PHYSICS. (c) m D D m. v vt. (c) 3. Width of the river = d = 800 m m/s. 600m. B 800m v m,r. (b) 1000m. v w

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1 PHYSICS T. T 8 k/hr = /s, ' Hz, ' 7 Hz T. r D D D,. r D D. D. D.... g/c. Width o the rier = d = 8 w /s B 8,r P O. sin t cos., on the copring this eqution, with sin t cos k =, k =. k. /s w Q. =, 9 9 = 8, 8,, 9,. sin t, sin t or sin t, cos t So Thus prticle describes circle o rdius

2 7. N =, s g. g g = g T cos g, T sin g g tn, tn g g tn ( s ) N g F.B.D. o kg Tcos Tsin g F.B.D. o kg 8. est be the ss per unit length o the chin nd be the length o hnging prt o the chin. In cse o liiting equilibriu o the chin, we he g l g l 9.. Distnce trelled b B 8 (pproitel). ppling consertion o liner oentu, we he cos ' ' cos sin sin equired distnce cos g g. gh k For the solid sphere, h = s sin I M MK g ssin 7 k gssin 7

3 . Fro the principle o consertion echnicl energ, we he L g g L Now, T L g L T L g N /s T L L. During the otion o the issile, its echnicl energ is lws consered. GM g GM ME g. We know, g ' g cos g For equtor, = For north pole, = 9 g' g g g equired weight, g' g g g 7 N. Let be the olue outside the wter For the equilibriu o the bod, we he L g L d g = c.c. h 7. We he, g gh lso, gh h g c, h = c When the hole is de t depth h below the ree surce, elocit o elu h ' g c/s olue o wter coing out/second = 9 = 7 c/s

4 8. We he V P gh B 9 N/ / / 9.. T U U I r, r Work done i.e. q U F U I q. U F q K J B, B B q, q 9 9 q 9 q q 9 K 9 q. Net electricl ield strength t L is = N/C orce on C chnge is = = N. Q, Q Distnce between two chnges = F 8.8.N. d E. dl 8 8

5 . = or iu power trnser P I wtt. Mgnetic lu densit due to single wire is B I sin sin B net = B =. T =.8 T 7 = 7 T =. T I. B t centre r, I r B t centre r i.e. independent o. So B t centre is zero 7. ccelertion o rod Distnce d BIL - /s 8. ev V F e FB Be B l l B. T outwrd 9. B tn l olt =. Energ LI L Energ I Diensions o sel inductnce = ML T

6 . V V V V L C L VC V V C 8 V V., u = c nd = c u Soling = 7 c. Such tringles (OBC) re round in nuber ech will he two new relections ecept the irst tringle in which we he three relections. Hence relections in ll C B.. D ' ' ' erction ro IInd surce eq eq eq 8 c 8 8

7 8. d = = c d conce cone 9. 8 e e 9 e 8, e L c c, e c, c. t t. u Becuse ige is erect nd ties ens lens is cone nd object between pole nd ocus. u u u c. (i) (ii) =.. Dn, n n, d n n 7, n, n c =. c.

8 . E E, E 8 E. P P E P, E P,, : : E, E, E = T TT T T in 8 ers, o the initill substnce will rein undec. rction which will dec is. Q = chnge in binding energ =. MeV dt dh H 7., C, s dt dt dt H H s, s dt dt L L C C 8 L L 8 s s 8. PV constnt, H r r P T C p 7 CP r = / C 7,, Q Q 7 J. H U nc nc p T T H 7, U J 7

9 . CHEMISTY C CH COCH H O Hg +. With increse in teperture; the interoleculr ttrction decreses.. C(s) + O (g) CO (g) ; H = 9. kcl (i) CO(g) C(s) + ½O (g) ; H = +. kcl (ii) CO(g) + ½ O (g) CO (g) ; H = 8. Eqution (ii) is reerse o the eqution o het o ortion o CO(g).. H = U + ng T H U = ng T = = 7.7 kj ol.. t node: ½H (g) H + (q) + e t cthode: gcl(s) + e g(s) + Cl (q) ½H (g) + gcl(s) H + (q) + Cl (q) + g(s).. te o rection = d [SO dt ] = d [O dt ] d[so ] = dt Hence, the rte o rection will be. ol/l sec.. B(OH) + H O B(OH) + H +. It s onobsic cid. 7. Becuse ter giing one electron, it gins inert gs conigurtion.

10 ubidiu belongs to I group nd th period, hence its outerost conigurtion is s, this iplies tht n =, l =, =, S = +, B C + D t =.. t equilib (. ) (. ) ccording to question [C] = =. =. K c = [C] [D] [] [B] = (.. ) (. ) = (. ) =.. 7. In I, =, or = Eleent belonging to groups nd rd period ust he e in rd orbit, hence electronic conigurtion is s, s, p, s, p. 79. In BCl, Boron hs onl si electrons in outer ost orbit There is no sp hbridised hdrogen in the copound O O. sp sp 8. CH =CH CH C CH

11 8. Copounds contining t lest hdrogens (in cse o crbonl copounds) or hdrogens (in cse o lcohols) will gie iodoor test For FCC, r = r = Closest distnce = r = 9. te = k[n O ].. [N O ] = =.8. toic weight 7 9. W = 7 ; E = = = 9 lenc E 9 Z = = = 9.. F 9 9. Milli equilent o NOH =. = Milli equilent o HCl = = Milli equilent o H SO = = Totl illi equilent o cid = + = Totl illi equilent o NOH = Let illi equilent o NOH = = i.e., lkline E n =. Z ev n [Z =, n = ] =. =. ev g(s) + ½O (g) g O(s) H = U + ng T H = U ½ T H < U.

12 log k T 98. E = ln F [H ] =.9 log =.9 log =.9. [H ] 99. ccording to rrhenius eqution log k = log E. T Plot o log k s T is stright line Slope = E. Intercept = log. /T. The bond between chlorine nd benzene hs prtil double bond chrcter.