[ All Batch ] ANSWER KEY WITH SOLUTION MATHEMATICS SECTION A 1. A 2. C 3. B 4. D 5. D 6. A 7. D 8. A 9. D 10. C 11. C 12. B 13. D 14. B 15.
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1 TAGET IIT-JEE CLASS XII [MAIN] DATE : -7-6 [ All Batch ] ANSWE KEY WITH SOLUTION MATHEMATICS SECTION A. A. C. B. D 5. D 6. A 7. D 8. A 9. D. C. C. B. D. B 5. C 6. B 7. B 8. C 9. B. C. C. C. D. A 5. A 6. A 7. B 8. B 9. C. D HYSICS SECTION A. B. A. A. A 5. C 6. B 7. B 8. D 9. B. C. C. D. D. A 5. D 6. A 7. B 8. B 9. A. D. C. B. D. C 5. D 6. B 7. A 8. B 9. C. A CHEMISTY SECTION A. A. C. A. D 5. B 6. A 7. B 8. C 9. D. A. C. A. B. D 5. A 6. D 7. D 8. C 9. B. D. C. A. D. D 5. D 6. C 7. A 8. B 9. B. B 9 - ajeev Ganhi Nagar Kota, h. No , 7-967
2 SOLUTIONS MATHEMATICS SECTION A Single Correct. A. C. B at an a t t a at t [Slope of tangent at the point (at, at)] Since tangent line is perpenicular to -ais, then t t Therefore require point (, ) (b + c) a bc (b + c) sina sin A b c bc bc (using AM GM equality hols if b c) A 9 an b c (C) < < y > + + > < < < <. D y 8 () Let slope of tangent is m tangent to the parabola is y m + a/m here a Equation of tangent is y m + /m () ut value of y from () in y (m + /m) m + m 5. D m + + m () touches the y D m m m m equation of tangent is y + < <. < < < <.8 [] / (tan / n tan / tan n n ( tan )( ) ) n tan n.sec ; t tan n n t t t n 6. A A A.. t t cm /sec n 7. D If Statement-I is false but Statement-II is true S n 7n S' n n 7 S-I is not true. S-II is true. T n 7(n ) 7 then is not T' (n ) 7 8. A Slope of C is sin an for > slope of C is. Thus for point of contact n 9 - ajeev Ganhi Nagar Kota, h. No , 7-967
3 sin or Hence point of contact is, or, For 9. D,, we get a For,, we get a g(), g (). g () 7 f () e g().g () f () e g().g () e. 7 7 slope of the normal (/) an the normal passes through the point (, ) the normal has the equation y (/) + y, require istance / 5.. D inepenent of a sin + cos perio /. B y a + y y Q. C y...(i) equation of tangent at (m, 8m ) is y m m...(ii) on solving (i) & (ii) (m m ) m, m ut m in equation (ii) y m so let Q(m, m ) so m (m, m ) & slope of normal at (m, m ) m since tangent at & normal at Q are same m 9m 7m 6 m. C OA HA cosa (istance of orthocentre from the verte A is cosa) cosa A (C). B The equation of the curve is y e +, when, y. e + at the point (, ) O y Equation of tangent at (, y) is 5. C an then Y y Y y (X ) y y X + X + Y y + y a [from Eq. (i)] X Y a + a y O a OQ a y ac b H.. a c O + OQ a + a y a( y) a a a. ac log (a + c) + log a c a c log (a + c) + log (a b) log (a + c) log (a c) log (c a) c > a 9 - ajeev Ganhi Nagar Kota, h. No , 7-967
4 6. B Equation of the joining the points (, ) an (5, ) is y. If this line is tangent to y a, then ( ) ( + ) a shoul ( ) have equal roots. Thus (a ) a ± 9. B area of heagon 9 absin C bcsin A ca sin 9 [] B 7. B (where is the area of the triangle ABC) area of heagon area of triangle lim h h z z z h ( h) lim h 8. C ( h) ( ) a + b + c...(i) Since, the curve y a + b + c + 5 touches -ais at (, ), the curve meets y-ais in (, 5). an an (,5) (,) + + c (given)...(ii) a b + c a b + [from Eq. (ii)]...(iii) (, ) lies on the curve, then 8a + b c + 5 8a + b [ c ] 8a b +...(iv). C. C 6 h I ; t e ( ) t t e ( t) t t e ( ( t)) e e( t) I I e t (by r property) h ; (where h is the altitue from A); r Also r h (using r s) r h 7 From Eqs. (iii) an (iv) we get a, b a b + c Now 8 So. now AQ an ABC are similiar h r Q r h 6 h Q 6 Q 5 Q Q Ans. 9 - ajeev Ganhi Nagar Kota, h. No , 7-967
5 . C ut + sin sin t cos sin f() sin cos f() (cos sin sin) (t ) sin (t ) t sin f() cot cosec f () cosec + c.. D S T + T + T 5 + T n & S T + T + T 6. + T n S rt + rt + rt rt n S rs S S r t t sin t t.. t sin. A t t / t t t t t t t t t t t t 5. A Here e f Now + e + f given t + f + f f utting in a + b + c we get a f + c b f a c b + f e t 6. A 7. B a, e b, f c e f,, a b c are in A.. are in H.. Here, f () (f()) >, f (g()) Lim a g() g(a) a f(g()) As f (g()) g() must be ifferentiable at a Clearly sin cos sin & y sin cos & z sin cos. y y z y yz y + z 8. B F(a) F(b) (Apply HVT) 9. C cos y a a... a f () n (a... a n ) /n 9 a a... a n Sum. D n at a 9 Also / 9. /9 8 a / 8 tan a a.sec tan tan (sin ) tan / / ( a) a /8. tan a / / / (a) a a a /.a a 9 a 6 a 9 - ajeev Ganhi Nagar Kota, h. No ,
6 HYSICS Section A. B E kq/r outsie a uniformly charge sphere an also for a point charge.. A F M m GM m GMm. A The given circuit is equivalent to 6. B GMm F Mg 8a Orbital velocity of earth. If h, g g h v g g g v If h, (B) where is raius v. 7. B. A As 5 8 this is balance Wheatstone network (5 ) ( 8) Therefore 6 ohm 9 8 Use C A A C 5. C The gravitational fiel at any point on the ring ue to the sphere is equal to the fiel ue to single particle of mass M lace at the centre of the sphere. Thus, the force on the ring ue to the sphere is also equal to the force on it by particle of mass M place at this point. By Newton's thir law it is equal to the force on the particle by the ring. Now the gravitational fiel ue to the ring at a istance Gm Gm g / (a ) 8a M a a on its ais is given as a The force on sphere of mass M place here is m a V V esistance of first bulb is, an resistance of the secon bulb is V In series same current will pass through each bulb ' ower evelope across first is I an that across secon is ' ' I ' V as V ' ' ' ' The bulb rate V & W will glow more. 8. D GMm r GM r v v e GM v v v e GMm mv 9 - ajeev Ganhi Nagar Kota, h. No ,
7 9. B V a V b V 8 V i c V g f 7 Along agbfc V a V V a + V V - V a V (B). C Fringe with. Therefore, an hence ecreases.5 times when immerse in liqui. The istance between central maima an th maima is cm in vacuum. When immerse in liqui it will reuce to cm. osition of central maima will not change while th maima will be obtaine at y cm.. C i 5 otential rop across otential rop across AB. A kq E (r) 5. D kq 6r y q + (, ) + + q + + r (r, ) After charging, isolate So Q constant increasing C V C V increasing 6. A We can observer that, an are shorte an bypasse, so no current will Alternative : flow in then an I 5V + A 9 5V 9.5 A. + 9 A I 5. 5V I.5A 9. D If sin ( - )t, central fringe is obtaine at O. If sin > ( - )t, central fringe is obtaine above O an if sin < ( - )t, central fringe is obtaine below O. (D). D At earth's equator effective value of gravity is g eq g s e If g eff at equator to be zero, we have g s e or g e e 7. B Shift 7 t. D t. D D S 5 t 8 mm 8 m 8. B From symmetry flu through each point of the sphere is same. Flu through whole sphere q q o 9 - ajeev Ganhi Nagar Kota, h. No ,
8 Total surface area m flu through. m is q / o q. o (B) 9. A. C there is zero potential ifference across an 6 resistance. i A power by battery b i W Ans.. B Given C A If separation is halve ' / C' A/' A C. D Efficiency output power input power So, eg. circuits iagrams (A) eq I 9 eq A Heat prouce in celli r9 (/) 6 W (B) I I So, (A) is correct. D q q 5q 6 (5/8) 7 5. A r r i i i r + r 5 or r 6 6 r.9 9% Ans.. C The potential on the surface of the sphere is given by q q v... (a) a b The potential on the surface of the sphere is given by, q q V b b v v v q q v a b v q a b (C) 9 - ajeev Ganhi Nagar Kota, h. No ,
9 5. D I I 8 I A net I + [I ] 7. A the ring can be treate as electric ipole. 8. B 9. C E j. A Energy store in capacitor is in the form of electric fiel or electric energy which is a potential energy I q + I B GMm E For escaping to infinity GMm Energy require E (B) v v v 5 ; v + v + (v 5) v v v i 5 / amp. Ans. 9 - ajeev Ganhi Nagar Kota, h. No ,
10 CHEMISTY. A SECTION - A. C. A. C Charge Transfer. A. D 5. B 6. A (SnCl ) n & (BeCl ) n 7. B. B. D 5. A 6. D 7. D 8. C 9. B 8. C. D 9. D. A (A) Cocl (B) Nicl (C) Co(CN) Ni(CN) (F) Buff. ppt Green ppt (D) K[Co(CN) 6] K [Ni(CN) ] (G) Brown Yellow [o] Eposure (E) KCN KCN (e) K [Co(CN) 6] Yellow & KCN KCN. C. A. D prouct A is obtaine by OMDM (i.e. aition of HOH acc. to morkovnikov s without rearrangement.) prouct B is obtaine by HBO (Hyroboration oiation) (i.e. final prouct is aition of HOH acc. to antimorkovnikov s without rearrangement.) H OH + + hmgbr Aci base reaction O MgBr Benzene. D 9 - ajeev Ganhi Nagar Kota, h. No , 7-967
11 5. D 6. C ate of ehyration stability of carbocation 7. A 8. B M. 5 M.5 9. B CH CH CCH A B K c K b K a a - b + Therefore, (A) option is correct. Ka Kb K K. B H /BaSO Cis--butene (acemic Miture) Br /CCl (Anti aition) 9 - ajeev Ganhi Nagar Kota, h. No , 7-967
12
1. The electron volt is a measure of (A) charge (B) energy (C) impulse (D) momentum (E) velocity
AP Physics Multiple Choice Practice Electrostatics 1. The electron volt is a measure of (A) charge (B) energy (C) impulse (D) momentum (E) velocity. A soli conucting sphere is given a positive charge Q.
ANSWER KEY 1. [A] 2. [C] 3. [B] 4. [B] 5. [C] 6. [A] 7. [B] 8. [C] 9. [A] 10. [A] 11. [D] 12. [A] 13. [D] 14. [C] 15. [B] 16. [C] 17. [D] 18.
![ANSWER KEY 1. [A] 2. [C] 3. [B] 4. [B] 5. [C] 6. [A] 7. [B] 8. [C] 9. [A] 10. [A] 11. [D] 12. [A] 13. [D] 14. [C] 15. [B] 16. [C] 17. [D] 18.](/thumbs/89/97771760.jpg "ANSWER KEY 1. [A] 2. [C] 3. [B] 4. [B] 5. [C] 6. [A] 7. [B] 8. [C] 9. [A] 10. [A] 11. [D] 12. [A] 13. [D] 14. [C] 15. [B] 16. [C] 17. [D] 18.")
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