NARAYANA I I T / N E E T A C A D E M Y. XI-REG (Date: ) Physics Chemistry Mathematics

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CPT-7 / XI-REG / 5..07 / Hints & Solution NARAYANA I I T / N E E T A C A D E M Y XI-REG (Date: 5..07) CODE XI-REG Physics Chemistry Mathematics.. 6... 6... 6... 6. 5. 5. 65. 6. 6. 66. 7. 7. 67. 8. 8. 68.

1 CPT-7 / XI-REG / / Hints & Solution NARAYANA I I T / N E E T A C A D E M Y XI-REG (Date: 5..07) CODE XI-REG Physics Chemistry Mathematics Page

2 CPT-7 / XI-REG / / Hints & Solution. w oil w oil Hints & Solution PART A: PHYSICS P h g h g h 500 g() g( 0.85) 6.5 gf / cm 9 F F F 0 A A A A F 00 0 cm 7 0 0cm g 8cm g cm g 0.6 8Y.6 Y 0.8 g / cc X Y Hg. L l r. Mgl l 7.50 m 0.075mm 6 AY 0.0. l 0. 0 l 00 Fl 00 Y 0 N / m 6 Al 0 0 l r l l cm r l r. Stress F (Strain) 6. Strain Y AY(Strain) 6 AY 0 0( 0) 8. W( l) 0.J L P hg K 0 N / m V V 0 V V F l 0. Y. F Y. l r r l l F = 9 0 ( 0 ) 0. 60N 00. OB = OC = 0 m BC BC 0 m OB BC 0 radian AB 80º º L W. Weight of length BC W L Effective downward force at B W W Stress at B W W / S. W. joule 6 6. Elastic potential energy is converted into kinetic energy of the ball Now elastic potential energy Page

3 CPT-7 / XI-REG / / Hints & Solution Page (strain). 6 Y volume J mv v 60 v = 0 ms m PART B: CHEMISTRY. KO is superoide K O BaO is peroide Ba + (O ) MnO and NO are dioides O=Mn=O N O O. The correct order is Li + > Na + > K + Lower the ionic radius higher is the hydration enthalpy.. Li gives Li O only.. All alkali metals show + oidation state. In KO, O.N. of K is + and that of O is Here, KO is actually K + O. Na y NH Na NH NH 5. y 5M Ammoniated electrons make the solution paramagnetic. These absorb red part of visible spectrum and blue (complementary) colour is observed. At higher concentration of NH electrons get paired up. Solution becomes diamagnetic and bronze in colour. On standing: Na + NH NaNH + H 6. (C) NH HCO + NaCl NaHCO + NH Cl 7. (D) CO HOH HCO OH PART C: MATHEMATICS 6. There are 0 chairs on each side of table. Number of ways of arranging 5 person on 0 chairs is 0 P 5 and persons sit on other side in 0 P ways and remaining person in remaining chairs in! Ways Hence total number of ways in which persons can be seated is 0 P 5 0 P! = 0!! 5! 7!. 6. Required number of ways = n C + n C + n C n C n = n n. 6 Required number = coefficient of n in ( ) = coefficient of n in ( 6 ) ( ) - = coefficient of n in ( )( ) - = coefficient of n in ( 6 ) ( ) - [9 n 6 n ] = n + C n - + n 9 C n 9

4 CPT-7 / XI-REG / / Hints & Solution = n C n - n 7 C n 9 = -n + n 8 6. (D) Terminal digits are the first and last digits Since terminals digits are even Since I st place can be filled in ways and last place can be filled in ways and remaining places can be filled in 5 P = 0. Hence the number of si digit numbers, the terminal digit are is 0 = We can first arrange the n gentlemen in (n )! ways. Between the n gentlemen there are n places and we can arrange the n ladies there in n! ways Total number of ways = n! (n )! = 5 7 For the divisors to be even and divisible by 5;, and 5 must occur atleast once. Therefore total number of required divisors are = = Page Required sum = Total ways of selection without restriction = 0 C number of ways of selection when two are adjacent = 0 6 C number of ways of selection when all the three are adjacent = 0 required number of ways = 0 C = = 0 [57 6 ] = 0 0 = Here = 7! 7!, y = 6! 7! y =. 70. If eaminer gives any 5 questions (each) marks. Then marks on remaining question = 0 5 = = 0 i 5, i =,,.6 number of ways = coefficient of 0 in ( ) 6 = coefficient of in ( ) 6 = coefficient of in ( ) -6 = 7 C 7. Total number of ways = 6!/ = We first choose the 6 cards which will go to the right places and dearrange the remaining. Number of ways to select the 6 = 0 C 6 and number of ways to dearrange =!!! Total number of ways = 0 C 6 9 = Number of 5 letters words which can be formed using ten letters when the letters can be repeated = 0 5 Number of 5 letters words which can be formed using ten letters when the letters are not repeated = 0 P5 5 0 Number of words which have at least one letter repeated 0 P 5. 7 Two women can sit on any of the chairs to in p ways. Now three men can sit on any of the remaining 6 chairs in 6 6 p ways. Total number of arrangements P P. 75 A number is divisible by if the sum of its digits is divisible by. Now =5, Which is divisible by Thus in order to form a five digit number divisible by we can left either 0 or. If 0 is ecluded then the number of numbers =5!=0! = 9

5 CPT-7 / XI-REG / / Hints & Solution If is ecluded then the number of numbers 5!! 96 (Total number of numbers =0+96=6 76 Let the number of newspaper is, then 60=00 5 =5 77 Using three digits,5 and 7 with repetition each position of the number can be filled in ways. So the n positions can be filled in. (n times) = n ways. According to the question n 900 n 00 n>6 Least value of n=7. 78 A nine digit number has 5 odd positions and even positions. Four odd digits 55 can be arranged in even positions in! 5! ways. Remaining digits 888 can be arranged in remaining positions in!!!! ways.! 5! Total number of ways 60!!!! 79 Number of triangles which can be formed using the vertices of a regular polygon of n vertices n C T n Thus T n T n n n C C n(n ) n 7 80 Total number of arrangements of the word BANANA= 6! 60!! n C Number of arrangement in which N s occur together 5! 0! Number of arrangement in which two N s do not appear adjacently =60-0=0 8 LCM of p, q will be r if eponent of r is in at least one of p and q, i.e., (0, ), (, 0), (,), (, ) and (, ), so eponent for r can be selected in +=5 way. Similarly eponent for s and t can be selected in +=9 and +=5 ways respectively. Hence, the number of selections=5 9 5=5. 8 Before COCHIN, the words starting with CC, CH, CI and CN will occur. Hence, the second place can be selected in ways and remaining four alphabets can be arranged in! ways. So the total number of such words =! =96 Clearly the net word will be COCHIN itself. 8 The required number of integers=number of integral solutions of , where i,i,, in( ) 7 in in in ( ) 7 ( )( ) C... C C C... 8 For any two subsets A and B of S we have (i) a A and a B (ii) a A and a B (iii) a A and a B (iv) a A and a B Where as C C Page 5

6 CPT-7 / XI-REG / / Hints & Solution Thus, total number of ways for A B. Thus, for each element of S there are three ways such that A B. Thus, total number of ways for A B is A B is one of the possibilities, otherwise in all other cases A and B are interchangeable. Thus, the required number of unordered pairs of disjoint subsets of S 85 The value of k 7 j C8 C7 C0 j k j j C8 C7 C7 C7 C7 C7 C7 C 7 C0 j j C C C C C C C C C C C C C C C C C C = 7 C a b c 0 a b c 6,7,8,9,0herea,b,c Hence required no. of ways = 5 C C C C C =0 87 Let a=+, b=y+, c=z+, d=u+, where,y,z,uw Then a+b+c+d=0 +y+z+u+=0 +y+z+u=8 No. of ordered quadruplet = 8 C C The number of the word in all formed by using the letters of the word SMALL = 5! 60! Let s count backwards. The 59 th word is SMALL 58 th word is SMALL 89. No. of words 6 = 6 C C Let A= {a a,...,a n } (i) ai P, ai a in! ( )!! in 6( ) ( ) in 6( ) ( ) ( ) in 6( ) ( )( ) 6 Q (ii) ai Q (iv) ai P, ai P, ai Q (iii) ai P, i Q P Q contains eactly two elements taking elements in (i) and (n-) elements in (ii) or (iii) or (iv) No. of ways = n C n Page 6

EXERCISE - 02 BRAIN TEASERS UNIT # 12 PERMUTATION & COMBINATION PERMUTATION & COMBINATION AND PROBABILITY EXERCISE - 01 CHECK YOUR GRASP

EXERCISE - 02 BRAIN TEASERS UNIT # 12 PERMUTATION & COMBINATION PERMUTATION & COMBINATION AND PROBABILITY EXERCISE - 01 CHECK YOUR GRASP EXERCISE - 01 CHECK YOUR GRASP 1. Total possible words words do not begi n or terminate with vowel Total words! 10 Words which do not begin and terminate with vowel 1 6 Desired words : 180 6 84 II-Method