SAFE HANDS & IIT-ian's PACE EDT-04 (JEE) Solutions

Transcription

1 ED- (JEE) Slutins Answer : Optin () ass f the remved part will be / I Answer : Optin () r L m (u csθ) (H) Answer : Optin () P 5 rad/s ms - because f translatin ωr ms - because f rtatin Cnsider a thin shell f radius r and thickness dr Vlume f shell dv r dr ass f shell dm ρ dv ρ r dr I f shell dm r ρ r 5 dr I f sphere dr Answer : Optin () Use parallel axes therem t calculate I f individual rds Answer : Optin () 5 Q Answer : Optin () 9 Assume the frce f frictin t be f in the same directin as that f F hen F f a and FR fr Iα F f Rα a Answer : Optin () hink abut the directin f tendency f relative slipping Answer : Optin () Nrmal will shift twards right t balance the trque f F Answer : Optin () x dx x L csθ Answer : Optin () 6 Answer : Optin () 7 H θ P L (-sinθ) L mv r sinθ mv x (Perpendicular distance between reference pint and line f actin f mmentum) At highest pint, velcity f prjectile will be hrizntal and it will be u csθ S perpendicular distance between reference pint and line f actin f mmentum will be H itself V u csθ Cnsider the trque abut center due t weight f the rd dτ W dm g By integrating frm x t x L, τ W his trque shuld be balanced by the cunterweight Hence, mgd τ W d L/ Answer : Optin () By parallel axes therem, I I C d I C I P a g e

2 ED- (JEE) Slutins Hence, I C I d Answer : Bnus As rd is in rtatinal equilibrium, trque abut any pint shuld be zer Equating trque abut pint where mass is attached t zer, > Answer : Optin () 5 Answer : Optin () 6 Answer : Optin () 7 he angular mmentum will be cnserved abut pint f suspensin mu L I ω Answer : Optin () As radius is halved, I becmes ne furth By law f cnservatin f angular mmentum, ω must becme fur times Answer : Optin () 9 Answer : Optin () ass f hllw sphere will be significantly smaller and s will be I Answer : Optin () Answer : Optin () As rd is heated its length will increase and hence will increase its mment f inertia Als, I m i r i Fr each particle, (r i ) new r i ( α Δθ) Hence, I new I ( α Δθ) By cnservatin f angular mmentum, Iω I new ω new ω new () Answer : Optin () As acceleratin f the center f mass is zer, mg Als, R α Acceleratin f the rpe will be Rα g Answer : Optin () Cnsider any pint n the surface he trque f frictin abut that pint will be zer and hence the angular mmentum will be cnserved abut that pint Answer : Optin () 5 Answer : Optin () 6 Let the mass f the ball be m By cnservatin f mmentum, mu V By cnservatin f angular mmentum, mu By cnservatin f energy, mu V Iω Put values f V and ω frm first tw equatins int third ne t get relatin between and m Answer : Optin () 7 Answer : Optin () Answer : Optin () 9 Assume axis is placed at distance x frm rigin hen, I mx m(d x) Fr minimum I, Hence, x d/ Answer : Optin () Given, α -kω ie ω -kω mx m(d x) dω -k dθ Nw simplify fr θ in terms f ω and put the limits Answer : Optin () P a g e

3 ED- (JEE) Slutins () KE R () () (KE) R (KE) R (KE) (KE) r (KE) (KE) Suppse w kg f each is taken n CH CH w w H 6 n p p mle fractin p CH p ttal ttal Slutin:(U AV ) A fr A (U AV ) pttal 9 w 6 w w 6 R A A B R A A B and (U rms ) B fr B V AV r < () P () Use relatin, P R B R B n a 5 () P (V nb) nr V At lw pressures, b can be ignred as the vl f the gas is very high At high temperatures a can be ignred as the press f the gas is high P (V b) R C PV - Pb R PV R Pb PV R () 5 Z Pb R P 6() r 66 (Fr natmic as CV He, Ne, Ar ) 7 () Higher the critical tempreture, mre easily gases will be liquefy s the sequence, O, N, H, He () he average kinetic energy f an atm is given as k It des nt depend n mass f the atm 9 () Slutin: he V rms at K is V V R At 5 K V R V nf lecules () 6 R 56 R V P a g e

4 ED- (JEE) Slutins () Gas equatin is PV m R ---() Again V V t V t ml () P N l fractin f N tal perssure atm atm P m V R ----() divide () by () m m m m Gas escaped is then () n N n O wt f mlecular wt N f N tal mles n N n O 5 Nw tal Pressure P 5 9 atm Partial pressure f nr V le f N N tal pressure tal mle 7 5 n a V () P ( V nb ) nr 9atm () () 7 P (5) m 65 atm 7 P ( 5 7 ) 66 6 () KE R ml r KE nr J 9 J 7 () P V P V If P mm, P will be mm Hence V V, V V, V Decrease in vlume V f V ie % () 9 () 5 () 5 () 5 () 5 () 5 () 6 5 W 6 (lecular weight f xalic acid 6) 6 5 gm 6 gm 6 gm 5 55 () Peptisatin: the cnversin ff ppt t cllidal slutin 56 () 57 () Negatively charged As S sl cagulated mst effectively by AlCl his is because ppsitely charged Al ins have maximum charge 5 () As > Ca > Na 59 () 6 () P a g e

5 ED- (JEE) Slutins ( a)sin(5 )sin(5 )sin(5 ) (sin(6 5 )sin(5 )sin(6 sin( 5 ) sin(75 ) ( a b c ) ( a)6sin (9 ) ( (9 )) sin ( cs( )) (a) 5 5 put the values and slve it 5 5 )) ( d)cs( α β ) cs(( θ β ) ( θ α)) cs( θ β )cs( θ α ) sin( θ β )sin( θ α) 5(a) It is given that sin A cs A a b a b ( cs A) ( cs A) a b a b b ( a b)( cs A cs A) a( a b)( cs A cs A) ab { b( a b) a( a b)} cs A ( a b)( a b) cs A a ( a b) b( a b) ab ( a b) cs A ( a b)( a b) cs A ( a b) {( a b)cs A ( a b)} Hence, r sin A cs A ( cs A) ( cs A) a b 6 a 6b 6a b a 6a 6a ( b a) b a 6b 6b 6b ( b a) ( a b) ( b a) ( a b) b a cs A b a b a b a 6(b) Let u cs θ sinθ sin θ sin α ( u sinθ cs θ ) cs θ(sin θ sin α) u tan θ u tan θ u sin α Since tan θ is real, therefre u u ( u sin α) u ( sin α) u sin α 7(c) cs 56 cs 5 cs 66 cs sin 6 sin cs cs sin cs (cs cs 6 ) cs cs 57 cs 9 cs cs 9 sin (b) A B C 7 A B C 9, then cs A cs B cs C sin A sin B sin C cs cs cs sin 9 sin 9 9(d) (b) sin 9 A cs θ sin θ A cs θ sin θsin θ A cs θ sin θ, [ Q sin θ ] A Again A cs θ sin θ ( sin θ ) sin A sin θ Hence, / A α β α β cs cs α β α β α β α β cs cs sin sin cs cs sin sin α β α β α β sin sin cs cs tan tan (b) he given expressin can be written as (cs 6 x cs x) 5 (cs x cs x) (cs x ) cs 5 x 5 cs x cs x After slving, we get the required result ie cs x θ P a g e 5

6 ED- (JEE) Slutins (b) cs( A C) cs A cs C sin A sin C cs B cs( A C) cs A cs C sin A sin C (b) sin tan tan B tan A tan C B tan A tan C tan B tan A tan C tan A tan C tan tan B tan A tan C tan A tan C tan B tan B tan A tan C tan B tan A tan C Hence, tan A, tan B and tanc will be in GP α sin ( α) 6 6 sin α sin (5 α) 6 6 {( cs α) ( sin α) } { cs α sin α} { cs α sin α) sin α cs α} { α sin α) cs α sin α(cs α sin α) } (cs 6 sin α cs α 6 sin α cs α (a) a sin θ b sin θ cs θ c cs θ ( a c) 5(b) tan( [ a cs θ b sin θ c cs θ ] [ b sin θ ( a c)cs θ ] Q b sin θ ( a c)cs θ b ( a c) { b sin θ ( a c)cs θ } b ( a c) a sin θ b sin θ cs θ c cs θ ( a c) b ( a c) B) A B A (i) and 6(b) We have sec( frm (i) and (ii), 9 B B) A B 6 A (ii) 5 7 k sin sin sin 5 cs cs cs 7 B sin 9 sin 9 7(a) If sin sin cs cs cs sin sin 9 9 L, then L L r L Bth L L as sec A tan A L (a) he slpe f line x y is It makes an angle f 5 with x-axis he equatin f line passing thrugh (, ) and making an angle f x y r cs 5 sin 5 x y r / / 5 is, C-rdinates f any pint n this line are r, then r If this pint lies n x y, r r r 9(a) Slving x y 9, y mx we get x is an integer if m,, 5, 5 5 x m m,,, S, m has tw integral values (b) he set f lines is ax by c, where a b c Eliminating c, we get ax by ( a b) a ( x ) b(y ) his passes thrugh the intersectin f the lines x and y ie x, y ie,, P a g e 6

7 ED- (JEE) Slutins (c) Given tangents drawn t the given circle, where and are the length f () (d) r 5 CP C (, ) ( ) () (7 ) k aximum distance 5 5 k (c) If d is the distance between the centres f tw circles f radii r and r, then they intersect in tw distinct pints, if r r < d < ( r ) r Here, r 5 6 r r r < 5 < r < r < (b) Equatin f radical axis is S S S x y x y 5 S 7 y x x y Radical axis is x y Hence gradient f radical axis 5(c) he c-rdinates f fci are ( ± ae, ) Here a, b 9 b a ( e ) 9 6( e ) e 6 7 e ± ; Pints are 7,) (± Radius ( 7 ) ( ) (a) 7(a) (b) It is given that the lines ax by p x cs α y sinα p are inclined at an angle a cs α herefre tan b sinα a cs α b sinα and a cs α b sin α a sin α b cs α (i) It is given that the lines ax by p, x cs α y sinα p and x sin α y cs α are cncurrent a b p cs α sinα p sinα cs α ap cs α bp sin α p a cs α b sin α cs α b sin α a (ii) Frm (i) and (ii), a sin α b cs α Frm (ii) and (iii), ( a cs α b sinα) ( a sin α b cs α) a b 9(b) he crdinates f A and B are (,) and (,) respectively he equatin f the perpendicular bisectr f AB is y 6 ( x ) r x y (i) Equatin f a line passing thrugh (, ) and parallel t x-axis is y his meets (i) at C, herefre the crdinates f C are Hence the area f the triangle ABC is 9 sq units, (c) he equatin can be written as ( ) ( x ) y Obviusly, it is a parabla whse fcus is 5, and directrix is x P a g e 7