UNIFIED COUNCIL. An ISO 9001: 2015 Certified Organisation NATIONAL LEVEL SCIENCE TALENT SEARCH EXAMINATION (UPDATED)

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UNIFIED COUNCIL A ISO 900: 05 Certified Orgaisatio NATIONAL LEVEL SCIENCE TALENT SEARCH EXAMINATION (UPDATED) CLASS - (PCM) Questio Paper Code : UN6 KEY A C A B 5 B 6 D 7 C 8 A 9 B 0 D A C B B 5 C 6

1 UNIFIED COUNCIL A ISO 900: 05 Certified Orgaisatio NATIONAL LEVEL SCIENCE TALENT SEARCH EXAMINATION (UPDATED) CLASS - (PCM) Questio Paper Code : UN6 KEY A C A B 5 B 6 D 7 C 8 A 9 B 0 D A C B B 5 C 6 D 7 C 8 D 9 C 0 C B B D A 5 D 6 C 7 C 8 A 9 B 0 A C A D B 5 D 6 C 7 C 8 A 9 B 0 A C D B B 5 A 6 C 7 B 8 A 9 A 50 C 5 D 5 A 5 C 5 B 55 D 56 C 57 D 58 Del 59 A 60 D SOLUTIONS MATHEMATICS (A) Give f () + (A) F The set of parallelograms F The set of rectagles F The set of rhombus F The set of squares For domai of f (), 0 ad > 0 ad > By defiitio of parallelogram, opposite ad (, ) (, ) sides are equal ad parallel (, ) (, ] (C) Give, f () + + Give < <, the f () (B) Give, 6 siθ, cosθ, taθ are i GP + + cos θ So, f () is oe-oe ad oto fuctio 6 siθ taθ si è f () is a bijective cos θ 6cosè

2 6 6 6 cos θ cos θ + cos θ + si θ 0 cos θ 0 cos θ 0 (cosè ) (cos è+cos è+) 0 For cos θ+cos θ + 0value of cosθ is imagiary Hece, cosider oly [ ta(80 è ) ta è ; ta(90 + è ) cot è ] λ+/ λ + λ + λ λ λ 7 (C) Let S upto terms cosθ 0 cosθ cosθ cos π θ π ± π 5 (B) ta 5 ta (60 5 ) ta60 ta 5 +ta60 ta5 ta ta y Q ta ( y) +ta tay ta ad cot 5 ta 5 ta 5 + cot ( ) + ( + ) ( + )( ) ( ) () (D) Give, ta0 λ ta60 ta0 +(ta60 ) (ta0 ) ta(80 0 ) ta( ) + (ta(80 0 ) (ta (90 +0 )) ta0 + cot0 + ta0 cot0 ( + 8 (A) We have, a + ib upto terms upto terms [ th term of, 5, 8, + () + ] + + ( + ) ( + i ) [( + i )] ( i) [( i) ] ( + i+i ) ( i+i ) 008 (i) ( )(i) (i) i 9 (B) We have, a + ib + i + i i + i + i i [multiplyig umerator ad deomiator by + i] + i+i + i + i + () [i ] + i + +

3 O comparig real ad imagiary parts both sides, we get a ad b + + Now, a + b (i) [from eq ()] ( ) ( +) ( +) ++ ( +) ( +) ( +) 0 (D) We have, 9 < + ad > 6 From iequality (i), we get < + < 6 7 < + 9 (i) (ii) [multiplyig both sides by ] < [addig 7 o both sides] 6 < < + 6 [subtractig from both sides] < 6 < 9 [dividig both sides by ] Thus, ay value of less tha 9 satisfies the iequality So, the solutio of iequality (i) is give by (,9) < 9 9 From iequality (ii), we get > > > 6 [multiplyig by 6 o both sides] 7 + > 6 + [addig o both sides] 7 > > [subtractig 6 from both sides] > Thus, ay value of greater tha satisfies the iequality So the solutio set is (, ) (iv) > The solutio set of iequalities (i) ad (ii) are represeted graphically o umber lie as give below Clearly, the commo value of satisfyig iequalities (iii) ad (iv) lie betwee ad 9 Hece, the solutio of the give system is < < 9 ie, (, 9) (A) We have 0 itermediate statios betwee A ad B There are 7 statios where the trai does ot stop ad the three statios where the trai stops should be ay three of the 8 places Total umber of ways 8 C (C) We have, P m m P m m! + P + P + P ++ P + +! +! +! ++! + rr! + [( r+ ) ] r! r r r + [( r+) r! r!] r + [( r+)! r!] +[(!!)+(!!)+(!!)++( +)!!] +[(+)!!] (+)! Hece proved

4 (B) We have 6 distict white roses ad 5 distict red roses Total umber of way makig a garlad such that o two red roses come together is (6 )! 6! (B) We have, 6! 5! (!) (!) (!) 5 (C) We have, the biomial of Clearly, the epasio cotais 7 terms, therefore 5th term from the ed is (6 5 + ) rd term from the begiig T T + 6 C ( ) ( ) 8 6! 6 5! 5!!! 6 (D) We have,, ad y, y, y are i GP with the same commo ratio Let r be the commo ratio 8 6 +!, r ad r Similarly, y y y yr ad y yr ! Area of y y y + + +! y r yr r yr + + +! / / + (+) ( ) y r r 0 0 r r [ C, C are idetical] The give poits are colliear 7 (C) I the epasio of ( + ), we have T 5 T + C, T 6 T 5+ C 5 5 ad T 7 T 6+ C 6 6 Give, coefficiets of T 5, T 6 ad T 7 ad i AP C, C 5 ad C 6 are i AP!!!,,!( )! 5!( 5)! 6!( 6)! are i AP

5 ( )( )( ) ( )( )( ) ( ), 5 ( )( )( ) ( )( 5) 65 O multiplyig each term by ( )( )( ), we get are i AP 9 +0,, are i AP 5 0 As we kow that if a, b ad c are i AP the b a c b So, here we get ( 9) 6 6(9) ( 7) ( ) 0 7 or 8 (D) Sice, + y 9 0, + by 0 ad y 0 are cocurret 9 b 0 (b)(6+)9(b) 0 b b 0 b 70 b 5 O solvig equatios + y 9 0 ad y 0, we het ad y Cocurrecy poit of lie is (, ) Equatio of lie passig through (5, 0) ad (, ) is y 0 0 (+5) + 5 y ( + 5) y + 5 y (C) Let the vertices of ABC be A (, ), B(, ) ad C (0, 0) The orthocetre is the poit of itersectio of the altitudes from the vertices to the opposite sides F O A(, ) (h, k) E B(, ) D C (0, 0) Let AD, BE ad CF be the altitudes ad O (h, k) be the orthocetre of ABC AO BC Slope of lie AO Slope of lie BC k [ m h + m ] k h + h k (i) Also, BO AC Slope of lie BO Slope of lie AC k + 0 [ m h 0 m ] k + ( ) h h k + h h 5 0 (ii) O solvig Eqs (i) ad (ii), we get h k

6 h k h, k Hece, orthocetre is (, ) 0 (C) Slopes of lies X y ad k k 5y 9 0 are ad, respectively 5 k Let m ad m 5 The, agle betwee two lies whose slopes are m ad m, is give by ta θ m m + mm 5 ta k 5 k k 5+k 0 k ± 5+k 0 k 0 k or 5+k 5+k 0 k 5 + k or 0 k 5 k 5 k or But k > 0 k (B) Give curve is y Also, poit (, 0) is the focus of the parabola It is clear from the graph that oly oe ormal is possible y y y (, 0) (B) Equatio of lie BC is y + z λ (say) 8 Thus, coordiates of ay poit o te lie BC are ( λ,8λ, λ + ) Coordiates of D are of form ( λ,8λ, λ + ) Now, DR s of BC are, 8, ad DR s of AD are λ,, 8λ 9, λ AD BC ( λ ) + 8(8λ 9) + ( )( λ) 0 λ +6λ 5+9 λ0 77 λ 5 λ Hece, the coordiates of D are (, 5, ) (D) Cosider, lim 0 ( cos)( + cos) ta ( + cos ) si lim 0 ta si lim lim lim ( + cos ) 0 0 ta 0 lim (+) 0 ta (A) To prove, dy d y ( y) We have, y +5 O differetiatig both sides wrt, we get dy d d d +5 d d ( + 5) ( + 5) d d ( +5)

7 du dv v u d u Q d d d v v ( + 5) () ( + 0) +5 ( +5) ( +5) dy 5 d ( +5) dy 5 Now, LHS d ( +5) RHS y ( y) RHS ( +5) From Eqs (i) ad (ii), we get dy y ( y) d 5 (D) We have, g() (f(f() + ) Hece proved g () (f(f()+)) f (f()+)f () g (0) (f(f(0) + ) f (f(0) + )f (0) g (0) (f(()+)f (()+() [ f(0), f (0) ] g (0) [f(+)f (+)] g (0) f(0)f (0) g (0) () () PHYSICS du CdT v Cv R / 6 (C) dw RdT R R 7 (C) p atmospheric pressure, p ecess pressure iside smaller bubble of radius r cm r r p p+p p p+p p ecess pressure iside bigger bubble of radius r cm σ σ Now, p ;p p r r σ σ σ p p + + r r r Also Hece p σ r + r r r r σ σ σ + r r r rr cm r r (A) Let R be the origial radius of a plaet The attractio o a body of mass m GMm placed o its surface will be F R If size of the plaet is made double ie, R' R, the mass of the plaet becomes M' π( R) ρ 8 πr ρ 8 M New force

8 GM' m G 8 M m F' F R' (R) ie, force of attractio icreases due to the icrease i mass of the plaet 9 (B) The mass of the complete disc will be m M ad its momet of iertia about mr the ais will be Therefore, the momet of iertia of the half disc about the ais will be mr MR MR 0 (A) Relative velocity m/s Time required to meet distace 0 0 s relative vel Distace from st poit vel of first t m (C) Eergy desity eergy volume ML T L [M L T ] Youg s modulus MLT / L stress strai [M L T ] (A) Mass of flywheel m 00 kg Radius r m I mr 00 I 00 kg m Iitial agular velocity 0 ω 0 π rad / s ω ω 0 96 α θ 8 π 0 99 rad / s 0 Torque required to stop the flywheel τ Iα Nm (D) [pc] (MLT ) (LT ) ML T [torque] (B) Istataeous velocity υ bt t s, υ 800 m/s t 0 s, υ 0 80 m/s Average velocity υ +υ m / s 5 (D) Apply the law of coservatio of 00 mometum V υ (C) The total work doe is (0 J) + (0 J) 0 J So, by the work-eergy theorem, W total K, we have 0 J K Sice K K f K i, we fid K f K i + K 0 J + 0 J 0 J 7 (C) Whe a metal wire elogates by hagig a load Mg o it, decrease i potetial eergy of the load Mgl (where l elogatio i metal wire) Elastic potetial eergy stored i stretched wire Mgl Differece of Mgl ad Mgl appears as heat eergy i the stretched wire Eergy appearig as heat Mgl Mgl Mgl Fial agular velocity ω 0 Agular displacemet i revolutios π 8 π radia

9 8 (A) The coefficiet of epasio of iro is less tha that of the water but its desity is more tha the liquids The relative decrease i the desity of water will be more tha that or iro As a result, the buoyat force will decrease ad the apparet weight will icrease 9 (B) m 05 kg u 5 km 5 ms o θ 5 Impulse imparted to the ball mu cos θ cos ( 5 o ) kg m s The impulse imparted to the ball is 57 kg m s directed alog the bisector of iitial ad fial directio 0 (A) elogatio i sprig due to mass 0 kg 0 0 m 00 W F 00 [() () ] J CHEMISTRY (C) I B H 6, the B atoms are liked through hydroge bridges The structure is ot similar to that of C H 6, there is o B B bod ad also all the atoms do ot lie i the same plae (D) I the st oide, oyge 76 parts, metal parts I the d oide, oyge 0 parts, metal parts As st oide is M O, 7 parts of M atoms of M ad 76 parts of O atoms of O 70 parts of M 70 atoms of M 7 9 atoms of M 0 parts of O 0 atoms of O 76 5 atoms of O Ratio of M : O i the d oide 9 : 5 : 5 : Hece, the formula is M O (B) I H O structure, two OH bods lie i differet plaes (B) Iitial pressure: p atm 5 (A) C(s) + CO (g) ƒ CO (g) Equilibrium pressure : (p5p) p As give: p 5p + p ; p 8 atm At equilibrium: Pco 8 atm P CO atm K p p CO 8 6 atm PCO 050 Mole of P Mole of Q Total mole 007 Total pressure 750 mm Partial pressure of P moles of P total pressure total moles mm 007 Partial pressure of Q mm 007

10 6 (C) The give ios of elemets belog to d period The atomic radius decreases from left to right (Li, B, O, F) i the period Catios are smaller, whereas aios are larger tha the correspodig atoms The aio carryig more egative charge is larger So, O has the highest value of ioic radius 7 (B) N (g) + H (g) NH (g) L of N reacts with L of H to form L of NH Thus, N is the limitig reactat 0 L N will react with 0 L H to form 0 L NH As actual yield is 50% of the epected value, NH formed 0 L, N reacted 5 L, H reacted 5 L Miture will cotai 0 L NH, 5 L N, 5 L H 8 (A) Elemet with At No 9 will lose oe electro ad elemet with AtNo 7 will gai that electro to form a ioic water soluble compoud 9 (A) KMO O o of M K MO MO M O Thus, the highest oidatio umber of + 7 for M is i KMO 50 (C) Give that, (i) C (graphite) + O (g) CO (g); H 905 kcal (ii) C (diamod) +O (g) CO (g); H 950 kcal Thus, applyig the ispectio method, [Eq (i) Eq (ii)] we get C(graphite) + O (g) C (diamod) O (g) CO (g) CO (g); H 905 (950) or C(graphite) C (diamod); H + 05 kcal Sice this ethalpy chage is oly for coversio of mole, ie, g of C (graphite) to C (diamod), therefore, for the coversio of 0 g of C (graphite) to C (diamod) 0 H kcal 5 (D) C has si, N has seve ad O has eight electros Thus, the total umber of electros ad their distributio for each species are give below : O Species CN NO + O Total o of electros 7 MO cofiguratio σs σ*s σs σ*s πp πp y σp z π*p π*p y CN NO + CN NO + O N b N a 7 Bod Order / So, the species CN ad NO + have the same bod order 5 (A) Lies caot be assiged quatum umbers 5 (C) a is directly related to forces of attractio Hece, greater the value of a, more easily the gas is liquefied

11 5 (B) H + I ƒ HI Applyig law of mass actio, [H I] K c [H ][ I ] Give [H ] 80 mole litre [I ] 0 mole litre [HI] 80 mole litre So, (80) K c 66 (80) (0) 55 (D) Mg + O MgO g 6 g ( + 6)g Thus, 8 g of Mg requires g of O to form 80 g of MgO Therefore, 0 g of Mg requires 0 g 8 of O to form 80 0 g 8 of MgO or 0 g of Mg requires 0 g of O to form 50 g of MgO The residual miture thus cotais 50 g of MgO ad 0 g 0f O CRITICAL THINKING 56 (C) 57 (D) 58 (Del) 59 (A) 60 (A) THE END

SAFE HANDS & IIT-ian's PACE EDT-10 (JEE) SOLUTIONS

SAFE HANDS & IIT-ian's PACE EDT-10 (JEE) SOLUTIONS . If their mea positios coicide with each other, maimum separatio will be A. Now from phasor diagram, we ca clearly see the phase differece. SAFE HANDS & IIT-ia's PACE ad Aswer : Optio (4) 5. Aswer : Optio