NARAYANA XII STD BATCHES [CF] (Hint & Solution) PART A : PHYSICS. When then material is removed, the force between them becomes

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NARAYANA I I T / N E E T A C A D E M Y Co m mo n Pr act i ce Te st Date: 6.04.7 XII STD BATCHES [CF] ANSWER.. 3. 4. 5. 6. 7. 8. 9. 0... 3. 4. 5. PHYSICS 6. 7. 8. 9. 0... 3. 4. 5. 6. 7. 8. 9. 30. 3. 3. 33.

80. 8. 8. 83. 84. 85. 86. 87. 88. 89. 90. (Hint & Solution) PART A : PHYSICS.. Since q = ne n.6 0 9 or n.

1 NARAYANA I I T / N E E T A C A D E M Y Co m mo n Pr act i ce Te st Date: XII STD BATCHES [CF] ANSWER PHYSICS CHEMISTRY MATHEMATICS (Hint & Solution) PART A : PHYSICS.. Since q = ne n or n.5 08 The force between two point charges q and q placed in a material medium of dielectric constant r at a distance r is qq qq 4 0 r r r 4 0 r When then material is removed, the force between them becomes qq F' rf 4 0 r Charge q will follow circular path if electrostatic force between q and q is attractive, F 3. i.e., qq mv qq v 4 0 r r 4 0 mr Also, v r r T ()

2 6 3 0 mr 3 r v qq Net electrical force on Q is zero FB FA FC FD 0 T 4. Now FA FC FD qq. 4 0 a Q. 4 0 a Resultant of FA and FC, FAC qq. 4 0 a qq Q a 4 0 a Q Q Or q or q By symmetry, the components of force on charge q0 due to charges at A and B along X-axis will cancel each other while along Y-axis will add. The resultant force on charge q0 is FB 5. F F cos q0 q 4 0 a y y a y q0 qy 0 a y 3/ Force on charge q0 will be maximum, when df 0 dy 3 y y q0 q a y 3/ a y 5/ Or a y 3/ 3y a 5/ y 0 or = 3 y a y ()

3 a q or y= The situation is as shown in the figure Or y 6. The origin is given a small displacement along the y-axis then the situation is as shown in the figure. By symmetry, the components of forces on the particle of charge q0 due to charges at A and B along x-axis will cancel each other where along y-axis will add up. The net force acting on the particle is Fnet F cos q q / 4 0 y a qq0 4 0 y a y y a y y a q y 4 0 y a 3/ As y a Fnet 7. q y or Fnet y 4 0 a 3 Initially, F 6 C 9 C 4 0 d When a charge of 3 C is given to both the spheres, then charges on spheres are 6 C 3 C 3 C and 9 C 3 C 6 C F ' 3 C 6 C 4 0 d Dividing eqn. (ii) by eqn. (i) we get F' F F ' F Fr Nm C Nm 4 0 qq C 9. Here q q 3 C C (3)

4 r iˆ ˆj kˆ; r iˆ 3 ˆj 3kˆ then r iˆ 3 ˆj 3kˆ iˆ ˆj kˆ iˆ ˆj kˆ r 3 According to Coulomb s law F qq N 4 0 r 3 0. F ; so when r is halved the force becomes four times. r. F q q. and F a 4 0 a F F3. 3. Conceptual 4. From principle of super position F Net FA F B F C F D 5. The resultant force at the centre will be 0 due to symmetry 6. As net external force along on the system is zero m a m a hence a cm =0 = m m a 8 m m a 0 = gm 7. Charge at rest produces only electric field 8. By using Q = ne Q = = +.6C In the LHS of q, the fields due to (-q) and (q) are opposite.. (4)

5 . At the present position all the charges are in equilibrium. But when they displaced slightly from their present position, they do not return back. So, they are in unstable equilibrium position. a a 4q 4q -q v KE qe mv m at m t m C is the mid point of AB. The field is directed to the right (positive) in the region between A and C, and to the left (negative) between C and B. The magnitude of E will increase sharply for x 0 and x l 5. In the following fig, in equilibrium Fe = T sin 300, r = m 0 m m T cos T 0 C Fk 0 r Q T r T T sin 30 Fe 0 C mg T.8N. Q x x d df 0 dx 7. F 30o +Q A F -Q +Q B C The net force normal to F cos 30 F cos 30 = 0, since F = F = Q 4 0 a 8. The net electrostatic force due to Cs ions at the centre (Cl ion) is zero, due to symmetry. 9. Force on each charge is zero, But if any of the charge is displaced, the net force starts acting on all of them. (5)

6 PART B : CHEMISTRY 3. Pyridine CH 3CH CH OH SOCl CH 3CH CH Cl SO HCl 3. CCl4 PhCH COOAg PhCH Br CO AgBr reflux 33. CCl4 CH 3CH COOAg I CH 3CH COOC H 5 CO AgI Silver propanoate Ethyl propanoate Conceptual Being strained cyclopropane ring readily opens up to form only n-propyl bromide. In contrast, reaction (a) gives a mixture of n-propyl and isopropyl bromides, reaction (b) gives isopropyl bromide with reaction (c) does not occur at all O O C H 5 C OAg I C H 5 C OC H 5 CO AgI It is Birnbaum Simonini reaction 39. CH3 CH3 0 HI CH 3 C CH OH CH 3 C CH H O, I H H CH3 CH3 I CH 3 C CH CH 3 C CH I Allylic substitution It is allylic substitution (6)

7 A, through halogen exchange reaction 54. Gattermann reaction occurs in presence of Cu-Powder 55. CCl4 R CH COOAg Br R Br CO AgBr reflux Inversion of configuration takes place in SN reaction CH 3O is a negative nucleophile and a much stronger base than CH 3COO 59. Viny halides usually fail in nucleophile substitution due to resonance phenomena leading to increase in C X bond strength. 60. PART C : MATHEMATICS R is reflexive and transitive but not symmetric R is symmetric and transitive but not reflexive a a is not true. So, R is not reflexive a b and b c does not imply a c. So, R is not transitive But, a b b a is always true. (i) a a = 0 is always true (ii) a R b a b -(a b) b a b R a. (iii) R and R But, is not related to So, R is not transitive. R is reflexive (3, 3), (6,6), (9, 9), (, ) R again (6, ) R but (, 6) R R is not symmetric R is transitive [ (3, 6) R, (6, ) R, (3, ) R others are clear Clearly, (x, y)r(x, y), (x, y) A, since xy = yx. This shows that R is reflexive. Further, (x, y)r(u, v) xv = yu uy vx and hence (u, v)r(x, y). This shows that R is symmetric. Similarly, (x, y)r(u, v) and (u, v)r(a, b) xv = yu and ub = va xv 69. a a b a yu xv yu xb ya and hence u u v u (x, y)r(a, b). Thus, R is transitive. Thus, R is an equivalent relation. Given, r = {(a, b) a, b R and a b 3 as an irrational number} (7)

8 (i) Reflexive ara a a 3 3, which is irrational number. (ii) Symmetric Now, r 3 3 3, which is not an irrational. Also, 3 r , which is an irrational. Which is not symmetric. (iii) Transitive 3r and r 4 5, i.e., Now, r 4 5 It is not transitive. 70. Given, R = {(, 3), (4, ), (, 4), (, 3), (3, )} is a relation on the set A = {,, 3, 4}, then since (, 4) R and (, 3) R, so R is not a function. since (3, ) R and (, 3) R but (, ) R, so R is not transitive. since (, 3) R but (3, ) R, so R is not symmetric. since (a, a) R a =,, 3, 4. not reflexive 7. Let W = {CAT, TOY, YOU,..) Clearly, R is reflexive and symmetric but not transitive. Since, CATRTOY, TOYRYOU CATRYOU 7. Conceptual Conceptual mrm, as m is multiple of m R is reflexive mrn nrm R is not symmetric. mrn and nrp mrp R is transitive 75. l Rl l is not perpendicular to l l Rl l l l l l Rl, Hence, R is symmetric. l Rl and l Rl3 l Rl3. ( R is not transitive) Reflexive but not Symmetric as, R but, R Not transitive as (, ) and, 3 R but, 3 R arb bra Since, R. So, R is not reflexive. Now (, ) R but, R, (8)

9 R is not symmetric. Clearly R is transitive. xry, yrz xrz R, 4, 4,,, 3, 3, a + b = b + a = Let a R a.a a 0 a, a R R is reflexive on R. Let a, b R. ab 0 ba 0 b, a R R is symmetric on R Since, R and, R but, R R is not transitive on R. We have ()² =, ()² = 4, (3)² = 9, (4)² = 6, (5)² = 5, (6)² = 36 R = {(, ), (4, ), (9, 3), (6, 4), (5, 5), (36, 6)} Domain of R = {x; x, y R} {, 4,9,6, 5,36) Range of R = { y; x, y R} (,,3, 4,5, 6) x R y iff x < y. We have < 3, < 5, < 3, < 5, 3 < 5 R = {(,3), (,5), (, 3), (, 5), (3, 5)} Let n N, then n is a factor of n nrn R is reflexive 3 R 6 because 3 is factor of 6 arc R is transitive R {(, ), (, 4), ( 3, 9)( 4, 6), ( 5, 5)} {(, ), (3,5), (4,7), (6, 6)} Domain of R {x : ( x, y) R} {,3, 4, 5, 6} Range of R { y : ( x, y ) R} {,5,0,7, 6} We have R = {(, 3)}. Here,3 R but 3, R. R is not symmetric. The relation R is also transitive i i R 3 is wrong R ( 3) is wrong i i R is wrong i 0 ir is correct (9)

10 Since R A B and S B C, we have SoR A C. SoR is a relation from A to C. R is reflexive because a a for a A. We have 4, R because 4 but, 4 R. R is not symmetric. Let a, b b, c R a b, b c. a c a, c R R is transitive TOPIC/SUB TOPIC Q. NOS. PHYSICS,7,3,7,8,6,30,3,4,5,,8,9,0,,,4,5,6,0,,5,7,8,9 9,,3,4 CHEMISTRY (Preparation of alkyle and aryl halide) 3 to 60 Alkyle and Aryl Halide MATHEMATICS Relation 6 to 90 Charge and its properties Coulomb s Law Electric Field due to Point charge (0)

CPT XII-NEW (REG) Batches (Date: ) ANSWERS PHYSICS CHEMISTRY MATHEMATICS 1. (D) 31. (D) 61. (D) 2. (B) 32. (D) 62. (B) 3. (B) 33.

CPT XII-NEW (REG) Batches (Date: ) ANSWERS PHYSICS CHEMISTRY MATHEMATICS 1. (D) 31. (D) 61. (D) 2. (B) 32. (D) 62. (B) 3. (B) 33. CPT-0/XII-S (NEW)/09.05.16 NARAYANA I I T / P M T A C A D E M Y CPT - 0 XII-NEW (REG) Batches (Date: 09.05.16) ANSWERS PHYSICS CHEMISTRY MATHEMATICS 1. (D) 31. (D) 61. (D). (B) 3. (D) 6. (B) 3. (B) 33.