SCORE I DATE : Test Pattern : JEE (Main)
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1 LEADER CURSE (PHASE ELS) TARGET : JEE (MAIN) 05 SCRE I DATE : MAJR TEST Test Pattern : JEE (Main) ANSWER KEY Que Ans. Que Ans. Que Ans. Que Ans. Que Ans. HINT SHEET. Due to 0. ev poton, one poton of energy 0. ev will be detected. Due to 5 ev poton te electron will coe out of te ato wit energy (5 -.6). ev.. I > I (given) Þ i > i (Q i I) and stopping potential does not depend upon intensity. So its value will be sae (V 0 )... l Since v is increasing in case (i), but it is v not canging in case (ii). Hence, in te first case de-broglie wavelengt will cange, but it second case, it reain te sae l p Þ Þ Kota/0CE l- l 00 p +D p 99l p +Dp 00 p 00 Þ p+d p p 99 Þ p 99 Dp 5. Lower NT gate inverts input to zero. NT gate fro NAND gate inverts tis output to upper NAND gate converts tis input and input 0 to. Tus A and B becoe inputs of NAND gate giving final output as zero. Coice A is correct. 6. Te P-N junction will conduct only wen it is forward biased i.e. wen 5V is fed to it, so it will conduct only for rd quarter part of signal sown and wen it conducts potential drop 5 volt will be across bot te resistors, so output voltage across R is.5 V. \ V 0.5V 7. In saple x no ipurity level seen, so it is undoped. In saple y ipurity energy level lies below te conduction band so it is doped wit fift group ipurity. In saple z, ipurity energy level lies above te valence band so it is doped wit tird group ipurity. (f) :(f) :(f) / 8. f c (N) Þ c E c F c F / / / ( 0 ) : (9 0 ) : (6 0 ) : : HS-/6
2 Target : JEE(Main) 05/ Here, fc.5 MHz 500 khz, HS-/6 f \ Low side band frequency c 0 khz f - f 500 khz - 0 khz 90 khz Upper side band frequency f + f 500 khz + 0 khz 50 khz c a 0.8Þb ( -0.8).. Also K Di b c D i i i b c b particle v also ÞD b D 6 A. l v æ ö n ç èlnø l Þ K particle. n poton...(i) c K...(ii) l 8 K particle n Kpoton c 0 8 \ l Þ E. Fro snell's law sin i sin r sin q.6 sin x 5 sin x sin q 8 l (E sae). Refraction fro plane surface - - v u / - 0 v (-0.) Þ v 0. For lens u ( ) 0.5 f v? - Þ - v u f v (-0.5) v Focal lengt of plano convex lens 0 c Fro lens aker forula 7. / æ ç - öæ ç - ö 0 è øè -Rø Þ R 5 Focal lengt of equi-concave lens of water æ / öæ ö -5 ç - ç - Þ f c f è øè-5 5ø So, net focal lengt Þ - + Þ fnet 5c f net 00 Pnet 6.67D 5 sini v.7 sin r v tan0º - - ic sin sin 8. N A N + N C x x x + ( l/.5) l/ l/. Þ.7 9. Eyes x - x 80 x 0 Here x 5.5 c. b r 0 H x x B 8p 0 r -7 H 0 p 0 0 Kota/0CE05
3 . For. T.I.R. Leader Course/Pase-ELS/Score-I/-0-05 i > q c sin i ³ sin q c Fro triangle sin i R ³ R+ d R R+ d sin Q c. ¾¾¾¾¾¾¾ C / Na Kolbe scidt reaction CH C C CH C Salicylic Acid Acetylation C CH C d R ³ For nd dark d sin q l -5 dsin q 0 sin 0º -5 l 6 0 c 6000Å 6. H ni - H H Hl 0 0 I n (N / l) N 60 Aspirin. Reducing sugar contain free group at Anoeric carbon.. Anoers are pair of disatereoers wic ave opposite configuration at carbon ato of functional group (Anoeric carbon). C NH is peptide bond 5. C CH H H H Conc. HS / D I 8A S N B G C CH p,p'-dicloro dipeny/ tricloroetane (DDT) 0. B Z 6. CH ¾ ¾ ¾ ¾ ¾ NaN+ H Diazotation CH net + B B B NH NN æ 0 Mö æm 0 ö net ç + ç B è p r ø è p r ø CH Hydrolysis H CH 0 M -7-7 Bnet T p r () CH ¾¾¾¾¾ CH+ K Reier-Tieann forlyation Kota/0CE05 HS-/6
4 Target : JEE(Main) 05/ Å Br Ha logenation ¾¾¾¾¾ Br Å N Nitration. ole product foration sould be fro eleents in ost stable allotropic for 5. Here, q 0 Þ tan q tan 0 0; æ x ö logç è ø log k + n log C N N Br ¾¾¾ Å Nitration N Br Q n tan q Þ n 0 NaBr N N CH Na CH N 8. nly ' Aide gives Hoffann Broaide reaction 9. Initial volue, V 0L V (final) nrt P W P DV (9 0) 9 L bar 9L 0. Nuber of oles of H MV n H Nuber of oles of Ca() MV n In te process of neutralisation ole H +, will be neutralised \ DH kcal kcal 5 cal. DS fusion D H fusion J ol T 7 p. DH reaction o o DH -DH (0) 605 kj ol. f(c H ) f(c H ) P (s) + (g) P 65 (l) ; DH kj P (l) + (g) P 5 (s); DH 7 kj n adding P (s) + 5 (g) P 5 (s) ; DH 5.5kJ 7. Since ferric ions can coagulate negatively carged blood solution, terefore ferric cloride ay be applied to stop bleeding. 8. Q Productive power gold nuber Q rder of productive power will be Gelatin > Haeoglobin > Sodiu acetate (0.005) (0.05) (0.7) 9. PV w RT 50. P 0.0 \ P 67.7 Pa u u é ë ù æ ö ú / ç Urs è M ø 5.. lit ixture. 0.ol. Since ixture is equi-olecular, Z 0.05olCH 0.ol ] 0.05olC H 6 ¾¾ 0 6 ¾ ¾ ¾ \ Mass of ixture.g 5. DS v D H v B.pt 50 7 ; DS f D H f F.pt \ D S DS 6. b a + c v f g.5g 80 7 ; Þ log 5 ( x æ7 x ö ) log 5 + log 5 ç + è ø Þ ( x ) æ7 x ö ç + è ø Þ x 7. x 8 0 Þ x HS-/6 Kota/0CE05
5 Leader Course/Pase-ELS/Score-I/ Tese are factors so æ 5 ç x+ öæ ç x+ ö è øè ø 6. æ 9 9 ç x+ ö æ ç x+ ö è ø è ø so coefficient of x æ ö ç è ø a+ b+ c+ d (abcd) ³ 5. Þ G A Þ a b c d \ a + b + c + 5d Þ a b c d So a + b + c + 5d f(0) C > Õ k é ù é ùé ù é ù k ú ú ú ú 0 ú 0 ú 0 ú 0 ú ë ë ë ë é ( ) + ù é 00ù Þ ë0 ú ë0 ú 7. A A T, B B T é ù A B ë ú...(i) Þ A T + B T é ù ë ú Þ (A + B)T é ù ë ú é Þ A + B ù ë ú...(ii) é 5ù ú Adding (i) & (ii) A 5 ú ë ú \ A d d d 0 Þ D D D 0 So f( ) a b + c > 0 Þ a + c > b 66. a < 0, b > 0, c > 0 Þ a > 0, b < 0, c < 0 So roots are of opposite sign and positive root is greater in agnitude. So b is positive root and a is negative root (Q a < b) 68. D > 0, Af() < 0 Kota/0CE05 a f() (i) D > 0 Þ 9 a > 0 Þ 9 > a (ii) Af() < 0 Þ (a ) < 0 Þ a < Þ a Î (, ) 69. A, B 9 AB A 9 B 70. D is ade up of cofactors of D. So by te property D D. \ D D D b D a (a + ) (a + ) a (a + ) (a + ) 0 Þ a 7. By properties of adjoint atrix options (), (), () all are correct. So option () is correct å å (r+ ) r+ (r+ ) r+ r 0 r å å å (r+ )r+ r+ r+ r 0 r 0 r 0 Þ ( ) ( ) Þ Put te value of n. 656 r r T r+ ( ) ( 9) C 7 r r r Cr 7 Þ r 0, 9, 8,..., 656 Þ T n a + (n )d Þ (n )9 Þ n 70 HS-5/6
6 Target : JEE(Main) 05/ () 507 (9 + ) 507 Reainder 507 ( ) 69 (6 + ) 69 so reainder 79. Let tere aer n candidates C + C + C + + C 6 n n n n n Þ n 6 Þ n 6 Þ n No. of ways!! 8. No. of words 5!!! 0 8. Considering cc as single object _U_CC_E_! Now at vacant places s can be arranged by so no. of words! 8. Required probability 7 56 P! æc 6 C ö + ç 7C 8 è C ø 8. x x 0 0 Þ (x + )(x 5) 0 Þ x 5 n 00 C, 5 C P! 86. Required probability {( 0.)( 0.) ( 0.)( 0.)} Let z x + iy ( ) ì( ) z + x + + iy y ixü í ý iz + ( y) + ix î( y) ix þ { } (x + )( y) + xy + i y( y) x(x + ) Þ ( y) + x y( y) x(x + ) So I.P. ( y) + x so locus is circle. 88. x.x.x... æ p pöæ p pö Þ ç cos isin ç cos isin... è øè 9 9ø æ Þ cos p p p ö æ i sin p p p ö ç ç è 9 7 ø è 9 7 ø p p Þ cos isin i 89. z Þ zz, z Þ zz, z 9 Þ zz 9 9z z + z z + z z zzzz + zzzz + zzzz Þ required probability 5 00 C C 0 Þ z z z z+ z + z Þ z + z + z Þ z + z + z 85. P() 6, P() 6, P() 6 {,, } or {,, } Probability æ ç ö! + æ ç ö è6 6 6ø è6 6 6ø 6 HS-6/6 Kota/0CE05
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