TEST - 4 (Paper-I) ANSWERS PHYSICS CHEMISTRY MATHEMATICS

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1 TEST - 4 (Pper-I) NSWERS PHYSICS CHEMISTRY MTHEMTICS. (4). (). () 4. () 5. () 6. (4) 7. () 8. () 9. (). (). (). (). () 4. () 5. () 6. (4) 7. () 8. (4) 9. (). (). (). (). () 4. (4) 5. (4) 6. () 7. () 8. (4) 9. (4). (). (). (4). () 4. () 5. (4) 6. () 7. () 8. () 9. () 4. () 4. () 4. () 4. () 44. () 45. () 46. () 47. () 48. (4) 49. () 5. () 5. () 5. () 5. () 54. () 55. (4) 56. () 57. (4) 58. (4) 59. () 6. () 6. () 6. () 6. () 64. () 65. () 66. () 67. () 68. () 69. (4) 7. () 7. () 7. () 7. (4) 74. () 75. () 76. () 77. (4) 78. () 79. () 8. () 8. () 8. () 8. () 84. () 85. () 86. (4) 87. (4) 88. () 89. () 9. () /

2 PRT - (PHYSICS). nswer (4) Spring force : k ( ) g R( ) R g Tol force long cenre of circle g cos + g cos g N B. nswer () ˆ R ˆ R i i ( jˆ) p. nswer () ˆ i ( j ˆ ) R l Ml u 4 l/ l/ M d d d l 4. nswer () 9u 8Ml Le poin P oes wih elociy nd is direcion kes ngle θ wih l. Now, sin θ nd sin(α θ) Soling (i) nd (ii), we ge 5. nswer () cos sin f f R R f f R f R or pure rolling, Now, s R s 6. nswer (4) BC C R 8 g OC sin OC cos O 9 θ f C R 45 θ...(iii) OC BC C sin cos lso, co lso, BC C B...(iii) Soling equions (i), (ii) nd (iii), we ge 7. nswer () u u ro equions (i) nd (ii), we ge u cos 9...(iii) Thus, 8. nswer () Velociy of BLL w.r.. ground (fer collision) ( u) + u ΔK K f K i [( u ) ] u ( u) B /

3 9. nswer () The ipc should e inelsic. u u KE ( ) 4 s 9. nswer () rel ( + M) M rel d M d M. nswer () y CM y rel N y s. nswer () WE heore, work done 4. nswer () K μ ( + )g N d Here, fricion (liiing nd kineic) eween kg nd horizonl surfce is N. 5. nswer () P gh W V gh gh P gh V V R 4R 5 R. nswer () The lock lees he surfce 5 s. Now, d d d dh d h 5 5 d 6. nswer (4) 5 kw [ ][ ] Hence, 4 d 7. nswer () orwrd journey : s y s, y 5 s /

4 Bckwrd journey : 5 5 s 4 e 5 e 8. nswer (4) 5 e C or no skidding : r f l N G N g g rg s s r Here, is he iu sfe speed for no skidding. or no oerurning : N + N g l l N f h N ro (ii) nd (iii), N fh g l l N When fh g g h r gr h f...(iii) is he iu sfe speed for no oppling. gr If, hen rg s h h 9. nswer () nswer () ω R f s α R f R f R ω ω + α...(iii) ωr ω R + αr...() f R R or pure rolling,. nswer () ω R f ω f f R R f f R 5 5 f 9 g g T I I T 6 Now, I...(iii) Soling 6 s. nswer () α I 5 α R 7 T αi g 7 4L rd/s 4/

5 . nswer () KE ou P T KE ou P + R KE ou P T KE ou poin P, ecuse poin P rnsles wih V CM R KE ou poin P 4. nswer (4) I 5 R IP R ICM 6 R R 6 9 R 8 I I I R CM CM 8 R I I R R CM 8 I R 5 5. nswer (4) p [ iˆ jˆ] [ iˆ jˆ] [ iˆ jˆ]ns ll p ( iˆ jˆ)ns surfce Coponen of ( iˆ j ˆ) long ( iˆ j ˆ), is 4 ˆ ˆ 4 [ i j] 5 5 Speed of pproch 4 5 Coponen of ( i ˆ j ˆ ) long ( iˆ j ˆ), is ˆ ˆ [ i j] 5 5 Speed of seprion e nswer () Ne work done y fricion is non-zero. 7. nswer () Liiing fricion is insufficien for rolling of disc nd ring. 8. nswer (4) orce of fricion ecoes he inernl force for he syse of ruck nd erh. 9. nswer (4) K ( )( e ) ( ). nswer () Tol work done upon spring is k. PRT - B (CHEMISTRY). nswer () ccording o Ben's rule, ore elecronegie o prefers o occupy il posiion so s o he iniu repulsion. I. P II. P μ μ III. CH C P CH CH C μ IV. CH C P CH CH C μ. nswer (4) YS hs les lue of K sp. Precipiion occurs when ionic produc > soluiliy produc.. nswer () KE per ole RT 9 cl 4. nswer () Enropy is se funcion ΔS (X P) ΔS (X Y) + ΔS (Y Z) + ΔS (Z Q) + ΔS (Q P) e.u. + 4 e.u. + ( e.u.) + e.u. 4 e.u. 5/

6 5. nswer (4) 6 On incresing eperure, wer dissocies ore nd hence lue of K w increses fro 4 o. Thus, ph scle now ecoes sller nd neurl ph is 6. lso, ph + poh So, poh lso ecoes 6. Bu, wer will sill rein neurl. 6. nswer () He gs shows posiie deiion only. high eperure nd low pressure, rel gs ehes like n idel gs. or oher gses odere pressure rnge, rcie forces doine. 7. nswer () E φ + KE hc hc hc hc 8. nswer () n + p 45 n.4p On soling, p nd n 4. So, oic nuer of o is. Elecronic configurion of ion ( ) s s p 6 s p 6 So, oueros shell of ion hs 8 elecrons. 9. nswer () 4. nswer () Spin qunu nuer is no deried fro soluion of Schrodinger we equion. Cr(Z 4) Tol spin 6 nuer of unpired e Oril ngulr oenu of n e in 5s 4. nswer () ll ( ) ( ) S.65 4 ol/l K sp 4S 4 (.65 4 ).8 poh [OH ] 5 [Mg + ][OH ] K sp.8 [Mg + ] ( ) 4. nswer () When CH OH is dded, equiliriu shifs ckwrd nd new equiliriu se is eslished. ddiion of clys or iner gs consn olue will he no effec. 4. nswer () H HO HO C HB H O H O B C Since α nd α re ery sll, so C α C nd C α C Xe I K (H) ( ) C Boh re liner olecules. K (HB) ( ) C 6/

7 Diide, C 4 C [ ]. [B ] nswer () In solid for, Xe 6 eis s Xe nswer () P (V ) RT V If '' is negligile, hen 5. nswer () P(V ) RT PV RT + P Xe d hs wo nodl cones. z Hyridision se is sp d. 5. nswer () 45. nswer () 54. nswer () I 8 I onds re presen. I 55. nswer (4) CCO CO CO (X) XisCO 46. nswer () N CO + H O + CO NHCO oiling poin, eperure rein consn. So, he ol energy rein se for he. 47. nswer () Since K p depends only on eperure, if he olue is douled, equiliriu is disured nd ore producs would e fored. So, α will chnge while K p will no chnge. 48. nswer (4) Pure eleen hs zero sndrd enhlpy of forion in is sndrd sle se. 49. nswer () Molr he cpciy is inensie propery. 5. nswer () HO(s) HO() l 7 K is eling poin of wer nd he process is equiliriu, so ΔG while ΔS +e, since solid is chnged o liquid phse. Moles of CCO oles of CO Moles of N CO ole Nuer of oles of NHCO fored is. ole ecuse CO is liiing regen. 56. nswer () I is he drwck of firs lw of herodynics h i fils o predic he direcion of recion king plce. 57. nswer (4) KE per olecule kt, where 'k' is Bolzn's consn. 58. nswer (4) Dipole oen of CH > CH. 59. nswer () 6. nswer () 7/

8 PRT - C (MTHEMTICS) 6. nswer () cos cos 4 4 n cos 4 6. nswer () sin sin y sin sin y ( )sin ( )sin, y sin sin y sin sin y co + co y 6. nswer () n ( )(cos cos y) sin sin y ( )co y n n sin θ + cos θ n ( ) ( ) ( ) ( ) 64. nswer () ( n θ ) sec θ n θ 8 n θ + n n 65. nswer () sin 5 cos ( sin 5 cos ) 4 < {} ( {}) (, ) is 66. nswer () s, ( + ) cos C + ( + c) cos + (c + ) cos B + + c (Using projecion forule) 67. nswer () Period of cos π{} Period of sin π{} lso, sin cos[] cos sin[] sin( []) sin{} Is period So, required period LCM of 68. nswer () α + β 5, αβ 5 ( ) ( ) ( ) ( ) [( )( )],, ( ) ( ) ( ) ( ) ( ) ( ) 6( ) ( ) ( ( ) ) 5 5( 5) ( 5) 6( 5) ( 5) ( 5 5) ( 9) nswer (4) sin cos (sin cos ) 7 sin cos sin cos (sin cos ) 7 sin cos sin cos 8/

9 ( sin ) 7 sin sin ( + )( + ) (7 ), where sin (sin ) 8 7 sin nswer (), 8 ±, 8 n( B B) n() n() n(b) n(b) 7. nswer () (4 ) y y + y y 6 + y ( 4 y + y 4 ) y y ( + y ) y ( + y ) y y y y y 7. nswer () sin θ + sin θ + sin θ sin θ cos θ sin θ cos θ (sin θ sin θ cos θ) + (sin θ sin θ cos θ) + (sin θ cos θ) (sin θ + sin θ + )(sin θ cos θ) Since sin θ + sin θ + sin θ cos θ n θ ro figure, he nuer of soluion is. π y π y 7. nswer (4) Siplifying he equion ecoes, ( + ) ( ) ( + ) The su of he coefficiens. α is roo. ( ) Produc of he roos αβ ( ) β 74. nswer () sec ( + ) + n ( + ) + n( + ) nd n,, (, ),,,, sisfying he equion. 75. nswer () Le he 7-digi nuer is cdefg 9 re he ordered pirs, c nd d cn e seleced in wys. in 9 wys. Now, when,, c nd d re fied, e, f nd g will heseles e fied nd cn e seleced in wy only. Nuer of 7-digi nuers nswer () Required diisors re, 5,, 7, 7, 5, 7 5, 5 7, 7, nswer (4) z z zz is purely iginry nuer. rg z z zz is Now, rg z is 5 9. or z lies in second qudrn. I(z) is posiie. z z rg zz 9/

10 78. nswer () cos45 + icos5 cos( ) icos cos(645 )... ers i ( i ) i ( i ) 79. nswer () + c c cos ( c) c cos + c ( c) c sin ( c) 4k sin 4k sin i.e., k sin 8. nswer () s s s s c s s s s c s, c ( s )( s c) s( s ) c n c 8. nswer ()... ers n, i( ) + ( ) + iy y ( ) 4 y [( ) + ( ) ( )] 4 + y (4) (4) 4 8. nswer ()!!!!!!! 9!! 7! 5! 5! 7!! 9!! [ C C C ] 5 C7 C9! 9! 9, n 8. nswer () 5 + n B C s, r 4R sin sin sin R B C 4R sin sin sin nswer () C B B sin cos cos 9 C B C sin cos sin 9 C C sin sin C 7 cos C sin 9 9 Le +, + ( )( ) i.e., + + So, ( )( )( ) (, ) (, ] [, ) 85. nswer () We re gien n By ril, 7, n 7n ( ) /

11 86. nswer (4) Vere of f() will e Cse : >, hen iniu lue of f() is. f() 5 + or which is no possile. Cse : <, hen he iniu lue is f(). f() 5 + or is possile Now, if < <, hen he iniu lue is f or 4 Su of lues of 87. nswer (4) is possile 4 f Seen- is correc s period of sin is π nd hence period of sin(π) is. Siilrly, period of cos is π nd hence period of. 88. nswer () Tol nuer of funcions n. 89. nswer () Seen- is correc nd seen- is flse. cos(sin ) cos cos sin cos n, n I sin n sin n RHS > for ll n I The equion does no possess rel roos. Seen- is rue, seen- is rue u i is no he correc eplnion of seen-. 9. nswer () z z z z k is () Hyperol if k < z z () Two rys if k z z (c) Epy se if k > z z cos( ) is. Hence, seen- is no rue s LCM of does no eis. nd Here, z z, k Seen- represens rnch of hyperol seen- is flse. /

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