IIT-JEE 2010 (Code-8) Answer Keys (Paper I)

Transcription

1 (ode-8) Answer Keys (Paper I) heistry. A. D. A 4. D 5. B B 9. A,. A,B. B,,D.,D. B,D 4. B 5. D B Matheatics 9. B. A. A.. 4. D 5. D 6. B 7. B 8. B, 9.,D 4. A 4. A,,D 4. D D 45. B 46. A Physics B D 6. B A 64. A 65. A, 66. A,B, 67. A,D 68. A,B 69. A, B 7. D 7. A 74. B /Paper I/Solutions Page

2 heistry 9. A,. A. D E a k Ae RT ; k increases eponentially with teperature(t). Na + Na SN + + Na. A 4. D NaN Na SN 5. B [r( ) 4 l ]N is ionization isoer of [r( ) 4 N l]l. 6. N N. A,B. B,,D a(),na, Nal can be used to reove teporary hardness..,d Na N NaN can act as buffer solution because weak acid ( ) and its canjugate base ( ) present in solution.. B,D 7. The single bond dissociation energy is approiately Kcal/ol. 8. B Standard olar enthalpy of foration of eleents in their eleental state or naturally occuring state is said to be zero. is liquid at roo teperature, and 4 are copounds B Roasting ufes us Fe us u Page /Paper I/Solution

3 5. D Fe Si FeSi 6. us u u S S is reducing species.. 4 Na Acid Salt If acid in above reaction is stronger acid than 7. B M(s) M (aq)(.5 M) M (aq)(m) M(s) ; ; ;.59 [M (aq)] LS Ecell E cell log [M (aq)] RS are stronger acid than. N( ).59.5 Ecell log 7 V So, E cell is positive G nfe G is ve Ecell log Ecell log 7 4 V. KN, K, LiN on dissolving in water gives basic solution which turns red litus into blue.. F F. Be Al Si 6 8 F F F.. Zn/ Na k dt.5 k k 5..9 k U 54 Xe 8 Sr n ; ; ; ; /Paper I/Solutions Page

4 9. B li 4 t 4 Matheatics tln t dt Use L ospital rule li li 4 4 for ln ln 4 4 li 4 4. A Three planes cannot cut at two distinct point.. A PQ 6i ˆ ˆj,QR ˆi j,rs ˆ 6i ˆ ˆj SP ˆi j ˆ Now PQ QR so it can not be a squre or rhobus. but PQ RS and QR SP so it can be a rectangle or a parallelogra. But PQ.QR so it a parallelogra.. Nuber of events in saple space = = 6 for r r Nuber of possible triplets are eight (,, ) (,, 4) (, 4, 5) (4, 5, 6) Each triplet gives 6 ways so possible favourable ways = 6 8 = 48 Required probability = Let the plane be a + by + cz = d as it contains y z 4 d = and a + b + 4c = (i) Further the plane is perpendicular to plane containing straight lines y z and y z 4 4 ence the noral to desired plane will be perpendicular to both the straight lines ence noral will be at 9 to the vector fored by i ˆ 4 ˆ j k ˆ 4i ˆ j ˆ k ˆ 8i ˆ ˆj kˆ 8a b c = (ii) Solving (i) and (ii) we get a = c and b = c required plane will be c cy + cz = or y + z = 4. D A, B, are in A.P. A + = B A + B + = 8 A + = and B = 6 a c sinc sina c a sin A sin.sincos.sin A cosa sin sin A sin A sin (by sine rule) 5. D All f(), g(), h() are onotonically increasing functions in [, ] a = f(), b = g(), c = h() 6. B p q Desired equation q p p q q p p p q p p q p p p q p q p q p p q q p Page 4 /Paper I/Solution

5 7. B A 8. B, f () sin B + /6 ++ cos ,D cos f () = sin f eist for all belonging to,, f does not eist for all, (t, t ) A (t, t ) y =4 B 4, ( ) as c will be negative 4. A t t Point,t t r t t AB t t r Siilarly a circle eist for lower lobe AB r I f() where f() f() f() f() I 4 7 I 7 /Paper I/Solutions Page 5

6 4. A,,D 4. D D z z z A is true as z z z z z z is true as this is nothing the equation of line D is true as Argz z z z B is not true as Argz z Argz z 45. B 4 y 4 4 y 4,y , y y A y , 6 For = 6, y = y 6 Equation of circle will be 6 6 y y y ˆi j ˆ a i ˆ ˆ j k b ˆ 5 4 since a b and a.b then let a b c Such that a,b and c for a right hand syste of unit vectors. a b c, b c a,c a b a b. a b a b a b. c a c b a b. b a 4 a b 4 5 Line y = + is tangent to a y b = c = fro condition of tangency c = a b = 4a b b 4a Now line pass through a, e a e Since b a e 4a a e a, e 4 e 4 e e e 5e 4 4 e or e for hyperbola e = Equation of plane containing lines y z and y z will be a ˆi j ˆ kˆ b i ˆ ˆj 4kˆ Page 6 /Paper I/Solution

7 5. 4 c i ˆ 4j ˆ 5kˆ r a. b c ˆi ˆj kˆ b c 4 ˆi j ˆ kˆ 4 5 a. b c 4 r. b c a b c r. ˆi j ˆ kˆ put r i ˆ 4j ˆ zkˆ y z, y + z = Distance between planes will be d 4 d = 6 6 {} if []is odd f if [] is even Now 9 9 f()cos cos f()cos 9 cos 9 f()cos f()cos f()cos I f() cos {}cos I f()cos cos cos cos sin sin sin cos cos I f()cos cos sin sin cos I 4 f()cos 4 5. Perforing z z z Perforing R R R and R R R = z z z = z z z = z z z a z z = zz = z k k! k k k Sk k k! k! k k k Sk = k k k k! k k k! k! k k k k k k k k k! k! k! k! /Paper I/Solutions Page 7

8 5. k k S k = k =!! k =!! 4 k = 4!! 4 5 k = 5! 4! 5 6 k = 6 4! 5! 99 98! 99! k Sk k 99!!! y z cos yzsin () z cos y sin yz sin () y zcos ysin yz sin () k = fro () and () yz sin zcos yzsin 4zcos yz sin y z cos fro () 4zcos 4 zsin cos z sin y z zcos 6zsin yz.z sin cos 6sin 4cos sin cos tan further ence three values of dy Y y X X = dy Y y dy y dy y I.F. e y. c y c = y = = c y tan cot5 tan tan 5 n 5 n 6 for n =, for n =, n =, 5 n =, n =, sin cos4 Page 8 /Paper I/Solution

9 cos 4 cos 4 4 n or n 6 for n = or 4 for n = or 4 for n = for n =, for n =, 5 or or 7 9 or 5 Now coon solution are,, 57. Physics T T onsider a sall segent of the circular wire Tsin Bi KQ T BiR as is very sall BiL L T R 56. For aiu value sin sin cos 5cos should be iniu sin cos 4cos.sin cos 4 sin cos sin Miniu value of sincos iniu value of denoinator 5 aiu value of epression = 58. B X and z R c c as increases z decreases ence potential drop across the capacitor decreases and that across R increases. Therefore the bulb glows brighter. 59. In the eperient one of the galvanoeters has to serve as a volteter connected in parallel to R T with a high resistance in series with itself. Moreover, the other galvanoeter serve as an aeter in series with R T with a low resistance in parallel to gavanoeter. 6. D For a bulb P rated v R Rated This iplies that the bulb with highest power rating should have least resistance Therefore R4 R6 R R R R B A real gas ehibits close to ideal behaviour at high teperature and low pressure. /Paper I/Solutions Page 9

10 6. as 6. A R l R A L Lt t hence error in the easureent of T.5 T.5s T l g Now T l g g T l g T l ence percentage error in deterination of g is 5%. 4R P 66. A,B, B 6 Potential due to the differential ring at P G 4r. M dv 6R 7R 6 5 Substituting 6r t dt Gr t E VP G t R 5R GM 4 5 7R A D Now V hence work required GM 4 5 7R 64. A The friction will vary as : V V p f gsinp tillp gsin f when P gsin f gsin Pwhen P gsin overall, f follows the sae straight line with negative slope. 65. A, As the least count of the stopwatch is s, the error in easuring total tie (4s) for oscillations is s. percentage error Now tie period.5% 4 4 s At E sinic sin sini > sini c i > i c TIR occurs at E, (A) is correct sin sini c, ray cannot get totally internally reflected (B) is correct. Angle between the incident and the eergent ray is 9, as shown above by geoetry. 67. A,D Electric field density is greater near Q than Q, so Q > Q harge Q is positive and Q is negative Electric field due to Q and Q will cancle out at a point to the right of Q Page /Paper I/Solution

11 68. A,B As teperature is sae at A and B, so the interval energies are sae at A and B. 69. A, 4V W nrt ln P V ln4 [at A P V = nrt ] AB V As the graph of the process B is not given to pass through the origin, so one cannot surely say any thing about the pressure and teperature at. Before u = u v =/s After v V K 4 {for sall aplitude} 7. B K T K T putting = A, T A T A kg 5 kg kg onserving the lienar oentu U 5 5 V u 5v () coefficient of restitution u v () Solving () and () u / s,v / s e v u 5 kg Moentu of the SP = + 5 = kg-/s V c 5 / s J KE of M KE of SP 4.5J 7. In the region of oscillation < E < V 7. D For > X V() = V = constant dv F acceleration = 7. A Since B > B T (B ) < T (B ) and beyond T, B doesn t affect R. 74. B As < 5 Tesla < 7.5 Tesla 75 K < T (5) < K certainly for series case, for parallel case, given J.5J solving R 4 T T E A T r T E A T r T solving we get E E i,j R R R E E E i,j R R R E 9 Region of oscillation V R a a a a cos 9 R = 5 /Paper I/Solutions Page

12 78. 4 Effective force constant of the wire K W W solving n = 4 K W YA l 8. 7 For reflected frequency fro car V V f V V f by differentiating w.r.t. V df V dv V V (V : velocity of the car) f A d 6 d B 5 LT.E..MS d MS d LB MS d 6 LT.E. 6 L B 6 gp ge cp pe v e gr M 4 R M R 4e / GM Also g R GM R 8. 6 Thus dv df Given V >> V so dv v u f V V V f V df 7 f When u = 5 c, v = c, = 4 when u = 5 c, v c, KEP KE qv qv P p p K p h/ K qv.4 h / K K qv p p p Using above we get v ep KM/ sec 8. 8 Let = ass of ice (in g) heat required to raise the teperature of ice fro 5 to is 5 heat required to elt g of ice =.6 5 z given + = 4 solving = 8 g Page /Paper I/Solution