NARAYANA IIT ACADEMY INDIA XI -REG_Adv. Dt: Time: 09:00 AM to 12:00 Noon Max.Marks: 180

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1 6..06 XI_Reg.ADV._Jee-Adv_03-Pkey & hits NARAYANA IIT ACADEMY INDIA XI -REG_Adv. Dt: Time: 09:00 AM to :00 Noo Max.Marks: 80 KEY & SOLUTIONS PHYSICS CHEMISTRY MATHS C C 4 B A C 4 D 3 A 3 D 43 A 4 A 4 C 44 A 5 D 5 C 45 A 6 B 6 A 46 A 7 B 7 B 47 A 8 C 8 B 48 D 9 D 9 B 49 C 0 C 30 A 50 A B,C 3 C,D 5 A,C A,D 3 A 5 B,D 3 C,D 33 A,B 53 A,B,C,D 4 A,C 34 B, C,D 54 A,B,C,D 5 A,D 35 C 55 A,B,C,D PHYSICS Page

2 6..06 XI_Reg.ADV._Jee-Adv_03-Pkey & hits P. B V / V Sice, for a icompressible liquid, V 0, B is equal to ifiity.. Stress, Y N / m 00 Eergy stored per uit volume, u J / m Average pressure o the side of the vessel is gh /. Sice force o the side = force o the bottom, we have gh rh gh r h = r 4. The average pressure exerted by water o the left side ad right side respectively are gh / ad gh /. Resultat thrust 3 gh gh gb bh bh h h N. p p gh ad p3 p wgh O addig the two equatio, we get p p g h h Pa 5. 0 w 3 0 This is the gauge pressure at the lower face of this block. 6. Termial velocity is give by r g v0 f 9 Here, v 3.5 mm / s, r cm, kg / m f 3 3 O solvig, we get 09kgm s r g 4 3 mass, m r 9 3 Sice, the other sphere has mass 8 m, its radius is double ad hece, its termial velocity is the same liquid shall be 4v 0 7. Termial velocity, v0 f Page

3 6..06 XI_Reg.ADV._Jee-Adv_03-Pkey & hits 0. Clearly, form figure, r cos R R r cos. For material to be ductile, it should be able to deform permaetly uder tesile stress. Sice Y >Y, is more ductile. For material to be malleable, it should be able to deform permaetly uder shear stress. Sice,B >B, is more malleable.. The desity of material of sphere is greater tha the desity of liquid ad F B F A < mg, otherwise the sphere would have moved up. Also, FB FA fvg is ot depedet o atmospheric pressure. 3. (C, D) Whe g = 9.8 m/s ad the sphere is movig with termial velocity. Weigth Buoyat force =Drag force i. e., mg Fb Fd I gravity free space, both mg ad F b are zero. Sice, its velocity ow also is 0 m/s, the same amout of drag force will act upwards. Now, ma Fd mg Fb F b a g upwards m (a) ad (b) are icorrect. As due to retardatio, speed will decreases F d will also decreases. Therefore, the magitude of acceleratio will decreases as time passes ad it will evetually stop. c ad d are icorrect. r g 4. (A,C) termial velocity, vt f is uiform ad is idepedet of height h Pressure iside the bubble is always greater tha the pressure outside. Also as it move up, outside pressure decreases resultig i decrees i iside pressure also. 6. Whe tap is closed. 5 P 3.50 Nm, v 0, h h Whe top is ope, 5 P 3.00 Nm, v?, h h Accordig to Beroulli s theorem, P v gh P v gh gh v v 0ms 500 v 4 50 F A N 3 x gh Page 3

4 6..06 XI_Reg.ADV._Jee-Adv_03-Pkey & hits S cos h rg The mass of water that rises is give by S cos Sr cos m r h r rg g m r. Hece, if r is doubled, m will also double. /3 P V /3 8 P V Let r be the radius of each droplet. The, the volume beig same, we have R 8 r 3 3 R Therefore, work doe W U fial Uiitial 8T 4r T 4 R 4T 8 R 4 R T CHEMISTRY. (C) O air -Ehtylathraquiol H O Ethylathraquioe H /Pd 6. (A) Ca + + Na CO 3 CaCO 3 + Na + Mg + + Na CO 3 MgCO 3 + Na + 7. (B) NaOH Z Na ZO H Sod.Zicate (B) KO is superoxide K O BaO is peroxide Ba + (O ) MO ad NO are dioxides O=M=O O N O 30. (A) Li gives Li O oly. 3. (C), (D) 3. (A) Calgo has formula (NaPO 3 ) 6 which is represeted as Na Na PO 33. (A), (B) All alkali metals show + oxidatio state. I KO, O.N. of K is + ad that of O is Here, KO is actually K + O. 39. (4) 4 i is aqueous solutio. It forms BeH O 40. (3) Basic salts solutio will have ph > 7, will chage colour of litmus paper red to blue. KCN, K CO 3 ad LiCN are the oly basic salts amog these. Page 4

5 6..06 XI_Reg.ADV._Jee-Adv_03-Pkey & hits Maths 0 4. sum=! d d d3... d 3! First arrage ay 3 cosoats at eve places i 4 p 3 ways. Now the ewly created four odd places ca be filled by the remaiig letters which icludes 3 vowels ad cosoats, which ca be doe i 4 p 4 ways. Hece the required umber of permutatio. 4 4 p3 p There are total 9 places out of which 4 are eve ad rest 5 place are odd.4 wome ca be arraged at 4 eve places i 4! Ways ad 5 me ca be placed i remaiig 5 places i 5 p 5 ways. Hece the required umber of permutatio. 5 4! P Total umber of arragemet =8!=4030 Number of arragemet whe best ad worst papers are together = 7!! 0080 Numbers of arragemet i which best ad worst papers are ot together = = T,T,T N,N,.N,G,G.G,R,I,I,I There are total 3 letters of which 3 are T s, 3 are G s, 3 are I s. Also there are 6 eve places ad 7 odd places. 3Is (i.e vowels) ca be arrages i the 6 eve places i 6 P3 3! 0 ways. 0! Now, we are left with with 0 places i which 0 letters ca be filled up i 3! 3! 3! 6800 The total umber of permutatios i which vowels occupy eve places= Total umber of 3digit umbers i which there is at least oe digit is Number of zeroes = = = 3. Hece (A) is the correct aswer. 48. We have two sets of parallel lies each cotaiig (m + ) lies. So umber of parallelograms = = m m m m C C Pr Pr Pr a b c Pr b Pr ad c Pr a P r b b c r ad r a b c b b b a b c. a 50. First we arrage B, A, A ad A i 4! 3! ca be arraged i 5 C ways. So total ways = 4 0 = 40. ways B A A A, Now deotes the place for N. So N 5. First we seat remaiig ( ) persos which ca be doe i ( )! ways. I ay such arragemet, the two particular persos ca be made to sit i ( ) gaps i P ways. Total ways = ( )! P Also, we may calculate the required ways as: Page 5

6 6..06 XI_Reg.ADV._Jee-Adv_03-Pkey & hits Total ways (ways i which the particular persos always sit together) Ways i which both are together = P P = ( )! Required ways =! ( )! 5. He ca ivite or ot ivite each of his frieds. Total ways i which this ca be doe = ( 8 times) = 8. But this also icludes oe way i which oe of these of his frieds is ivited. Rejectig that way, the umber of ways are 8. Also, he may ivite oe fried i 8 C way frieds i 8 C ways ad so o Required ways = 8 C + 8 C C No. of ways of arragig remaiig ( 3) boys = ( 3)! Here, there are ( ) gaps i which the 3 particular boys ca be seated i C 3 ways. The three boys ca rearrage amog themselves i 3! ways. Total ways k = ( 3)! C 3. 3!! = ( 3)! 3! 3! 5! = ( 3) ( 4) ( )! = ( 3) ( 4) ( 5)! 55. First ad secod prime i Mathematics (Physics) ca be awarded i 30 P ( 30 P ) ways. First prize i Chemistry (Biology) ca be awarded i 30 (30) ways. N = ( 30 P ). (30) = = sice 400 = = = N is divisible by each of 400, 600, 800. Also, N is divisible by four distict primes i.e., 3, 5, If a boy is selected the the o. of ways = 4 C 6 C If a boy is ot selected the umber of ways = 6 C 4 Captai ca be selected i 4 C ways Req o. of ways = 4 C C3 C C4 C4 C p r = p r+ -r =... () c r = c r- r =... () Solvig () & () we vet = 3, r = 3630 = 3 5. Now a divisor will be of the form (4+) if divisor is form the help of (4+) type umber or by (4+3) types umber take eve times. Hece divisors are, 5, 3,, 5, 5 3, i.e., C x > C x > 4x > 33 x 9, but x 0. x x So x = 9, 0. Hece there are two solutios 60. Each place of places ca e filled i 3 ways. Total o. of umbers = 3. 3 > 900. = 7 is least value. (B) is correct aswer. Page 6