ ÁŸ◊ʸáÊ ∑§ Á‹∞ ÁflÁ÷㟠environment or phases of life to form different
•flSÕÊ•Ê¥ ∑§Ê •ŸÈ‚⁄UáÊ ∑§⁄UÃ „Ò¥– ß‚ ˇÊ◊ÃÊ ∑§Ê ÄÿÊ ∑§„Ê kinds of structures. This ability is called :
¡ÊÃÊ „Ò?
(1) Elasticity
(1) ¬˝àÿÊSÕÃÊ
(2) Flexibility
(2) ŸêÿÃÊ
(3) ‚ÈÉÊ≈KÃÊ (3) Plasticity
(4) ¬Á⁄U¬ÄflÃÊ (4) Maturity

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109. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ª‹Ã „Ò? 109. Which of the following is an incorrect
(1) ¬Á⁄U¬Äfl øÊ‹ŸË ŸÁ‹∑§Ê ÃàflÊ¥ ◊¥ ∞∑§ S¬c≈U ∑§ãŒ˝∑§ statement ?
•ÊÒ⁄U ‚ÊœÊ⁄UáÊ ∑§ÊÁ‡Ê∑§ÊŒ˝√ÿËÿ ©¬Ê¥ª „ÊÃ „Ò¥– (1) Mature sieve tube elements possess a
(2) ‚͡◊∑§Êÿ, ¬ÊŒ¬Ê¥ •ÊÒ⁄U ¡ãÃÈ•Ê¥ ŒÊŸÊ¥ ◊¥ ¬ÊÿË ¡ÊÃË „Ò¥– conspicuous nucleus and usual cytoplasmic
(3) ¬Á⁄U∑§ãŒ˝∑§Ë •fl∑§Ê‡Ê, ∑§ãŒ˝∑§ ∑§ •ãŒ⁄U ©¬ÁSÕà ¬ŒÊÕÊZ organelles.
•ÊÒ⁄U ∑§ÊÁ‡Ê∑§ÊŒ˝√ÿ ◊¥ ©¬ÁSÕà ¬ŒÊÕÊZ ∑§ ’Ëø •fl⁄UÊœ (2) Microbodies are present both in plant and
∑§Ê ∑§Ê◊ ∑§⁄UÃÊ „Ò– animal cells.
(4) ∑§ãŒ˝∑§ Á¿UŒ˝, ∑§ãŒ˝∑§ •ÊÒ⁄U ∑§ÊÁ‡Ê∑§ÊŒ˝√ÿ ∑§ ’Ëø ŒÊŸÊ¥ (3) The perinuclear space forms a barrier
between the materials present inside the
ÁŒ‡ÊÊ•Ê¥ ◊¥ ¬˝Ê≈UËŸ •ÊÒ⁄U •Ê⁄U.∞Ÿ.∞. •áÊÈ•Ê¥ ∑§ Á‹∞,
nucleus and that of the cytoplasm.
∞∑§ ¬Õ ∑§Ë ÷Ê¥Áà ∑§Êÿ¸ ∑§⁄UÃ „¥Ò–
(4) Nuclear pores act as passages for proteins
110. ¬Á⁄U¬Äfl •flSÕÊ ◊¥ ∞∑§ ¬˝ÊM§¬Ë •ÊflÎÃ’Ë¡Ë ÷˝ÍáÊ∑§Ê· and RNA molecules in both directions
ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê „ÊÃÊ „Ò? between nucleus and cytoplasm.
(1) 8-∑§ãŒ˝∑§Ëÿ •ÊÒ⁄U 7-∑§ÊÁ‡Ê∑§Ëÿ
110. A typical angiosperm embryo sac at maturity is :
(2) 7-∑§ãŒ˝∑§Ëÿ •ÊÒ⁄U 8-∑§ÊÁ‡Ê∑§Ëÿ
(1) 8-nucleate and 7-celled
(3) 7-∑§ãŒ˝∑§Ëÿ •ÊÒ⁄U 7-∑§ÊÁ‡Ê∑§Ëÿ
(2) 7-nucleate and 8-celled
(4) 8-∑§ãŒ˝∑§Ëÿ •ÊÒ⁄U 8-∑§ÊÁ‡Ê∑§Ëÿ
(3) 7-nucleate and 7-celled
111. ∞∑§ πà ◊¥ ÉÊÊ‚¬Êà ∑§Ê ‚◊Êåà ∑§⁄UŸ ∑§ Á‹∞ ∑§ÊÒŸ ‚Ê ¬ÊŒ¬ (4) 8-nucleate and 8-celled
„Ê⁄U◊ÊŸ ©¬ÿÊª ◊¥ ‹ÊÿÊ ¡ÊÃÊ „Ò?
(1) •Êß.∞.∞. 111. The plant hormone used to destroy weeds in a field
(2) ∞Ÿ.∞.∞. is :
(3) 2, 4-«UË (1) IAA
(4) •Ê߸.’Ë.∞. (2) NAA
(3) 2, 4-D
112. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚ ‡ÊÒflÊ‹ ∑Ò§⁄UÊ¡ËŸ ©à¬ãŸ ∑§⁄UÃ „Ò¥? (4) IBA
(1) „Á⁄Uà ‡ÊÒflÊ‹
(2) ÷Í⁄U ‡ÊÒflÊ‹ 112. Which of the following algae produce Carrageen ?
(3) ‹Ê‹ ‡ÊÒflÊ‹ (1) Green algae
(4) ŸË‹-„Á⁄Uà ‡ÊÒflÊ‹ (2) Brown algae
(3) Red algae
113. ¬ÊŒ¬ ∑§ÊÁ‡Ê∑§Ê•Ê¥ ◊¥ Á∑§‚∑§ mÊ⁄UÊ ©à¬Á⁄UfløŸ ¬˝Á⁄Uà Á∑§ÿÊ ¡Ê (4) Blue-green algae
‚∑§ÃÊ „Ò?
(1) ∑§ÊߟÁ≈UŸ 113. Mutations in plant cells can be induced by :
(2) •fl⁄UÄà Á∑§⁄UáÊ¥ (1) Kinetin
(3) ªÊ◊Ê Á∑§⁄UáÊ¥ (2) Infrared rays
(4) Á¡∞Á≈UŸ (3) Gamma rays
(4) Zeatin
114. ‚ÍøË - I ∑§Ê ‚ÍøË - II ∑§ ‚ÊÕ ‚È◊Á‹Ã ∑§ËÁ¡∞–
114. Match List - I with List - II.
‚ÍøË - I ‚ÍøË - II
List - I List - II
(a) ¡ËflŒ˝√ÿ ‚¥‹ ÿŸ (i) ¬Íáʸ‡ÊÄÃÃÊ (a) Protoplast fusion (i) Totipotency
(b) ¬ÊŒ¬ ™§Ã∑§ ‚¥flœ¸Ÿ (ii) ¬Ê◊≈UÊ (b) Plant tissue culture (ii) Pomato
(c) ◊Á⁄US ≈U◊ ‚¥flœ¸Ÿ (iii) ‚Ê◊ÊÄ‹ÊŸ (c) Meristem culture (iii) Somaclones
(d) Micropropagation (iv) Virus free plants
(d) ‚͡◊¬˝flœ¸Ÿ (iv) Áfl·ÊáÊÈ ◊ÈÄà ¬ÊŒ¬
Choose the correct answer from the options given
ŸËø ÁŒÿ ªÿ Áfl∑§À¬Ê¥ ◊¥ ‚ ‚„Ë ©ûÊ⁄U øÈÁŸ∞– below.
(a) (b) (c) (d) (a) (b) (c) (d)
(1) (iii) (iv) (ii) (i) (1) (iii) (iv) (ii) (i)
(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)

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115. ÁŸêŸÁ‹Áπà ◊¥ ‚ •œ¸‚ÍòÊË Áfl÷Ê¡Ÿ ∑§Ë Á∑§‚ •flSÕÊ ◊¥ 115. Which of the following stages of meiosis involves
ªÈáÊ‚ÍòÊÁ’ãŒÈ ∑§Ê Áfl÷Ê¡Ÿ „ÊÃÊ „Ò? division of centromere ?
(1) ◊äÿÊflSÕÊ - I (1) Metaphase I
(2) ◊äÿÊflSÕÊ - II (2) Metaphase II

(3) ¬‡øÊflSÕÊ - II (3) Anaphase II

(4) •¥àÿÊflSÕÊ - II (4) Telophase II

116. ‚ÍøË - I ∑§Ê ‚ÍøË - II ∑§ ‚ÊÕ ‚È◊Á‹Ã ∑§ËÁ¡∞– 116. Match List - I with List - II.
List - I List - II
‚ÍøË - I ‚ÍøË - II
(a) Lenticels (i) Phellogen
(a) flÊÃ⁄¥Uœ ˝ (i) ∑§Êª¡Ÿ (b) Cork cambium (ii) Suberin deposition
(b) ∑§Ê∑¸§ ∑Ò¥§Á’ÿ◊ (ii) ‚È’Á⁄UŸ ÁŸˇÊ¬áÊ (c) Secondary cortex (iii) Exchange of gases
(d) Cork (iv) Phelloderm
(c) ÁmÃËÿ∑§ flÀ∑ȧ≈U (iii) ªÒ‚Ê¥ ∑§Ê •ÊŒÊŸ-¬˝ŒÊŸ
Choose the correct answer from the options given
(d) ∑§Ê∑¸§ (iv) ∑§Êª-•SÃ⁄U below.
ŸËø ÁŒÿ ªÿ Áfl∑§À¬Ê¥ ◊¥ ‚ ‚„Ë ©ûÊ⁄U øÈÁŸ∞– (a) (b) (c) (d)
(a) (b) (c) (d) (1) (iv) (i) (iii) (ii)
(1) (iv) (i) (iii) (ii) (2) (iii) (i) (iv) (ii)
(2) (iii) (i) (iv) (ii) (3) (ii) (iii) (iv) (i)
(3) (ii) (iii) (iv) (i) (4) (iv) (ii) (i) (iii)
(4) (iv) (ii) (i) (iii) 117. The site of perception of light in plants during
photoperiodism is :
117. ¬˝∑§Ê‡Ê∑§ÊÁ‹ÃÊ ∑§ ŒÊÒ⁄UÊŸ ¬ÊŒ¬Ê¥ ◊¥ ¬˝∑§Ê‡Ê ∑§ •flª◊ ∑§Ê SÕÊŸ
∑§ÊÒŸ ‚Ê „Ò? (1) Shoot apex
(1) ¬˝⁄UÊ„ ‡ÊË·¸ (2) Stem
(2) ÃŸÊ (3) Axillary bud
(3) ∑§ˇÊËÿ ∑§Á‹∑§Ê (4) Leaf
(4) ¬ûÊË
118. Which of the following plants is monoecious ?
118. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê ¬ÊŒ¬ ©÷ÿÁ‹¥ªÊüÊÿË „Ò? (1) Carica papaya
(1) ∑Ò§Á⁄U∑§Ê ¬¬ÊÿÊ (2) Chara
(2) ∑§Ê⁄UÊ (3) Marchantia polymorpha
(3) ◊Ê∑Z§Á‡ÊÿÊ ¬ÊÁ‹◊Ê»§Ê¸ (4) Cycas circinalis
(4) ‚Êß∑§‚ ‚Á‚¸ŸÁ‹‚
119. When the centromere is situated in the middle of
119. ¡’ ªÈáÊ‚ÍòÊÁ’ãŒÈ, ªÈáÊ‚ÍòÊ ∑§Ë ŒÊ ’⁄UÊ’⁄U ÷È¡Ê•Ê¥ ∑§ ◊äÿ ◊¥ two equal arms of chromosomes, the chromosome
ÁSÕà „ÊÃÊ „Ò Ã’ ÿ„ ÄÿÊ ∑§„‹ÊÃÊ „Ò? is referred as :
(1) ◊äÿ∑§ãŒ˝Ë (1) Metacentric
(2) •¥Ã∑§ãŒ˝Ë (2) Telocentric
(3) ©¬◊äÿ∑§ãŒ˝Ë (3) Sub-metacentric
(4) •ª˝Á’ãŒÈ∑§ (4) Acrocentric

120. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ∞∑§, ¬Ë.‚Ë.•Ê⁄U.(¬ÊÚÁ‹◊⁄U¡ o¥Îπ‹Ê 120. Which of the following is not an application of PCR
•Á÷Á∑˝§ÿÊ) ∑§Ê ∞∑§ •ŸÈ¬˝ÿÊª Ÿ„Ë¥ „Ò? (Polymerase Chain Reaction) ?
(1) •ÊáÊÁfl∑§ ÁŸŒÊŸ (1) Molecular diagnosis
(2) ¡ËŸ ¬˝flœ¸Ÿ (2) Gene amplification
(3) ¬ÎÕ∑§ Á∑§ÿ ªÿ ¬˝Ê≈UËŸ ∑§Ê ‡ÊÈhË∑§⁄UáÊ (3) Purification of isolated protein
(4) ¡ËŸ ©à¬Á⁄UfløŸ ∑§Ê ¬ÃÊ ‹ªÊŸÊ (4) Detection of gene mutation

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121. ∑ȧ¿U fl¥‡Ê ¡Ò‚ Á∑§ Á‚‹Á¡ŸÒ‹Ê •ÊÒ⁄ ‚ÊÁÀflÁŸÿÊ, ŒÊ ¬˝∑§Ê⁄U ∑§ 121. Genera like Selaginella and Salvinia produce two
’Ë¡ÊáÊÈ ©à¬ÊÁŒÃ ∑§⁄UÃ „Ò¥– ∞‚ ¬ÊŒ¬Ê¥ ∑§Ê ÄÿÊ ∑§„Ê ¡ÊÃÊ „Ò? kinds of spores. Such plants are known as :
(1) „Ê◊Ê‚Ê⁄U‚ (1) Homosorus

(2) „≈U⁄UÊ‚Ê⁄U‚ (2) Heterosorus

(3) ‚◊’Ë¡ÊáÊÈ∑§ (3) Homosporous
(4) Heterosporous
(4) Áfl·◊’Ë¡ÊáÊÈ∑§
122. ◊Í‹ Á‚hÊãà (‚ã≈˛U‹ «UÊÇ◊Ê) ∑§Ê ¬Íáʸ ¬˝flÊ„ ÁøòÊ „Ò — 122. Complete the flow chart on central dogma.

(1) (a)-¬˝ÁÃ∑ΧÁÃ∑§⁄UáÊ; (b)-•ŸÈ‹πŸ; (1) (a)-Replication; (b)-Transcription;
(c)-¬Ê⁄U∑§˝ ◊áÊ; (d)-¬˝Ê≈ UËŸ (c)-Transduction; (d)-Protein
(2) (a)-M§¬ÊãÃ⁄UáÊ; (b)-¬˝ÁÃ∑ΧÁÃ∑§⁄UáÊ; (2) (a)-Translation; (b)-Replication;
(c)-•ŸÈ‹π Ÿ; (d)-¬Ê⁄U∑§˝ ◊áÊ (c)-Transcription; (d)-Transduction
(3) (a)-¬˝ÁÃ∑ΧÁÃ∑§⁄UáÊ; (b)-•ŸÈ‹π Ÿ; (3) (a)-Replication; (b)-Transcription;
(c)-M§¬ÊãÃ⁄UáÊ; (d)-¬˝Ê≈ UËŸ (c)-Translation; (d)-Protein
(4) (a)-¬Ê⁄U∑§˝ ◊áÊ; (b)-M§¬ÊãÃ⁄UáÊ; (4) (a)-Transduction; (b)-Translation;
(c)-¬˝ÁÃ∑ΧÁÃ∑§⁄UáÊ; (d)-¬˝Ê≈ UËŸ (c)-Replication; (d)-Protein

123. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ‚„Ë Ÿ„Ë¥ „Ò? 123. Which of the following statements is not correct ?
(1) ‚◊ÈŒ˝ ◊¥ ¡Ëfl÷Ê⁄U ∑§Ê Á¬⁄UÁ◊«U ‚ÊœÊ⁄UáÊÃÿÊ ©À≈UÊ „ÊÃÊ (1) Pyramid of biomass in sea is generally
inverted.
„Ò–
(2) Pyramid of biomass in sea is generally
(2) ‚◊ÈŒ˝ ◊¥ ¡Ëfl÷Ê⁄U ∑§Ê Á¬⁄UÁ◊«U ‚ÊœÊ⁄UáÊÃÿÊ ‚ËœÊ „ÊÃÊ upright.
„Ò–
(3) Pyramid of energy is always upright.
(3) ™§¡Ê¸ ∑§Ê Á¬⁄UÁ◊«U ‚ŒÒfl ‚ËœÊ „ÊÃÊ „Ò–
(4) Pyramid of numbers in a grassland
(4) ∞∑§ ÉÊÊ‚ ÷ÍÁ◊ ¬Á⁄UÃãòÊ ◊¥ ‚¥ÅÿÊ ∑§Ê Á¬⁄UÁ◊«U ‚ËœÊ ecosystem is upright.
„ÊÃÊ „Ò–
124. Which of the following algae contains mannitol as
124. ÁŸêŸÁ‹Áπà ◊¥ ‚ Á∑§‚ ‡ÊÒflÊ‹ ◊¥ ‚¥øÿË πÊl ∑§ M§¬ ◊¥ reserve food material ?
◊ÒŸË≈UÊ‹ „ÊÃÊ „Ò? (1) Ectocarpus
(1) ∞Ä≈UÊ∑ §Ê¬¸‚ (2) Gracilaria
(2) ª˝ÊÁ‚‹Á⁄UÿÊ (3) Volvox
(3) flÊÚÀflÊÚÄ‚ (4) Ulothrix
(4) ÿÍ‹ÊÁÕ˝Ä‚
125. During the purification process for recombinant
125. ¬ÈŸÿÊ¸ª¡ «UË.∞Ÿ.∞. ¬˝ÊÒlÊÁª∑§Ë ∑§ Á‹∞ ‡ÊÈhË∑§⁄UáÊ ¬˝Á∑˝§ÿÊ ◊¥ DNA technology, addition of chilled ethanol
‡ÊËË ßÕŸÊÚ‹ ∑§Ê Á◊‹ÊŸ ‚ ÿ„ Á∑§‚ •flˇÊÁ¬Ã ∑§⁄UÃÊ „Ò? precipitates out :
(1) •Ê⁄U.∞Ÿ.∞. (1) RNA
(2) «UË.∞Ÿ.∞. (2) DNA
(3) Á„S≈UÊŸ (3) Histones
(4) ¬ÊÚÁ‹‚Ò∑§⁄UÊß«U (4) Polysaccharides

126. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§Ê⁄U∑§ ∞∑§ ‚◊Ác≈U »§Ê©¥«U⁄U ¬˝÷Êfl 126. The factor that leads to Founder effect in a
©àåÊ㟠∑§⁄UÃÊ „Ò? population is :
(1) ¬˝Ê∑ΧÁÃ∑§ øÿŸ (1) Natural selection
(2) •ÊŸÈfl¥Á‡Ê∑§ ¬ÈŸÿÊ¸ª¡Ÿ (2) Genetic recombination
(3) ©à¬Á⁄UfløŸ (3) Mutation
(4) •ÊŸÈfl¥Á‡Ê∑§ Áflø‹Ÿ (4) Genetic drift

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127. ‚ÍøË- I ∑§Ê ‚ÍøË - II ∑§ ‚ÊÕ ‚È◊Á‹Ã ∑§ËÁ¡∞– 127. Match List - I with List - II.
List - I List - II
‚ÍøË- I ‚ÍøË- II
More attraction in
Œ˝fl •flSÕÊ ◊¥ •¬ˇÊÊ∑Χà (a) Cohesion (i)
(a) ‚◊¥¡Ÿ (i) liquid phase
•Áœ∑§ •Ê∑§·¸áÊ Mutual attraction
¡‹ •áÊÈ•Ê¥ ∑§ ’Ëø (b) Adhesion (ii) among water
(b) •Ê‚¥¡Ÿ (ii)
¬Ê⁄US¬Á⁄U∑§ •Ê∑§·¸áÊ molecules
Water loss in liquid
(c) ¬Îc∆U ßÊfl (iii) Œ˝fl •flSÕÊ ◊¥ ¡‹ „ÊÁŸ (c) Surface tension (iii)
phase
œÈ̋flËÿ ¬Îc∆UÊ¥ ∑§Ë •Ê⁄U Attraction towards
(d) Á’ãŒÈ dÊfl (iv) (d) Guttation (iv)
•Ê∑§·¸áÊ polar surfaces
ŸËø ÁŒÿ ªÿ Áfl∑§À¬Ê¥ ◊¥ ‚ ‚„Ë ©ûÊ⁄U øÈÁŸ∞– Choose the correct answer from the options given
below.
(a) (b) (c) (d)
(a) (b) (c) (d)
(1) (ii) (iv) (i) (iii)
(1) (ii) (iv) (i) (iii)
(2) (iv) (iii) (ii) (i)
(2) (iv) (iii) (ii) (i)
(3) (iii) (i) (iv) (ii)
(3) (iii) (i) (iv) (ii)
(4) (ii) (i) (iv) (iii)
(4) (ii) (i) (iv) (iii)
128. Ám‚¥œË ¬È¥∑§‚⁄U Á∑§‚◊¥ ¬Êÿ ¡ÊÃ „Ò¥?
128. Diadelphous stamens are found in :
(1) øÊßŸÊ ⁄UÊ¡
(1) China rose
(2) ŸË¥’Í
(2) Citrus
(3) ◊≈U⁄U (3) Pea
(4) øÊßŸÊ ⁄UÊ¡ •ÊÒ⁄U ŸË¥’Í (4) China rose and citrus
129. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ, ¬ÊŒ¬Ê¥ ◊¥ ÁmÃËÿ∑§ ©¬Ê¬øÿ¡ Ÿ„Ë¥ 129. Which of the following are not secondary
„Ò¥? metabolites in plants ?
(1) ◊Ê»§Ë¸Ÿ, ∑§Ê«UËŸ (1) Morphine, codeine
(2) ∞◊ËŸÊ •ê‹, Ç‹Í∑§Ê¡ (2) Amino acids, glucose
(3) ÁflŸé‹S≈UËŸ, ∑§⁄UÄÿÍÁ◊Ÿ (3) Vinblastin, curcumin
(4) ⁄U’⁄U, ªÊ¥Œ (4) Rubber, gums

130. ‚ÍøË - I ∑§Ê ‚ÍøË - II ‚ ‚È◊Á‹Ã ∑§ËÁ¡∞– 130. Match List - I with List - II.

‚ÍøË - I ‚ÍøË - II List - I List - II
Primary constriction in
ªÈáÊ‚ÍòÊ ◊¥ ¬˝ÊÕÁ◊∑§ (a) Cristae (i)
(a) Á∑˝§S≈UË (i) chromosome
‚¥∑§ËáʸŸ Disc-shaped sacs in
ªÊÚÀ¡Ë ©¬∑§⁄UáÊ ◊¥ Á«US∑§- (b) Thylakoids (ii)
(b) ÕÊÿ‹∑§ÊÚÿ«U (ii) Golgi apparatus
•Ê∑§Ê⁄U ∑§Ë ÕÒÁ‹ÿÊ¥ (c) Centromere (iii)
Infoldings in
(c) ªÈáÊ‚ÍòÊÁ’¥ŒÈ (iii) ‚ÍòÊ∑§ÁáÊ∑§Ê ◊¥ •¥Ãfl¸Á‹Ã mitochondria
Flattened membranous
‹fl∑§ ∑§Ë ¬ËÁ∆U∑§Ê ◊¥ (d) Cisternae (iv) sacs in stroma of
(d) Á‚S≈UŸË¸ (iv)
ø¬≈UË Á¤ÊÀ‹ËŸÍ◊Ê ÕÒÁ‹ÿÊ¥ plastids
ŸËø ÁŒÿ ªÿ Áfl∑§À¬Ê¥ ◊¥ ‚ ‚„Ë ©ûÊ⁄U øÈÁŸ∞– Choose the correct answer from the options given
below.
(a) (b) (c) (d)
(a) (b) (c) (d)
(1) (iv) (iii) (ii) (i)
(1) (iv) (iii) (ii) (i)
(2) (i) (iv) (iii) (ii)
(2) (i) (iv) (iii) (ii)
(3) (iii) (iv) (i) (ii)
(3) (iii) (iv) (i) (ii)
(4) (ii) (iii) (iv) (i)
(4) (ii) (iii) (iv) (i)

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131. ⁄UÊª ∑§Ê ∆UË∑§ ∑§⁄UŸ ∑§ Á‹∞, ¡ËŸ ¬˝flœ¸Ÿ ∑§⁄UÃ „È∞ ∞∑§ ¡ËŸ 131. When gene targetting involving gene amplification
∑§Ê ‹ˇÿ Á∑§ÿÊ ¡ÊÃÊ „Ò Ã’ ÿ„ ¬˝Á∑˝§ÿÊ ÄÿÊ ∑§„‹ÊÃË „Ò? is attempted in an individual’s tissue to treat
disease, it is known as :
(1) ’ÊÿÊ¬Êß⁄U‚Ë
(1) Biopiracy
(2) ¡ËŸ ÁøÁ∑§à‚Ê
(2) Gene therapy
(3) •ÊáÊÁfl∑§ ÁŸŒÊŸ
(3) Molecular diagnosis
(4) ‚È⁄UˇÊÊ ¬⁄UˡÊáÊ
(4) Safety testing

132. ∞∑§ ¬ÊŒ¬ ∑§ ¬⁄Uʪ∑§Ê‡Ê ‚ ¬⁄Uʪ∑§áÊÊ¥ ∑§, ∞∑§ Á÷㟠¬ÊŒ¬ ∑§
flÁø∑§Êª˝ ¬⁄U SÕÊŸÊãÃ⁄UáÊ ∑§ Á‹∞ ∑§ÊÒŸ ‚Ê ‡ÊéŒ ¬˝ÿÈÄà „ÊÃÊ „Ò 132. The term used for transfer of pollen grains from
Á¡‚◊¥ ¬⁄UʪáÊ ∑§ ŒÊÒ⁄UÊŸ flÁø∑§Êª˝ ¬⁄U •ÊŸÈfl¥Á‡Ê∑§ M§¬ ◊¥ Á÷㟠anthers of one plant to stigma of a different plant
¬˝∑§Ê⁄U ∑§ ¬⁄Uʪ∑§áÊ ‹Êÿ ¡ÊÃ „Ò¥? which, during pollination, brings genetically
different types of pollen grains to stigma, is :
(1) ¬⁄UÁŸ·øŸ
(1) Xenogamy
(2) ‚¡ÊìÈc¬Ë ¬⁄UʪáÊ (2) Geitonogamy
(3) ©ã◊Ë‹ ¬⁄UʪáÊË (3) Chasmogamy
(4) •ŸÈã◊ËÀÿ ¬⁄UʪáÊË (4) Cleistogamy

133. ÁŸêŸÁ‹Áπà ◊¥ ‚, ¬ÊÚÁ‹◊⁄U¡ o¥ÎÅÊ‹Ê •Á÷Á∑˝§ÿÊ (¬Ë.‚Ë.•Ê⁄U.) 133. Which of the following is a correct sequence of
∑§ ø⁄UáÊÊ¥ ∑§Ê ∑§ÊÒŸ ‚Ê ‚„Ë •ŸÈ∑˝§◊ „Ò? steps in a PCR (Polymerase Chain Reaction) ?
(1) ÁŸÁc∑˝§ÿ∑§⁄UáÊ, ÃʬʟȇÊË‹Ÿ, ¬˝‚Ê⁄U (1) Denaturation, Annealing, Extension
(2) ÁŸÁc∑˝§ÿ∑§⁄UáÊ, ¬˝‚Ê⁄U, ÃʬʟȇÊË‹Ÿ (2) Denaturation, Extension, Annealing
(3) ¬˝‚Ê⁄U, ÁŸÁc∑˝§ÿ∑§⁄UáÊ, ÃʬʟȇÊË‹Ÿ (3) Extension, Denaturation, Annealing
(4) ÃʬʟȇÊË‹Ÿ, ÁŸÁc∑˝§ÿ∑§⁄UáÊ, ¬˝‚Ê⁄U (4) Annealing, Denaturation, Extension

134. ∑§Ê’¸Ÿ, ŸÊß≈˛UÊ¡Ÿ, »§ÊS»§Ê⁄U‚ •ÊÒ⁄U ∑Ò§ÁÀ‡Êÿ◊ ¡Ò‚ ¬Ê·∑§Ê¥ ∑§Ë 134. The amount of nutrients, such as carbon, nitrogen,
◊ÎŒÊ ◊¥ ◊ÊòÊÊ, Á∑§‚Ë ÁŒÿ ªÿ ‚◊ÿ ◊¥ ∑Ò§‚ ‚¥ŒÁ÷¸Ã ∑§Ë ¡ÊÃË phosphorus and calcium present in the soil at any
„Ò? given time, is referred as :

(1) ø⁄U◊ •flSÕÊ (1) Climax

(2) ø⁄U◊ ‚◊ÈŒÊÿ (2) Climax community

(3) Standing state
(3) SÕÊÿË •flSÕÊ
(4) Standing crop
(4) SÕÊÿË »§‚‹
135. The first stable product of CO2 fixation in sorghum
135. ‚ÊÚ⁄UÉÊ◊ ◊¥ CO2 ÁSÕ⁄UË∑§⁄UáÊ ◊¥ ¬„‹Ê SÕÊÿË ©à¬ÊŒ ÄÿÊ „Ò? is :
(1) ¬ÊßL§Áfl∑§ •ê‹ (1) Pyruvic acid
(2) •ÊÄ¡‹Ê∞Á‚Á≈U∑§ •ê‹ (2) Oxaloacetic acid

(3) ‚Ä‚ËÁŸ∑§ •ê‹ (3) Succinic acid

(4) »§ÊS»§ÊÇ‹Ë‚Á⁄U∑§ •ê‹ (4) Phosphoglyceric acid

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•ŸÈ÷ʪ - B (¡ËflÁflôÊÊŸ — flŸS¬ÁÃÁflôÊÊŸ) Section - B (Biology : Botany)
136. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ª‹Ã „Ò? 136. Which of the following statements is incorrect ?
(1) ATP •ÊÒ⁄U NADPH+H+ ŒÊŸÊ¥ ∑§Ê ªÒ⁄U ø∑˝§Ëÿ ¬˝∑§Ê‡Ê (1) Both ATP and NADPH+H + are
»§ÊS»§ÊÁ⁄U‹‡ÊŸ ∑§ ŒÊÒ⁄UÊŸ ‚¥‡‹·áÊ „ÊÃÊ „Ò– synthesized during non-cyclic
(2) ¬ËÁ∆U∑§Ê - ‹ÒÁ◊‹Ë ◊¥ ∑§fl‹ PS I „ÊÃÊ „Ò •ÊÒ⁄U NADP photophosphorylation.
Á⁄U«UÄ≈U¡ ∑§Ê •÷Êfl „ÊÃÊ „Ò– (2) Stroma lamellae have PS I only and lack
(3) ª˝ÊŸÊ ‹ÒÁ◊‹Ë ◊¥ PS I •ÊÒ⁄U PS II ŒÊŸÊ¥ „ÊÃ „Ò¥– NADP reductase.
(4) ø∑˝§Ëÿ ¬˝∑§Ê‡Ê »§ÊS»§ÊÁ⁄U‹‡ÊŸ ◊¥ PS I •ÊÒ⁄U PS II ŒÊŸÊ¥ (3) Grana lamellae have both PS I and PS II.
‡ÊÊÁ◊‹ „ÊÃ „Ò¥– (4) Cyclic photophosphorylation involves both
137. ÁŸêŸÁ‹Áπà ◊¥ ‚ ‚„Ë ∑§ÕŸ øÈÁŸ∞– PS I and PS II.
(1) ∑Ò§Á¬¥ª ◊¥ ◊ÁÕ‹ ÇflÊŸÊ‚ËŸ ≈˛UÊß»§ÊS»§≈U ∑§Ê hnRNA 137. Identify the correct statement.
∑§ 39 Á‚⁄U ‚ ¡Ê«∏Ê ¡ÊÃÊ „Ò– (1) In capping, methyl guanosine triphosphate
(2) ¡ËflÊáÊÈ•Ê¥ ◊¥ •ŸÈ‹πŸ ∑§Ë ¬˝Á∑˝§ÿÊ ∑§Ê ‚◊ʬŸ ∑§⁄UŸ is added to the 39 end of hnRNA.
∑§ Á‹∞ •Ê⁄U.∞Ÿ.∞. ¬ÊÚÁ‹◊⁄U¡, Rho ∑§Ê⁄U∑§ ∑§ ‚ÊÕ (2) RNA polymerase binds with Rho factor to
’ÁãœÃ „Ê ¡ÊÃÊ „Ò– terminate the process of transcription in
(3) ∞∑§ •ŸÈ‹πŸ ß∑§Ê߸ ◊¥ ∑ͧ≈U ⁄Uí¡È∑§, ∞∑§ mRNA ¬⁄U bacteria.
¬˝ÁÃ∑Χà „ÊÃÊ „Ò– (3) The coding strand in a transcription unit is
(4) Áfl÷Äà ¡ËŸ-√ÿflSÕÊ ¬˝Ê∑Ò§Á⁄UÿÊ≈UÊ¥ ∑§Ê ÁflÁ‡Êc≈U ‹ˇÊáÊ copied to an mRNA.
„Ò– (4) Split gene arrangement is characteristic of
138. SÃ¥÷ - I ∑§Ê SÃ¥÷ - II ‚ ‚È◊Á‹Ã ∑§ËÁ¡∞– prokaryotes.
SÃ¥÷ - I SÃ¥÷ - II 138. Match Column - I with Column - II.
(a) (i) ’˝ÊÁ‚∑§‚Ë Column - I Column - II

(a) (i) Brassicaceae
(b) (ii) Á‹Á‹∞‚Ë
(c) (iii) »Ò§’‚Ë (b) (ii) Liliaceae

(d) (iv) ‚Ê‹Ÿ‚Ë (c) (iii) Fabaceae

ŸËø ÁŒÿ ªÿ Áfl∑§À¬Ê¥ ◊¥ ‚ ‚„Ë ©ûÊ⁄U øÈÁŸ∞– (d) (iv) Solanaceae
(a) (b) (c) (d)
Select the correct answer from the options given
(1) (iii) (iv) (ii) (i) below.
(2) (i) (ii) (iii) (iv) (a) (b) (c) (d)
(3) (ii) (iii) (iv) (i) (1) (iii) (iv) (ii) (i)
(4) (iv) (ii) (i) (iii) (2) (i) (ii) (iii) (iv)
139. ÁŸêŸÁ‹Áπà ◊¥ ‚ ‚„Ë ÿÈÇ◊ ∑§Ê øÈÁŸ∞– (3) (ii) (iii) (iv) (i)
(1) ÉÊÊ‚ ∑§Ë ¬ÁûÊÿÊ¥ ∑§Ë ’ÊsàfløÊ - ‚„Êÿ∑§ ∑§ÊÁ‡Ê∑§Êÿ¥ (4) (iv) (ii) (i) (iii)
◊¥ ’«∏Ë ⁄¥Uª„ËŸ Á⁄UÄà 139. Select the correct pair.
∑§ÊÁ‡Ê∑§Êÿ¥ (1) Large colorless empty - Subsidiary cells
(2) Ám’Ë¡¬òÊË ¬ÁûÊÿÊ¥ ◊¥ - ÿÊÒÁª∑§ ™§Ã∑§ cells in the epidermis
‚¥fl„Ÿ ’á«U‹, of grass leaves
◊Ê≈UË Á÷ÁûÊ flÊ‹Ë ’«∏Ë (2) In dicot leaves, vascular - Conjunctive
∑§ÊÁ‡Ê∑§Ê•Ê¥ ‚ bundles are surrounded tissue
ÁÉÊ⁄U „ÊÃ „Ò¥ by large thick-walled
(3) ◊îÊÊ Á∑§⁄UáÊÊ¥ ∑§Ë ∑§ÊÁ‡Ê∑§Êÿ¥ ¡Ê - •¥Ã⁄Uʬ͋Ëÿ cells
(3) Cells of medullary rays - Interfascicular
∑Ò¥§Á’ÿ◊ fl‹ÿ ∑§ ÷ʪ ∑§Ê ∑Ò¥§Á’ÿ◊
that form part of cambium
ÁŸ◊ʸáÊ ∑§⁄UÃË „Ò¥ cambial ring
(4) ’ÊsàfløÊ ∑§Ê »§Ê«∏Ÿ flÊ‹Ë - S¬¥¡Ë ◊ÎŒÍÃ∑§ (4) Loose parenchyma cells - Spongy
Á‡ÊÁÕ‹ ◊ÎŒÈ ∑§ÊÁ‡Ê∑§Êÿ¥ ¡Ê rupturing the epidermis parenchyma
¿UÊ‹ ◊¥ ‹¥‚ ∑§ •Ê∑§Ê⁄U and forming a lens-
∑§Ë Á¿UŒ˝ ’ŸÊÃË „Ò¥ shaped opening in bark

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140. ø⁄ÉÊÊÃÊ¥∑§Ë flÎÁh ‚◊Ë∑§⁄UáÊ 140. In the exponential growth equation
Nt=Noert ◊¥, e Á∑§‚ ÁŸM§Á¬Ã ∑§⁄UÃÊ „Ò? Nt=Noert, e represents :
(1) ‚¥ÅÿÊ ‹ÉÊȪáÊ∑§ ∑§Ê •ÊœÊ⁄U (1) The base of number logarithms
(2) ø⁄UÉÊÊÃÊ¥∑§Ë ‹ÉÊȪáÊ∑§ ∑§Ê •ÊœÊ⁄U (2) The base of exponential logarithms
(3) ¬˝Ê∑ΧÁÃ∑§ ‹ÉÊȪáÊ∑§ ∑§Ê •ÊœÊ⁄U (3) The base of natural logarithms
(4) íÿÊÁ◊ÁÃ∑§ ‹ÉÊȪáÊ∑§ ∑§Ê •ÊœÊ⁄U (4) The base of geometric logarithms

141. ‚‚Ë◊∑¥§Œ˝∑§Ë ¡ËflÊ¥ ◊¥ •ŸÈ‹πŸ ∑§Ë ¬˝Á∑˝§ÿÊ ◊¥ •Ê⁄U.∞Ÿ.∞. 141. What is the role of RNA polymerase III in the
¬ÊÚÁ‹U◊⁄U¡-III ∑§Ë ÷ÍÁ◊∑§Ê ÄÿÊ „Ò? process of transcription in eukaryotes ?
(1) rRNA (28S, 18S •ÊÒ⁄U 5.8S) ∑§Ê •ŸÈ‹Áπà ∑§⁄UÃÊ (1) Transcribes rRNAs (28S, 18S and 5.8S)
„Ò (2) Transcribes tRNA, 5s rRNA and snRNA
(2) tRNA, 5s rRNA •ÊÒ⁄U snRNA ∑§Ê •ŸÈ‹Á πà ∑§⁄UÃÊ
(3) Transcribes precursor of mRNA
„Ò
(4) Transcribes only snRNAs
(3) mRNA ∑§ ¬Ífl¸flÃ˸ ∑§Ê •ŸÈ‹Áπà ∑§⁄UÃÊ „Ò
(4) ∑§fl‹ snRNAs ∑§Ê •ŸÈ‹Áπà ∑§⁄UÃÊ „Ò 142. Plasmid pBR322 has PstI restriction enzyme site
within gene amp R that confers ampicillin
142. å‹ÒÁí◊«U pBR322 ◊¥ ¡ËŸ ampR ∑§ •ãŒ⁄U PstI ¬˝ÁÃ’¥œŸ resistance. If this enzyme is used for inserting a
∞ã¡Êß◊ „Ò ¡Ê ∞ê¬ËÁ‚‹ËŸ ¬˝ÁÃ⁄UÊœ Œ‡ÊʸÃÊ „Ò– ÿÁŒ ß‚ gene for β-galactoside production and the
∞ã¡Êß◊ ∑§Ê ’Ë≈UÊ-ªÒ‹Ä≈UÊ‚Êß« ©à¬ÊŒŸ ∑§ Á‹∞, ∞∑§ ¡ËŸ ∑§ recombinant plasmid is inserted in an E.coli strain
Á‹∞ ÁŸflÁ‡Êà Á∑§ÿÊ ¡ÊÃÊ „Ò •ÊÒ⁄U ¬ÈŸÿÊ¸ª¡ å‹ÒÁí◊«U ∑§Ê (1) it will not be able to confer ampicillin
ß.∑§Ê‹Ë S≈˛Ÿ ◊¥ ÁŸflÁ‡Êà Á∑§ÿÊ ¡ÊÃÊ „Ò Ã’ — resistance to the host cell.
(1) ÿ„ ¬Ê·Ë ∑§ÊÁ‡Ê∑§Ê ∑§Ê ∞ê¬ËÁ‚‹ËŸ ¬˝ÁÃ⁄UÊœ Ÿ„Ë¥ ’ŸÊ (2) the transformed cells will have the ability
¬ÊÿªÊ to resist ampicillin as well as produce
β-galactoside.
(2) M§¬Ê¥ÃÁ⁄Uà ∑§ÊÁ‡Ê∑§Ê•Ê¥ ◊¥ ∞ê¬ËÁ‚‹ËŸ ¬˝ÁÃ⁄UÊœ ∑§Ë ˇÊ◊ÃÊ
„ÊªË •ÊÒ⁄U ‚ÊÕ „Ë ’Ë≈UÊ-ªÒ‹Ä≈UÊ‚Êß«U ©à¬ÊŒŸ ∑§⁄¥UªË (3) it will lead to lysis of host cell.

(3) ß‚‚ ¬Ê·Ë ∑§ÊÁ‡Ê∑§Ê ◊¥ ‹ÿŸ „ÊU ¡ÊÿªÊ (4) it will be able to produce a novel protein with
dual ability.
(4) ß‚◊¥ mÒà ˇÊ◊ÃÊ ∑§ ‚ÊÕ Ÿÿ ¬˝Ê≈UËŸ ©à¬ÊŒŸ ∑§Ë ˇÊ◊ÃÊ
„ÊªË 143. Match List - I with List - II.
143. ‚ÍøË - I ∑§Ê ‚ÍøË - II ∑§ ‚ÊÕ ‚È◊Á‹Ã ∑§ËÁ¡∞– List - I List - II
Proteins are
‚ÍøË - I ‚ÍøË - II (a) S phase (i)
synthesized
(a) S •flSÕÊ (i) ¬˝Ê≈UËŸÊ¥ ∑§Ê ‚¥‡‹·áÊ „ÊÃÊ „Ò (b) G2 phase (ii) Inactive phase
(b) G2 •flSÕÊ (ii) ÁŸÁc∑˝§ÿ •flSÕÊ
Interval between
‚ÍòÊË Áfl÷Ê¡Ÿ •ÊÒ⁄U (c) Quiescent stage (iii) mitosis and initiation
(c) ‡Êʥà •flSÕÊ (iii) «UË.∞Ÿ.∞. ¬˝ÁÃ∑ΧÁÃ∑§⁄UáÊ ∑§ of DNA replication
¬˝Ê⁄Uïê÷
’ „ÊŸ ∑§ ’Ëø •ãÃ⁄UÊ‹ (d) G1 phase (iv) DNA replication
(d) G1 •flSÕÊ (iv) «UË.∞Ÿ.∞. ∑§Ê ¬˝ÁÃ∑ΧÁÃ∑§⁄UáÊ
Choose the correct answer from the options given
ŸËø ÁŒÿ ªÿ Áfl∑§À¬Ê¥ ◊¥ ‚ ‚„Ë ©ûÊ⁄U øÈÁŸ∞– below.
(a) (b) (c) (d) (a) (b) (c) (d)
(1) (iii) (ii) (i) (iv) (1) (iii) (ii) (i) (iv)
(2) (iv) (ii) (iii) (i) (2) (iv) (ii) (iii) (i)
(3) (iv) (i) (ii) (iii) (3) (iv) (i) (ii) (iii)
(4) (ii) (iv) (iii) (i) (4) (ii) (iv) (iii) (i)

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144. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ‚„Ë „Ò? 144. Which of the following statements is correct ?
(1) ŒÊ ∑§ÊÁ‡Ê∑§Ê•Ê¥ ∑§ ‚¥‹ÿŸ ∑§Ê ∑§ãŒ˝∑§ ‚¥‹ÿŸ ∑§„Ê (1) Fusion of two cells is called Karyogamy.
¡ÊÃÊ „Ò– (2) Fusion of protoplasms between two motile
(2) •ø‹ ÿÈÇ◊∑§Ê¥ ¬⁄U ŒÊ ø‹ ÿÈÇ◊∑§Ê¥ ∑§ ’Ëø ¡ËflŒ˝√ÿ ∑§ on non-motile gametes is called plasmogamy.
‚¥‹ÿŸ ∑§Ê ∑§ÊÁ‡Ê∑§Ê Œ˝√ÿ-‹ÿŸ ∑§„Ê ¡ÊÃÊ „Ò– (3) Organisms that depend on living plants are
(3) ¡Ê ¡Ëfl ¡ËÁflà ¬ÊŒ¬Ê¥ ¬⁄U ÁŸ÷¸⁄U „ÊÃ „Ò¥ ©ã„¥ ◊ÎÃÊ¬¡ËflË called saprophytes.
∑§„Ê ¡ÊÃÊ „Ò– (4) Some of the organisms can fix atmospheric
(4) ∑ȧ¿U ¡Ëfl flÊÿÈ◊¥«U‹Ëÿ ŸÊß≈˛UÊ¡Ÿ ∑§ •Êë¿UÊŒ ∑§ÊÁ‡Ê∑§Ê nitrogen in specialized cells called sheath
cells.
∑§„Ë ¡ÊŸ flÊ‹Ë ÁflÁ‡Êc≈U ∑§ÊÁ‡Ê∑§Ê•Ê¥ ◊¥ ÁSÕÁ⁄U∑Χà ∑§⁄U
‚∑§Ã „Ò¥–
145. Which of the following statements is incorrect ?
145. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ª‹Ã „Ò? (1) During aerobic respiration, role of oxygen is
limited to the terminal stage.
(1) flÊÿflËÿ ‡fl‚Ÿ ∑§ ŒÊÒ⁄UÊŸ, •ÊÚÄ‚Ë¡Ÿ ∑§Ë ÷ÍÁ◊∑§Ê •¥ÁÃ◊
•flSÕÊ Ã∑§ ‚ËÁ◊à „Ò– (2) In ETC (Electron Transport Chain), one
molecule of NADH+H + gives rise to
(2) ß.≈UË.‚Ë.(ß‹ Ä ≈˛ U ÊÚ Ÿ ¬Á⁄Ufl„Ÿ o¥ Î π ‹Ê) ◊ ¥ , 2 ATP molecules, and one FADH2 gives rise
NADH+H+ ∑§ ∞∑§ •áÊÈ ‚ ∞.≈UË.¬Ë. ∑§ ŒÊ •áÊÈ to 3 ATP molecules.
’ŸÃ „Ò¥ •ÊÒ⁄U ∞∑§ FADH2 ‚ ÃËŸ ATP •áÊÈ ’ŸÃ „Ò¥–
(3) ATP is synthesized through complex V.
(3) ∞.≈UË.¬Ë. ∑§Ê ‚¥‡‹·áÊ ‚Áê◊üÊ V ∑§ mÊ⁄UÊ „ÊÃÊ „Ò–
(4) Oxidation-reduction reactions produce
(4) ©¬Ê¬øÿŸ •Á÷Á∑˝§ÿÊÿ¥, ‡fl‚Ÿ ◊¥ ¬˝Ê≈UÊŸ ¬˝fláÊÃÊ ©à¬ãŸ proton gradient in respiration.
∑§⁄UÃË „Ò¥–
146. In some members of which of the following pairs
146. ÁŸêŸÁ‹Áπà ◊¥ ‚ Á∑§Ÿ ∑ȧ‹Ê¥ ∑§ ÿÈÇ◊ ◊¥ ©Ÿ∑§ ∑ȧ¿U ‚ŒSÿÊ¥ ◊¥ of families, pollen grains retain their viability for
¬⁄Uʪ∑§áÊÊ¥ ∑§Ë ¡ËflŸˇÊ◊ÃÊ, ©Ÿ∑§ ◊ÈÄà „ÊŸ ∑§ ’ÊŒ ◊„ËŸÊ¥ Ã∑§ months after release ?
⁄U„ÃË „Ò? (1) Poaceae ; Rosaceae
(1) ¬Ê∞‚Ë, ⁄UÊ¡‚Ë (2) Poaceae ; Leguminosae
(2) ¬Ê∞‚Ë, ‹ÇÿÈÁ◊ŸÊ‚Ë
(3) Poaceae ; Solanaceae
(3) ¬Ê∞‚Ë, ‚Ê‹Ÿ‚Ë
(4) Rosaceae ; Leguminosae
(4) ⁄UÊ¡‚Ë, ‹ÇÿÈÁ◊ŸÊ‚Ë
147. Match Column - I with Column - II.
147. SÃ¥÷ - I ∑§Ê SÃ¥÷ - II ‚ ‚È◊Á‹Ã ∑§ËÁ¡∞–
Column - I Column - II
SÃ¥÷ - I SÃ¥÷ - II (a) Nitrococcus (i) Denitrification
(a) ŸÊß≈˛UÊ∑§Ê∑§‚ (i) ÁflŸÊß≈˛UË∑§⁄UáÊ Conversion of
(b) Rhizobium (ii)
ammonia to nitrite
•◊ÊÁŸÿÊ ∑§Ê ŸÊß≈˛UÊß≈U ◊¥
(b) ⁄UÊß¡ÊÁ’ÿ◊ (ii) Conversion of nitrite
¬Á⁄UfløŸ (c) Thiobacillus (iii)
to nitrate
ŸÊß≈˛UÊß≈U ∑§Ê ŸÊß≈˛≈U ◊¥ Conversion of
(c) ÁÕÿÊ’Ò‚Ë‹‚ (iii)
¬Á⁄UfløŸ (d) Nitrobacter (iv) atmospheric nitrogen
flÊÿÈ◊¥«U‹Ëÿ ŸÊß≈˛UÊ¡Ÿ ∑§Ê to ammonia
(d) ŸÊß≈˛UÊ’ÒÄ≈U⁄U (iv)
•◊ÊÁŸÿÊ ◊¥ ¬Á⁄UfløŸ Choose the correct answer from options given
ŸËø ÁŒÿ ªÿ Áfl∑§À¬Ê¥ ◊¥ ‚ ‚„Ë ©ûÊ⁄U øÈÁŸ∞– below.

(a) (b) (c) (d) (a) (b) (c) (d)

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

(2) (i) (ii) (iii) (iv) (2) (i) (ii) (iii) (iv)
(3) (iii) (i) (iv) (ii) (3) (iii) (i) (iv) (ii)
(4) (iv) (iii) (ii) (i) (4) (iv) (iii) (ii) (i)

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148. ∑§ÊÁ‡Ê∑§Ê•Ê¥ ∑§ ∞∑§ Ä‹ÊŸ ◊¥ ⁄UÁ«UÿÊœ◊˸ ¬˝Ê’ ‚ ß‚∑§ «UË.∞Ÿ.∞. 148. Now a days it is possible to detect the mutated
∑§Ê ‚¥∑§⁄UáÊ ∑§⁄U •ÊÒ⁄U ©‚∑§ ’ÊŒ •ÊÚ≈UÊ⁄UÁ«UÿÊª˝Ê»§Ë ¬˝ÿÈÄà ∑§⁄U gene causing cancer by allowing radioactive probe
to hybridise its complimentary DNA in a clone of
ß‚∑§Ë ¬„øÊŸ ∑§⁄U ∑Ò¥§‚⁄U ©à¬ãŸ ∑§⁄UŸ flÊ‹Ë ©à¬Á⁄UflÁøà ¡ËŸ
cells, followed by its detection using
∑§Ê ¬ÃÊ ‹ªÊŸÊ •Ê¡∑§‹ ‚¥÷fl „Ò ÄÿÊ¥Á∑§ — autoradiography because :

(1) ©à¬Á⁄UflÁøà ¡ËŸ »§Ê≈UÊª˝ÊÁ$»§∑§ Á»§À◊ ¬⁄U •Ê¥Á‡Ê∑§ M§¬ (1) mutated gene partially appears on a
◊¥ •ÊÃË „Ò– photographic film.

(2) ©à¬Á⁄UflÁøà ¡ËŸ »§Ê≈UÊª˝ÊÁ$»§∑§ Á»§À◊ ¬⁄U ¬Íáʸ •ÊÒ⁄U S¬c≈U (2) mutated gene completely and clearly
M§¬ ◊¥ •ÊÃË „Ò– appears on a photographic film.

(3) ©à¬Á⁄UflÁøà ¡ËŸ »§Ê≈UÊª˝ÊÁ$»§∑§ Á»§À◊ ¬⁄U Ÿ„Ë¥ •ÊÃË (3) mutated gene does not appear on a
photographic film as the probe has no
ÄÿÊ¥Á∑§ ¬˝Ê’ ◊¥ ß‚∑§Ê ∑§Ê߸ ¬Í⁄U∑§ Ÿ„Ë¥ „Ò–
complimentarity with it.
(4) ©à¬Á⁄UflÁøà ¡ËŸ »§Ê≈UÊª˝ÊÁ$»§∑§ Á»§À◊ ¬⁄U Ÿ„Ë¥ •ÊÃË (4) mutated gene does not appear on
ÄÿÊ¥Á∑§ ¬˝Ê’ ◊¥ ß‚∑§Ê ¬Í⁄U∑§ „ÊÃÊ „Ò– photographic film as the probe has
complimentarity with it.

149. «UË.∞Ÿ.∞. •¥ªÈ‹Ë¿UÊ¬Ë ◊¥ «UË.∞Ÿ.∞. •ŸÈ∑˝§◊ ◊¥ ∑ȧ¿U ÁflÁ‡Êc≈U
SÕÊŸÊ¥ ◊¥ Á÷ãŸÃÊ•Ê¥ ∑§Ë ¬„øÊŸ ∑§Ë ¡ÊÃË „Ò– ߟ ÁflÁ‡Êc≈U 149. DNA fingerprinting involves identifying differences
SÕÊŸÊ¥ ∑§Ê ÄÿÊ ∑§„Ê ¡ÊÃÊ „Ò? in some specific regions in DNA sequence, called
as :
(1) •ŸÈ·¥ªË «UË.∞Ÿ.∞.
(1) Satellite DNA
(2) ¬ÈŸ⁄UÊflÎÁûÊ «UË.∞Ÿ.∞.
(2) Repetitive DNA
(3) ∞∑§‹ ãÿÍÁÄ‹ÿÊ≈UÊß«U (3) Single nucleotides
(4) ’„ÈM§¬Ëÿ «UË.∞Ÿ.∞. (4) Polymorphic DNA

150. ‚ÍøË - I ∑§Ê ‚ÍøË - II ∑§ ‚ÊÕ ‚È◊Á‹Ã ∑§ËÁ¡∞– 150. Match List - I with List - II.

‚ÍøË - I ‚ÍøË - II List - I List - II
(a) ¬˝Ê≈UËŸ (i) C=C Ám’㜠(a) Protein (i) C=C double bonds
Unsaturated
(b) •‚¥ÃÎåà fl‚Êfl‚Ê•ï‹ •ê‹ (ii) »§ÊS»§Ê«UÊß∞S≈U⁄U ’㜠(b) (ii) Phosphodiester bonds
fatty acid
(c) ãÿÍÄ‹Ë∑§ •ï‹ •ê‹ (iii) Ç‹Êß∑§Ê‚ÊßÁ«U∑§ ’㜠(c) Nucleic acid (iii) Glycosidic bonds
(d) ¬ÊÚ‹ Ë‚Ò∑§⁄UÊß«U (iv) ¬å≈UÊß«U ’㜠(d) Polysaccharide (iv) Peptide bonds

ŸËø ÁŒÿ ªÿ Áfl∑§À¬Ê¥ ◊¥ ‚ ‚„Ë ©ûÊ⁄U øÈÁŸ∞– Choose the correct answer from the options given
below.
(a) (b) (c) (d)
(a) (b) (c) (d)
(1) (iv) (i) (ii) (iii) (1) (iv) (i) (ii) (iii)

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

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

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

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•ŸÈ÷ʪ - A (¡ËflÁflôÊÊŸ — ¬˝ÊÁáÊÁflôÊÊŸ) Section - A (Biology : Zoology)

151. ÁŸêŸ ◊¥ Á∑§‚ ¡Ëfl ◊¥ ‹ê’Ë •ÁSÕÿÊ° πÊπ‹Ë ∞fl¥ flÊÁË 151. Which one of the following organisms bears hollow
and pneumatic long bones ?
„ÊÃË „Ò¥? (1) Neophron
(1) ÁŸ•Ê»§˝ ÊÚŸ (2) Hemidactylus
(2) „Ò◊Ë«UÄ≈UÊÿ‹‚ (3) Macropus
(3) ◊Ò∑§˝ Ê¬‚ (4) Ornithorhynchus
(4) •ÊÒ⁄UÁŸÕÊÁ⁄¥U∑§‚ 152. Chronic auto immune disorder affecting neuro
152. ŒËÉʸ∑§Ê‹Ë Sfl¬˝ÁÃ⁄UˇÊÊ Áfl∑§Ê⁄U ¡Ê Ã¥ÁòÊ∑§Ê ¬‡ÊËÿ ‚¥Áœ ∑§Ê ¬˝÷ÊÁflà muscular junction leading to fatigue, weakening
∑§⁄UÃË „Ò ∞fl¥ Á¡‚∑§ ∑§Ê⁄UáÊ ∑¥§∑§Ê‹ ¬‡ÊË ◊¥ ∑§◊¡Ê⁄UË ∞fl¥ and paralysis of skeletal muscle is called as :
(1) Arthritis
¬ˇÊÉÊÊà „ÊÃÊ „Ò, ∑§„‹ÊÃË „Ò — (2) Muscular dystrophy
(1) ‚¥Áœ‡ÊÊÕ (3) Myasthenia gravis
(2) ¬‡ÊËÿ ŒÈc¬Ê·áÊ (4) Gout
(3) ◊ÊßSÕÁŸÿÊ ª˝Áfl‚ 153. Which one of the following is an example of
(4) ªÊ©≈U Hormone releasing IUD ?
153. ÁŸêŸ ∑§ ∑§ÊÒŸ „Ê◊Ê¸Ÿ ◊Êø∑§ •Ê߸ ÿÍ «UË ∑§Ê ©ŒÊ„⁄UáÊ „Ò? (1) CuT
(2) LNG 20
(1) CuT
(3) Cu 7
(2) LNG 20
(4) Multiload 375
(3) Cu 7
154. Which of the following characteristics is incorrect
(4) ◊À≈UË‹Ê«U 375 with respect to cockroach ?
154. ÁËø^ ∑§ ‚¥’¥œ ◊¥ ∑§ÊÒŸ ‚Ë ÁflÁ‡Êc≈UÃÊ∞° •ŸÈÁøà „Ò¥? (1) A ring of gastric caeca is present at the
(1) ◊äÿÊ¥òÊ ∞fl¥ ¬‡øÊ¥òÊ ∑§ ‚¥Áœ SÕ‹ ¬⁄U ¡∆U⁄UËÿ •¥œŸÊ‹ junction of midgut and hind gut.
∑§Ê ∞∑§ fl‹ÿ „ÊÃÊ „Ò– (2) Hypopharynx lies within the cavity enclosed
by the mouth parts.
(2) •œÊª˝‚ŸË ◊Èπ ÷ʪÊ¥ mÊ⁄UÊ ’ŸÊ߸ ªÿË ¬˝ªÈ„Ê ◊¥ „ÊÃË
(3) In females, 7th-9th sterna together form a
„Ò– genital pouch.
(3) ◊ÊŒÊ ◊¥ 7-9 Ã∑§ •œ⁄U∑§ Á◊‹∑§⁄U ¡ŸÁŸ∑§ ∑§Êc∆U ’ŸÊÃ (4) 10th abdominal segment in both sexes, bears
„Ò¥– a pair of anal cerci.
(4) ŒÊŸÊ¥ Á‹¥ªÊ¥ ◊¥ Œ‚fl¥ π¥«U ◊¥ ªÈŒËÿ ‹Í◊ ¬Êÿ ¡ÊÃ „Ò¥– 155. Select the favourable conditions required for the
155. ∑ͧÁ¬∑§Ê•Ê¥ ◊¥ •ÊÚÄ‚Ë„Ë◊ÊÇ‹ÊÁ’Ÿ ’ŸŸ ∑§ Á‹∞ •ŸÈ∑ͧ‹ formation of oxyhaemoglobin at the alveoli.
¬Á⁄UÁSÕÁÃÿÊ¥ ∑§Ê øÿŸ ∑§⁄UÊ– (1) High pO 2 , low pCO 2 , less H + , lower
(1) ©ëø pO2, ÁãÊêŸ pCO2, ÁŸêŸ H+, •¬ˇÊÊ∑Χà ∑§◊ temperature
(2) Low pO2, high pCO2, more H+, higher
Ãʬ◊ÊŸ temperature
(2) ÁŸêŸ pO2, ©ëø pCO2, •Áœ∑§ H+, •¬ˇÊÊ∑Χà ©ìÊ (3) High pO2, high pCO2, less H+, higher
Ãʬ◊ÊŸ temperature
(3) ©ëø pO2, ©ëø pCO2, ÁŸêŸ H+, •¬ˇÊÊ∑Χà ©ëø (4) Low pO2 , low pCO2, more H +, higher
Ãʬ◊ÊŸ temperature
(4) ÁŸêŸ pO2, ÁŸêŸ pCO2, •Áœ∑§ H+, •¬ˇÊÊ∑Χà ©ëø 156. Sphincter of oddi is present at :
Ãʬ◊ÊŸ (1) Ileo-caecal junction
156. •Ê«UË •fl⁄UÊÁœŸË ∑§„Ê° ©¬ÁSÕà „ÊÃË „Ò? (2) Junction of hepato-pancreatic duct and
duodenum
(1) ÁòÊ∑§Ê¥òÊ ‚¥Áœ (3) Gastro-oesophageal junction
(2) ÿ∑Χà •ÇŸÊ‡ÊÿË flÊÁ„ŸË ∞fl¥ ª˝„áÊË ‚¥Áœ (4) Junction of jejunum and duodenum
(3) ¡∆U⁄U-ª˝Á‚∑§Ê ‚¥Áœ 157. The organelles that are included in the
(4) •ª˝ˇÊÈŒ˝Ê¥òÊ ∞fl¥ ª˝„áÊË ‚¥Áœ endomembrane system are :
157. •¥Ã—Á¤ÊÁÀ‹∑§Ê Ã¥òÊ ◊¥ ∑§ÊÒŸ ‚ ∑§ÊÁ‡Ê∑§Ê¥ª ‚Áê◊Á‹Ã „ÊÃ „Ò¥? (1) Endoplasmic reticulum, Mitochondria,
(1) •¥ÃŒ˝√¸ ÿË ¡ÊÁ‹∑§Ê, ‚ÍòÊ∑§ÁáÊ∑§Ê, ⁄UÊß’Ê‚Ê◊ ∞fl¥ ‹ÿŸ∑§Êÿ Ribosomes and Lysosomes
(2) •¥ÃŒ˝¸√ÿË ¡ÊÁ‹∑§Ê, ªÊÚÀ¡Ë ‚Áê◊üÊ, ‹ÿŸ∑§Êÿ ∞fl¥ (2) Endoplasmic reticulum, Golgi complex,
⁄U‚œÊŸË Lysosomes and Vacuoles
(3) Golgi complex, Mitochondria, Ribosomes and
(3) ªÊÚÀ¡Ë ‚Áê◊üÊ, ‚ÍòÊ∑§ÁáÊ∑§Ê, ⁄UÊß’Ê‚Ê◊ ∞fl¥ ‹ÿŸ∑§Êÿ
Lysosomes
(4) ªÊÚÀ¡Ë ‚Áê◊üÊ, •¥ÃŒ˝¸√ÿË ¡ÊÁ‹∑§Ê, ‚ÍòÊ∑§ÁáÊ∑§Ê ∞fl¥ (4) Golgi complex, Endoplasmic reticulum,
‹ÿŸ∑§Êÿ Mitochondria and Lysosomes

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158. ‚ÍøË - I ∑§Ê ‚ÍøË - II ∑§ ‚ÊÕ Á◊‹ÊŸ ∑§⁄UÊ– 158. Match List - I with List - II.
‚ÍøË - I ‚ÍøË - II List - I List - II
(a) Metamerism (i) Coelenterata
(a) Áflπ¥«UÊflSÕÊ (i) ‚Ë‹ã≈U⁄U≈UÊ
(b) Canal system (ii) Ctenophora
(b) ŸÊ‹-Ã¥òÊ (ii) ≈UËŸÊ»§Ê⁄UÊ
(c) Comb plates (iii) Annelida
(c) ∑¥§∑§Ã ¬Á^∑§Ê (iii) ∞ŸÁ‹«UÊ (d) Cnidoblasts (iv) Porifera
(d) Œ¥‡Ê ∑§ÊÁ‡Ê∑§Ê (iv) ¬Ê⁄UË»§⁄UÊ Choose the correct answer from the options given
below.
ÁŸêŸ Áfl∑§À¬Ê¥ ◊¥ ‚ ©Áøà ©ûÊ⁄U ∑§Ê øÿŸ ∑§⁄UÊ– (a) (b) (c) (d)
(a) (b) (c) (d)
(1) (iv) (iii) (i) (ii) (1) (iv) (iii) (i) (ii)
(2) (iii) (iv) (i) (ii) (2) (iii) (iv) (i) (ii)
(3) (iii) (iv) (ii) (i) (3) (iii) (iv) (ii) (i)
(4) (iv) (i) (ii) (iii) (4) (iv) (i) (ii) (iii)
159. ∞∑§ ÁflÁ‡Êc≈U ¬„øÊŸ •ŸÈ∑˝§◊ ¡Ê ∞¥«UÊãÿÍÁÄ‹∞¡ mÊ⁄UÊ 159. A specific recognition sequence identified by
«UË ∞Ÿ ∞ ∑§Ê ÁflÁ‡Êc≈U ÁSÕÁà ¬⁄U ∑§Ê≈UŸ ∑§ Á‹∞ ¬„øÊŸÊ ¡ÊÃÊ endonucleases to make cuts at specific positions
„Ò, ÄÿÊ ∑§„‹ÊÃÊ „Ò? within the DNA is :
(1) •¬OÊÁ‚à ¬˝Ê⁄Uê÷∑§ •ŸÈ∑˝§◊ (1) Degenerate primer sequence
(2) •Ê∑§Ê$¡Ê∑§Ë •ŸÈ∑˝§◊ (2) Okazaki sequences
(3) ¬Ò‹Ëã«˛UÊÁ◊∑§ ãÿÍÁÄÀÊ•Ê≈UÊß«U •ŸÈ∑˝§◊ (3) Palindromic Nucleotide sequences
(4) ’„È (∞) ¬Èë¿UŸ •ŸÈ∑˝§◊ (4) Poly(A) tail sequences
160. ÁŸêŸ ◊¥ ¬˝Ê≈UËŸ ‚¥‡‹·áÊ ∑§ Á‹∞ ∑§ÊÒŸ ‚Ê RNA •Êfl‡ÿ∑§ 160. Which of the following RNAs is not required for
Ÿ„Ë¥ „Ò? the synthesis of protein ?
(1) mRNA (1) mRNA
(2) tRNA (2) tRNA
(3) rRNA (3) rRNA
(4) siRNA (4) siRNA
161. ∑ͧÁ¬∑§Ê•Ê¥ (Áfl‚⁄UáÊ SÕ‹) ◊¥ •ÊÚÄ‚Ë¡Ÿ (O2) ∞fl¥ ∑§Ê’¸Ÿ 161. The partial pressures (in mm Hg) of oxygen (O2)
«UÊß•ÊÚÄ‚Êß«U (CO2) ∑§Ê •Ê¥Á‡Ê∑§ Œ’Êfl (mm Hg ◊¥) „ÊÃÊ and carbon dioxide (CO2) at alveoli (the site of
„Ò — diffusion) are :
(1) pO2=104 ∞fl¥ pCO2=40 (1) pO2=104 and pCO2=40
(2) pO2=40 ∞fl¥ pCO2=45 (2) pO2=40 and pCO2=45
(3) pO2=95 ∞fl¥ pCO2=40 (3) pO2=95 and pCO2=40
(4) pO2=159 ∞fl¥ pCO2=0.3 (4) pO2=159 and pCO2=0.3
162. ÁŸêŸ ∑§ÕŸÊ¥ ◊¥ ∑§ÊÒŸ Áø∑§ŸË ¬‡ÊË ∑§Ë ¬˝flÎÁûÊ ∑§Ê •ŸÈÁøà M§¬ 162. Which of the following statements wrongly
‚ Œ‡ÊʸÃÊ „Ò? represents the nature of smooth muscle ?
(1) ߟ ¬Á‡ÊÿÊ¥ ◊¥ œÊÁ⁄UÿÊ° Ÿ„Ë¥ „ÊÃË– (1) These muscle have no striations
(2) ÿ •ŸÒÁë¿U∑§ ¬Á‡ÊÿÊ° „ÊÃË „Ò¥– (2) They are involuntary muscles
(3) ∑§ÊÁ‡Ê∑§Ê•Ê¥ ∑§ ◊äÿ ‚¥ø⁄UáÊ •¥ÃÁfl¸c≈U Á«US∑§ mÊ⁄UÊ „ÊÃÊ (3) Communication among the cells is
„Ò– performed by intercalated discs
(4) ÿ ¬Á‡ÊÿÊ° ⁄UÄà flÊÁ„∑§Ê ∑§Ë Á÷ÁûÊ ◊¥ ©¬ÁSÕà „ÊÃË „Ò– (4) These muscles are present in the wall of
163. ‚ÍøË - I ∑§Ê ‚ÍøË - II ∑§ ‚ÊÕ Á◊‹ÊŸ ∑§⁄UÊ– blood vessels
163. Match the following :
‚ÍøË - I ‚ÍøË - II List - I List - II
(a) »§Êß‚Á‹ÿÊ (i) ◊ÈÄÃʇÊÈÁÄà (a) Physalia (i) Pearl oyster
(b) Á‹◊Í‹ ‚ (ii) ¬Èøª Ê‹Ë ÿÈh ◊ÊŸfl (b) Limulus (ii) Portuguese Man of War
(c) Ancylostoma (iii) Living fossil
(c) ∞Ÿ‚Êß‹ÊS ≈UÊ◊Ê (iii) ¡ËÁflà ¡Ëflʇ◊
(d) Pinctada (iv) Hookworm
(d) Á¬¥∑§≈UÊ«UÊ (iv) •¥∑ȧ‡Ê∑ΧÁ◊ Choose the correct answer from the options given
ÁŸêŸ Áfl∑§À¬Ê¥ ‚ ©Áøà ©ûÊ⁄U ∑§Ê øÿŸ ∑§⁄UÊ– below.
(a) (b) (c) (d) (a) (b) (c) (d)
(1) (ii) (iii) (i) (iv) (1) (ii) (iii) (i) (iv)
(2) (iv) (i) (iii) (ii) (2) (iv) (i) (iii) (ii)
(3) (ii) (iii) (iv) (i) (3) (ii) (iii) (iv) (i)
(4) (i) (iv) (iii) (ii) (4) (i) (iv) (iii) (ii)

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164. ‚Ä∑§‚ ∞¥≈UÁ⁄U∑§‚ ∑§Ê ‚¥Œ÷¸ „Ò — 164. Succus entericus is referred to as :
(1) •ÇŸÊ‡ÊÿË ⁄U‚ ‚ (1) Pancreatic juice
(2) •Ê¥òÊ ⁄U‚ ‚ (2) Intestinal juice
(3) ¡∆U⁄U ⁄U‚ ‚ (3) Gastric juice
(4) ∑§Êß◊ ‚ (4) Chyme
165. ÁŸêŸ ◊¥ ∑§ÊÒŸ »§‚‹Ê¥ ∑§ ’ÊÿÊ»§ÊÁ≈¸UÁ»§∑§‡ÊŸ ∑§Ê ©g‡ÿ Ÿ„Ë¥ „Ò? 165. Which of the following is not an objective of
(1) ¬˝Ê≈UËŸ ∑§Ë ◊ÊòÊÊ ◊¥ ‚ÈœÊ⁄U ∑§⁄UŸÊ Biofortification in crops ?
(2) ⁄UÊªÊ¥ ∑§Ë ¬˝ÁÃ⁄UÊœÃÊ ◊¥ ‚ÈœÊ⁄U ∑§⁄UŸÊ (1) Improve protein content
(2) Improve resistance to diseases
(3) Áfl≈UÊÁ◊Ÿ ∑§Ë ◊ÊòÊÊ ◊¥ ‚ÈœÊ⁄U ∑§⁄UŸÊ
(3) Improve vitamin content
(4) ‚͡◊-¬Ê·∑§Ê¥ ∞fl¥ πÁŸ¡Ê¥ ∑§Ë ◊ÊòÊÊ ◊¥ ‚ÈœÊ⁄U ∑§⁄UŸÊ (4) Improve micronutrient and mineral content
166. ⁄UÁá ⁄UÊª Á∑§‚∑§ mÊ⁄UÊ ‚¥øÁ⁄Uà „ÊÃ „Ò¥? 166. Veneral diseases can spread through :
(a) ⁄UÊªÊáÊÈ⁄UÁ„à ‚ÍßÿÊ¥ ∑§ ©¬ÿÊª ‚ (a) Using sterile needles
(b) ‚¥∑˝§Á◊à √ÿÁÄà ∑§ ⁄UÄà ø…∏UÊŸ ‚ (b) Transfusion of blood from infected person
(c) ‚¥∑˝§Á◊à ◊ÊÃÊ ‚ ÷˝ÍáÊ ◊¥ (c) Infected mother to foetus
(d) øÈê’Ÿ ‚ (d) Kissing
(e) fl¥‡ÊªÁà ‚ (e) Inheritance
ÁŸêŸ Áfl∑§À¬Ê¥ ‚ ‚„Ë ©ûÊ⁄U ∑§Ê øÿŸ ∑§⁄UÊ– Choose the correct answer from the options given
(1) ∑§fl‹ (a), (b) ∞fl¥ (c) below.
(2) ∑§fl‹ (b), (c) ∞fl¥ (d) (1) (a), (b) and (c) only
(3) ∑§fl‹ (b) ∞fl¥ (c) (2) (b), (c) and (d) only
(4) ∑§fl‹ (a) ∞fl¥ (c) (3) (b) and (c) only
(4) (a) and (c) only
167. ߥ‚È‹ËŸ ∑§ ‚¥Œ÷¸ ◊¥ ©Áøà Áfl∑§À¬ ∑§Ê øÿŸ ∑§⁄UÊ–
167. With regard to insulin choose correct options.
(a) ¬Á⁄U¬Äfl ߥ‚È‹ËŸ ◊¥ ‚Ë-¬å≈UÊß«U Ÿ„Ë¥ „ÊÃË– (a) C-peptide is not present in mature insulin.
(b) •Ê⁄U «UË ∞Ÿ ∞ ¬˝ÊÒlÊÁª∑§Ë ‚ ©à¬ÊÁŒÃ ߥ‚È‹ËŸ ◊¥ (b) The insulin produced by rDNA technology
‚Ë-¬å≈UÊß«U „ÊÃÊ „Ò– has C-peptide.
(c) ¬˝Ê∑˜§-ߥ‚È‹ËŸ ◊¥ ‚Ë-¬å≈UÊß« „ÊÃÊ „Ò– (c) The pro-insulin has C-peptide.
(d) ߥ‚È‹ËŸ ∑§ ∞-¬å≈UÊß«U ∞fl¥ ’Ë-¬å≈UÊß«U «UÊß‚À»§Êß« (d) A-peptide and B-peptide of insulin are
’¥œÊ¥ mÊ⁄UÊ ¬Ê⁄US¬Á⁄U∑§ ¡È«∏ „ÊÃ „Ò¥– interconnected by disulphide bridges.
Choose the correct answer from the options given
ŸËø ÁŒ∞ ªÿ Áfl∑§À¬Ê¥ ◊¥ ‚ ©Áøà ©ûÊ⁄U ∑§Ê øÿŸ ∑§⁄UÊ– below.
(1) ∑§fl‹ (b) ∞fl¥ (d) (1) (b) and (d) only
(2) ∑§fl‹ (b) ∞fl¥ (c) (2) (b) and (c) only
(3) ∑§fl‹ (a), (c) ∞fl¥ (d) (3) (a), (c) and (d) only
(4) ∑§fl‹ (a) ∞fl¥ (d) (4) (a) and (d) only
168. »§‹◊ÄπË ∑§Ë ¬˝àÿ∑§ ∑§ÊÁ‡Ê∑§Ê ◊¥ 8 ªÈáÊ‚ÍòÊ (2n) „ÊÃ „Ò¥– 168. The fruit fly has 8 chromosomes (2n) in each cell.
ÿÁŒ ‚ÍòÊË Áfl÷Ê¡Ÿ ∑§Ë •¥Ã⁄UÊflSÕÊ ∑§Ë G1 ¬˝ÊflSÕÊ ◊¥ ªÈáÊ‚ÍòÊÊ¥ During interphase of Mitosis if the number of
∑§Ë ‚¥ÅÿÊ 8 „Ò Ã’ S-¬˝ÊflSÕÊ ∑§ ’ÊŒ ªÈáÊ‚ÍòÊÊ¥ ∑§Ë ‚¥ÅÿÊ ÄÿÊ chromosomes at G1 phase is 8, what would be the
„ÊªË? number of chromosomes after S phase ?
(1) 8 (1) 8
(2) 16 (2) 16
(3) 4 (3) 4
(4) 32 (4) 32
169. •‚Ë◊∑§ãŒ˝∑§Ë ∑§Ë •ŸÈ‹πŸ ¬˝Á∑˝§ÿÊ ◊¥ ∑§ÊÒŸ ‚Ê ∞¥¡Êß◊ ¬˝Ê⁄¥U÷Ÿ, 169. Which is the “Only enzyme” that has “Capability”
ŒËÉÊ˸∑§⁄UáÊ ∞fl¥ ‚◊ʬŸ ∑§Ê ©à¬˝Á⁄Uà ∑§⁄UÃÊ „Ò? to catalyse Initiation, Elongation and Termination
in the process of transcription in prokaryotes ?
(1) «UË ∞Ÿ ∞ ¬⁄U ÁŸ÷¸⁄U «UË ∞Ÿ ∞ ¬ÊÚ‹Ë◊⁄U¡ (1) DNA dependent DNA polymerase
(2) «UË ∞Ÿ ∞ ¬⁄U ÁŸ÷¸⁄U •Ê⁄U ∞Ÿ ∞ ¬ÊÚ‹Ë◊⁄U¡ (2) DNA dependent RNA polymerase
(3) «UË ∞Ÿ ∞ ‹Êߪ¡ (3) DNA Ligase
(4) «UË ∞Ÿ ∞¡ (4) DNase
170. ∞∑§ Ÿ⁄U •ÊÒ⁄U ◊ÊŒÊ ŒÊŸÊ¥ ŒÊòÊ ∑§ÊÁ‡Ê∑§Ê •⁄UÄÃÃÊ ∑§ ¡ËŸ ∑§ Á‹∞ 170. In a cross between a male and female, both
Áfl·◊ÿÈÇ◊¡Ë „Ò¥ ∑§ ‚¥∑§⁄UáÊ ‚ ©à¬ãŸ ‚¥ÃÁà ∑§Ê Á∑§ÃŸÊ ¬˝ÁÇÊà heterozygous for sickle cell anaemia gene, what
⁄UÊªÿÈÄà „ÊªÊ ? percentage of the progeny will be diseased ?
(1) 50%
(1) 50%
(2) 75%
(2) 75% (3) 25%
(3) 25% (4) 100%
(4) 100%

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171. Á∑§‚Ë ⁄UÊª ∑§ ¬˝÷ÊflË ©¬øÊ⁄U ∑§ Á‹∞ ß‚∑§ •Ê⁄¥UÁ÷∑§ ÁŸŒÊŸ 171. For effective treatment of the disease, early
∞fl¥ ⁄UÊª-Á∑˝§ÿÊ ÁflôÊÊŸ ∑§Ê ‚◊¤ÊŸÊ ’„Èà ◊„àfl¬Íáʸ „Ò– ÁŸêŸ ◊¥ diagnosis and understanding its pathophysiology
∑§ÊÒŸ ‚Ë •ÊÁáfl∑§ ÁŸŒÊŸ Ã∑§ŸË∑§ •Ê⁄¥UÁ÷∑§ ¬„øÊŸ ∑§ Á‹∞ is very important. Which of the following
’„Èà ©¬ÿÊªË „Ò? molecular diagnostic techniques is very useful for
early detection ?
(1) flÒS≈UŸ¸ é‹ÊÁ≈¥Uª Ã∑§ŸË∑§ (1) Western Blotting Technique
(2) ‚ŒŸ¸ é‹ÊÁ≈¥Uª Ã∑§ŸË∑§ (2) Southern Blotting Technique
(3) ELISA Ã∑§ŸË∑§ (3) ELISA Technique
(4) ‚¥∑§⁄UáÊ Ã∑§ŸË∑§ (4) Hybridization Technique
172. •œ¸‚ÍòÊË ¬ÍflʸflSÕÊ ∑§Ë ∑§ÊÒŸ ‚Ë ¬˝ÊflSÕÊ ∑§Ê ÁflÁ‡Êc≈U ‹ˇÊáÊ 172. Which stage of meiotic prophase shows
∑§Êß∞í◊≈UÊ ∑§Ê ©¬Ê¥ÃË÷flŸ „Ò? terminalisation of chiasmata as its distinctive
(1) ßȬ^ feature ?
(1) Leptotene
(2) ÿÈÇ◊¬^ (2) Zygotene
(3) ¬Ê⁄UªÁÃ∑˝§◊ (3) Diakinesis
(4) SÕÍ‹¬^ (4) Pachytene
173. ÁŸêŸ ∑§ÕŸÊ¥ ∑§Ê ¬…∏Ê– 173. Read the following statements.
(a) ◊≈UÊ¡ŸÁ‚‚ ∑ΧÁ◊ÿÊ¥ ◊¥ ¬ÊÿÊ ¡ÊÃÊ „Ò– (a) Metagenesis is observed in Helminths.
(b) ∞∑§ÊߟÊ«U◊¸U ÁòÊ∑§Ê⁄U∑§ ∞fl¥ ªÈ„Ëÿ ¡¥ÃÈ „ÊÃ „Ò¥– (b) Echinoderms are triploblastic and coelomate
(c) ªÊ‹∑ΧÁ◊ÿÊ¥ ◊¥ ‚¥ª∆UŸ ∑§Ê SÃ⁄U •¥ª Ã¥òÊ „ÊÃÊ „Ò– animals.
(c) Round worms have organ-system level of
(d) ≈UËŸÊ»§Ê⁄U ◊¥ ©¬ÁSÕà ∑¥§∑§Ã ¬Á^∑§Ê∞° ¬ÊøŸ ◊¥ ‚„ÊÿÃÊ body organization.
∑§⁄UÃË „Ò¥–
(d) Comb plates present in ctenophores help in
(e) ¡‹ ‚¥fl„Ÿ ∑§Ê ∞∑§ÊߟÊ«U◊¸ ∑§Ë ÁflÁ‡Êc≈UÃÊ „ÊÃË „Ò– digestion.
ÁŸêŸ Áfl∑§À¬Ê¥ ‚ ©Áøà ©ûÊ⁄U ∑§Ê øÿŸ ∑§⁄UÊ– (e) Water vascular system is characteristic of
(1) (c), (d) ∞fl¥ (e) ‚„Ë „Ò¥– Echinoderms.
(2) (a), (b) ∞fl¥ (c) ‚„Ë „Ò¥– Choose the correct answer from the options given
below.
(3) (a), (d) ∞fl¥ (e) ‚„Ë „Ò¥–
(1) (c), (d) and (e) are correct
(4) (b), (c) ∞fl¥ (e) ‚„Ë „Ò¥–
(2) (a), (b) and (c) are correct
174. «UÊÚ’‚Ÿ ß∑§Ê߸ Á∑§‚ ∑§Ë ◊Ê≈UÊ߸ ◊ʬŸ ∑§ Á‹∞ ©¬ÿÊª ∑§Ë ¡ÊÃË (3) (a), (d) and (e) are correct
„Ò? (4) (b), (c) and (e) are correct
(1) ‚Ë ∞»§ ‚Ë 174. Dobson units are used to measure thickness of :
(2) ‚◊Ãʬ◊¥«U‹ (1) CFCs
(3) •Ê¡$ ÊŸ (2) Stratosphere
(3) Ozone
(4) ˇÊÊ÷◊¥«U‹
(4) Troposphere
175. ¬Ë ‚Ë •Ê⁄U ∑§ ©¬ÿÊª ‚ ¡ËŸ ¬˝flœ¸Ÿ ∑§Ë •Á÷Á∑˝§ÿÊ ∑§ ŒÊÒ⁄UÊŸ 175. During the process of gene amplification using
ÿÁŒ •Ê⁄¥U÷ ◊¥ ©ëø Ãʬ◊ÊŸ ’ŸÊ Ÿ„Ë¥ ⁄U„ÃÊ Ã’ ÁŸêŸ ◊¥ PCR, if very high temperature is not maintained
¬Ë ‚Ë •Ê⁄U ∑§Ê ∑§ÊÒŸ ‚Ê ø⁄UáÊ ¬„‹ ¬˝÷ÊÁflà „ÊªÊ? in the beginning, then which of the following steps
(1) ÃʬʟȇÊË‹Ÿ of PCR will be affected first ?
(2) ¬˝‚Ê⁄U (1) Annealing
(2) Extension
(3) ÁŸÁc∑˝§ÿ∑§⁄UáÊ
(3) Denaturation
(4) ‹Êߪ‡ÊŸ
(4) Ligation
176. •ŸÈÁøà ÿÈÇ◊ ∑§Ê ¬„øÊÁŸ∞– 176. Identify the incorrect pair.
(1) ∞À∑Ò§‹ÊÚß«U˜‚ - ∑§Ê«UËŸ (1) Alkaloids - Codeine
(2) Áfl· - ∞’Á⁄UŸ (2) Toxin - Abrin
(3) ‹ÒÁÄ≈UŸ - ∑§ÊŸ∑Ò§ŸÊflÁ‹Ÿ A (3) Lectins - Concanavalin A
(4) «˛Uª - Á⁄UÁ‚Ÿ (4) Drugs - Ricin

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177. ‘AB’ ⁄UÄà ‚◊Í„ ∑§ √ÿÁÄÃÿÊ¥ ∑§Ê ‚fl¸ ª˝Ê„Ë ÄÿÊ¥ ∑§„Ã „Ò¥? 177. Persons with ‘AB’ blood group are called as
(1) RBC ∑§Ë ‚Ä ¬⁄U ¬˝Á០A ∞fl¥ B ∑§Ë •ŸÈ¬ÁSÕÁà “Universal recipients”. This is due to :
∑§ ∑§Ê⁄UáÊ (1) Absence of antigens A and B on the surface
(2) å‹Êï◊Ê ◊¥ ¬˝Á០A ∞fl¥ B ∑§Ë •ŸÈ¬ÁSÕÁà ∑§ ∑§Ê⁄UáÊ of RBCs
(3) RBC ◊¥ ∞¥≈UË-A ∞fl¥ ∞¥≈UË-B ¬˝ÁÃ⁄UˇÊË ∑§Ë ©¬ÁSÕÁà ∑§ (2) Absence of antigens A and B in plasma
∑§Ê⁄UáÊ (3) Presence of antibodies, anti-A and anti-B,
on RBCs
(4) å‹Êï◊Ê ◊¥ ∞¥≈UË-A ∞fl¥ ∞¥≈UË-B ¬˝ÁÃ⁄UˇÊË ∑§Ë •ŸÈ¬ÁSÕÁÃ
(4) Absence of antibodies, anti-A and anti-B, in
∑§ ∑§Ê⁄UáÊ
plasma
178. ÁŸêŸ ◊¥ ∑§ÊÒŸ ◊ÁS∑§«UË ∑ȧ‹ ◊¥ •ÊÃÊ „Ò?
178. Which one of the following belongs to the family
(1) ¡ÈªŸÍ Muscidae ?
(2) Á≈UaÊ (1) Fire fly
(3) ÁËø^Ê (2) Grasshopper
(4) ◊ÄπË (3) Cockroach
179. ÃÊ⁄U∑§∑¥§Œ˝ ∑§’ ÁmªÈÁáÊà „ÊÃÊ „Ò? (4) House fly
(1) S-¬˝ÊflSÕÊ 179. The centriole undergoes duplication during :
(2) ¬ÍflʸflSÕÊ (1) S-phase
(3) ◊äÿÊflSÕÊ (2) Prophase
(4) G2 ¬˝ÊflSÕÊ (3) Metaphase
(4) G2 phase
180. SßœÊÁ⁄UÿÊ¥ ◊¥ ‡ÊÈ∑˝§ÊáÊÈ ’¥œŸ ∑§ ª˝Ê„Ë ∑§„Ê° ©¬ÁSÕà „ÊÃ „Ò¥?
(1) ∑§Ê⁄UÊŸÊ ⁄UÁ«U∞≈UÊ 180. Receptors for sperm binding in mammals are
present on :
(2) ¬ËÃ∑§ Á¤ÊÀ‹Ë (1) Corona radiata
(3) ¬Á⁄U¬ËÃ∑§ •fl∑§Ê‡Ê (2) Vitelline membrane
(4) ¡ÊŸÊ ¬ÀÿÈÁ‚«UÊ (3) Perivitelline space
181. ∑§ÊÒŸ ‚Ê ∞¥¡Êß◊ ÁŸÁc∑˝§ÿ »§ÊßÁ’˝ŸÊ¡Ÿ ∑§Ê »§ÊßÁ’˝Ÿ ◊¥ ’Œ‹Ÿ (4) Zona pellucida
∑§ Á‹∞ ©ûÊ⁄UŒÊÿË „Ò? 181. Which enzyme is responsible for the conversion of
(1) Õ˝ÊÁ ê’Ÿ inactive fibrinogens to fibrins ?
(2) ⁄UÁŸŸ (1) Thrombin
(3) ∞Á¬Ÿ»§˝ ËŸ (2) Renin
(4) Õ˝Êê ’Ê∑§Êߟ¡ (3) Epinephrine
182. ‚ÍøË - I ∑§Ê ‚ÍøË - II ∑§ ‚ÊÕ Á◊‹ÊŸ ∑§⁄UÊ– (4) Thrombokinase
182. Match List - I with List - II.
‚ÍøË - I ‚ÍøË - II
List - I List - II
(a) ∞S¬⁄UÁ¡‹‚ ŸÊߪ⁄U (i) ∞‚ËÁ≈U∑§ •ï‹•ê‹
(a) Aspergillus niger (i) Acetic Acid
(b) ∞‚Ë≈UÊ’ÒÄ≈U⁄U ∞Á‚≈UÊ߸ •ê‹
(ii) ‹ÒÁÄ≈U∑§ •ï‹
(b) Acetobacter aceti (ii) Lactic Acid
(c) Ä‹ÊS ≈˛U ËÁ«Uÿ◊ éÿÍ≈UÊÿÁ‹∑§◊ (iii) Á‚Á≈˛U∑§ •ï‹
•ê‹ (c) Clostridium butylicum (iii) Citric Acid
(d) ‹ÒÄ≈UÊ’ÒÁ‚‹‚ •ê‹
(iv) éÿÍÁ≈Á⁄U∑§ •ï‹ (d) Lactobacillus (iv) Butyric Acid

ÁŸêŸ Áfl∑§À¬Ê¥ ‚ ©Áøà ©ûÊ⁄U ∑§Ê øÿŸ ∑§⁄UÊ — Choose the correct answer from the options given
below.
(a) (b) (c) (d)
(a) (b) (c) (d)
(1) (iii) (i) (iv) (ii)
(2) (i) (ii) (iii) (iv) (1) (iii) (i) (iv) (ii)
(3) (ii) (iii) (i) (iv) (2) (i) (ii) (iii) (iv)
(4) (iv) (ii) (i) (iii) (3) (ii) (iii) (i) (iv)
183. ßÁ⁄UÕ˝Ê¬Ê߸Á≈UŸ „Ê◊Ê¸Ÿ ¡Ê •Ê⁄U.’Ë.‚Ë. ∑§ ÁŸ◊ʸáÊ ∑§Ê ¬˝Á⁄Uà ∑§⁄UÃÊ (4) (iv) (ii) (i) (iii)
„Ò ©‚∑§Ê ©à¬ÊŒŸ ∑§ÊÒŸ ∑§⁄UÃÊ „Ò? 183. Erythropoietin hormone which stimulates R.B.C.
(1) •ÇŸÊ‡Êÿ ∑§Ë α-∑§ÊÁ‡Ê∑§Ê∞° formation is produced by :
(1) Alpha cells of pancreas
(2) ⁄UÊS≈˛U‹ ∞Á«UŸÊ„Ê߬Ê»§ÊßÁ‚‚ ∑§Ë ∑§ÊÁ‡Ê∑§Ê∞° (2) The cells of rostral adenohypophysis
(3) •ÁSÕ ◊í¡Ê ∑§Ë ∑§ÊÁ‡Ê∑§Ê∞° (3) The cells of bone marrow
(4) flÎÄ∑§ ∑§Ë ¡ÄS≈UÊÇ‹Ê◊L§‹⁄U ∑§ÊÁ‡Ê∑§Ê∞° (4) Juxtaglomerular cells of the kidney

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184. ‚ÍøË - I ∑§Ê ‚ÍøË - II ∑§ ‚ÊÕ Á◊‹ÊŸ ∑§ËÁ¡∞– 184. Match List - I with List - II.

‚ÍøË - I ‚ÍøË - II List - I List - II

ª÷ʸ‡Êÿ ª˝ËflÊ ◊¥ ‡ÊÈ∑˝§ÊáÊÈ ∑§ Entry of sperm through
(a) Vaults (i)
(a) flÊÚÀ ≈U (i) Cervix is blocked
¬˝fl‡Ê ∑§Ê ⁄UÊ∑§Ã „Ò¥–
(b) IUDs (ii) Removal of Vas deferens
(b) •Ê߸ ÿÍ «UË (ii) ‡ÊÈ∑˝§flÊ„∑§ ∑§Ê „≈UÊŸÊ
Phagocytosis of sperms
‡ÊÈ∑˝§flÊ„∑§ ª÷ʸ‡Êÿ ◊¥ ‡ÊÈ∑˝§ÊáÊÈ•Ê¥ ∑§Ë (c) Vasectomy (iii)
within the Uterus
(c) (iii)
©ë¿UŒŸ ÷ˇÊ∑§ÊáÊÈÁ∑˝§ÿÊ
(d) Tubectomy (iv) Removal of fallopian tube
Á«Uêï’flÊÁ„ŸË ŸÁ‹∑§Ê ∑§Ê
(d) ŸÁ‹∑§Ê ©ë¿UŒŸ (iv)
„≈UÊŸÊ Choose the correct answer from the options given
below.
ÁŸêŸ Áfl∑§À¬Ê¥ ‚ ‚„Ë ©ûÊ⁄U ∑§Ê øÿŸ ∑§⁄UÊ– (a) (b) (c) (d)
(a) (b) (c) (d) (1) (iv) (ii) (i) (iii)
(1) (iv) (ii) (i) (iii) (2) (i) (iii) (ii) (iv)
(2) (i) (iii) (ii) (iv) (3) (ii) (iv) (iii) (i)
(3) (ii) (iv) (iii) (i) (4) (iii) (i) (iv) (ii)
(4) (iii) (i) (iv) (ii) 185. If Adenine makes 30% of the DNA molecule, what
185. ÿÁŒ ∞∑§ «UË ∞Ÿ ∞ •áÊÈ ◊¥ ∞«UŸËŸ ∑§Ë ◊ÊòÊÊ 30% „Ò Ã’ will be the percentage of Thymine, Guanine and
ÕÊÿ◊ËŸ, ÇflÊŸËŸ ∞fl¥ ‚Êß≈UÊ‚ËŸ Á∑§ÃŸ ¬˝ÁÇÊà „Ê¥ª? Cytosine in it ?
(1) T : 20 ; G : 30 ; C : 20
(1) T : 20 ; G : 30 ; C : 20
(2) T : 20 ; G : 20 ; C : 30
(2) T : 20 ; G : 20 ; C : 30 (3) T : 30 ; G : 20 ; C : 20
(3) T : 30 ; G : 20 ; C : 20 (4) T : 20 ; G : 25 ; C : 25
(4) T : 20 ; G : 25 ; C : 25
Section - B (Biology : Zoology)
•ŸÈ÷ʪ - B (¡ËflÁflôÊÊŸ — ¬˝ÊÁáÊÁflôÊÊŸ) 186. Which of the following secretes the hormone,
186. ‚ª÷¸ÃÊ ∑§ ©ûÊ⁄UÊœ¸ ∑§Ë •flÁœ ◊¥ ∑§ÊÒŸ Á⁄U‹ÒÁÄ‚Ÿ „Ê◊Ê¸Ÿ ∑§Ê relaxin, during the later phase of pregnancy ?
(1) Graafian follicle
dÊÁflà ∑§⁄UÃÊ „Ò? (2) Corpus luteum
(1) ª˝Ê»§Ë ¬È≈U∑§ (3) Foetus
(2) ¬Ëà Á¬¥«U (4) Uterus
(3) ÷Í̋áÊ 187. Identify the types of cell junctions that help to stop
the leakage of the substances across a tissue and
(4) ª÷ʸ‡Êÿ facilitation of communication with neighbouring
187. ∑§ÊÁ‡Ê∑§Ê ‚¥ÁœÿÊ¥ ∑§Ê ¬„øÊÁŸ∞ ¡Ê ¬ŒÊÕÊZ ∑§Ê ™§Ã∑§ ‚ ’Ê„⁄U cells via rapid transfer of ions and molecules.
ÁŸ∑§‹Ÿ ‚ ⁄UÊ∑§ÃË „Ò¥ ∞fl¥ •ÊÿŸÊ¥ ∞fl¥ •áÊÈ•Ê¥ ∑§ ÃÈ⁄U¥Ã SÕÊŸÊ¥Ã⁄UáÊ (1) Gap junctions and Adhering junctions,
‚ ¬«UÊ‚Ë ∑§ÊÁ‡Ê∑§Ê•Ê¥ ∑§ ◊äÿ ‚¥ø⁄UáÊ SÕÊÁ¬Ã ∑§⁄UÃË „Ò¥ — respectively.
(2) Tight junctions and Gap junctions,
(1) ∑˝§◊‡Ê— •¥Ã⁄UÊ‹Ë ‚¥Áœ ∞fl¥ •Ê‚¥¡Ë ‚¥Áœ respectively.
(2) ∑˝§◊‡Ê— ŒÎ…∏ ‚¥Áœ ∞fl¥ •¥Ã⁄UÊ‹Ë ‚¥Áœ (3) Adhering junctions and Tight junctions,
(3) ∑˝§◊‡Ê— •Ê‚¥¡Ë ‚¥Áœ ∞fl¥ ŒÎ…∏ ‚¥Áœ respectively.
(4) ∑˝§◊‡Ê— •Ê‚¥¡Ë ‚¥Áœ ∞fl¥ •¥Ã⁄UÊ‹Ë ‚¥Áœ (4) Adhering junctions and Gap junctions,
respectively.
188. ÁŸêŸ ◊¥ Á„S≈UÊŸ ∑§ Áfl·ÿ ◊¥ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ª‹Ã „Ò?
188. Which one of the following statements about
(1) Á„S≈UÊŸ ‚¥ªÁ∆Uà „Ê∑§⁄U 8 •áÊÈ•Ê¥ ∑§Ë ∞∑§ ß∑§Ê߸ ’ŸÊÃ Histones is wrong ?
„Ò¥– (1) Histones are organized to form a unit of
(2) Á„S≈UÊŸ ∑§Ë pH Á∑¥§Áøà •ê‹Ëÿ „ÊÃË „Ò– 8 molecules.
(3) Á„S≈UÊŸ ◊¥ ‹ÊßÁ‚Ÿ ∞fl¥ •ÊÁ¡¸ÁŸŸ ∞◊ËŸÊ •ê‹ ¬˝øÈ⁄U (2) The pH of histones is slightly acidic.
(3) Histones are rich in amino acids - Lysine
„ÊÃ „Ò¥– and Arginine.
(4) Á„S≈UÊŸ ∑§Ë ¬Ê‡fl¸ oÎ¥π‹Ê ◊¥ œŸÊà◊∑§ •Êfl‡Ê „ÊÃÊ „Ò– (4) Histones carry positive charge in the side
chain.
189. ◊ÊŸfl ◊¥ ¬˝‚fl ∑§ •Ê⁄¥U÷ ∑§ Á‹∞ ÁŸêŸ ◊¥ ∑§ÊÒŸ ◊„àfl¬Íáʸ 189. Which of these is not an important component of
•flÿfl Ÿ„Ë¥ „Ò? initiation of parturition in humans ?
(1) ∞S≈˛UÊ¡Ÿ ∞fl¥ ¬˝Ê¡S≈U⁄UÊŸ ∑§ •ŸÈ¬Êà ◊¥ flÎÁh (1) Increase in estrogen and progesterone ratio
(2) ¬˝ÊS≈UÊÇ‹Ò¥Á«UŸ ∑§Ê ‚¥‡‹·áÊ (2) Synthesis of prostaglandins
(3) •ÊÚÄ‚Ë≈UÊÁ‚Ÿ ∑§Ê ◊ÊøŸ (3) Release of Oxytocin
(4) ¬˝Ê‹ÒÁÄ≈UŸ ∑§Ê ◊ÊøŸ (4) Release of Prolactin

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190. ‚ÍøË - I ∑§Ê ‚ÍøË - II ∑§ ‚ÊÕ Á◊‹ÊŸ ∑§⁄UÊ– 190. Match List - I with List - II.
List - I List - II
‚ÍøË - I ‚ÍøË - II
(a) Allen's Rule (i) Kangaroo rat
(a) ∞‹Ÿ ∑§Ê ÁŸÿ◊ (i) ∑¥§ªÊM§ øÍ„Ê Physiological
(b) (ii) Desert lizard
‡Ê⁄UË⁄U Á∑˝§ÿÊà◊∑§ adaptation
(b) (ii) ◊L§SÕ‹Ë Á¿U¬∑§‹Ë Behavioural
•ŸÈ∑ͧ‹Ÿ (c) (iii) Marine fish at depth
adaptation
√ÿfl„ÊÁ⁄U∑§
(c) (iii) ª„⁄UÊ߸ ◊¥ ‚◊ÈŒ˝Ë ◊¿U‹ Ë (d)
Biochemical
(iv) Polar seal
•ŸÈ∑ͧ‹Ÿ adaptation
¡Òfl⁄U‚ÊÿÁŸ∑§ Choose the correct answer from the options given
(d) (iv) œÈ̋flËÿ ÷Ê‹Í
•ŸÈ∑ͧ‹Ÿ below.
(a) (b) (c) (d)
ŸËø ÁŒ∞ ªÿ Áfl∑§À¬Ê¥ ‚ ©Áøà ©ûÊ⁄U ∑§Ê øÿŸ ∑§⁄UÊ– (1) (iv) (ii) (iii) (i)
(a) (b) (c) (d) (2) (iv) (i) (iii) (ii)
(1) (iv) (ii) (iii) (i) (3) (iv) (i) (ii) (iii)
(2) (iv) (i) (iii) (ii) (4) (iv) (iii) (ii) (i)
(3) (iv) (i) (ii) (iii) 191. Match List - I with List - II.
(4) (iv) (iii) (ii) (i)
List - I List - II
191. ‚ÍøË - I ∑§Ê ‚ÍøË - II ∑§ ‚ÊÕ Á◊‹ÊŸ ∑§⁄UÊ– Haemophilus
(a) Filariasis (i)
‚ÍøË - I ‚ÍøË - II influenzae
(a) »§Êß‹Á⁄U∞Á‚‚ (i) „Ë◊ÊÁ»§À‚ ߥç (b) Amoebiasis (ii) Trichophyton
ï‹È∞°¡Ë
(c) Pneumonia (iii) Wuchereria bancrofti
(b) •◊Ë’ÃÊ (ii) ≈˛UÊß∑§Ê»§Êß≈UÊÚŸ
(d) Ringworm (iv) Entamoeba histolytica
(c) ãÿÍ◊ÊÁŸÿÊ (iii) flÈø⁄UÁ⁄UÿÊ ’Ò¥∑˝§Êï
ç≈UÊ߸ Choose the correct answer from the options given
(d) Á⁄U¥ª fl◊¸ (iv) ∞¥≈U•◊Ë’Ê Á„S≈UÊÁ‹Á≈U∑§Ê below.
(a) (b) (c) (d)
ÁŸêŸ Áfl∑§À¬Ê¥ ‚ ©Áøà ©ûÊ⁄U ∑§Ê øÿŸ ∑§⁄UÊ– (1) (iv) (i) (iii) (ii)
(a) (b) (c) (d) (2) (iii) (iv) (i) (ii)
(1) (iv) (i) (iii) (ii) (3) (i) (ii) (iv) (iii)
(2) (iii) (iv) (i) (ii) (4) (ii) (iii) (i) (iv)
(3) (i) (ii) (iv) (iii) 192. Which of the following is not a step in Multiple
(4) (ii) (iii) (i) (iv) Ovulation Embryo Transfer Technology
(MOET) ?
192. ÁŸêŸ ◊¥ ∑§ÊÒŸ ◊À≈Uˬ‹ •ÊflÍ‹‡ÊŸ ∞ê’˝ÿÊ ≈˛UÊ¥‚»§⁄U Ã∑§ŸË∑§ (1) Cow is administered hormone having LH
(MOET) ∑§Ê ø⁄UáÊ Ÿ„Ë¥ „Ò? like activity for super ovulation
(2) Cow yields about 6-8 eggs at a time
(1) ªÊÿ ◊¥ ©ëø •¥«UÊà‚¡¸Ÿ ∑§ Á‹∞ LH ¡Ò‚Ë Á∑˝§ÿÊ (3) Cow is fertilized by artificial insemination
flÊ‹Ê „Ê◊Ê¸Ÿ ÁŒÿÊ ¡ÊÃÊ „Ò– (4) Fertilized eggs are transferred to surrogate
(2) ªÊÿ ∞∑§ ‚◊ÿ ◊¥ ‹ª÷ª 6-8 •¥«U ©àåÊ㟠∑§⁄UÃË „Ò– mothers at 8-32 cell stage
(3) ªÊÿ ∑§Ê ∑ΧÁòÊ◊ flËÿ¸‚øŸ mÊ⁄UÊ ÁŸ·Áøà Á∑§ÿÊ ¡ÊÃÊ „Ò– 193. Statement I :
(4) ÁŸ·Áøà •¥«U 8-32 ∑§ÊÁ‡Ê∑§Ê •flSÕÊ ◊¥ ¬˝ÁÃÁŸÿÈÄà The codon ‘AUG’ codes for methionine and
◊ÊŒÊ ◊¥ SÕÊŸÊ¥ÃÁ⁄Uà Á∑§ÿ ¡ÊÃ „Ò¥– phenylalanine.
Statement II :
193. ∑§ÕŸ I : ¬˝∑ͧ≈U ‘AUG’ ◊ËÁÕÿÊŸËŸ ∞fl¥ »§ÁŸ‹-∞‹ÊÁŸŸ ‘AAA’ and ‘AAG’ both codons code for the amino
∑§Ê ∑ͧ≈U ‹πŸ ∑§⁄UÃÊ „Ò– acid lysine.
In the light of the above statements, choose the
∑§ÕŸ II : ‘AAA’ ∞fl¥ ‘AAG’ ŒÊŸÊ¥ ¬˝∑ͧ≈U ∞◊ËŸÊ •ê‹ correct answer from the options given below.
‹ÊßÁ‚Ÿ ∑§Ê ∑ͧ≈U ‹πŸ ∑§⁄UÃ „Ò¥– (1) Both Statement I and Statement II are
©¬ÿȸÄà ∑§ÕŸÊ¥ ∑§ •ÊœÊ⁄U ¬⁄U ÁŸêŸ Áfl∑§À¬Ê¥ ‚ ‚„Ë ©ûÊ⁄U ∑§Ê true
øÿŸ ∑§⁄UÊ– (2) Both Statement I and Statement II are
(1) ŒÊŸÊ¥ ∑§ÕŸ I ∞fl¥ ∑§ÕŸ II ‚„Ë „Ò¥– false
(3) Statement I is correct but Statement II
(2) ŒÊŸÊ¥ ∑§ÕŸ I ∞fl¥ ∑§ÕŸ II ª‹Ã „Ò¥– is false
(3) ∑§ÕŸ I ‚„Ë „Ò ‹Á∑§Ÿ ∑§ÕŸ II ª‹Ã „Ò– (4) Statement I is incorrect but
(4) ∑§ÕŸ I ª‹Ã „Ò ‹Á∑§Ÿ ∑§ÕŸ II ‚„Ë „Ò– Statement II is true

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194. ‚ÍøË - I ∑§Ê ‚ÍøË - II ∑§ ‚ÊÕ Á◊‹ÊŸ ∑§⁄UÊ– 194. Match List - I with List - II.

‚ÍøË - I ‚ÍøË - II List - I List - II
(a) Scapula (i) Cartilaginous joints
(a) S∑Ò§¬È‹ Ê (i) ©¬ÊÁSÕ ÿÈÄà ¡Ê«∏
(b) Cranium (ii) Flat bone
(b) ∑§¬Ê‹ (ii) ø¬≈UË •ÁSÕ (c) Sternum (iii) Fibrous joints
(c) ©⁄UÊÁSÕ (iii) ⁄U‡ÊËÿ ¡Ê«∏ (d) Vertebral column (iv) Triangular flat bone
Choose the correct answer from the options given
(d) ∑§‡ÊL§∑§ Œ¥«U (iv) ÁòÊ÷È¡Ê∑§Ê⁄U ø¬≈UË •ÁSÕ
below.
ÁŸêŸ Áfl∑§À¬Ê¥ ◊¥ ‚ ©Áøà ©ûÊ⁄U ∑§Ê øÿŸ ∑§⁄UÊ– (a) (b) (c) (d)
(a) (b) (c) (d) (1) (i) (iii) (ii) (iv)
(1) (i) (iii) (ii) (iv) (2) (ii) (iii) (iv) (i)
(2) (ii) (iii) (iv) (i) (3) (iv) (ii) (iii) (i)
(3) (iv) (ii) (iii) (i) (4) (iv) (iii) (ii) (i)
(4) (iv) (iii) (ii) (i) 195. The Adenosine deaminase deficiency results into :
195. ∞«UËŸÊÁ‚Ÿ Á«U∞◊ËŸ¡ ∑§Ë ∑§◊Ë ‚ ÄÿÊ „ÊÃÊ „Ò? (1) Dysfunction of Immune system
(1) ¬˝ÁÃ⁄UˇÊË Ã¥òÊ ∑§Êÿ¸„ËŸ „ÊŸÊ (2) Parkinson’s disease
(2) ¬ÊÁ∑¸§ã‚Ÿ ⁄UÊª (3) Digestive disorder
(3) ¬ÊøŸ Áfl∑§Ê⁄U (4) Addison’s disease
(4) ∞«UË‚Ÿ ⁄UÊª 196. Assertion (A) :
196. ∑§ÕŸ (A) : ¡’ ∞∑§ √ÿÁÄà ©ëø ÃÈ¥ªÃÊ ¬⁄U ¡ÊÃÊ „Ò Ã’ A person goes to high altitude and experiences
fl„ ÃÈ¥ªÃÊ ’Ë◊Ê⁄UË ∑§ ‹ˇÊáÊ ¡Ò‚ ‚Ê¥‚ ‹Ÿ ◊¥ ‘altitude sickness’ with symptoms like breathing
∑§Á∆UŸÊ߸ ∞fl¥ NŒÿ ∑§Ë œ«∏∑§Ÿ ’…∏ŸÊ ◊„‚Í‚ difficulty and heart palpitations.
Reason (R) :
∑§⁄UÃÊ „Ò–
Due to low atmospheric pressure at high altitude,
∑§Ê⁄UáÊ (R) : ©ëø ÃÈ¥ªÃÊ ¬⁄U ÁŸêŸ flÊÿÈ◊¥«U‹Ëÿ ŒÊ’ ∑§ ∑§Ê⁄UáÊ the body does not get sufficient oxygen.
‡Ê⁄UË⁄U ∑§Ê ¬ÿʸåà •ÊÚÄ‚Ë¡Ÿ Ÿ„Ë¥ Á◊‹ ¬ÊÃË– In the light of the above statements, choose the
©¬ÿ¸ÈÄà ∑§ÕŸÊ¥ ∑§ ¬˝∑§Ê‡Ê ◊¥ ÁŸêŸ Áfl∑§À¬Ê¥ ◊¥ ‚ ‚„Ë ©ûÊ⁄U ∑§Ê correct answer from the options given below.
øÿŸ ∑§⁄UÊ– (1) Both (A) and (R) are true and (R) is the
(1) ŒÊŸÊ¥ (A) ∞fl¥ (R) ‚àÿ „Ò¥ ∞fl¥ (R), (A) ∑§Ë ©Áøà correct explanation of (A)
(2) Both (A) and (R) are true but (R) is not the
√ÿÊÅÿÊ „Ò–
correct explanation of (A)
(2) ŒÊŸÊ¥ (A) ∞fl¥ (R) ‚àÿ „Ò¥ ‹Á∑§Ÿ (R), (A) ∑§Ë (3) (A) is true but (R) is false
©ÁøÃ √ÿÊÅÿÊ Ÿ„Ë¥ „Ò– (4) (A) is false but (R) is true
(3) (A) ‚àÿ „Ò ‹Á∑§Ÿ (R) •‚àÿ „Ò–
197. Match List - I with List - II.
(4) (A) •‚àÿ „Ò ‹Á∑§Ÿ (R) ‚àÿ „Ò–
List - I List - II
197. ‚ÍøË - I ∑§Ê ‚ÍøË - II ∑§ ‚ÊÕ Á◊‹ÊŸ ∑§⁄UÊ–
Selection of resistant
‚ÍøË - I ‚ÍøË - II (a)
Adaptive
(i)
varieties due to excessive
‡ÊÊ∑§ŸÊ‡ÊË ∞fl¥ ¬Ë«∏∑§ŸÊ‡ÊË radiation use of herbicides and
pesticides
(a) A•ŸÈ∑ͧ‹Ë ÁflÁ∑§⁄UáÊ (i) ∑§ •àÿÁœ∑§ ©¬ÿÊª ‚
Convergent Bones of forelimbs in Man
¬˝ÁÃ⁄UÊœË Á∑§S◊Ê¥ ∑§Ê øÿŸ (b) (ii)
evolution and Whale
◊ŸÈcÿ ∞fl¥ √„‹ ∑§ •ª˝¬ÊŒ Divergent Wings of Butterfly and
(b) •Á÷‚Ê⁄UË Áfl∑§Ê‚ (ii) (c) (iii)
∑§Ë •ÁSÕÿÊ° evolution Bird
Evolution by
(c) •¬‚Ê⁄UË Áfl∑§Ê‚ (iii) ÁÃÃ‹Ë ∞fl¥ ¬ˇÊË ∑§ ¬¥π
(d) anthropo- (iv) Darwin Finches
◊ÊŸflËÿ Á∑˝§ÿÊ•Ê¥ genic action
(d) (iv) «UÊÁfl¸Ÿ ∑§Ë Á»¥§ø
∑§ mÊ⁄UÊ Áfl∑§Ê‚ Choose the correct answer from the options given
ÁŸêŸ Áfl∑§À¬Ê¥ ‚ ©Áøà ©ûÊ⁄U ∑§Ê øÿŸ ∑§⁄UÊ– below.
(a) (b) (c) (d) (a) (b) (c) (d)
(1) (iv) (iii) (ii) (i) (1) (iv) (iii) (ii) (i)
(2) (iii) (ii) (i) (iv) (2) (iii) (ii) (i) (iv)
(3) (ii) (i) (iv) (iii) (3) (ii) (i) (iv) (iii)
(4) (i) (iv) (iii) (ii) (4) (i) (iv) (iii) (ii)

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198. Á‹Á¬«U ‚ ‚¥’¥ÁœÃ ∑§ÕŸ ŸËø ÁŒ∞ ª∞ „Ò– 198. Following are the statements with reference to
‘lipids’.
(a) ∞‚ Á‹Á¬«U Á¡Ÿ◊¥ ∑§fl‹ ∞∑§‹ •Ê’¥œ „ÊÃ „Ò¥ ©ã„¥
•‚¥ÃÎåà fl‚Ê •ê‹ ∑§„Ã „Ò¥– (a) Lipids having only single bonds are called
unsaturated fatty acids.
(b) ‹Á‚ÁÕŸ »§ÊÚS»§ÊÁ‹Á¬«U „Ò– (b) Lecithin is a phospholipid.
(c) ≈˛UÊß„Êß«˛UÊÚÄ‚Ë ¬˝ÊÚ¬Ÿ ÁÇ‹‚⁄UÊÚ‹ „Ò– (c) Trihydroxy propane is glycerol.
(d) ¬ÊÁÀ◊Á≈U ∑ § •ê‹ ◊ ¥ ∑§Ê’Ê ¸ Á Ä‚‹ ∑§Ê’¸ Ÿ ‚Á„à (d) Palmitic acid has 20 carbon atoms including
20 ∑§Ê’¸Ÿ ∑§ ¬⁄U◊ÊáÊÈ „ÊÃ „Ò¥– carboxyl carbon.
(e) ∞⁄UÁ∑§«UÊÁŸ∑§ •ê‹ ◊¥ 16 ∑§Ê’¸Ÿ ¬⁄U◊ÊáÊÈ „ÊÃ „Ò¥– (e) Arachidonic acid has 16 carbon atoms.
ÁŸêŸ Áfl∑§À¬Ê¥ ‚ ©Áøà ©ûÊ⁄U ∑§Ê øÿŸ ∑§⁄UÊ– Choose the correct answer from the options given
below.
(1) ∑§fl‹ (a) ∞fl¥ (b)
(1) (a) and (b) only
(2) ∑§fl‹ (c) ∞fl¥ (d)
(2) (c) and (d) only
(3) ∑§fl‹ (b) ∞fl¥ (c) (3) (b) and (c) only
(4) ∑§fl‹ (b) ∞fl¥ (e) (4) (b) and (e) only

199. ¬‡ÊË ‚¥∑ȧøŸ ◊¥ ∑§ÊÒŸ ‚Ë ÉÊ≈UŸÊ∞° „ÊÃË „Ò¥? 199. During muscular contraction which of the
following events occur ?
(a) ‘H’ ˇÊòÊ Áfl‹Èåà „Ê ¡ÊÃÊ „Ò
(a) ‘H’ zone disappears
(b) ‘A’ ’Ò¥«U øÊÒ«∏Ê „Ê ¡ÊÃÊ „Ò
(b) ‘A’ band widens
(c) ‘I’ ’Ò¥«U ∑§Ë øÊÒ«∏Ê߸ ∑§◊ „Ê ¡ÊÃË „Ò (c) ‘I’ band reduces in width
(d) ◊ÊÿÊÁ‚Ÿ ATP ∑§Ê ¡‹•¬ÉÊÁ≈Uà ∑§⁄UU ADP ∞fl¥ Pi (d) Myosine hydrolyzes ATP, releasing the ADP
∑§Ê ◊ÊøŸ ∑§⁄UÃÊ „Ò and Pi
(e) ∞ÁÄ≈UŸ ‚ ‚¥‹ÁÇŸÃ Z-⁄UπÊ •ãŒ⁄U ∑§Ë Ã⁄U»§ πË¥ø (e) Z-lines attached to actins are pulled inwards
¡ÊÃË „Ò Choose the correct answer from the options given
ÁŸêŸ Áfl∑§À¬Ê¥ ‚ ‚„Ë ©ûÊ⁄U ∑§Ê øÿŸ ∑§⁄UÊ– below.
(1) ∑§fl‹ (a), (c), (d), (e) (1) (a), (c), (d), (e) only
(2) ∑§fl‹ (a), (b), (c), (d) (2) (a), (b), (c), (d) only
(3) (b), (c), (d), (e) only
(3) ∑§fl‹ (b), (c), (d), (e)
(4) (b), (d), (e), (a) only
(4) ∑§fl‹ (b), (d), (e), (a)
200. Following are the statements about prostomium
200. ∑¥§øÈ∞ ∑§ ¬È⁄UÊ◊Èπ ‚ ‚¥’¥ÁœÃ ∑§ÕŸÊ¥ ◊¥ ÁŒ∞ ª∞ „Ò– of earthworm.
(a) ÿ„ ◊Èπ ∑§Ê …U∑§Ÿ ∑§Ê ∑§Êÿ¸ ∑§⁄UÃÊ „Ò– (a) It serves as a covering for mouth.
(b) ÿ„ »§ÊŸ ◊ÎŒÊ Œ⁄UÊ⁄UÊ¥ ∑§Ê πÊ‹Ÿ ◊¥ ◊ŒŒ ∑§⁄UÃÊ „Ò Á¡‚◊¥ (b) It helps to open cracks in the soil into which
it can crawl.
ÿ„ ⁄¥Uª ‚∑§ÃÊ „Ò–
(c) It is one of the sensory structures.
(c) ÿ„ ∞∑§ ‚¥flŒË ‚¥⁄UøŸÊ „Ò–
(d) It is the first body segment.
(d) ÿ„ ‡Ê⁄UË⁄U ∑§Ê ¬˝Õ◊ π¥«U „Ò– Choose the correct answer from the options given
ÁŸêŸ Áfl∑§À¬Ê¥ ‚ ©Áøà ©ûÊ⁄U ∑§Ê øÿŸ ∑§⁄UÊ– below.
(1) (a), (b) ∞fl¥ (c) ‚„Ë „Ò¥– (1) (a), (b) and (c) are correct
(2) (a), (b) and (d) are correct
(2) (a), (b) ∞fl¥ (d) ‚„Ë „Ò¥–
(3) (a), (b), (c) and (d) are correct
(3) (a), (b), (c) ∞fl¥ (d) ‚„Ë „Ò¥–
(4) (b) and (c) are correct
(4) (b) ∞fl¥ (c) ‚„Ë „Ò¥–

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Hindi+English M4
47
⁄»§ ∑§Êÿ¸ ∑§ Á‹∞ ¡ª„ / Space For Rough Work

M4 Hindi+English
48

ÁŸêŸÁ‹Áπà ÁŸŒ¸‡Ê äÿÊŸ ‚ ¬…∏¥ — Read carefully the following instructions :

6. ¬⁄ˡÊÊ ‚ê¬ÛÊ „ÊŸ ¬⁄, ¬⁄ˡÊÊÕ˸ ∑§ˇÊ/„ÊÚ‹ ¿Ê«∏Ÿ ‚ ¬Ífl¸ ©ûÊ⁄ 6. On completion of the test, the candidate must
hand over the Answer Sheet (ORIGINAL and
¬òÊ (◊Í‹ ¬˝ÁÃÁ‹Á¬ ∞fl¥ ∑§Êÿʸ‹ÿ ¬˝ÁÃÁ‹Á¬) ∑§ˇÊ ÁŸ⁄ˡÊ∑§ OFFICE Copy) to the Invigilator before leaving
∑§Ê •fl‡ÿ ‚ÊÒ¥¬ Œ¥– ¬⁄ˡÊÊÕ˸ •¬Ÿ ‚ÊÕ ¬˝‡Ÿ ¬ÈÁSÃ∑§Ê ‹ ¡Ê the Room/Hall. The candidates are allowed to take
‚∑§Ã „Ò¥– away this Test Booklet with them.

7. ß‚ ¬ÈÁSÃ∑§Ê ∑§Ê ‚¥∑§Ã „Ò M4– ÿ„ ‚ÈÁŸÁ‡øà ∑§⁄ ‹¥ Á∑§ 7. The CODE for this Booklet is M4. Make
sure that the CODE printed on the Original
ß‚ ¬ÈÁSÃ∑§Ê ∑§Ê ‚¥∑§Ã, ©ûÊ⁄ ¬òÊ ∑§ ◊Í‹ ¬˝ÁÃÁ‹Á¬ ¬⁄U ¿Uʬ
Copy of the Answer Sheet is the same as that
ªÿ ‚¥∑§Ã ‚ Á◊‹ÃÊ „Ò– •ª⁄ ÿ„ Á÷ÛÊ „Ê ÃÊ ¬⁄ˡÊÊÕ˸ ŒÍ‚⁄Ë on this Test Booklet. In case of discrepancy, the
¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê •ÊÒ⁄ ©ûÊ⁄ ¬òÊ ‹Ÿ ∑§ Á‹∞ ÁŸ⁄ˡÊ∑§ ∑§Ê ÃÈ⁄¥Uà candidate should immediately report the matter to
•flªÃ ∑§⁄Ê∞¥– the Invigilator for replacement of both the Test
Booklet and the Answer Sheet.
8. ¬⁄ˡÊÊÕ˸ ‚ÈÁŸÁ‡øà ∑§⁄¥ Á∑§ ß‚ ©ûÊ⁄ ¬òÊ ∑§Ê ◊Ê«∏Ê Ÿ ¡Ê∞ ∞fl¥ 8. The candidates should ensure that the Answer
©‚ ¬⁄ ∑§Ê߸ •ãÿ ÁŸ‡ÊÊŸ Ÿ ‹ªÊ∞¥– ¬⁄ˡÊÊÕ˸ •¬ŸÊ •ŸÈ∑˝§◊Ê¥∑§ Sheet is not folded. Do not make any stray marks
¬˝‡Ÿ ¬ÈÁSÃ∑§Ê/©ûÊ⁄ ¬òÊ ◊¥ ÁŸœÊ¸Á⁄à SÕÊŸ ∑§ •ÁÃÁ⁄Äà •ãÿòÊ on the Answer Sheet. Do not write your Roll No.
anywhere else except in the specified space in the
ŸÊ Á‹π¥– Test Booklet/Answer Sheet.
9. ©ûÊ⁄ ¬òÊ ¬⁄ Á∑§‚Ë ¬˝∑§Ê⁄ ∑§ ‚¥‡ÊÊœŸ „ÃÈ √„Êß≈ $ç‹Íß« ∑§ 9. Use of white fluid for correction is NOT permissible
¬˝ÿÊª ∑§Ë •ŸÈ◊Áà Ÿ„Ë¥ „Ò– on the Answer Sheet.
10. Each candidate must show on-demand his/her
10. ¬Í¿ ¡ÊŸ ¬⁄ ¬˝àÿ∑§ ¬⁄ˡÊÊÕ˸, ÁŸ⁄ˡÊ∑§ ∑§Ê •¬ŸÊ ¬˝fl‡Ê-¬òÊ Admit Card to the Invigilator.
ÁŒπÊ∞¥– 11. No candidate, without special permission of the
11. ∑¥§Œ˝ •œËˇÊ∑§ ÿÊ ÁŸ⁄ˡÊ∑§ ∑§Ë Áfl‡Ê· •ŸÈ◊Áà ∑§ Á’ŸÊ ∑§Ê߸ centre Superintendent or Invigilator, would leave
his/her seat.
¬⁄ˡÊÊÕ˸ •¬ŸÊ SÕÊŸ Ÿ ¿Ê«∏¥–
12. The candidates should not leave the Examination
12. ∑§Êÿ¸ ⁄ à ÁŸ⁄ˡÊ∑§ ∑§Ê •¬ŸÊ ©ûÊ⁄ ¬òÊ ÁŒ∞ Á’ŸÊ ∞fl¥ Hall without handing over their Answer Sheet to
©¬ÁSÕÁÃ-¬òÊ∑§ ¬⁄ ŒÈ’Ê⁄UÊ „SÃÊˇÊ⁄ (‚◊ÿ ∑§ ‚ÊÕ) Á∑§∞ the Invigilator on duty and sign (with time) the
Attendance Sheet twice. Cases, where a
Á’ŸÊ ∑§Ê߸ ¬⁄ˡÊÊÕ˸ ¬⁄ˡÊÊ „ÊÚ‹ Ÿ„Ë¥ ¿Ê«∏¥ª– ÿÁŒ Á∑§‚Ë
candidate has not signed the Attendance
¬⁄ˡÊÊÕ˸ Ÿ ŒÍ‚⁄Ë ’Ê⁄ ©¬ÁSÕÁÃ-¬òÊ∑§ ¬⁄ „SÃÊˇÊ⁄ Ÿ„Ë¥ Sheet second time, will be deemed not to
Á∑§∞ ÃÊ ÿ„ ◊ÊŸÊ ¡Ê∞ªÊ Á∑§ ©‚Ÿ ©ûÊ⁄ ¬òÊ Ÿ„Ë¥ ‹ÊÒ≈ÊÿÊ „Ò have handed over the Answer Sheet and
•ÊÒ⁄ ÿ„ •ŸÈÁøà ‚ÊœŸ ∑§Ê ◊Ê◊‹Ê ◊ÊŸÊ ¡Ê∞ªÊ– dealt with as an Unfair Means case.
13. Use of Electronic/Manual Calculator is prohibited.
13. ß‹Ä≈˛ÊÁŸ∑§/„SÃøÊÁ‹Ã ¬Á⁄∑§‹∑§ ∑§Ê ©¬ÿÊª flÁ¡¸Ã „Ò– 14. The candidates are governed by all Rules and
14. ¬⁄ˡÊÊ-∑§ˇÊ/„ÊÚ‹ ◊¥ •Êø⁄áÊ ∑§ Á‹∞ ¬⁄ˡÊÊÕ˸, ¬⁄ˡÊÊ ∑§ Regulations of the examination with regard to their
ÁŸÿ◊Ê¥ ∞fl¥ ÁflÁŸÿ◊Ê¥ mÊ⁄Ê ÁŸÿÁ◊à „Ò¥– •ŸÈÁøà ‚ÊœŸ ∑§ conduct in the Examination Room/Hall. All cases
of unfair means will be dealt with as per the Rules
‚÷Ë ◊Ê◊‹Ê¥ ∑§Ê »Ò§‚‹Ê ß‚ ¬⁄ˡÊÊ ∑§ ÁŸÿ◊Ê¥ ∞fl¥ ÁflÁŸÿ◊Ê¥ ∑§ and Regulations of this examination.
•ŸÈ‚Ê⁄ „ÊªÊ– 15. No part of the Test Booklet and Answer
15. Á∑§‚Ë „Ê‹Êà ◊¥ ¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê •ÊÒ⁄ ©ûÊ⁄ ¬òÊ ∑§Ê ∑§Ê߸ Sheet shall be detached under any
circumstances.
÷ʪ •‹ª Ÿ ∑§⁄¥–
16. The candidates will write the Correct Test Booklet
16. ¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê / ©ûÊ⁄ ¬òÊ ◊¥ ÁŒ∞ ª∞ ¬⁄ˡÊÊ ¬ÈÁSÃ∑§Ê ‚¥∑§Ã ∑§Ê Code as given in the Test Booklet/Answer Sheet
¬⁄ˡÊÊÕ˸ ‚„Ë Ã⁄Ë∑§ ‚ ©¬ÁSÕÁÃ-¬òÊ∑§ ◊¥ Á‹π¥– in the Attendance Sheet.