Narayana IIT Academy

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1 INDIA Sec: Sr. IIT_IZ Jee-Advnced Dte: --7 Time: 09:00 AM to :00 Noon 0_P Model M.Mrks: 0 KEY SHEET CHEMISTRY C D 3 D B 5 A 6 D 7 B 8 AC 9 BC 0 ABD ABD A 3 C D 5 B 6 B PHYSICS B 5 D 6 B 7 A 8 B 9 C 30 A 3 BCD 3 ABCD 33 ABC 3 ABC 35 B 36 C 37 C 38 D 39 C MATHS 7 A 8 C 9 D 50 A 5 D 5 C 53 C 5 ABCD 55 ABD 56 AC 57 ABCD 58 B 59 B 60 C 6 A 6 B

2 CHEMISTRY.. () Acid strength H SO > H CO 3. Bsic strength HCO 3 > HSO. (b) Acid strength HCl > HF. Bsic strength F > Cl. (d) Acid strength HSO > HCN. 3. Sol. Initil formul º AB3 lter the formul becomes A /8 B3 i.e. A7/8 B3 i.e. A7B HCOOH H + + HCOO C K neglect w.r.t. 0.0 due to common ion effect K = X= [HCOO ] = 0 Sec: Sr.IIT_IZ Pge --7_Sr.IIT_IZ_JEE-Adv_(0_P)_CTA-_Key & Sol s 8. A) As ([H + ] = [OH ]), Hence (ph = poh) for pure wter (B) ph = ph = 3 [H + ] = 0 [H + ] = 0 3 = 00 (D) Acording to Lewis concept wter cts only bse 9. KSP for AgCl = 0 9 (0.) = Cl / AgCl / Ag E = log0 0 = 0. V.. Wter softened by ion echnge resin is completely free form minerls nd is not useful for drinking purpose.. Mg OH MgOH

3 _Sr.IIT_IZ_JEE-Adv_(0_P)_CTA-_Key & Sol s poh 6 log 5. ph y.60 y 00 6 y 0 0 poh ph moles of NOH left Mg OH Mg OH z 0. z z poh ph= Ans: B o Pb/ PbSO / SO (A) E = E (B) E = E E o 0.06 log K Pb Pb sp = ( 6) = = 0.60V Pb/ PbSO / SO Pb Pb E o Pb Pb = log 0. E o Pb + Pb 0.06 log [ Pb ] Sec: Sr.IIT_IZ Pge 3

4 = 0.03( 5) 0.03 = = 0.39 Ans.] --7_Sr.IIT_IZ_JEE-Adv_(0_P)_CTA-_Key & Sol s 0.0 log Ans. 9 Pressure t the surfce of lke in terms of H O = tm = 0 m of wter P t the bottom = = 90 m of H O Now pplying boyle's lw 0 V = 90 V V V 8. Ans = /9 V - initil volume Sol. CuSO. 5HO(s) CuSO(s) + 5HO(g) Kp = 0 0 (tm) cm = P HO P HO = 0 tm. n = PV RT = 0.5 = Ans. 3 Sol. When NCl is electrolysed, Q = it =» 0. At node : Cl Cl + e At cthode : HO + e H (g) + OH 0. F 0. Mole So, [OH ] = 0. poh = Sol. (D) t0 = log log [ 3 log ] t50 = log log t50 log 3 t 3log 0 or t50 = 3 t0 = 3t Sec: Sr.IIT_IZ Pge

5 PHYSICS. At the initil moment, ngulr velocity of rod is zero. --7_Sr.IIT_IZ_JEE-Adv_(0_P)_CTA-_Key & Sol s Accelertion of end B of rod with respect to end A is shown in figure. Centripetl ccelertion of point B with respect to A is zero l 0 So t the initil moment, ccelertion of end B with respect to end A is perpendiculr to the rod which is equl to b rel l b l where is ngulr ccelertion 5. T. cos0 l g h g T. g l h l h T g 6. Friction force between wedge nd block is internl i.e. will not chnge motion of COM. Friction force on the wedge by ground is eternl nd cuses COM to move towrds right. Grvittionl force (mg) on block brings it downwrd hence COM comes down. 7 When coches (m ech) re ttched with engine (m) ccording to question P = K 6mgv () (constnt power), (K being proportionlity constnt) Since resistive force is proportionl to weight Now if coches re ttched P K.mg.v () Sec: Sr.IIT_IZ Pge 5

6 Since engine power is constnt So by eqution () nd () 6Kmgv Kmgv 6 0 v v 8.5m / sec 6 v Similr for 6 coches K6mgv K8mgv 6 v m / sec 8. Force on tble due to collision of blls: dp dt 3 Fdynmic N 0. 0 N Net force on one leg --7_Sr.IIT_IZ_JEE-Adv_(0_P)_CTA-_Key & Sol s 9. During st collision perpendiculr component of v, v becomes e times, while nd component v II remin unchnged nd similrly for second collision. The end result is tht both v II nd v becomes e times their initil vlue nd hence v" ev(the (-) sign indictes the reversl of direction). 30. Consider prllel is pssing through center of bse. The two es re equidistnt from the C.M. Sec: Sr.IIT_IZ Pge 6

7 --7_Sr.IIT_IZ_JEE-Adv_(0_P)_CTA-_Key & Sol s 3. At mimum etension both should move with equl velocity. By momentum conservtion, V V = 5 m/sec Now, by energy conservtion V k Put V nd k m m 5 cm. Also first mimum compression occurs t; 3T 3 t k sec mm (where reduced mss, m m ). 3. The bll hs v', component of its velocity perpendiculr to the length of rod immeditely fter the collision. u is velocity of COM of the rod nd is ngulr velocity of the rod, just fter collision. The bll strikes the rod with speed v cos53 in perpendiculr direction nd its component long the length of the rod fter the collision is unchnged. Using for the point of collision. Sec: Sr.IIT_IZ Pge 7

8 --7_Sr.IIT_IZ_JEE-Adv_(0_P)_CTA-_Key & Sol s Velocity of seprtion = Velocity of pproch 3v l u v ' 5 () Conserving liner momentum (or rod + prticle), in the direction of the rod. 3 mv. mu mv ' 5 () Conserving ngulr moment bout point D s shown in the figure 0 0 mu u 3 l ml l (3) By solving v 7v u, w l Time tken to rotte by ngle t In the sme time, distnce trvelled l u.t 3 Using ngulr impulse-ngulr momentum eqution. l ml 7v N.dt l mv N.dt 55 or using impulse momentum eqution on Rod mv Ndt mu As V v V 30 m v 30 first Resonnce depth (from upper end) R m 5 cm Sec: Sr.IIT_IZ Pge 8

9 3. U 3 y F m U / 3 m y y F U / y 5m / s m m Let t time t prticle crosses y-is t sec Then 6 3 t Along y-direction : y 8 prticle crosses y-is t y = - At (6, ) : U = 3 & KE = 0 At (0, -) : U = -6 KE = 50 mv 50 or v = 0m/s while crossing y-is PASSAGE-_(35, 36, 37) --7_Sr.IIT_IZ_JEE-Adv_(0_P)_CTA-_Key & Sol s Upon plcing on the step due to symmetry B, D should be overll eqully compressed. Net torque must be zero bout ny is, including digonls, so chnges in tension t the opposite ends of digonl must be equl. This is why springs A, C should be eqully further compressed by n mount nd B, D should eqully lengthen by sme mount. This equlity of chnges in length ensures tht net force provided by springs to support weight of the tble doesn t chnge. Denoting upwrds s positive. Verticl displcement t A, B, C nd D will be 8,, - nd respectively. Displcement of midpoint of AC = displcement of midpoint of PASSAGE-_(38, 39) S cos S S l sl s 0. & be the displcement from equilibrium position 8 BD cm. Sec: Sr.IIT_IZ Pge 9

10 Now for hollow sphere, pplying A 5 k r m r 3 r () --7_Sr.IIT_IZ_JEE-Adv_(0_P)_CTA-_Key & Sol s By ngulr momentum conservtion (bout A) of the system, m v r 5 v v () 5 (3) I 5 7 m v r Using () nd (3) we get, k m 3 6 k 35 m 6k f 35m. v gh H / v g h By continuity eqution dh A v v dt dh A g h H / g h dt or H/ t A dh dt t A g h h H / 3 H 0 H g. Work done by force F ˆ ˆ ˆ ˆ W F.d yi j. di dyj () y d y dy 0 ydy y W y dy dy 0 y dy Sec: Sr.IIT_IZ Pge 0

11 --7_Sr.IIT_IZ_JEE-Adv_(0_P)_CTA-_Key & Sol s It cn be observed tht the force is tngentil to the curve t ech point nd the mgnitude is constnt. The direction of force is opposite to the direction of motion of the prticle. work done = (force) (distnce) y J 3. Mimum kinetic energy is gined when the semicirculr disc hs minimum potentil energy. By energy conservtion, r mg I 3 0 ; here I 0 M.I bout point of contct r mr r mr r r 8r 0 cm cm 0 I I m r I m ; I m mr m m I 0 mr Due to error in g 9 6 r m rd/sec. l l l 00 l l 00 l l00 l When l00 l is mimum then m will be minimum, tht mens l 50 cm mg 5. Y d / l Y mgl () / d dy m l d Y m l d m m 0.0 kg m 0.kg l = 5m l cm d = cm d 0.00cm = 0.00 cm 0.00cm dy 0.kg cm 0.00cm 0.00cm 00%.3% Y 0.0 kg 5cm 0.05cm 0.00cm m 6. Effective re should be tken. Sec: Sr.IIT_IZ Pge

12 --7_Sr.IIT_IZ_JEE-Adv_(0_P)_CTA-_Key & Sol s MATHS 7. I.cos d 3 by prts cos 0 8. Let 3 sin. 3 sin d cos cos 0 I I I 0 0 cos d sin d But sin cos 9. / d GI y z0 z y / z z 3 0 z 3 3 y 0 z 0 3, 0, 0,, R. D,, 6 5. P.I =(,3,5) eqution of plne +3y+5z= ,, G. E (,,) 0 y 3 L : Infinite solutions 3 7 rectngulr hyper y e Rd= e 7 re = k 5. Split res into trpeziums ppro vlue of ln d Sec: Sr.IIT_IZ Pge

13 --7_Sr.IIT_IZ_JEE-Adv_(0_P)_CTA-_Key & Sol s f f f f 3 f f Is f lnk lnk! k ln d ln k ln k! k ln k k ln k ln k! k re we get k! e k e n n n k k / n k / n k / n d) proceed j S sin e d G. E e j k e 56. 0,0,0,,, re vertices third verte (l,l,3l) Are = 56 l (,,6) G,, OG GE r, y, z z y c, z c, y c Also +y+z= solving r 0,, y / 3c 3 z / Solving y Given y d d 5 R.A= 0 p, q, 6. f Sec: Sr.IIT_IZ Pge 3

14 --7_Sr.IIT_IZ_JEE-Adv_(0_P)_CTA-_Key & Sol s R.A= d d /8 y p q p q 3 f p= q=- 6.,,, f f, f or f 3 No solutions / b b 63. GS k b k f d f d k R You get minimum looking it s qudrtic in k, y y y clerly d d y d y y y ydy d nd K=0 GE 5 k f R f f Sec: Sr.IIT_IZ Pge

15 0 0 f d c da f f R k f 67. r y b r y b 3 3 6y r b Let P y y, PA PB y 5 cos 3,5 y sin GE sum of digit A= tn A tn B tnc (tna+tnb+tnc) 3 58, 8.6 {9,50,5 } 30 K 9 --7_Sr.IIT_IZ_JEE-Adv_(0_P)_CTA-_Key & Sol s Sec: Sr.IIT_IZ Pge 5

FULL MECHANICS SOLUTION

FULL MECHANICS SOLUTION FULL MECHANICS SOLUION. m 3 3 3 f For long the tngentil direction m 3g cos 3 sin 3 f N m 3g sin 3 cos3 from soling 3. ( N 4) ( N 8) N gsin 3. = ut + t = ut g sin cos t u t = gsin cos = 4 5 5 = s] 3 4 o