PART - I : PHYSICS. SECTION I : Single Correct Answer Type ALLEN. n=1.5. n=1.2. R=14cm. (A) cm (B) 40.0 cm (C) 21.5 cm (D) 13.

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1 IIT-JEE EXAMINATIN TARGET : IIT-JEE 3 (eld on 8-4-) AER - ( ANSWERS & SLUTINS)

2 IIT-JEE EXAMINATIN (ELD N : 8-4-) ART - I : YSICS SECTIN I : Single Correct Answer Type This section contains multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which NLY NE is correct.. A biconvex lens is formed with two thin plano-convex lenses as shown in the figure, Refractive index n of the first lens is.5 and that of the second lens is.. Both the curved surfaces are of the same radius of curvature R = 4 cm. For this biconvex lens, for an object distance of 4 cm, the image distance will be :- Ans. (B) n=.5 R=4cm n=. (A) 8. cm (B) 4. cm (C).5 cm (D) 3.3 cm Equivalent focal length of system is ( ) æ ö.5 (. ) æ ö = - ç + - f è ç 4ø è4ø = from lens equation - = we get v u f = + f f f v - ( 4) = ; v = + 4 cm -. A thin uniform rod, pivoted at, is rotating in the horizontal plane with constant angular speed w. as shown in the figure. At time t =, a small insect starts from and moves with constant speed v with respect to the rod towards the other end. If reaches the end of the rod at t= T and stops. The angular speed of the system remains w throughout. The magnitude of the torque ( t ) on the system about, as a function of time is best represented by which plot? z w v t t t t (A) T t (B) T t (C) T (D) t T t Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: / 33

3 Ans. (B) IIT-JEE EXAMINATIN (ELD N : 8-4-) Net torque on a system is rate of change of angular momentum. Let insect be at a radial distance r from axis at an instant. Angular momentum of system as function of r is given by r æ ML ö ç è 3 ø L = + mr w As w is of constant magnitude, angular momentum varies due to variation in r. dl di t = = w dt dt d ML w æ mr ö = ç + dt è 3 ø æ dr ö = w ç + mr ( ) dr = m r è dt ø w æ ç ö è dt ø = mw( vt)( v) = ( mwv t) t µ t Thus graph is linear passing through origin. 3. Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at temperatures T and 3T respectively. The temperature of the middle (i.e. second) plate under steady state condition is (A) æ65ö ç è ø /4 T (B) æ97ö ç è 4 ø /4 T (C) æ97ö ç è ø /4 T (D) ( ) /4 97 T Ans. (C) T q 3T p received p emitted p recieved p emitted For steady state, net received = net emitted 4 ( ( ) 4 ) ( ) 4 4 sa q - T = saéë 3T -q ù û q - 6T = 8T - q 97 q =ç æ ö è ø /4 T / 33 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website:

4 IIT-JEE EXAMINATIN (ELD N : 8-4-) 4. Consider a thin spherical shell of radius R with its centre at the origin, carrying uniform positive surface r charge density. The variation of the magnitude of the electric field Er ( ) and the electric potential V(r) with the distance r from the centre, is best represented by which graph? E(r) V(r) E(r) V(r) (A) (B) (C) R r E(r) V(r) (D) R E(r) R r R Ans. (D) Electric field inside a uniformly charged spherical shell : E E= r=r KQ R V kq Eµ /r E inside = ; r < R E outside = r r r=r vµ /r r KQ Electric potential : Vinside = ; R r R V r V(r) r ; r ³ R outside KQ = ; r r ³ R æ 4MLgö 5. In the determination of Young's modulus ç Y = è pld ø by using Searle's method, a wire of length L = m and diameter d =.5 mm is used. For a load M =.5 kg, an extension l =.5 mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They have the same pitch of.5 mm. The number of divisions on their circular scale is. The contributions to the maximum probable error of the Y measurement (A) due to the errors in the measurements of d and l are the same. (B) due to the error in the measurement of d is twice that due to the error in the measurement of l. (C) due to the error in the measurement of l is twice that due to the error in the measurement of d. (D) due to the error in the measurement of d is four times that due to the error in the measurement of l. Ans. (A) Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: 3 / 33

5 IIT-JEE EXAMINATIN (ELD N : 8-4-) 4MLg y ( ) ( d) y = d d d p d y = l l l + d d l [.5 /] d ( d) [.5 /] = = and = = l.5 5 d A small block is connected to one end of a massless spring of un-stretched length 4.9 m. The other end of the spring (see the figure) is fixed. They system lies on a horizontal frictionless surface. The block is stretched by. m and released from rest at t =. It then executes simple harmonic motion Ans. (A) with angular frequency p w = rad/s. Simultaneously at t =, a small pebble is projected with speed 3 v from point at an angle of 45 as shown in the figure. oint is at a horizontal distance of m from. If the pebble hits the block at t = s, the value of v is (take g = m/s ) z m (A) 5 m/s (B) 5 m/s (C) 5 m/s (D) 53 m/s Time of flight = v sin 45 = g 45 g v = = 5 m/s 7. Young's double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are b G, b and R B v b, respectively. Then (A) bg > bb > br (B) bb > bg > br (C) br > bb > bg (D) br > bg > bb Ans. (D) Dl Fringe width is given by ( b ) = and order of wavelength in electromagnetic spectrum is given d by lr > lg > lb 8. A small mass m is attached to a massless string whose other end is fixed at as shown in the figure. The mass is undergoing circular motion in the x-y plane with centre at and constant angular speed w. If the angular momentum of the system, calculated about and are denoted by L r and L r respectively, then z x m w 4 / 33 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website:

6 (A) (B) (C) (D) L r and L r IIT-JEE EXAMINATIN (ELD N : 8-4-) do not vary with time L r varies with time while L r remains constant L r remains constant while L r varies with time L r and L r both vary with time Ans. (C) L L v v r r Angular momentum of particle about is vertical as shown in figure. Angular momentum L L r r v about point is as shown in figure at an instant w L L L L L r is constant & L r is changing in direction. 9. A mixture of moles of helium gas (atomic mass = 4 amu) and mole of argon gas (atomic mass = 4 amu) is kept at 3 K in a container. The ratio of the rms speeds (A).3 (B).45 (C).4 (D) 3.6 Ans. (D) v rms 3RT = M ( v ) ( v ) M 4 = = = M 4 = 3.6 rms e Ar rms Ar e æv ( ) rms helium ö ç è v argon ø is rms ( ) Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: 5 / 33

7 IIT-JEE EXAMINATIN (ELD N : 8-4-). Two large vertical and parallel metal plates having a separation of cm are connected to a DC voltage source of potential difference X. A proton is released at rest midway between the two plates. It is found to move at 45 to the vertical JUST after release. Then X is nearly (A) 5 V (B) 7 V (C) 9 V (D) V Ans. (C) E mg As proton moves at 45 in vertical plane it implies magnitude of electric force and gravitational force is same. for 45 direction mg = qe mg V = E = q d mgd V = = q qe -7 - (.6 kg)( m) -9 ( ) = 9 volt.6 C Section-II : Multiple Correct Answer(s) Type This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which NE or MRE are correct.. A cubical region of side a has its centre at the origin. It encloses three fixed point charges, q at (, a/4, ), +3q at (,,) and q at (, +a/4, ). Choose the correct option(s). a z -q 3q -q y 6 / 33 x (A) The net electric flux crossing the plane x = +a/ is equal to the net electric flux crossing the plane x = a/ (B) The net electric flux crossing the plane y = +a/ is more than the net electric flux crossing the plane y = a/. (C) The net electric flux crossing the entire region is (D) The net electric flux crossing the plane z = +a/ is equal to the net electric flux crossing the plane x = +a/. Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: q e

8 Ans. (A,C,D) IIT-JEE EXAMINATIN (ELD N : 8-4-) For (A): osition of charges is symmetrical with respect to planes at x = + a/ and x = a/ For (C) : Net charge enclosed by cube is q. For (D) : Charges are symmetrically placed iwth given planes.. For the resistance network shown in the figure, choose the correct option(s). I W 4W W W 4W Q V (A) the current through Q is zero (B) I = 3A (C) The potential at S is less than that at Q Ans. (A,B,C,D) I W W W 4W 4W Q V W S T W W W 4W 4W S T (D) I = A Since resitances in two arms upper and lower are in same ratio. Current through Q is zero. I W 4W W S W W W 4W 4W Q T V I = = A, I = + = 3A and potential of S is less than that at Q. 6 6 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: 7 / 33

9 IIT-JEE EXAMINATIN (ELD N : 8-4-) 3. A small block of mass of. kg lies on a fixed inclined plane Q which makes an angle q with the horizontal. A horizontal force of N acts on the block through its center of mass as shown in the figure. The block remains stationary if (take g = m/s ) Q N (A) q = 45 (B) q > 45 and a frictional force acts on the block towards (C) q > 45 and a frictional force acts on the block towards Q (D) q < 45 and a frictional force acts on the block towards Q Ans. (A,C) Q cosq N q N sinq q () If sinq = cosq q = 45 no friction will act and the block will remain at rest. () If sinq > cosq q > 45 friction will act towards Q (3) If sinq < cosq q < 45 friction will act towards 4. Consider the motion of a positive point charge in a region where there are simultaneous uniform electric r and magnetic fields E= Ej ˆ r and ˆ B= Bj. At time t =, this charge has velocity v r in the x-y plane, making an angle q with the x-axis. Which of the following option (s) is (are) correct for time t >? (A) If q =, the charge moves in a circular path in the x-z plane. (B) If q =, the charge undergoes helical motion with constant pitch along the y-axis. (C) If q =, the charge undergoes helical motion with its pitch increasing with time, along the y- axis (D) If q = 9, the charge undergoes linear but accelerated motion along the y-axis Ans. (C,D) y q B,E q vcosq is perpendicular to magnetic field and it plays role of tangential velocity (responsible for circular motion). vsinq is axial velocity parallel to magnetic field responsible for axial motion. If q =, the path of the charge particle will be helical with increasing pitch. If q =, the path of the charge particle will be helical with increasing pitch. If q = 9, the path of the charge particle will be linear and accelerating. x 8 / 33 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website:

10 IIT-JEE EXAMINATIN (ELD N : 8-4-) 5. A person blows into open-end of a long pipe. As a result, a high-pressure pulse of air travels down Ans. (B,D) the pipe. When this pulse reaches the other end of the pipe. (A) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is open (B) a low -pressure pulse starts travelling up the pipe, if the other end of the pipe is open (C) a low pressure pulse starts travelling up the pipe, if the other end of the pipe is closed (D) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed Section-III : Integer Answer Type This section contains 5 questions. The answer to each question is a single digit Integer, ranging from to 9 (both inclusive) 6. An infinitely long solid cylinder of radius R has a uniform volume charge density r. It has a spherical cavity of radius R/ with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point, which is at a distance R from the axis of the cylinder, is given by the expression 3rR 6ke. The value of k is x R/ z y Ans.6 Electric field due to cylinder = Kl R KQ Electric field due to sphere = ( ) R From principle of superposition. = ( rp ) [where K = 4p Î ] 3 æ 4p æ R öö K( l) KQ ( ) K r E = - R ( R) = K( rpr ) ç è ç 3 è 8 ø ø - R 4R K( )( R) 3 R 3 R K R - r p 6 4 = rp r k 4pe 4 = 6 6e = Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: 9 / 33

11 IIT-JEE EXAMINATIN (ELD N : 8-4-) 7. A cylindrical cavity of diameter a exists inside a cylinder of diameter a as shown in the figure. Both the cylinder and the cavity are infinitely long. A uniform current density J flows along the length. N If the magnitude of the magnetic field at the point is given by m aj, then the value of N is a Ans.5 Magnetic field on the surface of cylinder is given by mja m ja Magnetic field at radial distance r outside is given by r From principle of superposition we get r ( ) J a J a 5 aj B = m 5 - m p m N æ3aö 4 = = pç è ø 8. A lamina is made by removing a small disc of diameter R from a bigger disc of uniform mass density and radius R, as shown in the figure. The moment of inertia of this lamina about axes passing through and is I and I respectively. Both these axes are perpendicular to the plane of the lamina. The I ratio I to the nearest integer is Ans.3 C / 33 Let mass of compelte disc of radius R is 4M & mass of removed part is M. ( ) 4M R MR 3 I é ù = - MR MR ê + = ë ú û é4m( R) ù ém( R) ù 37 I 4 ( ) ( 5 ) = ê + M R ú- ê + M R ú= MR ë û ë û I I o 37 =» 3 3 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website:

12 IIT-JEE EXAMINATIN (ELD N : 8-4-) 9. A circular wire loop of radius R is placed in the x-y plane centred at the origin. A square loop of side a (a<<r) having two turns is placed with its center at z = 3 R along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of 45 with respect to m the z-axis. If the mutual inductance between the loops is given by /, then the value of p is ar p 45 z Ans.7 x m Magnetic field at a distance x along axis of a circular coil is given by ( ) ir B x = ( R + x ) Flux linked with N coils of square loop inclined at q = 45 with vertical is given by é m ( ) ir ù f x = ê N( a )( cos45 ) 3/ ú ë( R + x ) û As = Mi so M = m R ( ) 3/ ç 7/ ( ) R é è ø R + 3R ù a æ a ö = m a thus p = 7 ë û. A proton is fired from very far away towards a nucleus with charge Q = e, where e is the electronic charge. It makes a closest approach of fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is (take the proton mass, m = (5/3) 7 kg; h/e = 4. 5 J.s/C; Ans.7 4pe = 9 9 m/f; fm = 5 m) +e +Q proton r min 3/ From work energy theorem we get and de-broglie wavelength is given by h l = = p h mk Welectric =D KE ; h h = = æ Qq ö m( e)( e) mç è4p Î r ø 4p Î r min min on substituting numerical values we get l = 7 5 m ( Qq) = mv 4p Î r min Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: / 33

13 IIT-JEE EXAMINATIN (ELD N : 8-4-) ART-II : CEMISTRY SECTIN I : Single correct Answer Type This section contains multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which NLY NE is correct.. As per IUAC nomenclature, the name of the complex [Co( ) 4 (N 3 ) ]Cl 3 is : Ans. (D) (A) Tetraaquadiaminecobalt(III) chloride (B) Tetraaquadiamminecobalt(III) chloride (C) Diaminetetraaquacobalt(III) chloride (D) Diamminetetraaquacobalt(III) chloride [Co( ) 4 (N 3 ) ]Cl 3 Diamminetetraaquacobalt(III) chloride. In allene (C 3 4 ), the type(s) of hybridisation of the carbon atoms is (are) Ans. (B) (A) sp and sp 3 (B) sp and sp (C) only sp (D) sp and sp 3 Allene structure C C C sp sp sp The type of hybridisation of carbon atoms are sp & sp 3. For one mole of a van der Waals gas when b = and T=3K, the V vs. /V plot is shown below. The value of the van der Waals constant a (atm. liter mol ) is V (liter-atm mol ) [Graph not to scale]». 3. V (mol liter ) (A). (B) 4.5 (C).5 (D) 3. / 33 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website:

14 IIT-JEE EXAMINATIN (ELD N : 8-4-) Ans.(C) æ a ö ç + (V - ) = RT è V ø as (n = ) V = RT a V n comparing slope from graph.-.6 a = 3- - a =.5 = a =.5 4. The number of optically active products obtained from the complete ozonolysis of the given compound is : C 3 C C = C C C = 3 C C C = C C 3 C 3 (A) (B) (C) (D) 4 Ans. (A) C 3 C C = C C C = 3 C C C = C C 3 C 3 ¾ ¾ 3 Zn¾ C 3 C = + All are optically inactive products No. of optically active product = C 3 = C C C = + C 3 = C C C = + C 3 C = 5. A compound M p X q has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below. The empirical formula of the compound is : M = X = (A) MX (B) MX (C) M X (D) M 5 X 4 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: 3 / 33

15 Ans.(B) 'X' is forming fcc. Effective "X" atoms per unit cell = 4 IIT-JEE EXAMINATIN (ELD N : 8-4-) Effective "M" atoms per unit cell = = Formula = MX. 6. The number of aldol reaction(s) that occurs in the given transformation is Ans. (C) C 3 C + 4C conc. aq. Na (A) (B) (C) 3 (D) 4 C 3 C = + C C C C C = C C C C = C = C C C C C C = ( time aldol) C C = ( time aldol) C C C C = C (3 times aldol) C C + C C C C + C C (crossed cannizaro product) the carboxyl functional group ( C ) is presen in. 7. The colour of light absorbed by an aqueous solution of CuS 4 is - Ans. (A) (A) orange-red (B) blue-green (C) yellow (D) violet 4 / 33 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website:

16 V I B G Y R l n E IIT-JEE EXAMINATIN (ELD N : 8-4-) R V,I B Y G The colour of aqueous solution of CuS 4 is blue so colour of light absorbed by an aqueous solution of CuS 4 is orange-red. 8. The carboxyl functional group ( C) is present in - (A) picric acid (B) barbituric acid (C) ascorbic acid (D) aspirin Ans. (D) N N N icric acid : Barbituric acid : N N C C 3 C Ascorbic acid : Aspirin : (Acetyl salicylic acid) 9. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a is Bohr radius] (A) Ans. (C) h 4p ma (B) h 6p ma (C) h 3p ma (D) h 3p ma v = nh pmr nh K.E. = m. 4p mr for second orbit K.E. = 4h 8p mr since r = a r = 4a h K.E. = 3p ma Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: 5 / 33

17 IIT-JEE EXAMINATIN (ELD N : 8-4-) 3. Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen- (A) N 3, N, N 4 Cl, N (B) N 3, N, N, N 4 Cl (C) N 3, N 4 Cl, N, N (D) N, N 3, N 4 Cl, N Ans. (B) + N 3 N N N Cl.S + X + 3 ( ) =.S X + ( ) =.S. = zero.s. X + 4 = + X = 6 X = + X = 3 X = +5 so decreasing order is N 3, N, N N 4 Cl SECTIN-II : Multiple Correct Answser(s) Type This section contains 5 multiple choice question. Each question has four choices (A), (B), (C) and (D) out of which NE or MRE are correct. 3. Identify the binary mixtures (s) that cna be separated into the individual compounds, by differential Ans (B, D) extraction, as shown in the given scheme - Binary mixture containing Compund and compound (A) C 6 5 and C 6 5 C (C) C 6 5 C and C 6 5 Na(aq) NaC (aq) Acidic structure order of some organic compounds 3 Compound + Compound Compound + Compound (B) C 6 5 C and C 6 5 C (D) C 6 5 C and C 6 5 C C 4 C65C, CCC > C3 > C > > C C (A) C 6 5, C 6 5 C ¾ ¾ separated by NaC 3 only but not by Na (B) C 6 5 C, C 6 5 C ¾ ¾ separated by NaC 3 & Na both. (C) C 6 5 C, C 6 5 ¾ ¾ separated by Na only but not by NaC 3 (D) C 6 5 C, C 6 5 C C ¾ ¾ separated by NaC 3 & Na both. 3. Choose the correct reason(s) for the stability of the lyophobic colloidal particle. (A) referential adsorption of ions on their surface from the solution (B) referential adsorption of solvent on their surface from the solution (C) Attraction between different particles having opposite charges on their surface (D) otential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles 6 / 33 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website:

18 Ans. (A,D) IIT-JEE EXAMINATIN (ELD N : 8-4-) 33. Which of the following molecules, in pure from, is (are) unstable at room temperature? (A) (B) (C) (D) Ans (B, C) Both are antiaromatic. So are unstable. Non aromatic stable at room temperature. Aromatic, stable at room temperature. 34. Which of the following hydrogen halides react(s) with AgN 3 (aq) to give a precipitate that dissolves in Na S 3 (aq): (A) Cl (C) Br Ans. (A, C, D) AgN 3 + Cl ¾¾ AgCl (white ppt.) (B) F (D) I AgCl + Na S 3 ¾¾ [Ag(S 3 ) ] 3 soluble complex AgN 3 + Br ¾¾ AgBr pale yellow ppt. AgBr + Na S 3 ¾¾ [Ag(S 3 ) ] 3 soluble complex AgN 3 + I ¾¾ AgI yellow ppt. AgI + Na S 3 ¾¾ [Ag(S 3 ) ] 3 soluble complex AgN 3 + F ¾¾ No precipitation Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: 7 / 33

19 IIT-JEE EXAMINATIN (ELD N : 8-4-) 35. For an ideal gas, consider only -V work in going from an initial state X to the final state Z. The final state Z can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct? [take DS as change in entropy and w as work done] (A) DS x z = DS x y + DS y z (C) W x y z = W x y Ans.(A,C) (atmosphere) For option A : As entropy is state function. DS x z = DS x y + DS y z For option C : since W y z = as (V constant) W x y z = W x y + W y z X Y Z V(L) (B) W x z = W x y + W y z (D) DS x y z = DS x y W x y z = W x y SECTIN III : Integer Answer Type This Section contains 5 questions. The answer to each question is a single digit integer, ranging from to 9 (both inclusive) 36. The substitutes R and R for nine peptides are listed in the table given below. ow many of these peptides are positively changed at p = 7.? + 3N C C N C C N C C N C C R R eptide R R I II C 3 III C C IV C CN (C ) 4 N V C CN C CN VI (C ) 4 N (C ) 4 N VII C C C CN VIII C (C ) 4 N IX (C ) 4 N C 3 8 / 33 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website:

20 IIT-JEE EXAMINATIN (ELD N : 8-4-) Ans.4 (I) Neutral peptide (II) C 3 Neutral peptide (III) C C Acidic peptide ( )ve charge (IV) C CN (C ) 4 N Basic peptide (+)ve charge (V) C CN C CN Neutral peptide (VI) (C ) 4 N (C ) 4 N Basic peptide (+)ve charge (VII) C C C CN Acidic peptide ( )ve charge (VIII) C (C )N Basic peptide (+)ve charge (IX) (C ) 4 N C 3 Basic peptide (+)ve charge ence IV, VI, VIII & IX C, amides C C N are neutral groups. 37. The periodic table consists of 8 groups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below. To which group, element X belongs in the periodic table? Ans.8 Cu+ 6 n+a+ +X 63 9 Cu+ 6 n+ e+ + X X has atomic number = 6 = atomic number of iron group number = When the following aldohexose exists in its d-configuration, the total number of stereoisomers in its pyranose form is - C C C C C C Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: 9 / 33

21 IIT-JEE EXAMINATIN (ELD N : 8-4-) Ans.8 C= 6 C C C C Fixed configuration Fixed configuration D-configuration (Four stereoisomers w.r.t. to C & C chiral centres) C Fixed configuration New chiral centre generated at C carbon in pyranose form hence total 8 pyranose form w.r.t. to C & C variable configuration centres C % (w/w) Cl stock solution has a density of.5 g ml. The molecular weight of Cl is 36.5g mol. The volume (ml) of stock solution required to prepare a ml solution of.4 M Cl is. Ans.8 æw ç % ö density w 9..5 Molarity of given solution = è ø = = mol. wt during dilution, M V = M V V =.4 V = 8mL 4. An organic compound undergoes first-order decomposition. The time taken for its decomposition Ans.9 [t /8] to /8 and / of its initial concentration are t /8 and t / respectively. What is the value of? t/ (take log =.3) Q t =.33 log a k (a - x) t /8 =.33 log a k a/8 t /8 =.33 log8 k t / =.33 log a k a / 6 \ / 33 t / =.33 log k t/8 log8 = = 3log = 3.3 =.9 t log t t / /8 / =.9 = 9 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website:

22 IIT-JEE EXAMINATIN (ELD N : 8-4-) ART - III : MATEMATICS SECTIN I : Single Correct Answer Type This section contains multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which NLY NE is correct. 4. The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line 4x 5y = to the circle x + y = 9 is- (A) (x + y ) 36x + 45y = (B) (x + y ) + 36x 45y = (C) 36(x + y ) x + 45y = (D) 36(x + y ) + x 45y = Ans. (A) Let mid point be (h, k), then chord of contact : hx + ky = h + k Let any point on the line 4x 5y = be \ Chord of contact : 5x x + (4x )y = 45...(ii) (i) and (ii) are same \ 5x 4x- 45 = = h k h + k x 9h = h + k and 9h 45k + (h + k ) = h + k 4(h + k ) (h + k ) 36h + 45k = x...(i) + + 4(h + k) 45k (h k ) = \ Locus is (x + y ) 36x + 45y = æ 4x - ö ç x, è 5 ø 4. The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is - (A) 75 (B) 5 (C) (D) 43 Ans. (B) Balls can be distributed as,, 3 or,, to each person. When,, 3 balls are distributed to each person, then total number of ways : 5! =..3! = 6!! 3!! When,, balls are distributed to each person, then total number of ways : 5! =..3! = 9!!!! \ total = = 5 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: / 33

23 IIT-JEE EXAMINATIN (ELD N : 8-4-) 43. Let ì p ïx cos, x¹ ƒ(x) = í x, x ÎIR, then ƒ is - ï î, x = (A) differentiable both at x = and at x = (B) differentiable at x = but not differentiable at x = (C) not differentiable at x = but differentiable at x = (D) differentiable neither at x = nor at x = Ans. (B) At x = R..D = p - ( + h) -() h lim = lim h h h h p = lim h cos = cos( ) h h = finite = ƒ( -h) -ƒ() LD : = lim h -h = = lim h cos h h Q LD = RD at x = h cos æ p ö ç - è-hø æpö = lim-h cosç -h èhø ƒ(x) is differentiable at x = At x = ƒ( + h) -ƒ() RD = lim h h = lim h æ p ö ç è+ hø h (+ h).cos - æ p ö cosç h = 4lim è + ø h h æ p ö æ p ö sin ç. h ç - ( h) è + ø + =- 4lim è ø = p h / 33 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website:

24 IIT-JEE EXAMINATIN (ELD N : 8-4-) ƒ( -h) -ƒ() LD : lim h -h = lim h æ æ p öö - ç - ç è è øø -h ( h) cos - h æ p öæ p ö 4ç sin ( h) ( h) - ç - = lim è øè ø = p. h - LD ¹ RD at x = \ Not differentiable at x =. 44. The function ƒ : [,3] [,9], defined by ƒ(x) = x 3 5x + 36x +, is : (A) one-one and onto (B) onto but not one-one (C) one-one but not onto (D) neither one-one nor onto Ans. (B) ƒ(x) = x 3 5x + 36x + ƒ'(x) = 6(x 5x + 6) = 6(x )(x 3) Q ƒ(x) is non monotonic in x Î [,3] ƒ(x) is not one-one ƒ(x) is increasing in x Î[,) and decreasing in xî(,3] ƒ() =, ƒ() = 9 & ƒ(3) = 8 \ Range of ƒ(x) is [,9] ƒ(x) is onto. 45. æx + x+ ö If lim ç - ax - b = 4, then - x è x+ ø (A) a =, b = 4 (B) a =, b = 4 (C) a =, b = 3 (D) a =, b = 3 Ans. (B) æx + x+ ö lim ç - ax - b = 4 x è x+ ø ( ) ( ) æ - a x + x -a- b + -bö lim = 4 x ç x+ è ø Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: 3 / 33

25 IIT-JEE EXAMINATIN (ELD N : 8-4-) æ æ-böö ç ( - a ).x+ -a- b+ ç x lim è è øø = 4 x + x for limit to exist finitely a = and a b = 4 a = and b = Let z be a complex number such that the imaginary part of z is nonzero and a = z + z + is real. Ans. (D) Then a cannot take the value - (A) (B) 3 z + z + a = Q z is imaginary D < 4( a) < 4a < 3 3 a <. 4 Aliter : a = z + z + Q a = a (given a is real) \ z + z = z + z z - z = z-z z+ z =- (Q Im(z) is non zero) (C) (D) 3 4 Re(z) = \ z can be taken as + iy where y Î R æ ö æ ö \ a = ç + iy + + iy + è ç ø è ø a = 4 + iy + iy y a = 3 4 y a < 3 4 \ a ¹ / 33 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website:

26 IIT-JEE EXAMINATIN (ELD N : 8-4-) x x 47. The ellipse E : + = is inscribed in a rectangle R whose sides are parallel to the coordinate 9 4 axes. Another ellipse E passing through the point (,4) circumscribes the rectangle R. The eccentricity of the ellipse E is - (A) (B) 3 (C) (D) 3 4 Ans. (C) Let equation of E be x y + = (Q E passes through (, 4)) a 6 Q E passes through (3,) \ = a 6 a = \ a 3 e = - = - e= 6 4 y (,4) (,) ( 3,) (3,) x (3,) ( 3, ) (3, ) 48. Let =[a ij ] be a 3 3 matrix and let Q = [b ij ], where b ij = i+j a ij for < i, j < 3. If the determinant Ans. (D) of is, then the determinant of the matrix Q is - (A) (B) (C) (D) 3 a a a Q = a a a a a a a a a a3 a3 a33 Q =...a a a = = = 3 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: 5 / 33

27 IIT-JEE EXAMINATIN (ELD N : 8-4-) 49. The integral sec x dx ò 9/ equals (for some arbitrary constant K) (sec x + tan x) (A) ì ( sec x tan x ) ü - K / í - + ý+ î 7 þ ( sec x + tan x) ì ü - / sec x + tan x ý+ K secx+ tanx î 7 þ (B) í ( ) ( ) (C) (D) Ans. (C) Let I = = ò ì ( sec x tan x ) ü - K / í + + ý+ î 7 þ ( sec x + tan x) ( sec x + tan x) ò ì ( sec x tan x ) ü K / í + + ý+ î 7 þ sec x (sec x + tan x) 9/ dx sec x(sec x + tan x)sec x / dx (secx + tan x) ut secx + tanx = t (secx tanx + sec x) dx = dt Also Q sec x tan x = secx tanx = t \ secx = æ t ö ç + è tø \ æ ö dt ç + t I = è ø / ò t I. t t dt ò -9/ -3/ = ( + ) -7/ -/ æ t t ö I= ç K è 7 ø æ t ö I = - K / ç + + t è 7 ø ì ü - / í + sec x + tan x ý+ K sec x + tan x î 7 þ I = ( ) ( ) 6 / 33 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website:

28 IIT-JEE EXAMINATIN (ELD N : 8-4-) 5. The point is the intersection of the straight line joining the points Q(,3,5) and R(,,4) with the plane 5x 4y z =. If S is the foot of the perpendicular drawn from the point T(,,4) to QR, then the length of the line segment S is - (A) Ans. (A) Line QR : (B) (C) (D) x- y-3 z-5 = = =l 4 Any point on line QR : (l +, 4l + 3, l + 5) \ oint of intersection with plane : 5l + 6l l 5 = \ Also l=- 3 æ4 3 ö ç,, è3 3 3 ø Q TQ = TR = 5 S is the mid-point of QR æ3 9ö S ç,, è ø S = units Q (,3,5) T(,,4) R (,,4) SECTIN II : Multiple Correct Answer Type This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which NE or MRE are correct. 5. Let q, j Î [,p] be such that æ q qö ç è ø cos q( -sin j ) = sin q tan + cot cosj-, S tan( p-q ) > and Then j cannot satisfy- (A) p <j< 3 - < sin q<-. (B) p 4p <j< (C) 4 p 3 p <j< (D) 3 p <j< p 3 3 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: 7 / 33

29 Ans. (A,C,D) tan(p q) > p - q lies in I on III quadrant IIT-JEE EXAMINATIN (ELD N : 8-4-) Q lies in II or IV Quadrant...(i) Q < sinq < \ By (i) and (ii) : Also, given 3 æ4p 5pö - qî ç, è 3 3 ø æ3p 5pö qî ç, è 3 ø...(ii) æ q qö ç è ø cos q( -sin f ) = sin q tan + cot cos f- sin qcos f cos q-cos qsin f= - q q sin cos cosq cosqsinf = sinq cosf + cos q sin( q+f ) = sin( ) < q+f < Q p 5p 3p 7p <q+f< or <q+f< æ3 p 5 3 7, p ö qîç p <q+f< p è 3 ø 6 6 p 4p <f< 3 5. Let S be the area of the region enclosed by y= e -x, y =, x =, and x =. Then - (A) Ans. (A,B,D) S ³ (B) e Area (ABC) = S³ - (C) e y æ ö S 4 ç + è e ø (D) æ ö S + ç - e è ø Shaded area is S. Clearly S < and -x ò e dx > ò e -x dx C(, ) (, /) B(, ) (, /e) A x= y = e x x S > (\ (B) is correct e 8 / 33 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website:

30 Again S ³ Area (trapezium ACD) S ³ æ ç + öæ ö è ç e øè ø S ³ \ C is wrong æ ö ç + è e ø IIT-JEE EXAMINATIN (ELD N : 8-4-) A y B C E D F x = / Ö x = x Also S Sum of areas of rectangles ABD and CEFD S S æ öæ ö + - è ç ç øè eø + e æ ö ç - è ø (\ (D) is correct) 53. A ship is fitted with three engines E, E and E 3. The engines function independently of each other with respective probabilities, 4 and 4. For the ship to be operational at least two of X denotes 3 respectively the events that the engines E, E and E 3 are functioning. Which of the following is (are) true? (A) 3 éx Xù = 6 c ë û (B) [Exactly two engines of ship are functioning 5 7 (C) [X X] = (D) [X X] = 6 6 Ans. (B,D) (X) = E E E 3 + E E E 3 + EEE 3 + EEE = X] = 8 (X) = 4 C ( ÇX) c æx ö X /3 ç = = = è X ø ( X ) /4 8 (Exactly two engines are functioning x) 7/3 7 = = /4 8 ( ÇX ) ( ) æ X ö X 5/3 5 ç = = = è X ø X /4 8 ( ÇX ) ( ) æ X ö X 7/3 7 ç = = = è X ø X / 6 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: 9 / 33

31 IIT-JEE EXAMINATIN (ELD N : 8-4-) x y 54. Tangents are drawn to the hyperbola - =, parallel to the straight line x y =. The points 9 4 of contact of the tangents on the hyperbola are (A) Ans. (A,B) æ 9 ö ç, è ø (B) æ 9 ö ç -,- è ø (C) ( 3 3, - ) (D) (-3 3, ) Let parametric coordinates be (3secq, tanq) Equation of tangent at point will be xsec q y tan q - = 3 Q tangent is parallel to x y = sec q = 3tan q sin q= 3 \ coordinates are æ 9 ö ç, è ø and æ 9 ö ç -, - è ø 55. If y(x) satisfies the differential equation y' ytanx = x sec x and y() =, then (A) æpö p yç = è 4 ø 8 (B) æpö p y' ç = è4 ø 8 æpö p (C) yç = è 4 ø 9 Ans. (A,D) (D) æpö 4p p y' ç = + è3ø dy y tan x x sec x dx - = I.F. ò - tanxdx = e = cosx \ Equation reduces to y.cos x = x.sec x.cos xdx ò 3 / 33 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website:

32 y cosx = x + C Q y() = = + C \ y cosx = x y(x) = x secx IIT-JEE EXAMINATIN (ELD N : 8-4-) \ æpö p p yç 4 = = è ø 6 8 (\ (A) is correct) æpö p p y ç. 3 = = è ø 9 9 (\ (C) is wrong) Also y'(x) = x secx + x secx tanx and æpö p p y' ç =. + (\ (B) is wrong) è4 ø 6 æpö p p y' ç = + è ø 3 9 4p p = + (\ (D) is correct) SECTIN III : Integer Answer Type This section contains 5 questions. The answer to each question is a single digit integer, ranging from to 9 (both inclusive) 56. Let ƒ : IR IR be defined as ƒ(x) = x + x. The total number of points at which ƒ attains either a local maximum or a local minimum is Ans.5 f(x) = x + (x + ) (x ) f(x) = x x x x + x + x + x + x x < x < x < x ³ y x \ f has 5 points where it attains either a local maximum or local minimum. Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: 3 / 33

33 IIT-JEE EXAMINATIN (ELD N : 8-4-) æ ö 57. The value of 6 + log ç ç è ø Ans. 4 is Let y = y = 4 3 y 3 y = y 3 y + y = (3y 4 ) ( y + 3) = y = 4 3 ; y = 3 (reject) æ ö \ V = 6 + log y 3/ ç è 3 ø æ 4 ö = 6 + log. 3/ ç è3 3 ø = 6 + log æö 3/ ç è 3ø = 6 = Let p(x) be a real polynomial of least degree which has a local maximum at x = and a local minimum Ans. 9 at x = 3. If p() = 6 and p(3) =, then p'() is Let '(x) = k(x ) (x 3) = k(x 4x + 3) (x) = æx ö ç è 3 ø 3 k - x + 3x + c Q () = 6 4k c =...() (3) = c =...() by (i) and (ii) k = 3 \ '(x) = 3(x ) (x 3) '() = 9 3 / 33 Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website:

34 r 59. If a,b r IIT-JEE EXAMINATIN (ELD N : 8-4-) and c r r r r r r r r r r are unit vectors satisfying a- b + b- c + c- a = 9, then a + 5b + 5c is Ans. 3 r r r r r r a- b + b- c + c- a = 9 rr 6-S a.b= 9 r r 3 S a.b =- r r r a+ b+ c ³ r rr S + S ³ a a.b r r 3 S a.b ³-...() r r r for equality a+ b+ c = r r r a+ b+ c= r r r 5b + 5c =-5a r r r r a + 5b + 5c =-3a a r r + 5b + 5c r = 3 a r = 3 6. Let S be the focus of the parabola y = 8x and let Q be the common chord of the circle x + y x 4y = and the given parabola. The area of the triangle QS is Ans.4 Focus of parabola S(,) points of intersection of given curves : (,) and (,4). y Q (,4) (,) S(,) x Area (DSQ) =..4= 4 sq. units Corporate ffice : CAREER INSTITUTE, SANKAL, C-6, INDRA VIAR, KTA-345 NE : , Fax : , info@allen.ac.in Website: 33 / 33