ANSWERKEY SOLUTION = X O2

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1 PPER CODE. NCERT XIII Pat Page 7.. In ƒcc unit cell a [ adius of Cu atom, a edge length] So, a SCORE-I 6 7 pm Sb S Sb + S ;K ( x ).( x) sp x x K sp 08x 5 ; K sp 08 ( 0 5 ) It is buffe solution so, + log Salt cid log cid Salt log 0 æ [ cid] ö ç Given 0 : [ Salt] è ø 6. NCERT-XII, Pat-I, Page-8(fomula-.6) P p 78 X ; \ P 75; \ P to NSWERKEY SOLUTION TEST # 06 PTTERN : JEE (Main) Date : Q Q Q Q Q C E 0 5 Copoate Office : LLEN CREER INSTITUTE, SNKLP, CP-6, INDR VIHR, KOT-005 PHONE : , ax : , info@allen.ac.in Website: 8. NCERT XI th Pat-I Page No.. Let the mass of methane and oxygen is w mole faction of oxygen + 6 w w w + 6 Let the total pessue be P The pessue exeted by oxygen (patial pessue) X O P total Þ P. Hence () is coect. 9. NCERT XI Pat Page # 96 + C ( 0.75) ( 0.75).5 Path to success [C] (.5).5 KOT (RJSTHN) K.0 [][] NCl 5, PH 5, CuI exist ugha djs gs a. PCl is sp hybidised and pyamidal in shape. PCl sp ladj.k dk gksk gs ;g fijkehmh; gksk gs 6. [X] is MnO x + ( ) x + 6 KOT / HS - /7 TM

2 8. SO, Cl O 7, N O 5 ¾ cidic natue oxides N O ¾ Neutal oxide. 9. ClO does not dispopotionate because in this oxoanion chloine is pesent in its highest oxidation state. 0. CCl + H O. Diffeence of oots R.T. ¾¾¾ No eaction a b (a + h) (b + h) D D D a a p D p. a(p + x) + bpx + c 0 oots ae, p 9. ab. a b cos a + b + c I.6. a + b + C I. 0 0 Þ LHS is intege if c 0 then a ai ˆ+ bj ˆ which lies in xy plane 0. Let P(h, k) be the point fom which two tangents ae dawn to y x. ny tangent to the paabola y x is poduct of oots p + c a. a(bc ) (c ) + ( 6) > 0 abc a b c + > 0...() > G a b c on adding e n () and () + + > / (abc) Þ (a + b + c) (abc) / >0..() abc (abc) / + > 0 x x + > 0 (x ) (x + ) > 0 x > (abc) / > abc > 8 y mx+ m If it passes though P(h, k), then k mh + m Þ m h mk + 0 Let m, m be the oots of this euation. Then, m + m k h and m m h Þ m k h and m h [Q m m (given)] æ k ö Þ ç Þ k 9h è hø h Hence, P(h, k) lies on y 9x KOT / HS - /7 ' Ç' n(' Ç ') Since a, b, c ae the oots of the euation x x + x + l 0 Þ a + b + c Þ a + b c Now aea of the tiangle will be a + b (a + b) ( -c) d > 0 dc ( - c) Þ s is an inceasing function & c Î [,] \ max s. units... (-6,0) dy dx 6 (xy) 5 x y 0 P(,5) p'(, 5) Þ y x (x, y ) lies on x y P'7 Þ x y ± \ Euation of tangent lines with slope ae y x + 6 & y x 6 Þ c c 0CE05

3 6. E of plane O Q R is x y z Since the aea is symmetical about x-axis, so aea ò 9 x dx y Þ x y + z 0 distance of P fom plane O x X x 7. P( - ) x 9 x + 9sin x/ù úû é êë 9p/ 8 9 sin (/) 9 (p/ sin /) 8 9 sec 8 (a) - MP ^ M (a+b) - - (b) 5. Solving the euation of the given cuves fo x, we get x x + Þ (x ) (x + ) 0 Þ x, Y Þ MP. O Q æ a+ b ö -. b- a 0 è ø Þ ç ( ) P x 0 x X é ù ê- a+ b a- b 0 ú ë û Þ ( ) ( ) So ed. aea ò éx+ x ù ê údx ë û 9. Statement- is clealy tue. uuuu uuu Since OM l OD ld Now points,, C and M ae coplana Þ [a b c] [m a b] + [m b c] + [m c a] Þ [ l d a b] +l [ d b c] +l [ d c a] [a b c] l [d a b] + [d b c] + [d c a] éx x ù ê + x ú ë û 9 [( + 8 / ) (/ + / )] 8 5. Given euation can be witten as é æ ö ù 5 dy æ m ö ædyö ê+ ç ú ç ç ê è dx ø ú èm + ë û ø èdx ø 5 This shows that its ode, degee 5 and theefoe Statement- is also tue. 0CE05 KOT / HS - /7

4 a a+b a-b a+b cos cos - cos + a+bæ a-b a+bö cos ç cos - cos + è ø a+b a b cos sin sin + b+ Þ a b 55. sin x cos x ( sin x) sin x cos x No value of x fo L.H.S. R.H.S. 56. Given, 0 99 s 59. f(x) x x + 00x + 00 f (x) x x + 00 > 0 " Î R \ f(x) is inceasing (stictly) \ f æ ç ö > f æ ö 999 ç 000 è ø è ø f(x + ) > f(x ) 60. Let f(x) x ln x x ln x \ f (x) x ln x x 0 Þ x e -/ lim x 0 f(x) 0, lim f ( x) x and - f(e ) e Þ s 0 \ Maximum value of f(x) is e SD of euied seies s Ù ( Ú ) is when Ú ( Ù ) is when, We have [ Ù ( )] º [ Ù (~ Ú )] º [( Ù (~)) Ú ( Ù )] º Ù º ~( Ù ) Ú º [(~) Ú (~)] Ú º (~) Ú [(~) Ú ] º (~) Ú T º T \ [ Ù ( )] is a tautology X C / D D X C D D + D - D - D X C D % + % - % -5% % + % - % -5% The pecentage eo contibuted by C is which is minimum among,, C and D. KOT / HS - /7 E a b C Statement- is Tue. In statement- if the height of the pole is h, then the length of the adjacent sides of the field ae h cot a and h cot b and the aea if h cot a cot b h as a + b p/ Þ cot a cot b. So h 500 Þ h 5 units. and the statement- is tue using statement-. h D 6. (i) [(a cos ) î + (b sin ) ĵ]. [(b sin ) î (a cos ) ĵ] ab sin cos ba sin cos 0 (ii) [a(cos ) î + (b sin ) ĵ]. éæ öˆ æ ö sin i cos ˆù êç -ç j a a ú ëè ø è ø û sin cos sin cos 0 (iii) [(a cos ) î + (b sin ) ĵ]. 5 ˆk 0 Hence, all the thee options ae coect because the dot poduct of two pependicula vectos is zeo. 0CE05

5 Using v u gh (fo vetically upwad motion unde gavity) 0CE05 0 u 0 5 u 0 ms lso using u u gt t t s which means the each ball is thown afte sec. Theefoe the numbe of balls thown up pe minute is Velocity of the stone elative to the gound m/s (upwads) Velocity of the stone afte s, elative to the gound v 5 0 v 5 ms o v +5 ms 65. Given x ct + bt ct dx velocity v a + bt -ct dt dv and acceleation f b -6ct dt cceleation is zeo at time b 6ct 0 (using, v u gt) b t c Putting this value of t in e. (i), we get æ b ö æ b ö v a + bç -cç èc ø èc ø b b b o v a+ - a+ c c c 66. Given, R H u sin u sin g g o sin cos sin o cos sin tan Þ 0º Mv 67. Mgcos- R when the body leaves the suface, R 0 Mv Mgcos Mg cos Mg R v (5) cos g º 68. Time taken to each the bag on the gound 69. h 90 t 0s g 9.8 \ R m u sin Hmax g u sin R g Dividing both the euations, we get tan o sin 5 Putting the value of sin in en. (i), we get, 8g u 6 / 5 o u 5 g/ 70. µ (m + M) g + (M + m)a...(i) µ mg µ (m+m)g m M Now, since the fiction foce between m and M is µ mg, the acceleation. oce µ mg a µg mass m Putting the value of a in en. (i), we get µ (m + M)g + µ (m + M)g (µ + µ ) (m + M)g ( ) ( + 5) 0 96 N KOT / HS - 5/7

6 7. mv + 0 mv + mg( l) l If v v then v v (given) \ m(v) mv + mgl O mv mg l v gl d(mv) dm 7. v dt dt Wok Powe (P) Time kg-m/s oce Displacement oce velocity Time J/s o watt 7. Initial hoizontal momentum inal hoizontal momentum kg /m/s kg v cos m/s+ kg v cosm/s 0 0 n amount of 8 J is impated by explosion. Thus, the total enegy of the fagments is 6 J, i.e. each fagment has J kinetic enegy. kg v J v 6 v 8 m/s and v cos cos Þ 60º 7. s 6t t ds dt v 8t -t t t 0; v 0 t t s ; v m/s lso W m(v -v) (56 0) J 75. Let the sping constant of sping Q is k and that of P is k. The extensions poduced by applying eual foces on them ae x p and x Q, espectively. Since kx (numeically) o x k v v and o U kx æö U kç èk ø k 8 v cos v cos m/s...(i) The initial kinetic enegy of the shell kg (m / s) 6 J o Thus, o U µ k UQ k k U k k P P Q U Q U P E KOT / HS - 6/7 0CE05

7 76. Relative velocity afte collision e Relative velocity befoe collision o the collision between the blocks and v -v v -v e 0.5 v -v 0-0 (Q u 0 and u 0 m/s) o v v 5...(i) lso fom the pinciple of consevation of linea momentum, mu + mu mv + mv o v + v o v + v 0...(ii) om es. (i) and (ii) v 7.5 m/s and v.5 m/s Now fo the collision between the blocks and C. vc -v vc -v e 0.5 u -u o v C v (iii) and also fom momentum consevation pinciple, mu + mu C mv + mv C v + v C o v C + v (iv) om es. (iii) and (iv), we get v C 5.6 m/s u u u 77. V Vi ˆ+wR( - ˆi);V Vi;V ˆ C Vi ˆ+wRi ˆ u u V ˆ C - V wri u u [V - V C] [V(i) -V(i) -w R(i)] -w R(i) u u u u Hence VC - V -(V -V C) C ngula momentum of a body of mass m moving with velocity v in cicula obit of adius about the cente of the obit is mv om the angula momentum consevation pinciple, if v' is the linea velocity of the coment when it is fathest distance ' fom the sun, then mv' ' mv 0 v v' 0 m/s ' t. ^. R t Ia whee I MR (Moment of inetia of the disc about axis passing though its cente and ^ to its plane face) \ Ia. R R.R a I MR MR Thus, aµ R R C w V wr wr V V 0CE05 KOT / HS - 7/7

Physics 181. Assignment 4

Physics 181. Assignment 4 Physics 181 Assignment 4 Solutions 1. A sphee has within it a gavitational field given by g = g, whee g is constant and is the position vecto of the field point elative to the cente of the sphee. This