Narayana IIT/NEET Academy INDIA IIT_XI-IC_SPARK 2016_P1 Date: Max.Marks: 186

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1 Narayaa IIT/NEET Academy INDIA IIT_XI-IC_SPARK 6_P Date: 5--8 Max.Marks: 86 KEY SHEET PHYSICS B B c 4 B 5 c 6 ac 7 ac 8 ac 9 ad abc bc acd ad CHEMISTRY 9 c b c a a 4 bc 5 ab 6 abcd 7 ab 8 cd 9 abc acd abd MATHS 7 a 8 c 9 d 4 d 4 c 4 abc 4 ad 44 ab 45 cd 46 abcd 47 bc 48 abc 49 abcd

2 Narayaa IIT Academy SOLUTIONS IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & So s ' V + V = V Vs + = 8 5 = 98 + Hz f V V = = f V V V + V V = m / s s s ( θ ) V = V + r s ( α θ )( α θ ) L = L + + θ ( α θ )( α θ ) + r = + + r = α + α PV P = P V PV PV V PV T = = R RV dt V = V = dv 9PV T = 4R PHYSICS 5. Fractio of Eergy supplied for icremet i iteral eergy r = 5 5 (ad r= ) Therefore eergy supplied = = 6% 5 Ad eergy supplied for exteral work doe = 4% 6. Frequecy received by A ad B are v + u = v v u = v + = 7. Page

3 Narayaa IIT Academy. V + V wall f ' = f V Vwall f ' f = % f θ C= T θ It is possible to obtai almost ay set of values. A r. = A r % icrease i area =.4% V..6% V = = l = lα T 4 α =.5 % C. Vibratio eergy of mooatomic gas = At all temperatures Ad, for diatomic gas 5R At high temperature, C v> 4. f = 9 + v 5. v f = 9 + So, f = 9 v ad f = 9 V also, f f = 9 v v = m / s ( C) ( C) ( C) ( C) E E water water E = MC T = 5 = kcal Ice Ice E =.5 = kcal So, 8 kcal of eergy left wire melt kg of ice at C 5 = α t At C = α ( t) At 5 C, 7. IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & So s Page

4 Narayaa IIT Academy R C = 6 Cv = R w ad = C Cv = T ( + ) R R = 5 8. = 6 R ( P P )( V V ) = ( V V )( P P ) RT Also, P = V P V VV T = V R ( + 4 ) dt ad = V = V dv 4PV Tmax = R IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & So s Page 4

5 Narayaa IIT Academy IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & So s CHEMISTRY 9. Alkoxymercuratio-demercuratio-additio of CH OH. Hydroboratio-oxidatio-additio of water.. Formatio of CCl (dichlorocarbee) which act as attackig aget ad added to multiple bod.. CH N : CH + N Carbee. Reductive ozoolysis 4. Cis alkee sy additio meso product CSM Tras alkee ati additio meso product TAM 6. Hoffma s rule applicable i all case. Due to presece of bulky leavig group.. HI give markoikov s product eve i presece of peroxide.. H Br H Br CH H 5. 4 (reagets ii, iii, v & vi, give sy additio with alkee) C H, C H, C H N, C H, C H Degree of usaturatio = ( ) = o. of carbo = o. of hydroge + Page 5

6 Narayaa IIT Academy IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & So s MATHEMATICS 7. p,q,r lie o the circle z =, whose circumcetre is origi. Also, (p+q+r)/=. Hece the ceroid coicides with circumcetre. So the triagle is equilateral. Now, ( p + q + r) = p + q + r = pqr + + p q r pqr + + = aω + bω + c ω ω 8. Let α ( ) be th roots of uity, the the vertices of the regular -go iscribed i the uit circle of radius with cetre of O are determied by the umbers ( )( α )( α )...( α ). α, α, α,... α, where z = z z z z let M be a poit o the ray OA, lyig outside the polygo, the usig A. M G. M., we have = k MA r α k= k k= r > r = r = OM = = MA r r OM MA OM k= k k= 9. ( ) ( )( )( )...( ) 4. Z Z Z Z Z Z Z Z Takig logarithms & o differetiatio. Z = Z Z Z Z Z Z For Z= = Z Z Z + + t t t t ( + t ) A k B E D C cosθ cosθ cosθ cosθ + + siθ cosθ siθ siθ + cotθ Page 6

7 Narayaa IIT Academy si ( θ + θ ) si ( θ + θ ) siθ cos ecθ siθ siθ si ( θ + θ ) si ( θ + θ ) siθ siθ Apply sie rule i ABD ad ACD AD a AD a = ; = si C siθ si B siθ + θ si C siθ = si B si ( θ + θ )... ( i) Apply sie rule i ABE ad AEC AE = a ; AE = a si C siθ + θ si B siθ si B siθ = si C si ( θ + θ ) From (i) ad (ii) siθ + θ siθ + θ siθ siθ... = 4 4. A A= [ OA A is equilibrium] A A = A A4 = x le OA A, apply cosie rule. I x = x = x = A A A A A A = 4 4. W is colliear with z ad z. m 4+ m rπ π i m m e = ( i) = m= r= m= 45. We have x + x + x ω x ω Sice f(x) is divisible by x + x + f ω =, f ω = ( ω ) + ω ( ω ) = + ω = ( ω ) + ω ( ω ) = + ω =... P Q P Q P Q P Q Solvig () ad () P = ad Q = We obtai f x = P x + xq x we get f(x) is divisible by x By simple geometry, i AFE, we get AF=AE. IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & So s Page 7

8 Narayaa IIT Academy IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & So s Therefore, AFE, is a isosceles triagle Now area ( ABC) = are ( ABD) + area A ( ADC) bc si A = cadsi + b A bc cos A ADsi AD b + c A AD = AE cos 47. A cos B + cosc = 4si B + C B C 4 cos cos = 4si B C cos = B + C cos By compoets ad dividedo, we get B C B + C cos + cos = B C B + C cos cos B C s a ta ta = = s s = s + a b + c = a AB + AC > BC Therefore, locus of A is ellipse. a b c ap + bp + cp = K y = + + P P P is miimum whe a b c y = ( ap + bp + cp ) + + K P P P 48. Give ( a + b + c) y K P P P P P P = = = = = = P P P P P P P is icetre of ABC 49. For a right agled triagle iscribed i a circle of radius R the legth of the hypotheuse is R. The area is maximum whe its is isosceles triagle Page 8

9 Narayaa IIT Academy With each side= R S = ( + ) R = ( + ) R, = R R = R = = = r r r r R R π π ω = cos + si = + i ω is oe of cue root of uity. z + ω ω ω z ω ω + = R R + R + R z z z ω z ω ω z + ω + = + + = z = = z= ω ω z + ω ( α)( α ) ( α8 ) 9 ( α )( α ) ( α ) x x x x = x = 8 Sice α = α whe i+j=9 ( α )( α )( α )( α ) = 5 = C C absi C = 6b si + 6asi C C C C absi cos = 6bsi + 6a si + = a b 9 λ = 9 π π cos + cos = + k k π π + π cos + cos = Let = k k k + t + t = where cos θ / = t ( ) IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & So s Page 9

10 Narayaa IIT Academy 54. =, 4 θ π = k = 6 = absi C 5 si C = = = C = ab cos = 4 ow r r = s 5 = = IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & So s Page

SINGLE CORRECT ANSWER TYPE QUESTIONS: TRIGONOMETRY 2 2

Class-Jr.X_E-E SIMPLE HOLIDAY PACKAGE CLASS-IX MATHEMATICS SUB BATCH : E-E SINGLE CORRECT ANSWER TYPE QUESTIONS: TRIGONOMETRY. siθ+cosθ + siθ cosθ = ) ) ). If a cos q, y bsi q, the a y b ) ) ). The value