JEE Main 2018 Question Paper with Answers (15 Apr Evening Shift)

Set - 03 1
PART A — PHYSICS ÷ʪ A — ÷ÊÒÁÃ∑§ ÁflôÊÊŸ cpN A — cp¥rsL$ rhop_
ALL THE GRAPHS/DIAGRAMS GIVEN ARE
ÁŒ∞ ªÿ ‚÷Ë ª˝Ê»§/⁄UπÊ∑ΧÁÃÿÊ° •Ê⁄UπËÿ „Ò¥ sdpd Apg¡M/rQÓp¡ õL$ud¡V$uL$ R>¡ A_¡ õL¡$g âdpZ¡ v$p¡fpe¡gp
SCHEMATIC AND NOT DRAWN TO SCALE.
•ÊÒ⁄U S∑§‹ ∑§ •ŸÈ‚Ê⁄U ⁄UπÊ¥Á∑§Ã Ÿ„Ë¥ „Ò– _\u.
1. The characteristic distance at which
quantum gravitational effects are 1. å‹Ê¥∑§ ŒÍ⁄UË fl„ ÁflÁ‡Êc≈U ŒÍ⁄UË „Ò Á¡‚ ¬⁄U ÄflÊã≈U◊ ªÈL§àflËÿ 1. Äep„ L$¹hp¡ÞV$d-Nyê$ÐhpL$j}e Akfp¡ ANÐe_u lp¡e s¡hy„
significant, the Planck length, can be ¬˝÷Êfl ◊„àfl¬Íáʸ „ÊÃ „Ò¥– ß‚∑§Ê ÁŸœÊ¸⁄UáÊ ◊Í‹÷Íà gpnrZL$ A„sf, àgpÞL$ A„sf; G,
A_¡ c S>¡hp cp¥rsL$
determined from a suitable combination
of the fundamental physical constants G,
÷ÊÒÁÃ∑§ ⁄UÊÁ‡ÊÿÊ¥ G,
ÃÕÊ c ‚ „Ê ‚∑§ÃÊ „Ò– ÁŸêŸ ◊¥ AQmp„L$p¡_p ep¡Áe k„ep¡S>__u dv$v$\u v$ip®hu iL$pe R>¡.

and c. Which of the following correctly ‚ ∑§ÊÒŸ ‚Ê ‚ÍòÊ å‹Ê¥∑§ ŒÍ⁄UË ∑§Ê ‚„Ë ÁŒπÊÃÊ „Ò? _uQ¡ Ap`¡gpdp„\u L$ep¡ rhL$ë` àgpÞL$ A„sf_¡ kpQu
gives the Planck length ?
(1) G
2 c3 fus¡ v$ip®h¡ R>¡ ?
(1) G
2 c3 (1) G
2 c3
(2) G2
c
(2) G2
c 1 (2) G2
c
2
1
(3) G 2
c
1
(3) G 2
2 c (3) G 2
2 c
1
1
 G
 2
 G
 2
(4)  3  1
c   G
 2
(4)  3  (4)  3 
c  c 

SET - 03 ENGLISH PHYSICS SET - 03 HINDI PHYSICS SET - 03 GUJARATI PHYSICS

Set - 03 2
2. A man in a car at location Q on a straight 2. ∞∑§ •ÊŒ◊Ë ∑§Ê⁄U ◊¥ SÕÊŸ Q ‚ ∞∑§ ‚ËœË ‚«∏∑§ ¬⁄U 2. A¡L$ kyf¡M lpCh¡ `f Q õ\p_¡ L$pfdp„ fl¡g A¡L$ ìe[¼s
highway is moving with speed v. He
ªÁà v ‚ ¡Ê ⁄U„Ê „Ò– fl„ πà ∑§ ∞∑§ Á’ãŒÈ P ¬⁄U, ¡Ê v S>¡V$gu TX$`\u Nrs L$f¡ R>¡. ApL©$rsdp„ v$ip®ìep âdpZ¡
decides to reach a point P in a field at a
distance d from the highway (point M) as ÁŒπÊÿ ªÿ ÁøòÊÊŸÈ‚Ê⁄U ‚«∏∑§ ‚ d ŒÍ⁄UË ¬⁄U „Ò s¡ lpCh¡ (tbvy$ M) \u d A„sf¡ fl¡gp M¡sf_p tbvy$ P
shown in the figure. Speed of the car in (Á’¥ŒÈ M), ¬„°ÈøŸ ∑§Ê ÁŸ‡øÿ ∑§⁄UÃÊ „Ò– ∑§Ê⁄U ∑§Ë ApNm `lp¢Qhp_y„ _½$u L$f¡ R>¡. M¡sfdp„ L$pf_u TX$`
the field is half to that on the highway. øÊ‹ πà ◊¥, ‚«∏∑§ ∑§Ë øÊ‹ ∑§Ë •ÊœË „Ò– fl„ ŒÍ⁄UË lpCh¡ `f s¡_u TX$` L$fsp AX$^u R>¡. RM A„sf L¡$V$gy„
What should be the distance RM, so that
RM ÄÿÊ „ÊªË Á¡‚‚ Á∑§ P Ã∑§ ¬„°ÈøŸ ∑§Ê ‚◊ÿ li¡ L¡$ S>¡\u P ky^u `lp¢Qsp gpNsp¡ kde gOyÑd
the time taken to reach P is minimum ?
ãÿÍŸÃ◊ „Ò? \pe ?

(1) d
(1) d
d
(2) d (1) d
2 (2)
2 d
d (2)
(3) 2
2 d
(3) d
2
d (3)
(4) 2
3 d
(4) d
3
(4)
3

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Set - 03 3
3. A body of mass 2 kg slides down with an 3. 30
∑§ÊáÊ ‚ ¤ÊÈ∑§ „È∞ ∞∑§ ÉÊ·¸áÊÿÈÄà ‚◊Ë ¬⁄U ∞∑§ 3. 30
S>¡V$gp¡ Y$pm ^fphsp A¡L$ Y$m¡gp fa¹ kdsg `f\u
acceleration of 3 m/s2 on a rough inclined
2 kg Œ˝√ÿ◊ÊŸ ∑§Ê ∞∑§ Á¬á«U àfl⁄UáÊ 3 m/s2 ‚ ŸËø 2 kg v$m ^fphsp¡ `v$p\® 3 m/s2 _p âh¡N\u _uQ¡
plane having a slope of 30
. The external
force required to take the same body up ∑§Ë •Ê⁄U Á»§‚‹ÃÊ „Ò– ©‚ Á¬á«U ∑§Ê ß‚ ‚◊Ë ¬⁄U sfa kfL¡$ R>¡. Ap S> `v$p\®_¡ Ap S> kdp_ `°h¡N\u
the plane with the same acceleration will ©‚Ë àfl⁄UáÊ ‚ ™§¬⁄U ‹ ¡ÊŸ ∑§ Á‹ÿ ’Ês ’‹ ∑§Ë D`f sfa gC S>hp dpV¡$ gNphhy„ `X$sy„ bpü bm
be : (g=10 m/s2) •Êfl‡ÿ∑§ÃÊ „ÊªË — (ÁŒÿÊ „Ò g=10 m/s2) __________ \i¡. (g=10 m/s2)
(1) 14 N (1) 14 N
(2) 20 N (2) 20 N (1) 14 N
(3) 6 N (3) 6N (2) 20 N
(4) 4 N (4) 4N (3) 6N
(4) 4N
4. A proton of mass m collides elastically with
4. m Œ˝√ÿ◊ÊŸ ∑§Ê ∞∑§ ¬˝Ê≈UÊÚŸ Á∑§‚Ë •ôÊÊà Œ˝√ÿ◊ÊŸ ∑§
a particle of unknown mass at rest. After
Áfl⁄UÊ◊ÊflSÕÊ ◊¥ ⁄Uπ „È∞ ∞∑§ ∑§áÊ ‚ ¬˝àÿÊSÕ ‚¥ÉÊ^ 4. A¡L$ Aops v$m ^fphsp [õ\f L$Z kp\¡ m v$m ^fphsp¡
the collision, the proton and the unknown
particle are seen moving at an angle of 90
∑§⁄UÃÊ „Ò– ‚¥ÉÊ^ ∑§ ¬‡øÊØ, ¬˝Ê≈UÊÚŸ •ÊÒ⁄U •ôÊÊà ∑§áÊ âp¡V$p¡_ [õ\rsõ\p`L$ A\X$pdZ A_ych¡ R>¡. Ap k„Ops
with respect to each other. The mass of ¬⁄US¬⁄U 90
∑§Ê ∑§ÊáÊ ’ŸÊÃ „È∞ ø‹ ¡ÊÃ „Ò¥– •ôÊÊà bpv$, Ap âp¡V$p¡_ A_¡ Aops L$Z A¡L$buÅ_¡ kp`¡n¡
unknown particle is : 90
_p L$p¡Z¡ Nrs L$fsp dpg|d `X¡$ R>¡. Aops L$Z_y„
∑§áÊ ∑§Ê Œ˝√ÿ◊ÊŸ „Ò —
m v$m :
(1) m
2 (1)
2 m
(2) m (1)
(2) m 2
m (2) m
(3) m
3 (3)
3 m
(4) 2m (3)
(4) 2m 3
(4) 2m

SET - 03 ENGLISH PHYSICS SET - 03 HINDI PHYSICS SET - 03 GUJARATI PHYSICS

Set - 03 4
5. A disc rotates about its axis of symmetry 5. ∞∑§ Á«S∑§ •¬Ÿ ‚◊Á◊à •ˇÊ ∑§ ¬Á⁄U× ˇÊÒÁá ‚◊Ë 5. A¡L$ s[¼s s¡_u k„rdrs An_¡ A_ygnu_¡ kdrnrsS>
in a horizontal plane at a steady rate of
◊¥ 3.5 øÄ∑§⁄U ¬˝Áà ‚∑§á«U ∑§Ë ÁSÕ⁄U ªÁà ‚ ÉÊÍáʸŸ kdsgdp„ 3.5 `qfc°dZ ârs k¡L$ÞX$_p [õ\f v$f¡
3.5 revolutions per second. A coin placed
at a distance of 1.25 cm from the axis of ∑§⁄U ⁄U„Ë „Ò– ÉÊÍáʸŸ •ˇÊ ‚ 1.25 cm ∑§Ë ŒÍ⁄UË ¬⁄U ⁄UπÊ `qfc°dZ L$f¡ R>¡. s¡_u c°dZpn \u 1.25 cm A„sf¡
rotation remains at rest on the disc. The ∞∑§ Á‚Ä∑§Ê Á«S∑§ ¬⁄U ÁSÕ⁄U ⁄U„ÃÊ „Ò– Á‚Ä∑§ •ÊÒ⁄U d|L¡$g A¡L$ rk½$p¡ s[¼s `f [õ\f [õ\rsdp„ fl¡ R>¡.
coefficient of friction between the coin and Á«S∑§ ∑§ ’Ëø ◊¥ ÉÊ·¸áÊ ªÈáÊÊ¥∑§ ∑§Ê ◊ÊŸ „ÊªÊ — rk½$p A_¡ s[¼s hÃQ¡_p¡ Oj®Zp„L$ __________ \i¡.
the disc is : (g=10 m/s2) (g=10 m/s2)
(1) 0.5
(ÁŒÿÊ „Ò — g=10 m/s2)
(2) 0.3 (1) 0.5
(2) 0.3 (1) 0.5
(3) 0.7 (2) 0.3
(4) 0.6 (3) 0.7
(4) 0.6 (3) 0.7
(4) 0.6

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Set - 03 5
6. A thin uniform bar of length L and mass 6. L ‹ê’Ê߸ ÃÕÊ 8 m Œ˝√ÿ◊ÊŸ ∑§Ê ∞∑§ ∞∑§‚◊ÊŸ ¬Ã‹Ë 6. A¡L$ L g„bpC_p¡ A_¡ 8 m v$m ^fphsy„ `psmy„ kdp„Nu
8 m lies on a smooth horizontal table. Two
¿U«∏ ∞∑§ Áø∑§Ÿ ˇÊÒÁá ◊$¡ ¬⁄U ⁄UπÊ „Ò– ŒÊ Á’ãŒÈ Qp¡kgy„ A¡L$ Oj®Zfrls kdrnrsS> V¡$bg `f `X¡$gy„ R>¡.
point masses m and 2 m are moving in the
same horizontal plane from opposite sides Œ˝√ÿ◊ÊŸ m ÃÕÊ 2 m ©‚Ë ˇÊÒÁá ‚◊Ë ◊¥ ¿U«∏ ∑§ Qp¡ k gp_u rhê$Ý^ qv$ipdp„ \ u kdp_ kdrnrsS>
of the bar with speeds 2v and v Áfl¬⁄UËà Ã⁄U»§ ‚ ∑˝§◊‡Ê— 2v ÃÕÊ v øÊ‹ ‚ •ÊÃ „Ò¥– kdsgdp„ b¡ qb„vy$hs v$mp¡ m A_¡ 2 m A_y¾$d¡ 2v
respectively. The masses stick to the bar ŒÊŸÊ¥ Œ˝√ÿ◊ÊŸ ‚¥ÉÊ^ ∑§ ’ÊŒ ¿U«∏ ∑§ ∑§ãŒ˝ ‚ ∑˝§◊‡Ê— A_¡ v S>¡V$gu TX$`p¡ kp\¡ Nrs L$f¡ R>¡. Ap k„Ops_¡ A„s¡
L L L L L L
after collision at a distance and
3 6 3
ÃÕÊ 6
ŒÍ⁄UË ¬⁄U Áø¬∑§ ¡ÊÃ „Ò¥– ‚¥ÉÊ^U ∑§ Ap v$mp¡ Qp¡kgp_p L¡$ÞÖ\u A_y¾$d¡ 3
A_¡ 6
A„sf¡
respectively from the centre of the bar. If
the bar starts rotating about its center of
»§‹SflM§¬ ÿÁŒ ¿U«∏ •¬Ÿ Œ˝√ÿ◊ÊŸ ∑¥§Œ˝ ∑§ ‚ʬˇÊ Qp¡kgp_¡ Qp¢V$u Åe R>¡. Ap k„Ops_¡ L$pfZ¡ Å¡ Qp¡kgy„
mass as a result of collision, the angular ÉÊÍ◊ŸÊ ‡ÊÈM§ ∑§⁄U ŒÃË „Ò ÃÊ ¿U«∏ ∑§Ë ∑§ÊáÊËÿ øÊ‹ „ÊªË — s¡_p Öìedp_ L¡$ÞÖ_¡ A_ygnu_¡ `qfc°dZ L$fhp_y„ Qpgy„
speed of the bar will be : L$f¡, sp¡ Qp¡kgp_u L$p¡Zue TX$` __________ \i¡.

v
v (1) 5L v
(1) 5L (1) 5L
6v
6v (2) 5L 6v
(2) 5L (2)
3v 5L
3v (3) 5L 3v
(3) 5L (3)
v 5L
v (4) 6L v
(4) 6L (4) 6L

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Set - 03 6
7. A thin rod MN, free to rotate in the vertical 7. ∞∑§ ¬Ã‹Ë ¿U«∏ MN, ¡Ê Á∑§ ™§äflʸœ⁄U ‚◊Ë ◊¥ 7. A¡L$ kdrnrsS> fpM¡g `psmp¡ krmep¡ MN s¡_p S>qX$s
plane about the fixed end N, is held
ÁSÕ⁄U Á‚⁄U N ∑§ ‚ʬˇÊ ÉÊÍ◊Ÿ ∑§ Á‹∞ SflÃ¥òÊ „Ò, ∑§Ê R>¡X$p N \u DÝh® kdsgdp„ dy¼s fus¡ `qfc°dZ L$fu iL¡$
horizontal. When the end M is released
the speed of this end, when the rod makes ˇÊÒÁá ÁSÕÁà ◊¥ ⁄UÊ∑§Ê ªÿÊ „Ò– ¡’ Á‚⁄U M ∑§Ê ¿UÊ«∏Ê R>¡. Äepf¡ M R>¡X$p_¡ dy¼s L$fhpdp„ Aph¡ R>¡, sp¡ Äepf¡
an angle α with the horizontal, will be ¡ÊÃÊ „Ò ÃÊ ß‚ Á‚⁄U ∑§Ë øÊ‹, ¡’ ¿U«∏ ˇÊÒÁá ‚ α krmep¡ kdrnrsS> kp\¡ α L$p¡Z b_ph¡ Ðepf¡ Ap R>¡X$p_u
proportional to : (see figure) ∑§ÊáÊ ’ŸÊÃË „Ò, ‚◊ʟȬÊÃË „ÊªË — (ÁøòÊ Œπ¥) TX$` ________ _p kdâdpZdp„ li¡. (Sy>Ap¡ ApL©$rs)

(1) sin α (1) sin α (1) sin α
(2) sin α sin α sin α
(2) (2)
(3) cos α (3) cos α (3) cos α
(4) cosα cosα cosα
(4) (4)

8. As shown in the figure, forces of 105 N 8. ÁøòÊÊŸÈ‚Ê⁄U, 10 cm ÷È¡Ê flÊ‹ ∞∑§ ÉÊŸ ∑§ ™§¬⁄U •ÊÒ⁄U 8. ApL©$rsdp„ bspìep âdpZ¡, 10 cm _u bpSy> ^fphsp
each are applied in opposite directions, on ŸËø flÊ‹ »§‹∑§ ¬⁄U 105 N ∑§ ’⁄UÊ’⁄U ’‹Ê¥ ∑§Ê O__u D`f_u A_¡ _uQ¡_u v$f¡L$ bpSy> D`f rhê$Ý^
the upper and lower faces of a cube of side
Áfl¬⁄UËà ÁŒ‡ÊÊ ◊¥ ‹ªÊÿÊ ¡ÊÃÊ „Ò Á¡‚‚ ™§¬⁄UË »§‹∑§ qv$ipdp„ 105 N S>¡V$gy„ bm gNpX$hpdp„ Aphsp s¡_u
10 cm, shifting the upper face parallel to
itself by 0.5 cm. If the side of another cube •¬Ÿ ‚◊ÊãÃ⁄U 0.5 cm ‚ ÁflSÕÊÁ¬Ã „Ê ¡ÊÃË „Ò– D`f_u bpSy s¡_¡ kdp„sf 0.5 cm S>¡V$gu Mk¡ R>¡.
of the same material is 20 cm, then under ÿÁŒ ‚◊ÊŸ ¬ŒÊÕ¸ ∑§ ŒÍ‚⁄U 20 cm ÷È¡Ê flÊ‹ ÉÊŸ ∑§Ê buÅ Ap S> Öìe_p b_¡gp O__u bpSy> 20 cm lp¡e
similar conditions as above, the ’ÃÊÿ ªÿ •flSÕÊ ◊¥ ⁄UπÊ ¡Êÿ ÃÊ ÁflSÕʬŸ ∑§Ê ◊ÊŸ sp¡ D`f S>Zph¡g kdp_ [õ\rs dpV¡$ dmsy„ bpSy>_y„
displacement will be :
„ÊªÊ — õ\p_p„sf __________.

(1) 0.25 cm
(2) 0.37 cm (1) 0.25 cm (1) 0.25 cm
(3) 0.75 cm (2) 0.37 cm (2) 0.37 cm
(4) 1.00 cm (3) 0.75 cm (3) 0.75 cm
(4) 1.00 cm (4) 1.00 cm
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Set - 03 7
9. When an air bubble of radius r rises from 9. ¡’ r ÁòÊíÿÊ ∑§Ê „flÊ ∑§Ê ∞∑§ ’È‹’È‹Ê ∞∑§ ¤ÊË‹ ∑§ 9. Äepf¡ r rÓÄep ^fphsp¡ lhp_p¡ `f`p¡V$p¡ smph_p srme¡
the bottom to the surface of a lake, its
ÁŸø‹ ‚Ä ‚ ©∆U ∑§⁄U ™§¬⁄UË ‚Ä Ã∑§ •ÊÃÊ „Ò, ÃÊ 5r
5r \u k`pV$u D`f Aph¡ R>¡ Ðepf¡ s¡_u rÓÄep S>¡V$gu
radius becomes . Taking the 5r 4
4 ©‚∑§Ë ÁòÊíÿÊ ’…∏ ∑§⁄U „Ê ¡ÊÃË „Ò– flÊÿÈ◊¥«U‹Ëÿ
4 \pe R>¡. hpsphfZ_y„ v$bpZ `pZu_p õs„c_u 10 m
atmospheric pressure to be equal to 10 m
height of water column, the depth of the ŒÊ’ ∑§Ê 10 m ™°§øÊ߸ ∑§ ¡‹ SÃ¥÷ ∑§ ’⁄UÊ’⁄U ◊ÊŸ¥, ÃÊ KQpC S>¡V$gy„ g¡sp„, smph_u k„r_L$¹V$ (gNcN) KX$pC $
lake would approximately be (ignore the ¤ÊË‹ ∑§Ë ª„⁄UÊ߸ ∑§Ê ‚ÁÛÊ∑§≈U ◊ÊŸ „ÊªÊ (¬Îc∆U ßÊfl __________.
surface tension and the effect of ÃÕÊ Ãʬ◊ÊŸ ∑§Ê ¬˝÷Êfl Ÿªáÿ „Ò) — (`©›$spZ A_¡ sp`dp__u Akf AhNZp¡)
temperature) :
(1) 11.2 m
(1) 11.2 m (1) 11.2 m (2) 8.7 m
(2) 8.7 m (2) 8.7 m (3) 9.5 m
(3) 9.5 m (3) 9.5 m (4) 10.5 m
(4) 10.5 m (4) 10.5 m
10. b¡ L$p_p£ A¡[ÞS>_ A A_¡ B A¡L$buÅ kp\¡ î¡Zudp„
10. Two Carnot engines A and B are operated
10. ŒÊ ∑§ÊŸÊ¸ ߥ¡Ÿ A ÃÕÊ B ∑§Ê üÊáÊË’h ∑˝§◊ ◊¥ ø‹ÊÿÊ L$pe®fs R>¡. A¡[ÞS>_ A, 600 K sp`dp_¡ fl¡gp k„N°plL$
in series. Engine A receives heat from a
reservoir at 600 K and rejects heat to a ¡ÊÃÊ „Ò– ߥ¡Ÿ A 600 K ∑§ ÷¥«Ê⁄U ‚ ™§c◊Ê •fl‡ÊÊÁ·Ã `pk¡\u Dódp âpá L$f¡ R>¡ A_¡ T sp`dp_¡ fl¡gp k„N°plL$_¡
reservoir at temperature T. Engine B ∑§⁄UÃÊ „Ò ÃÕÊ T Ãʬ◊ÊŸ ∑§ ÷¥«Ê⁄U ∑§Ê ™§c◊Ê ©à‚Á¡¸Ã `pR>u Ap`¡ R>¡. A¡[ÞS>_ B A¡ A¡[ÞS>_ A A¡ a¢L¡$g
receives heat rejected by engine A and in ∑§⁄UÃÊ „Ò– ߥ¡Ÿ B, ߥ¡Ÿ A mÊ⁄UÊ ©à‚Á¡¸Ã ™§c◊Ê ∑§Ê
turn rejects it to a reservoir at 100 K. If
EÅ®_¡ âpá L$f¡ R>¡ A_¡ 100 K sp`dp_¡ fl¡g k„N°plL$_¡
the efficiencies of the two engines A and B •fl‡ÊÊÁ·Ã ∑§⁄UÃÊ „Ò •ÊÒ⁄U Á»§⁄U 100 K ∑§ ÷¥«Ê⁄U ∑§Ê `pR>u Ap`¡ R>¡. Å¡ Ap b¡ A¡[ÞS>_p¡ A A_¡ B _u
are represented by ηA and ηB, respectively, ©à‚Á¡¸Ã ∑§⁄UÃÊ „Ò– ÿÁŒ ŒÊŸÊ¥ ߥ¡Ÿ A ÃÕÊ B ∑§Ë ηB
L$pe®ndsp A_y¾$d¡ ηA A_¡ ηB lp¡e sp¡ _u
η η ηA
then what is the value of B ? ŒˇÊÃÊ ∑˝§◊‡Ê— ηA ∞fl¥ ηB „Ê¥ ÃÊ B ∑§Ê ◊ÊŸ „ÊªÊ —
ηA ηA qL„$ds L¡$V$gu \i¡ ?
12 12 12
(1) (1) (1)
7 7 7
7 7 7
(2) (2) (2)
12 12 12
12 12 12
(3) (3) (3)
5 5 5
5 5 5
(4) (4) (4)
12 12 12

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Set - 03 8
11. The value closest to the thermal velocity 11. ∑§ˇÊ Ãʬ◊ÊŸ (300 K) ¬⁄U „ËÁ‹ÿ◊ ¬⁄U◊ÊáÊÈ ∑§ ÃʬËÿ 11. Ap¡fX$p_p sp`dp_¡ (300 K) rlrged `fdpÏ dpV¡$_y„
of a Helium atom at room temperature
flª ∑§Ê ms−1 ◊¥ ÁŸ∑§≈UÃ◊ ◊ÊŸ „ÊªÊ, Dódue h¡N_y„ _ÆL$sd d|ëe ms−1 dp„ __________
(300 K) in ms−1 is : [ kB=1.4×10−23 J/K;
[kB=1.4×10−23 J/K; mHe=7×10−27 kg] [ kB=1.4×10−23 J/K; mHe=7×10−27 kg ]
mHe=7×10−27 kg ]
(1) 1.3×10 4
(1) 1.3×10 4
(2) 1.3×10 3 (1) 1.3×10 4
(2) 1.3×10 3
(3) 1.3×10 5 (2) 1.3×10 3
(3) 1.3×10 5
(4) 1.3×10 2 (3) 1.3×10 5
(4) 1.3×10 2
(4) 1.3×10 2
12. Two simple harmonic motions, as shown 12. ŸËø Œ‡Êʸ߸ „È߸ ŒÊ ‚⁄U‹ •Êflø ªÁÃÿÊ° ∞∑§ ŒÍ‚⁄U ∑§
12. _uQ¡ v$ i p® h ¡ g b¡ kfm Aphs® NrsAp¡ (SHM)
below, are at right angles. They are ‹ê’flà „Ò ¥ – ©Ÿ∑§Ê ‚¥ ÿ È Ä Ã ∑§⁄U ∑  § Á‹‚Ê¡È ‚
combined to form Lissajous figures. A¡L$buÅ_¡ L$pV$L$p¡Z¡ \pe R>¡. s¡Ap¡ guk¡ÅDk ApL©$rsAp¡
(Lissajous) ÁøòÊ ’ŸÊÃ „Ò¥–
x(t)=A sin (at+δ) x(t)=A sin (at+δ) b_ph¡ R>¡.
y(t)=B sin (bt) y(t)=B sin (bt) x(t)=A sin (at+δ)
Identify the correct match below. y(t)=B sin (bt)
Parameters Curve
ÁŸêŸ ◊¥ ‚ ‚„Ë ◊‹ ∑§Ë ¬„øÊŸ ∑§ËÁ¡ÿ–
_uQ¡_p `¥L$u L$ep¡ rhL$ë` kpQp¡ rhL$ë` R>¡ s¡ ip¡^p¡.
(1) A ≠ B, a=b ; δ=0 Parabola ⁄UÊÁ‡ÊÿÊ° fl∑˝§
âpQgp¡ h¾$
(1) A ≠ B, a=b ; δ=0
A=B, a=b ; δ= π 2
(2) Line
¬⁄Ufl‹ÿ
(1) A ≠ B, a=b ; δ=0
A=B, a=b ; δ= π 2
`fhge
(2) ⁄UπÊ
(3) A ≠ B, a=b ; δ= π 2 Ellipse (2) A=B, a=b ; δ= π 2 f¡Mp
(3) A ≠ B, a=b ; δ= π 2 ŒËÉʸflÎûÊ
(4) A=B, a=2b ; δ= π 2 Circle (3) A ≠ B, a=b ; δ= π 2 D`hge
(4) A=B, a=2b ; δ= π 2 flÎûÊ
13. 5 beats/second are heard when a tuning (4) A=B, a=2b ; δ= π 2 hsym
®
fork is sounded with a sonometer wire 13. ÿÁŒ ‚ÊŸÊ◊Ë≈U⁄U ∑§ ÃÊ⁄U ∑§Ë ‹¥’Ê߸ 0.95 m ÿÊ 1 m „Ê
under tension, when the length of the 13. Äepf¡ f¡Tp¡_¡V$f (A_y_pv$us) spf_u g„bpC 0.95 m
sonometer wire is either 0.95 m or 1 m. ÃÊ ¡’ ∞∑§ SflÁ⁄UòÊ Ám÷È¡ ∑§Ê ‚ÊŸÊ◊Ë≈U⁄U ∑§ ßÊfl A\hp 1 m fpMhpdp„ Aph¡ R>¡ Ðepf¡ Ýhr_ Qu`uep_¡
The frequency of the fork will be : flÊ‹ ÃÊ⁄U ∑§ ‚ÊÕ ’¡ÊÿÊ ¡ÊÃÊ „Ò ÃÊ 5 ÁflS¬¥Œ ¬˝Áà kp¡_p¡duV$f spf kp\¡ Aamphhp\u 5 õ`„v$/k¡L$ÞX$
(1) 195 Hz ‚∑§á«U ‚ÈŸÊÿË ¬«∏Ã „Ò¥– SflÁ⁄UòÊ Ám÷È¡ ∑§Ë •ÊflÎÁûÊ
(2) 150 Hz k„cmpe R>¡. Ýhr_ rQ`uep_u Aph©rÑ __________
„ÊªË —
(3) 300 Hz \i¡.
(1) 195 Hz
(4) 251 Hz (1) 195 Hz
(2) 150 Hz
(3) 300 Hz (2) 150 Hz
(4) 251 Hz (3) 300 Hz
(4) 251 Hz

SET - 03 ENGLISH PHYSICS SET - 03 HINDI PHYSICS SET - 03 GUJARATI PHYSICS

Set - 03 9
14. A solid ball of radius R has a charge 14. R ÁòÊíÿÊ ∑§ ∞∑§ ∆UÊ‚ ªÊ‹ ∑§ •Êfl‡Ê ÉÊŸàfl ρ ∑§Ê 14. R rÓÄep_p bp¡ g dpV¡ $ rhÛy s cpf O_sp ρ A¡
density ρ given by ρ=ρ o 1 − r (
R
for) 0 ≤ r ≤ R ∑§ Á‹∞ ρ=ρo 1 − r
R ( ) mÊ⁄UÊ ¬˝∑§≈U (
ρ=ρo 1 − r )
; Äep„ 0 ≤ r ≤ R hX¡$ A`pe R>¡.
0 ≤ r ≤ R. The electric field outside the R
Á∑§ÿÊ ¡ÊÃÊ „Ò– ªÊ‹ ∑§ ’Ê„⁄U ÁfllÈà ˇÊòÊ „ÊªÊ — bp¡g_u blpf_p cpNdp„ rhÛysn¡Ó __________.
ball is :
3
ρo R ρo R 3
ρo R 3 (1) 2
(1) o r (1)
o r 2
o r 2
ρo R 3 ρo R 3
ρo R 3 (2)
(2) 12 o r 2 (2)
12 o r 2
12 o r 2
4 ρo R 3 4 ρo R 3
4 ρo R 3 (3)
(3) 3 o r 2 (3)
3 o r 2
3 o r 2
3 ρo R 3 3 ρo R 3
3 ρo R 3 (4)
(4) 4 o r 2 (4)
4 o r 2
4 o r 2

15. A parallel plate capacitor with area 15. ˇÊòÊ»§‹ 200 cm2 ÃÕÊ å‹≈UÊ¥ ∑§ ’Ëø ∑§Ë ŒÍ⁄UË 15. 200 cm2 S>¡V$gy„ n¡Óam A_¡ b¡ àg¡V$p¡ hÃQ¡_y„ A„sf
200 cm2 and separation between the plates 1.5 cm, flÊ‹ ∞∑§ ‚◊ÊãÃ⁄U å‹≈U ‚¥œÊÁ⁄UòÊ ∑§Ê ÁfllÈà 1.5 cm lp¡e s¡hy„ kdp„sf àg¡V$ L¡$`¡kuV$f (k„OpfL$) V
1.5 cm, is connected across a battery of flÊ„∑§ ’‹ V flÊ‹Ë ∞∑§ ’Ò≈U⁄UË ‚ ¡Ê«∏Ê ªÿÊ „Ò– ÿÁŒ
emf V. If the force of attraction between hp¡ëV$ S>¡V$gy„ emf ^fphsu b¡V$fu_¡ kdp„sf Å¡X¡$g R>¡. Å¡
the plates is 25×10−6 N, the value of V is
å‹≈UÊ¥ ∑§ ’Ëø •Ê∑§·¸áÊ ’‹ 25×10−6 N „Ê ÃÊ, V àg¡V$p¡ hÃQ¡_y„ ApL$j®Z 25×10−6 N lp¡e sp¡ V _y„
approximately : ∑§Ê ‹ª÷ª ◊ÊŸ „ÊªÊ — gNcN __________ S>¡V$gy„ d|ëe \i¡.
 2 
 2  −12 C  2 
−12 C 
o = 8.85 × 10  −12 C

o = 8.85 × 10  N.m 2  
o = 8.85 × 10 
 N.m 2    N.m 2 
(1) 250 V (1) 250 V
(2) 100 V (1) 250 V
(2) 100 V (2) 100 V
(3) 300 V (3) 300 V
(4) 150 V (3) 300 V
(4) 150 V (4) 150 V

SET - 03 ENGLISH PHYSICS SET - 03 HINDI PHYSICS SET - 03 GUJARATI PHYSICS

Set - 03 10
16. A copper rod of cross-sectional area A 16. A •ŸÈ¬˝SÕ ∑§Ê≈U ∑§ ˇÊòÊ»§‹ ∑§Ë ∞∑§ ∑§ÊÚ¬⁄U ∑§Ë ¿U«∏ 16. A S>¡V$gy„ ApX$R>¡v$_y„ n¡Óam ^fphsp sp„bp_p„ krmepdp„\u
carries a uniform current I through it. At
‚ „Ê∑§⁄U I œÊ⁄UÊ ’„ÃË „Ò– T Ãʬ◊ÊŸ ¬⁄U ÿÁŒ ¿U«∏ ∑§Ê I S>¡V$gp¡ kdp_ rhÛysâhpl `kpf \pe R>¡. Å¡ T
temperature T, if the volume charge
density of the rod is ρ, how long will the •ÊÿÃŸË •Êfl‡Ê ÉÊŸàfl ρ „Ê ÃÊ •Êfl‡ÊÊ¥ ∑§Ê d ŒÍ⁄UË Ãÿ sp`dp_¡, _mpL$pf_u L$v$ rhÛyscpf O_sp ρ lp¡e sp¡
charges take to travel a distance d ? ∑§⁄UŸ ◊¥ Á∑§ÃŸÊ ‚◊ÿ ‹ªªÊ? huS>cpfp¡_¡ d A„sf L$p`hp dpV¡$ gpNsp¡ kde L¡$V$gp¡
2ρdA 2ρdA \i¡ ?
(1) (1)
I I 2ρdA
(1)
2ρdA 2ρdA I
(2) (2)
IT IT 2ρdA
(2) IT
ρdA ρdA
(3) (3)
I I ρdA
(3)
ρdA ρdA I
(4) (4)
IT IT ρdA
(4) IT
17. A capacitor C1=1.0 µF is charged up to a 17. ∞∑§ ‚¥œÊÁ⁄UòÊ C1=1.0 µF ∑§Ê ∞∑§ ÁSflø (1) mÊ⁄UÊ
voltage V=60 V by connecting it to battery
B through switch (1). Now C 1 is
’Ò≈U⁄UË B ‚ ¡Ê«∏ ∑§⁄U V=60 V Áfl÷fl Ã∑§ •ÊflÁ‡Êà 17. A¡L$ k„OpfL$ C1=1.0 µF _¡ B b¡V$fu kp\¡ L$m (1)
disconnected from battery and connected Á∑§ÿÊ ¡ÊÃÊ „Ò– •’ C1 ∑§Ê ’Ò≈U⁄UË ‚ ÁflÿÊÁ¡Ã ∑§⁄U kp\¡ Å¡X$u V=60 V S>¡V$gy„ rhÛyscpqfs L$fhpdp„ Aph¡
to a circuit consisting of two uncharged ÁŒÿÊ ¡ÊÃÊ „Ò ÃÕÊ ÁSflø (2) ∑§ mÊ⁄UÊ ŒÊ •ŸÊflÁ‡Êà R>¡. lh¡, C1 _¡ b¡V$fu\$u R|>Vy„$ L$fu ApL©$rsdp„ bspìep
capacitors C 2 =3.0 µF and C 3 =6.0 µF ‚¥œÊÁ⁄UòÊÊ¥ C2=3.0 µF ÃÕÊ C3=6.0 µF ∑§ ‚ÊÕ âdpZ¡ L$m (2) _u dv$v$\u b¡ rhÛyscpf frls k„OpfL$p¡
through switch (2), as shown in the figure.
The sum of final charges on C2 and C3 is : ∞∑§ ¬Á⁄U¬Õ ◊¥ ¡Ê«∏ ÁŒÿÊ ¡ÊÃÊ „Ò ¡Ò‚Ê Á∑§ ÁøòÊ ◊¥ C2=3.0 µF A_¡ C3=6.0 µF ^fphsp `qf`\_¡
Œ‡ÊʸÿÊ ªÿÊ „Ò– C2 ÃÕÊ C3 ¬⁄U •¥ÁÃ◊ •Êfl‡ÊÊ¥ ∑§Ê Å¡X$hpdp„ Aph¡ R>¡. sp¡ C2 A_¡ C3 `f_p¡ A„rsd Ly$g
ÿÊª „ÊªÊ — rhÛyscpf __________ R>¡.

(1) 40 µC
(2) 36 µC
(3) 20 µC (1) 40 µC (1) 40 µC
(4) 54 µC (2) 36 µC (2) 36 µC
(3) 20 µC (3) 20 µC
(4) 54 µC (4) 54 µC
SET - 03 ENGLISH PHYSICS SET - 03 HINDI PHYSICS SET - 03 GUJARATI PHYSICS

Set - 03 11
18. A current of 1 A is flowing on the sides of 18. 4.5×10−2 m ÷È¡Ê ∑§ ∞∑§ ‚◊’Ê„È ÁòÊ÷È¡ ◊¥ 1 A 18. 4.5×10−2 m _u g„bpC ^fphsp kdbpSy> rÓL$p¡Z_u
an equilateral triangle of side 4.5×10−2 m.
∑§Ë œÊ⁄UÊ ¬˝flÊÁ„à „Ê ⁄U„Ë „Ò– ß‚ ÁòÊ÷È¡ ∑§ ∑§ãŒ˝ ¬⁄U bpSy>dp„\u 1 A S>¡V$gp¡ rhÛysâhpl `kpf \pe R>¡.
The magnetic field at the centre of the
triangle will be : øÈê’∑§Ëÿ ˇÊòÊ ∑§Ê ◊ÊŸ „ÊªÊ — rÓL$p¡Z_p L¡$ÞÖ ApNm Qy„bL$ue n¡Ó __________
(1) 2×10−5 Wb/m2 (1) 2×10−5 Wb/m2 li¡.
(2) Zero (2) ‡ÊÍãÿ
(3) 8×10−5 Wb/m2 (3) 8×10−5 Wb/m2 (1) 2×10−5 Wb/m2
(4) 4×10−5 Wb/m2 (4) 4×10−5 Wb/m2
(2) i|Þe
19. At the centre of a fixed large circular coil (3) 8×10−5 Wb/m2
19. R ÁòÊíÿÊ ∑§Ë ∞∑§ ÁSÕ⁄U ∞fl¥ ’«∏Ë ªÊ‹Ê∑§Ê⁄U ∑ȧá«U‹Ë (4) 4×10−5 Wb/m2
of radius R, a much smaller circular coil of
radius r is placed. The two coils are
∑§ ∑§ãŒ˝ ¬⁄U •àÿÁœ∑§ ¿UÊ≈UË r ÁòÊíÿÊ ∑§Ë ∞∑§ ªÊ‹Ê∑§Ê⁄U
concentric and are in the same plane. The ∑ȧá«U‹Ë ⁄UπË „Ò– ŒÊŸÊ¥ ∑ȧá«UÁ‹ÿÊ° ‚¥∑§ãŒ˝Ë ÃÕÊ ∞∑§ 19. A¡L$ S>qX$s A_¡ R rÓÄep_u dp¡V$p hsy®mpL$pf N|„Qmp_p
larger coil carries a current I. The smaller „Ë ‚◊Ë ◊¥ „Ò¥– ’«∏Ë ∑È§á«‹Ë ◊¥ I œÊ⁄UÊ ’„ÃË „Ò, L¡$ÞÖ ApNm M|b S> _p_u r rÓÄep ^fphsy„ hsy®mpL$pf
coil is set to rotate with a constant angular ŒÊŸÊ¥ ∑ȧá«Á‹ÿÊ¥ ∑§ ©÷ÿÁŸc∆U √ÿÊ‚ ‚ „Ê∑§⁄U ¡ÊŸ N|„Qmy„ d|L¡$g R>¡. Ap b„_¡ N|„QmpAp¡ kdL¡$[ÞÖe A_¡
velocity ω about an axis along their
common diameter. Calculate the emf flÊ‹ •ˇÊ ∑§ ‚ʬˇÊ ¿UÊ≈UË ∑È§á«‹Ë ∑§Ê ∞∑§ ∞∑§‚◊ÊŸ kdp_ kdsgdp„ R>¡. dp¡Vy„$ N|„Qmy„ I âhpl ^fph¡ R>¡.
induced in the smaller coil after a time t of ∑§ÊáÊËÿ flª ω ‚ ÉÊÈ◊ÊÿÊ ¡ÊÃÊ „Ò– ÉÊÍáʸŸ ‡ÊÈM§ „ÊŸ ∑§ s¡d_p kpdpÞe ìepk_u qv$ip_¡ A_ygnu_¡ An_¡ afs¡
its start of rotation. t ‚◊ÿ ©¬⁄UÊãà ¿UÊ≈UË ∑ȧá«U‹Ë ◊¥ ¬˝Á⁄Uà ÁfllÈà flÊ„∑§ _p_p N|„Qmp_¡ AQm L$p¡Zue h¡N ω \u `qfc°dZ
’‹ ∑§Ë ªáÊŸÊ ∑§⁄¥U — L$fphhpdp„ Aph¡ R>¡. `qfc°dZ iê$ \ep bpv$ t kde¡
µo I
ω π r 2 sin ωt µo I _p_p N|„Qmpdp„ â¡qfs \sy„ emf NZp¡.
(1) 2R (1) ω π r 2 sin ωt
2R µo I
(1) ω π r 2 sinωt
µo I 2R
ω π r 2 sin ωt µo I
(2) 4R (2) ω π r 2 sin ωt
4R µo I
(2) ω π r 2 sin ωt
µo I 4R
ω r 2 sin ωt µo I
(3) 4R (3) ω r 2 sin ωt
4R µo I
(3) ω r 2 sin ωt
µo I 4R
ω r 2 sin ωt µo I
(4) 2R (4) ω r 2 sin ωt
2R µo I
(4) ω r 2 sin ωt
2R

SET - 03 ENGLISH PHYSICS SET - 03 HINDI PHYSICS SET - 03 GUJARATI PHYSICS

Set - 03 12
20.
20. 20.

m v$m ^fphsp¡ L$p¡`f_p¡ A¡L$ krmep¡ Nyê$ÐhpL$j®Z bm_u
A copper rod of mass m slides under ˇÊÒÁá ‚ θ ∑§ÊáÊ ¬⁄U ÁSÕà ŒÊ Áø∑§ŸË ‚◊ÊŸÊ¥Ã⁄U ¿U«∏Ê¥, Akf l¡W$m l A„sf¡ fpM¡g b¡ kdp„sf `pV$p L¡$ S>¡
gravity on two smooth parallel rails, with
separation l and set at an angle of θ with
Á¡Ÿ∑§ ’Ëø ∑§Ë ŒÍ⁄UË l „Ò, ∑§ ™§¬⁄U m Œ˝√ÿ◊ÊŸ ∑§Ë kdrnrsS>\u θ L$p¡Z¡ Np¡W$h¡g R>¡, `f kfL¡$ R>¡. ApL©$rsdp„
the horizontal. At the bottom, rails are ÃÊ°’ ∑§Ë ∞∑§ ¿U«∏ ªÈL§àfl ∑§ •¥Ãª¸Ã Á»§‚‹ÃË „Ò– bspìep âdpZ¡ Ap `pV$p_¡ srme¡ R Ahfp¡^ hX¡$
joined by a resistance R. There is a ¿U«∏Ê¥ ∑§ ÁŸø‹ Á‚⁄UÊ¥ ∑§Ê ∞∑§ ¬˝ÁÃ⁄UÊœ R mÊ⁄UÊ ¡Ê«∏Ê Å¡X$hpdp„ Aph¡g R>¡. `pV$p_p kdsg_¡ g„b kdp„N
uniform magnetic field B normal to the ªÿÊ „Ò– ‚◊ÊŸÊ¥Ã⁄U ¿U«∏Ê¥ ∑§ ‚◊Ë ∑§ ‹ê’flà ÁŒ‡ÊÊ Qy„bL$ue n¡Ó B R>¡. L$p¡`f_p Ap krmep_u A„rsd
plane of the rails, as shown in the figure.
The terminal speed of the copper rod is :
◊¥ ∞∑§ ∞∑§‚◊ÊŸ øÈê’∑§Ëÿ ˇÊòÊ B „Ò ¡Ò‚Ê Á∑§ ÁøòÊ ◊¥ (V$rd®_g) TX$` __________ R>¡.
Œ‡ÊʸÿÊ ªÿÊ „Ò– ÃÊ¥’ ∑§ ¿U«∏ ∑§Ë ‚Ë◊Êãà øÊ‹
mg R tan θ mg R tan θ
(1) 2 2 „ÊªË — (1)
B l B2 l 2
mg R tan θ
mg R cot θ (1) mg R cot θ
(2) 2 2 B2 l 2 (2)
B l B2 l 2
mg R cot θ
mg R sin θ (2) mg R sin θ
(3) 2 2 B2 l 2 (3)
B l B2 l 2
mg R sin θ
mg R cos θ (3) mg R cos θ
(4) B2 l 2
B2 l 2 (4)
B2 l 2
mg R cos θ
(4)
B2 l 2

SET - 03 ENGLISH PHYSICS SET - 03 HINDI PHYSICS SET - 03 GUJARATI PHYSICS

Set - 03 13
21. A plane polarized monochromatic 21. ∞∑§ ‚◊Ë œ˝ÈÁflà ∞∑§fláÊ˸ÿ ÁfllÈÃøÈê’∑§Ëÿ Ã⁄¥Uª 21. sg ^y°hue A¡L$hZ} EM sf„N z qv$ipdp„ i|ÞephL$pidp„
EM wave is traveling in vacuum along z
ÁŸflʸà ◊¥ z-ÁŒ‡ÊÊ ∑§ ‚¥ªÃ ß‚ Ã⁄U„ ø‹ ⁄U„Ë „Ò Á∑§ A¡hu fus¡ Nrs L$f¡ R>¡ L¡$ t=t1 A¡ AhL$piue tbvy$ z1
direction such that at t=t1 it is found that
the electric field is zero at a spatial point Á∑§‚Ë SÕÊÁŸ∑§ Á’¥ŒÈ z1 ¬⁄U ‚◊ÿ t=t1 ¬⁄U ÁfllÈà ApNm rhÛys n¡Ó i|Þe R>¡. s¡_u _ÆL$dp„ Ðepf `R>u_u
z 1 . The next zero that occurs in its ˇÊòÊ ‡ÊÍãÿ „Ò– ß‚∑§ ‚◊ˬ ÁfllÈà ˇÊòÊ ∑§Ê •ª‹Ê i|Þe suh°sp A¡ z2 ApNm dm¡ R>¡. rhÛysQy„bL$ue sf„N_u
neighbourhood is at z2. The frequency of ‡ÊÍãÿ z2 ¬⁄U ¬ÊÿÊ ¡ÊÃÊ „Ò– ß‚ ÁfllÈà øÈê’∑§Ëÿ Ã⁄¥Uª Aph©rÑ __________.
the electromagnetic wave is :
3×108
∑§Ë •ÊflÎÁûÊ „ÊªË —
3×108 (1)
(1) 3×108
z2 −z1


z2 −z1
(1)

z2 −z1

8 1.5×108
1.5×10 (2)
(2) 1.5×108
z2 −z1


z2 −z1
(2)

z2 −z1

6×108
6×108 (3)
(3) 6×108
z2 −z1


z2 −z1
(3)

z2 −z1

1
1 (4)
(4) 1
z −z


z −z
(4) t 1+ 2 81
t 1+ 2 81
z −z
3×10
3×10 t 1+ 2 81
3×10 22. Np¡gue rh`\_ (spherical aberration) ky^pfhp
22. A convergent doublet of separated lenses, dpV¡$ R|>V$p `pX¡$gp g¡Þkp¡_p k„L¡$[ÞÖ Å¡X$L$p_u `qfZpdu
corrected for spherical aberration, has 22. Á∑§ã„Ë¥ ŒÊ •Á÷‚Ê⁄UË ‹ã‚Ê¥ ‚ ’Ÿ ‚¥ÿÊª ∑§Ë, ªÊ‹Ëÿ L¡$ÞÖ g„bpC 10 cm R>¡. b¡ g¡Þk_p Å¡X$L$p„ hÃQ¡_y„ A„sf
resultant focal length of 10 cm. The ŒÊ· ŒÍ⁄U ∑§⁄UŸ ∑§ ’ÊŒ, ¬˝÷ÊflË »§Ê∑§‚ ŒÍ⁄UË 10 cm „Ò– 2 cm R>¡. Ap OV$L$ g¡Þkp¡_u L¡$ÞÖg„bpCAp¡ ________.
separation between the two lenses is 2 cm. ŒÊŸÊ¥ ‹ã‚Ê¥ ∑§ ’Ëø ∑§Ë ŒÍ⁄UË 2 cm „Ò– ŒÊŸÊ¥ ‹ã‚Ê¥ ∑§Ë
The focal lengths of the component lenses
•‹ª-•‹ª »§Ê∑§‚ ŒÍÁ⁄UÿÊ¥ „Ò¥ — (1) 10 cm, 12 cm
are :
(1) 10 cm, 12 cm (2) 12 cm, 14 cm
(1) 10 cm, 12 cm
(2) 12 cm, 14 cm (3) 16 cm, 18 cm
(2) 12 cm, 14 cm
(3) 16 cm, 18 cm (4) 18 cm, 20 cm
(3) 16 cm, 18 cm
(4) 18 cm, 20 cm (4) 18 cm, 20 cm

SET - 03 ENGLISH PHYSICS SET - 03 HINDI PHYSICS SET - 03 GUJARATI PHYSICS

Set - 03 14
23. A plane polarized light is incident on a 23. ∞∑§ ‚◊Ë œ˝ÈÁflà ¬˝∑§Ê‡Ê Á∑§‚Ë ∞∑§ œ˝Èfl∑§ Á¡‚∑§Ê 23. ApL©$rsdp„ bspìep A_ykpf A¡L$ sg ^°yhue âL$pi_¡ L¡$
polariser with its pass axis making angle θ
¬ÊÁ⁄UÃ-•ˇÊ x-•ˇÊ ‚ θ ∑§ÊáÊ ’ŸÊÃÊ „Ò, ¬⁄U •Ê¬ÁÃà S>¡_y„ v$L$¹ -An kdsg x-An_¡ kp`¡n θ L$p¡Z¡ b_ph¡
with x-axis, as shown in the figure. At
four different values of θ, θ=8
, 38
, 188
„ÊÃÊ „Ò, ¡Ò‚Ê ÁøòÊ ◊¥ ÁŒπÊÿÊ ªÿÊ „Ò– ∑§ÊáÊ θ ∑§ øÊ⁄U R>¡. θ _p Qpf Sy>v$p-Sy>v$p d|ëep¡ `f θ=8
, 38
, 188

and 218
, the observed intensities are same. ÁflÁ÷ÛÊ ◊ÊŸÊ¥, θ=8
, 38
, 188
ÃÕÊ 218
¬⁄U ÃËfl˝ÃÊÿ¥ A_¡ 218
fpMsp suh°sp kdp_ Å¡hp dm¡ R>¡. ^y°huc|s_u
What is the angle between the direction of ’⁄UÊ’⁄U ¬ÊÿË ¡ÊÃË „Ò¥– œ˝ÈfláÊ ÁŒ‡ÊÊ ÃÕÊ x-•ˇÊ ∑§ qv$ip A_¡ x-An_u hÃQ¡_p¡ L$p¡Z L¡$V$gp¡ li¡ ? $
polarization and x-axis ?
’Ëø ∑§Ê ∑§ÊáÊ „ÊªÊ —

(1) 98
(1) 98

(2) 128
(1) 98
(2) 128

(3) 203
(2) 128
(3) 203

(4) 45
(3) 203
(4) 45

(4) 45

24. If the de Broglie wavelengths associated 24. Å¡ âp¡V$p¡_ A_¡ α-L$Z kp\¡ k„L$mpe¡g X$u-b°p¡Águ
with a proton and an α-particle are equal, 24. ÿÁŒ ∞∑§ ¬˝Ê≈UÊÚŸ ∞fl¥ ∞∑§ α-∑§áÊ ∑§Ë Á«U-’˝ÊÇ‹Ë
then the ratio of velocities of the proton
sf„Ng„bpCAp¡ kdp_ lp¡e sp¡ âp¡V$p¡_ A_¡ α-L$Z_u
and the α-particle will be :
Ã⁄¥UªŒÒäÿ¸ ’⁄UÊ’⁄U „Ò¥ ÃÊ ß‚ ¬˝Ê≈UÊÚŸ ÃÕÊ α-∑§áÊ ∑§ flªÊ¥ h¡Np¡_p¡ NyZp¡Ñf __________ li¡.
(1) 4 : 1 ∑§Ê •ŸÈ¬Êà „ÊªÊ —
(2) 2 : 1 (1) 4:1 (1) 4:1
(3) 1 : 2 (2) 2:1 (2) 2:1
(4) 1 : 4 (3) 1:2 (3) 1:2
(4) 1:4 (4) 1:4

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Set - 03 15
25. Muon (µ − ) is a negatively charged 25. êÿÍ•ÊÚŸ (Muon) (µ−) ∞∑§ ´§áÊÊà◊∑§ •ÊflÁ‡Êà 25. çe|Ap¡_ (µ−) A¡ F>ZrhÛyscpqfs ( q = e ) L$Z R>¡
( q = e ) particle with a mass mµ=200 me,
( q = e ) ∑§áÊ „Ò Á¡‚∑§Ê Œ˝√ÿ◊ÊŸ mµ=200 me „Ò L¡$ S>¡_y„ v$m mµ=200 me, Äep„ me, A¡ Cg¡¼V²$p¡__y„
where me is the mass of the electron and e
is the electronic charge. If µ− is bound to ¡„Ê° me ß‹Ä≈˛UÊÚŸ ∑§Ê Œ˝√ÿ◊ÊŸ ÃÕÊ e ß‹Ä≈˛UÊÚŸ ∑§Ê v$m R>¡ A_¡ e Cg¡¼V²$p¡r_L$ QpS>® R>¡. Å¡ µ− A¡ âp¡V$p¡_
a proton to form a hydrogen like atom, •Êfl‡Ê „Ò– „Êß«˛UÊ¡Ÿ ¡Ò‚Ê ¬⁄U◊ÊáÊÈ ’ŸÊŸ ∑§ Á‹∞ ÿÁŒ kp\¡ Å¡X$pC (b„^pC) lpCX²$p¡S>_ S>¡hp¡ `fdpÏ b_ph¡
identify the correct statements. êÿÍ•ÊÚŸ ´§áÊÊà◊∑§ ∞∑§ ¬˝Ê≈UÊÚŸ ∑§ ‚ÊÕ ¬Á⁄U’h „ÊÃÊ „Ò, R>¡, sp¡ kpQp„ rh^p_p¡ ip¡^p¡.
(A) Radius of the muonic orbit is 200
ÃÊ ‚„Ë ∑§ÕŸ „Ê¥ª¥ — (A) çeyAp¡r_L$¹ L$np_u rÓÄep Cg¡¼V²$p¡__u rÓÄep L$fsp
times smaller than that of the
electron. (A) êÿÍ•ÊÚŸ ∑§ ∑§ˇÊ ∑§Ë ÁòÊíÿÊ ß‹Ä≈˛UÊÚŸ ∑§ ∑§ˇÊ 200 NZu _p_u li¡.
(B) The speed of the µ− in the nth orbit ∑§Ë ÁòÊíÿÊ ‚ 200 ªÈŸÊ ¿UÊ≈UË „Ò– (B) µ−_u nth L$npdp„ TX$` A¡ nth L$npdp„_p
1 (B) n fl¥ ∑§ˇÊ ◊¥ µ− ∑§Ë øÊ‹, n fl¥ ∑§ˇÊ ◊¥ 1
is times that of the electron in Cg¡¼V²$p¡__u TX$` L$fsp„ 200 NZu li¡.
200 1
the nth orbit. ß‹Ä≈˛UÊÚŸ ∑§Ë øÊ‹ ∑§Ë 200 ªÈŸÊ „ÊªË– (C) çey A p¡ r _L$ `fdpÏ_u Ape_uL$fZ EÅ®
(C) The ionization energy of muonic
atom is 200 times more than that of (C) êÿÍ•ÊÚÁŸ∑§ ¬⁄U◊ÊáÊÈ ∑§Ë •ÊÿŸŸ ™§¡Ê¸, „Êß«˛UÊ¡Ÿ lpCX²$p¡S>_ `fdpÏ„ L$fsp„ 200 NZu h^pf¡ li¡.
an hydrogen atom. ¬⁄U◊ÊáÊÈ ∑§ •ÊÿŸŸ ™§¡Ê¸ ‚ 200 ªÈŸÊ íÿÊŒÊ (D) nth L$ n pdp„ çey A p¡ _ _y „ h¡ N dp_ Cg¡ ¼ V² $ p ¡ _ _p
(D) The momentum of the muon in the „Ò– h¡Ndp_ L$fsp„ 200 NÏ„ h^pf¡ li¡.
nth orbit is 200 times more than that
(D) n fl¥ ∑§ˇÊ ◊¥ êÿÍ•ÊÚŸ ∑§Ê ‚¥flª, n fl¥ ∑§ˇÊ ◊¥ (1) (A), (B), (D)
of the electron.
(1) (A), (B), (D) ß‹Ä≈˛UÊÚŸ ∑§ ‚¥flª ‚ 200 ªÈŸÊ íÿÊŒÊ „Ò– (2) (A), (C), (D)
(2) (A), (C), (D) (1) (A), (B), (D) (3) (B), (D)
(3) (B), (D) (2) (A), (C), (D) (4) (C), (D)
(4) (C), (D) (3) (B), (D)
(4) (C), (D) 26. A¡L$ A[õ\f cpf¡ Þey[¼gek b¡ Þey[¼gekdp„ M„X$_
26. An unstable heavy nucleus at rest breaks `pd¡ R>¡ A_¡ Ap b„_¡ cpNp¡ 8 : 27 S>¡V$gp h¡Np¡_p
into two nuclei which move away with 26. ∞∑§ ÁSÕ⁄U •flSÕÊ ∑§Ê •SÕÊÿË ÷Ê⁄UË ŸÊÁ÷∑§, ŒÊ NyZp¡Ñf\u v|$f a¢L$pe R>¡. Ap Þey[¼gek (_¡ Np¡mpL$pf
velocities in the ratio of 8 : 27. The ratio of ŸÊÁ÷∑§Ê¥ ◊¥ ≈ÍU≈U ¡ÊÃÊ „Ò ¡Ê 8 : 27 ∑§ flª •ŸÈ¬Êà ‚
the radii of the nuclei (assumed to be ^pfsp„) _u rÓÄepAp¡_p¡ NyZp¡Ñf __________.
spherical) is :
ŒÍ⁄U ¡ÊÃ „Ò¥– ≈ÍU≈U „È∞ ŸÊÁ÷∑§Ê¥ ∑§Ë ÁòÊíÿÊ•Ê¥ (◊ÊŸÊ fl
(1) 8 : 27
(1) 8 : 27 ªÊ‹Ê∑§Ê⁄U „Ò¥) ∑§Ê •ŸÈ¬Êà „ÊªÊ — (2) 4:9
(2) 4 : 9 (1) 8 : 27 (3) 3:2
(3) 3 : 2 (2) 4:9 (4) 2:3
(4) 2 : 3 (3) 3:2
(4) 2:3

SET - 03 ENGLISH PHYSICS SET - 03 HINDI PHYSICS SET - 03 GUJARATI PHYSICS

Set - 03 16
27. Truth table for the following digital circuit 27. _uQ¡ Ap`¡ g X$uTuV$g `qf`\ dpV¡ $ _y „ V³ \ V¡ b g
27. ÁŒÿ ªÿ •¥∑§∑§ ¬Á⁄U¬Õ ∑§ Á‹∞ ‚àÿ◊ÊŸ ‚ÊÁ⁄UáÊË
will be : __________.
„ÊªË —

x y z
x y z
x y z 0 0 0
0 0 0
0 0 0 0 1 0
0 1 0 (1)
(1) 0 1 0 1 0 0
1 0 0 (1) 1 0 0 1 1 1
1 1 1
1 1 1
x y z
x y z
x y z 0 0 1
0 0 1
0 0 1 0 1 1
0 1 1 (2)
(2) 0 1 1 1 0 1
1 0 1
(2) 1 0 1 1 1 0
1 1 0
1 1 0

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Set - 03 17

x y z x y z x y z
0 0 1 0 0 1 0 0 1
0 1 1 0 1 1 0 1 1
(3) 1 0 1 (3) 1 0 1 (3) 1 0 1
1 1 1 1 1 1 1 1 1

x y z x y z x y z
0 0 0 0 0 0 0 0 0
0 1 1 0 1 1 0 1 1
(4) 1 0 1 (4) 1 0 1 (4) 1 0 1
1 1 1 1 1 1 1 1 1

28. 49 µH ApÐdâ¡fL$Ðh ^fphsp„ N|„Qmp_p A_¡ 2.5 nF
28. The carrier frequency of a transmitter is 28. ∑ȧá«U‹Ë ∑§ ¬˝⁄U∑§àfl 49 µH ÃÕÊ œÊÁ⁄UÃÊ 2.5 nF
provided by a tank circuit of a coil of k„OpfL$sp ^fphsp k„OpfL$\u b_¡gp A¡L$ V¡$ÞL$ `qf`\\u
flÊ‹ ∞∑§ ≈Ò¥U∑§ ¬Á⁄U¬Õ mÊ⁄UÊ ∞∑§ ¬˝·∑§ ∑§Ë flÊ„∑§
inductance 49 µH and a capacitance of V²$pÞkduV$f dpV¡$_p L¡$fuef sf„N_u Aph©rÑ Ap`hpdp„
2.5 nF. It is modulated by an audio signal •ÊflÎ Á ûÊ ©à¬ÛÊ ∑§Ë ¡ÊÃË „Ò – ß‚ •ÊflÎ Á ûÊ ∑§Ê
Aph¡ R>¡. s¡_¡ 12 kHz _p Ýhr_ rkÁ_g\u dp¡X$éyg¡V
of 12 kHz. The frequency range occupied 12 kHz ∑§ ∞∑§ äflÁŸ ‚¥∑§Ã (audio signal) ‚
by the side bands is : L$fhpdp„ Aph¡ R>¡. kpCX$ b¡ÞX$dp„ kpd¡g Aph©rÑ Npmp¡
◊Ê«ÈUÁ‹Ã ∑§⁄UÃ „Ò¥– ¬Ê‡fl¸ ’Ò¥«U ∑§Ë •ÊflÎÁûÊ ∑§Ê ¬⁄UÊ‚ __________.
(1) 13482 kHz − 13494 kHz
(2) 442 kHz − 466 kHz
„ÊªÊ —
(1) 13482 kHz − 13494 kHz
(3) 63 kHz − 75 kHz (1) 13482 kHz − 13494 kHz
(2) 442 kHz − 466 kHz
(4) 18 kHz − 30 kHz (2) 442 kHz − 466 kHz
(3) 63 kHz − 75 kHz
(3) 63 kHz − 75 kHz
(4) 18 kHz − 30 kHz
(4) 18 kHz − 30 kHz

SET - 03 ENGLISH PHYSICS SET - 03 HINDI PHYSICS SET - 03 GUJARATI PHYSICS

Set - 03 18
29. A constant voltage is applied between two 29. ∞∑§ œÊÃÈ ∑§ ÃÊ⁄U ∑§ ŒÊŸÊ¥ Á‚⁄UÊ¥ ∑§ ’Ëø ∞∑§ ÁSÕ⁄U 29. A¡L$ ^pÐhue spf_p b¡ R>¡X$p hÃQ¡ AQm hp¡ëV¡$S> gNpX¡$g
ends of a metallic wire. If the length is
Áfl÷fl ‹ªÊÿÊ ¡ÊÃÊ „Ò– ÿÁŒ ÃÊ⁄U ∑§Ë ‹ê’Ê߸ •ÊœË R>¡. Å¡ spf_u g„bpC AX$^u A_¡ rÓÄep bdZu \pe
halved and the radius of the wire is
doubled, the rate of heat developed in the ÃÕÊ ÁòÊíÿÊ ŒÊªÈŸË ∑§⁄U ŒË ¡Êÿ ÃÊ ÃÊ⁄U ◊¥ ©à¬ÛÊ ™§c◊Ê sp¡ spfdp„ DÐ`Þ_ Dódp_p¡ v$f __________.
wire will be : Œ⁄U — (1) bdZp¡
(1) Doubled (1) ŒÈªÈŸË „Ê ¡ÊÿªË
(2) Halved (2) AX$^p¡
(2) •ÊœË „Ê ¡ÊÿªË (3)
(3) Unchanged bv$gpi¡ _l]
(4) Increased 8 times (3) fl„Ë ⁄U„ªË (4) 8 hMs h^i¡
(4) 8 ªÈŸÊ ’…∏ ¡ÊÿªË
30. A body takes 10 minutes to cool from 60
C
30. A¡L$ `v$p\® 60
C \u 50
C W„$X$p¡ `X$hp dpV¡$ 10 du_uV$
to 50
C. The temperature of surroundings 30. ∞∑§ Á¬á«U 60
C ‚ 50
C Ã∑§ ∆¥U«UÊ „ÊŸ ◊¥ 10 Á◊Ÿ≈U
is constant at 25
C. Then, the temperature g¡ R>¡. Apk`pk_p hpsphfZ_y„ sp`dp_ 25
C AQm
of the body after next 10 minutes will be ∑§Ê ‚◊ÿ ‹ÃÊ „Ò– flÊÃÊfl⁄UáÊ ∑§Ê Ãʬ◊ÊŸ 25
C ¬⁄U lp¡e sp¡ `v$p\®_y„ 10 du_uV$ `R>u_y„ sp`dp_ gNcN
approximately : ÁSÕ⁄U „Ò– ©‚∑§ 10 Á◊Ÿ≈U ’ÊŒ Á¬á«U ∑§ Ãʬ◊ÊŸ ∑§Ê __________ \i¡ .
(1) 47
C ∑§⁄UË’Ë ◊ÊŸ „ÊªÊ —
(2) 41
C (1) 47
C
(1) 47
C
(3) 45
C (2) 41
C
(2) 41
C
(4) 43
C (3) 45
C
(3) 45
C
(4) 43
C
(4) 43
C

SET - 03 ENGLISH PHYSICS SET - 03 HINDI PHYSICS SET - 03 GUJARATI PHYSICS

Set - 03 19
PART B — CHEMISTRY ÷ʪ B — ⁄U‚ÊÿŸ ÁflôÊÊŸ cpN B — fkpeZ ipõÓ
31. For per gram of reactant, the maximum
31. ÁŸêŸ ÃʬËÿ ÁflÉÊ≈UŸ •Á÷Á∑˝§ÿÊ•Ê¥ ◊¥, ¬˝Áê˝Ê◊ 31. _uQ¡ Ap`¡gp Dódue rhOV$_ âq¾$epAp¡dp„, `°rsN°pd
quantity of N2 gas is produced in which
of the following thermal decomposition •Á÷∑§Ê⁄U∑§ ‚ Á∑§‚◊¥ N2 ªÒ‚ ∑§Ë ◊ÊòÊÊ ‚flʸÁœ∑§ âq¾$eL$ \u L$epdp„ N2 hpey_u dpÓp khp®r^L$ âpá \pe
reactions ? ¬˝Êåà „ÊªË? R>¡ ?
(Given : Atomic wt. - Cr=52 u, Ba=137 u) (ÁŒÿÊ ªÿÊ „Ò — ¬⁄U◊ÊáÊÈ ÷Ê⁄U - Cr=52 u, Ba=137 u) (Ap`¡g `fdpÎhue hS>_ Cr=52 u, Ba=137 u)
(1) (NH4)2Cr2O7(s) → N2(g)+4H2O(g) (1) (NH4)2Cr2O7(s) → N2(g)+4H2O(g)
+Cr2O3(s) (1) (NH4)2Cr2O7(s) → N2(g)+4H2O(g)
+Cr2O3(s) +Cr2O3(s)
(2) 2NH4NO3(s) → 2 N2(g)+4H2O(g) (2) 2NH4NO3(s) → 2 N2(g)+4H2O(g)
+O2(g) (2) 2NH4NO3(s) → 2 N2(g)+4H2O(g)
+O2(g) +O2(g)
(3) Ba(N3)2(s) → Ba(s)+3N2(g) (3) Ba(N3)2(s) → Ba(s)+3N2(g)
(4) 2NH3(g) → N2(g)+3H2(g) (3) Ba(N3)2(s) → Ba(s)+3N2(g)
(4) 2NH3(g) → N2(g)+3H2(g)
(4) 2NH3(g) → N2(g)+3H2(g)
32. ÁŸêŸ ◊¥ ‚ ∞∑§ ∑§ •ÁÃÁ⁄UÄà ‚’∑§Ë Á∑˝§S≈U‹ ‚¥⁄UøŸÊ
32. All of the following share the same crystal
∞∑§ ¡Ò‚Ë „Ò, fl„ ∞∑§ ∑§ÊÒŸ „Ò? 32. _uQ¡ Ap`¡gpdp„\u A¡L$ rkhpe b^p A¡L$S>¡hp õaqV$L$
structure except :
(1) LiCl (1) LiCl b„^pfZ ^fph¡ R>¡ :
(2) NaCl (2) NaCl (1) LiCl
(3) RbCl (3) RbCl (2) NaCl
(4) CsCl (4) CsCl (3) RbCl
(4) CsCl
33. The de-Broglie’s wavelength of electron 33. „Êß«˛UÊ¡Ÿ ¬⁄U◊ÊáÊÈ ∑§ ¬˝Õ◊ ’Ê⁄U ∑§ˇÊÊ ◊¥ ©¬ÁSÕÃ
present in first Bohr orbit of ‘H’ atom is : 33. ‘H’ `fdpÏ_u â\d ålp¡ f L$ n pdp„ lpS>f fl¡ g p
ß‹Ä≈˛UÊÚŸ ∑§Ê «UË-’˝ÊÇ‹Ë Ã⁄¥UªŒÒÉÿ¸ „ÊªÊ —
(1) 0.529 Å Cg¡¼V²$p¡__u X$u-b°p¡Ngu sf„N g„bpC ip¡^p¡.
(1) 0.529 Å
(2) 2π×0.529 Å (1) 0.529 Å
(2) 2π×0.529 Å
0.529 (2) 2π×0.529 Å
(3) Å 0.529
2π (3) Å 0.529
2π (3) Å
(4) 4×0.529 Å
(4) 4×0.529 Å 2π
(4) 4×0.529 Å

SET - 03 ENGLISH CHEMISTRY SET - 03 HINDI CHEMISTRY SET - 03 GUJARATI CHEMISTRY

Set - 03 20
34. ∆fG at 500 K for substance ‘S’ in liquid 34. ¬ŒÊÕ¸ ‘S’ ∑§ Á‹ÿ, Œ˝fl •flSÕÊ ÃÕÊ ªÒ‚Ëÿ •flSÕÊ 34. 500 K `f, ∆ f G `v$p\® ‘S’ dpV¡ $ , âhplu
state and gaseous state are
◊ ¥ , ∆ f G ∑§Ê ◊ÊŸ 500 K ¬⁄U ∑˝ § ◊‡Ê— Ahõ\p A_¡ hpey d e Ahõ\pdp„ A_y ¾ $d¡
+100.7 kcal mol−1 and +103 kcal mol−1,
respectively. Vapour pressure of liquid ‘S’ +100.7 kcal mol−1 ÃÕÊ +103 kcal mol−1 +100.7 kcal mol−1 A_¡ +103 kcal mol−1
at 500 K is approximately equal to : „Ò¥– 500 K ¬⁄U Œ˝fl ‘S’ ∑§Ê flÊc¬ ŒÊ’ ‹ª÷ª ÁŸêŸ R>¡. 500 K `f âhplu ‘S’ _y„ bpó`v$bpZ Apif¡
(R=2 cal K−1 mol−1) ∑§ ’⁄UÊ’⁄U „ÊªÊ — _uQ¡_pdp„\u L$p¡C A¡L$_¡ bfpbf \i¡ S>¡ ip¡^p¡.
(1) 0.1 atm (R=2 cal K−1 mol−1) (R=2 cal K−1 mol−1)
(2) 1 atm (1) 0.1 atm (1) 0.1 atm
(3) 10 atm (2) 1 atm (2) 1 atm
(4) 100 atm (3) 10 atm (3) 10 atm
(4) 100 atm (4) 100 atm
35. Given
(i) 2Fe 2 O 3 (s) → 4Fe(s)+3O 2 (g) ; 35. Ap`¡g
35. ÁŒÿÊ ªÿÊ „Ò,
∆rG=+1487.0 kJ mol−1 (i) 2Fe 2 O 3 (s) → 4Fe(s)+3O 2 (g) ;
(ii) 2CO(g)+O2(g) → 2CO2(g) ; (i) 2Fe 2 O 3 (s) → 4Fe(s)+3O 2 (g) ;
∆rG=+1487.0 kJ mol−1 ∆rG=+1487.0 kJ mol−1
∆rG= −514.4 kJ mol−1 (ii) 2CO(g)+O2(g) → 2CO2(g) ;
Free energy change, ∆rG for the reaction (ii) 2CO(g)+O2(g) → 2CO2(g) ;
∆rG= −514.4 kJ mol−1 ∆rG= −514.4 kJ mol−1
2Fe 2O 3 (s)+6CO(g) → 4Fe(s)+6CO 2 (g) 2Fe 2O 3 (s)+6CO(g) → 4Fe(s)+6CO 2 (g)
will be : •Á÷Á∑˝§ÿÊ, 2Fe2O3(s)+6CO(g) → 4Fe(s)+
(1) −112.4 kJ mol−1
âq¾$ep dpV¡$ dy¼s EÅ® a¡fapf ∆rG _uQ¡_pdp„\u iy„
6CO2(g) ∑§ Á‹∞ ◊ÈÄà ™§¡Ê¸ ¬Á⁄UfløŸ, ∆rG „ÊªÊ —
(2) −56.2 kJ mol−1 (1) −112.4 kJ mol−1 li¡ ?
(3) −168.2 kJ mol−1 (2) −56.2 kJ mol−1 (1) −112.4 kJ mol−1
(4) −208.0 kJ mol−1 (3) −168.2 kJ mol−1 (2) −56.2 kJ mol−1
(4) −208.0 kJ mol−1 (3) −168.2 kJ mol−1
(4) −208.0 kJ mol−1

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Set - 03 21
36. Two 5 molal solutions are prepared by 36. X ÃÕÊ Y Áfl‹Êÿ∑§Ê ¥ ◊ ¥ ÁfllÈ Ã •Ÿ¬ÉÊ≈K ÃÕÊ 36. rhÛys ArhcpÄe Abpó`iug Öpìe_¡ ÖphL$p¡ X A_¡
dissolving a non-electrolyte non-volatile
•flÊc¬‡ÊË‹ Áfl‹ÿ ∑§Ê ÉÊÊ‹∑§⁄U •‹ª-•‹ª 5 ◊Ê‹‹ Y dp„ Ap¡Npmu_¡ AgN AgN 5 dp¡gg_p b¡ ÖphZp¡
solute separately in the solvents X and Y.
The molecular weights of the solvents are Áfl‹ÿŸ ÃÒÿÊ⁄U Á∑§ÿ ¡ÊÃ „Ò¥– Áfl‹Êÿ∑§Ê¥ ∑§ •áÊÈ÷Ê⁄U b_phhpdp„ Apìep. ÖphL$p¡_p ApÎhue v$m A_y¾$d¡
M X and M Y , respectively where 3 3
∑˝§◊‡Ê— MX ÃÕÊ MY „Ò¥ ¡„Ê° M X = M Y . X ◊¥ MX A_¡ MY R>¡, Äep„ M X = M Y . ÖphZ Y
3 4 4
MX= M Y . The relative lowering of
4 ’ŸÊÿ „È∞ Áfl‹ÿŸ ∑§ flÊc¬ŒÊ’ ∑§Ê ‚ʬˇÊ •flŸ◊Ÿ L$fsp„ ÖphZ X _p bpó`v$bpZdp„ \sp¡ kp`¡n OV$pX$p¡
vapour pressure of the solution in X is “m” Y ◊¥ ’ŸÊÿ „È∞ Áfl‹ÿŸ ∑§ ‚ʬˇÊ flÊc¬ŒÊ’ •flŸ◊Ÿ “m” NZp¡ R>¡ . Öpìe_p dp¡ g _u k„ ¿ ep ÖphL$ _ u
times that of the solution in Y. Given that
the number of moles of solute is very small
∑§Ê “m” ªÈŸÊ „Ò– ÁŒÿÊ ªÿÊ „Ò Á∑§ Áfl‹ÿ∑§ ∑§Ë kfMpdZudp„ M|b S> Ap¡R>u R>¡ S>¡ Ap`¡g R>¡, sp¡
in comparison to that of solvent, the value ÃÈ‹ŸÊ ◊¥ Áfl‹ÿ ∑§ ◊Ê‹Ê¥ ∑§Ë ‚¥ÅÿÊ ’„Èà ∑§◊ „Ò– “m” _u qL„$ds ip¡^p¡.
of “m” is : “m” ∑§Ê ◊ÊŸ „ÊªÊ — 4
(1)
4 4 3
(1) (1)
3 3 3
(2)
3 3 4
(2) (2)
4 4 1
(3)
1 1 2
(3) (3)
2 2 1
(4)
1 1 4
(4) (4)
4 4

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Set - 03 22
37. Following four solutions are prepared by 37. •‹ª-•‹ª ‚ÊãŒ˝ÃÊ•Ê¥ ∑§ NaOH ÃÕÊ HCl ∑§ 37. Sy>v$p Sy>v$p L$v$ A_¡ Sy>v$u Sy>v$u kp„Ösp ^fphsp NaOH
mixing different volumes of NaOH and
•‹ª-•‹ª •ÊÿßÊ¥ ∑§Ê Á◊‹Ê∑§⁄U øÊ⁄U Áfl‹ÿŸ ÃÒÿÊ⁄U A_¡ HCl _p ÖphZp¡_¡ rdî L$fu_¡ Qpf ÖphZp¡ Ap`¡gp
HCl of different concentrations, pH of
which one of them will be equal to 1 ? Á∑§ÿ ¡ÊÃ „Ò¥– ÁŸêŸ ◊¥ ‚ Á∑§‚ ’ŸÊÿ „È∞ Áfl‹ÿŸ ∑§Ê R>¡. _uQ¡ Ap`¡gpdp„\u L$ep A¡L$_u pH 1 _¡ bfpbf
M M pH, ∞∑§ (1) „ÊªÊ? \i¡ ?
(1) 100 mL HCl+100 mL NaOH
10 10 M M M M
(1) 100 mL HCl+100 mL NaOH (1) 100 mL HCl+100 mL NaOH
M M 10 10 10 10
(2) 75 mL HCl+25 mL NaOH
5 5 M M M M
(2) 75 mL HCl+25 mL NaOH (2) 75 mL HCl+25 mL NaOH
M M 5 5 5 5
(3) 60 mL HCl+40 mL NaOH
10 10 M M M M
(3) 60 mL HCl+40 mL NaOH (3) 60 mL HCl+40 mL NaOH
M M 10 10 10 10
(4) 55 mL HCl+45 mL NaOH
10 10 M M M M
(4) 55 mL HCl+45 mL NaOH (4) 55 mL HCl+45 mL NaOH
10 10 10 10
38. At a certain temperature in a 5 L vessel,
2 moles of carbon monoxide and 3 moles 38. ∞∑§ ÁŒÿ „È∞ Ãʬ ¬⁄U, 5 L ∑§ ¬ÊòÊ ◊¥ 2 ◊Ê‹ ∑§Ê’¸Ÿ 38. 5 L `pÓdp„ r_es sp`dp_¡ 2 dp¡g L$pb®_ dp¡_p¡¼kpCX$
of chlorine were allowed to reach
equilibrium according to the reaction,
◊ÊŸÊÄ‚Êß«U ÃÕÊ 3 ◊Ê‹ Ä‹Ê⁄UËŸ ∑§Ê •Á÷Á∑˝§Áÿà A_¡ 3 dp¡g ¼gp¡qf__¡ _uQ¡ Ap`¡g âq¾$ep âdpZ¡ k„syg_
CO+Cl2
COCl2
∑§⁄UÊ∑§ ÁŸêŸ ¬˝∑§Ê⁄U ‚ ‚Êêÿ ¬⁄U ‹ÊÿÊ ¡ÊÃÊ „Ò, âpá L$f¡ Ðep„ ky^u c¡Np L$fhpdp„ Apìep -
At equilibrium, if one mole of CO is present CO+Cl2
COCl2 CO+Cl2
COCl2
then equilibrium constant (K c ) for the ‚Êêÿ ¬⁄U ÿÁŒ CO ∑§Ê ∞∑§ ◊Ê‹ ©¬ÁSÕà „Ê ÃÊ k„syg_¡, Å¡ A¡L$ dp¡g CO lpS>f lp¡e sp¡ âq¾$ep dpV¡$_p¡
reaction is : •Á÷Á∑˝§ÿÊ ∑§Ê ‚Êêÿ ÁSÕ⁄UÊ¥∑§ (Kc) „ÊªÊ — k„syg_ AQmp„L$ (Kc) ip¡^p¡.
(1) 2
(1) 2 (1) 2
(2) 2.5
(2) 2.5 (2) 2.5
(3) 3
(3) 3 (3) 3
(4) 4
(4) 4 (4) 4

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Set - 03 23
39. If x gram of gas is adsorbed by m gram of 39. ÿÁŒ P ŒÊ’ ¬⁄U, Á∑§‚Ë ªÒ‚ ∑§Ê x ª˝Ê◊ Á∑§‚Ë ∞∑§ 39. v$bpZ P `f, Å¡ x N°pd hpey A¡ m N°pd Ar^ip¡jL$
x x x
adsorbent at pressure P, the plot of log m ª˝Ê◊ •Áœ‡ÊÊ·∑§ ‚ •Áœ‡ÊÊÁ·Ã „ÊÃÊ „Ò, ÃÊ log `f Ar^ip¡rjs \pe R>¡. log rhê$Ý^ log P _p¡
m m m
versus log P is linear. The slope of the plot
∑§Ê log P ∑§ ÁflL§h å‹Ê≈U ⁄UπËÿ „ÊªÊ– å‹Ê≈U ∑§Ë Apg¡M f¡Mue R>¡. Apg¡M_p¡ Y$pm iy„ R>¡ ?
is :
(n and k are constants and n > 1) ¬˝fláÊÃÊ (S‹Ê¬) ÁŸêŸ „ÊªË — (n A_¡ k AQmp„L$p¡ R>¡ A_¡ n > 1)
(1) 2 k (n ÃÕÊ k ÁSÕ⁄UÊ¥∑§ „Ò¥ ÃÕÊ n > 1) (1) 2k
(2) log k (1) 2k (2) log k
(3) n (2) log k (3) n
1 (3) n 1
(4) (4)
n 1 n
(4)
n
40. For a first order reaction, A → P, t½ (half- 40. â\d ¾$d_u âq¾$ep A → P dpV¡$, t½ (A^®-Apeyóe)
1 th 40. ¬˝Õ◊ ∑§ÊÁ≈U ∑§Ë •Á÷Á∑˝§ÿÊ A → P, ∑§ Á‹∞, t½ 1
life) is 10 days. The time required for 10 qv$hkp¡ R>¡. A _y„
4 1 4
ê$`p„sfZ \hp dpV¡$_p¡ S>ê$fu
(•h¸ •ÊÿÈ) 10 ÁŒŸ „Ò– A ∑§ 4 ¬Á⁄UfløŸ ∑§ Á‹∞
conversion of A (in days) is : kde (qv$hkp¡dp„) ip¡^p¡.
(ln 2=0.693, ln 3=1.1) (ÁŒŸÊ¥ ◊¥) ‹ªŸ flÊ‹Ê ‚◊ÿ „ÊªÊ — (ln 2=0.693, ln 3=1.1)
(1) 5 (ln 2=0.693, ln 3=1.1)
(2) 3.2 (1) 5
(1) 5
(3) 4.1 (2) 3.2
(2) 3.2
(4) 2.5 (3) 4.1
(3) 4.1
(4) 2.5
(4) 2.5

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Set - 03 24
41. Which of the following best describes the 41. •áÊÈ ∑§ˇÊ∑§ ∑§ ÁŒÿ ªÿ ÁøòÊ ∑§Ê, ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ 41. _uQ¡ Ap`¡ g ApÎhue L$nL$_u ApL© $ rs_¡ _uQ¡
diagram below of a molecular orbital ?
‚flÊ¸ûÊ◊ …¥ª ‚ ‚◊¤ÊÊÃÊ „Ò? Ap`¡gpdp„\u L$ep hX¡$ kp¥\u kpfu fus¡ hZ®hu iL$pe ?

(1) A non-bonding orbital
(2) An antibonding σ orbital (1) ∞∑§ •ŸÊ’¥œË ∑§ˇÊ∑§ (1) Ab„^L$pfL$ L$nL$
(3) A bonding π orbital (2) ∞∑§ ¬˝ÁÕʒ¥œË σ ∑§ˇÊ∑§ (2) b„^ârsL$pfL$ σ L$nL$
(4) An antibonding π orbital (3) ∞∑§ •Ê’¥œË π ∑§ˇÊ∑§ (3) b„^L$pfL$ π L$nL$
42. Biochemical Oxygen Demand (BOD) value
(4) ∞∑§ ¬˝ÁÕʒ¥œË π ∑§ˇÊ∑§ (4) b„^ârsL$pfL$ π L$nL$
can be a measure of water pollution caused
by the organic matter. Which of the 42. ¡Òfl ⁄UÊ‚ÊÿÁŸ∑§ •ÊÚÄ‚Ë¡Ÿ •Êfl‡ÿ∑§ÃÊ (BOD) ∑§Ê 42. L$pb®r_L `v$p\p£ Üpfp \sy„ `pZu_y„ âv|$jZ S>¥hfpkperZL$
following statements is correct ? ◊ÊŸ ∑§Ê’¸ÁŸ∑§ ¬ŒÊÕÊZ mÊ⁄UÊ Á∑§ÿ ªÿ ¡‹ ¬˝ŒÍ·áÊ ∑§Ê Ap¡[¼kS>_ X$udpÞX$ (BOD) _p d|ëe Üpfp Ap`u iL$pe
(1) Aerobic bacteria decrease the BOD ◊ʬ „Ê ‚∑§ÃÊ „Ò– ÁŸêŸ ∑§ÕŸÊ¥ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ‚„Ë „Ò? R>¡. _uQ¡ Ap`¡gp rh^p_p¡dp„\u L$ey„ kpQy„ R>¡ ?
value.
(2) Anaerobic bacteria increase the BOD (1) flÊÿÈ¡ËflË ’ÒÄ≈UËÁ⁄UÿÊ BOD ∑§Ê ◊ÊŸ ÉÊ≈UÊÃ „Ò¥– (1) hpeyÆhu$ b¡¼V¡$fuep BOD _y„ d|ëe OV$pX¡$ R>¡.
value. (2) •flÊÿflËÿ ’ÒÄ≈UËÁ⁄UÿÊ BOD ∑§Ê ◊ÊŸ ’…∏ÊÃ (2) Ahpehue$ b¡¼V¡$fuep BOD _y„ d|ëe h^pf¡$ R>¡.
(3) Clean water has BOD value higher „Ò–¥ (3) iyÙ `pZu_y„ BOD d|ëe 10 ppm \u h^pf¡
than 10 ppm.
(3) ‚Ê»§ ¡‹ ∑§ BOD ∑§Ê ◊ÊŸ 10 ppm ‚ lp¡e R>¡.
(4) Polluted water has BOD value higher
than 10 ppm. íÿÊŒÊ „ÊÃÊ „Ò– (4) âv|$rjs `pZu_y„ BOD d|ëe 10 ppm \u h^pf¡
(4) ¬˝ŒÍÁ·Ã ¡‹ ∑§ BOD ∑§Ê ◊ÊŸ 10 ppm ‚ lp¡e R>¡.
43. In KO2, the nature of oxygen species and íÿÊŒÊ „ÊÃÊ „Ò–
the oxidation state of oxygen atom are, 43. KO2 dp„, Ap¡[¼kS>_ õ`ukuT_u âL©$rs A_¡ Ap¡[¼kS>_
respectively :
43. KO2 ◊¥ •ÊÚÄ‚Ë¡Ÿ S¬Ë‡ÊË$¡ ∑§Ë ¬˝∑ΧÁà ÃÕÊ •ÊÚÄ‚Ë¡Ÿ `fdpÏ_u Ap¡[¼kX¡$i_ Ahõ\p A_y¾$d¡ ip¡^p¡.
(1) Oxide and −2
(2) Superoxide and −1/2 ¬⁄U◊ÊáÊÈ ∑§Ë •ÊÚÄ‚Ë∑§⁄UáÊ •flSÕÊ ∑˝§◊‡Ê— „Ò¥ — (1) Ap¡¼kpCX$ A_¡ −2
(3) Peroxide and −1/2 (1) •ÊÚÄ‚Êß«U ÃÕÊ −2 (2) ky`fAp¡¼kpCX$ A_¡ −1/2
(4) Superoxide and −1 (2) ‚Ȭ⁄U•ÊÚÄ‚Êß«U ÃÕÊ −1/2 (3) `fAp¡¼kpCX$ A_¡ −1/2
(3) ¬⁄U•ÊÚÄ‚Êß«U ÃÕÊ −1/2 (4) ky`fAp¡¼kpCX$ A_¡ −1
(4) ‚Ȭ⁄U•ÊÚÄ‚Êß«U ÃÕÊ −1

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Set - 03 25
44. The number of P−O bonds in P4O6 is : 44. P4O6 ◊¥ P−O •Ê’ãœÊ¥ ∑§Ë ‚¥ÅÿÊ „Ò — 44. P4O6 dp„ P−O b„^p¡_u k„¿ep ip¡^p¡.
(1) 6 (1) 6 (1) 6
(2) 9 (2) 9 (2) 9
(3) 12 (3) 12 (3) 12
(4) 18 (4) 18 (4) 18

45. Lithium aluminium hydride reacts with
45. ‹ËÁÕÿ◊ ∞ ‹ È ◊ ËÁŸÿ◊ „Êß«˛ U Êß«U, Á‚Á‹∑§ÊÚ Ÿ 45. rgr\ed A¡ëeyrdr_ed lpCX²$pCX$ kuguL$p¡_ V¡$V²$p¼gp¡fpCX$
silicon tetrachloride to form :
(1) LiCl, AlH3 and SiH4 ≈U≈˛UÊÄ‹Ê⁄UÊß«U ∑§ ‚ÊÕ •Á÷Á∑˝§ÿÊ ∑§⁄U∑§ ’ŸÊÃÊ „Ò — kp\¡ âq¾$ep L$fu_¡ S>¡ b_ph¡ s¡ ip¡^p¡.
(2) LiCl, AlCl3 and SiH4 (1) LiCl, AlH3 ÃÕÊ SiH4 (1) LiCl, AlH3 A_¡ SiH4
(3) LiH, AlCl3 and SiCl2 (2) LiCl, AlCl3 ÃÕÊ SiH4 (2) LiCl, AlCl3 A_¡ SiH4
(4) LiH, AlH3 and SiH4
(3) LiH, AlCl3 ÃÕÊ SiCl2 (3) LiH, AlCl3 A_¡ SiCl2
46. The correct order of spin-only magnetic (4) LiH, AlH3 ÃÕÊ SiH4 (4) LiH, AlH3 A_¡ SiH4
moments among the following is :
(Atomic number : Mn=25, Co=27, 46. ÁŸêŸ ∑§ ’Ëø ¬˝ø∑˝§áÊ ◊ÊòÊ øÈê’∑§Ëÿ •ÊÉÊÍáʸ ∑§Ê ‚„Ë 46. _uQ¡ Ap`¡gpdp„\u a¼s õ`u_ Qy„bL$ue QpL$dpÓp_p¡ kpQp¡
Ni=28, Zn=30) ∑˝§◊ „Ò — ¾$d ip¡^p¡.
(1) [ZnCl4]2− > [NiCl4]2− > [CoCl4]2−
> [MnCl4]2− (¬⁄U◊ÊáÊÈ ‚¥ÅÿÊ — Mn=25, Co=27, Ni=28, (`fdpÎhue k„¿ep : Mn=25, Co=27, Ni=28,
(2) [CoCl4]2− > [MnCl4]2− > [NiCl4]2− Zn=30 ) Zn=30 )
> [ZnCl4]2− (1) [ZnCl4]2− > [NiCl4]2− > [CoCl4]2− (1) [ZnCl4]2− > [NiCl4]2− > [CoCl4]2−
(3) [NiCl4]2− > [CoCl4]2− > [MnCl4]2− > [MnCl4]2− > [MnCl4]2−
> [ZnCl4]2− (2) [CoCl4]2− > [MnCl4]2− > [NiCl4]2− (2) [CoCl4]2− > [MnCl4]2− > [NiCl4]2−
(4) [MnCl4]2− > [CoCl4]2− > [NiCl4]2− > [ZnCl4]2− > [ZnCl4]2−
> [ZnCl4]2− (3) [NiCl4]2− > [CoCl4]2− > [MnCl4]2− (3) [NiCl4]2− > [CoCl4]2− > [MnCl4]2−
> [ZnCl4]2− > [ZnCl4]2−
47. The correct order of electron affinity is : (4) [MnCl4]2− > [CoCl4]2− > [NiCl4]2− (4) [MnCl4]2− > [CoCl4]2− > [NiCl4]2−
(1) F > Cl > O > [ZnCl4]2− > [ZnCl4]2−
(2) F > O > Cl
(3) Cl > F > O 47. ß‹Ä≈˛UÊÚŸ ’¥œÈÃÊ ∑§Ê ‚„Ë ∑˝§◊ „Ò —
(4) O > F > Cl (1) F > Cl > O 47. Cg¡¼V²$p¡_ b„^ysp_p¡ kpQp¡ ¾$d ip¡^p¡.
(2) F > O > Cl (1) F > Cl > O
(3) Cl > F > O (2) F > O > Cl
(4) O > F > Cl (3) Cl > F > O
(4) O > F > Cl

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Set - 03 26
48. In XeO3F2, the number of bond pair(s), 48. XeO3F2 ◊ ¥ , •Ê’¥ œ -ÿÈ Ç ◊ (ÿÈ Ç ◊Ê ¥ ) , π -•Ê’¥ œ 48. XeO3F2 dp„ b„^ eyÁd(dp¡), π-b„^(^p¡) _u k„¿ep
π-bond(s) and lone pair(s) on Xe atom
(•Ê’¥œÊ¥) ÃÕÊ Xe ¬⁄U◊ÊáÊÈ ¬⁄U ∞∑§Ê∑§Ë ÿÈÇ◊ (ÿÈÇ◊Ê¥) A_¡ Xe D`f Ab„^L$pfL$ eyÁd(dp¡) _u k„¿ep A_y¾$d¡
respectively are :
(1) 5, 2, 0 ∑§Ë ‚¥ÅÿÊ ∑˝§◊‡Ê— „Ò¥ — ip¡^p¡.
(2) 4, 2, 2 (1) 5, 2, 0 (1) 5, 2, 0
(3) 5, 3, 0 (2) 4, 2, 2 (2) 4, 2, 2
(4) 4, 4, 0 (3) 5, 3, 0 (3) 5, 3, 0
(4) 4, 4, 0 (4) 4, 4, 0
49. In the leaching method, bauxite ore is 49. r_npg_ `Ý^rsdp„, bp¡¼kpCV$ AeõL$_¡ kp„Ö NaOH
digested with a concentrated solution of 49. ÁŸˇÊÊ‹Ÿ ÁflÁœ ◊¥ ’ÊÚÄ‚Êß≈U •ÿS∑§ ∑§Ê NaOH ∑§
NaOH that produces ‘X’. When CO2 gas _p ÖphZ kp\¡ Qe_ L$fsp„ ‘X’ DÐ`Þ_ \pe R>¡. ‘X’
‚ÊãŒ˝ Áfl‹ÿŸ ◊¥ ¬ÊÁøà Á∑§ÿÊ ¡ÊÃÊ „Ò Á¡‚‚ ‘X’
is passed through the aqueous solution of _p S>gue ÖphZdp„\u CO2 hpey `kpf L$fhpdp„ Aphsp
‘X’, a hydrated compound ‘Y’ is
¬˝Êåà „ÊÃÊ „Ò– ¡’ CO2 ∑§Ê ‘X’ ∑§ ¡‹Ëÿ Áfl‹ÿŸ ‚
`pZuey¼s k„ep¡S>_ ‘Y’ Ahn¡r`s \pe R>¡. ‘X’ A_¡
precipitated. ‘X’ and ‘Y’ respectively ¬˝flÊÁ„à Á∑§ÿÊ ¡ÊÃÊ „Ò Ã’ ∞∑§ ¡‹ÿÊÁ¡Ã ÿÊÒÁª∑§
‘Y’ A_y¾$d¡ ip¡^p¡.
are : ‘Y’ •flˇÊÁ¬Ã „ÊÃÊ „Ò– ‘X’ ÃÕÊ ‘Y’ ∑˝§◊‡Ê— „Ò¥ —
(1) NaAlO2 and Al2(CO3)3⋅x H2O (1) NaAlO2 A_¡ Al2(CO3)3⋅x H2O
(1) NaAlO2 ÃÕÊ Al2(CO3)3⋅x H2O
(2) Al(OH)3 and Al2O3⋅x H2O (2) Al(OH)3 A_¡ Al2O3⋅x H2O
(3) Na[Al(OH)4] and Al2O3⋅x H2O (2) Al(OH)3 ÃÕÊ Al2O3⋅x H2O
(3) Na[Al(OH)4] A_¡ Al2O3⋅x H2O
(4) Na[Al(OH)4] and Al2(CO3)3⋅x H2O (3) Na[Al(OH)4] ÃÕÊ Al2O3⋅x H2O
(4) Na[Al(OH)4] A_¡ Al2(CO3)3⋅x H2O
(4) Na[Al(OH)4] ÃÕÊ Al2(CO3)3⋅x H2O
50. The total number of possible isomers for 50. kdsgue kdQp¡fk
square-planar [Pt(Cl)(NO2)(NO3)(SCN)]2−
50. flª¸ ‚◊Ã‹Ë [Pt(Cl)(NO2)(NO3)(SCN)]2− ∑§ [Pt(Cl)(NO 2 )(NO 3 )(SCN)] 2− dpV¡ $ i¼e
is :
(1) 8 Á‹∞ ‚ê÷fl ‚◊ÊflÿÁflÿÊ¥ ∑§Ë ∑ȧ‹ ‚¥ÅÿÊ „Ò — kdOV$L$p¡_u Ly$g k„¿ep ip¡^p¡.
(2) 12 (1) 8 (1) 8
(3) 16 (2) 12 (2) 12
(4) 24 (3) 16 (3) 16
(4) 24 (4) 24

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51. Two compounds I and II are eluted by 51. SÃê÷fláʸ ‹πŸ mÊ⁄UÊ ŒÊ ÿÊÒÁª∑§Ê¥ I ÃÕÊ II (•Áœ‡ÊÊ·áÊ 51. b¡ k„ep¡S>_p¡ I A_¡ II _¡ õs„c ¾$p¡d¡V$p¡N°pau (Ar^ip¡jZ
column chromatography (adsorption of
I > II) ∑§Ê ˇÊÊÁ‹Ã Á∑§ÿÊ– ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ ∞∑§ ‚„Ë I > II) Üpfp r_npgus (eluted) L$fhpdp„ Aph¡ R>¡.
I > II). Which one of following is a correct
statement ? ∑§ÕŸ „Ò? _uQ¡ Ap`¡gpdp„\u L$ey„ A¡L$ rh^p_ kpQy„ R>¡ ?
(1) I moves faster and has higher R f (1) I Ã¡ ø‹ÃÊ „Ò ÃÕÊ ©‚∑§ Rf ∑§Ê ◊ÊŸ II ∑§Ë (1) I TX$`\u Qpg¡ R>¡ A_¡ s¡_y„ Rf d|ëe II L$fsp„
value than II ÃÈ‹ŸÊ ◊¥ ©ëøÃ⁄U „Ò– h^pf¡ R>¡.
(2) II moves faster and has higher R f
(2) II Ã¡ ø‹ÃÊ „Ò ÃÕÊ ©‚∑§ Rf ∑§Ê ◊ÊŸ I ∑§Ë (2) II TX$`\u Qpg¡ R>¡ A_¡ s¡_y„ Rf d|ëe I L$fsp„
value than I
(3) I moves slower and has higher Rf ÃÈ‹ŸÊ ◊¥ ©ëøÃ⁄U „Ò– h^pf¡ R>¡.
value than II (3) I œË◊Ê ø‹ÃÊ „Ò ÃÕÊ ©‚∑§ Rf ∑§Ê ◊ÊŸ II ∑§Ë (3) I ^ud¡ Qpg¡ R>¡ A_¡ s¡_y„ Rf d|ëe II L$fsp h^pf¡
(4) II moves slower and has higher Rf ÃÈ‹ŸÊ ◊¥ ©ëøÃ⁄U „Ò– R>¡.
value than I
(4) II œË◊Ê ø‹ÃÊ „Ò ÃÕÊ ©‚∑§ Rf ∑§Ê ◊ÊŸ I ∑§Ë (4) II ^ud¡ Qpg¡ R>¡ A_¡ s¡_y„ Rf d|ëe I L$fsp„ h^pf¡
52. Which of the following statements is not ÃÈ‹ŸÊ ◊¥ ©ëøÃ⁄U „Ò– R>¡.
true ?
(1) Step growth polymerisation requires 52. ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ‚àÿ Ÿ„Ë¥ „Ò? 52. _uQ¡ Ap`¡gp rh^p_p¡dp„\u L$ey„ kpQy„ _\u ?
a bifunctional monomer.
(1) ‚Ê¬ÊŸflÎÁh ’„È‹∑§Ÿ ∑§ Á‹∞ Ám•Á÷‹ˇÊ∑§ (1) sb½$phpf h©[Ý^ blºguL$fZ dpV¡$ b¡ q¾$epiug
(2) Nylon 6 is an example of step-
growth polymerisation. ∞∑§‹∑§ ∑§Ë •Êfl‡ÿ∑§ÃÊ „ÊÃË „Ò– dp¡_p¡df S>ê$fu R>¡.
(3) Chain growth polymerisation (2) ŸÊÿ‹ÊŸ - 6 ‚Ê¬ÊŸflÎÁh ’„È‹∑§Ÿ ∑§Ê ∞∑§ (2) _pegp¡_ 6 A¡ sb½$phpf h©[Ý^ blºguL$fZ_y„
includes both homopolymerisation ©ŒÊ„⁄UáÊ „Ò– Dv$plfZ R>¡.
and copolymerisation.
(3) oÎ¥π‹Ê flÎÁh ’„È‹∑§Ÿ ◊¥ ‚◊’„È‹∑§Ÿ ÃÕÊ (3) kp„L$m h©[Ý^ blºguL$fZ A¡ lp¡dp¡blºguL$fZ A_¡
(4) Chain growth polymerisation
involves homopolymerisation only. ‚„’„È‹∑§Ÿ ŒÊŸÊ¥ „ÊÃ „Ò¥– L$p¡blºguL$fZ bÞ_¡ _¡ kdph¡ R>¡.
(4) oÎ¥π‹Ê flÎÁh ’„È‹∑§Ÿ ◊¥ ◊ÊòÊ ‚◊’„È‹∑§Ÿ (4) kp„L$m h©[Ý^ blºguL$fZdp„ a¼s lp¡dp¡blºguL$fZ
„ÊÃÊ „Ò– ipd¡g R>¡.

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53. When 2-butyne is treated with H 2 / 53. ¡’ 2-éÿÍ≈UÊߟ ∑§Ê H2/Á‹ã«U‹⁄U ©à¬˝⁄U∑§ ∑§ ‚ÊÕ 53. Äepf¡ 2-åe|V$pC__u H2/guÞX$gf DØu`L$ kp\¡ âq¾$ep
Lindlar’s catalyst, compound X is
•Á÷Á∑˝§Áÿà Á∑§ÿÊ ¡ÊÃÊ „Ò ÃÊ ÿÊÒÁª∑§ X ∞∑§ ◊ÈÅÿ L$fsp k„ep¡S>_ X A¡ dy¿e _u`S> sfuL¡$ âpá \pe R>¡
produced as the major product and when
treated with Na/liq. NH3 it produces Y as ©à¬ÊŒ ∑§ M§¬ ◊¥ Á◊‹ÃÊ „Ò •ÊÒ⁄U ¡’ ©‚ Na/Œ˝fl A_¡ Äepf¡ s¡_u Na/liq. NH3 kp\¡ âq¾$ep L$fsp„ Y
the major product. Which of the following NH3 ∑§ ‚ÊÕ •Á÷Á∑˝§Áÿà Á∑§ÿÊ ¡ÊÃÊ „Ò Ã’ fl„ Y A¡ dy¿e _u`S> sfuL¡$ âpá \pe R>¡.
statements is correct ? ∞∑§ ◊ÈÅÿ ©à¬ÊŒ ∑§ M§¬ ◊¥ ŒÃÊ „Ò– ÁŸêŸ ∑§ÕŸÊ¥ ◊¥ ‚ _uQ¡ Ap`¡gp rh^p_p¡dp„\u L$ey„ rh^p_ kpQy„ R>¡ ?
(1) X will have higher dipole moment
∑§ÊÒŸ ‚Ê ∑§ÕŸ ‚„Ë „Ò? (1) X _u Y L$fsp„ qÜ^y°h QpL$dpÓp A_¡ DÐL$g_tbvy$
and higher boiling point than Y.
(2) Y will have higher dipole moment (1) X ∑§Ê, Y ∑§Ë ÃÈ‹ŸÊ ◊¥, ©ëøÃ⁄U Ámœ˝Èfl •ÊÉÊÍáʸ dlÑd \i¡.
and higher boiling point than X. ÃÕÊ ©ëøÃ⁄U ÄflÕŸÊ¥∑§ „ÊªÊ– (2) Y _u X L$fsp„ qÜ^y°h QpL$dpÓp A_¡ DÐL$g_tbvy$
(3) X will have lower dipole moment (2) Y ∑§Ê, X ∑§Ë ÃÈ‹ŸÊ ◊¥, ©ëøÃ⁄U Ámœ˝Èfl •ÊÉÊÍáʸ dlÑd \i¡.
and lower boiling point than Y.
(4) Y will have higher dipole moment ÃÕÊ ©ëøÃ⁄U ÄflÕŸÊ¥∑§ „ÊªÊ– (3) X _u Y L$fsp„ qÜ^y°h QpL$dpÓp A_¡ DÐL$g_tbvy$
and lower boiling point than X. (3) X ∑§Ê, Y ∑§Ë ÃÈ‹ŸÊ ◊¥, ÁŸêŸÃ⁄U Ámœ˝Èfl •ÊÉÊÍáʸ _uQy„ \i¡.
ÃÕÊ ÁŸêŸÃ⁄U ÄflÕŸÊ¥∑§ „ÊªÊ– (4) Y _u X L$fsp„ qÜ^°yh QpL$dpÓp dlÑd \i¡ A_¡
54. The increasing order of the acidity of the
(4) Y ∑§Ê, X ∑§Ë ÃÈ‹ŸÊ ◊¥, Ámœ˝Èfl •ÊÉÊÍáʸ ©ëøÃ⁄U DÐL$g_tbvy$ _uQy„ \i¡.
following carboxylic acids is :
ÃÕÊ ÄflÕŸÊ¥∑§ ÁŸêŸÃ⁄U „ÊªÊ–
54. _uQ¡ Ap`¡ g p L$pbp£ [ ¼krgL$ A¡ r kX$p¡ d p„ \ u s¡ _ u
54. ÁŸêŸ ∑§Ê’Ê¸ÁÄ‚Á‹∑§ •ê‹Ê¥ ∑§Ë •ê‹ËÿÃÊ ∑§Ê ’…∏ÃÊ A¡rkqX$L$sp_p¡ QY$sp¡ ¾$d ip¡^p¡.
∑˝§◊ „Ò —

(1) I < III < II < IV
(2) IV < II < III < I
(3) II < IV < III < I
(4) III < II < IV < I
(1) I < III < II < IV (1) I < III < II < IV
(2) IV < II < III < I (2) IV < II < III < I
(3) II < IV < III < I (3) II < IV < III < I
(4) III < II < IV < I (4) III < II < IV < I

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Set - 03 29
55. The major product formed in the following 55. ÁŸêŸ •Á÷Á∑˝§ÿÊ ◊¥ ’ŸŸflÊ‹Ê ◊ÈÅÿ ©à¬ÊŒ „Ò — 55. _uQ¡ Ap`¡g âq¾$epdp„ b_su dy¿e _u`S> ip¡^p¡.
reaction is :

(1)
(1) (1)

(2)
(2) (2)

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(3) (3) (3)

(4) (4) (4)

56. On treatment of the following compound 56. ÁŸêŸ ÿÊÒÁª∑§ ∑§Ê ∞∑§ ¬˝’‹ •ê‹ ‚ •Á÷Á∑˝§Áÿà 56. _uQ¡ Ap`¡gp k„ep¡S>__u âbm A¡rkX$ kp\¡ âq¾$ep
with a strong acid, the most susceptible
∑§⁄UŸ ¬⁄U •Ê’㜠≈ÍU≈UŸ ∑§Ê ‚flʸÁœ∑§ ‚ª˝Ês SÕÊŸ L$fsp„, b„^ s|V$hp_y„ kp¥\u h^y k„crhs õ\p_ ip¡^p¡.
site for bond cleavage is :
„ÊªÊ —

(1) C1−O2 (1) C1−O2
(2) O2−C3 (1) C1−O2 (2) O2−C3
(3) C4−O5 (2) O2−C3 (3) C4−O5
(4) O5−C6 (3) C4−O5 (4) O5−C6
(4) O5−C6

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57. The increasing order of diazotisation of the 57. ÁŸêŸ ÿÊÒÁª∑§Ê¥ ∑§ «UÊß∞¡Ê≈UË∑§⁄UáÊ ∑§Ê ’…∏ÃÊ „È•Ê ∑˝§◊ 57. _uQ¡ Ap`¡gp k„ep¡S>_p¡_p¡ X$peA¡Tp¡V$pCT¡i__p¡ QY$sp¡
following compounds is :
„Ò — ¾$d ip¡^p¡.
(a)
(a) (a)

(b)
(b) (b)

(c)
(c) (c)

(d)
(d) (d)

(1) (a) < (b) < (c) < (d)
(1) (a) < (b) < (c) < (d) (1) (a) < (b) < (c) < (d)
(2) (a) < (d) < (b) < (c)
(2) (a) < (d) < (b) < (c) (2) (a) < (d) < (b) < (c)
(3) (a) < (d) < (c) < (b)
(3) (a) < (d) < (c) < (b) (3) (a) < (d) < (c) < (b)
(4) (d) < (c) < (b) < (a)
(4) (d) < (c) < (b) < (a) (4) (d) < (c) < (b) < (a)

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58. The dipeptide, Gln-Gly, on treatment with 58. «UÊ߬å≈UÊß«U, Gln-Gly ∑§Ê CH3COCl ∑§ ‚ÊÕ 58. X$pe`¡àV$pCX$ Gln-Gly, _u CH3COCl kp\¡ âq¾$ep
CH3COCl followed by aqueous work up
•Á÷Á∑˝§Áÿà ∑§⁄UŸ ∑§ ÃଇøÊØ ¡‹Ëÿ ∑§◊¸áÊ (work L$ep® bpv$ S>gue L$pe® L$fsp (aqueous work up)
gives :
up) ¬⁄U ¬˝Êåà „ÊªÊ — âpá \pe s¡ ip¡^p¡.
(1)
(1) (1)

(2)
(2) (2)

(3)
(3) (3)

(4)
(4) (4)

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59. The total number of optically active 59. ÁŸêŸ •Á÷Á∑˝§ÿÊ ◊¥ ’Ÿ œÈ˝fláÊ ÉÊÍáʸ∑§ÃÊ flÊ‹ ÿÊÒÁª∑§Ê¥ 59. _uQ¡ Ap`¡ g âq¾$ e pdp„ \ u b_sp âL$ p i q¾$ e piug
compounds formed in the following
∑§Ë ∑ȧ‹ ‚¥ÅÿÊ „Ò — k„ep¡S>_p¡_u Ly$g k„¿ep ip¡^p¡.
reaction is :

(1) Two (1) ŒÊ (1) b¡
(2) Four (2) øÊ⁄U (2) Qpf
(3) Six (3) ¿U—
(4) Zero (3) R>
(4) ‡ÊÍãÿ (4) i|Þe
60. The major product formed in the following
reaction is : 60. ÁŸêŸ •Á÷Á∑˝§ÿÊ ◊¥ ’ŸŸflÊ‹Ê ◊ÈÅÿ ©à¬ÊŒ „Ò, 60. _uQ¡ Ap`¡g âq¾$epdp„\u b_su dy¿e _u`S> ip¡^p¡.

(1)
(1) (1)

(2)
(2) (2)

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(3) (3) (3)

(4) (4) (4)

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PART C — MATHEMATICS ÷ʪ C — ªÁáÊà cpN C — NrZs
^pfp¡ L¡ $ f : A → B A¡ x −1
61. Let f : A → B be a function defined as 61. f (x) = Üpfp
61. ÿÁŒ A=R−{2}, B=R−{1} „Ò ¥ ÃÕÊ »§‹Ÿ x −2
x − 1 , where A=R−{2} and
f (x) =
x −2 f : A → B; f ( x ) = x − 1 mÊ⁄UÊ ¬Á⁄U÷ÊÁ·Ã „Ò, ìep¿epres rh^¡ e R>¡ , Äep„ A=R−{2} A_¡
x −2 B=R−{1}. sp¡ f _y„ __________
B=R−{1}. Then f is :
ÃÊ f — (1) ârsrh^¡e A[õsÐh ^fph¡ R>¡ A_¡
−1 3y − 1
(1) invertible and f (y ) = 3y − 1
y −1 √ÿÈà∑˝§◊áÊËÿ „Ò ÃÕÊ f −1 ( y ) = 3y − 1
(1)
y −1 f −1 ( y ) =
y −1
−1 2y − 1
(2) invertible and f (y) = 2y − 1 (2)
√ÿÈà∑˝§◊áÊËÿ „Ò •ÊÒ⁄U f −1 ( y ) =
y −1 ârsrh^¡e A[õsÐh ^fph¡ R>¡ A_¡
(2)
y −1 2y − 1
−1 2y + 1 f −1 ( y ) =
(3) invertible and f (y) = 2y + 1 y −1
y −1 (3) √ÿÈà∑˝§◊áÊËÿ „Ò •ÊÒ⁄U f −1 ( y ) =
y −1 (3) ârsrh^¡e A[õsÐh ^fph¡ R>¡ A_¡
(4) not invertible
(4) √ÿÈà∑˝§◊áÊËÿ Ÿ„Ë¥ „Ò– 2y + 1
62. If f (x) is a quadratic expression such that f −1 ( y ) =
y −1
f (1)+f (2)=0, and −1 is a root of f (x)=0, 62. ÿÁŒ f (x) ∞∑§ ÁmÉÊÊà √ÿ¥¡∑§ „Ò, Á¡‚∑§ Á‹ÿ
then the other root of f (x)=0 is : (4) ârsrh^¡e A[õsÐh ^fphsy„ _\u
f (1)+f (2)=0 ÃÕÊ f (x)=0 ∑§Ê ∞∑§ ◊Í‹ −1, „Ò,
5
(1) − ÃÊ f (x)=0 ∑§Ê ŒÍ‚⁄UÊ ◊Í‹ „Ò — 62. Å¡ f (x) A¡ h u qÜOps `v$phrg lp¡ e L¡ $ S>¡ \ u
8
5 f (1)+f (2)=0 A_¡ −1 A¡ f (x)=0 _y„ A¡L$ buS>
8 (1) −
(2) − 8 lp¡e, sp¡ f (x)=0 _y„ AÞe buS> __________ R>¡.
5
8 5
5 (2) − (1) −
(3) 5 8
8
5 8
8 (3) (2) −
(4) 8 5
5
8 5
(4) (3)
5 8
8
(4)
5

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63. If
z−3+2i
≤ 4 then the difference 63. ÿÁŒ
z−3+2i
≤ 4 „Ò, ÃÊ
z
∑§ •Áœ∑§Ã◊ ÃÕÊ 63. Å¡
z−3+2i
≤ 4 sp¡
z
_u Ar^L$Ñd qL„$ds A_¡
between the greatest value and the least
ãÿÍŸÃ◊ ◊ÊŸÊ¥ ∑§Ê •¥Ã⁄U „Ò — gOyÑd qL„$ ds hÃQ¡_p¡ saphs __________ R>¡.
value of
z
is :
(1) 2 13 (1) 2 13
(1) 2 13
(2) 8 (2) 8
(2) 8
(3) 4 + 13 (3) 4 + 13
(3) 4 + 13
(4) 13 (4) 13
(4) 13

64. Suppose A is any 3×3 non-singular 64. ◊ÊŸÊ Á∑§ A ∑§Ê߸ 3×3 √ÿÈà∑˝§◊áÊËÿ •Ê√ÿÍ„ „Ò Á¡‚∑§ 64. ^pfp¡ L¡$ A A¡ A¡L$ 3×3 kpdpÞe î¡rZL$ R>¡ A_¡
matrix and (A−3I)(A−5I)=O, where Á‹∞ (A−3I)(A−5I)=O, ¡„Ê° I=I 3 ÃÕÊ (A−3I)(A−5I)=O, Äep„ I=I3 A_¡ O=O3.
I=I3 and O=O3. If αA+βA−1=4I, then O=O3 „Ò– ÿÁŒ αA+βA−1=4I „Ò, ÃÊ α+β Å¡ αA+βA −1 =4I, sp¡ α+β bfpbf
α+β is equal to :
’⁄UÊ’⁄U „Ò — __________ R>¡.
(1) 8
(2) 7 (1) 8 (1) 8
(3) 13 (2) 7 (2) 7
(4) 12 (3) 13 (3) 13
(4) 12 (4) 12
65. If the system of linear equations
x+ay+z=3 65. ÿÁŒ ÁŸêŸ ⁄ÒUÁπ∑§ ‚◊Ë∑§⁄UáÊ ÁŸ∑§Êÿ 65. Å¡ kyf¡M kduL$fZ k„lrs
x+2y+2z=6 x+ay+z=3 x+ay+z=3
x+5y+3z=b x+2y+2z=6 x+2y+2z=6
has no solution, then : x+5y+3z=b x+5y+3z=b
(1) a=−1, b=9 ∑§Ê ∑§Ê߸ „‹ Ÿ„Ë¥ „Ò, ÃÊ — _¡ A¡L$ `Z DL¡$g _ lp¡e, sp¡
(2) a=−1, b ≠ 9 (1) a=−1, b=9
(3) a ≠−1, b=9 (2) a=−1, b ≠ 9 (1) a=−1, b=9
(4) a=1, b ≠ 9 (3) a ≠−1, b=9 (2) a=−1, b ≠ 9
(4) a=1, b ≠ 9 (3) a ≠−1, b=9
(4) a=1, b ≠ 9

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Set - 03 37
66. The number of four letter words that can 66. ‡ÊéŒ BARRACK ∑§ •ˇÊ⁄UÊ¥ ∑§Ê ¬˝ÿÊª ∑§⁄U∑§ ’ŸÊ∞ 66. BARRACK iåv$_p d|mpnfp¡_p¡ D`ep¡N L$fu Qpf
be formed using the letters of the word
¡Ê ‚∑§Ÿ flÊ‹ øÊ⁄U •ˇÊ⁄UÊ¥ ∑§ ‚÷Ë ‡ÊéŒÊ¥ ∑§Ë ‚¥ÅÿÊ d|mpnfp¡ hpmp L¡$V$gp„ iåv$p¡ b_phu iL$pe ?
BARRACK is :
(1) 120 „Ò — (1) 120
(2) 144 (1) 120 (2) 144
(3) 264 (2) 144 (3) 264
(4) 270 (3) 264 (4) 270
(4) 270
67. The coefficient of x10 in the expansion of 67. (1+x)2(1+x2)3(1+x3)4 _p„ rhõsfZdp„ x10 _p¡
(1+x)2(1+x2)3(1+x3)4 is equal to : 67. (1+x)2(1+x2)3(1+x3)4 ∑§ ¬˝‚Ê⁄U ◊¥ x10 ∑§Ê klNyZL$ __________ R>¡.
(1) 52 ªÈáÊÊ¥∑§ ’⁄UÊ’⁄U „Ò —
(2) 56 (1) 52
(1) 52
(3) 50 (2) 56
(2) 56
(4) 44 (3) 50
(3) 50
(4) 44
(4) 44
68. If a, b, c are in A.P. and a2, b2, c2 are in
3 68. Å¡ a, b, c kdp„sf î¡Zu (A.P.) dp„ A_¡ a2, b2, c2
G.P. such that a < b < c and a + b + c = , 68. ÿÁŒ a, b, c ∞∑§ ‚◊Ê¥Ã⁄U üÊ…∏Ë ◊¥ „Ò¥ ÃÕÊ a2, b2, c2
4 kdNyZp¡Ñf î¡Zu (G.P.) dp„ lp¡e s\p a < b < c
∞∑§ ªÈáÊÊûÊ⁄U üÊ…∏Ë ◊¥ „Ò¥, ¡’Á∑§ a < b < c ÃÕÊ
A_¡ a + b + c = 3 lp¡e, sp¡ a _u qL„$ds .... R>¡.
then the value of a is :
3
1 1 a + b + c = „Ò, ÃÊ a ∑§Ê ◊ÊŸ „Ò — 4
(1) − 4
4 4 2 1 1
1 1 (1) −
1 1 (1) − 4 4 2
(2) − 4 4 2
4 3 2 1 1
1 1 (2) −
1 1 (2) − 4 3 2
(3) − 4 3 2
4 2 2 1 1
1 1 (3) −
1 1 (3) − 4 2 2
(4) − 4 2 2
4 2 1 1
1 1 (4) −
(4) − 4 2
4 2

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Set - 03 38
2 3 2 3 2 3
3 3 3 3 3 3 3 3 3
69. Let A n =   −   +   − ..... + 69. ÿÁŒ A n =   −   +   − ..... + 69. ^pfp¡ L¡$ A n =   −   +   − ..... +
4 4 4 4 4 4 4 4 4
n n n
3 3 3
(−1)n−1   and B n = 1 − A n . Then, (−1)n−1   ÃÕÊ Bn = 1 − A n „Ò, ÃÊ ãÿÍŸÃ◊ (−1)n−1   A_¡ Bn = 1 − A n . sp¡ b^pS>
4 4 4
the least odd natural number p, so that Áfl·◊ ¬ÍáÊÊZ∑§ p, Á¡‚∑§ Á‹∞ ‚÷Ë n  p ∑§ Á‹∞ n  p dpV¡$, Bn > An \pe s¡hu Þe|_sd AeyÁd
Bn > An, for all n  p, is :
Bn > An „Ò, „Ò — âpL©$rsL$ k„¿ep p L$C R>¡ ?
(1) 9
(1) 9 (1) 9
(2) 7
(2) 7 (2) 7
(3) 11
(3) 11 (3) 11
(4) 5
(4) 5 (4) 5
lim x tan 2 x − 2x tan x equals :
70. x→0 lim x tan 2 x − 2x tan x ’⁄UÊ’⁄U „Ò — lim x tan 2 x − 2x tan x =__________
(1 − cos 2x )2 70. x→0
(1 − cos 2x )2 70. x→0
(1 − cos 2x )2
1
(1) 1 1
4 (1)
4 (1)
4
(2) 1
(2) 1 (2) 1
1
(3) 1 1
2 (3) (3)
2 2
1
(4) − 1 1
2 (4) − (4) −
2 2
(x − 1) 2−1 x , x > 1, x ≠ 2
71. Let f ( x ) =  (x − 1) 2−1 x , x > 1, x ≠ 2 1
71. ◊ÊŸÊ, f ( x ) =  „Ò–  , x > 1, x ≠ 2
 k , x =2
^pfp¡ L¡$ f ( x ) = (x − 1)
2−x

 k , x = 2 71.
The value of k for which f is continuous at  k , x =2
x=2 is : ÃÊ k ∑§Ê fl„ ◊ÍÀÿ, Á¡‚∑§ Á‹ÿ f, x=2 ¬⁄U ‚¥Ãà „Ò,
x=2 ApNm f kss \pe s¡ dpV¡$_u k _u qL„$ds
(1) 1 „Ò —
(2) e (1) 1 __________ R>¡.
(3) e−1 (2) e (1) 1
(4) e−2 (3) e−1 (2) e
(4) e−2 (3) e−1
(4) e−2
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Set - 03 39

 2 × 3x   1  x  x
72. If f ( x ) = sin−1 
 1+9 x  , then f
−  72. ÿÁŒ f ( x ) = sin−1  2 × 3x  „Ò, ÃÊ f
− 1  72. Å¡ f ( x ) = sin−1  2 × 3x  , sp¡  1
f
−  =
   2  1+9   2  1+9   2
equals : ’⁄UÊ’⁄U „Ò — __________.
(1) − 3 log e 3 (1) − 3 log e 3 (1) − 3 log e 3
(2) 3 log e 3 (2) 3 log e 3 (2) 3 log e 3
(3) − 3 log e 3 (3) − 3 log e 3 (3) − 3 log e 3
(4) 3 log e 3 (4) 3 log e 3 (4) 3 log e 3

73. Let f (x) be a polynomial of degree 4 having
73. ◊ÊŸÊ f (x) ÉÊÊà 4 ∑§Ê ∞∑§ ’„ȬŒ „Ò Á¡‚∑§ x=1 ÃÕÊ 73. ^pfp¡ L¡$ f (x) A¡ 4 Opshpmu A¡L$ A¡hu blº`v$u R>¡ L¡$
extreme values at x=1 and x=2.
x=2 ¬⁄U ŒÊ ø⁄U◊ ◊ÊŸ (Extreme Values) „Ò¥– S>¡_p„ ApÐe„rsL$ d|ëep¡ x=1 A_¡ x=2 ApNm dm¡
 f (x) 
If xlim
→0  2
+ 1  = 3 then f (−1) is equal  f (x)  f (x)
 x ÿÁŒ xlim  
 →0  2
+ 1  = 3 , ÃÊ f (−1) ’⁄UÊ’⁄U R>¡ . Å¡ xlim
→0  2
+ 1  = 3 , sp¡ f (−1) =
to :  x   x 
9 „Ò — __________.
(1)
2 9 9
(1) (1)
5 2 2
(2)
2 5 5
(2) (2)
3 2 2
(3)
2 3 3
(3) (3)
1 2 2
(4)
2 1 1
(4) (4)
2 2

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Set - 03 40
2x + 5 2x + 5 74. Å¡
74. If
∫ 7 − 6x − x 2
dx = A 7 − 6x − x 2 + 74. ÿÁŒ
∫ 7 − 6x − x 2
dx = A 7 − 6x − x 2 +

∫
2x + 5
dx = A 7 − 6x − x 2 +
2
7 − 6x − x
−1  x + 3  −1  x + 3 
B sin   +C B sin   +C
 4   4   x +3 
B sin−1   +C
(where C is a constant of integration), then (¡„Ê° C ∞∑§ ‚◊Ê∑§‹Ÿ •ø⁄U „Ò), ÃÊ ∑˝§Á◊à ÿÈÇ◊  4 
the ordered pair (A, B) is equal to :
(A, B) ’⁄UÊ’⁄U „Ò — (Äep„ C A¡ k„L$g__p¡ AQmp„L$ R>¡), sp¡ ¾$dey¼s Å¡X$
(1) (2, 1)
(1) (2, 1) (A, B)=__________.
(2) (−2, −1)
(2) (−2, −1) (1) (2, 1)
(3) (−2, 1)
(3) (−2, 1) (−2, −1)
(4) (2, −1) (2)
(4) (2, −1) (3) (−2, 1)
(4) (2, −1)
3π
4 3π

∫
x 4
3π

∫
75. The value of integral dx is : x 4
1 + sinx dx ∑§Ê ◊ÊŸ „Ò —
∫
75. ‚◊Ê∑§‹ x
π 1 + sinx 75. dx _u k„L$g qL$„ds __________
4 π
4
1 + sinx
π
(1) 4
π 2 (1) π 2
(2) π ( 2 − 1) (2) π ( 2 − 1)
R>¡.
(1) π 2
π
(3)
2
( 2 + 1) (3)
π
( 2 + 1) (2) π ( 2 − 1)
2
(4) 2π ( 2 − 1) (4) 2π ( 2 − 1) (3)
π
( 2 + 1)
2

(4) 2π ( 2 − 1)

SET - 03 ENGLISH MATHS SET - 03 HINDI MATHS SET - 03 GUJARATI MATHS

Set - 03 41
1 1 1

76. If I 1 =
∫
0
e−x cos 2 x dx , 76. ÿÁŒ I1 =
∫
0
e−x cos 2 x dx , 76. Å¡ I1 =
∫
0
e−x cos 2 x dx ,

1 1 1

∫ −x 2
∫ −x 2
∫
2
2 2
I2 = e cos x dx and I2 = e cos x dx ÃÕÊ I 2 = e−x cos 2 x dx A_¡
0 0 0

1 1 1

∫ −x 3
∫ −x 3
∫
3
I3 = e dx ; then : I3 = e dx „Ò, ÃÊ — I 3 = e−x dx ; sp¡ :
0 0 0

(1) I2 > I3 > I1 (1) I 2 > I3 > I 1 (1) I2 > I3 > I1
(2) I2 > I1 > I3 (2) I 2 > I1 > I 3 (2) I2 > I1 > I3
(3) I3 > I2 > I1 (3) I 3 > I2 > I 1 (3) I3 > I2 > I1
(4) I3 > I1 > I2 (4) I 3 > I1 > I 2 (4) I3 > I1 > I2

77. The curve satisfying the differential 77. fl„ fl∑˝§ ¡Ê •fl∑§‹ ‚◊Ë∑§⁄UáÊ 77. rhL$g kduL$fZ (x 2 −y 2 )dx+2xydy=0 _¡
equation, (x 2 −y 2 )dx+2xydy=0 and
(x2−y2)dx+2xydy=0 ∑§Ê ‚¥ÃÈc≈U ∑§⁄UÃÊ „Ò ÃÕÊ k„sp¡jsp¡ A_¡ tbvy$ (1, 1) dp„\u `kpf \sp¡ h¾$ :
passing through the point (1, 1) is :
(1) a circle of radius one. Á’ãŒÈ (1, 1) ‚ „Ê∑§⁄U ¡ÊÃÊ „Ò, „Ò — (1) rÓÄep A¡L$ lp¡e s¡hy L$p¡C A¡L$ hsy®m R>¡.
(2) a hyperbola. (1) ∞∑§ flÎûÊ Á¡‚∑§Ë ÁòÊíÿÊ ∞∑§ ∑§ ’⁄UÊ’⁄U „Ò– (2) L$p¡C A¡L$ Arshge R>¡.
(3) an ellipse. (2) ∞∑§ •Áì⁄Ufl‹ÿ–
(4) a circle of radius two. (3) L$p¡C A¡L$ D`hge R>¡.
(3) ∞∑§ ŒËÉʸflÎûÊ– (4) rÓÄep b¡ lp¡e s¡hy L$p¡C A¡L$ hsy®m R>¡.
(4) ∞∑§ flÎûÊ Á¡‚∑§Ë ÁòÊíÿÊ ŒÊ ∑§ ’⁄UÊ’⁄U „Ò–

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Set - 03 42
78. The sides of a rhombus ABCD are parallel 78. ∞∑§ ‚◊øÃÈ÷Ȩ̀¡ ABCD ∞‚Ê „Ò Á¡‚∑§Ë ÷È¡Êÿ¥ ⁄UπÊ•Ê¥ 78. kdbpSy> QsyóL$p¡Z ABCD _u bpSy>Ap¡ f¡MpAp¡
to the lines, x−y+2=0 and 7x−y+3=0.
x−y+2=0 ÃÕÊ 7x−y+3=0 ∑§ ‚◊Ê¥Ã⁄U „Ò¥– x−y+2=0 A_¡ 7x−y+3=0 _¡ kdp„sf R>¡.
If the diagonals of the rhombus intersect
at P(1, 2) and the vertex A (different from ÿÁŒ ß‚ ‚◊øÃÈ÷ȸ¡ ∑§ Áfl∑§áʸ Á’ãŒÈ P(1, 2) ¬⁄U Å¡ Ap kdbpSy> QsyóL$p¡Z_p rhL$Zp£ tbvy$ P(1, 2)
the origin) is on the y-axis, then the ∑§Ê≈UÃ „Ò¥ ÃÕÊ ∞∑§ ‡ÊË·¸ A(A ≠ O, ◊ÍÀÊ Á’ãŒÈ) ApNm R>¡v$¡ A_¡ rifp¡tbvy$ A (ENdtbvy$ \u rcÞ_)
ordinate of A is : y-•ˇÊ ¬⁄U „Ò, ÃÊ A ∑§Ë ∑§ÊÁ≈U (ordinate) „Ò — y-An `f lp¡e, sp¡ A _p¡ epd __________ R>¡.
5 5 5
(1)
2 (1)
2
(1)
2
7 7 7
(2)
4 (2)
4
(2)
4
(3) 2 (3) 2 (3) 2
7 7 7
(4)
2 (4)
2
(4)
2

79. The foot of the perpendicular drawn from 79. ENdtbvy$dp„\u f¡Mp 3x+y=λ(λ ≠ 0) `f v$p¡f¡gp
the origin, on the line, 3x+y=λ(λ ≠ 0) is
79. ◊Í‹ Á’ãŒÈ ‚ ⁄UπÊ, 3x+y=λ(λ ≠ 0) ¬⁄U «UÊ‹ ª∞
P. If the line meets x-axis at A and y-axis ‹ê’ ∑§Ê ¬ÊŒ P „Ò– ÿÁŒ ÿ„ ⁄UπÊ x-•ˇÊ ∑§Ê A ÃÕÊ g„b_p¡ g„b`pv$ P R>¡. Å¡ Ap`¡gu f¡Mp x-An_¡ A dp„
at B, then the ratio BP : PA is : y-•ˇÊ ∑§Ê B ¬⁄U ∑§Ê≈UÃË „Ò, ÃÊ •ŸÈ¬Êà BP : PA „Ò — A_¡ y-An_¡ B dp„ dm¡, sp¡ NyZp¡Ñf BP : PA=
(1) 1 : 3 (1) 1:3 __________.
(2) 3 : 1 (2) 3:1 (1) 1:3
(3) 1 : 9 (3) 1:9 (2) 3:1
(4) 9 : 1 (4) 9:1 (3) 1:9
(4) 9:1
80. The tangent to the circle 80. ÿÁŒ flÎûÊ C1 : x2+y2−2x−1=0 ∑§ Á’ãŒÈ (2, 1)
C1 : x2+y2−2x−1=0 at the point (2, 1) 80. hsy ® m C 1 : x 2 +y 2 −2x−1=0 `f_p tbvy $
cuts off a chord of length 4 from a circle ¬⁄U πË¥øË ªß¸ S¬‡Ê¸ ⁄UπÊ, ∞∑§ ŒÍ‚⁄U flÎûÊ C2, Á¡‚∑§Ê
(2, 1) ApNm_p¡ õ`i®L$ A¡ (3, −2) L¡$ÞÖhpmp hsy®m
C2 whose centre is (3, −2). The radius of ∑§ãŒ˝ (3, −2) „Ò, ‚ ‹ê’Ê߸ øÊ⁄U ∑§ ’⁄UÊ’⁄U ∞∑§ ¡ËflÊ
C2 is : C2 dp„\u 4 g„bpChpmp A¡L$ Æhp L$p`¡ R>¡. sp¡ C2 _u
∑§Ê≈UÃË „Ò, ÃÊ C2 ∑§Ë ÁòÊíÿÊ „Ò —
(1) 2 (1) 2 rÓÄep __________ R>¡.
(2) 2 (2) (1) 2
2
(3) 3 (2)
(3) 3 2
(4) 6 (4) 6 (3) 3
(4) 6

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Set - 03 43
81. Tangents drawn from the point (−8, 0) to 81. Á’ãŒÈ (−8, 0) ‚ ¬⁄Ufl‹ÿ, y2=8x ¬⁄U πË¥øË ªß¸ 81. tbvy$ (−8, 0) dp„\u `fhge y2=8x _¡ v$p¡f¡gp õ`i®L$p¡
the parabola y2=8x touch the parabola at
S¬‡Ê¸ ⁄UπÊ∞° ¬⁄Ufl‹ÿ ∑§Ê P ÃÕÊ Q ¬⁄U S¬‡Ê¸ ∑§⁄UÃË „Ò¥– `fhge_¡ P A_¡ Q ApNm õ`i£ R>¡. Å¡ Ap `fhge_u
P and Q. If F is the focus of the parabola,
then the area of the triangle PFQ ÿÁŒ F ß‚ ¬⁄Ufl‹ÿ ∑§Ë ŸÊÁ÷ „Ò, ÃÊ ∆PFQ ∑§Ê ˇÊòÊ»§‹ _prc F lp¡e, sp¡ rÓL$p¡Z PFQ _y„ n¡Óam (Qp¡. A¡L$ddp„)
(in sq. units) is equal to : (flª¸ ß∑§ÊßÿÊ¥ ◊¥) ’⁄UÊ’⁄U „Ò — __________ R>¡.
(1) 24 (1) 24 (1) 24
(2) 32 (2) 32 (2) 32
(3) 48 (3) 48 (3) 48
(4) 64 (4) 64 (4) 64
82. A normal to the hyperbola, 4x2−9y2=36 82. •Áì⁄Ufl‹ÿ, 4x2−9y2=36, ¬⁄U ∞∑§ •Á÷‹ê’ 82. Arshge 4x2−9y2=36 _p¡ L$p¡C Arcg„b epdpnp¡
meets the co-ordinate axes x and y at A
and B, respectively. If the parallelogram
ÁŸŒ¸‡ÊÊ¥∑§ •ˇÊÊ¥ x ÃÕÊ y ∑§Ê ∑˝§◊‡Ê— A ÃÕÊ B ¬⁄U x A_¡ y _¡ A_y¾$d¡ A A_¡ B dp„ dm¡ R>¡. Å¡ OABP
OABP (O being the origin) is formed, then ∑§Ê≈UÃÊ „Ò– ÿÁŒ ‚◊ÊãÃ⁄U øÃÈ÷ȸ¡ OABP (O, ◊Í‹ (Äep„ O ENdtbv$y R>¡) kdp„sfbpSy> QsyóL$p¡Z b_sp¡
the locus of P is : Á’ãŒÈ „Ò) ’ŸÊÿÊ ¡ÊÃÊ „Ò, ÃÊ P ∑§Ê Á’ãŒÈ¬Õ „Ò — lp¡e, sp¡ P _p¡ tbvy$`\ __________ R>¡.
(1) 4x 2+9y 2=121 (1) 4x 2+9y 2 =121
(2) 9x 2+4y 2=169 (1) 4x 2+9y 2=121
(2) 9x 2+4y 2 =169
(3) 4x 2−9y 2=121 (2) 9x 2+4y 2=169
(3) 4x 2−9y 2 =121
(4) 9x 2−4y 2=169 (3) 4x 2−9y 2=121
(4) 9x 2−4y 2 =169
(4) 9x 2−4y 2=169
83. An angle between the lines whose 83. ⁄ U πÊ•Ê ¥ , Á¡Ÿ∑§Ë ÁŒ∑˜ § ∑§Ê í ÿÊ∞° ‚◊Ë∑§⁄UáÊÊ ¥
direction cosines are given by the 83. S>¡ f¡MpAp¡_p qv$L$¹L$p¡kpC_, kduL$fZp¡ l+3m+5n=0
equations, l+3m+5n=0 and l+3m+5n=0 ÃÕÊ 5lm−2mn+6nl=0 mÊ⁄UÊ
A_¡ 5lm−2mn+6nl=0 Üpfp Ap`¡g lp¡e, s¡
5lm−2mn+6nl=0, is : ¬˝ŒûÊ „Ò¥, ∑§ ’Ëø ∑§Ê ∞∑§ ∑§ÊáÊ „Ò — f¡MpAp¡ hÃQ¡_p¡ A¡L$ M|Zp¡ __________ R>¡.
1 1
cos−1   cos−1   1
(1)
3
(1)
3 (1) cos−1  
3
1 1
cos−1   cos−1   1
(2)
4
(2)
4 (2) cos−1  
4
1 1
cos−1   cos−1   1
(3)
6
(3)
6 (3) cos−1  
6
 1  1
(4) cos−1   (4) cos−1    1
8 8 (4) cos−1  
8

SET - 03 ENGLISH MATHS SET - 03 HINDI MATHS SET - 03 GUJARATI MATHS

Set - 03 44
84. A plane bisects the line segment joining 84. ∞∑§ ‚◊Ë Á’ãŒÈ•Ê¥ (1, 2, 3) ÃÕÊ (−3, 4, 5) ∑§Ê 84. tbvy$Ap¡ (1, 2, 3) A_¡ (−3, 4, 5) _¡ Å¡X$sp„ f¡MpM„X$_¡
the points (1, 2, 3) and (−3, 4, 5) at right
Á◊‹ÊŸ flÊ‹ ⁄UπÊπ¥«U ∑§Ê ‚◊∑§ÊáÊ ¬⁄U ‚◊Ám÷ÊÁ¡Ã L$p¡C A¡L$ kdsg L$pV$M|Z¡ vy$cpN¡ R>¡. sp¡ Ap kdsg
angles. Then this plane also passes through
the point : ∑§⁄UÃÊ „Ò, ÃÊ ÿ„ ‚◊Ë Á¡‚ •ÊÒ⁄U Á’ãŒÈ ‚ ªÈ$¡⁄UÃÊ „Ò, __________ tbvy$dp„\u `Z `kpf \i¡.
(1) (−3, 2, 1) fl„ „Ò — (1) (−3, 2, 1)
(2) (3, 2, 1) (1) (−3, 2, 1) (2) (3, 2, 1)
(3) (−1, 2, 3) (2) (3, 2, 1) (3) (−1, 2, 3)
(4) (1, 2, −3) (3) (−1, 2, 3) (4) (1, 2, −3)
(4) (1, 2, −3)
85. If the position vectors of the vertices A, B 85. Å¡ ∆ABC _p rifp¡tbvy$Ap¡ A, B A_¡ C _p õ\p_
and C of a ∆ABC are respectively 85. ÿÁŒ ∞∑§ ÁòÊ÷È¡ ABC ∑§ ‡ÊË·ÊZ A, B ÃÕÊ C ∑§ ∧ ∧ ∧ ∧ ∧ ∧
∧ ∧ ∧ ∧ ∧ ∧ kqv$ip¡ A_y¾$d¡ 4 i + 7 j + 8 k , 2 i + 3 j + 4 k
∧ ∧ ∧
4 i +7 j +8k , 2 i +3 j +4k and SÕÊŸËÿ ‚ÁŒ‡Ê ∑˝ § ◊‡Ê— 4 i +7 j +8k ,
∧ ∧ ∧
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ A_¡ 2 i + 5 j + 7 k lp¡e, sp¡ ∠A _p¡ qÜcpS>L$
2 i + 5 j + 7 k , then the position vector of 2 i + 3 j + 4 k ÃÕÊ 2 i + 5 j + 7 k „Ò¥, ÃÊ ©‚
the point, where the bisector of ∠A meets BC _¡ S>¡ tbvy$dp„ dm¡ s¡_p¡ õ\p_ kqv$i __________
BC is :
Á’ãŒÈ, ¡„Ê° ∠A ∑§Ê ‚◊Ám÷Ê¡∑§ BC ¬⁄U Á◊‹ÃÊ „Ò, R>¡.

( ) ( )
1 ∧ ∧ ∧
∑§Ê SÕÊŸËÿ ‚ÁŒ‡Ê „Ò — 1 ∧ ∧ ∧
4 i + 8 j + 11 k
( ) 2 4 i + 8 j + 11 k
(1) 1 ∧ ∧ ∧ (1)
2
2 4 i + 8 j + 11 k
(1)
1
( ∧ ∧
) ∧ 1
( ∧ ∧
) ∧
3 6 i + 11 j + 15 k
( ) 3 6 i + 11 j + 15 k
(2) 1 ∧ ∧ ∧ (2)
3 6 i + 11 j + 15 k
(2)
1
( ∧ ∧
) ∧ 1
( ∧ ∧
) ∧
3 6 i + 13 j + 18 k
( ) 3 6 i + 13 j + 18 k
(3) 1 ∧ ∧ ∧ (3)
3 6 i + 13 j + 18 k
(3)
1
( ∧ ∧
) ∧ 1
( ∧ ∧
) ∧
4 8 i + 14 j + 19 k
( ) 4 8 i + 14 j + 19 k
(4) 1 ∧ ∧ ∧ (4)
4 8 i + 14 j + 19 k
(4)

SET - 03 ENGLISH MATHS SET - 03 HINDI MATHS SET - 03 GUJARATI MATHS

Set - 03 45
86. A player X has a biased coin whose 86. ∞∑§ Áπ‹Ê«∏Ë X ∑§ ¬Ê‚ ∞∑§ •Á÷ŸÃ Á‚Ä∑§Ê „Ò 86. M¡gpX$u X `pk¡ A¡L$ Arc_s (biased) rk½$p¡ R>¡ S>¡_u
probability of showing heads is p and a
Á¡‚∑§Ë ÁøûÊ (Heads) Œ‡ÊʸŸ ∑§Ë ¬˝ÊÁÿ∑§ÃÊ ‘p’ „Ò R>p` v$¡MpX$hp_u k„cph_p p R>¡ A_¡ M¡gpX$u Y `pk¡
player Y has a fair coin. They start playing
a game with their own coins and play ÃÕÊ Áπ‹Ê«∏Ë Y ∑§ ¬Ê‚ ∞∑§ •ŸÁ÷ŸÃ Á‚Ä∑§Ê „Ò– A¡L$ kdsp¡g rk½$p¡ R>¡. s¡Ap¡ `p¡s-`p¡sp_p rk½$pAp¡
alternately. The player who throws a head fl„ •¬Ÿ •¬Ÿ Á‚Ä∑§Ê¥ mÊ⁄UÊ ∞∑§ π‹ ‡ÊÈM§ ∑§⁄UÃ „Ò¥ kp\¡ A¡L$ fds fdhp_u iê$Aps L$f¡ R>¡ s\p hpfpaf\u
first is a winner. If X starts the game, and Á¡‚◊¥ ©ã„¥ ’Ê⁄UË ’Ê⁄UË ‚ π‹ŸÊ „Ò– fl„ Áπ‹Ê«∏Ë ¡Ê fd¡ R>¡. S>¡ M¡gpX$u rk½$p¡ DR>pmu â\d R>p` d¡mh¡ s¡
the probability of winning the game by both
the players is equal, then the value of ‘p’
¬„‹ ÁøûÊ »¥§∑§ÃÊ „Ò¥, Áfl¡ÃÊ ◊ÊŸÊ ¡ÊÃÊ „Ò– ÿÁŒ X, rhS>¡sp b_¡. Å¡ X fds_u iê$Aps L$f¡ A_¡ bÞ_¡
is : π‹ ‡ÊÈM§ ∑§⁄UÃÊ „Ò ÃÕÊ ŒÊŸÊ¥ ∑§ ¡Ëß ∑§Ë ¬˝ÊÁÿ∑§ÃÊ∞° M¡gpX$uAp¡_u rhS>¡sp b_hp_u k„cph_p kdp_ lp¡e,
1 ‚◊ÊŸ „Ò¥, ÃÊ ‘p’ ∑§Ê ◊ÊŸ „Ò — sp¡ ‘p’ _u qL„$ds __________ R>¡.
(1)
5 1 1
(1) (1)
1 5 5
(2)
3 1 1
(2) (2)
2 3 3
(3)
5 2 2
(3) (3)
1 5 5
(4)
4 1 1
(4) (4)
4 4
87. If the mean of the data : 7, 8, 9, 7, 8, 7, λ, 8
is 8, then the variance of this data is : 87. ÿÁŒ •Ê°∑§«∏Ê¥ 7, 8, 9, 7, 8, 7, λ, 8 ∑§Ê ◊Êäÿ 8 „Ò, ÃÊ 87. Å¡ dprlsu : 7, 8, 9, 7, 8, 7, λ, 8 _p¡ dÝeL$ 8 lp¡e,
7 ߟ •Ê°∑§«∏Ê¥ ∑§Ê ¬˝‚⁄UáÊ „Ò — sp¡ Ap dprlsu_p¡ rhQfZ __________ R>¡.
(1)
8
7 7
(2) 1 (1) (1)
8 8
9
(3) (2) 1 (2) 1
8
9 9
(4) 2 (3) (3)
8 8
(4) 2 (4) 2

SET - 03 ENGLISH MATHS SET - 03 HINDI MATHS SET - 03 GUJARATI MATHS

Set - 03 46
88. The number of solutions of sin 3x=cos 2x,
π  π 
88. sin 3x=cos 2x ∑§ •¥Ã⁄UÊ‹  , π  ◊¥ „‹Ê¥ ∑§Ë 88. sin 3x=cos 2x _p  , π  A„sfpgdp„ DL¡$gp¡_u
π  2  2 
in the interval  , π  is :
2  ‚¥ÅÿÊ „Ò — k„¿ep __________ R>¡.
(1) 1 (1) 1
(2) 2 (1) 1
(2) 2 (2) 2
(3) 3 (3) 3
(4) 4 (3) 3
(4) 4 (4) 4
89. A tower T1 of height 60 m is located exactly
89. ∞∑§ ‚ËœË ‚«∏∑§ ¬⁄U 60 ◊Ë. ™°§øË ∞∑§ ◊ËŸÊ⁄U T1, 89. 60 duV$f KQpC ^fphsy„ A¡L$ V$phf T1, 80 duV$f KQpC
opposite to a tower T2 of height 80 m on a
straight road. From the top of T1, if the 80 ◊Ë. ™°§øË ∞∑§ ◊ËŸÊ⁄U T2 ∑§ ∆UË∑§ ‚Ê◊Ÿ SÕÊÁ¬Ã ^fphsp„ A¡L$ V$phf T2 _u bfpbf kpd¡ S> A¡L$ ku^p
angle of depression of the foot of T2 is twice „Ò– ÿÁŒ T1 ∑§ Á‡Êπ⁄U ‚ T2 ∑§ ¬ÊŒ ∑§Ê •flŸ◊Ÿ fõsp `f Aph¡gy„ R>¡. T1 _u V$p¡Q\u, T2 _p srmep_p¡
the angle of elevation of the top of T2, then ∑§ÊáÊ, T2 ∑§ Á‡Êπ⁄U ∑§ ©ÛÊÿŸ ∑§ÊáÊ ∑§Ê ŒÈªÈŸÊ „Ò, ÃÊ
the width (in m) of the road between the Ahk¡^L$p¡Z Å¡ T2 _p V$p¡Q_p DÐk¡^L$p¡Z\u bdZp¡
feet of the towers T1 and T2 is :
◊ËŸÊ⁄UÊ¥ T1 ÃÕÊ T2 ∑§ ¬ÊŒÊ¥ ∑§ ’Ëø ‚«∏∑§ ∑§Ë øÊÒ«∏Ê߸ lp¡e, sp¡ V$phf T1 A_¡ T2 _p„ srmepAp¡ hÃQ¡_p fõsp_u
(◊Ë≈U⁄UÊ¥ ◊¥) „Ò — `lp¡mpC (duV$f dp„) __________ R>¡.
(1) 10 2
(1) 10 2
(2) 10 3 (1) 10 2
(2) 10 3
(3) 20 3 (2) 10 3
(3) 20 3
(4) 20 2 (3) 20 3
(4) 20 2
(4) 20 2

SET - 03 ENGLISH MATHS SET - 03 HINDI MATHS SET - 03 GUJARATI MATHS

Set - 03 47
90. Consider the following two statements : 90. ÁŸêŸ ŒÊ ∑§ÕŸÊ¥ ¬⁄U ÁfløÊ⁄U ∑§ËÁ¡∞ — 90. _uQ¡_p„ b¡ rh^p_p¡ rhQpfp¡ :
Statement p :
∑§ÕŸ p : rh^p_ p :
The value of sin 120 can be derived by
taking θ=240 in the equation sin 120 ∑§Ê ◊ÊŸ, ‚◊Ë∑§⁄UáÊ sin 120 _u qL„$ds kduL$fZ
θ θ θ
2 sin = 1 + sin θ − 1 − sin θ . 2 sin = 1 + sin θ − 1 − sin θ ◊¥ 2 sin = 1 + sin θ − 1 − sin θ .
2 2 2
Statement q : θ=240 ‹Ÿ ‚ ôÊÊà Á∑§ÿÊ ¡Ê ‚∑§ÃÊ „Ò– dp„ θ=240 g¡sp d¡mhu iL$pe.
The angles A, B, C and D of any ∑§ÕŸ q : rh^p_ q :
quadrilateral ABCD satisfy the equation
Á∑§‚Ë øÃÈ÷ȸ¡ ABCD ∑§ ∑§ÊáÊ A, B, C ÃÕÊ D, L$p¡C`Z QsyóL$p¡Z ABCD _p M|ZpAp¡ A, B, C
1  1 
cos  (A + C)  + cos  (B + D)  = 0 ‚◊Ë∑§⁄UáÊ A_¡ D kduL$fZ
 2   2 
1  1  1  1 
Then the truth values of p and q are cos  (A + C)  + cos  (B + D)  = 0 cos  (A + C)  + cos  (B + D)  = 0
respectively : 2  2  2  2 
(1) F, T ∑§Ê ‚¥ÃÈc≈U ∑§⁄UÃ „Ò¥– _y„ kdp^p_ L$f¡.
(2) T, F ÃÊ p ÃÕÊ q ∑§ ‚àÿ◊ÊŸ, ∑˝§◊‡Ê— „Ò¥ — sp¡ p A_¡ q _p kÐep\®sp d|ëep¡ A_y¾$d¡ __________
(3) T, T
(1) F, T R>¡.
(4) F, F
(2) T, F
(3) T, T (1) F, T
-o0o- (2) T, F
(4) F, F
(3) T, T
-o0o- (4) F, F

-o0o-

SET - 03 ENGLISH MATHS SET - 03 HINDI MATHS SET - 03 GUJARATI MATHS

AglaSem Admission

JEE Main Official Answer Key 2018 for online exam held on 15 April
: Evening Key

Qs Ans Qs Ans Qs Ans
Q1 4 Q31 4 Q61 2
Q2 4 Q32 4 Q62 4
Q3 2 Q33 2 Q63 3
Q4 2 Q34 1 Q64 1
Q5 4 Q35 2 Q65 2
Q6 2 Q36 2 Q66 4
Q7 1 Q37 2 Q67 1
Q8 1 Q38 2 Q68 3
Q9 3 Q39 4 Q69 2
Q10 1 Q40 3 Q70 3
Q11 2 Q41 4 Q71 3
Q12 3 Q42 4 Q72 2
Q13 1 Q43 2 Q73 1
Q14 2 Q44 3 Q74 2
Q15 1 Q45 2 Q75 2
Q16 3 Q46 4 Q76 3
Q17 1 Q47 3 Q77 1
Q18 4 Q48 3 Q78 1
Q19 1 Q49 3 Q79 4
Q20 3 Q50 2 Q80 4
Q21 2 Q51 2 Q81 3
Q22 4 Q52 4 Q82 4
Q23 3 Q53 1 Q83 3
Q24 1 Q54 4 Q84 1
Q25 2 Q55 4 Q85 3
Q26 3 Q56 2 Q86 2
Q27 3 Q57 3 Q87 2
Q28 2 Q58 3 Q88 1
Q29 4 Q59 2 Q89 3
Q30 4 Q60 3 Q90 1