JEE-Main. Practice Test-3 Solution. Physics

Transcription

1 JEE-Main Practice Test- Sutin Phsics. (d) Differentia area d = rdr b b q = d 0 rdr g 0 e r a a. (d) s per Gauss aw eectric fu f eectric fied is reated with net charge encsed within Gaussian surface. Which is sphere here. S b shifting q wi nt change eectric fied at S.. q (c) Ging fr 8V end t V end 8 q () c c q q ging fr v end t 8v end 8 0 C C ( ) t junctin net charge = 0 q q q () Sving (), (), () q 0. (a) E E ess current wi fw in circuit in secnd case. S nu pint wi be shifted eft. 5. (c) Charge n differentia area dq = d d Current, di = d T 0dI r db = / ( ) (Bit sa vart aw) B db 0 r d 0 r / r. (c) K. E. = W eectric [n eectric fied changes speed] 5 qe 0 0 v 0 0 = qe0 7. (a) Here ptentia energ is due t rtatin f disc. U K.E. nw a w a V q q ne e ac I bd Ib a c 8. (c) Q Bd 0 0 B n c c d B 0 b c a d t induced n I0 dt c dt I b c a Q I /R 0 0 n R c I 9. (c) B 0 I B 0 a ( r a) r r d d B 0 I a r 0 Ia dr 0 Ia v induced. = dt dt r dt r B- /- M, Brahanand Cn, Durgakund, Varanasi; Page

2 dv dv d dv dv 0. (c) a vv dt d dt d d v dv Since bd started in directin s a > 0 hence wi be psitive d. (b) puing has a cpnent ang nra which reduces N. Hence frictin wi be reduced.. (d) s is increased trque increases thus anguar acceeratin increases. s F is cnstant cpnent f reactin frce n hinge des nt change.. (c) apping parae ais there I I 0 M S graph wi be (c). (b) In first fa f 00 air frictin was ess than gravitatina frce. In secnd fa after 00 frictin f air baances weight. S air frictin wi be re than first 00 s des wrk. 5. (a) In secnd case kinetic and ptentia energ bth is cnverted int thera energ. In third case kinetic energ is cnverted int ptentia and thera energ. Where as in first case kinetic energ is cnverted int thera energ. dq dt. (b) K since rds are cnnected in series sae heat current wi fw. dt d ( 00 T) (T T) K(T 8) k kq ( 00 T). 5 (T 8) (T T ). 5 (T 8) T T. 5T 7. 5 T T 7 T 7 T. 5. ( T).T T T 7 00 T = T - T T.. 5 T 0 C dq dt 7. (c) K dt d rea acrss B is re than acrss CD s teperature difference acrss B is ess than that acrss CD t aintain sae heat current. v v 8. (a) w v g vg = g T w g 9. (c) Fr n sipping a bcks ve tgether that is with sae acceeratin. f N,f N here f is frictin between bck and and f frictin between bck and. F.B.D. f k f f f f k = a = a (bck and ves tgether) k k = (d) When pseud frce w equas weight the cnditin f questin is fu fied. S g T g. (d) Since O is a rigid supprt s right haf part f puse wi be refected with ppsite phase. B- /- M, Brahanand Cn, Durgakund, Varanasi; Page

3 v u. (b) s per Dpper s aw 0 v us 0 v 5.5 = v = v 5 0 v. (b) sin i, sin r h 5 h h h 5 sin sin. (d) 0 sin sin cs sin = (c) nd (Distance f bright fringe fr id pint) d hc. (a) in E h. 0 Js (after cacuatin) 7. (b) Fr secnd segent width is re than first s re phtn is present in secnd segent. 8. (c) N.N. M 9. (c) Effective area after bending beces cs and change in entu wi be ang nra. 0. (d) New intensit wi be I cs and new frce I cs = I cs. cs = F.cs c c P n s graph wi be hperba B- /- M, Brahanand Cn, Durgakund, Varanasi; Page

4 Cheistr. (a) CH C C C g CH C C CH gc C. (c) NO is paraagnetic and having ne unpaired e.. (b) buk grup increases reactivit decreases due t steric hinderence.. a (d) P (V b) RT V t w P, vue bece ver high is, V b V a a S P V RT PV RT V V Q PV a PV RT K V RT VRT 5. (c). (d) kai and akaine earth eta can nt be btained b eectrsis in an aqueus sutin f its sat. 7. (c) Fr Rb s s p s p s d p 5 s n = 5, = 0, = 0, s = 8. (a) 9. (c) Fr First sutin 5s Specific cnductance (k) =. S, R 50,M 0. k R Fr nd Sutin R 80, k 0. (c). (b) S K New ar cnductivit ( ) C S e. (c) Effective arit = i arit H OH C 5 Mg(PO) 5 0. KBr NaPO Effective arit f each sutin are sae. S sae stic pressure.. (c). (d) Ligand fied strength Energ f ight absrbed. Energ. bue ( L ) > Green ( L ) > Yew ( L ) > red ( L ) S rder is - L L L L B- /- M, Brahanand Cn, Durgakund, Varanasi; Page

5 5. (d) Fe (OH) is psitive s s caguated b negative charge eectrte. Greater the agnitude f ive charge, greater wi be caguating pwer.. (d).e. f NH = eq. Of H SO.e. f NaOH = 0 0 = N. f equivaent f NH Me f N = 0.0 Wt. f N = 0.0 = 0. g 0. % f N = (b) 0 P the pairing wi nt ccur fr d. S t g eg. LiH c.koh 5 8. (a) CHCOOH CHCH OH CHCH C CH CH 9. (d) Ksp s s fr CuS = CuS 0 7 Ksp g S s s fr 0 gs Ksp s s fr HgS = Hgs S subiit rder. 0. (c) 0 5 g S CuS HgS PC. (c) Miniu difference f ph and pk a is better buffer sutin. (b) Carb aine test RNH CHC KOH RNC KC HO. (d) CH O CO H (g) (g) (g) ( ) O E H n RT. 78. ( R 98) = ( 59R 7.8) kj.. (d) CsI g Cs I 5. (c) CH5OH() O(g) CO (g) HO() ng H E n g RT =.7 ( ) E.7 kj e =. 95 kj e. (b) Rate = 7. c 8. b K [] [B]. 0 K(0.) (0.) ----(). 0 K(0.) (0.) ----(). 0 K(0.) (0.) () Dividing eq. () b eq. () B- /- M, Brahanand Cn, Durgakund, Varanasi; Page 5 n g = Dividing Eq. () b Eq. () () () = 0 S, Rate = k []

6 9. (c) Standard eectrde ptentia f reactin (E ) E ce E ce E i E red If = psitive then reactin is spntaneus because Other wise nn spntaneus ce can be cacuated as G ne ce F is negative. E =.5 (.8).9V Reactin wi nt ccur. 0. (c) Basicit f aines in aqueus sutin des nt n depend n inductive effect but n cbined infuence f inductive effect, svent effect and steric effect s rder is CH5NH (CH) N CHNH (CH) NH. (a) Since, z and Matheatics z w ; w w z [ + w w Re(w) w w Re(w) ; Using z z z z z Re(z z )] w Re w 0Re(w) 0 c. (d) a b c 0 has rts and, then + = - b/a and =. Find the vaues f + a and and then put in ( ) ( ) t get required vaue. 5 ( ) ( ) ; ( ) 8 9. (b) Perpendicuar fr the center t ine wi be greater than radius Center : - (0, 0) Radius : (c) 5. (b). (a) k k 0 a a a a c c c c 9 b 5 S equatin (c) Cefficient f in ( ) ( ) ( ) Nw, cnsider the fwing cases fr 7 n 5 (r ) n r 5 7 ( ) ( ) ( ) and r r 5 n 5n 8 n + = 5n + 8 n = 8. (a) P().P(B).P(C).P(D) 7 5,... ; (b) 0 in B- /- M, Brahanand Cn, Durgakund, Varanasi; Page 7

7 0. (b) Given, P P ( ) ( ) ( ) P adj. (c) Given, f(n) n n,f(),f(), f(),f() Let f() f() f() f() f() f() f() f(). k k (d) Given, fk () (sin cs ), where R and k > k f() f() (sin cs s) (sin cs ); ( sin.cs ), ( sin.cs ) 0, If n a n n 0 a, If n. (b) 0 a... an i, b0 b0 b... b, If n anda0b0 0, If n anda0b0 0 ( a b) ( b) a = 0 and i a b = b = - [ (-a) = 0]. (c) 5. (c) f() = g, then f() f( ) 0 f() 0,. (a) t f(t )t t f(t ) t f f (b) Equatin f B 0 0 ( 0 ) 8. (c) h d h d 0 h h 0 h h 0 h 0 h 0 0 d 0 B- /- M, Brahanand Cn, Durgakund, Varanasi; Page 8

8 9. (b) (~ p q) verba fr If Ra des nt wrk hard then Ra gets gd grade (c) h,k h, k, Substituting in,(k) (h) k h is required cus (c), r a Eccentricit f a b a. (c) Case I : 0 r Then a e, b E r 0 r, S (, ) () Case II : r Then r r S, (, ] (, ).. () Fr () and (), (, ] (, ) (, ).. (d) g( h ) We have i f ( ) i f ( h) i and i f ( ) i f ( h) h0 h0 g( h) h0 g( h ) i S, f ( ) and f ( ) d nt eist. h 0 g ( h). (d) Given, Trignetrica Equatin (sin sin ) + sin = C D C D - cs sin + sin cs = [ sinc sind cs sin and sin = sin cs ] sin ( cs cs ) = sin ( cs cs ) = 5. (d) I Let t d d r I t ) dt ( dt d t t C t ( t ) C ( ) C. (b) r e B- /- M, Brahanand Cn, Durgakund, Varanasi; Page 9

9 7. (a) n d [ n ] n Required area = ( n ) = n sq. units. 8. (d) S, the equatin f pane is + + = 0 r + + = (a) The equatin f the tangent at the pint R (, f ( )) is Y f ( ) f ( )( X ) The crdinates f the pint P are ( 0, f ( ) f ( )). The spe f the perpendicuar ine thrugh f ( ) f ( ) P is r ( ) ( ) ( ( )) d d f f f r f ( ) d d Which is the required differentia equatin t the curve at f (). ( ).( ) ( ) ( ) 0. (a) ( ) ( ) ( ) z z 0 r 5 z 7 0 = B- /- M, Brahanand Cn, Durgakund, Varanasi; Page 0