MUMBAI / AKOLA / DELHI / KOLKATA / LUCKNOW / NASHIK / GOA / BOKARO / PUNE / NAGPUR IIT JEE: 09 TW TEST (ADV) DATE: 04/03/8 ANSWER KEY PHYSICS : GRAVITATION. (D). (A) 3. (B) 4. (B) 5. (C) 6. (ACD 7. (AD) 8. (AD) 9. (AD) 0. (BD). (BCD). (C) 3. (ABCD) 4. (C) 5. (ABD) 6. () 7. (0) 8. (3) 9. (8) 0. () CHEMISTRY : VOLUMETRIC. (C). (A) 3. (C) 4. (B) 5. (A) 6. (CD 7. (ACD) 8. (AC) 9. (ABC) 30. (CD) 3. (AB) 3. (BCD) 33. (ACD) 34. (ABD) 35. (BCD) 36. (4) 37. (5) 38. () 39. () 40. (3) MATHEMATICS : LIMIT 4. (B) 4. (D) 43. (A) 44. (B) 45. (A) 46. (ABCD) 47. (BC) 48. (ABC) 49. (AD) 50. (BC) 5. (AC) 5. (BC) 53. (ACD) 54. (AB) 55. (BCD) 56. (3) 57. (0) 58. (4) 59. () 60. () CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI #
MUMBAI / AKOLA / DELHI / KOLKATA / LUCKNOW / NASHIK / GOA / BOKARO / PUNE / NAGPUR IIT JEE: 09 TW TEST (ADV) DATE: 04/03/8 TOPIC: GRAVITATION. (D) From modified Gauss s theorem for gravitatio. (A) 3. (B) 4. (B) 5. (C) SOLUTION r r r r k E.ds 4G dv E4r 4G 4 r dr r r 0 r 0 get E = costat as E is costat, so the Potetial V Edr U m V V f i will be proportioal to r GM GM 3 GM 3 GM 3 U m m mr mrg 4R R 4 R 4 R 4 6. (ACD The readig of the beam balace will be idepedet of effective g, so W b = W' b but the readig of the sprig balace will Proportioal to g effactive At equator due to cetrifugal force of earth, g effactive is less so Readig of sprig ballace is less W s < W' b 7. (AD) 8. (AD) 9. (AD) 0. (BD). (BCD) CENTERS : MUMBAI / DELHI / AKOLA / LUCKNOW / NASHIK / PUNE / NAGPUR / BOKARO / DUBAI #
. (C) 3. (ABCD) 4. (C) 5. (ABD) 6. () GM g R dr R 7. (0) dg g dr R dg % % g 8. (3) The acceleratio due to gravity at earth s surface is g ad at a distace R from earth s surface it is g/9. Hece 9. (8) The gravitatioal field at ay poit o the rig due to the sphere is equal to the field due to a sigle particle of mass M placed at the cetre of the sphere. Thus, the force o the rig due to the sphere is also equal to the force o it by a particle of mass M placed at this poit. By Newto s third law it is equal to the force o the particle by the rig. Now the gravitatioal field due to the rig at a distace d = 3a o its ais is 0. () E Gmd 3Gm a d 3/ 8a The force o a particle of mass M placed here is F = ME 3GMm 8a This is also the force due to the sphere o the rig. Gm( M m) F r df For maimum force 0 dm d GmM Gm dm r r 0 M M m 0 m CENTERS : MUMBAI / DELHI / AKOLA / LUCKNOW / NASHIK / PUNE / NAGPUR / BOKARO / DUBAI #
CENTERS : MUMBAI / DELHI / AKOLA / LUCKNOW / NASHIK / PUNE / NAGPUR / BOKARO / DUBAI # 3
MUMBAI / AKOLA / DELHI / KOLKATA / LUCKNOW / NASHIK / GOA / BOKARO / PUNE / NAGPUR IIT JEE: 09 TW TEST (ADV) DATE: 04/03/8 TOPIC: LIMIT SOLUTION 4. (B) lim lim r r r r d ta 0 4 0 4. (D) ta left eighbourhood of 0 f si right eighbourhood of a lim f 0 & lim f 43. (A) 44. (B) 0 0 Limit does ot eist. si k k 3 3 lim lim si k lim lim si k 0 0 k k si k k 3 lim 0 si k lim e k lim... e 3 lim e 6 3 e lim a 0 b cos b CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI #
Ad lim 0 a cos 45. (A) a a 4 cos 4 cos cos / 8 4 cos cos 8 6... cos L.cos.cos...cos Multiply divide by si si si L If. L si L 46. (ABCD) k a k lim a lim.lim k a lim.lim l a.lim k CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI #
47. (BC) lim f 0 if k 0,, l a if k 3 if k 4ad a 0, if k 4ad a 0 0 f lim 0 f Ad lim 0 f lim 0 lim f 0 0 Ad lim f 0 48. (ABC) (A) limit does t eist at =,, 3, 4, 5 0,6 (B) f()= for all 5 3 7 5 9 (C) limit does t eist at,,,,, 6 6 6 6 6 6 (D) if a=0 fuctio is discotiuous at may poit but limit still eist 49. (AD) si lim si lim 0 0 si 0 is ot i domai ad lim si e 0 0 &0 both are ot i domai lim F lim l F e l Pa P a... P a lim e P a l a... P a l a lim Pa P a... P a e l a e a Similarly 50. (BC) 5. (AC) lim F a CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI # 3
(A) e lim sec lim cos ( ) ( ).cos lim cos lim cos 0 4. ( ).cos e e e / lim () 4 lim (C) e e 4 (D) e e e 5. (BC) lim 3 (cos )(cos e ) 0....... lim! 4! 3! 5! 0 4 5 lim... 4 0 4, l / 53. (ACD) lim f ( ) 0 4 3 5 3 3 4 lim a0 a a a3 a4... 3 0 If limits eists the a 0, a 0, a, a 0 3 54. (AB) lim l({ sih}.{cosh} ) lim l(( sih).(cosh) ) f (0 ) l h 0 { sih}.{cosh} h 0 ( sih).(cosh) lim l({sih}.{cosh} ) lim l((sih.cosh) ) f (0 ) h 0 {sih}.{cosh} h 0 (sih.cosh) l si h. cos h lim lim l({ sih}.{cosh} ) f h 0 h 0 { sih}.{cosh} si h. cos h l si h. cos h lim lim l({sih}.{cosh} ) f h 0 h 0 {sih}.{cosh} si h. cos h f (0 ) f (0 ) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI # 4
55. (BCD) lim lim a / f ( ) b / a lim a if 0 a b,, f ( ) 0 b b a lim a fa b 0,, f ( ) b b lim a lim lim ifa b, f ( ) e e a 56. (3) 4 5 6 7 5 4 3 lim....... 7 8 9 0 3 lim0 57. (0) k 0 / lim cos lim cos 0 lim cos 0 0 58. (4) 3 k k 4k k k k k k k k k () k k Usig partial fractios ad summig the telescopig series, we get k k k k k 59. () k k k / a () Therefore, the Squeeze Theorem ad () yield lim k k 60. () Telescopig, we have CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI # 5
3 k 8k 6k lim lim k k k 4k 3 k k k k3 k 3 lim k k k k 3 5 lim 3 5 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI # 6