UNIFIED COUNCIL. An I SO : 2015 Cer t i f i ed Or g a n i sa t i o n. Test Assess Achieve

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UNIFIED CUNCIL An I S 9 : 5 Ce t i f i ed g a n i sa t i o n Test Assess Achieve NATINAL LEVEL SCIENCE TALENT SEARCH EXAMINATIN (UPDATED) CLASS - (PCM) Question Pape Code : UN449 KEY. A. C. C 4. A 5.

1 UNIFIED CUNCIL An I S 9 : 5 Ce t i f i ed g a n i sa t i o n Test Assess Achieve NATINAL LEVEL SCIENCE TALENT SEARCH EXAMINATIN (UPDATED) CLASS - (PCM) Question Pape Code : UN449 KEY. A. C. C 4. A 5. B 6. B 7. D 8. C 9. D. B. B. B. B 4. D 5. A 6. D 7. C 8. A 9. D. A. D. B. B 4. B 5. C 6. D 7. B 8. D 9. C. D. C. A. B 4. C 5. C 6. A 7. C 8. B 9. D 4. D 4. D 4. A 4. D 44. C 45. A 46. C 47. A 48. A 49. C 5. C 5. A 5. B 5. B 54. A 55. D 56. D 57. D 58. B 59. D 6. Del SLUTINS MATHEMATICS. (A) Given, f() = (p n ) /n, p > Now, f[f()] = f[(p n ) /n ] = {p(p n ) /n n } /n = ( n ) /n =. (C) Given, f() = 4 Now, f() = f() () () 4 = () () =. (C) 4 = 4 = = {} 5 cos cos cos cos,. cos cos. cos cos cos 65

2 4. (A) Given equation is (tan ) (cot ) = (tan cot ) tan.cot = tan cot = 4 tan. tan ( / tan ) = (tan ) tan = 5. (B) We have, (tan ) = tan tan = = 4 4 =, = tan 4, tan 4 =, A 4 T A 4 T T A A 4 AA 8 T T A (A A ) (A A ) 4 A A = B C 6. (B) Given, 7. (D) C.5 = C T.5 A A A... () Now, A = ( ) () = C =, C =, C = C = () =, C =, C = C =, C = (), C = A... () Fom Eq. () & () A A = A 5 Let Apply opeations R R R, R R R, we get 5 Let Again, apply opeation C C C C C C, we get

3 8. (C) 6 4 Epand along R, we get = () (44) (8 ) (46) = 4 () = 4 6 = b ab b c bc ac Let ab a a b b ab bc ac c a ab a b(b a) b c c (b a) a(b a) a b b (b a) c(b a) c a a (b a) Taking common (b a) fom C and C, espectively b b c c (b a) (b a) a a b b b c (b a) a b = 9. (D) We have, c a 4 D c c a a = 4 (5 9) (7 7) ( 45) = 4 84 = 5 D 5 7 = 4(7 6)5(7 7) ( 9) = 4 4 = 4 5 and D 5 9 = 4( ) ( 9) 5( 45) = 8 7 = So, the given system of equations has infinite numbe of solutions.. (B) Given, f() = sin, if, if sin, if Since, f is continuous at =, LHL = f() = RHL Now, LHL = f() RHL = sin sin h h h h and f() = sin sin h h Fom eq. (i), we get = 5 (5 9) (7 6) ( ) = 4 = 4 5 D 9 7

4 . (B) LHL RHL f() ( h) ( h) h h LHL = RHL = f() Hence, f() is continuous at =. LHD f() f() f() f() ( ). (B) Given, LHD RHD RHD Hence, f() is not diffeentiable at = 5cos sin y = tan cos 5sin y = tan 5 tan 5 tan [divide by cos in denominato and numeato] 5 tan y = tan 5 y = Diffeentiating both sides w..t. dy = d. (B) Given, dv dt = cm /s Volume of sphee, V = 4 n diffeentiating w..t. t, we get dv 4 d dt dt d 4 dt 4 4. (D) Given, d cm / s dt V Volume of sphee 4 V = n diffeentiating w..t., we get du 4 4 d du V 4 d Relative pe cent eo V 4 = V 4 = = (.) 6 =.6 5. (A) Let V be the volume, be the adius and h be the height of cone at any time t. Then A B 4 m We have, dv dt = m /min We have to find : dh dt Clealy, C h D 6 m when h = m V h... (i) Since, BD BAC

5 [By AA-similaity citeion] B D BA AC 6 4 h h Now, fom Eq. (), we get 4 V h h h 7 dv 4 dh h dt 7 dt h 4 dh 7 dt dh 7 dt 4 h 6. (D) Let i = = = dh 7 m / min dt h e sin d cos sin e d cos cos sin cos e d cos cos = e sec tan d = e sec d e tan d = e tan e tan d e tan d = e tan C 7. (C) Let f() = Now, f() = = So, f() is an odd function f() f() d d 8. (A) Let I = Put sin d d cos d I cos d sin cos cos d d = C sin C 9. (D) Given cuve is y = = y X and line y = Y y = y = = Y = The intesection point of cuve and line is = X = Now, b 4ac = 4 < Hence, thee is no point of intesection Requied aea = (y y ) d [( ) ( )] d [ ] 5

6 . (A) I = 8 [4 4 ( )] 4 4 [ ] = 6 = 9 /4 tan cot d /4 /4 sin cos sin cos d sin cos /4 cos d sin (sin cos ) d sin cos /4 sin cos d (sin cos ) Let sin cos = t I (cos sin ) d = dt /4 dt t /4 [sin t] /4 [sin (sin cos ] sin sin cos sin (sin cos ) 4 4 [sin sin ( )]. (D) Given diffeential equation is dy y e d dy y y/ e d y/ It is a homogeneous diffeential equation. Put y = v dy v dv d d dv v v/ v X e d 6 v X dv X e d dv v e d v v e dv d n integating both sides, we get e v = log c e y/ = log c Given, y() = e / = log c = c c = e y/ = log = log e y/. (B) We have, dy d a by c d a by c dy d a = by c dy Hence, the above equation is a linea diffeential euqation in.. (B) Let the coodinates of fou points P, Q, R and S be (, 4, 5), (,, 4), (4, 5, ) and (, 4, ) espectively. Now, equation of line PQ is y 4 z y 4 z 5 4 Equation of line RS is 4 y 4 z y 5 z v (say)... (i) (say)...(ii) Let (, 4 4, 5) and ( 4, 5, ) be the points on line (i) and (ii), espectively. Since, both lines intesect at a common point, then = 4 = 7... (iii)

7 and 5 = = 9... (iv) n subtacting Eq. (iv) fom Eq. (iii), we get = n putting the value of in Eq. (iii), we get () = 7 = So, equied point of intesection is (, 4, ) i.e., i 4j k 4. (B) Given, a = i j k [(a i) j] = a = i j k = ( 4) 6 [(a.j)i] 5. (C) Let a = i j 4k = A b = i 4j k = B and c = 4i j k = C and AB = B A = i j k BC = C B = i j k CA = A C = i j k Now, AB = 4 6 BC = 4 6 and CA = 4 6 Since, the length of all thee sides ae equal. So, the taingle is an equilateal tiangle. PHYSICS 6. (D) The points, and 4 have same potential and the points 5 and 6 have same potential. Theefoe, the cicuit may be educed as shown below. 4 5 Thus, R 6 = 4 7. (B) Magnitude of the othe field o. sin 5 o sin T 8. (D) Speed of infaed adiation in vacuum = 8 m s v = fl 8 f( 5 ) f =.5 Hz 9. (C) Intensity at the cente of bight finge, I I I I I cos I I I I 4 I Intensity at a point distant b/4 (with a phase diffeence = p/4 = p/) is I' I I I I cos o I' I I I I' I I 4I I' I 7

8 . (D) The enegy of the incident photons is E = hf = (4.4 5 ev. s) (7. 5 Hz) = ev Since E > f, photoelectons will be poduced, with maimum kinetic enegy K ma = E f = ev 6 ev = 4 ev. (C) No.emf is induced in the paallel hoizontal wies. Equal emf of same polaity is induced in the two paallel vetical wies. Hence, induced cuent is zeo as two equal and opposite emf s ae pesent in the loop.. (A) R = W, R = 7 W E = V, E I E ms ms 4.4 V Ems Total esis tance A 7 Powe developed acoss R = I ms R (.44) 7 =.4 W.. (B) Magnetic field due to ADB is 4. (C) B i a (Pependicula to pape outwads) and magnetic field due to ACB is B i a (Pependicula to pape inwads) i Bnet B B a (Pependicula to pape inwads) q q F 4 o q q = 4 F Setting F =.75 N and = m, we get q q = (i) Also, q q = mc = 6 C..(ii) Fom the equations (i) and (ii), it can be obtained that q = 5 6 C and q = 5 6 C 5. (C) m =.54 kg, I = A, t =? m t, z E M z. I F pf Atomic mass of coppe = M = 6.5 kg Valency = p = Faaday = F = 965 C M 6.5 z.9 kg C pf z. I.9 7 m.54 t 77. second. 6. (A) R (4 l / d ) Theefoe, R µ l/d Hence, R :R :R 4 4 : : : : : Cuents ae in the invese atio of esistances i : i : i : : 54 : 64 : 75 4 o i = 54k, i = 64k and i = 75k Whee k is the common atio. But i i i = 5 54k 64k 75k = 5 o k = 5/ i = A, i = A, i = A (C) In steady state the following chages will appea on diffeent faces of the plates. q q q q Netchage on cental plate is q. Thus, q chage will flow though the switch. 8

9 8. (B) The distance of closest appoach is given by Z e 4 m u Hee, Z = 79; m u = 5 MeV = 5.6 J We know, e =.6 9 C 79 (.6 ) = m 9. (D) The lens make s fomula is : nl f n R R m Whee n L =Refactive inde of lens and n m = Refactive inde of medium. In case of double concave lens, R is negative and R is positive. Theefoe, R R 4. (D) Seies will be negative. Fo the lens to be diveging in natue, focal length f should be negative o nl n m should be positive o n > n L m but since n > n (given), theefoe the lens should be filled with L and immesed in L C C C S Effective capacitance C 5 F CS F n 5 Paallel C P = C C.... Effective capacitance = C P = nc = 5 5 mf = 5 mf. CHEMISTRY 4. (D) 4Zn HN 4Zn (N ) NH 4 N H. 4. (A) Hee, Mn is in 7 state. The atomic numbe of Mn is 5. So, the electonic configuation of Mn 7 is s s s p 6. Thus, thee is no electon in its d obital. 4. (D) Difluooacetic acid is the stongest acid out of the given acids. Hence, it ionizes maimum and theefoe, has highest electical conductivity. 44. (C) Ethyl chloide and acetyl chloide eact with alc. KCN by nucleophilic substitution eaction while benzaldehyde undegoes benzoin condensation. KCN C Cl alc. C CN KCl KCN CH CCl alc. CH CCN KCl KCN C 6 CH alc. C 6 CHHCC 6 Thus, only chloobenzene does not eact. 45. (A) NaCl changes into CsCl type (6:6 to 8:8) on applying pessue. 46. (C) Addition of an electon to a negatively chaged species is not a favouable pocess. So, enegy is absobed in such a step. 47. (A) nly Na eacts with both ethanol and phenol. In contast, NaH/I eacts only with ethanol while neutal FeCl and B /H eact in the pesence of conc. H S 4 with phenol only. 48. (A) Numbe of moles of acetic acid.6 ml.6g ml = 6 g mol =.6 mol = n.6 mol Molality = ml g ml =.6 mol kg DT f =.86 K kg mol.6 mol kg =.97 K van t Hoff facto (i) bseved feezing point.5 K Calculated feezing point.97 K =.4 9

10 49. (C) Heats of adsoption in physisoption lie in the ange 4 kj mol. Condensation 5. (C) C 6 CH = H NC 6 Reaction C 6 CH = NC 6 H (Benzylideneaniline) 5. (A) Both Zn and Hg eact with ai (oygen) on heating to fom Zn and Hg. All othe popeties ae shown by Zn but not Hg. 5. (B) S N eaction leads to acemisation. 5. (B) Rate = ate of disappeaance of A pe mole. A t The negative sign simply indicates the fall in concentation of A. Thus ignoing the negative sign, the ate of the eaction is.5 mole/lite/minute. 56. (D) CRITICAL THINKING 57. (D) If Geetha is olde than Manish and Rohan is olde than Geetha, then Manish has to be the youngest of the thee. Choice b is clealy wong because Rohan is the oldest. Thee is no infomation in the paagaph to suppot eithe choice a o choice c. 58. (B) Clealy, damage to cops due to high tempeatue may have esulted in a shot supply fo vegetables and hence an incease in thei pices. 59. (D) The women is the mothe of shaikh ganddaughte. Hence, the woman is the daughte-in-law of shaukh. 6. (Delete) 54. (A) S H N H S 4 N Catalyst Catalyst 55. (D) CH C HNa C H Sod. bicabonate CH C Na H C H C H Thus, C of C comes fom bicabonate. THE END

Physics 2212 GH Quiz #2 Solutions Spring 2016

Physics 2212 GH Quiz #2 Solutions Sping 216 I. 17 points) Thee point chages, each caying a chage Q = +6. nc, ae placed on an equilateal tiangle of side length = 3. mm. An additional point chage, caying