12th IIT 3rd May Test Paper-1 Solution

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1 th IIT 3rd May Test Paper- Solution Subject-PHYSICS Part : SECTIN-A (MULTIPLE CHICE NE R MRE THAN NE CRRECT ANSWER) Q Choose the correct statements if the given network is in steady state The P.D. across all capacitors equal 0 V The P.D. across all resistance equals zero The charges on all the three capacitors are equal The change on 8µF capacitors is maximum among the three capacitors (a, b, d) Apply KVL in all the three loops Q Bodies of mass 0.5 kg resting on a horizontal frictionless tabletop are connected with an unstretched spring of length L = 0 cm and of spring constant 6 N/m. The mass of the spring is negligible. At a certain moment the bodies are given an initial speed of v0 = 0.36 m/s towards the wall on the right. The body at the right collides with the wall totally elastically There will be collisions with the wall After st collision centre of mass comes to rest After nd collision centre of mass moves to left with speed v0 After all collisions are over, the system oscillates about the centre of mass (a, b, c) After st collision two are moving with speeds v0 in opposite direction. Hence centre of mass is at rest & two start oscilating with initial position as mean position. So mass on right again collides with wall when at mean position & both masses have same speed towards left after collision. 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

2 3 th IIT 3rd May Test Paper- Solution Q3 Figure shows a system of three concentric metal shells A, B and C with radii a, a and 3a respectively. Shell B is earthed and shell C is given a charge Q. Now if shell C is connected to shell A Charge Potential of C is Charge on shell B is 4Q / 3 on shell B is 8Q / KQ a Potential of C is 3 KQ a (b, c) K 8 KQ Q Q 3 a a Q4 In the circular region with centre magnetic field is into the plane of paper. Two conductor ABC and DEF VC are placed as shown. As magnetic field is increasing with time, potentials at A, B, D, E are related as VA > VB VD > VE VB < VE As A and D are joined VB > VE (b, d) 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

3 4 th IIT 3rd May Test Paper- Solution Q5 Mark the correct statement(s). For an exothermic reaction, if Q value is +.56 MeV and the KE of incident particle is.44 MeV, then the total KE of products of reaction is 5.00 MeV For an exothermic reaction, if Q value is +.56 MeV and the KE of incident particle is.44 MeV, then the total KE of products of reaction is.56 MeV For an endothermic reaction, if we give the energy equal to Q value of reaction, then the reaction will be carried out For an exothermic reaction, the BE per nucleon of products should be greater than the BE per nucleon of reactants (a, d) For an exothermic reaction X x Y y If K i is the kinetic energy of incident particle x, then from energy conservation, K i (m x M x ) c K Y K y ( M Y m y ) c K Y K y K i (m x M X M Y m y ) c KY K y K i Q In any exothermic reaction, mass of the products is less than the mass of reactants, i.e., in products, the nucleons are more tightly bound and hence have greater BE per nucleon as compared to BE per nucleon of reactants. For endothermic reaction to be carried out, minimum energy given to the reactant must be greater than Q value. Q6 In an experiment to determine the focal length ( f ) of a concave mirror by the u-v method, a student places the object pin A on the principal axis at a distance x from the pole P. The student looks at the pin and its inverted image from a distance keeping his/her eye in line with PA. When the student shifts his/her eye towards left, the image appears to the left of the object pin. Then Q7 x f f x f x f x f (d) A solid sphere is of mass 5 m and radius R. There is a small frictionless hole passing through the centre of the sphere and there are two small particles each of mass m situated at the centre of solid sphere as shown in figure and are free to move in the hole. If at t = 0, an angular velocity 0 is given to the solid sphere in gravity free space, then 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

4 5 th IIT 3rd May Test Paper- Solution Angular velocity of the sphere when both the particles just approach the middle of radius is 3 0 / 5 Angular velocity of the sphere when both the particles just approach the surface of sphere (just inside) is 0 / Angular velocity of the sphere when both the particles just outside the surface of sphere, is 0 / Angular velocity of the sphere when both the particles just approach the surface of sphere (just inside) is 3 0 / Q8 (b, c) Function x log e ( D sin t E cos t ) A sin t B cos t C sin t cos t represents SHM For If Q9 any value of A, B, C, D and E (except C = 0) Lf A = -B, C = B, D = E ; amplitude = B If A = B; C = 0; D = E = 0 A = B ; C = B ; D = E ; amplitude = B (b, d) ne end of a copper rod of uniform cross section and length.5 m is kept in contact with ice and the other end with water at 00 C. At what point along its length should a temperature of 00 C be maintained so that in the steady state, the mass of ice melting be equal to that of the steam produced in same interval of time. Assume that the whole system is insulated from surroundings [ Lice 80 cal / g, Lsteam 540 cal / g ] 0.34 cm from the end at 00 C 0.34 mm from the end at 00 C.034 cm from the end at 00 C.034 m from the end at 00 C (a) If the point is at a distance x from water at00 C, heat transferred to ice in time t to melt it is m L KA (00 0) t (.5 x) or m KA 00t 80 (.5 x) Similarly, heat conducted by the rod to water at 00 C in time t is Q KA (00 00) t m s Ls x ms KA (00 00)t KA 00t xls x 540 According to problem, ml ms 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

5 6 th IIT 3rd May Test Paper- Solution i.e., KA 00t KA 00t 80 (.5 x) x (.5 x) 54 x or Solving it, we get, x m or 0.34 cm Q0 The photon radiated from hydrogen corresponding to nd line of Lyman series is adsorbed by a hydrogen like atom X in nd excited state. As a result the hydrogen like atom X makes a transition to n th orbit X X He Li n n6 9 (b, d) Energy (nd line of lyman, H) =. ev n 3 Now,. Z (3.6) Checking values Z 3 and n 9 SECTIN-D Integer Answer Type This section Contains 8 questions. The answer to each of the questions is a single digit integer, ranging from 00 to 99. The appropriate bubbles below the respective question numbers in the MR have to be darkened. Q The heat generated in 5Ω resistor due to current flowing through it is 0 cal/sec. Find the heat generated in 4Ω resistance (in cal/sec). Q () The path of photoelectrons emitted due to electromagnetic radiation incident on a sample of material A, is found to have a maximum bending radius of 0. m in a magnetic field of / T. When the radiation is incident normally on a double slit having a slit separation of 0. mm, it is observed that there are 0 fringes in a width of 3. cm on a screen placed at a distance of m from the double slit. Find the work function of the material in ev. Given that mass of the electron = 0.5 MeV/c, hc = 400 ev - A () The bending radius r of a particle of mass m and charge e moving with a velocity v in a uniform magnetic field is given by mv evb, e being the charge and B the strength of the field. r Momentum, mv Ber (i) 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

6 7 th IIT 3rd May Test Paper- Solution When the em radiation is incident normally on a double slit. W Q3 A student throws tennis balls out of the window at different angles to the horizontal. All tennis balls have the same initial speed v 0 3 m/s. It turns out that all tennis balls landing velocities make angles 30 or greater with the horizontal. Find the height h (in m) of the window above the ground. (5) gh tan 0 3 Given that 30 tan tan 30 gh h 5m Q4 Find the maximum value of mass of block C (in x 0 ) so that neither A nor B moves. (m A 00 kg, m B 00 kg, Pulley and strings are massless. Coefficient of friction between A and B and between B and horizontal surface is 0.4) 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

7 8 th IIT 3rd May Test Paper- Solution () Maximum friction that can be obtained between A and B is f m A g (0.4) (00) (0) 400 N and maximum friction between B and ground is f (m A m B ) g (0.4) (00 00) (0) 00 N. Drawing diagrams of A, B and C in limiting case, T f 400 N Equilibrium of A gives (i) Equilibrium of B gives and equilibrium of C gives mc g T Q5 or 0 mc 000 or mc 00 kg 0 kg Two small blocks A and B of same mass m are placed at the mid point of a smooth horizontal track of length as shown. A massless spring of force constant k and free length is touching the block A but is not attached to it. The block A is displaced towards left by a small distance d = AP there by causing a compression in the spring. The block is then let free to move. What is the minimum time after which the block A again K d m reaches its starting position P. (Assume elastic collision) () Ttotal d m K m K From P to A speed at A equilibrium potition V d vb d K collision is elastic it will transfer to B m K m 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

8 9 th IIT 3rd May Test Paper- Solution Time for B for half oscillation For A for half oscillation Ttotal Q6 d m K d m K m K m K In a circus act, a 4 kg dog is trained to jump from B cart to A cart and then immediately back to the B cart. The carts each have a mass of 0 kg and they are initially at rest. In both cases the dog jumps at 6 m/s relative to the cart. If the cart moves along the same line with negligible friction, If the final magnitude of velocity of cart B with respect to the floor is x /36, find x. (55) 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

9 0 th IIT 3rd May Test Paper- Solution Q7 A 40 cm long wire having a mass 3. gm and area of cross section mm is stretched between the support cm apart. In its fundamental mode. It vibrate with frequency 000/64 Hz. The young s modulus of the wire is 0 x N/m. Find the value of x. Q8 (9) When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which does external work is. Find P. P (7) SECTIN-E (CLUMN MATCHIMG TYPE) Q. ne mole of a monatomic ideal gas is taken along two cyclic processes E F G E and E F H E as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

10 th IIT 3rd May Test Paper- Solution Match the paths in Column A with the magnitudes of the work done in Column B and select the correct answer using the codes given below the lists Column A Q. Column B (p) G E 60 P0V0 ln (q) G H 36 P0V0 (r) F H 4 P0V0 (s) F G 3P0V0 p d, q c, r b, s a. Some physical quantities are given in column I and some possible SI units in which these quantities may be expressed are given in column II. Match the physical quantities in column I with the units in column II. Column A (p) 0 E 0 = Permitivity of free space E = Column B (Kilogram) (Meter) (Second) (Kilogram) (Meter) (Kelvin) Electric field (q) nrt n = number of Mole R = Gas V constant T = Absolute Temperature V = Volume (r) mb m = mass b = Wein s constant (Kilogram) (Meter) (Second) (s) LtT = Coefficent of viscosity (Kilogram) (Meter) (Second) (Kelvin) L = Length 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

11 th IIT 3rd May Test Paper- Solution t = time T = Absolute temperature p c, q c, r b, s b. 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

12 3 th IIT 3rd May Test Paper- Solution Subject-CHEMISTRY Part : SECTIN-A (MULTIPLE CHICE NE R MRE THAN NE CRRECT ANSWER) Q Which statement(s) is (are) correct regarding to above observations: P & Q are stereo isomers P & Q are enantiomers P & Q are Diastereomers P & Q can be separated by physical methods (a, c, d) P & Q are diastereomers. P & Q are stereo isomers As P & Q are diastereomers they can be separated by physical methods. Q Which of the following statement(s) are correct regarding catalyst? It increases the value of rate constant It does not alter enthalpy of reaction It alter the mechanism of reaction It shortens the time for a reversible reaction to reach equilibrium (a, b, c, d) 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

13 4 th IIT 3rd May Test Paper- Solution Q3 The quantum number of last electron added are 4, 0,0, 9 Q4 0, the atomic number of the element could be 9 30 (a, b, c, d) Z electron configuration 9 [Ar] 4s 0 [Ar] 4s 9 [Ar] 3d 0 4s 30 [Ar] 4s 8 g glucose is taken in 8. litre vessel at 7 C. Enough oxygen is added so that total pressure is 3 atm at same T. The mixture is allowed to react by sparking. After reaction, the total pressure at 373 k is atm (c) C 6 H 6( s ) 6( g ) 6C ( g ) 6 H Initial mole 0. After reaction nc no PC At 373 k (normal boiling point of water) PH is atm and cannot exceed atm. Water exists partly as vapor and partly as liquid PToTal 4.73atm ption (c) Q5 Final product Z in the given reaction sequence. 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

14 5 th IIT 3rd May Test Paper- Solution (d) no change in configuration of product. Q6 Which of the following order s are correct. Acidity order : -salicyldehyde > p-hydroxy benzal dehyde > heat of hydrogenation Basicity order Heat of hydrogenation Q7 (b, c) Which of the following will exhibit square planar geometry? K [Pt Cl3 CH4] Rh (P Ph3)3 Cl K [Mn Cl4] Na (a, b, d) Q8 Which of the following statements is / are true for the following reactions 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783 [AuCl4]

15 6 th IIT 3rd May Test Paper- Solution Product B is CH3 CH = CH Br Product E give +ve halo form test Product B is Product Q9 E give ve halo form test (b, d) The disaccharide cellobiose (X) has the following structure. CH H H H CH H H H H H H H H Which of the following statements is/are correct for (X) Disaccharide (X) is reducing sugar. Two molecules of glucose are formed on hydrolysis Disaccharide (X) is non reducing sugar. The monosaccharide (s) formed on hydrolysis of (X) are ketohexose. (a, b) Q0 Choose the correct statement(s): Ksp of AgCl is at a given temperature. Then It s Ksp in KCl solution will be less than It s solubility in CaCl solution will be less than 0 5M but more than 0 5 M in NH3 solution due to complex formation. The solubility of AgCl in 0.5M KCl (80% dissociated) is equal to 4 0 9M 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

16 7 th IIT 3rd May Test Paper- Solution If G0 = 0 for a reaction: A( g ) B( g ) then a mixture of A and B in 5 : molar proportion at a total pressure of 30 atm will be in equilibrium at 5 C. (b, c, d) The value of Ksp is independent of other ions present. Solubility decreases due to common ion effect & increases due to formation of complex. Go = 0, Kp = ; PA = 5 atm, PB = 5 atm. SECTIN-D Integer Answer Type This section Contains 8 questions. The answer to each of the questions is a single digit integer, ranging from 00 to 99. The appropriate bubbles below the respective question numbers in the MR have to be darkened. Q Count the total number of S bounds are having equal length in HS4 and S36 respectively. (If the answer is 5 and 3 then represent as 53) (36) Count the total X bonds Ans. 3 & 6 Q Due to Resonance How many geometrical isomers are possible for (A) CH3 CH CH CH CH CH C H A 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

17 8 th IIT 3rd May Test Paper- Solution Q3 () If conductance of X+ is 60 ohm cm equivalent and Y is 40 ohm cm mole then find molar conductance of XY. If answer is X, fill answer X/0. (6) Molar conductance of XY = Molar conductance of X+ + Molar conductance of Y = 60 ohm cm mole. = Q4 st membrane only allow to pass He atoms. nd membrane allow to pass N and H. Find the total pressure in nd compartment. Q5 () BH6 + 3H + A B Find the number of hybridised orbitals used by boron in products. (3) B H 6 3 H H 3 B 3 3H Hybridization of boron in H3B3 is sp Q6 How many chiral carbons are there in Reserpine (an antipsychotic drug)? CH3 H3C N N H CH3 CH3 CH3 H3C (8) N H3C CH3 N H CH3 CH3 CH3 H3C Q7 A g B g The rate constant of the above first order reaction is min. After how many minutes from the commencement of the reaction, a : molar ratio of A and B is attained in the reaction mixture? 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

18 9 th IIT 3rd May Test Paper- Solution log 3 = 0.76 (8) A B t 0, a 0 t t, a a 3 a 3 t Q8 a log a The ph of 0.M solution of a weak monobasic acid (HA) is 3.5. Then the value of pkb of its conjugate base would be (8) pka log C 3.5 pka pka 6; pkb 8 ph SECTIN-E (CLUMN MATCHIMG TYPE) Q. Column A (p) Column B CuS4 in its of molecular weight 6 Equivalent weight of Na S 3 in its of molecular weight 3 Equivalent weight of reaction with KI (q) reaction with I (r) Equivalent weight of K Cr 7 as an of molecular weight oxidizing agent in acid solution (s) Equivalent weight of H 3 P3 in complete Equal to molecular weight neutralization with NaH Q. p d; q d; r a; s c. Column (A) is reaction & Column (B) is reaction intermediates involved in the reaction Column A (p) Column B Carbon anion 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

19 0 th IIT 3rd May Test Paper- Solution (q) Carbocation (r) Carbene (s) Benzyne p b; q a, c; r b; s a, d. 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

20 th IIT 3rd May Test Paper- Solution Subject-Mathematics Part : 3 SECTIN-A (MULTIPLE CHICE NE R MRE THAN NE CRRECT ANSWER) Q The points A(0, 0), B(cos, sin ) and C (cos, sin ) are the vertices of a right angled triangle if sin cos cos sin (a, b, c, d) tan tan Q cos( ) 0 Consider the circle x y 0 x 6 y Let be the centre of the circle and tangent at A (7, 3) and B (5, ) meet at C. Let S 0 represents family of circles passing through A and B, then- Area of quadrilateral ACB = 4 The radical axis for the family of circles S 0 is x y 0 The smallest possible circle of the family S 0 is x y x 4 y 38 0 The coordinates of point C are (7, ) (a, c, d) Coordinates of are (5, 3) and radius Equation of tangent at A (7, 3) is 7 x 3 y 5 ( x 7) 3 ( y 3) 30 0 i.e. x 4 0 i.e. x 7 Equation of tangent at B (5, ) is 5 x y 5 ( x 5) 3 ( y ) 30 0 i.e. y 0 i.e. y coordinate of C are (7, ) area of ACB = 4 Equation of AB is x y 4 (radical axis) Equation of the smallest circle is ( x 7) ( x 5) ( y 3) ( y ) 0 i.e. x y x 4 y 38 0 Q3 A differentiable function satisfies f ( x) f (x) has a minimum value e. x 0 ( f (t ) cos t cos (t x )) dt. Which of the following hold good? f (x) has a maximum value e. 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

21 th IIT 3rd May Test Paper- Solution f ' (0). (a, b, c) Q4 If foci of x y x y concide with the focii of and eccentricity of the hyperbola is, then 5 9 a b There is no directosr circle to the hyperbola a b 6 Centre f ' ' e. of the director circle is (0, 0) Length of latus rectum of the hyperbola = (a, b, d) For the ellipse : e focii are ( 4, 0) and (4, 0) For the hyperbola ae 4, e a 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

22 3 th IIT 3rd May Test Paper- Solution b 4 (4 ) b Q5 The equation of the line x y z 0, 4 x y z 0 written in the symmetrical form is x y z 0 x y z x / y z / x y z (a, b, c) x y z 0 4x y z 0 direction ratios of the line are 3, 6, 3 i.e.,, Let z k, then x k, y k i.e. (k, k, k ) is any point on the line (,, 0), (0, 0, ) and,, are points on the line (A), (B) and (C) are correct options Q6 Identify the statement(s) which is/are INCRRECT? a [a (a b )] (a b ) (a ) If a, b, c are non-zero non coplanar vectors, and v. a v. b v. c 0 they v must be a null vector If a and b lie in a plane normal to the plane containing the vectors a b, c d ; a, b, c, d are non zero vector, then (a b ) (c d ) 0 If a, b, c and a ', b ', c ' are reciprocal system of vectors then a. b ' b. c ' c. a ' 3 (a, c, d) (a) a [a (a b )] a [(a. b ) a (a. a ) b ] (a. a ) (a b ) (A) is not correct (b) v. a 0 v 0 or v a v. b 0 v 0 or v b v. c 0 v 0 or v c 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

23 4 th IIT 3rd May Test Paper- Solution v 0 or v a, b, c v0 (c) ( a b ). (c d ) 0 statement is incorrect (d) a b ' b. c ' c. a ' 0. (Property of reciprocal system) Q7 Which of the following statement(s) is/are correct? Rolle s theorem is applicable to the function F ( x ) 5 x 6 on the interval [, ] The slope of the tangent to the curve represented parametrically by the equations x t 3t and y t 3t 4 at the point M (, 0) is The value of a for which the function F ( ) a sin 00 The value of k {x k} is {x} (where {*} denotes fractional part function) 00 (a, b, d) (a) Apply rolle s theorem (b) dy 4t 3 dx t t 3 t 00 {x k} k 00 (d) Q8 00 {x} 00 {x} k If z 5 i and z 4 then Maximum z iz 7 Minimum z ( i) z 3 4 Minimum z z sin 3 has an extremum at is z 3 4 Maximum z z 4 z 3 3 (a, b, d) (a) z i z z z (b) z ( i ) z z i (c) z z 4 4 z z z 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

24 5 th IIT 3rd May Test Paper- Solution z (d) Q9 Let f ( x) 4 4 z z z ex and g ( x) f ' ( x) then x g (x ) has two local maxima and two local minima points g (x ) has exactly one local maxima and one local minima point x is a point of local maxim for g (x ) There is a point of local maxima for g (x ) in the interval (, 0) (b, d) ( x ) ( x 3 3 x 5 x ). e x g ' ( x) ( x ) 3 Now, ' x 3 3 x 5 x ' is strictly increasing and has a root in (, 0). Q0 If the equation cos x cos x b 0 has two distinct roots in [0, ] then all values of b must lie in [, [, 0) 0] (, (, 0] 0) (a, b) SECTIN-D Integer Answer Type This section Contains 8 questions. The answer to each of the questions is a single digit integer, ranging from 00 to 99. The appropriate bubbles below the respective question numbers in the MR have to be darkened. Q Let G be the centroid of the ΔABC whose sides are of lengths a, b, c. If P be a point in the plane of ΔABC such that PA =, PB = 3, PC = 4, and PG =, then the value of a b c is (4) As a b c a b b c c a 3( a b c ) So, 9 (PG) (AB) (BC) (CA) 3 (PA) (PB) (PC) ) Q 9 (a b c ) 3 ( 3 4 ) a b c 4. The number of complex numbers z which satisfy arg ( z 4) 5 / 6 and arg ( z 4) / 3 is () We get one value of z corresponding to point of intersection of these lines then z can have two values. 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

25 6 th IIT 3rd May Test Paper- Solution Q3 log 5 (log (log 4 x )) 0 has a solution for x when y b, y a. If b is as large log 7 (log (log y (log 4 x))) The equation as possible, then the value of a b is. () log 5 (log (log 4 x)) 0 log 4 x log (log y (log 4 x)) log (log y ) y log y Also, log (log y (log 4 x)) 0 log y (log 4 6) y From () and (), a Q4 (), b The area bounded by the curves x y and x y is k. Then find integral part of k. () 0 Q5 If f ( x) , then the value of ' a ' which satisfies f log (3 x) ( a 4) is (3) Given f Q6 () log y Ar 4 Ar in first quadrant ( x x ) dx x 6 (a 4) f a 4 Put x 9 in f ( x ) 3 log (3 x) a 4 a3 If tan cot and tan 4 cot 4 07, then the number of real value(s) of is () 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

26 7 th IIT 3rd May Test Paper- Solution tan cot tan 4 cot 4 ( ) 07 Q7 Q Find number of point(s) of intersection of curve sin x cos y and the circle x y. (0) x y z Let matrix A 3, where x, y, z N. If adj (adj (adj (adja))) , then the number of such matrix is. (36) adj (adj (adj (adja))) A A 0 x y z 0 SECTIN-E (CLUMN MATCHIMG TYPE) Q. Column A (p) Column B There are five points in a plane. The 8 maximum number of points of intersection of the perpendiculars drawn from each point to the lines joining the other points (q) The number of ordered pairs 35 (m, n) {,,..., 0} Such that m n is divisible by 7 (r) The total number of positive integral 7 solutions of the equation xyz = 539 is equal to (s) A dictionary comprising of four letter 34 (e.) 335 words is formed by arranging individually. The letters of words SLW and FAST. Rank of the word SLW is p b, q d, r a, s c. 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

27 8 th IIT 3rd May Test Paper- Solution (p) The total number of perpendiculars 4 C 5 30 points of intersection of 30 lines (i) 30 C Perpendiculars to the same line segment are parallel giving zero points of intersection number of line segments 5 C number that has to be subtracted from 30 C (ii) Altitudes of a triangle are concurrent number that has to be subtracted 5 C 3 0. (iii) Perpendiculars from a single point to 4 C lines are concurrent. 6 perpendiculars are given one point of intersection instead of 6 C points. Number to be subtracted (q) C m n ( m ) ( n ) Case-I : m 3 p m ( 3 ) p ( 7) p 7 k Case-II: m 3 p m 7k ' Case-III: m 3 p m 7k" 3 for m n to be divisible by 7 both m and n are to be of the same form. Thus required number of ordered pairs n (A) n (B) n (C) A {, 4, 7}, B{, 5, 8}, C {3, 6, 9} (r) xyz (s) C 3C 8 Letters starting with A 3! Letters starting with F 3! Letters starting with SA! Letters starting with SF! Next word is SLW Rank (3!) (!) 7. Q. Column A (p) Number of solutions of the equation Column B cos (cos x ) [ x ] where [ ] denotes the greatest integer function, is 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783

28 9 th IIT 3rd May Test Paper- Solution (q) The number of real common tangents to the 5 x 5 y 6 circle and the hyperbola 3 x y 48, is (r) The number of real common normals to parabola y 4x and the circle ( x ) ( y ), is (s) Number of positive integer solutions of xyz 0800 is N then N 300 p d, q c, r a, s b. (p) x 3 x 3 Hence the solutions are {0,,, 3, 3} S : x y (q) H x y 6 48 y mx 6m 48 Tangent to H is, 6m 48 m (r) m (c) 4 5 5(6m 48) 6( m ) 4 common tangent m 4 [for each value of m there are two parallel tangents] P : y 4x S : ( x ) ( y ) Normal to P : y mx m m 3 passes through the centre of the circle (, ) m 3 m 0 f ' ( m) 0 one real root one common normal (a) (s) 4 C 5C 6 C 900 0, Blue Diamond Complex, Fatehgunj, Vadodara-0.Tel. No / / 783