Delete Chemistry Time: 2 hours Marks: 60 [10 ´ 2 = 20] 1. Calculate the molarity of water if its density is 1000 kg/m3. 2. The average velocity of gas molecules is 400 m/sec. Calculate its rms velocity at the same temperature. 3. Write down the heterogeneous catalyst involved in the polymerisation of ethylene. 4. Which one is more soluble in diethyl ether anhydrous AlCl3 or hydrous AlCl3? Explain in terms of bonding. 5. Using VSEPR theory, draw the shape of PCl5 and BrF 5. 6. A racemic mixture of (±) 2-phenyl propanoic acid on esterification with (+) 2 -butanol gives two esters. Mention the stereochemistry of the two esters produced. 7. Wavelength of high energy transition of H-atoms is 91.2nm. Calculate the corresponding wavelength of He atoms. 8. Match the Ka values Ka a) Benzoic acid 3.3 ´ 10–5 b) O 2 N COOH 6.3 ´ 10–5 c) Cl COOH 30.6 ´ 10–5 d) H3CO COOH 6.4 ´ 10–5 e) C H3 COOH 4.2 ´ 10–5 9. Write down reactions involved in the extraction of Pb. What is the oxidation number of lead in litharge? 10. Following two amino acids lionise and glutamine form dipeptide linkage. What are two possible dipeptides? N H2 COOH NH2 HOOC COOH NH2 + [10 ´ 4 = 40] 11. a) You are given marbles of diameter 10 mm. They are to be placed such that their centres are lying in a square bound by four lines each of length 40 mm. What will be the arrangements of marbles in a plane so that maximum number of marbles can be placed inside the area? Sketch the diagram and derive expression for the number of molecules per unit area. b) 1 gm of charcoal adsorbs 100 ml 0.5 M CH3COOH to form a monolayer, and thereby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01 ´ 102 m2/gm. 12. a) Will the pH of water be same at 4°C and 25°C? Explain. b) Two students use same stock solution of ZnSO4 and a solution of CuSO4. The emf of one cell is 0.03 V higher than the other. The conc. of CuSO4 in the cell with higher emf value is 0.5 M. Find out the conc. of CuSO4 in the other cell (2.203 RT/F = 0.06). IIT – JEE 2003(MAINS SOLIIIIUTIONS) 13. Convert a) COOH COOH F(in not more than 3 steps) b) OH Asprin 14. There is a solution of p-hydroxy benzoic acid and p-amino benzoic acid. Discuss one method by which we can separate them and also write down the confirmatory tests of the functional groups present. 15. ) Cl H C ( HCl 12 6 13 6 C B ) H C ( A + ¾ ¾ ® ¾ D B KOH . alc ¾ ¾ ¾ ® ¾ (isomer of A) E D ozonolysis ¾ ¾ ¾ ¾ ® ¾ (it gives negative test with Fehling solution but responds to iodoform test). G F A Ozonolysis + ¾ ¾ ¾ ¾ ® ¾ (both give positive Tollen’s test but do not give iodoform test). alcohol primary a HCOONa G F NaOH . conc + ¾ ¾ ¾ ¾ ® ¾ + Identify from A to G. 16. Identify the following: D C B A CO Na 2 3 2 2 I S Elemental CO Na SO 3 2 ¾® ¾ ¾¾ ¾ ¾ ¾ ® ¾ ¾ ¾ ¾ ® ¾ ¾ ¾ ® ¾ D Also mention the oxidation state of S in all the compounds. 17. Write the IUPAC nomenclature of the given complex along with its hybridisation and structure. K2 [Cr (NO)(NH3)(CN)4], m = 1.73 BM 18. A mixture consists A (yellow solid) and B (colourless solid) which gives lilac colour in flame. a) Mixture gives black precipitate C on passing H2S(g). b) C is soluble in aqua-regia and on evaporation of aqua-regia and adding SnCl2 gives greyish black precipitate D. The salt solution with NH4OH gives a brown precipitate. i) The sodium extract of the salt with CCl4/FeCl3 gives a violet layer. ii) The sodium extract gives yellow precipitate with AgNO3 solution which is insoluble in NH3. Identify A and B, and the precipitates C and D. 19. a) Match the following if the molecular weights of X, Y and Z are same. Boiling Point Kb X 100 0.68 Y 27 0.53 Z 253 0.98 b) Cv value of He is always 3R/2 but Cv value of H2 is 3R/2 at low temperature and 5R/2 at moderate temperature and more than 5R/2 at higher temperature explain in two to three lines. 20. a) C H3 CH2 OH Write resonance structure of the given compound. b) Compound A of molecular formula C9 H7O2Cl exists in ketoform and predominantly in enolic form ‘B’. On oxidation with KMnO4, ‘A’ gives m- chlorobenzoic acid. Identify ‘A’ and ‘B’. ************************************************************************************************* Chemistry Solutions 1. 1 litre water = 1kg i.e. 1000 g water (Q d = 1000 kg/m3) º 18 1000 = 55.55 moles of water So, molarity of water = 55.55M 2. Crms = M RT 8 C , MRT 3 av p = 8 3 RT 8 M MRT 3 C C av rms p = p ´ = = 1.085 Crms = 1.085 ´ Cav = 1.085 ´ 400 = 434 ms–1 3. Ziggler Natta catalyst (R3Al + TiCl4) 4. Oxygen atom of diethyl ether by donation of its lone pair to vacant 3p orbitals of Al in anhydrous AlCl3 solvates it more compared to hydrous AlCl3. 5. PCl5 : sp3d Trigonal bipyramid BrF5 : sp3d2 Square pyramidal Cl Cl Cl PCl F F F F Br F Cl 6. ) (+ Ph COOH H C H3 H C2H5 OH C H3 C H3 O CH3 H Ph O H5C2 H C H3 O CH3 H Ph O H5C2 H + ) (+ ) (- (Racemic mixture) ) (+ ) (± ) (+ The bonds attached to the chiral carbon in both the molecules are not broken during the esterification reaction. (+) Acid reacts with (+) alcohol to give an (++) ester while (–) acid reacts with (+) alcohol to give (+ –) ester. These two esters are diastereoisomers. 7. ÷ ÷ ø ö ç ç è æ - = l 2 2 2 1 2 H n 1 n 1 Z . R 1 RH is a constant and transition remaining the same, 2 Z 1 µ l 41 ZZ2He 2H H He = = l l So, lHe = 2 . 91 41 ´ = 22.8 nm 8. Ka value a) Benzoic acid 6.3 ´ 10–5 b) p-NO2 - C6H4 - COOH 30.6 ´ 10–5 c) p-Cl - C6H4 - COOH 6.4 ´ 10–5 d) p-CH3 -C6H4 – COOH 4.3 ´ 10–5 e) p-OCH3 – C6H4 – COOH 3.3 ´ 10–5 9. 2PbS + 3O2 ¾® 2PbO + 2SO2 PbS + 2O2 ¾® PbSO4 PbS + 2PbO ¾® 3Pb + SO2 PbS + PbSO4 ¾® 2Pb + 2SO2 Oxidation number of Pb in litharge (PbO) is +2. 10. NH N H OO N H2 CO2H N H NH OO O OH N H2 11. a) Area of quadrilateral = q ´ ´ ´ sin a a 21 2 = a2´sinq Where a = length of the side of the quadrilateral To have the maximum area, i.e. sinq = 1 or q = 90°. In other words, the quadrilateral must be a square q a fig.1 Area of square = 4 ´ 4 = 16 cm2 Again to have the maximum no. of spheres the packing must be hcp Maximum no. of sphere s= 18 (see fig. 2) Area = 16 sq. cm \ No. of spheres per cm2 = 16 18 = 1.125 fig.2 b) No. of m mole of CH3COOH initially taken = 100 ´ 0.5 = 50 Since concentration reduces to 0.49 M \ Final no. of m mole of CH3COOH = 100 ´ 0.49 = 49 \ No. of m mole of CH3COOH get adsorbed = 50 – 49 = 1 \ No. of molecules of CH3COOH get adsorbed = 6.02 ´ 1020 Since 1g charcoal has area = 3.01 ´ 102 m2 Q 6.02 ´ 1020 molecules of acetic acid gets adsorbed in 3.01 ´ 102 m2 area \ 1 molecule of acetic acid gets adsorbed = 20 2 10 02 . 6 10 01 . 3 ´´ = 18 10 21 - ´ = 5 ´ 10–19 m2 12. a) At 25°C: Kw = 10–14 pKw= 14 \ pH + pOH = 14 Pure water being neutral, pH = pOH = 7 As temperature decreases, K w decreases and hence pKw increases. Obviously, pH of water which is pKw/2 will increase. Thus, pH of water at 4°C will be more than that at 25°C. b) Daniel cell is: Zn | Zn2+ || Cu2+ | Cu Let there be two Daniel cells with their Ecell as given below: Zn | Zn2+ (C1) || Cu2+ (C = ?) | Cu, Ecell = E1 Zn | Zn2+ (C2) || Cu2+ (C = 0.5 M) | Cu Ecell = E2 where E2 > E1 From question E2 – E1= 0.03 and C 2 = C 1 The cell reaction is Zn + Cu2+ ¾® Zn2+ + Cu, Q = ] Cu [ ] Zn [ 22++ So, E cell = ] Cu [ ] Zn [ log 206 . 0 E 22 0cell ++ - Thus, E1 = C C log 206 . 0 E 1 0cell - E2 = 0.5 C log 2 0.06 E 2 0cell - So, E2 – E1 = úû ù êë é ´ 1 2 C5 . 0 C C log 206 . 0 Þ 0.03 = C5 . 0 log 206 . 0 Þ C5 . 0 log = 1 C = 0.05 M 13. a) COOH D ¾ ¾ ¾ ¾ ¾ ® ¾ 4 2SO H . conc COOH SO3H COOH F fusion KHF2 ¾ ¾ ® ¾ Alternatively COOH 4 2 3 SO HHNO¾ ¾ ® ¾ 4 2 HBF ) ii C 5 0 HCl /NaNO ) i ¾ ¾ ¾ ¾ ¾ ® ¾ ° - COOH NO2 COOH NH2 ¾ ¾ ¾ ® ¾ HCl /Sn COOH F b) OH + ¾ ¾ ¾ ¾ ¾ ® ¾H ) ii temperture and pressure High CO /NaOH ) i 2 OH COOHpyridine AcOH O Ac2 ¾ ¾ ¾ ¾ ¾ ® ¾ + O COOH O CH3 14. filtre HCl ¾ ¾ ® ¾ OH COOH NH2 COOH [NH3+]Cl COOH OH COOH in aqueous solution in non-soluble form + p-NH2 COOH & p-OH COOH both will give effervesences of CO2 with NaHCO3 p-NH2 COOH will give positive azo dye test p-OH COOH will give positive liebermann nitroso test 15. ¾® ¾ C H2 CH3 CH3 C H3 C H3 Cl C H3 CH3 CH3 ¾ ¾ ® ¾HCl C H3 CH3 Cl CH3 CH3 O3/H2O KOH H O C H3 CH3 CH3 C H3 CH3 C H3 CH3 ¾ ¾ ¾ ® ¾ O H /O 2 3 C H3 OCH3 + HCOONaconc. NaOH (cross cannizaro) H OH C H3 CH3 CH3 + HCHO (A) (B) (F) (G) (C) 16. Na2CO3 + 2SO2 + H2O ¾® ) A ( 3 NaHSO 2 + CO2 2NaHSO3 + Na2CO3 ¾® ) B ( 3 2SO Na 2 + H2O + CO2 Na2SO3 + S ¾® ) C ( 3 2 2 O S Na 2 2Na2S2O3 + I2 ¾® ) D ( 6 4 2 O S Na 2 + 2NaI Oxidation states of ‘S’ are in (A) and (B) (+4) in (C) (+6, –2) in (D) (+5, 0) 17. Potassiumamminetetracyano(C)nitrosoniumchromium(I) Cr is in + 1 state and d2sp3 hybridisation m = 3 ) 2 n ( n = + = 1.73 B.M. 18. [A = KI, B = HgI2] HgI2 + H2S ¾® HgS + 2HI HgI2 ¾ ¾ ¾ ¾ ® ¾ -regia aqua HgCl2 HgCl2 + SnCl2 ¾® ) black greyish ( Hg ¯ + SnCl4 2KI + HgI2 ¾® orange 4 2 ] HgI [ K ; 2K2HgI4 + NH3+ 3KOH ¾® base s ' Milon of Iodide 2 I ] HgOHgNH [ + 7KI + 2H2O Na2CO3 + HgI2 ¾® Hg¯ + NaI + CO2 + O 2 NaI + Fe3+ ¾® ) violet layer CCl ( 24 Na I + + + Fe2+ AgNO3 + I– ¾® ) NH in le lub inso ( Yellow 3 AgI 19. a) Kb = v 2 b v 2 b H 1000 M RT , l 1000 RT D = = S 1000M RTBD úû ù êë é D = MH l v v Q A change from liquid to vapour at boiling point is accompanied by increase in disorderness and hence increase in entropy. However, since a vapour is highly disordered state the difference of the extent of disorderness between vapour and liquid is so high that even if the extent of disorderness varies from liquid to liquid, the same may be considered to be almost equal. In otherwords DS may be considered to be constant. M is the same and R is constant so, Kb µ Tb. Thus, K b (x) = 0.68, Kb(y) = 0.53 and Kb(z) = 0.98. b) Helium being mono-atomic it has only translational degrees of freedom. The contribution of transitional degree of freedom towards Cv being R/2 so Cv = 3 ´ R/2 = 3R/2. Hydrogen molecule is diatomic. However, at low temperature rotational and vibrational contribution are also zero so Cv is 3R/2. At moderate temperature rotational contribution (C = 2 ´ R/2) also becomes dominant and at even higher temperature vibrational contribution (1 ´ R) also becomes significant. 20. a) C H3 CH2 OH C H3 CH2 OH b) O CHOCl (A) O H CHOCl (B) stable form KMnO 4 Metachlorobenzoic acid Physics 1. If nth division of main scale coincides with (n+1)th divisions of vernier scale. Given one main scale division is equal to ‘a’ units. Find the least count of the vernier. 2. Find the focal length of the lens shown in the figure. The radii of curvature of both the surfaces are equal to R. m1 m2 m3 R R m1 < m2 < m3 3. Frequency of a photon emitted due to transition of electron of a certain element from L to K shell is found to be 4.2 ´ 1018 Hz. Using Moseley’s law, find the atomic number of the element, given that the Rydberg’s constant R = 1.1 ´ 107 m-1. 4. An insulated container containing mono atomic gas of molar mass m is moving with a velocity v0. If the container is suddenly stopped, find the change in temperature. 5. A particle of mass m, moving in a circular path of radius R with a constant speed v2 is located at point (2R, 0) at time t = 0 and a man starts moving with a velocity v1 along the +ve y-axis from origin at time t = 0. Calculate the linear momentum of the particle w.r.t. the man as a function of time. R m v2 v1 y x (0, 0) 6. A tuning fork of frequency 480 Hz resonates with a tube closed at one end of length 16 cm and diameter 5 cm in fundamental mode. Calculate velocity of sound in air. 7. How a battery is to be connected so that the shown rheostat will behave like a potential divider? Also indicate the points about which output can be taken. A B C R 8. Charges +q and -q are located at the corners of a cube of side a as shown in the figure. Find the work done to separate the charges to infinite distance. -q +q -q +q +q -q +q -q 9. A radioactive sample emits n b-particles in 2 sec. In next 2 sec it emits 0.75 n b-particle, what is the mean life of the sample? 10. In a photoelectric experiment set up, photons of energy 5 eV falls on the cathode having work function 3 eV. (a) If the saturation current is iA = 4mA for intensity 10-5 W/m2, then plot a graph between anode potential and current. (b) Also draw a graph for intensity of incident radiation 2 ´ 10-5 W/m2. 11. Hot oil is circulated through an insulated container with a wooden lid at the top whose conductivity K = 0.149 J/(m-°C-sec), thickness t = 5 mm, emissivity = 0.6. Temperature of the top of the lid is maintained at Tl = 127°. If the ambient temperature T a = 27°C. Calculate (a) rate of heat loss per unit area due to radiation from the lid. (b) temperature of the oil. (Given s = 3 17 ´ 10-8) Tl= 127°C Ta = 27°C Hot Oil To 12. Two masses m1 and m2 connected by a light spring of natural length l0 is compressed completely and tied by a string. This system while moving with a velocity v0 along +ve x-axis pass through the origin at t= 0. At this position the string snaps. Position of mass m1 at time t is given by the equation x1(t) = v0t - A(1 - coswt) Calculate (a) position of the particle m2 as a function of time. (b) l0 in terms of A. 13. Shown in the figure is a prism of an angle 30° and refractive index mp = 3 . Face AC of the prism is covered with a thin film of refractive index mf = 2.2. A monochromatic light of wavelength l = 550 nm fall on the face AB at an angle of incidence of 60°. Calculate (a) angle of emergence. (b) minimum value of thickness t so that intensity of emergent ray is maximum. 30° 60° mp = Ö3 A B C mf = 2.2 14. A body is projected vertically upwards from the bottom of a crater of moon of depth 100 R where R is the radius of moon with a velocity equal to the escape velocity on the surface of moon. Calculate maximum height attained by the body from the surface of the moon. 15. A square loop of side ‘a’ with a capacitor of capacitance C is located between two current carrying long parallel wires as shown. The value of I in the wires is given as I = I0 sinwt. (a) Calculate maximum current in the square loop. (b) Draw a graph between charges on the upper plate of the capacitor vs time. I a a a I 16. A charge +Q is fixed at the origin of the co-ordinate system while a small electric dipole of dipole-moment P r pointing away from the charge along the x-axis is set free from a point far away from the origin. (a) Calculate the K.E. of the dipole when it reaches to a point (d, 0). (b) Calculate the force on the charge +Q at this moment. 17. A wheel of radius R having charge Q, uniformly distributed on the rim of the wheel is free to rotate about a light horizontal rod. The rod is suspended by light inextensible strings and a magnetic field B is applied as shown in the figure. The initial tensions in the strings are T 0. If the breaking tension of the strings are 2T 3 0 , find the maximum angular velocity w0 with which the wheel can be rotated. T0 T0 w0 d B 18. A string tied between x = 0 and x = l vibrates in fundamental mode. The amplitude A, tension T and mass per unit length m is given. Find the total energy of the string. x = 0 x = l 19. A bubble having surface tension T and radius R is formed on a ring of radius b (b << R). Air is blown inside the tube with velocity v as shown. The air molecule collides perpendicularly with the wall of the bubble and stops. Calculate the radius at which the bubble separates from the ring. R b v 20. Shown in the figure is a container wh ose top and bottom diameters are D and d respectively. At the bottom of the container, there is a capillary tube of outer radius b and inner radius a. The volume flow rate in the capillary is Q. If the capillary is removed the liquid comes out with a velocity of v0. The density of the liquid is given as r. Calculate the coefficient of viscosity h. d D P h r SOLUTIONS 1. (n + 1) division of vernier scale = n division of main scale \ one Vernier division = 1 nn+ main scale division Least count = 1 M.S.D. – 1 V.D. = 1 n 1+ M.S. D. = 1 na+ 2. For an object placed at infinity the image after first refraction will be formed at v1 R v 1 2 1 12 + m - m = ¥ -m - m … (i) The image after second refraction will be found at v2 R v v 2 3 12 23 + m - m = m - m … (ii) adding (i) and (ii) R v 1 3 23 m - m = m Þ 1 3 3 2 R v m - mm = Therefore focal length will be 1 3 3Rm - mm 3. (Z – 1)2 Rhc n = úû ù êë é - h 41 11 (Z – 1)2 = RC 3 4 n Þ Z = 42 4. Loss in K.E. of the gas DE = 21 (nm) 20 v , where n = number of moles. If its temperature change by DT. Then n 23 RDT = 21 (nm) 20 v Þ DT = R 3 mv20 . v1 v2 5. w = Rv2 ) jˆ t cos v iˆ t sin v ( v 2 2 2 w + w - = j ˆ v v 1 1 = v PM = jˆ ) v t cos v ( iˆ t sin v v v 1 2 2 1 2 - w + w - = - p r PM = m v PM = -mv2 sin wt jˆ ) v t cos v ( m iˆ 1 2 - w + . wt v2 wt 6. (l + 0.6r) = f 4v 4 = l v = 4f (l + 0.6 r) = 336 m/s. 7. Battery should be connected across A and B. Out put can be taken across the terminals A and C or B and C. A B C R 8. Wexternal = DPE = úû ù êë é - + - pe 3 1 2 3 13 a q 4 1 2 0 ´ 28 = ] 2 6 3 3 3 [ 6 4 . a q 4 1 2 0 - - pe . -q +q -q +q +q -q +q -q 9. Let N be the number of active nuclei at time t = 0. Hence n = N0 (1 – e-2l) 1.75 n = N0 (1 – e-4l) Þ l - l - -- = 42 e 1 e 1 75 . 1 1 Þ e-4l -1.75 e-2l + 0.75 = 0 Þ ) 3 /4 ln( 2 1 = l sec. 10. Vs = -2V 4mA 8mA (a) (b) 11. (a) dt dQ = seA[(T l)4 - (Ta)4], Rate of heat loss per unit area = 595 watt /m2. (b) Let T o be the temperature of the hot oil Þ ( ) A 595 t T T KA o = - l Þ To » 420 K 12. (a) x1 = v0t - A(1 - coswt) xcm = t v m m x m x m 0 2 1 2 2 1 1 = ++ Þ x2 = v0t + ( ) t cos 1 A mm21 w - (b) a1 = t cos A dtx d 2 21 2 w w - = The separation (x2 - x1) between the two blocks will be equal to l0 when the acceleration will be equal to zero. x2 - x1 = ( ) ( ) t cos 1 A t cos 1 A mm21 w - + w - for a1 = 0 x2 – x1 = ÷ ÷ø ö ç çè æ + = 1 mm21 0 l A Alternate (b) In center of mass reference frame, maximum separation of the blocks = 2l0 (using conservation of energy). If x1 and x2 be the separation of the blocks from center of mass at the moment of maximum separation x1 + x2 = 2l0 and m1x1 = m2x2 Þ x1 = ÷ ÷ø ö ç çè æ + 21 0 mm 1 2l but x1 = 2A Þ l0 = A ÷ ÷ø ö ç çè æ + 21 mm 1 13. (a) mairsin60° = mpsinr Þ 3 23 = sinr Þ r = 30° The refracted ray inside the prism hits the other face at 90°; hence deviation produced by this face is zero and hence angle of emergence is zero. 60° (b) Multiple reflection occurs between the surfaces of the film for minimum thickness Dx = 2mt = l, where t = thickness Þ t = m l 2 = 125 nm 14. For escape velocity from the surface of moon Þ ve = RGM 2 where M is the mass of the moon. P.E. inside the crater of moon will be = - ò + 100 R 99 R 3 dx R GMmx R GMm = R 20000 GMm 199 - K.E. of the body at the highest point will be zero R 20000 GMm 199 - + 2e mv 21 = h RGMm + - + 0 Þ h » 100 R 15. (a) Flux through the square loop = ò - + p m a 2a 0 adx x a 3 1 x1 I 2 4 = p m40 Ia4ln2 Induced emf e = - dt df = - p m 0 aI0w ln2 coswt I x I dx Charge on the capacitor Q = Ce = -C p m0 aI0w ln2 coswt = -Q0coswt (say) Current in the loop = dt dQ = p m 0 CI0w2a ln2 sinwt Imax = p m 0 CI0 aw2 ln2 (b) p/2w p/w 3p/2w 2p/w Q0 Q -Q0 t 16. (a) K.E.final = -DPE. = E . P r r = 22 0 dQ 4Ppe (b) F = E Q r = 3 0d 2 QP pe along positive x-axis. 17. 2T0 = mg . . . (1) For moment of forces about P q p w2max ´pR2 ´ B + mg 2d = d 2T 3 0 ÷ ÷ø ö ç çè æ Þ wmax = 2 0 qBRd T T2 T1 w0 O d B P 18. Equation of the standing wave in the string is y = A sin kx cos wt where k = p/l and w = m p T l w - = A dt dy sin kx sin wt Þvmax (x) = Aw sin kx E = ò w m l0 2 2 2 kx sin dxA 21 = l 4 T A 2 2p . 19. 2pb ´ 2Tsin q = rAv2 Þ 4pbT ´ Rb = rpb2v2 Þ R = 2 vT 4 r T q T Tsinq Tsinq 20. When the tube is not there. P + hrg + 21 r 21 v = 21 r 20 v d D P h r 0 2 1 2 v 4b v 4D p = p By Poiseuille’s equation the rate of flow of liquid in the capillary tube Q = l h D p8Pa4 where DP = P + rgh = 21 r 20 v úû ù êë é - 22 Db 1 Q = 21 r 20 v úû ù êë é - 22 Db 1 l h p8a 4 h = l Q 16 v a 20 4r p úû ù êë é - 22 Db 1 MATHEMATICS Time: 2 hours Marks: 60 1. If z1 and z2 are two complex numbers such that |z1| < 1 < |z2| then prove that 1 z z z z 1 2 1 2 1 < - - . [2] 2. Find a point on the curve x2 + 2y2 = 6 whose distance from the line x + y = 7, is minimum. [2] 3. If matrix A = ú ú ú ûù ê ê ê ëé b a c a c b c b a where a, b, c are real positive numbers, abc = 1 and ATA = I, then find the value of a3 + b3 + c3. [2] 4. Prove that 2 k ( ) ÷ ÷ø ö ç çè æ = ÷ ÷ø ö ç çè æ - ÷ ÷ø ö ç çè æ - - ÷ ÷ø ö ç çè æ -- ÷ ÷ø ö ç çè æ + ÷ ÷ø ö ç çè æ -- ÷ ÷ø ö ç çè æ - ÷ ÷ø ö ç çè æ ÷ ÷ø ö ç çè æ - - kn 0 k n kn 1 ..... 2 k 2 n 2n 2 1 k 1 n 1n 2 kn 0n k 2 k 1 k . [2] 5. If f is an even function then prove that ( ) ( ) ò ò p p = 4 /0 2 /0 dx x cos x 2 sin f 2 dx x cos x 2 cos f . [2] 6. For a student to qualify, he must pass at least two out of three exams. The probability that he will pass the 1st exam is p. If he fails in one of the exams then the probability of his passing in the next exam is 2p otherwise it remains the same. Find the probability that he will qualify. [2] 7. For the circle x2 + y2 = r2, find the value of r for which the area enclosed by the tangents drawn from the point P(6, 8) to the circle and the chord of contact is maximum. [2] 8. Prove that there exists no complex number z such that |z| < 31 and 1 z a n1 r r r = å= where |ar| < 2. [2] 9. A is targeting to B, B and C are targeting to A. Probability of hitting the target by A, B and C are 31 and 21 , 32 respectively. If A is hit then find the probability that B hits the target and C does not. [2] 10. If a function f : [-2a, 2a] ® R is an odd function such that f(x) = f(2a -x) for x Î[a, 2a] and the left hand derivative at x = a is 0 then find the left hand derivative at x = -a. [2] 11 Using the relation 2(1 -cos x) < x2, x ¹ 0 or otherwise, prove that sin (tan x) ³ x, " x Î úû ù êë é p4 , 0 [4] 12. If a, b, c are in A.P., a2, b2, c2 are in H.P., then prove that either a = b = c or a, b, -2c form a G.P. [4] 13. If x2 + (a -b)x + (1 -a -b) =0 where a, b Î R then find the values of a for which equation has unequal real roots for all values of b. [4] 14. Normals are drawn from the point P with slopes m1, m2, m3 to the parabola y2 = 4x. If locus of P with m1 m2 = a is a part of the parabola itself then find a. [4] 15. If the function f : [0, 4] ® R is differentiable then show that (i). For a, bÎ (0, 4) , (f(4))2 -(f(0))2 = 8f¢(a) f(b) (ii). ( ) ( ) ( ) [ ] 2 2 40 f f 2 dt t f b b + a a = ò " 0 < a, b < 2 [4] 16. (i). Find the equation of the plane passing through the points (2, 1, 0), (5, 0, 1) and (4, 1, 1). (ii). If P is the point (2, 1, 6) then find the point Q such that PQ is perpendicular to the plane in (i) and the mid point of PQ lies on it. [4] 17. If P(1) = 0 and ) x ( P dx ) x ( dP > for all x ³ 1 then prove that P(x) > 0 for all x > 1. [4] 18. If In is the area of n sided regular polygon inscribed in a circle of unit radius and On be the area of the polygon circumscribing the given circle, prove that In = ÷ ÷ ÷ øö ç ç ç èæ ÷ø ö çè æ - + 2 2 1 1 2 nI O n n . [4] 19. If w , v , u r r r are three non-coplanar unit vectors and a, b, g are the angles between v and u r r , w and v r r , u and w r r respectively and z , y , x r r r are unit vectors along the bisectors of the angles a, b, g respectively. Prove that [ ] [ ] 2 sec 2 sec 2 sec w v u 16 1 x z z y y x 2 2 2 2 g b a = ´ ´ ´ r r r r r r r r r . [4] 20. A right circular cone with radius R and height H contains a liquid which evaporates at a rate proportional to its surface area in contact with air (proportionality constant = k > 0). Find the time after which the cone is empty. [4] SOLUTIONS 1. To prove 2 1 2 1 z z z z 1 - < - Û ( )( ) ( )( ) 2 1 2 1 2 1 2 1 z z z z z z 1 z z 1 - - < - - Û 2 2 2 1 1 2 2 1 2 2 2 1 2 1 2 1 z z z z z z z z z z z z 1 + - - < + - - Û (1 -|z1|2) -|z2|2 (1 -|z1|2) < 0 Û (1 -|z2|2) (1 -|z1|2) < 0 Which is obvious as |z1| < 1 < |z2|. 2. ( ) q q sin 3 , cos 6 P shortest distance exists along the common normal Slope of normal at P = 1 tan 2 ec cos 3 sec 6 = q = q q so cosq = 32 and sinq = 3 1 Hence P º (2, 1). x + y = 7 N P 3. ATA = I ú ú ú ûù ê ê ê ëé b a c a c b c b a ú ú ú ûù ê ê ê ëé b a c a c b c b a = ú ú ú ûù ê ê ê ëé 1 0 0 0 1 0 0 0 1 Þ ú ú ú ûù ê ê ê ëé = ú ú ú ûù ê ê ê ëé + + + + + + + + + + + + + + + + + + 1 0 0 0 1 0 0 0 1 c b a ca bc ab ca bc ab ca bc ab c b a ca bc ab ca bc ab ca bc ab c b a 2 2 2 2 2 2 2 2 2 Þ a2 + b 2 + c2 = 1 ….(1) and ab + bc + ca = 0 ….(2) Now a3 + b3 + c3 = (a + b + c) (a2 + b2 + c2 -ab -bc -ca) + 3abc = (a + b + c) + 3 …(3) Now (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 1 + 2. 0 = 1 Þ a + b + c = 1 (since a, b, c are real positive number) Now from (3) a3 + b3 + c3 = 1 + 3 = 4. Alternate: ATA = I Þ |ATA| = | I | Þ |A|2 = 1 Þ (a3 + b3 + c3 -3abc)2 = 1 Þ a3 + b 3 + c3 -3abc = 1 (since a, b, c are positive real number Þ a3 + b 3 + c3 ³ 3abc from AM ³ GM) Þ a3 + b 3 + c3 = 4 4. ( ) ( ) ( ) ( ) ! r k ! k n ! r n ! r ! r n ! n 2 ) 1 ( C C 2 ) 1 ( r k k0 r r r k r n r n r k k0 r r - - - - - = - - = - - - = å å = ( ) ( ) ! r k ! r ! k ! k ! k n ! n 2 ) 1 ( r k k0 r r - - - - = å = r k k0 r r k k n C 21 2 C å= ÷ø ö çè æ- = nCk 2k k 21 1 ÷ø ö çè æ - = nCk. 5. ò p 2 /0 xdx cos ) x 2 (cos f = ò p ú úû ù ê êë é ÷ø ö çè æ - p ÷ ÷ø ö ç çè æ ÷ø ö çè æ - p + 4 /0 dx x 2 cos x 2 2 cos f x cos ) x 2 (cos f = ( ) [ ] ò p - + 4 /0 dx x sin x 2 cos f x cos ) x 2 (cos f = ( )[ ] ò p + 4 /0 dx x sin x cos x 2 cos f = ( ) ò p ÷ø ö çè æ - p 4 /0 dx x 4 cos x 2 cos f 2 = ò p ÷ ÷ø ö ç çè æ ÷ø ö çè æ - p - p ÷ ÷ø ö ç çè æ ÷ø ö çè æ - p 4 /0 dx x 4 4 cos x 4 2 cos f 2 = 2 ò p 4 /0 xdx cos ) x 2 (sin f 6. Let Ei : d enotes the event that the student will pass the ith exam, i = 1, 2, 3 E: denotes the event that the student will qualify. P (E) = P (E1) ´ P ÷ ÷ø ö ç çè æ 12 EE + P (E1) ´ P ÷ ÷ø ö ç çè æ ¢12 EE ´ P ÷ ÷ø ö ç çè æ ¢23 EE + P (E¢1) ´ P ÷ ÷ø ö ç çè æ ¢12 EE ´ P ÷ ÷ø ö ç çè æ 23 EE = p2 + p ´ (1 – p) 2p + (1 – p) ´ 2p ´ p Þ P (E) = 2 p -p p -p 2p 3 2 3 2 2 + + = 2p2 – p3 7. Since OP = 10, sinq = 10 r where q Î ÷ø ö çè æ p2 , 0 A = 21 ´ 2r cosq (10 -r sinq) = 10 sinq cosq (10 -10 sin2q) Þ A = 100 cos2q sinq cosq [ ] q q - q = q 2 2 4 cos sin 3 cos 100 d dA O(0, 0) P(6, 8) r q q = 300 cos4q ÷ ÷ø ö ç çè æ q + ÷ ÷ø ö ç çè æ q - tan 3 1 tan 3 1 Þ A is maximum at q = 6p Þ r = 10´ 21 = 5 units. 8 a1z + a2z2 + …. + anzn = 1 Þ | a1z + a2z2 + …. + anzn| = 1 Þ | a1z| + |a 2z2 | + …. + |anzn| ³ 1 Þ 2[ |z| + |z|2 + …. + |z|n] > 1 Þ ( ) 1 z 1 z 1 z 2 n > - ÷ø öçè æ- as |z| < 31 , |z| > 1 n z 32 31 + + Þ |z| > 31 This is a contradiction. Hence there exists no such complex number. 9. P(A) = Probability that A will hit B P(B) = Probability that B will hit A P(C) = Probability that C will hit A P(E) = Probability that A will be hit Then P(E) = ( ) ( ) ( ) 32 32 . 21 1 C P . B P 1 C B P 1 = - = - = Ç - ( ) ( ) ( ) 21 3232 . 21 E P C P . B P E C B P = = = ÷ ÷øö ç çèæ Ç 10. L.H.D. at x = a - ¢f (a) = h ) a ( f ) h a ( f lim0 h - + - ® = 0 (given) - ¢f (-a) = h ) a ( f ) h a ( f lim0 h - - + - - ® = h ) a ( f ) a h ( f lim0 h + + - - - ® (As f is odd) = h ) a ( f ) a h a 2 ( f lim0 h + - + - - ® = -- ¢f (a) = 0. 11. Let f (x) = sin (tan x) – x f¢ (x) = cos (tan x) sec2 x – 1 = tan2x cos(tanx) + cos(tanx) -1 > tan2x cos(tanx) -2 x tan2 Þ f¢(x) > tan2x(cos(tanx) -cos3p ) > 0 (since 0 £ tanx £ 1 < 3p ) Þ f(x) is an increasing function " x Î úû ù êë é p4 , 0 As f(0) = 0 Þ f(x) ³ 0 " x Î úû ù êë é p4 , 0 Þ sin(tan x) ³ x. 12. 2b = a + c …(1) b2 = 2 2 2 2 2 2 c a c a c a 2 ÷ø ö çè æ + = + Þ (a2 + c2)2 + 2ac(a2 + c2) = 8a2c2 Þ (a2 + c2 + ac)2 = 9a2c2 Þ a2 + c2 + ac = ± 3ac …(2) Þ a2 + c2 -2ac = 0 Þ a = c Þ b = c or, a 2 + c2 = -4ac Þ (a + c)2 = -2ac Þ 4b2 = -2ac Þ b2 = -2 ac Þ a, b, -2c are in G.P. 13. For unequal real roots D > 0 Þ (a - b)2 - 4 (1 - a - b) > 0 Þ b2 + b (4 - 2a) + a2 + 4a - 4 > 0 For the above quadratic expression to be true " b Î R Discriminant of its corresponding equation should be less than zero i.e. (4 - 2a)2 - 4 (a 2 + 4a - 4) < 0 Þ - 32a + 32 < 0 Þ a > 1 . 14. Let the point P be (h, k) Þ k = mh – 2m – m3 or, m3 + m(2 – h) + k = 0 Þ m1m2m3 = – k Þ m3 = – ak Þ ( ) 0 k h 2 k k 3 = + - a - ÷ø ö çè æ a - Þ k2 = a2h -2a2 + a3 Þ y2 = a2x -2 a2 + a3 Comparing it with y2 = 4x, we get a2 = 4 and -2 a2 + a3 = 0 Þ a = 2. Alternate: Since locus of P is a part of the parabola Þ normals at any two points t1 and t2 meet at P Þ t1t2 = 2 Þ (-m1) (-m2) = 2 Þ a = 2 m2 m3 m1 P(h, k) t1 t2 15(i). From Lagrange’s mean value theorem ( ) ( ) ( ) a f 0 4 0 f 4 f ¢ = -- for a Î (0, 4) …(1) Also from Intermediate mean value theorem ( ) ( ) ( ) b f 2 0 f 4 f = + for b Î (0, 4) …(2) From (1) and (2), we get ( ) ( ) ( ) ( ) ( ) ( ) b f a f 8 0 f 4 f 2 2 ¢ = - . for a, b Î (0, 4) (ii). Replacing t by z2, we get ( )dt t f 40ò = ( ) dz z f z 2 20 2 ò From Lagrange’s mean value theorem ( ) ( ) ( ) 2 00 2 20 2 f 2 0 2 dz z zf 2 dz z zf 2 g g = -- ò ò for g Î (0, 2) Þ ( ) ( ) ( 2 20 2 f 2 2 dz z zf 2 g g = ò = 2 ( ) ( )÷ ÷øö ç çèæ b b + a a 2 f 2 f 2 2 2 ( where 0< a < g Þ 0 ) x ( P dx ) x ( dP > - Þ ) e ) x ( P ( dx d x - > 0 Þ P (x) . e–x is an increasing function. Þ P (x) e–x > P (1) e–1 " x ³ 1 Þ P (x) e–x > 0 " x > 1 (since P (1) = 0) Þ P (x) > 0 " x > 1. 18. In = 2n r2 sin n 2p Þ n 2 sin nI 2 n p = .…(1) On = nr2 tan np ….(2) From (1) and (2), we get n tan n 2 sin 21 OI n n pp = = cos2 2 1 n 2 cos n + p = p = 2 nI 2 1 1 2 n ÷ø ö çè æ - + ( using (1) ) In = ÷ ÷ ÷ øö ç ç ç èæ ÷ø ö çè æ - + 2 n n nI 2 1 1 2 O 19. ( ) v u 2 sec 21 v u v u x r r r r r r r + a = ++ = , Similarly for vectors z and y r r As ( ) ( ) ( ) [ ] x z z y y x r r r r r r ´ ´ ´ = [ ]2 z y x r r r = [ ]2 2 2 2 u w w v v u 2 sec 2 sec 2 sec 64 1 r r r r r r + + + g b a = [ ]2 2 2 2 w v u 2 sec 2 sec 2 sec 64 4 r r r g b a As [ ] [ ] w v u 2 u w w v v u r r r r r r r r r = + + + = [ ]2 2 2 2 w v u 2 sec 2 sec 2 sec 16 1 r r r g b a 20. Let the semi vertical angle of the cone be q = tan–1 ÷ø ö çè æ HR Let height of the liquid at time ‘t’ be ‘h’ from the base BC and radius r. Volume of liquid at time ‘t’ = V = 31 pr2h = 31 pr3 cot q S = Surface area in contact with air at time ‘t’ = pr2 Given that – dt dV µ S Þ – dt dV = kS = kpr2 R r B C H h q Þ 3 cot q p 3r2 dt dr = – kpr2 Þ cot q ò ò - = T0 0R dt k dr (where T is the required time) Þ RH R = kT Þ T = kH *********************************************************************************************************