JEE Main 2016 Question Paper 10 Apr B.Tech

Set - 03 1
1. A, B, C and D are four different physical 1. A, B, C ÃÕÊ D øÊ⁄U Á÷ÛÊ ◊ÊòÊÊ∞° „Ò¥ Á¡Ÿ∑§Ë Áfl◊Ê∞¥ 1. A, B, C A_¡ D A¡ Qpf Sy>v$p-Sy>v$p `qfdpZ ^fphsu
quantities having different dimensions.
Á÷㟠„Ò¥– ∑§Ê߸ ÷Ë ◊ÊòÊÊ Áfl◊Ê-⁄UÁ„à ◊ÊòÊÊ Ÿ„Ë¥ „Ò¥, Sy>v$u-Sy>v$u cp¥rsL$fpriAp¡ R>¡. s¡dp„_u L$p¡C`Z `qfdpZ
None of them is dimensionless. But we
know that the equation AD=C ln(BD) ‹Á∑§Ÿ AD=C ln(BD) ‚àÿ „Ò– Ã’ ÁŸêŸ ◊¥ ‚ frls _\u. Å¡ AD=C ln(BD) kduL$fZ kpQy„ lp¡e
holds true. Then which of the combination ∑§ÊÒŸ •Ê‡Êÿ-⁄UÁ„à ◊ÊòÊÊ „Ò? sp¡ _uQ¡ Ap`¡g `¥L$u L$ey„ k„ep¡S>_ A¡ A\®kcf (kpQu)
is not a meaningful quantity ? (1) A2 − B2C2 fpri v$ip®hsy„ _\u ?
(1) A2 − B2C2 (1) A2 − B2C2
(A − C )
(A − C ) (2)
(2) D (A − C )
D (2)
A D
A (3) −C
(3) −C B A
B (3) −C
B
C AD2
AD2 (4) −
(4)
C
− BD C C AD2
BD C (4) −
BD C
2. Œ˝√ÿ◊ÊŸ M ∑§Ê ∞∑§ ∑§áÊ ÁŸÁ‡øà ÁòÊíÿÊ R ∑§ flÎûÊËÿ
2. A particle of mass M is moving in a circle
of fixed radius R in such a way that its
¬Õ ¬⁄U-ß‚ ¬˝∑§Ê⁄U ø‹ ⁄U„Ê „Ò Á∑§ ‚◊ÿ ‘t’ ¬⁄U •Á÷∑§ãŒ˝Ë 2. M v$m ^fphsp¡ L$Z R S>¡V$gu AQm rÓÄep ^fphsp
centripetal acceleration at time t is given àfl⁄UáÊ n2 R t2 mÊ⁄UÊ ÁŒÿÊ ¡Ê ‚∑§ÃÊ „Ò, ÿ„Ê° ‘n’ •ø⁄U hsy®mpL$pf dpN® `f A¡hu fus¡ Nrs L$f¡ R>¡ L¡$ t kde¡ s¡_p¡
by n2 R t2 where n is a constant. The power „Ò– Ã’ ∑§áÊ ¬⁄U ‹ª ⁄U„ ’‹ mÊ⁄UÊ ©‚∑§Ê ŒË ªß¸ ‡ÊÁÄà L¡$ÞÖNpdu âh¡N n2 R t2 hX¡$ Ap`u iL$pe; Äep„ n A¡
delivered to the particle by the force acting „Ò — AQmp„L$ R>¡. sp¡ L$Z `f gpNsp bm hX¡$ L$Z_¡ dmsp¡
on it, is :
(1) M n2 R2 t `phf (L$pe®v$nsp) __________ \i¡.
(1) M n2 R2 t
(2) M n R2 t (1) M n2 R2 t
(2) M n R2 t
(3) M n R2 t2 (2) M n R2 t
(3) M n R2 t2
1 (3) M n R2 t2
1 (4) M n2 R2 t2
(4) M n2 R2 t2 2 1
2 (4) M n2 R2 t2
2

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Set - 03 2
3. Concrete mixture is made by mixing 3. ∑¥§∑˝§Ë≈U Á◊Ä‚ø⁄U ’ŸÊŸ ∑§ Á‹ÿ ‚Ë◊¥≈U, ⁄Uà ÃÕÊ ⁄UÊ«∏Ë 3. A¡L$ QpL$Nrs L$fsp _mpL$pfue X²$ddp„ rkd¡ÞV$, `Õ\f
cement, stone and sand in a rotating
∑§Ê ∞∑§ ÉÊÍáÊ˸ÿ ’‹ŸÊ∑§Ê⁄U «˛U◊ ◊¥ «UÊ‹Ê ¡ÊÃÊ „Ò– ÿÁŒ A_¡ f¡su_¡ c¡Np L$fu_¡ L$p¢q¾$V$ rdîZ s¥epf L$fhpdp„
cylindrical drum. If the drum rotates too
fast, the ingredients remain stuck to the «˛U◊ ∑§Ë ÉÊÍáʸŸ-ªÁà ’„Èà Ã¡ „Ê ÃÊ ‚¥ÉÊ≈U∑§ «˛U◊ ∑§Ë Aph¡ R>¡. lh¡ Å¡ X²$d M|b S> TX$`\u QpL$Nrs L$f¡ sp¡
wall of the drum and proper mixing of ŒËflÊ⁄U ‚ Áø¬∑§ ⁄U„Ã „Ò¥ •ÊÒ⁄U Á◊Ä‚ø⁄U ∆UË∑§ ‚ Ÿ„Ë¥ rdîZ L$f¡gp sÒhp¡ X²$d_u qv$hpg_¡ Qp¢V$u Åe R>¡ A_¡
ingredients does not take place. The ’ŸÃÊ– ÿÁŒ «˛U◊ ∑§Ë ÁòÊíÿÊ 1.25 m „Ò •ÊÒ⁄U ß‚∑§Ë s¡d_y„ ep¡Áe rdîZ b_phu iL$psy„ _\u. sp¡ `qfc°dZ
maximum rotational speed of the drum in
revolutions per minute(rpm) to ensure
œÈ⁄UË ˇÊÒÁá „Ò, Ã’ •ë¿UË Ã⁄U„ Á◊Ä‚ „ÊŸ ∑§ Á‹ÿ ârs rd_uV$ (rpm) _p `v$dp„, ep¡Áe rdîZ b_hp
proper mixing is close to : ¡M§⁄UË •Áœ∑§Ã◊ ÉÊÍáÊ˸ÿ-ªÁà rpm ◊¥ „Ò — dpV¡$_u X²$d_u dlÑd L$p¡Zue TX$` __________ _u
(Take the radius of the drum to be 1.25 m (1) 0.4 _ÆL$_u li¡. (X²$d_u rÓÄep 1.25 m A_¡ s¡_u An
and its axle to be horizontal) : (2) 1.3 kdrnrsS> R>¡ s¡d ^pfp¡).
(1) 0.4 (3) 8.0 (1) 0.4
(2) 1.3 (4) 27.0 (2) 1.3
(3) 8.0 (3) 8.0
(4) 27.0 (4) 27.0

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Set - 03 3
4. Velocity-time graph for a body of mass 4. 10 kg Œ˝√ÿ◊ÊŸ ∑§ Á¬¥«U ∑§ Á‹ÿ flª-‚◊ÿ ª˝Ê»§ ÁøòÊ 4. 10 kg v$m ^fphsp `v$p\® dpV¡$_p¡ h¡N-kde_p¡ Apg¡M
10 kg is shown in figure. Work-done on
◊¥ ÁŒÿÊ „Ò– Á¬¥«U ¬⁄U ¬„‹ 2 ‚. ◊¥ Á∑§ÿÊ ªÿÊ ∑§Êÿ¸ ApL©$rÑdp„ v$ip®h¡g R>¡. `v$p\® `f â\d 2 k¡L$ÞX$dp„ \su
the body in first two seconds of the motion
is : „Ò — Nrs v$fçep_ \sy„ L$pe® __________\i¡.

(1) 12000 J (1) 12000 J (1) 12000 J
(2) −12000 J (2) −12000 J (2) −12000 J
(3) −4500 J (3) −4500 J (3) −4500 J
(4) −9300 J (4) −9300 J (4) −9300 J

5. In the figure shown ABC is a uniform wire. 5. ÁŒÿ ªÿ ÁøòÊ ◊¥ ÃÊ⁄U ABC ∞∑§ ‚◊ÊŸ „Ò– ÿÁŒ 5. ApL©$rÑdp„ v$ip®ìep dyS>b ABC A¡ kdp_ spf R>¡. Å¡
If centre of mass of wire lies vertically below ‚¥„ÁÃ-∑§ãŒ˝ Á’¥ŒÈ A ∑ ™§äflʸœ⁄U ŸËø ÁSÕà „Ò, Ã’ spf_y„ Öìedp_ L¡$ÞÖ tbvy$ A _u bfp¡bf _uQ¡ Aphsy„
BC BC
point A, then is close to : BC
AB ‹ª÷ª „Ò — lp¡e sp¡ _y„ d|ëe _________ _u _ÆL$_y„ \i¡.
AB AB

(1) 1.85 (1) 1.85 (1) 1.85
(2) 1.37 (2) 1.37 (2) 1.37
(3) 1.5 (3) 1.5 (3) 1.5
(4) 3 (4) 3 (4) 3

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Set - 03 4
6. An astronaut of mass m is working on a 6. ¬ÎâflË ∑§Ë ‚Ä ‚ ‘h’ ŒÍ⁄UË ¬⁄U ÁSÕà ∞∑§ ©¬ª˝„ ¬⁄U 6. `©Õhu_u k`pV$u\u h KQpC A¡ `qfc°dZ L$fsp
satellite orbiting the earth at a distance h
∞∑§ ‘m’ Œ˝√ÿ◊ÊŸ ∑§Ê •¥ÃÁ⁄UˇÊ-ÿÊòÊË ∑§Ê◊ ∑§⁄U ⁄U„Ê „Ò– k¡V¡gpCV$dp„ m v$m ^fphsp¡ MNp¡mipõÓu L$pe® L$f¡ R>¡.
from the earth’s surface. The radius of the
earth is R, while its mass is M. The ¬ÎâflË ∑§Ê Œ˝√ÿ◊ÊŸ ‘M’ ÃÕÊ ÁòÊíÿÊ ‘R’ „Ò– Ã’ ©‚ `©Õhu_u rÓÄep R R>¡ Äepf¡ s¡_„y v$m M R>¡. MNp¡mipõÓu
gravitational pull FG on the astronaut is : ÿÊòÊË ¬⁄U ‹ª ⁄U„Ê ªÈL§àflËÿ ’‹ FG „Ò — `f gpNsy„ NyfyÐhpL$j} M¢QpZ FG __________ li¡.
(1) Zero since astronaut feels weightless (1) ‡ÊÍãÿ, ÄÿÊ¥Á∑§ fl„ ÿÊòÊË ÷Ê⁄U„ËŸÃÊ ◊„‚Í‚ ∑§⁄UÃÊ (1) i| Þ e, L$ p fZ L¡ $ MNp¡ m ipõÓu hS>_frls

0 < FG <
GMm „Ò– [õ\rsdp„ R>¡.
(2)
R2 GMm GMm
(2) 0 < FG < (2) 0 < FG <
GMm GMm R2 R2
(3) < FG <
(R + h )2 R2 GMm GMm GMm GMm
(3) < FG < (3) < FG <
GMm (R + h )2 R2 (R + h )2 R2
(4) FG =
(R + h )2 GMm GMm
(4) FG = (4) FG =
(R + h )2 (R + h )2

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Set - 03 5
7. A bottle has an opening of radius a and 7. ∞∑§ ’ÊË ∑§ ◊È°„ ∑§Ë ÁòÊíÿÊ ‘a’ „ÒÒ ÃÕÊ ‹ê’Ê߸ ‘b’ „Ò– 7. A¡L$ bp¡V$g_p Nmp_u rÓÄep a A_¡ g„bpC b R>¡.
length b. A cork of length b and radius
∞∑§ ‘b’ ‹ê’Ê߸ •ÊÒ⁄U (a+∆a) ÁòÊíÿÊ (∆a<<a) flÊ‹ ApL©$rÑdp„ v$ip®ìep dyS>b, b g„bpC_p A_¡ (a+∆a)
(a+∆a) where (∆a<<a) is compressed to fit
into the opening completely (See figure). ∑§Ê∑¸§ ∑§Ê ©‚∑§ ◊È°„ ◊¥ ¬Í⁄UË Ã⁄U„ ∆Í°U‚ ÁŒÿÊ ªÿÊ „Ò (ÁøòÊ Äep„ (∆a<<a) _u rÓÄep ^fphsp b|Q_¡ v$bpZ`|h®L$
If the bulk modulus of cork is B and ŒÁπÿ)– ÿÁŒ ∑§Ê∑¸§ ∑§Ê •Êÿß ¬˝àÿÊSÕÃÊ ªÈáÊÊ¥∑§ B hX¡$ bp¡V$g_p Nmphpmp cpN_¡ k„`|Z® b„^ L$fhpdp„ Aph¡
frictional coefficient between the bottle and „Ò ÃÕÊ ’ÊË •ÊÒ⁄U ∑§Ê∑¸§ ∑§ ’Ëø ÉÊ·¸áÊ-ªÈáÊÊ¥∑§ µ „Ò, R>¡. Å¡ b|Q_p¡ L$v$ [õ\rsõ\p`L$spA„L$ B lp¡e A_¡
cork is µ then the force needed to push the
cork into the bottle is :
Ã’ ∑§Ê∑¸§ ∑§Ê ◊È°„ ◊¥ ÉÊÈ‚ÊŸ ∑§ Á‹ÿ •Êfl‡ÿ∑§ ’‹ „Ò — bp¡V$g A_¡ b|Q hÃQ¡_p¡ OjZp¯L$ µ lp¡e sp¡ b|Q_¡
bp¡V$gdp„ O|kpX$hp S>ê$fu bm ________ \i¡.

(1) (πµB b) ∆a (1) (πµB b) ∆a (1) (πµB b) ∆a
(2) (2πµB b) ∆a (2) (2πµB b) ∆a (2) (2πµB b) ∆a
(3) (πµB b) a (3) (πµB b) a (3) (πµB b) a
(4) (4πµB b) ∆a (4) (4πµB b) ∆a (4) (4πµB b) ∆a

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Set - 03 6
8. A Carnot freezer takes heat from water at 8. ∞∑§ ∑§ÊŸÊ¸≈U »˝§Ë¡⁄U •¬Ÿ •¥Œ⁄U 08C ¬⁄U ⁄Uπ „È∞ ¡‹ ‚ 8. A¡L$ L$p_p£V$ f¡qäS>f¡V$f `pZudp„\u 08C sp`dp_¡ Dódp
08C inside it and rejects it to the room at a
™§c◊Ê ‹∑§⁄U ©‚ ∑§◊⁄U ∑§ Ãʬ◊ÊŸ 278C ¬⁄U ÁŸc∑§ÊÁ‚à ip¡ju 278C sp`dp_¡ fl¡gp hpsphfZdp„ a¢L¡$ R>¡. bfa_u
temperature of 278C. The latent heat of ice
is 336×103 J kg−1. If 5 kg of water at 08C is ∑§⁄UÃÊ „Ò– ’»¸§ ∑§Ë ªÈåà ™§c◊Ê 336×103 J kg−1 „Ò– g¡V¡$ÞV$ (Nyá) Dódp 336×103 J kg−1 R>¡. Å¡
converted into ice at 08C by the freezer, then ÿÁŒ »˝§Ë¡⁄U ◊¥ ⁄UπÊ 08C ¬⁄U 5 kg ¡‹, 08C ¬⁄U ’»¸§ ◊¥ f¡qäS>f¡V$fdp„ 08C _y„ 5 kg `pZu 08C bfadp„ ê$`p„sfus
the energy consumed by the freezer is close ’Œ‹ÃÊ „Ò Ã’ »˝§Ë¡⁄U mÊ⁄UÊ π¬ÊßZ ªß¸ ™§¡Ê¸ ‹ª÷ª „Ò — L$fhy„ lp¡e sp¡ f¡qäS>f¡V$f¡ hp`f¡g EÅ® __________ _u
to :
(1) 1.67×105 J _ÆL$_y„ d|ëe ^fphi¡.
(1) 1.67×105 J
(2) 1.68×106 J (1) 1.67×105 J
(2) 1.68×106 J
(3) 1.51×105 J (2) 1.68×106 J
(3) 1.51×105 J
(4) 1.71×107 J (3) 1.51×105 J
(4) 1.71×107 J
(4) 1.71×107 J

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Set - 03 7
9. Which of the following shows the correct 9. Á∑§‚Ë •ÊŒ‡Ê¸ ªÒ‚ ∑§ Á‹ÿ ÁSÕ⁄U Ãʬ◊ÊŸ ¬⁄U ©‚∑§ ŒÊ’ 9. _uQ¡ Ap`¡g ApL©$rsAp¡ `¥L$u L$C ApL©$rÑ AQm sp`dp_¡
relationship between the pressure ‘P’ and
‘P’ ÃÕÊ ÉÊŸàfl ‘ρ’ ∑§ ’Ëø ‚¥’¥œ ∑§ Á‹ÿ ÁŸêŸ ◊¥ ‚ fl¡g Apv$i® hpey dpV¡$ v$bpZ ‘P’ A_¡ O_sp ρ hÃQ¡_p¡
density ρ of an ideal gas at constant
temperature ? ∑§ÊÒŸ-‚Ê ÁøòÊ ‚„Ë „Ò? kpQp¡ k„b„^ v$ip®h¡ R>¡.

(1) (1) (1)

(2) (2) (2)

(3) (3) (3)

(4) (4) (4)

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Set - 03 8
10. In an engine the piston undergoes vertical 10. ∞∑§ ߥ¡Ÿ ∑§Ê Á¬S≈UŸ 7 cm •ÊÿÊ◊ ∑§Ë ‚⁄U‹-•Êflø- 10. A¡L$ A¡[ÞS>_dp„ r`õV$_ DÝh® qv$ipdp„ 7 cm _p
simple harmonic motion with amplitude
ªÁà ™§äflʸœ⁄U ◊¥ ∑§⁄U ⁄U„Ê „Ò– Á¬S≈UŸ ∑§ ™§¬⁄U ∞∑§ L„$`rhõspf kp\¡ SHM (k.Ap.N) L$f¡ R>¡. r`õV$__u
7 cm. A washer rests on top of the piston
and moves with it. The motor speed is flʇÊ⁄U ⁄UπÊ „Ò ¡Ê ©‚∑§ ‚ÊÕ ø‹ÃÊ „Ò– ◊Ê≈U⁄U ∑§Ë ªÁà D`f fl¡g hp¸if s¡_u kp\¡ S> Nrs L$f¡ R>¡. r`õV$__¡
slowly increased. The frequency of the œË⁄U-œË⁄U ’…∏Ê߸ ¡ÊÃË „Ò ÃÊ Á¬S≈UŸ ∑§Ë •ÊflÎÁûÊ Á¡‚ ¬⁄U Nrs L$fphsu dp¡V$f_u TX$` ^ud¡\u h^pfhpdp„ Aph¡ R>¡.
piston at which the washer no longer stays flʇÊ⁄U Á¬S≈UŸ ∑§Ê ‚ÊÕ ¿UÊ«∏ ŒÃÊ „Ò, fl„ ‹ª÷ª „Ò — hp¸if r`õV$__u k`pV$u_¡ R>p¡X$u v$¡ s¡ r`õV$__u S>ê$fu
in contact with the piston, is close to :
(1) 0.1 Hz Aph©rÑ __________ _u _ÆL$_y„ d|ëe li¡.
(1) 0.1 Hz
(2) 1.2 Hz (1) 0.1 Hz
(2) 1.2 Hz
(3) 0.7 Hz (2) 1.2 Hz
(3) 0.7 Hz
(4) 1.9 Hz (3) 0.7 Hz
(4) 1.9 Hz
(4) 1.9 Hz

11. A toy-car, blowing its horn, is moving with
11. ∞∑§ Áπ‹ÊÒŸÊ ∑§Ê⁄U „ÊŸ¸ ’¡ÊÃË „È߸ 5 m/s ∑§Ë ÁSÕ⁄U
a steady speed of 5 m/s, away from a wall. ªÁà ‚ ∞∑§ ŒËflÊ⁄U ‚ ŒÍ⁄U ÃÕÊ ∞∑§ √ÿÁÄà ∑§Ë •Ê⁄U ¡Ê 11. A¡L$ lp¡_® hNpX$su fdL$X$p„_u L$pf qv$hpg \u v|$f sfa
An observer, towards whom the toy car is ⁄U„Ë „Ò– ©‚ √ÿÁÄà ∑§Ê 5 ’Ë≈U/‚¥. ‚ÈŸÊ߸ ŒÃË „Ò¥– ÿÁŒ 5 m/s _u AQm TX$`\u Nrs L$f¡ R>¡ . A¡ L $
moving, is able to hear 5 beats per second. „flÊ ◊¥ äflÁŸ ∑§Ë ªÁà 340 m/s „Ò, Ã’ „ÊŸ¸ ∑§Ë Ahgp¡L$_L$pf, L¡$ S>¡_u sfa fdL$X$p„_u L$pf Nrs L$f¡ R>¡,
If the velocity of sound in air is 340 m/s,
the frequency of the horn of the toy car is •ÊflÎÁûÊ ‹ª÷ª „Ò — A¡L$ k¡L$ÞX$dp„ 5 õ`„v$ (beats) kp„cm¡ R>¡. Å¡ lhpdp„
close to : (1) 680 Hz Ýhr__p¡ h¡N 340 m/s lp¡e sp¡ L$pf_p lp¡_®_u Aph©qÑ
(1) 680 Hz (2) 510 Hz _______ _u _ÆL$_u li¡.
(2) 510 Hz (3) 340 Hz (1) 680 Hz
(3) 340 Hz (4) 170 Hz (2) 510 Hz
(4) 170 Hz (3) 340 Hz
(4) 170 Hz

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Set - 03 9
12. Within a spherical charge distribution of 12. •Êfl‡Ê-ÉÊŸàfl ρ (r) ∑§ Á∑§‚Ë ªÊ‹Ëÿ-•Êfl‡Ê-ÁflÃ⁄UáÊ, 12. ρ(r) S>¡V$gu Np¡gue rhÛyscpf rhsfZ dpV¡$, A_y¾$d¡
charge density ρ(r), N equipotential
∑§ •ãŒ⁄U N ‚◊Áfl÷fl-¬Îc∆U, Á¡Ÿ∑§Ë Áfl÷fl „Ò r 0 , r 1 , r 2 ,..........r N rÓÄep ^fphsu A_¡
surfaces of potential V0, V0+∆V, V0+2∆V,
V 0 , V 0 +∆V, V 0 +2∆V, ........ V 0 +N∆V V0, V0+∆V, V0+2∆V, .........., V0+N∆V
.......... V0+N∆V (∆V>0), are drawn and
have increasing radii (∆V>0), •Ê⁄UÁπà Á∑§ÿ ªÿ „Ò¥ •ÊÒ⁄U ©Ÿ∑§Ë ÁòÊíÿÊ∞¥ (∆V>0) S>¡V$gy„ [õ\rsdp_ ^fphsu N kd[õ\rsdp_
r 0 , r 1 , r 2 ,..........r N , respectively. If the ∑˝§◊‡Ê— r0, r1, r2,..........rN „Ò¥– ÿÁŒ ÁòÊíÿÊ•Ê¥ ∑§Ê k`pV$u (`©›$) v$p¡fhpdp„ Aph¡ R>¡. V0 A_¡ ∆V _p
difference in the radii of the surfaces is •ãÃ⁄UÊ‹, ‚÷Ë V0 ÃÕÊ ∆V ∑§ ◊ÊŸÊ¥ ∑§ Á‹ÿ, ÁSÕ⁄U „Ò b^pS> d|ëe dpV¡$ Å¡ k`pV$uAp¡_u rÓÄep hÃQ¡_p¡ saphs
constant for all values of V0 and ∆V then :
Ã’ — AQm fl¡sp¡ lp¡e sp¡ ________.
(1) ρ (r) α r
(1) ρ (r) α r (1) ρ (r) α r
(2) ρ (r) = constant
(2) ρ (r)=•ø⁄U (2) ρ (r) = AQm
1
(3) ρ (r ) α 1 1
r (3) ρ (r ) α (3) ρ (r ) α
r r
1
(4) ρ (r ) α 1 1
r2 (4) ρ (r ) α (4) ρ (r ) α
2
r r2

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Set - 03 10
13. Figure shows a network of capacitors where 13. ÁøòÊ ‚¥œÊÁ⁄UòÊÊ¥ ∑§Ê ÁŸ∑§Êÿ Œ‡ÊʸÃÊ „Ò, ¡„Ê° •¥∑§ µF ◊¥ 13. ApL©$rÑdp„ k„OpfL$p¡_p¡ b_¡gy„ _¡V$hL®$ v$ip®h¡ R>¡ Äep„
the numbers indicates capacitances in
œÊÁ⁄UÃÊ Œ‡ÊʸÃ „Ò¥– A fl B ∑§ ’Ëø ¬˝÷ÊflË œÊÁ⁄UÃÊ L¡ $ ` ¡ k uV$ f _u bpSy > dp„ gM¡ g _„ b f s¡ L¡ $ ` ¡ k uV$ f _y „
micro Farad. The value of capacitance C if
the equivalent capacitance between point 1 µF „ÊŸ ∑§ Á‹ÿ C ∑§Ë œÊÁ⁄UÃÊ „ÊŸË øÊÁ„ÿ — dpC¾$p¡a¡fpX$dp„ d|ëe v$ip®h¡ R>¡. Å¡ A A_¡ B hÃQ¡_y„
A and B is to be 1 µF is : `qfZpdu L¡$`¡kuV$Þk (k„OpfL$sp) 1 µF Å¡Csy„ lp¡e sp¡
k„OpfL$ C _y„ d|ëe ________ Å¡Ci¡.

31
(1) µF
31 23 31
(1) µF (1) µF
23 32 23
(2) µF
32 23 32
(2) µF (2) µF
23 33 23
(3) µF
33 23 33
(3) µF (3) µF
23 34 23
(4) µF
34 23 34
(4) µF (4) µF
23 23

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Set - 03 11
14. The resistance of an electrical toaster has a 14. Á’¡‹Ë ‚ ø‹Ÿ flÊ‹ ≈UÊS≈U⁄U ∑§ ¬˝ÁÃ⁄UÊœ ∑§Ê Ãʬ◊ÊŸ ‚ 14. A¡L$ Cg¡¼V²$uL$ V$p¡õV$f_p¡ s¡_p D`ep¡N v$fçep_, sp`dp_
temperature dependence given by
’Œ‹Êfl R(T)=R0 [1+α(T−T0)] mÊ⁄UÊ ÁŒÿÊ ªÿÊ kp\¡ _ p¡ k„ b „ ^ R(T)=R0 [1+α(T−T0)] hX¡ $
R(T)=R 0 [1+α(T−T 0 )] in its range of
operation. At T0=300 K, R=100 Ω and at „Ò – T 0 =300 K ¬⁄U R=100 Ω „Ò ÃÕÊ Ap`hpdp„ Aph¡ R>¡ . T 0 =300 K dpV¡ $
T=500 K, R=120 Ω. The toaster is T=500 K ¬⁄U R=120 Ω „Ò– ≈UÊS≈U⁄U 200 VU ∑§ R=100 Ω A_¡ T=500 K dpV¡$ R=120 Ω R>¡.
connected to a voltage source at 200 V and dÊà ‚ ¡È«∏Ê „Ò, ÃÕÊ ©‚∑§Ê Ãʬ◊ÊŸ 300 K ‚ ∞∑§ V$p¡õV$f_¡ A¡L$ 200 V _p hp¡ëV¡$S> Dv¹$Nd kp\¡ Å¡X$u s¡_y„
its temperature is raised at a constant rate
from 300 to 500 K in 30 s. The total work
‚◊ÊŸ Œ⁄U ¬⁄U ’…∏∑§⁄U 30 s ◊¥ 500 K „Ê ¡ÊÃÊ „Ò– Ã’ sp`dp_ AQm v$f¡ 300 \u 500 K ky^u gC S>hpdp„
done in raising the temperature is : ß‚ ¬˝∑˝§◊ ◊¥ Á∑§ÿÊ ªÿÊ ∑ȧ‹ ∑§Êÿ¸ „Ò — Aph¡ R>¡. s¡ dpV¡$ gpNsp¡ kdeNpmp¡ 30 s R>¡. sp¡
1.5 1.5 sp`dp__p¡ h^pfp¡ L$fhp dpV¡ $ S>ê$fu Ly $ g L$pe®
(1) 400 l n J (1) 400 l n J __________.
1.3 1.3
1.5
2 2 (1) 400 l n J
(2) 200 l n J (2) 200 l n J 1.3
3 3
2
5 5 (2) 200 l n J
(3) 400 l n J (3) 400 l n J 3
6 6
(4) 300 J (4) 300 J 5
(3) 400 l n J
6
(4) 300 J

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Set - 03 12
15. Consider a thin metallic sheet 15. ∞∑§ ¬Ã‹Ë œÊÃÈ ‡ÊË≈U ¬Îc∆U ∑§ ‹ê’flà ⁄UπË „Ò •ÊÒ⁄U 15. kdp„Nu Qy„bL$ue n¡Ó B dp„, `yõsL$_p `¡S>_¡ g„b A_¡
perpendicular to the plane of the paper
ÁøòÊ ◊¥ ÁŒπÊ߸ ÁŒ‡ÊÊ ◊¥ flª ‘v’ ‚ ∞∑§ ‚◊ÊŸ øÈê’∑§Ëÿ- ‘v’ S>¡V$gu AQm TX$`\u `¡S>_u A„v$f v$pMg \su
moving with speed ‘v’ in a uniform
magnetic field B going into the plane of the ˇÊòÊ B ◊¥ ø‹ ⁄U„Ë „Ò– øÈê’∑§Ëÿ-ˇÊòÊ ß‚ ‚◊Ë ¬Îc∆U ^psy _ u `psmu s[¼s (`© › $) _¡ Ýep_dp„ gp¡ .
paper (See figure). If charge densities σ1 and ◊¥ ¬˝fl‡Ê ∑§⁄U ⁄U„Ê „Ò– ÿÁŒ ß‚ ‡ÊË≈U ∑§Ë ’ÊßZ •ÊÒ⁄U ŒÊßZ (ApL©$rÑ Sy>Ap¡) Å¡ s¡_u X$pbu A_¡ S>dZu k`pV$u `f
σ2 are induced on the left and right surfaces, ‚ÄÊ¥ ¬⁄U ∑˝§◊‡Ê— ¬Îc∆U-•Êfl‡Ê-ÉÊŸàfl σ1 ÃÕÊ σ2 ¬˝Á⁄Uà A_y¾$d¡ σ1 A_¡ σ2 S>¡V$gu `©›$rhÛyscpf O_sp Dv¹$ch¡
respectively, of the sheet then (ignore
fringe effects) :
„ÊÃ „Ò¥, Ã’ ©¬Ê¥Ã-¬˝÷Êfl ∑§Ê Ÿªáÿ ◊ÊŸÃ „È∞ sp¡ __________ \i¡. (qäÞS> Akf AhNZp¡.)
σ1 ÃÕÊ σ2 ∑§ ◊ÊŸ „Ê¥ª —

(1) σ1=e0 v B, σ2=−e0 v B
(1) σ1=e0 v B, σ2=−e0 v B

vB −
0 vB
(1) σ1=e0 v B, σ2=−e0 v B (2) σ1 = 0 ,σ2 =

vB −
0 vB 2 2
(2) σ1 = 0 ,σ2 =
2 2
vB −
0 vB (3) σ1=σ2=e0 vB
(2) σ1 = 0 ,σ2 =
(3) σ1=σ2=e0 vB 2 2
−
0 v B
vB
(3) σ1=σ2=e0 vB (4) σ1 = , σ2 = 0
−
0 v B
vB 2 2
(4) σ1 = , σ2 = 0
2 2 −
0 v B
vB
(4) σ1 = , σ2 = 0
2 2

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

Set - 03 13
16. A fighter plane of length 20 m, wing span 16. ∞∑§ ‹«∏Ê∑ͧ ¡„Ê¡ ∑§Ë ‹ê’Ê߸ 20 m, ¬¥πÊ¥ ∑§ Á‚⁄UÊ¥ ∑§ 16. 20 m g„bpC ^fphsy„, 15 m `p„Muep_u `lp¡mpC (A¡L$
(distance from tip of one wing to the tip of
’Ëø ŒÍ⁄UË 15 m ÃÕÊ ™°§øÊ߸ 5 m „Ò, •ÊÒ⁄U ÿ„ ÁŒÀ‹Ë bpSy>_u `p„Muep_p R>¡X$p\u buÆ bpSy>_p `p„rMep_p
the other wing) of 15 m and height
5 m is flying towards east over Delhi. Its ∑§ ™§¬⁄U ¬Ífl¸-ÁŒ‡ÊÊ ◊¥ 240 ms−1 ªÁà ‚ ©«∏ ⁄U„Ê „Ò– R>¡X$p ky^u) A_¡ 5 m S>¡V$gu KQpC ^fphsy„ A¡L$ gX$peL$
speed is 240 ms−1. The earth’s magnetic ÁŒÀ‹Ë ∑§ ™§¬⁄U ¬ÎâflË ∑§Ê øÈê’∑§Ëÿ-ˇÊòÊ 5×10−5 T rhdp_ qv$ëlu_p `|h® sfa EX$u füy„ R>¡. s¡_u TX$`
field over Delhi is 5×10 −5 T with the „Ò, Á«UÁÄ‹Ÿ‡ÊŸ ∑§ÊáÊ ~08 „Ò, ÃÕÊ Á«U¬ ∑§ÊáÊ θ ∑§ Á‹ÿ 240 ms−1 R>¡. qv$ëlu D`f `©Õhu_y„ Qy„bL$ue n¡Ó_y„
declination angle ~08 and dip of θ such that
2 d|ëe 5×10−5 T d¡Á_¡V$uL$ X¡$[¼g_¡i_ ~08 A_¡ X$u`
2 sin θ =
3
„Ò– ÿÁŒ ¬˝Á⁄UÃ-Áfl÷fl „Ò¥ — VB ¡„Ê¡ ∑§
sin θ = . If the voltage developed is VB 2
3 A¡ÞNg θ A¡hp¡ R>¡ L¡$ S>¡\u sin θ = \pe. Å¡
between the lower and upper side of the
™§¬⁄U fl ŸËø ∑§ ’Ëø ; VW ¬¥πÊ¥ ∑§ Á‚⁄UÊ¥ ∑§ ’Ëø– 3
plane and VW between the tips of the wings Ã’ — àg¡__p _uQ¡_p A_¡ D`f_p R>¡X$p hÃQ¡ VB S>¡V$gp¡
then VB and VW are close to : (1) VB=45 mV ; VW=120 mV ŒÊÿÊ¥ ¬¥π- hp¡ëV¡$S> A_¡ VW S>¡V$gp¡ hp¡ëV¡$S> b¡ `p„Mp¡_p R>¡X$p hÃQ¡
(1) VB=45 mV; VW=120 mV with right Á‚⁄UÊ+ve DÑ`Þ_ \sp¡ lp¡ e sp¡ V B A_¡ V W _y „ d| ë e
side of pilot at higher voltage
(2) VB=45 mV ; VW=120 mV ’ÊÿÊ¥ ¬¥π- __________ _ÆL$_y„ li¡..
(2) VB=45 mV; VW=120 mV with left
side of pilot at higher voltage Á‚⁄UÊ−ve (1) VB=45 mV A_¡ VW= `pegp¡V$_u S>dZu
(3) VB=40 mV; VW=135 mV with right (3) VB=40 mV ; VW=135 mV ŒÊÿÊ¥ ¬¥π- bpSy> KQp hp¡ëV¡$S>¡ lp¡e s¡ fus¡ 120 mV
side of pilot at high voltage
Á‚⁄UÊ+ve (2) VB=45 mV; VW= `pegp¡V$_u X$pbu bpSy>
(4) VB=40 mV; VW=135 mV with left
side of pilot at higher voltage (4) VB=40 mV ; VW=135 mV ’ÊÿÊ¥ ¬¥π- KQp hp¡ëV¡$S>¡ lp¡e s¡ fus¡ 120 mV
Á‚⁄UÊ−ve (3) VB=40 mV; VW= `pegp¡V$_u S>dZu bpSy>
KQp hp¡ëV¡$S>¡ lp¡e s¡ fus¡ 135 mV
(4) VB=40 mV; VW= `pegp¡V$_u X$pbu bpSy>
KQp hp¡ëV¡$S>¡ lp¡e s¡ fus¡ 135 mV

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

Set - 03 14
17. A conducting metal circular-wire-loop of 17. ‘r’ ÁòÊíÿÊ ∑ § œÊÃÈ flÎ û ÊËÿ-ÃÊ⁄U - ‹Í ¬ ∑§Ê ¬Î c ∆U , 17. A¡L$ kyhplL$ ^psy_y„ hsy®mpL$pf N|„Qmy„ L¡$ S>¡_u rÓÄep r
radius r is placed perpendicular to a
B = B0 e
−t
mÊ⁄UÊ ’Œ‹Ã „È∞ øÈê’∑§Ëÿ-ˇÊòÊ ∑§ lp¡e s¡_¡ Qy„bL$ue n¡Ó_¡ g„bê$`¡ d|L$hpdp„ Aph¡ R>¡. Ap
magnetic field which varies with time as τ

−t ‹ê’flà ⁄UπÊ „Ò– ¡„Ê° ‚◊ÿ t=0 ¬⁄U B0 ÃÕÊ τ •ø⁄U Qy„bL$ue n¡Ó kde kp\¡ B = B0 e−t τ âdpZ¡ bv$gpe
0 and τ are constants,
B = B0 e τ , where B

at time t=0. If the resistance of the loop is „Ò¥– ÿÁŒ ‹Í¬ ∑§Ê ¬˝ÁÃ⁄UÊœ R „Ò, Ã’ ∑§Ê»§Ë íÿÊŒÊ ‚◊ÿ R>¡, Äep„ B0 A_¡ τ t=0 kde¡ AQmp„L$ R>¡. Å¡ N|Q
„ mp_p¡
R then the heat generated in the loop after (t →∞) ªÈ¡⁄UŸ ∑§ ’ÊŒ ©‚ ‹Í¬ ◊¥ ¬ÒŒÊ „È߸ ™§¡Ê¸ „Ò — Ahfp¡^ R lp¡e sp¡ M|b S> gp„bp kde (t →∞) _¡
a long time (t →∞) is : A„s¡ N|„Qmpdp„ DÐ`Þ_ Dódp _______ \i¡.
π 2 r 4 B04
(1)
π 2 r 4 B04 2τ R π 2 r 4 B04
(1) (1)
2τ R 2τ R
π 2 r 4 B02
(2)
π 2 r 4 B02 2τ R π 2 r 4 B02
(2) (2)
2τ R 2τ R
π 2 r 4 B02 R
(3)
π 2 r 4 B02 R τ π 2 r 4 B02 R
(3) (3)
τ τ
π 2 r 4 B02
(4)
π 2 r 4 B02 τR π 2 r 4 B02
(4) (4)
τR τR

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Set - 03 15
18. Consider an electromagnetic wave 18. √ÿÊ◊ ◊¥ ø‹ ⁄U„Ë flÒlÈÃ-ëÊÈê’∑§Ëÿ Ã⁄¥Uª ∑§ Á‹∞ ‚„Ë 18. i|ÞephL$pidp„ âkfsp rhÛysQy„bL$ue sf„N_¡ Ýep_dp„
propagating in vacuum. Choose the correct
Áfl∑§À¬ øÈÁŸ∞– gp¡. _uQ¡dp„\u kpQy„ rh^p_ `k„v$ L$fp¡.
statement :
(1) For an electromagnetic wave (1) +x ÁŒ‡ÊÊ ◊¥ øÊÁ‹Ã flÒlÈÃ-øÈê’∑§Ëÿ Ã⁄¥Uª ∑§ (1) +x -qv$ipdp„ Nrs L$fsp rhÛysQy„bL$ue sf„N
propagating in +x direction the dpV¡$ rhÛys n¡Ó
electric field is
→
Á‹ÿ E = 1 Eyz (x, t ) y − z , ( ∧ ∧
)
2 →
E=
1
(
∧ ∧
Eyz ( x, t ) y − z ) A_¡
→
E=
1
2
( ∧
Eyz ( x, t ) y − z and the
∧
) →
B=
1
( ∧
Byz ( x, t ) y + z
∧
) 2
2 Qy„bL$ue n¡Ó
magnetic field is
→
B=
1 ∧ ∧
Byz ( x, t ) y + z( ) (2) +x ÁŒ‡ÊÊ ◊¥ øÊÁ‹Ã flÒlÈÃ-øÈê’∑§Ëÿ Ã⁄¥Uª ∑§
→
B=
1
2
(∧ ∧
Byz ( x, t ) y + z ) \i¡.
(2)
2
For an electromagnetic wave
Á‹ÿ
→
E=
1
2
∧ ∧
Eyz ( y, z, t ) y + z , ( ) (2) +x -qv$ipdp„ Nrs L$fsp rhÛysQy„bL$ue sf„N
propagating in +x direction the dpV¡$ rhÛys n¡Ó
electric field is
→
B=
1
2
∧ ∧
Byz ( y, z, t ) y + z ( ) → 1 ∧
( ∧
) A_¡
E= Eyz ( y, z, t ) y + z
→
E=
1
2
∧
Eyz ( y, z, t ) y + z
∧
( ) and (3) +y ÁŒ‡ÊÊ ◊¥ ø‹ ⁄U„Ë flÒlÈÃ-øÈê’∑§Ëÿ Ã⁄¥Uª ∑§
2
Qy„bL$ue n¡Ó
the magnetic field is → 1 ∧
Á‹ÿ E= Eyz ( x, t ) y ,
→ 1
( ∧ ∧
) 2
→
B=
1
2
∧ ∧
Byz ( y, z, t ) y + z( ) \i¡.
B= Byz ( y, z, t ) y + z
2 → 1 ∧
B= Byz ( x, t ) z (3) +y - qv$ipdp„ Nrs L$fsp rhÛysQy„bL$ue sf„N
(3) For an electromagnetic wave 2
→ ∧
propagating in +y direction the dpV¡$ rhÛys n¡Ó E = 1 Eyz ( x, t ) y
→ 1 ∧ 2
electric field is E = Eyz ( x, t ) y
2 → 1 ∧
and the magnetic field is
A_¡ Qy„bL$ue n¡Ó B = Byz ( x, t ) z
2
→ 1 ∧ \i¡.
B= Byz ( x, t ) z
2

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

Set - 03 16
(4) For an electromagnetic wave (4) +y ÁŒ‡ÊÊ ◊¥ ø‹ ⁄U„Ë flÒlÈÃ-øÈê’∑§Ëÿ Ã⁄¥Uª ∑§ (4) +y -qv$ipdp„ Nrs L$fsp rhÛysQy„bL$ue sf„N
propagating in +y direction the
→ ∧ → ∧
→ 1 ∧ Á‹ÿ E = 1 Eyz ( x, t ) z , dpV¡ $ rhÛy s n¡ Ó E = 1 Eyz ( x, t ) z
electric field is E = Eyz ( x, t ) z 2 2
2
→ 1 ∧ → 1 ∧
and the magnetic field is B= Bz ( x, t ) y A_¡ Qy„bL$ue n¡Ó B = Bz ( x, t ) y
→ ∧ 2 2
1
B= Bz ( x, t ) y \i¡.
2
19. ∞∑§ ∑§Ê°ø ∑§ •h¸ªÊ‹Ëÿ ∆UÊ‚ ∑§Ë ÁòÊíÿÊ 10 cm ÃÕÊ
19. A hemispherical glass body of radius •¬fløŸÊ¥∑§ 1.5 „Ò– ©‚∑§Ë fl∑˝§Ëÿ ‚Ä ¬⁄U øÊ°ŒË ∑§Ë 19. 10 cm _u rÓÄep ^fphsu A_¡ 1.5 h¾$uch_p„L$ ^fphsp
10 cm and refractive index 1.5 is silvered ¬⁄Uà ø…∏Ê߸ ªß¸ „Ò– ‚◊Ë ¬Îc∆U ∑§ 6 cm ŸËø ÃÕÊ Ágpk_p A¡L$ A^®Np¡mpL$pf cpN_u h¾$ k`pV$u `f Qp„v$u_p¡
on its curved surface. A small air bubble is •ˇÊ ¬⁄U, ∞∑§ ‚͡◊ „flÊ ∑§Ê ’È‹’È‹Ê ÁSÕà „Ò– Ã’ Y$p¡m QY$phhpdp„ Aph¡g R>¡. s¡_u ku^u k`pV$u\u 6 cm
6 cm below the flat surface inside it along
the axis. The position of the image of the
fl∑˝§Ëÿ-Œ¬¸áÊ ‚ ’Ÿ ⁄U„ ’È‹’È‹ ∑§Ë ¬˝ÁÃÁ’ê’ ŒÍ⁄UË „Ò — _uQ¡, A^®Np¡mpL$pf_u A„v$f_p cpNdp„, A¡L$ lhp_p¡ _p_p¡
air bubble made by the mirror is seen : `f`p¡V$p¡ s¡_u An `f fl¡g R>¡. Afukp Üpfp lhp_p
`f`p¡V$p_p„ ârstbb_y„ õ\p_ _________ v$¡Mpi¡.

(1) ‚◊Ë ‚Ä ‚ 14 cm ŸËø
(1) 14 cm below flat surface (2) ‚◊Ë ‚Ä ‚ 30 cm ŸËø
(2) 30 cm below flat surface
(3) ‚◊Ë ‚Ä ‚ 20 cm ŸËø (1) ku^u k`pV$u_u _uQ¡ 14 cm A„sf¡
(3) 20 cm below flat surface
(4) ‚◊Ë ‚Ä ‚ 16 cm ŸËø (2) ku^u k`pV$u_u _uQ¡ 30 cm A„sf¡
(4) 16 cm below flat surface
(3) ku^¡ k`pV$u_u _uQ¡ 20 cm A„sf¡
(4) ku^u k`pV$u_u _uQ¡ 16 cm A„sf¡

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Set - 03 17
20. Two stars are 10 light years away from the 20. ŒÊ ÃÊ⁄U ¬ÎâflË ‚ 10 ¬˝∑§Ê‡Ê-fl·¸ ∑§Ë ŒÍ⁄UË ¬⁄U „Ò¥– ©Ÿ∑§Ê 20. b¡ spfpAp¡ `©Õhu\u 10 âL$pihj® v|$f R>¡. s¡Ap¡_¡
earth. They are seen through a telescope
∞∑§ ≈UÁ‹S∑§Ê¬ mÊ⁄UÊ ŒπÊ ¡ÊÃÊ „Ò, Á¡‚∑§Ê •Á÷ŒÎ‡ÿ∑§ 30 cm ìepk ^fphsp Ap¡bS>¡¼$¹ V$uh (hõsy-g¡Þk) hpmp
of objective diameter 30 cm. The
wavelength of light is 600 nm. To see the 30 cm √ÿÊ‚ ∑§Ê „Ò– ¬˝∑§Ê‡Ê ∑§Ë Ã⁄¥UªŒÒÉÿ¸ 600 nm V¡$guõL$p¡`\u Å¡hpdp„ Aph¡ R>¡. âL$pi_u sf„Ng„bpC
stars just resolved by the telescope, the „Ò– (1 ¬˝∑§Ê‡Ê-fl·¸ =9.46× 1015m) „Ò– ≈UÁ‹S∑§Ê¬ 600 nm R>¡ . Ap b¡ spfp_¡ just R| > V$ p `X¡ $ g p
minimum distance between them should •ª⁄U ©Ÿ ÃÊ⁄UÊ¥ ∑§Ê ‹ª÷ª Áfl÷ÁŒÃ Œπ ¬Ê ⁄U„Ê „Ò, Ã’ (rhc¡qv$s \e¡gp) Å¡hp dpV¡$ s¡d_u hÃQ¡_y„ A„sf
be (1 light year=9.46× 1015m) of the order
of :
©Ÿ∑§ ’Ëø ∑§Ë ŒÍ⁄UË ∑§Ê order „Ò — _________ _p ¾$d_y„ lp¡hy„ Å¡CA¡.

(1) 106 km (1) 106 km (1 âL$pi hj®=9.46× 1015m)
(2) 108 km (2) 108 km (1) 106 km
(3) 1011 km (3) 1011 km (2) 108 km
(4) 1010 km (4) 1010 km (3) 1011 km
(4) 1010 km
21. A photoelectric surface is illuminated 21. ∞∑§ ¬˝∑§Ê‡Ê-flÒlÈà ‚Ä ¬⁄U ¬„‹Ë ’Ê⁄U λ ÃÕÊ ŒÍ‚⁄UË
successively by monochromatic light of λ λ
λ
’Ê⁄U 2 Ã⁄¥UªŒÒÉÿ¸ ∑§Ê ¬˝∑§Ê‡Ê «UÊ‹Ê ¡ÊÃÊ „Ò– ÿÁŒ 21. A¡L$ ap¡V$p¡Cg¡¼V²$uL$ k`pV$u_¡ hpfpasu A_y¾$d¡ λ A_¡ 2
wavelengths λ and . If the maximum
2 ©à‚Á¡¸Ã ¬˝∑§Ê‡Ê-ß‹Ä≈˛UÊÚŸ ∑§Ë •Áœ∑§Ã◊ ªÁá-™§¡Ê¸ sf„Ng„bpC ^fphsp A¡L$f„Nu âL$pi\u âL$pris L$fhpdp„
kinetic energy of the emitted ŒÍ‚⁄UË ’Ê⁄U ◊¥ ¬„‹Ë ’Ê⁄U ∑§Ë ÁÃªÈŸË „Ê, Ã’ ©‚ ‚Ä
photoelectrons in the second case is
Aph¡ R>¡. buÅ qL$õkpdp„ Å¡ DÐkÅ®sp ap¡V$p¡Cg¡¼V²$p¡__u
3 times that in the first case, the work ∑§Ê ∑§Êÿ¸-»§‹Ÿ „Ò — dlÑd Nrs EÅ® â\d qL$õkp L$fsp„ ÓZ NZu dmsu
function of the surface is : hc lp¡e sp¡ k`pV$u_y„ hL®$a„¼i_ _________ \i¡.
(1)
hc 3λ hc
(1) (1)
3λ hc 3λ
(2)
hc 2λ hc
(2) (2)
2λ hc 2λ
(3)
hc λ hc
(3) (3)
λ λ
3 hc
(4)
3 hc λ 3 hc
(4) (4)
λ λ

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Set - 03 18
22. A neutron moving with a speed ‘v’ makes 22. ªÁà ‘v’ ‚ ø‹ÃÊ „È•Ê ∞∑§ ãÿÍ≈˛UÊÚŸ ∞∑§ ÁSÕ⁄U „Ê߸«˛UÊ¡Ÿ 22. ‘v’ S>¡V$gu TX$`\u Nrs L$fsp¡ ÞeyV²$p¡_ A¡L$ [õ\f A_¡
a head on collision with a stationary
¬⁄U◊ÊáÊÈ, ¡Ê •¬ŸË •Êl-•flSÕÊ ◊¥ „Ò, ‚ ‚ê◊Èπ ^fp[õ\rsdp„ fl¡gp lpCX²$p¡S>_ `fdpÏ kp\¡ ku^u
hydrogen atom in ground state. The
minimum kinetic energy of the neutron for ≈UÄ∑§⁄U ∑§⁄UÃÊ „Ò– ãÿÈ≈˛UÊÚŸ ∑§Ë fl„ ãÿÍŸÃ◊ ªÁá ™§¡Ê¸ A\X$pdZ A_ych¡ R>¡. ÞeyV²$p¡__u A[õ\rsõ\p`L$
which inelastic collision will take place is : ’ÃÊÿ¥ Á¡‚ ∑§ „ÊŸ ¬⁄U ÿ„ ≈UÄ∑§⁄U •¬˝àÿÊSÕ „ÊªË — A\X$pdZ \pe s¡ dpV¡ $ _u dlÑd NrsEÅ®
(1) 10.2 eV (1) 10.2 eV __________ \i¡.
(2) 16.8 eV (2) 16.8 eV (1) 10.2 eV
(3) 12.1 eV (3) 12.1 eV (2) 16.8 eV
(4) 20.4 eV (4) 20.4 eV (3) 12.1 eV
(4) 20.4 eV
23. To get an output of 1 from the circuit shown 23. ÁŒÿ ªÿ ¬Á⁄U¬Õ ‚ 1 ÁŸª¸◊ ¬˝Êåà ∑§⁄UŸ ∑§ Á‹ÿ •Êfl‡ÿ∑§
in figure the input must be :
ÁŸfl‡Ê „ÊŸÊ øÊÁ„ÿ — 23. ApL©$rÑdp„ v$ip®h¡g `qf`\ dpV¡$ ApDV$`yV$ 1 dm¡ s¡ dpV¡$
S>ê$fu C_`yV$ __________ \i¡.

(1) a=0, b=1, c=0
(1) a=0, b=1, c=0
(2) a=1, b=0, c=0
(2) a=1, b=0, c=0 (1) a=0, b=1, c=0
(3) a=1, b=0, c=1
(3) a=1, b=0, c=1 (2) a=1, b=0, c=0
(4) a=0, b=0, c=1
(4) a=0, b=0, c=1 (3) a=1, b=0, c=1
(4) a=0, b=0, c=1

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Set - 03 19
24. A modulated signal Cm(t) has the form 24. C m (t)=30 sin 300πt+10 (cos 200πt 24. dp¡X$éygf (Ar^rdrîs) \e¡gp rkÁ_g Cm(t) _uQ¡
C m (t)=30 sin 300πt+10 (cos 200πt −cos 400πt) ∞∑§ ◊Ê«ÈUÁ‹Ã Á‚ÇŸ‹ ∑§Ê Œ‡ÊʸÃÊ „Ò– dyS>b Ap`u iL$pe R>¡. Cm(t)=30 sin 300πt+10
−cos 400πt). The carrier frequency fc, the
Ã’ flÊ„∑§ •ÊflÎÁûÊ fc, ◊Ê«ÈU‹∑§ •ÊflÎÁûÊ fω ÃÕÊ ◊Ê«ÈU‹∑§ (cos 200πt−cos 400πt) L¡$fuef Aph©rÑ fc , k„v$¡ip
modulating frequency (message frequency)
f ω , and the modulation index µ are ߟ«UÄ‚ µ ∑˝§◊‡Ê— „Ò¥ — (dp¡X$éyg¡V$]N) Aph©rÑ fω, A_¡ dp¡X$éyg¡i_ A„L$
respectively given by : 1 (index) µ, A_y¾$d¡ ______ hX¡$ Ap`u iL$pe.
(1) fc=200 Hz ; fω=50 Hz ; µ =
1 2
(1) fc=200 Hz ; fω=50 Hz ; µ = 1
2 2 (1) fc=200 Hz ; fω=50 Hz ; µ =
2
(2) fc=150 Hz ; fω=50 Hz ; µ =
2 3
(2) fc=150 Hz ; fω=50 Hz ; µ = 2
3 1 (2) fc=150 Hz ; fω=50 Hz ; µ =
3
(3) fc=150 Hz ; fω=30 Hz ; µ =
1 3
(3) fc=150 Hz ; fω=30 Hz ; µ = 1
3 1 (3) fc=150 Hz ; fω=30 Hz ; µ =
3
(4) fc=200 Hz ; fω=30 Hz ; µ =
1 2
(4) fc=200 Hz ; fω=30 Hz ; µ = 1
2 (4) fc=200 Hz ; fω=30 Hz ; µ =
2
25. m Œ˝√ÿ◊ÊŸ ∑§ ∑§áÊ ¬⁄U F ’‹ ‹ª ⁄U„Ê „Ò, •ÊÒ⁄U ©‚∑§
25. A particle of mass m is acted upon by a force
R
R Á‹ÿ •ÊŸÈ÷Áfl∑§ ‚ê’¥œ „Ò F = 2 v(t) ß‚ ‚ê’¥œ 25.
R
m v$m ^fphsp¡ L$Z F = 2 v(t) _¡ A_ykfsp
F given by the empirical law F = 2 v(t) . t
t t
If this law is to be tested experimentally by ∑§ ‚àÿʬŸ ∑§ Á‹∞ ÁSÕ⁄U •flSÕÊ ‚ ∑§áÊ ∑§Ë ªÁà ∑§Ê F bm_u Akf l¡W$m R>¡. Å¡ Ap bm-r_ed_¡ âpep¡rNL$
observing the motion starting from rest, the ¬˝ˇÊáÊ (Observation) ∑§⁄U ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ fus¡, L$Z_u Nrs [õ\f [õ\rsdp„\u iê$ \pe s¡ fus¡
best way is to plot : ‚Ê ª˝Ê»§ ‚flÊ¸ûÊ◊ „ÊªÊ? QL$pkhp¡ lp¡e sp¡ __________ _p¡ N°pa A¡ kp¥\u kpfp¡
(1) v(t) against t2
(1) t2 ∑§ ÁflL§h v(t) rhL$ë` \i¡.
1
(2) log v(t) against 2 1 (1) v(t) rhfyÝ^ t2
t (2) ∑§ ÁflL§h log v(t)
t2 1
(3) log v(t) against t (2) log v(t) rhfyÝ^ 2
(3) t ∑§ ÁflL§h log v(t) t
1
(4) log v(t) against 1 (3) log v(t) rhfyÝ^ t
t (4)
t
∑§ ÁflL§h log v(t)
1
(4) log v(t) rhfyÝ^
t

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26. A thin 1 m long rod has a radius of 5 mm. 26. 1 m ‹ê’Ë ¬Ã‹Ë ¿U«∏ ∑§Ë ÁòÊíÿÊ 5 mm „Ò– ÿ¥ª 26. 1 m gp„bp A_¡ 5 mm rÓÄep ^fphsp¡ A¡L$ `psmp¡
A force of 50 πkN is applied at one end to
◊Ê«U‹‚ ÁŸ∑§Ê‹Ÿ ∑§ Á‹ÿ ß‚ ∑§ Á‚⁄U ¬⁄U 50 πkN krmep¡ R>¡ . s¡ _ p¡ ×Y$ [õ\rsõ\p`L$sp A„ L $
determine its Young’s modulus. Assume
that the force is exactly known. If the least ∑§Ê ’‹ ‹ªÊÿÊ ªÿÊ– ◊ÊŸ¥ Á∑§ ’‹ Á’‹∑ȧ‹ ∆UË∑§ ‚ (e„N dp¡X$éygk) ip¡^hp dpV¡$ s¡_p A¡L$ R>¡X$p D`f
count in the measurement of all lengths is ôÊÊà „Ò– ÿÁŒ ‹ê’ÊßÿÊ¥ ∑§ ◊ʬŸ ∑§ •À¬Ê¥‡Ê 50 πkN S>¡V$gy„ bm gNpX$hpdp„ Aph¡ R>¡. A¡hy„ ^pfp¡ L¡$
0.01 mm, which of the following statements 0.01 mm „Ò¥– Ã’ ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ ‚Ê ∑§ÕŸ ª‹Ã Ap bm_y„ d|ëe kQp¡V$ fus¡ dpg|d R>¡. Å¡ g„bpC_p
is false ?
„Ò? b^p S> dp`_dp„ gOyÑd dp` i[¼s 0.01 mm lp¡e
∆Y sp¡ _uQ¡ `¥L$u_y„ L$ey„ rh^p_ Mp¡Vy„$ li¡ ?
(1) gets minimum contribution ∆Y
Y (1) ◊¥ ‹ê’Ê߸ ∑§Ë •ÁŸÁ‡øÃÃÊ ∑§Ê ÿÊªŒÊŸ
Y ∆Y
from the uncertainty in the length. (1) g„bpC_u AQp¡½$kpC_¡ L$pfZ¡ dp„ gOyÑd
(2) The figure of merit is the largest for ãÿÍŸÃ◊ „Ò– Y
the length of the rod. (2) ¿U«∏ ∑§Ë ‹ê’Ê߸ ∑§ Á‹ÿ ŒˇÊÃÊ¥∑§ ‚’‚ ’«∏Ê „Ò– apmp¡ Aphsp¡ li¡.
(3) The maximum value of Y that can be (3) Y ∑§Ê •Áœ∑§Ã◊ ¬˝Ê# „Ê ‚∑§Ÿ flÊ‹Ê ◊ÊŸ (2) krmep_u g„bpC dpV¡$ v$nsp„L$ (figure of
determined is 1014 N/m2.
1014 N/m2. merit) kp¥\u h^pf¡ \i¡.
∆Y (3) d¡ m hu iL$ p e s¡ h u Y _u dlÑd qL„ $ d s
(4) gets its maximum contribution ∆Y
Y (4) ◊¥ Áfl∑ΧÁà ∑§Ë •ÁŸÁ‡øÃÃÊ ∑§Ê ÿÊªŒÊŸ 1014 N/m2 \i¡.
Y
from the uncertainty in strain.
•Áœ∑§Ã◊ „Ò– ∆Y
(4) dp„ dlÑd apmp¡ rhL© $ rsdp„ fl¡ g
Y
AQp¡½$kpC_¡ L$pfZ¡ li¡.

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27. A galvanometer has a 50 division scale. 27. ∞∑§ ªÀflŸÊ◊Ë≈U⁄U ∑§Ë S∑§‹ 50 ÷ʪÊ¥ ◊¥ ’¥≈UË „Ò– ’Ò≈U⁄UË 27. N¡ëh¡_p¡duV$fdp„ 50 L$p`p R>¡. b¡V$fu_p¡ Ap„sqfL$ Ahfp¡^
Battery has no internal resistance. It is
∑§Ê •Ê¥ à Á⁄U∑§ ¬˝ Á Ã⁄UÊ œ ‡ÊÍ ã ÿ „Ò – ÿÁŒ i|Þe R>¡. Äepf¡ =2400 Ω lp¡e R>¡ Ðepf¡ N¡ëh¡_p¡duV$f
found that there is deflection of 40 divisions
when R=2400 Ω. Deflection becomes R=2400 Ω „Ò ÃÊ ÁflˇÊ¬ =40 ÷ʪ „Ò– ÿÁŒ dp„ 40 L$p`p (divisions) ky^u_y„ Aphs®_ dm¡ R>¡.
20 divisions when resistance taken from R=4900 Ω „Ò ÃÊ ÁflˇÊ¬ =20 ÷ʪ „Ò– Ã’ „◊ Äepf¡ Ahfp¡^ `¡V$udp„\u 4900 Ω S>¡V$gp¡ L$pY$hpdp„ Aph¡
resistance box is 4900 Ω. Then we can ÁŸœÊ¸Á⁄Uà ∑§⁄U ‚∑§Ã „Ò¥ Á∑§ — R>¡ Ðepf¡ 20 divisions S>¡V$gy„ Aphs®_ \pe R>¡. sp¡
conclude :
Ap`Z¡ spfZ Ap`u iL$uA¡ L¡$ __________.

(1) ªÀflŸÊ◊Ë≈U⁄U ∑§Ê ¬˝ÁÃ⁄UÊœ 200 Ω „Ò–
(1) Resistance of galvanometer is 200 Ω. (1) N¡ëh¡_p¡duV$f_p¡ Ahfp¡^ 200 Ω li¡.
(2) Full scale deflection current is 2 mA. (2) »È§‹-S∑§‹ ÁflˇÊ¬ ∑§ Á‹ÿ œÊ⁄UÊ 2 mA „Ò–
(2) `|Z®-õL¡$g Aphs®_ dpV¡$ âhpl 2 mA li¡.
(3) Current sensitivity of galvanometer (3) ªÀflŸÊ◊Ë≈U⁄U ∑§Ë œÊ⁄UÊ-‚¥flŒŸ‡ÊË‹ÃÊ 20 µA
is 20 µA/division. (3) N¡ ë h¡ _ p¡ d uV$f_u `° h pl k„ h ¡ q v$sp
¬˝Áà ÷ʪ „Ò–
(4) Resistance required on R.B. for a 20 µA/division \i¡.
(4) ÁflˇÊ¬ =10 ÷ʪ ∑§ Á‹ÿ R=9800 Ω.
deflection of 10 divisions is 9800 Ω. (4) 10 divisions S>¡V$gy„ Aphs®_ d¡mhhp dpV¡$
Ahfp¡^ `¡V$udp„ Ahfp¡^_y„ d|ëe 9800 Ω Å¡Ci¡.
28. To determine refractive index of glass slab 28. ∑§Ê°ø ∑§Ë S‹Ò’ ∑§Ê ø‹-◊Ê߸∑˝§ÊS∑§Ê¬ mÊ⁄UÊ •¬fløŸÊ¥∑§
using a travelling microscope, minimum ÁŸ∑§Ê‹Ÿ ∑§ Á‹ÿ ¡M§⁄UË ¬Ê∆˜ÿÊ¥∑§Ê¥ ∑§Ë ãÿÍŸÃ◊ ‚¥ÅÿÊ
number of readings required are : 28. V²$ph¡g]N dpC¾$p¡õL$p¡`_u dv$v$\u Ágpk_p Qp¡kgp_p¡
„Ò —
(1) Two h¾$ u ch_p„ L $ dp`hp dpV¡ $ Ap¡ R >pdp„ Ap¡ R >p S>ê$ f u
(1) ŒÊ Ahgp¡L$_p¡_u k„¿ep __________ li¡.
(2) Three
(3) Four (2) ÃËŸ (1) b¡
(4) Five (3) øÊ⁄U (2) ÓZ
(4) ¬Ê°ø (3) Qpf
(4) `p„Q

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29. A realistic graph depicting the variation of 29. ∞∑§ ≈˛ U Ê ¥ Á ¡S≈U ⁄ U ∑§Ê common-emitter (CE) 29. A¡L$ V²$p[ÞTõV$f_p common emitter (CE) k„fQ_pdp„
the reciprocal of input resistance in an input
•Á÷ÁflãÿÊ‚ ◊¥ ÁŸfl‡Ê-•Á÷‹ÊˇÊÁáÊ∑§ ◊ʬŸ Á∑§ÿÊ ªÿÊ– C_`yV$ gpnrZL$sp dpV¡$ C_`yV$ Ahfp¡^_p ìeõs_p¡ kp¥\u
characteristics measurement in a common-
emitter transistor configuration is : Ã’ ÁŸfl‡Ê-¬˝ÁÃ⁄UÊœ ∑§ √ÿÈà∑˝§◊ ∑§Ê ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ- h^y hpõsrhL$ N°pa __________ \i¡.
‚Ê ª˝Ê»§ ©Áøà „Ò?

(1) (1)
(1)

(2)
(2) (2)

(3)
(3) (3)

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Set - 03 23

(4) (4) (4)

30. The ratio (R) of output resistance r0, and 30. Á∑§‚Ë ≈˛UÊ¥Á¡S≈U⁄U ∑§Ë ÁŸfl‡Ê-ÁŸª¸◊ •Á÷‹ÊˇÊÁáÊ∑§ 30. V²$p[ÞTõV$f_u C_`yV$ A_¡ ApDV$`yV$ gpnrZL$sp_p
the input resistance ri in measurements of
◊ʬŸ ∑ § Á‹ÿ ¬˝ ÿ È Ä Ã ÁŸª¸ ◊ -¬˝ Á Ã⁄U Ê  œ (r0) fl dp`_dp„ ApDV$`yV$ Ahfp¡^ r0 A_¡ C_`yV$ Ahfp¡^ ri
input and output characteristics of a
transistor is typically in the range : ÁŸfl‡Ê-¬˝ÁÃ⁄UÊœ (ri) ∑§ •ŸÈ¬Êà (R) ∑§Ê •ÊÿÊ◊ _p¡ NyZp¡Ñf (R) __________ S>¡V$gu f¡ÞS>dp„ lp¡e R>¡.
(1) R~102−103 (range) „ÊªÊ? (1) R~102−103
(2) R~1−10 (1) R~102−103 (2) R~1−10
(3) R~0.1−0.01 (2) R~1−10 (3) R~0.1−0.01
(4) R~0.1−1.0 (3) R~0.1−0.01 (4) R~0.1−1.0
(4) R~0.1−1.0
31. The volume of 0.1N dibasic acid sufficient 31. 0.1N X$peb¡rTL$ A¡rkX$_y„ L$v$ iy„ li¡ L¡$ S>¡ 1 g b¡CT_p
to neutralize 1 g of a base that furnishes 0.04 31. 0.1N ÁmˇÊÊ⁄UËÿ •ê‹ ∑§Ê •Êÿß ÄÿÊ „ÊªÊ ¡Ê 1 ª˝Ê◊ sV$õ\uL$fZ L$fhp dpV¡$ `ep®á lp¡e L¡$ S>¡_p S>gue ÖphZdp„
mole of OH− in aqueous solution is :
ˇÊÊ⁄U∑§ Á¡‚∑§ ¡‹Ëÿ Áfl‹ÿŸ ◊¥ 0.04 ◊Ê‹ OH− „Ò 0.04 dp¡g OH− Aph¡gp R>¡ ?
(1) 200 mL
∑§Ê ©ŒÊ‚ËŸ ∑§⁄UŸ ∑§ Á‹ÿ ¬ÿʸ# „Ò? (1) 200 mL
(2) 400 mL (2) 400 mL
(1) 200 mL
(3) 600 mL (3) 600 mL
(2) 400 mL
(4) 800 mL (4) 800 mL
(3) 600 mL
(4) 800 mL

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Set - 03 24
32. Initially, the root mean square (rms) velocity 32. ∞∑§ Áfl‡Ê· Ãʬ ¬⁄U ¬˝Ê⁄Uê÷ ◊¥ ŸÊß≈˛UÊ¡Ÿ •áÊÈ•Ê¥ (N2) 32. r_ròs sp`dp_ `f, iê$Apsdp„, N2 AÏAp¡_p¡
of N2 molecules at certain temperature is u.
∑§Ê flª¸ ◊Êäÿ ◊Í‹ flª u „Ò– ÿÁŒ ß‚ Ãʬ ∑§Ê ŒÈªÈŸÊ kf¡fpi hN®d|m (root mean square) (rms) h¡N u
If this temperature is doubled and all the
nitrogen molecules dissociate into nitrogen ∑§⁄U ÁŒÿÊ ¡Êÿ •ÊÒ⁄U ‚÷Ë ŸÊß≈˛UÊ¡Ÿ •áÊÈ ÁflÿÊÁ¡Ã R>¡. Å¡ Ap sp`dp_ bdÏ„ (Doubled) L$fhpdp„ Aph¡
atoms, then the new rms velocity will be : „Ê∑§⁄U ŸÊß≈˛UÊ¡Ÿ ¬⁄U◊ÊáÊÈ ’Ÿ ¡Ê∞ ÃÊ ŸÿÊ flª¸ ◊Êäÿ sp¡ b^p S> _pCV²$p¡S>_ AÏAp¡_y„ rhep¡S>_ _pCV²$p¡S>_
(1) u/2 ◊Í‹ flª „ÊªÊ — `fdpÏAp¡dp„ \pe sp¡, `R>u _hy„ kf¡fpi hN®d|m (rms)
(2) 2u (1) u/2 h¡N iy„ \i¡ ?
(3) 4u (2) 2u (1) u/2
(4) 14u (3) 4u (2) 2u
(4) 14u (3) 4u
33. Aqueous solution of which salt will not (4) 14u
contain ions with the electronic
configuration 1s22s22p63s23p6 ?
33. Á∑§‚ ‹fláÊ ∑§ ¡‹Ëÿ Áfl‹ÿŸ ◊¥ 1s22s22p63s23p6
33. Cg¡¼V²$p¡r_L$ rhÞepk 1s22s22p63s23p6 ^fphsp L$ep
(1) NaF ß‹Ä≈˛UÊÚÁŸ∑§ ÁflãÿÊ‚ ∑§ •ÊÿŸ Ÿ„Ë¥ „Ê¥ª?
npf_y„ S>gue ÖphZ Ape_p¡ ^fphsp„ _\u ?
(2) NaCl (1) NaF
(1) NaF
(3) KBr (2) NaCl (2) NaCl
(4) CaI2 (3) KBr (3) KBr
(4) CaI2 (4) CaI2
34. The bond angle H-X-H is the greatest in the
compound : 34. Á∑§‚ ÿÊÒÁª∑§ ◊¥ H-X-H •Ê’㜠∑§ÊáÊ ‚flʸÁœ∑§ „Ò? 34. L$ep k„ep¡S>_dp„ H-X-H b„^M|Zp¡ khp®r^L$ R>¡ ?
(1) CH4 (1) CH4 (1) CH4
(2) NH3 (2) NH3 (2) NH3
(3) H2O (3) H2O (3) H2O
(4) PH3 (4) PH3
(4) PH3

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35. If 100 mole of H2O2 decompose at 1 bar and 35. ÿÁŒ H2O2 ∑§ 100 ◊Ê‹ 1 bar ÃÕÊ 300 K ¬⁄U 35. Å¡ H2O2 _p 100 dp¡g 1 bpf (bar) A_¡ 300 K `f
300 K, the work done (kJ) by one mole of
ÁflÿÊÁ¡Ã „Ê ÃÊ 1 bar ŒÊ’ ∑§ ÁflL§h 1 ◊Ê‹ •ÊÚÄ‚Ë¡Ÿ rhOV$us \pe sp¡, 1 bpf (bar) v$bpZ_p rhfyÝ^ 1
O2(g) as it expands against 1 bar pressure
is : ∑§ ÁflSÃÊÁ⁄Uà „ÊŸ ¬⁄U Á∑§ÿÊ „È•Ê ∑§Êÿ¸ (kJ ◊¥) „ÊªÊ — dp¡g O2(g) _p rhõsfZ \hp_¡ gu^¡ \e¡g L$pe® (kJ)
2H2 O 2 (l) ⇌ 2H2 O(l) + O 2 (g) ip¡^p¡.
2H2 O 2 (l) ⇌ 2H2 O(l) + O 2 (g)
2H2 O 2 (l) ⇌ 2H2 O(l) + O 2 (g)
(R = 8.3 J K−1 mol−1) (R = 8.3 J K−1 mol−1)
(1) 62.25 (1) 62.25 (R = 8.3 J K−1 mol−1)
(2) 124.50 (2) 124.50 (1) 62.25
(3) 249.00 (3) 249.00 (2) 124.50
(4) 498.00 (4) 498.00 (3) 249.00
(4) 498.00
36. An aqueous solution of a salt MX2 at certain 36. Á∑§‚Ë Áfl‡Ê· Ãʬ ¬⁄U, ∞∑§ ‹fláÊ MX2 ∑§ ¡‹Ëÿ
temperature has a van’t Hoff factor of 2. 36. A¡L$ r_ròs sp`dp_ `f, npf MX2 _p S>gue
The degree of dissociation for this solution
Áfl‹ÿŸ ∑§Ê flÊã≈U •ÊÚ»§ »Ò§Ä≈U⁄U 2 „Ò– ‹fláÊ ∑§ ß‚
ÖphZ_p¡ hpÞV$-lp¡a Aheh 2 R>¡. sp¡ npf_p Ap
of the salt is : Áfl‹ÿŸ ∑§ Á‹∞ ÁflÿÊ¡Ÿ ◊ÊòÊÊ „ÊªË —
ÖphZ_p¡ rhep¡S>_A„i ip¡^p¡.
(1) 0.33 (1) 0.33
(1) 0.33
(2) 0.50 (2) 0.50
(2) 0.50
(3) 0.67 (3) 0.67
(3) 0.67
(4) 0.80 (4) 0.80
(4) 0.80

37. A solid XY kept in an evacuated sealed 37. ∞∑§ ’¥Œ (‚ËÀ«U) ÁŸflʸÁÃà ¬ÊòÊ ◊¥ ⁄UπÊ ªÿÊ ∆UÊ‚ XY 37. A¡L$ b„^ (sealed) r_hp®rss (evacuated) `pÓdp„
container undergoes decomposition to ÁflÉÊÁ≈Uà „Ê∑§⁄U Ãʬ T ¬⁄U ŒÊ ªÒ‚¥ X ÃÕÊ Y ∑§Ê Á◊üÊáÊ
form a mixture of gases X and Y at fpMhpdp„ Aph¡g O_ (solid) XY rhOV$us \C_¡ T
temperature T. The equilibrium pressure ’ŸÊÃÊ „Ò– ß‚ ¬ÊòÊ ◊¥ ‚Êêÿ ŒÊ’ 10 bar „Ò– ß‚ sp`dp_¡ hpeyAp¡_y„ rdîZ X A_¡ Y b_¡ R>¡. Ap `pÓdp„
is 10 bar in this vessel. Kp for this reaction •Á÷Á∑˝§ÿÊ ∑§ Á‹ÿ Kp „ÊªÊ — k„syg_ v$bpZ 10 bar (bpf) R>¡. Ap âq¾$ep dpV¡$ Kp
is : (1) 5 iy„ R>¡ ?
(1) 5 (2) 10 (1) 5
(2) 10 (3) 25 (2) 10
(3) 25 (4) 100 (3) 25
(4) 100
(4) 100

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38. Oxidation of succinate ion produces 38. ‚ÁÄ‚Ÿ≈U •ÊÿŸ ∑§ •ÊÚÄ‚Ë∑§⁄UáÊ ‚ ∞ÁÕ‹ËŸ ÃÕÊ ∑§Ê’¸Ÿ 38. k[¼k_¡V$ Ape__p Ap¡[¼kX¡$i_\u Cr\gu_ A_¡ L$pb®_
ethylene and carbon dioxide gases. On
«UÊß•ÊÚÄ‚Êß«U ªÒ‚¥ ’ŸÃË „Ò¥– ¬Ê≈Ò˜UÁ‡Êÿ◊ ‚ÁÄ‚Ÿ≈U ∑§ X$pep¡¼kpCX$ hpeyAp¡ DÐ`Þ_ \pe R>¡. `p¡V¡$ried
passing 0.2 Faraday electricity through an
aqueous solution of potassium succinate, ¡‹Ëÿ Áfl‹ÿŸ ‚ 0.2 »Ò§⁄UÊ«U ÁfllÈà ¬˝flÊÁ„à ∑§⁄UŸ ¬⁄U k[¼k_¡V$_p S>gue ÖphZdp„ 0.2 a¡fpX¡$ rhÛysâhpl `kpf
the total volume of gases (at both cathode ªÒ‚Ê¥ ∑§Ê ∑ȧ‹ •Êÿß (∑Ò§ÕÊ«U ÃÕÊ ∞ŸÊ«U ŒÊŸÊ¥ ¬⁄U) L$fsp, STP `f (1 atm A_¡ 273 K) A¡ hpey_p¡ _y„
and anode) at STP (1 atm and 273 K) is : STP (1 atm ÃÕÊ 273 K) ¬⁄U „ÊªÊ — Ly$g L$v$ (L¡$\p¡X$ A_¡ A¡_p¡X$ b„_¡ `f) ip¡^p¡.
(1) 2.24 L (1) 2.24 L (1) 2.24 L
(2) 4.48 L (2) 4.48 L (2) 4.48 L
(3) 6.72 L (3) 6.72 L (3) 6.72 L
(4) 8.96 L (4) 8.96 L (4) 8.96 L

39. The rate law for the reaction below is given 39. ŸËø ŒË ªß¸ •Á÷Á∑˝§ÿÊ ∑§ Á‹∞ Œ⁄U ÁŸÿ◊ k [A][B] 39. _uQ¡ Ap`¡g âq¾$ep dpV¡$_p¡ h¡N-r_ed k [A][B] ìe„S>L$
by the expression k [A][B] √ÿ¥¡∑§ ‚ √ÿÄà Á∑§ÿÊ ¡ÊÃÊ „Ò \u ìe¼s (expression) L$f¡g R>¡.
A+B g Product A+B g ©à¬ÊŒ A+B g _u`S>
If the concentration of B is increased from A _u kp„Ösp 0.1 mole (dp¡g) fpMuA¡ A_¡ Å¡ B _u
0.1 to 0.3 mole, keeping the value of A at A ∑§Ë ‚ÊãŒ˝ÃÊ ∑§Ê ◊ÊŸ 0.1 ◊Ê‹ ¬⁄U ⁄UπÃ „È∞ ÿÁŒ B
∑§Ë ‚ÊãŒ˝ÃÊ 0.1 ‚ ’…∏Ê∑§⁄U 0.3 ◊Ê‹ ∑§⁄U ŒË ¡ÊÃË „Ò ÃÊ kp„Ösp 0.1 \u h^pfu_¡ 0.3 dp¡g L$fhpdp„ Aph¡ sp¡
0.1 mole, the rate constant will be :
(1) k Œ⁄U ÁSÕ⁄UÊ¥∑§ „ÊªÊ — h¡NAQmp„L$ iy„ li¡ ?
(1) k
(2) k/3 (1) k
(2) k/3
(3) 3k (2) k/3
(3) 3k
(4) 9k (3) 3k
(4) 9k
(4) 9k
40. Gold numbers of some colloids 40. L¡$V$gpL$ L$rggp¡_p õhZp¯L$ (Np¡ëX$ _„bf) Ap âdpZ¡ R>¡.
are : Gelatin : 0.005 - 0.01, Gum 40. ∑È § ¿U ∑§Ê ‹ Êß«U Ê  ¥ ∑ § SfláÊÊZ ∑ § „Ò ¥ , Á¡‹ Á ≈U Ÿ —
Æg¡V$u_ (Gelatin) : 0.005 - 0.01, Nd Af¡rbL$
Arabic : 0.15 - 0.25; Oleate : 0.04 - 1.0; 0.005 - 0.01, ª◊ ∞⁄ U Á’∑§ — 0.15 - 0.25;
(Gum Arabic) : 0.15 - 0.25; Ap¡ r gA¡ V $$
Starch : 15 - 25. Which among these is a •ÊÁ‹∞≈U — 0.04 - 1.0; S≈UÊø¸ — 15 - 25, ߟ◊¥ ∑§ÊÒŸ-
better protective colloid ? (Oleate) : 0.04 - 1.0; õV$pQ® (Starch) : 15 - 25.
‚Ê ’„Ã⁄U ⁄UˇÊË ∑§Ê‹Êÿ«U „ÊªÊ?
(1) Gelatin Apdp„\u L$ep¡ kp¥\u h^pfp¡ kpfp¡ frns L$rgg R>¡ ?
(1) Á¡‹Á≈UŸ
(2) Gum Arabic (1) Æg¡V$u_ (Gelatin)
(3) Oleate (2) ª◊ ∞⁄UÁ’∑§ (2) Nd Af¡rbL$ (Gum Arabic)
(4) Starch (3) •ÊÁ‹∞≈U (3) Ap¡rgA¡V$ (Oleate)
(4) S≈UÊø¸ (4) õV$pQ® (Starch)
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41. The following statements concern elements 41. ÁŸêŸ ∑§ÕŸ •Êflø ÃÊÁ‹∑§Ê ◊¥ ©¬ÁSÕà ÃàflÊ¥ ‚ ‚ê’¥ÁœÃ 41. _uQ¡_p rh^p_p¡ Aphs®L$p¡ô$dp„_p sÒhp¡_¡ k„b„r^s R>¡.
in the periodic table. Which of the
„Ò¥– ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ-‚Ê ‚àÿ „Ò? _uQ¡_pdp„\u L$ey„ kpQy„ R>¡ ?
following is true ?
(1) All the elements in Group 17 are (1) ª˝È¬ 17 ◊¥ ‚÷Ë Ãàfl ªÒ‚ „Ò¥– (1) kd|l 17 dp„ b^p S> sÒhp¡ hpeyAp¡ R>¡.
gases. (2) ª˝È¬ 13 ◊¥ ‚÷Ë Ãàfl œÊÃÈ „Ò¥– (2) kd|l 13 dp„ b^p S> sÒhp¡ ^psyAp¡ R>¡.
(2) The Group 13 elements are all metals.
(3) ª˝È¬ 15 ∑§ ÃàflÊ¥ ∑§Ë ÃÈ‹ŸÊ ◊¥ ‚¥ªÃ •Êflø ∑§ (3) Aphs®_¡ A_ygnu_¡ kd|l 15 _p sÒhp¡_u
(3) Elements of Group 16 have lower syg_pdp„ kd|l 16 _p sÒhp¡_u â\d Ape_uL$fZ
ionization enthalpy values compared
ª˝È¬ 16 ∑§ ÃàflÊ¥ ◊¥ •ÊÿŸŸ ∞ãÕÒÀ¬Ë ∑§Ê ◊ÊŸ
to those of Group 15 in the ∑§◊ ⁄U„ÃÊ „Ò– A¡Þ\pë`u Ap¡R>u R>¡.
corresponding periods. (4) kd|l 15 _p sÒhp¡ dpV¡$, kd|ldp„ _uQ¡ S>CA¡
(4) ª˝È¬ 15 ∑§ ÃàflÊ¥ ∑§ Á‹∞, ª˝È¬ ◊¥ ŸËø ¡ÊŸ ¬⁄U
(4) For Group 15 elements, the stability +5 •ÊÚÄ‚Ë∑§⁄UáÊ •flSÕÊ ∑§Ê SÕÊÁÿàfl ’…∏ÃÊ
s¡d +5 Ap¡[¼kX¡$i_ Ahõ\p_u [õ\fsp h^¡
of +5 oxidation state increases down
„Ò– R>¡.
the group.

42. âÖphZ (smelting) Üpfp L$p¡`f_y„ r_óL$j®Z L$fsu hMs¡
42. Extraction of copper by smelting uses silica 42. S◊ÁÀ≈¥Uª mÊ⁄UÊ ∑§ÊÚ¬⁄U ∑§ ÁŸc∑§·¸áÊ ◊¥ Á‚Á‹∑§Ê ÿÊíÿ ∑§
Dd¡f¡g rkrgL$p_p¡ D`ep¡N _uQ¡_pdp„\u iy„ v|$f L$fhp
as an additive to remove : M§¬ ◊¥ ÁŸêŸ ◊¥ ‚ Á∑§‚∑§Ê „≈UÊŸ ∑§ Á‹∞ ∑§Ë ¡ÊÃË „Ò ?
(1) Cu2S
\pe R>¡ ?
(1) Cu2S
(1) Cu2S
(2) FeO (2) FeO
(2) FeO
(3) FeS (3) FeS
(3) FeS
(4) Cu2O (4) Cu2O
(4) Cu2O
43. Identify the reaction which does not
43. ©‚ •Á÷Á∑˝§ÿÊ ∑§Ê ’ÃÊß∞ Á¡‚◊¥ „Êß«˛UÊ¡Ÿ ©à‚Á¡¸Ã 43. âq¾$ep Ap¡mMu bsphp¡ L¡$ S>¡dp„ lpCX²$p¡S>_ DÐkS>®_
liberate hydrogen :
(1) Reaction of zinc with aqueous alkali.
Ÿ„Ë¥ „ÊÃË „Ò — \sp¡ _\u.
(2) Electrolysis of acidified water using (1) ¡‹Ëÿ ˇÊÊ⁄U ∑§ ‚ÊÕ Á¡¥∑§ ∑§Ë •Á÷Á∑˝§ÿÊ (1) S>gue ApëL$gu_u T]L$ kp\¡_u âq¾$ep
Pt electrodes. (2) å‹Á≈UŸ◊ ß‹Ä≈˛UÊ«UÊ¥ ∑§Ê ¬˝ÿÊª ∑§⁄U∑§ •ê‹Ë∑Χà (2) Pt Cg¡¼V²$p¡X$p¡_p¡ D`ep¡N L$fu_¡ A¡rkqX$L$ `pZu_y„
(3) Allowing a solution of sodium in ¡‹ ∑§Ê ÁfllÈà •¬ÉÊ≈UŸ rhÛysrhcpS>_
liquid ammonia to stand.
(4) Reaction of lithium hydride with (3) Œ˝fl •◊ÊÁŸÿÊ ◊¥ ‚ÊÁ«Uÿ◊ ∑§ Áfl‹ÿŸ ∑§Ê ÁSÕ⁄U (3) âhplu A¡dp¡r_epdp„ kp¡qX$ed_p ÖphZ_¡ fpMu_¡
B2H6. „ÊŸ ∑§ Á‹∞ ¿UÊ«∏ ŒŸÊ R>p¡X$u v$¡hy„
(4) B 2H 6 ∑ § ‚ÊÕ ‹ËÁÕÿ◊ „Êß«˛ U Ê ß«U ∑§Ë (4) B2H6 kp\¡ rgr\ed lpCX²$pCX$_u âq¾$ep
•Á÷Á∑˝§ÿÊ
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44. The commercial name for calcium oxide 44. ∑Ò§ÁÀ‡Êÿ◊ •ÊÚÄ‚Êß«U ∑§Ê √ÿÊfl‚ÊÁÿ∑§ ŸÊ◊ „Ò — 44. L¡$ëiued Ap¡¼kpCX$_y„ ìep`pfu ^p¡fZ¡ _pd Ap`p¡.
is :
(1) Á◊À∑§ •ÊÚ»§ ‹Êß◊ (1) v|$r^ep¡ Q|_p¡ (rdëL$ Ap¡a gpCd) (Milk of
(1) Milk of lime
lime)
(2) Slaked lime (2) S‹ÒÄ«U ‹Êß◊
(2) ap¡X¡$gp¡ Q|_p¡ (Slaked lime)
(3) Limestone (3) ‹Êß◊S≈UÊŸ
(3) Q|_p_p¡ `Õ\f (Limestone)
(4) Quick lime (4) ÁÄfl∑§ ‹Êß◊
(4) L$mu Q|_p¡ (Quick lime)
45. Assertion : Among the carbon
allotropes, diamond is an 45. ∑§ÕŸ — ∑§Ê’¸Ÿ ∑§ •¬⁄UM§¬Ê¥ ◊,¥ «UÊÿ◊¥«U ∑ȧøÊ‹∑§ 45. L$\_ : L$pb®__p blºê$`p¡dp„, lufp¡ A¡ AhplL$ R>¡
insulator, whereas,
„Ò ¡’ Á∑§ ª˝»§Êß≈U ∞∑§ ÁfllÈà ‚ÈøÊ‹∑§ Äepf¡ N°¡apCV$ A¡ rhÛys_p kyhplL$ R>¡.
graphite is a good
conductor of electricity. „Ò– L$pfZ : lufpdp„ A_¡ N°¡apCV$dp„ L$pb®__y„ k„L$fZ
Reason : Hybridization of carbon ∑§Ê⁄UáÊ — «UÊÿ◊ã«U ÃÕÊ ª˝»§Êß≈U ◊¥ ∑§Ê’¸Ÿ ∑§Ê A_y¾d¡ sp3 A_¡ sp2 R>¡.
in diamond and graphite ‚¥∑§⁄UáÊ ∑˝§◊‡Ê— sp3 ÃÕÊ sp2 „Ò–
are sp 3 and sp 2 , (1) L$\_ A_¡ L$pfZ b„_¡ kpQp R>¡, s\p L$pfZ A¡
respectively. (1) ∑§ÕŸ ÃÕÊ ∑§Ê⁄UáÊ ŒÊŸÊ¥ ‚„Ë „Ò¥ ÃÕÊ ∑§Ê⁄UáÊ ∑§ÕŸ L$\_ dpV¡$_u kpQu kdS|>su R>¡.
(1) Both assertion and reason are correct, ∑§Ë ‚„Ë √ÿÊÅÿÊ „Ò– (2) L$\_ A_¡ L$pfZ b„_¡ kpQp R>¡, `Z L$pfZ A¡
and the reason is the correct (2) ∑§ÕŸ ÃÕÊ ∑§Ê⁄UáÊ ŒÊŸÊ¥ ‚„Ë „Ò¥ ¬⁄UãÃÈ ∑§Ê⁄UáÊ, L$\_ dpV¡$_u kpQu kdS|>su _\u.
explanation for the assertion.
∑§ÕŸ ∑§Ë ‚„Ë √ÿÊÅÿÊ Ÿ„Ë¥ „Ò– (3) L$\_ A¡ AkÐe rh^p_ R>¡. `Z L$pfZ kÐe
(2) Both assertion and reason are correct,
but the reason is not the correct (3) ∑§ÕŸ •‚àÿ „Ò ¬⁄UãÃÈ ∑§Ê⁄UáÊ ‚àÿ „Ò– R>¡.
explanation for the assertion. (4) ∑§ÕŸ ÃÕÊ ∑§Ê⁄UáÊ ŒÊŸÊ¥ „Ë •‚àÿ „Ò¥– (4) L$\_ A_¡ L$pfZ b„_¡ AkÐe R>¡.
(3) Assertion is incorrect statement, but
the reason is correct. 46. Mp¡Vy„$ rh^p_ ip¡^u bsphp¡.
(4) Both assertion and reason are 46. •‚àÿ ∑§ÕŸ ∑§Ê ¬„øÊÁŸ∞ —
(1) S2 A¡ Ap¡[¼kS>__u dpaL$ A_yQy„bL$ue R>¡.
incorrect. (1) •ÊÚÄ‚Ë¡Ÿ ∑§Ë Ã⁄U„ S2 •ŸÈøÈê’∑§Ëÿ „Ò–
(2) f¹ l p¡ [ çbL$ A_¡ dp¡ _ p¡ [ ¼gr_L$ këafdp„ S8
46. Identify the incorrect statement : (2) ⁄UÊÁê’∑§ (Áfl·◊‹¥’ÊˇÊ) ÃÕÊ ◊ÊŸÊÄ‹ËÁŸ∑§
AÏAp¡ R>¡.
(1) S2 is paramagnetic like oxygen. ‚À»§⁄U ◊¥ S8 •áÊÈ „Ò¥–
(3) S8 hge_p¡ ApL$pf dyNV$ (¾$pD_) S>¡hp¡ R>¡.
(2) Rhombic and monoclinic sulphur (3) S8 fl‹ÿ ∑§Ê •Ê∑§Ê⁄U ◊È∑ȧ≈U ∑§Ë Ã⁄U„ „Ò–
have S8 molecules. (4) S8 A_¡ S6 hgep¡dp„ S-S-S b„^ M|ZpAp¡
(4) S8 ÃÕÊ S6 fl‹ÿÊ¥ ◊¥ S-S-S •Ê’㜠∑§ÊáÊ ∞∑§ A¡L$kfMp R>¡.
(3) S8 ring has a crown shape.
¡Ò‚ „Ò¥–
(4) The S-S-S bond angles in the S8 and
S6 rings are the same.
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Set - 03 29
47. Identify the correct statement : 47. ‚„Ë ∑§ÕŸ ∑§Ê ¬„øÊÁŸÿ — 47. kpQy„ rh^p_ ip¡^p¡.
(1) Iron corrodes in oxygen-free water. (1) Ape_® (gp¡M„X$) Ap¡[¼kS>_-dy¼s `pZudp„
(1) •Êÿ⁄UŸ •ÊÚÄ‚Ë¡Ÿ-◊ÈÄà ¡‹ ◊¥ ‚¥ˇÊÊÁ⁄Uà „ÊÃÊ
(2) Iron corrodes more rapidly in salt npqfs \pe R>¡.
water because its electrochemical
„Ò–
potential is higher. (2) ‹fláÊËÿ ¡‹ ◊¥ •Êÿ⁄UŸ ¡ÀŒË ‚ ‚¥ˇÊÊÁ⁄Uà „ÊÃÊ (2) npfhpmp `pZudp„ Ape_® (gp¡M„X$) M|b S>
(3) Corrosion of iron can be minimized „Ò ÄÿÊ¥Á∑§ ß‚∑§Ê ÁfllÈà ⁄UÊ‚ÊÿÁŸ∑§ Áfl÷fl ©ìÊ TX$`\u npqfs \pe R>¡ L$pfZ L¡ $ s¡ _ p¡
by forming a contact with another „Ò– rhÛysfpkperZL$ `p¡V¡$[Þieg KQp¡ R>¡.
metal with a higher reduction
(3) •Êÿ⁄UŸ ∑§Ê ‚¥ˇÊÊ⁄UáÊ ß‚∑§Ê ©ìÊ •¬øÿŸ Áfl÷fl (3) Ape_® (gp¡M„X$) _y„ npfZ, Äepf¡ s¡_p¡ buÆ
potential.
(4) Corrosion of iron can be minimized flÊ‹ œÊÃÈ ∑§ ‚ê¬∑¸§ ◊¥ ‹ÊŸ ¬⁄U ∑§◊ Á∑§ÿÊ ¡Ê KQp fuX$n_ `p¡V¡$[Þieg ^fphsu ^psy kp\¡
by forming an impermeable barrier ‚∑§ÃÊ „Ò– k„`L®$ L$fsp Ap¡Ry>„ L$fu iL$pe R>¡.
at its surface. (4) Ape_® (gp¡M„X$) _y„ npfZ, s¡_u k`pV$u `f
(4) •Êÿ⁄UŸ ∑§Ê ‚¥ˇÊÊ⁄UáÊ ß‚∑§ ‚Ä ¬⁄U •¬Ê⁄Uªêÿ
•fl⁄UÊœ ’ŸÊ∑§⁄U ∑§◊ Á∑§ÿÊ ¡Ê ‚∑§ÃÊ „Ò– A`pfNçe Ahfp¡^ b_phu_¡ Ap¡Ry>„ L$fu iL$pe
48. Which of the following is an example of R>¡.
homoleptic complex ?
(1) [Co(NH3)6]Cl3 48. ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ „Ê◊Ê‹Áå≈U∑§ (homoleptic) ‚¥∑ȧ‹
48. _uQ¡ Ap`¡ g pAp¡ d p„ \ u L$ey „ A¡ L $ lp¡ d p¡ g ¡ à V$uL$
(2) [Pt(NH3)2Cl2] ∑§Ê ∞∑§ ©ŒÊ„⁄UáÊ „Ò?
(homoleptic) kdp_uL©$s k„L$uZ®_y„ Dv$plfZ R>¡ ?
(3) [Co(NH3)4Cl2] (1) [Co(NH3)6]Cl3
(1) [Co(NH3)6]Cl3
(4) [Co(NH3)5Cl]Cl2 (2) [Pt(NH3)2Cl2]
(2) [Pt(NH3)2Cl2]
(3) [Co(NH3)4Cl2]
(3) [Co(NH3)4Cl2]
49. The transition metal ions responsible (4) [Co(NH3)5Cl]Cl2
for color in ruby and emerald are, (4) [Co(NH3)5Cl]Cl2
respectively :
(1) Cr3+ and Co3+
49. M§’Ë ∞fl¥ ß◊⁄UÊÀ«U ◊¥ Á¡Ÿ ‚¥∑˝§◊áÊ œÊÃÈ•Ê¥ ∑§ •ÊÿŸÊ¥ 49. dpZ¡L$ (Ruby) A_¡ `Þ_p (emerald) dp„ f„Np¡ dpV¡$
(2) Co3+ and Cr3+
∑§Ë ©¬ÁSÕÁà ∑§ ∑§Ê⁄UáÊ ⁄¥Uª „ÊÃÊ „Ò, fl ∑˝§◊‡Ê— „Ò¥ — S>hpbv$pf k„¾$p„rs ^psy Ape_p¡ A_y¾$d¡ _uQ¡_pdp„\u
(3) Co3+ and Co3+ (1) Cr3+ ÃÕÊ Co3+ ip¡^p¡.
(4) Cr3+ and Cr3+ (2) Co3+ ÃÕÊ Cr3+ (1) Cr3+ A_¡ Co3+
(3) Co3+ ÃÕÊ Co3+ (2) Co3+ A_¡ Cr3+
(4) Cr3+ ÃÕÊ Cr3+ (3) Co3+ A_¡ Co3+
(4) Cr3+ A_¡ Cr3+

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50. Which one of the following substances used 50. «˛UÊ߸ċËÁŸ¥ª ◊¥ ¬˝ÿÈÄà ÁŸêŸ ¬ŒÊÕÊZ ◊¥ ‚ Á∑§‚∑§Ê ¬˝ÿÊª 50. X²$pe¼gu_]Ndp„ _uQ¡ Ap`¡gp `v$p\p£dp„\u L$ep¡ A¡L$_p¡
in dry cleaning is a better strategy to control
flÊÃÊfl⁄UáÊ ¬˝ŒÍ·áÊ ∑§ ÁŸÿ¥òÊáÊ ∑§Ë ’„Ã⁄U ∑§Êÿ¸ ŸËÁà „Ò? D`ep¡N hpsphfZue âv|$jZ_¡ r_e„ÓZ L$fhpdp„ h^y
environmental pollution ?
(1) Tetrachloroethylene (1) ≈U≈˛UÊÄ‹Ê⁄UÊ∞ÁÕ‹ËŸ L$pe®nd R>¡ ?
(2) Carbon dioxide (2) ∑§Ê’¸Ÿ «UÊß•ÊÚÄ‚Êß«U (1) V¡$V²$p¼gp¡fp¡Cr\rg_
(3) Sulphur dioxide (2) L$pb®_ X$pep¡¼kpCX
(3) ‚À»§⁄U «UÊß•ÊÚÄ‚Êß«U
(4) Nitrogen dioxide (3) këaf X$pep¡¼kpCX$
(4) ŸÊß≈˛UÊ¡Ÿ «UÊß•ÊÚÄ‚Êß«U
(4) _pCV²$p¡S>_ X$pep¡¼kpCX$
51. Sodium extract is heated with concentrated
HNO 3 before testing for halogens 51. „Ò‹Ê¡Ÿ˜‚ ∑§Ë ¡Ê°ø ∑§ ¬„‹ ‚ÊÁ«Uÿ◊ ∞ÄS≈˛ÒUÄ≈U ∑§Ê
51. l¡gp¡S>_p¡_u L$kp¡V$u L$fsp `l¡gp kp¡qX$ed AL®$ (extract)
because : ‚ÊãŒ˝ HNO3 ∑§ ‚ÊÕ ª◊¸ Á∑§ÿÊ ¡ÊÃÊ „Ò ÄÿÊ¥Á∑§ —
(1) Silver halides are totally insoluble in
_¡ kp„Ö HNO3 _u kp\¡ Nfd L$fhpdp„ Aph¡ R>¡ L$pfZ
(1) Á‚Àfl⁄U „Ò‹Êß«U ŸÊßÁ≈˛U∑§ •ê‹ ◊¥ ¬ÍáʸM§¬áÊ L¡$...
nitric acid.
•ÉÊÈ‹Ÿ‡ÊË‹ „Ò¥– (1) rkëhf l¡gpCX$p¡ _pCV²$uL$ A¡rkXdp„ k„`|Z® fus¡
(2) Ag2S and AgCN are soluble in acidic
medium. (2) •ê‹Ëÿ ◊Êäÿ◊ ◊ ¥ Ag 2 S ÃÕÊ AgCN AÖpìe R>¡.
(3) S 2− and CN − , if present, are ÉÊÈ‹Ÿ‡ÊË‹ „Ò¥– (2) A¡rkqX$L$ dpÝeddp„ Ag2S A_¡ AgCN Öpìe
decomposed by conc. HNO 3 and
(3) ÿÁŒ S2− ÃÕÊ CN− ©¬ÁSÕà „Ò¥ ÃÊ ‚ÊãŒ˝ \pe R>¡.
hence do not interfere in the test.
HNO3 ‚ ÁflÉÊÁ≈Uà „Ê ¡ÊÃ „Ò¥ ß‚Á‹ÿ ¬⁄UˡÊáÊ (3) Å¡ S2− A_¡ CN− lpS>f lp¡e sp¡ kp„Ö
(4) Ag reacts faster with halides in acidic
medium. ◊¥ „SÃˇÊ¬ Ÿ„Ë¥ ∑§⁄UÃ– HNO3 \u rhOV$_ \pe R>¡. s¡\u `funZdp„
(4) •ê‹Ëÿ ◊Êäÿ◊ ◊¥ Á‚Àfl⁄U, „Ò‹Êß«UÊ¥ ∑§ ‚ÊÕ (L$kp¡V$udp„) v$MgNufu L$fsp„ _\u.
Ã¡ •Á÷Á∑˝§ÿÊ ∑§⁄UÃÊ „Ò– (4) A¡rkqX$L$ dpÝeddp„ Ag _u âq¾$ep l¡gpCX$p¡ kp\¡
TX$`u \pe R>¡.

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52. Bromination of cyclohexene under 52. ŸËø ÁŒÿ ªÿ ¬˝ÁÃ’ãœÊ¥ ◊¥ ‚ÊßÄ‹Ê„Ä‚ËŸ ∑§Ê ’˝Ê◊ËŸ‡ÊŸ 52. kpe¼gp¡ l ¡ ¼ Tu__y „ b° p ¡ r d_¡ i _ Ap`¡ g _uQ¡ _ u
conditions given below yields :
ŒÃÊ „Ò — `qf[õ\rsAp¡dp„ L$C _u`S> Ap`i¡ ?
Br2 /hν
Br2 /hν Br2 /hν

Br
Br Br

(1)
(1) (1)

Br
Br Br

Br
Br Br

(2)
(2) (2)

Br Br
Br
(3)
(3) (3)
Br Br
Br
Br
Br Br
(4)
(4) (4)
Br Br
Br

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Set - 03 32
53. Consider the reaction sequence below : 53. ŸËø ŒË ªß¸ •Á÷Á∑˝§ÿÊ ∑˝§◊ ¬⁄U ÁfløÊ⁄U ∑§ËÁ¡∞ — 53. _uQ¡ Ap`¡g âq¾$ep_p¡ ¾$d Ýep_dp„ gp¡.
OCH 3 OCH 3 OCH 3

X X

is : X iy„ R>¡ ?
„Ò —
OH
OH
H3CO OH
H3CO
H3CO
(1)
(1)
(1)
OH
OH
OH

OCH3 OH
OCH3 OH
OCH3 OH

(2)
(2)
(2)
OH
OH
OH
OCH3
OCH3
OCH3
(3)
(3)
(3)

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Set - 03 33
H 3CO H 3CO H 3CO

(4) (4) (4)

54. Which one of the following reagents is not 54. Áfl‹Ê¬Ÿ •Á÷Á∑˝§ÿÊ ∑§ Á‹∞ ߟ •Á÷∑§Ê⁄U∑§Ê¥ ◊¥ ‚ 54. _uQ¡ Ap`¡gpdp„\u L$ep¡ âq¾$eL$ A¡ rhgp¡`_ âq¾$ep dpV¡$
suitable for the elimination reaction ? kyk„Ns _\u ?
∑§ÊÒŸ-‚Ê ©¬ÿÈÄà Ÿ„Ë¥ „Ò?
Br Br
Br
(1) NaOH/H2O (1) NaOH/H2O
(1) NaOH/H2O
(2) NaOEt/EtOH (2) NaOEt/EtOH
(2) NaOEt/EtOH
(3) NaOH/H2O-EtOH (3) NaOH/H2O-EtOH
(3) NaOH/H2O-EtOH
(4) NaI (4) NaI
(4) NaI

55. The correct statement about the synthesis 55. PETN _u b_phV$dp„ h`fpsp¡ Cfu\° u V$p¡ g
of erythritol (C(CH 2OH) 4 ) used in the 55. PETN ∑§ ’ŸÊŸ ◊¥ ¬˝ÿÈÄà ß⁄UËÁÕ˝≈UÊÚ‹ (C(CH2OH)4) (C(CH2OH)4) _p k„ïg¡jZ dpV¡$ L$ey„ rh^p_ kpQy„
preparation of PETN is : ∑§ ‚¥‡‹·áÊ ∑§ ‚ê’㜠◊¥ ‚„Ë ∑§ÕŸ „Ò — R>¡ ?
(1) The synthesis requires four aldol (1) ‚¥‡‹·áÊ ◊¥ ◊ÕŸÊÚ‹ ÃÕÊ ∞ÕŸÊÚ‹ ∑§ ’Ëø øÊ⁄U (1) k„ïg¡jZ dpV¡$ rd\¡_p¡g A_¡ C\¡_p¡g_u hÃQ¡
condensations between methanol
and ethanol.
∞À«UÊ‹ ‚¥ÉÊŸŸ ∑§Ë •Êfl‡ÿ∑§ÃÊ „ÊÃË „Ò– Qpf ApëX$p¡g k„O___u AphíeL$sp lp¡e R>¡.
(2) The synthesis requires two aldol (2) ‚¥‡‹·áÊ ◊¥ ŒÊ ∞À«UÊ‹ ‚¥ÉÊŸŸ ÃÕÊ ŒÊ ∑Ò§ÁŸ¡Ê⁄UÊ (2) k„ïg¡jZdp„ b¡ ApëX$p¡g k„O__ A_¡ b¡ L¡$r_Tpfp¡
condensations and two Cannizzaro •Á÷Á∑˝§ÿÊ ∑§Ë ¡M§⁄Uà „ÊÃË „Ò– âq¾$ep_u AphíeL$sp R>¡.
reactions.
(3) ‚¥‡‹·áÊ ◊¥ ÃËŸ ∞À«UÊ‹ ‚¥ÉÊŸŸ ÃÕÊ ∞∑§ (3) k„ïg¡jZdp„ ÓZ ApëX$p¡g k„O__ A_¡ A¡L$
(3) The synthesis requires three aldol
condensations and one Cannizzaro ∑Ò§ÁŸ¡Ê⁄UÊ •Á÷Á∑˝§ÿÊ ∑§Ë •Êfl‡ÿ∑§ÃÊ „ÊÃË „Ò– L¡$r_Tpfp¡ âq¾$ep_u AphíeL$sp R>¡.
reaction. (4) ß‚ •Á÷Á∑˝§ÿÊ ◊¥ ∞ÕŸÊÚ‹ ∑§ •À»§Ê „Êß«˛UÊ¡Ÿ (4) Ap âq¾$epdp„ C\¡_p¡g_p¡ Apëap lpCX²$p¡S>_ A_¡
(4) Alpha hydrogens of ethanol and ÃÕÊ ◊ÕŸÊÚ‹ ÷ʪ ‹Ã „Ò¥– rd\¡_p¡g cpN g¡ R>¡.
methanol are involved in this
reaction.

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56. Fluorination of an aromatic ring is easily 56. Á∑§‚Ë ∞⁄UÊÒ◊ÒÁ≈U∑§ fl‹ÿ ∑§Ê ç‹È•Ê⁄UË∑§⁄UáÊ •Ê‚ÊŸË ‚ 56. L$p¡C A¡fp¡d¡qV$L$ hge_y„ ãgp¡qf_¡i_ M|b S> kl¡gpC\u
accomplished by treating a diazonium salt
‚¥÷fl „ÊÃÊ „Ò ÿÁŒ ©‚∑§ «UÊß∞¡ÊÁŸÿ◊ ‹fláÊ ∑§Ê k„ch lp¡e R>¡, Äepf¡ s¡_p X$peTp¡_ued npf_¡ HBF4
with HBF 4 . Which of the following
conditions is correct about this reaction ? HBF4 ∑§ ‚ÊÕ ©¬øÊÁ⁄Uà Á∑§ÿÊ ¡Êÿ– ß‚ •Á÷Á∑˝§ÿÊ kp\¡ âq¾$ep L$fhpdp„ Aph¡ R>¡. D`f_u `°q¾$ep dpV¡$ _uQ¡_p
(1) Only heat ∑§ ‚ê’㜠◊¥ ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ‚Ë ¬Á⁄UÁSÕÁà Ap`¡gpdp„\u L$C `qf[õ\rs kpQu R>¡ ?
(2) NaNO2/Cu ©¬ÿÈÄà „Ò? (1) a¼s Dódp
(3) Cu2O/H2O (1) ∑§fl‹ ™§c◊Ê (2) NaNO2/Cu
(4) NaF/Cu (2) NaNO2/Cu (3) Cu2O/H2O
(3) Cu2O/H2O (4) NaF/Cu
57. Which of the following polymers is (4) NaF/Cu
synthesized using a free radical 57. _uQ¡ Ap`¡gpdp„\u L$ep blºgL$_y„ k„ïg¡jZ A¡ dy¼s dygL$
polymerization technique ? 57. ÁŸêŸÁ‹Áπà ◊¥ ‚ ∑§ÊÒŸ-‚Ê ’„È‹∑§ ◊ÈÄà ◊Í‹∑§ blºguL$fZ sL$_uL$_p¡ D`ep¡N L$fu_¡ L$fhpdp„ Aph¡ R>¡
(1) Teflon
’„È‹∑§Ë∑§⁄UáÊ ÁflÁœ mÊ⁄UÊ ‚¥‡‹Á·Ã Á∑§ÿÊ ¡ÊÃÊ „Ò? ?
(2) Terylene
(1) ≈Uç‹ÊÚŸ (1) V¡$agp¡_
(3) Melamine polymer
(2) ≈ÒU⁄UË‹ËŸ (2) V¡$qfrg_
(4) Nylon 6,6
(3) ◊Ò‹Ò◊Êߟ ’„È‹∑§ (3) d¡g¡dpC_ blºgL$
58. The “N” which does not contribute to the (4) _pegp¡_ 6,6
(4) ŸÊÿ‹ÊÚŸ 6,6
basicity for the compound is :
58. fl„ “N” ¡Ê ÁŸêŸ ÿÊÒÁª∑§ ∑§Ë ˇÊÊ⁄UËÿ ¬˝flÎÁûÊ ◊¥ ÿÊªŒÊŸ 58. “N” L¡$ S>¡ Ap`¡g k„ep¡S>__u b¡rTL$spdp„ cpN g¡sp¡
6 7
5 N Ÿ„Ë¥ ŒÃÊ „Ò, fl„ „Ò — _\u s¡ bsphp¡.
1N
8
6 7 6 7
2 N9 5 N
N 4 H 5 N 1N
3 1N
8 8
2 2 N9
(1) N7 N 4 N9 N 4 H
3 H 3
(2) N9
(3) N1 (1) N7
(1) N7
(4) N3 (2) N9
(2) N9
(3) N1 (3) N1
(4) N3 (4) N3

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Set - 03 35
59. Which of the following is a bactericidal 59. ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ-‚Ê ’ÒÄ≈UËÁ⁄UÿʟʇÊË ¬˝ÁáÒÁfl∑§ „Ò? 59. _uQ¡_pdp„\u L$ey„ A¡L$ b¡¼V¡$qfep_piL$ ârsÆhuAp¡
antibiotic ?
(1) ∞Á⁄UÕ˝Ê◊Êß‚ËŸ (antibiotic) R>¡ ?
(1) Erythromycin
(2) ≈U≈˛UÊ‚ÊÿÄ‹ËŸ (1) Cqf\°p¡dpeku_
(2) Tetracycline
(2) V¡$V²$pkpe¼gu_
(3) Chloramphenicol (3) Ä‹Ê⁄U∞ê»Ò§ÁŸ∑§Ê‹
(4) Ofloxacin (3) ¼gp¡fA¡ça¡r_L$p¡g
(4) •ÊÚ»˜§‹ÊÄ‚Ò‚ËŸ
(4) Ap¡ãgp¡¼k¡rk_
60. Observation of “Rhumann’s purple” is a
60. L§„◊ÒŸ ŸË‹ ‹ÊÁ„à (¬¬¸‹) ∑§Ê ¬˝∑§≈U „ÊŸÊ ÁŸêŸÁ‹ÁπÃ
confirmatory test for the presence of : 60. ""fyldp_ Å„byX$uep¡''(``®g) (Rhumann’s purple)
(1) Reducing sugar ◊¥ ‚ Á∑§‚∑§Ê ‚¥¬ÈÁc≈U ¬⁄UˡÊáÊ „Ò?
Ahgp¡L$_ A¡ r_Zp®eL$ L$kp¡V$u _uQ¡_pdp„\u L$p¡_u lpS>fu
(2) Cupric ion (1) •¬øÊÿ∑§ ‡Ê∑¸§⁄UÊ k|Qh¡ R>¡ ?
(3) Protein (2) ÄÿͬÁ⁄U∑§ •ÊÿŸ (1) fuX$éyk]N iL®$fp
(4) Starch (3) ¬˝Ê≈UËŸ (2) ¼eyâuL$ Ape_
(4) S≈UÊø¸ (◊¥«U) (3) âp¡V$u_
61. Let P = {θ : sinθ − cosθ = 2 cosθ } and (4) õV$pQ®

Q = {θ : sinθ + cosθ = 2 sinθ } be two 61. ◊ÊŸÊ P = {θ : sinθ − cosθ = 2 cosθ } ÃÕÊ
sets. Then : 61. ^pfp¡ L¡ $ P = {θ : sinθ − cosθ = 2 cosθ }
Q = {θ : sinθ + cosθ = 2 sinθ } ŒÊ
(1) P ⊂ Q and Q−P ≠ φ
(2) Q ⊄ P
‚◊Èëøÿ „Ò¥, ÃÊ — A_¡ Q = {θ : sinθ + cosθ = 2 sinθ } b¡
(3) P ⊄ Q (1) P ⊂ Q ÃÕÊ Q−P ≠ φ NZp¡ R>¡. sp¡ :
(4) P = Q (2) Q⊄P (1) P ⊂ Q A_¡ Q−P ≠ φ
(3) P⊄Q (2) Q⊄P
(4) P=Q (3) P⊄Q
(4) P=Q

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Set - 03 36
62. If x is a solution of the equation, 62. ÿÁŒ ‚◊Ë∑§⁄UáÊ 62. Å¡ kduL$fZ
 1  1  1
2 x + 1 − 2x − 1 = 1,  x   , then 2 x + 1 − 2x − 1 = 1,  x   _p¡ A¡L$
 2 2 x + 1 − 2x − 1 = 1,  x   ,
 2
∑§Ê x  2
4x 2 − 1 is equal to : DL¡$g x R>¡, sp¡
∞∑§ „‹ „Ò, ÃÊ 4x 2 − 1 ’⁄UÊ’⁄U „Ò — 4x 2 − 1 =__________ \pe.
3 3 3
(1) (1) (1)
4 4 4
1 1 1
(2) (2) (2)
2 2 2
(3) 2 (3) 2 (3) 2
(4) 2 2 (4) (4) 2 2
2 2

63. Let z=1+ai be a complex number, a > 0, ^pfp¡ L¡$ z=1+ai, a > 0, A¡L$ A¡hu k„L$f k„¿ep R>¡ L¡$
such that z3 is a real number. Then the sum
63. ◊ÊŸÊ z=1+ai, a > 0 ∞∑§ ∞‚Ë ‚Áê◊üÊ ‚¥ÅÿÊ „Ò, 63.

1+z+z2+.....+z11 is equal to : Á∑§ z 3 ∞∑§ flÊSÃÁfl∑§ ‚¥ Å ÿÊ „Ò , ÃÊ ÿÊ ª S>¡ \ u z 3 hpõsrhL$ k„ ¿ ep \pe. sp¡ kfhpmp¡
1+z+z2+.....+z11 ’⁄UÊ’⁄U „Ò — 1+z+z2+.....+z11 =__________ \pe.
(1) −1250 3i
(1) −1250 3i (1) −1250 3i
(2) 1250 3i
(2) 1250 3i (2) 1250 3i
(3) 1365 3i
(3) 1365 3i (3) 1365 3i
(4) −1365 3i
(4) −1365 3i (4) −1365 3i

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64. Let A be a 3×3 matrix such that 64. ◊ÊŸÊ A, 3×3 ∑§Ê ∞∑§ ∞ ‚ Ê •Ê√ÿÍ „ „Ò Á∑§ 64. ^pfp¡ L¡ $ A A¡ L $ 3×3 î¡ r ZL$ R>¡ L¡ $ S>¡ \ u
A2−5A+7I=O.
A2−5A+7I=O „Ò– A2−5A+7I=O.
1
Statement - I : A−1 = (5I − A ). −1 1 1
7 ∑§ÕŸ - I :A = (5I − A ). rh^p_ - I : A−1 = (5I − A ).
7 7
Statement - II : The polynomial
A3−2A2−3A+I can be ∑§ÕŸ - II : ’„È ¬ Œ A 3 −2A 2 −3A+I ∑§Ê rh^p_ - II : blº ` v$u A 3 −2A 2 −3A+I _¡
reduced to 5(A−4I). 5(A−4I) ◊¥ ¬Á⁄UflÁøà Á∑§ÿÊ ¡Ê ‚∑§ÃÊ 5(A−4I) dp„ ê$`p„sqfs L$fu iL$pe R>¡.
Then : „Ò– sp¡ __________.
(1) Statement-I is true, but Statement-II ÃÊ, (1) rh^p_-I kÐe R>¡, `f„sy rh^p_-II AkÐe R>¡.
is false.
(2) Statement-I is false, but Statement-II (1) ∑§ÕŸ - I ‚àÿ „Ò ‹Á∑§Ÿ ∑§ÕŸ - II •‚àÿ „Ò– (2) rh^p_-I AkÐe R>¡, `f„sy rh^p_-II kÐe R>¡.
is true. (2) ∑§ÕŸ - I •‚àÿ „Ò ‹Á∑§Ÿ ∑§ÕŸ - II ‚àÿ „Ò– (3) b„_¡ rh^p_p¡ kÐe R>¡.
(3) Both the statements are true. (3) ŒÊŸÊ¥ ∑§ÕŸ ‚àÿ „Ò¥– (4) b„_¡ rh^p_p¡ AkÐe R>¡.
(4) Both the statements are false.
(4) ŒÊŸÊ¥ ∑§ÕŸ •‚àÿ „Ò¥–
−4 −1
−4 −1 65. Å¡ A = , sp¡ î¡ r ZL$
65. If A = , then the determinant of −4 −1 3 1 
 3 1  65. ÿÁŒ A = „Ò , ÃÊ •Ê√ÿÍ „
 3 1  (A 2016 −2A 2015 −A 2014 ) _p¡ r_òpeL$
the matrix (A2016−2A2015−A2014) is :
(A2016−2A2015−A2014) ∑§Ê ‚Ê⁄UÁáÊ∑§ „Ò — __________ R>¡.
(1) 2014
(1) 2014 (1) 2014
(2) −175
(2) −175 (2) −175
(3) 2016
(3) 2016 (3) 2016
(4) −25
(4) −25 (4) −25

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Set - 03 38
n+2 n+2 n+2
C6 C6
66. If
n−2
= 11, then n satisfies the 66. ÿÁŒ n−2
= 11, „Ò, ÃÊ n ÁŸêŸ ◊¥ ‚ Á∑§‚ 66. Å¡ n−2C6 = 11, sp¡ n _uQ¡_pdp„\u L$ey„ kduL$fZ
P2 P2 P2
equation : ‚◊Ë∑§⁄UáÊ ∑§Ê ‚¥ÃÈc≈U ∑§⁄UÃÊ „Ò? k„`p¡j¡?
(1) n2+3n−108=0 (1) n2+3n−108=0 (1) n2+3n−108=0
(2) n2+5n−84=0 (2) n2+5n−84=0 (2) n2+5n−84=0
(3) n2+2n−80=0 (3) n2+2n−80=0 (3) n2+2n−80=0
(4) n2+n−110=0 (4) n2+n−110=0 (4) n2+n−110=0

67. If the coefficients of x−2 and x−4 in the
1 18 1 18
ÿÁŒ  x 3 + Å¡  x 3 +
1 1
 13 1 18 67. 1 
, (x > 0), ∑§ ¬˝‚Ê⁄U ◊¥ x−2 67. 1 
, (x > 0) _p rhõsfZdp„ x−2
expansion of + , (x > 0), are  2x 3   2x 3 
x 1 
 2x 3 
m A_¡ x−4 _p klNyZL$p¡ A_y¾$d¡ m A_¡ n lp¡e sp¡,
m ÃÕÊ x−4 ∑§ ªÈáÊÊ¥∑§ ∑˝§◊‡Ê— m ÃÕÊ n „Ò¥, ÃÊ n
m and n respectively, then is equal to : m
n =__________ \pe.
’⁄UÊ’⁄U „Ò — n
(1) 182
(1) 182 (1) 182
4
(2) 4 4
5 (2) (2)
5 5
5
(3) 5 5
4 (3) (3)
4 4
(4) 27
(4) 27 (4) 27

68. Let a1, a2, a3, ......, an, ..... be in A.P. If
a3+a7+a11+a15=72, then the sum of its 68. ◊ÊŸÊ a1, a2, a3, ......, an, ..... ∞∑§ ‚◊Ê¥Ã⁄U üÊ…∏Ë ◊¥ „Ò¥– 68. ^pfp¡ L¡$ a1, a2, a3, ......, an, ..... kdp„sf î¡Zudp„ R>¡.
first 17 terms is equal to : ÿÁŒ a3+a7+a11+a15=72 „Ò, ÃÊ ©‚∑§ ¬˝Õ◊ 17 Å¡ a3+a7+a11+a15=72, sp¡ s¡_p â\d 17 `v$p_¡ p¡
(1) 306 ¬ŒÊ¥ ∑§Ê ÿÊª ’⁄UÊ’⁄U „Ò — kfhpmp¡ __________ \pe.
(2) 153 (1) 306 (1) 306
(3) 612 (2) 153 (2) 153
(4) 204 (3) 612 (3) 612
(4) 204 (4) 204
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Set - 03 39
10 10 10
The sum ∑ ( r + 1 ) × (r!) is equal to : ÿÊª»§‹ ∑ ( r 2+ 1 ) × (r!) ’⁄UÊ’⁄U „Ò — ∑ ( r 2+ 1) × (r!) =__________.
2
69. 69. 69.
r =1 r =1 r =1

(1) (11)! (1) (11)! (1) (11)!
(2) 10×(11!) (2) 10×(11!) (2) 10×(11!)
(3) 101×(10!) (3) 101×(10!) (3) 101×(10!)
(4) 11×(11!) (4) 11×(11!) (4) 11×(11!)

( 1 − cos2 x )2 ( 1 − cos2 x )2 ( 1 − cos2 x )2
70. lim is : 70. lim ’⁄UÊ’⁄U „Ò — 70. lim =__________.
x →0 2 x tan x − x tan 2 x x →0 2 x tan x − x tan 2 x x →0 2 x tan x − x tan 2 x
(1) −2 (1) −2 (1) −2
1 1 1
(2) − (2) − (2) −
2 2 2
1 1 1
(3) (3) (3)
2 2 2
(4) 2 (4) 2 (4) 2

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Set - 03 40
71. Let a, b e R, (a ≠ 0). If the function f defined 71. ◊ÊŸÊ a, b e R, (a ≠ 0)– ÿÁŒ »§‹Ÿ f ¡Ê, ÁŸêŸ mÊ⁄UÊ 71. ^pfp¡ L¡$ a, b e R, (a ≠ 0).
as
¬Á⁄U÷ÊÁ·Ã „Ò —
 2x2
 2x2  , 0≤x <1
 , 0≤x <1  2x2  a
 a  , 0≤x <1 
  a Å¡ rh^¡e f (x) =  a , 1≤x < 2
f (x) =  a , 1≤x < 2   2
 2 f (x) =  a , 1≤x < 2
 2b −4b ,
 2b −4b ,  2  x 3
2 ≤x < ∞
2 ≤x < ∞  2b −4b ,
 x 3 2 ≤x < ∞
 x 3
A„sfpg [0, ∞) dp„ kss lp¡e, sp¡ ¾$dey¼s Å¡X$
is continuous in the interval [0, ∞), then an
•¥Ã⁄UÊ‹ [0, ∞) ◊¥ ‚Ãà „Ò, ÃÊ ∞∑§ ∑˝§Á◊à ÿÈÇ◊ (a, b) = __________ R>¡.
ordered pair (a, b) is :
(a, b) „Ò — (1) ( 2, 1− 3)
(1) ( 2, 1− 3)
(1) ( 2, 1− 3) (2) (− 2, 1+ 3)
(2) (− 2, 1+ 3)
(2) (− 2, 1+ 3) (3) ( 2 , −1 + 3)
(3) ( 2 , −1 + 3)
(3) ( 2 , −1 + 3) (4) (− 2, 1− 3)
(4) (− 2, 1− 3)
(4) (− 2, 1− 3)
72. Let f(x)=sin 4 x+cos 4 x. Then f is an 72. ^pfp¡ L¡$ f(x)=sin4x+cos4x. sp¡ _uQ¡_pdp„\u L$ep
increasing function in the interval : 72. ◊ÊŸÊ f(x)=sin4x+cos4x „Ò, ÃÊ ÁŸêŸ ◊¥ ‚ Á∑§‚ A„sfpgdp„ f h^sy„ rh^¡e R>¡ ?
•¥Ã⁄UÊ‹ ◊¥ f ∞∑§ flœ¸◊ÊŸ »§‹Ÿ „Ò?
 π  π
(1)  0, 4   π (1)  0, 4 
(1)  0, 4 
π π  π π 
(2)  4 , 2  π π  (2)  4 , 2 
(2)  4 , 2 
 π 5π   π 5π 
(3)  2 , 8   π 5π  (3)  2 , 8 
(3)  2 , 8 
 5π 3π   5π 3π 
(4)  8 , 4   5π 3π  (4)  8 , 4 
(4)  8 , 4 

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Set - 03 41
73. Let C be a curve given by
73. ◊ÊŸÊ C ∞∑§ fl∑˝§ „Ò ¡Ê y ( x ) = 1 + 4x − 3 , 73. ^pfp¡ L¡$ A¡L$ h¾$ C A¡ y ( x) = 1 + 4x − 3,
3
y ( x) = 1 + 4x − 3, x > . If P is a point 3 3
4 x> mÊ⁄UÊ ¬˝ŒûÊ „Ò– ÿÁŒ C ¬⁄U P ∞∑§ ∞‚Ê Á’¥ŒÈ „Ò x> Üpfp v$ip®h¡g R>¡. Å¡ C `f A¡L$ A¡hy„ tbvy$ P R>¡
on C, such that the tangent at P has slope 4 4
2 , then a point through which the normal Á∑§ P ¬⁄U πË¥øË ªß¸ S¬‡Ê¸ ⁄UπÊ ∑§Ë …Ê‹ 2 „Ò, ÃÊ fl„ L¡$ S>¡\u tbvy$ P ApNm_p õ`i®L$_p¡ Y$pm 2 \pe, sp¡ P
3 3 3
at P passes, is : Á’¥ŒÈ Á¡‚‚ P ¬⁄U πË¥øÊ ªÿÊ •Á÷‹¥’ ªÈ$¡⁄UÃÊ „Ò, „Ò — ApNm_p¡ Arcg„b __________ tbvy$dp„\u `kpf \pe
(1) (2, 3) (1) (2, 3) R>¡.
(2) (4, −3) (2) (4, −3) (1) (2, 3)
(3) (1, 7) (3) (1, 7) (2) (4, −3)
(4) (3, −4) (4) (3, −4) (3) (1, 7)
(4) (3, −4)

dx dx
74. The integral is equal 74. ‚◊Ê∑§‹ ’⁄UÊ’⁄U „Ò —
(1 + x) x− x 2 (1 + x) x − x2 dx
74. =__________.
(1 + x) x − x2
to : (¡„Ê° C ∞∑§ ‚◊Ê∑§‹Ÿ •ø⁄U „Ò–)
(where C is a constant of integration.)
(Äep„ C k„L$g__p¡ AQmp„L$ R>¡.)
1+ x
1+ x (1) −2 +C
(1) −2 +C 1− x 1+ x
1− x (1) −2 +C
1− x
1− x
1− x (2) −2 +C
(2) −2 +C 1+ x 1− x
1+ x (2) −2 +C
1+ x
1− x
1− x (3) − +C
(3) − +C 1+ x 1− x
1+ x (3) − +C
1+ x
1+ x
1+ x (4) 2 +C
(4) 2 +C 1− x 1+ x
1− x (4) 2 +C
1− x

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Set - 03 42
75. The value of the integral
 x 2  dx  x 2  dx
75. ‚◊Ê∑§‹   , ¡„Ê° 75.   =_________.
 x 2  dx
  , where [x]  x 2− 28x + 196  +  x 2   x 2− 28x + 196  +  x 2 
       
 x 2− 28x + 196  +  x 2 
   
[x], x ‚ ∑§◊ ÿÊ x ∑§ ’⁄UÊ’⁄U ◊„ûÊ◊ ¬ÍáÊÊZ∑§ „Ò, ∑§Ê Äep„ [x] A¡ x \u _p_p A\hp x _¡ kdp_ sdpd
denotes the greatest integer less than or ◊ÊŸ „Ò — `|Zp¯L$p¡dp„ kp¥\u dp¡V$p¡ `|Zp¯L$ v$ip®h¡ R>¡.
equal to x, is :
(1) 6 (1) 6
(1) 6
(2) 3 (2) 3
(2) 3
(3) 7 (3) 7
(3) 7
1 1
1 (4) (4)
(4) 3 3
3
76. x e R, x ≠ 0, ∑§ Á‹∞, ÿÁŒ y(x) ∞∑§ ∞‚Ê •fl∑§‹ŸËÿ
76. For x e R, x ≠ 0, if y(x) is a differentiable 76. Å¡ x e R, x ≠ 0 dpV¡$, y(x) A¡L$ rhL$g_ue rh^¡e R>¡ L¡$
function such that »§‹Ÿ „Ò Á∑§
x x
x x x x S>¡ \ u x ∫ y (t ) dt = ( x + 1) ∫ t y (t ) dt , sp¡
x ∫ y (t ) dt = ( x + 1) ∫ t y (t ) dt , then y(x) x ∫ y (t ) dt = ( x + 1) ∫ t y (t ) dt „Ò, ÃÊ y (x) 1 1
1 1 1 1
y(x)=__________.
equals : ’⁄UÊ’⁄U „Ò — (Äep„ C AQm R>¡.)
(where C is a constant.) (¡„Ê° C ∞∑§ •ø⁄U „Ò–)
1
1 C −x
C − 1 (1) e
(1) e x C −x x
x (1) e
x
1
1
1
C −x
C −
C
(2) e
(2) x
x2
e −
2 (2) e x
x2
x
1 1
C −x 1 C −x
(3) e C − (3) e
x3 (3) 3
e x
x3
x
1 1
(4) 1
3 x
Cx e (4) (4) 3
3 x
Cx e Cx e x

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Set - 03 43
77. The solution of the differential equation dy y dy y
tanx tanx
dy y tanx 77. •fl∑§‹ ‚◊Ë∑§⁄UáÊ + secx = , ¡„Ê° 77. rhL$g kduL$fZ + secx = , Äep„
+ secx = , where 0 ≤ x < π , dx 2 2y dx 2 2y
dx 2 2y 2
and y(0)=1, is given by : 0≤x <
π
„Ò ÃÕÊ y(0)=1 „Ò, ∑§Ê „‹ „Ò — 0≤x <
π
, A_¡ y(0)=1, _p¡ DL¡$g _________
2 2

(1) y = 1−
x
x R>¡.
sec x + tan x (1) y = 1−
sec x + tan x x
(1) y = 1−
2 x sec x + tan x
(2) y = 1+ x
sec x + tan x (2) y2 = 1 +
sec x + tan x x
(2) y2 = 1 +
2 x sec x + tan x
(3) y = 1− x
sec x + tan x (3) y2 = 1 −
sec x + tan x x
(3) y2 = 1 −
x sec x + tan x
(4) y = 1+ x
sec x + tan x (4) y = 1+
sec x + tan x x
(4) y = 1+
sec x + tan x
78. A ray of light is incident along a line which
meets another line, 7x−y+1=0, at the 78. ¬˝∑§Ê‡Ê ∑§Ë ∞∑§ Á∑§⁄UáÊ ∞∑§ ⁄UπÊ ∑§Ë ÁŒ‡ÊÊ ◊¥ •Ê¬ÁÃÃ
point (0, 1). The ray is then reflected from „Ò ¡Ê ∞∑§ •ãÿ ⁄UπÊ 7x−y+1=0 ∑§Ê Á’¥ŒÈ 78. âL$pi_y„ A¡L$ qL$fZ f¡Mp 7x−y+1=0 `f Ap`ps
this point along the line, y+2x=1. Then (0, 1) ¬⁄U Á◊‹ÃË „Ò– fl„ Á∑§⁄UáÊ Á»§⁄U ß‚ Á’¥ŒÈ ‚ ⁄UπÊ \pe R>¡ S>¡ tbv$y$ (0, 1) ApNm dm¡ R>¡. Ðepfbpv$ Ap
the equation of the line of incidence of the
y+2x=1 ∑§Ë ÁŒ‡ÊÊ ◊¥ ¬Á⁄UflÁøà „ÊÃË „Ò, ÃÊ •Ê¬ÁÃà qL$fZ Ap tbvy$dp„\u `fphrs®s \C f¡Mp y+2x=1
ray of light is :
(1) 41x−38y+38=0 ¬˝∑§Ê‡Ê ∑§Ë Á∑§⁄UáÊ ∑§Ê ‚◊Ë∑§⁄UáÊ „Ò — `f fl¡ R>¡. sp¡ Ap`ps qL$fZ_y„ kduL$fZ __________
(2) 41x+25y−25=0 (1) 41x−38y+38=0 R>¡.
(3) 41x+38y−38=0 (2) 41x+25y−25=0 (1) 41x−38y+38=0
(4) 41x−25y+25=0 (3) 41x+38y−38=0 (2) 41x+25y−25=0
(4) 41x−25y+25=0 (3) 41x+38y−38=0
(4) 41x−25y+25=0

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79. A straight line through origin O meets the 79. ◊Í‹ Á’¥ŒÈ O ‚ „Ê∑§⁄U ¡ÊŸ flÊ‹Ë ∞∑§ ‚⁄U‹ ⁄UπÊ ⁄UπÊ•Ê¥ 79. ENdtbvy$ O dp„\u `kpf \su A¡L$ f¡Mp, f¡MpAp¡
lines 3y=10−4x and 8x+6y+5=0 at
3y=10−4x ÃÕÊ 8x+6y+5=0 ∑§Ê ∑˝ § ◊‡Ê— 3y=10−4x A_¡ 8x+6y+5=0 _¡ A_y ¾ $ d ¡
points A and B respectively. Then O
divides the segment AB in the ratio : Á’¥ŒÈ•Ê¥ A ÃÕÊ B ¬⁄U Á◊‹ÃË „Ò¥, ÃÊ Á’¥ŒÈ O ⁄UπÊπ¥«U tbvy$Ap¡ A A_¡ B dp„ dm¡ R>¡. sp¡ O A¡ f¡MpM„X$ AB _¡
(1) 2 : 3 AB ∑§Ê Á¡‚ •ŸÈ¬Êà ◊¥ Áfl÷ÊÁ¡Ã ∑§⁄UÃÊ „Ò, fl„ „Ò — __________ NyZp¡Ñfdp„ rhcpS>_ L$f¡ R>¡.
(2) 1 : 2 (1) 2:3 (1) 2:3
(3) 4 : 1 (2) 1:2 (2) 1:2
(4) 3 : 4 (3) 4:1 (3) 4:1
(4) 3:4 (4) 3:4
80. Equation of the tangent to the circle, at the
point (1, −1), whose centre is the point of 80. ©‚ flÎûÊ Á¡‚∑§Ê ∑§ãŒ˝ ‚⁄U‹ ⁄UπÊ•Ê¥ x−y=1 ÃÕÊ 80. S>¡_„y L¡$ÞÖ f¡MpAp¡ x−y=1 A_¡ 2x+y=3 _y„ R>¡v$tbvy$
intersection of the straight lines x−y=1
2x+y=3 ∑§Ê ¬˝ÁÃë¿UŒ Á’¥ŒÈ „Ò, ∑§ Á’¥ŒÈ (1, −1) ¬⁄U lp¡e s¡hp hsy®m_¡ tbvy$ (1, −1) ApNm_p õ`i®L$_y„
and 2x+y=3 is :
(1) 4x+y−3=0 πË¥øË ªß¸ S¬‡Ê¸ ⁄πÊ ∑§Ê ‚◊Ë∑§⁄UáÊ „Ò — kduL$fZ __________ R>¡.
(2) x+4y+3=0 (1) 4x+y−3=0 (1) 4x+y−3=0
(3) 3x−y−4=0 (2) x+4y+3=0 (2) x+4y+3=0
(4) x−3y−4=0 (3) 3x−y−4=0 (3) 3x−y−4=0
(4) x−3y−4=0 (4) x−3y−4=0
81. P and Q are two distinct points on the
parabola, y2=4x, with parameters t and t1 81. P ÃÕÊ Q ¬⁄Ufl‹ÿ y2=4x ¬⁄U ÁSÕà ŒÊ Á÷㟠Á’¥ŒÈ „Ò 81. P A_¡ Q `fhge y2=4x `f Aph¡gp b¡ rcÞ_
respectively. If the normal at P passes Á¡Ÿ∑§ ¬˝Êø‹ ∑˝§◊‡Ê— t ÃÕÊ t1 „Ò¥– ÿÁŒ P ¬⁄U πË¥øÊ tbv$y$Ap¡ R>¡ S>¡_p âpQgp¡ A_y¾$d¡ t A_¡ t1 R>¡. Å¡ P
through Q, then the minimum value of t12
ªÿÊ •Á÷‹¥’ Q ‚ „Ê∑§⁄U ¡ÊÃÊ „Ò, ÃÊ t12 ∑§Ê ãÿÍŸÃ◊ ApNm_p¡ Arcg„b Q dp„\u `kpf \sp¡ lp¡e, sp¡ t12
is :
◊ÊŸ „Ò — _y„ Þe|_sd d|ëe __________ R>¡.
(1) 2
(1) 2 (1) 2
(2) 4
(2) 4 (2) 4
(3) 6
(3) 6 (3) 6
(4) 8
(4) 8 (4) 8

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82. A hyperbola whose transverse axis is along 82. ∞∑§ •Áì⁄Ufl‹ÿ, Á¡‚∑§Ê •ŸÈ¬˝SÕ •ˇÊ ‡ÊÊ¥∑§fl 82. A¡ L $ Arshge S>¡ _ p¡ dy ¿ e An ip„ L $h
x2 y2 x2 y2 x2 y2
the major axis of the conic, + =4 + = 4 ∑§ ŒËÉʸ •ˇÊ ∑§Ë ÁŒ‡ÊÊ ◊¥ „Ò ÃÕÊ + = 4 _p â^p_ An `f R>¡ s\p s¡_p
3 4 3 4 3 4
and has vertices at the foci of this conic. If
Á¡‚∑§ ‡ÊË·¸ ß‚ ‡ÊÊ¥∑§fl ∑§Ë ŸÊÁ÷ÿÊ¥ ¬⁄U „Ò– ÿÁŒ rifp¡ t bvy $ A p¡ Ap ip„ L $ h _u _prcAp¡ `f R>¡ . Å¡
3
the eccentricity of the hyperbola is , then 3 3
2 •Áì⁄Ufl‹ÿ ∑§Ë ©à∑§ãŒ˝ÃÊ 2 „Ò, ÃÊ ÁŸêŸ ◊¥ ‚ ∑§ÊÒŸ Arshge_u DÐL¡$ÞÖsp 2 lp¡e, sp¡ _uQ¡_pdp„\u L$ey„
which of the following points does NOT
lie on it ? ‚Ê Á’¥ŒÈ ß‚ ¬⁄U ÁSÕà Ÿ„Ë¥ „Ò? tbvy$ s¡_p `f Aph¡gy„ _l] lp¡e ?
(1) (0, 2) (1) (0, 2) (1) (0, 2)

(2) ( 5, 2 2 ) (2) ( 5, 2 2 ) (2) ( 5, 2 2 )
(3) ( 10 , 2 3 ) (3) ( 10 , 2 3 ) (3) ( 10 , 2 3 )
(4) (5, 2 3 ) (4) (5, 2 3 ) (4) (5, 2 3 )

83. ABC is a triangle in a plane with vertices 83. ∞∑§ ‚◊Ë ◊¥ ∞∑§ ÁòÊ÷È¡ ABC „Ò Á¡‚∑§ ‡ÊË·¸ 83. A¡L$ kdsgdp„ A¡L$ rÓL$p¡Z ABC R>¡. S>¡_p rifp¡tbvy$Ap¡
A(2, 3, 5), B(−1, 3, 2) and C(λ, 5, µ). If the A(2, 3, 5), B(−1, 3, 2) ÃÕÊ C(λ, 5, µ) „Ò¥– ÿÁŒ A(2, 3, 5), B(−1, 3, 2) A_¡ C(λ, 5, µ) R>¡. Å¡ A
median through A is equally inclined to the
A ‚ „Ê∑§⁄U ¡ÊÃË ◊ÊÁäÿ∑§Ê ÁŸŒ¸‡ÊÊ¥∑§ •ˇÊÊ¥ ¬⁄U ‚◊ÊŸ dp„\u _uL$msu dÝeNp epdpnp kp\¡ kdp_ fus¡ Y$m¡g
coordinate axes, then the value of
(λ3+µ3+5) is : M§¬ ‚ ¤ÊÈ∑§Ë „Ò, ÃÊ (λ3+µ3+5) ∑§Ê ◊ÊŸ „Ò — R>¡, sp¡ (λ3+µ3+5) _u qL„$ds __________ R>¡.
(1) 1130 (1) 1130 (1) 1130
(2) 1348 (2) 1348 (2) 1348
(3) 676 (3) 676 (3) 676
(4) 1077 (4) 1077 (4) 1077

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Set - 03 46
84. The number of distinct real values of λ for 84. λ ∑§ fl„ Á÷㟠flÊSÃÁfl∑§ ◊ÊŸÊ¥ ∑§Ë ‚¥ÅÿÊ Á¡Ÿ∑§
x− 1 y− 2 z+ 3
84. S>¡_p dpV¡$ f¡MpAp¡ x − 1 = y − 2 = z +2 3
which the lines x− 1 y− 2 z+ 3 1 2 λ
1
=
2
=
λ2
Á‹∞ ⁄ π Ê∞° = = ÃÕÊ
1 2 λ2
A_¡ x − 3 = y −2 2 = z − 1 kdsgue \pe
x− 3 y− 2 z− 1 x− 3 y− 2 z − 1 ‚◊ËËÿ „Ò¥, „Ò — 1 λ 2
and = = are
1 2 2 = 2
=
λ 1 λ 2 s¡ h u λ _u rcÞ_ hpõsrhL$ qL„ $ dsp¡ _ u k„ ¿ ep
coplanar is :
(1) 4 __________ R>¡.
(1) 4
(2) 1 (1) 4
(2) 1
(3) 2 (2) 1
(3) 2
(4) 3 (3) 2
(4) 3
(4) 3
85. ◊ÊŸÊ ABC ∞∑§ ÁòÊ÷È¡ „Ò Á¡‚∑§Ê ¬Á⁄U∑§ãŒ˝ P ¬⁄U „Ò–
85. Let ABC be a triangle whose circumcentre
is at P. If the position vectors of A, B, C ÿÁŒ Á’¥ŒÈ•Ê¥ A, B, C ÃÕÊ P ∑§ ÁSÕÁà ‚ÁŒ‡Ê ∑˝§◊‡Ê— 85. ^pfp¡ L¡$ ABC A¡L$ rÓL$p¡Z R>¡ S>¡_y„ `qfL¡$ÞÖ P ApNm
→ → → → → → R>¡. Å¡ A, B, C A_¡ P _p õ\p_kqv$ip¡ A_y¾$d¡
→ → → → → →
and P are a , b , c and a + b + c a, b, c ÃÕÊ a + b + c „Ò¥, ÃÊ ß‚ ÁòÊ÷È¡ → → →
→ → →
4 4 a, b, c A_¡ a + b + c lp¡e, sp¡ Ap
respectively, then the position vector of the ∑§ ‹¥’ - ∑§ãŒ˝ ∑§Ê ÁSÕÁà ‚ÁŒ‡Ê „Ò — 4
orthocentre of this triangle, is :
→ → → rÓL$p¡Z_p g„bL¡$ÞÖ_p¡ õ\p_kqv$i __________ R>¡.
→ → → (1) a + b + c
→ → →
(1) a + b + c (1) a + b + c
→ → →
a b c 
−
→ → → (2) + +
a + b + c    → → →
(2) − 2 (2)  a + b + c 
 2    −
 
2
   
→
→ (3) 0
→
(3) 0 (3) 0
→ → →
→ → (4) a + b + c
→
  → → →
(4) a + b + c (4) a + b + c
  2  
2 2

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Set - 03 47
86. The mean of 5 observations is 5 and their 86. 5 ¬˝ˇÊáÊÊ¥ ∑§Ê ◊Êäÿ 5 „Ò ÃÕÊ ©Ÿ∑§Ê ¬˝‚⁄UáÊ 124 „Ò– 86. 5 Ahgp¡L$_p¡_p¡ dÝeL$ 5 A_¡ rhQfZ 124 R>¡. Å¡
variance is 124. If three of the observations
ÿÁŒ ©Ÿ◊¥ ‚ ÃËŸ ¬˝ˇÊáÊ 1, 2 ÃÕÊ 6 „Ò¥, ÃÊ ߟ •Ê°∑§«∏Ê¥ s¡dp„\u ÓZ Ahgp¡L$_p¡ 1, 2 A_¡ 6 lp¡e, sp¡ dprlsu_y„
are 1, 2 and 6 ; then the mean deviation
from the mean of the data is : ∑§Ê ◊Êäÿ ‚ ◊Êäÿ Áflø‹Ÿ „Ò — dÝeL$\u kf¡fpi rhQg_ __________ \pe.
(1) 2.4 (1) 2.4 (1) 2.4
(2) 2.8 (2) 2.8 (2) 2.8
(3) 2.5 (3) 2.5 (3) 2.5
(4) 2.6 (4) 2.6 (4) 2.6

87. An experiment succeeds twice as often as 87. ∞∑§ ¬˝ÿÊª ∑§ ‚»§‹ „ÊŸ ∑§Ê ‚¥ÿÊª ©‚∑§ Áfl»§‹ 87. A¡L$ âep¡N S>¡V$gp hMs Akam \pe R>¡ s¡_p\u b¡
it fails. The probability of at least 5 „ÊŸ ∑§ ‚¥ÿÊª ∑§Ê ŒÈªÈŸÊ „Ò– ß‚ ¬˝ÿÊª ∑§ 6 ¬⁄UˡÊáÊÊ¥ NZp¡ kam \pe R>¡. Ap âep¡N_p R> âeÐ_p¡dp„\u
successes in the six trials of this experiment ◊¥ ‚ ∑§◊ ‚ ∑§◊ ¬Ê°ø ∑§ ‚»§‹ „ÊŸ ∑§Ë ¬˝ÊÁÿ∑§ÃÊ
is :
Ap¡ R >pdp„ Ap¡ R >u 5 kamsp dmhp_u k„ c ph_p
„Ò — __________ R>¡.
240
(1) 240 240
729 (1) (1)
729 729
192
(2) 192 192
729 (2) (2)
729 729
256
(3) 256 256
729 (3) (3)
729 729
496
(4) 496 496
729 (4) (4)
729 729

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Set - 03 48
π π π
88. If A>0, B>0 and Α + B = , then the 88. ÿÁŒ A>0, B>0 ÃÕÊ Α+B= „Ò , ÃÊ 88. Å¡ A>0, B>0 A_¡ Α + B = lp¡ e , sp¡
6 6 6
minimum value of tanA+tanB is : tanA+tanB ∑§Ê ãÿÍŸÃ◊ ◊ÊŸ „Ò — tanA+tanB _y„ Þe|_sd d|ëe __________ R>¡.
(1) 3− 2 (1) 3− 2 (1) 3− 2
(2) 2− 3 (2) 2− 3 (2) 2− 3
(3) 4 −2 3 (3) 4 −2 3 (3) 4 −2 3
2 2 2
(4) (4) (4)
3 3 3

89. The angle of elevation of the top of a 89. Á’¥ŒÈ A ‚, ¡Ê ∞∑§ ™§äflʸœ⁄U ◊ËŸÊ⁄U ∑§ ¬Ífl¸ ∑§Ë •Ê⁄U
vertical tower from a point A, due east of it „Ò, ◊ËŸÊ⁄U ∑§ ‡ÊË·¸ ∑§Ê ©ãŸÿŸ ∑§ÊáÊ 458 „Ò– Á’¥ŒÈ B, 89. A¡L$ rifp¡g„b V$phf_u `|h® sfa_p tbvy$ A \u V$phf_u
is 458. The angle of elevation of the top of
¡Ê Á’¥ŒÈ A ∑§ ŒÁˇÊáÊ ◊¥ „Ò, ‚ ©‚Ë ◊ËŸÊ⁄U ∑§ ‡ÊË·¸ ∑§Ê V$p¡Q_p¡ DÐk¡^L$p¡Z 458 R>¡. tbvy$ A _u v$rnZ¡ Aph¡g
the same tower from a point B, due south
©ãŸÿŸ ∑§ÊáÊ 308 „Ò– ÿÁŒ A ÃÕÊ B ∑§ ’Ëø ∑§Ë ŒÍ⁄UË tbvy$ B \u V$phf_u V$p¡Q_p¡ DÐk¡^L$p¡Z 308 R>¡. Å¡ A
of A is 308. If the distance between A and
B is 54 2 m , then the height of the tower 54 2 ◊Ë. „Ò, ÃÊ ◊ËŸÊ⁄U ∑§Ë ™§°øÊ߸ (◊Ë. ◊¥) „Ò — A_¡ B hÃQ¡_y„ A„sf 54 2 m lp¡e, sp¡ V$phf_u
(in metres), is : KQpC (duV$fdp„) __________ R>¡.
(1) 36 3
(1) 36 3 (2) 54 (1) 36 3
(2) 54 (2) 54
(3) 54 3
(3) 54 3 (4) 108 (3) 54 3
(4) 108 (4) 108

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Set - 03 49
90. The contrapositive of the following 90. ÁŸêŸ ∑§ÕŸ ∑§Ê ¬˝ÁÃœŸÊà◊∑§ (contrapositive) 90. _uQ¡_p rh^p__y„ kdp_p\} â¡fZ L$ey„ R>¡ ?
statement,
„Ò — ""Å¡ L$p¡C Qp¡fk_u bpSy> bdZu L$fhpdp„ Aph¡, sp¡ s¡_y„
“If the side of a square doubles, then its area
increases four times”, is : ““ÿÁŒ Á∑§‚Ë flª¸ ∑§Ë ÷È¡Ê ŒÈªÈŸË „Ê ¡Ê∞, ÃÊ ©‚∑§Ê n¡Óam QpfNÏ„ h^¡.''
(1) If the side of a square is not doubled, ˇÊòÊ»§‹ øÊ⁄U ªÈŸÊ ’…∏ ¡ÊÃÊ „Ò”” — (1) Å¡ L$p¡C Qp¡fk_u bpSy> bdZu L$fhpdp„ _ Aph¡,
then its area does not increase four (1) ÿÁŒ ∞∑§ flª¸ ∑§Ë ÷È¡Ê ŒÈªÈŸË Ÿ ∑§Ë ¡Ê∞, ÃÊ sp¡ s¡_y„ n¡Óam QpfNÏ„ h^i¡ _rl.
times. ©‚∑§Ê ˇÊòÊ»§‹ øÊ⁄U ªÈŸÊ Ÿ„Ë¥ ’…∏ÃÊ– (2) Å¡ L$pC ¡ Qp¡fk_y„ n¡Óam QpfNÏ„ h^pfhpdp„ Aph¡,
(2) If the area of a square increases four
(2) ÿÁŒ Á∑§‚Ë flª¸ ∑§Ê ˇÊòÊ»§‹ øÊ⁄U ªÈŸÊ ’…∏ ¡Ê∞, sp¡ s¡_u bpSy> bdZu \pe.
times, then its side is doubled.
(3) If the area of a square increases four
ÃÊ ©‚∑§Ë ÷È¡Ê ŒÈªÈŸË „Ê ¡ÊÃË „Ò– (3) Å¡ L$p¡C Qp¡fk_y„ n¡Óam QpfNÏ„ h^pfhpdp„
times, then its side is not doubled. (3) ÿÁŒ Á∑§‚Ë flª¸ ∑§Ê ˇÊòÊ»§‹ øÊ⁄U ªÈŸÊ ’…∏ ¡ÊÃÊ Aph¡, sp¡ s¡_u bpSy> bdZu _ \pe.
(4) If the area of a square does not „Ò, ÃÊ ©‚∑§Ë ÷È¡Ê ŒÈªÈŸË Ÿ„Ë¥ „ÊÃË–
increase four times, then its side is not (4) Å¡ L$p¡C Qp¡fk_y„ n¡Óam QpfNÏ„ h^pfhpdp„ _
(4) ÿÁŒ Á∑§‚Ë flª¸ ∑§Ê ˇÊòÊ»§‹ øÊ⁄U ªÈŸÊ Ÿ„Ë¥ ’…∏ÃÊ,
doubled. Aph¡, sp¡ s¡_u bpSy> bdZu _ \pe.
ÃÊ ©‚∑§Ë ÷È¡Ê ŒÈªÈŸË Ÿ„Ë¥ „ÊÃË–
-oOo- -oOo-
-oOo-

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Set - 03 50

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