FIITJEE Solutions to JEE(Main)-2017

Transcription

1 Note: For the benefit of the students, specially the aspiring ones, the question of JEE(Main), 07 are also given in this booklet. Keeping the interest of students studying in class XI, the questions based on topics from class XI have been marked with *, which can be attempted as a test. FIITJEE Solutions to JEE(Main)-07 Instructions as printed on question paper. PAPER MATHEMATICS, PHYSICS & CHEMISTRY Test Booklet Code D Important Instructions:. Immediately fill in the particulars on this page of the Test Booklet with only Black Ball Point Pen provided in the examination hall.. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars carefully.. The test is of hours duration. 4. The Test Booklet consists of 90 questions. The maximum marks are There are three parts in the question paper A, B, C consisting of Mathematics, Physics and Chemistry having 0 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response. 6. Candidates will be awarded marks as started above in instruction No. 5 for correct response of each question. ¼ (one fourth) marks of the total marks allotted to the question (i.e. mark) will be deducted for indicating incorrect response of each question. No deduction from that total score will be made if no response is indicated for an item in the answer sheet. 7. There is only one correct response for each question. Filling up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above. 8. For writing particulars / marking responses on Side and Side of the Answer Sheet use only Black Ball Point Pen provided in the examination hall. 9. No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc. except the Admit Card inside the examination room/hall. 0. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the bottom of each page and in four pages (Page 0-0) at the end of the booklet.. n completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room / Hall. However, the candidates are allowed to take away this Test Booklet with them.. The CDE for this Booklet is D. Make sure that the CDE printed on Side of the Answer Sheet and also tally the serial number of the Test Booklet and Answer Sheet are the same as that on this booklet. In case of discrepancy, the candidate should immediately report the matter to the Invigilator for replacement of both the Test Booklet and the Answer Sheet.. Do not fold or make any stray mark on the Answer Sheet. Name of the Candidate (in Capital letters): Roll Number : in figures Examination Centre Number : : in words Name of Examination Centre (in Capital letters): Candidate's Signature :. Invigilator's Signature :. Invigilator's Signature : FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

2 JEE-MAIN-MPC-07- PART A - MATHEMATICS. If S is the set of distinct values of 'b' for which the following system of linear equations x + y + z = x + ay + z = ax + by + z = 0 has no solution, then S is: () an empty set () an infinite set () a finite set containing two or more elements a singleton Here, D a 0 a = a b For a =, the equations become x + y + z = x + y + z = x + by + z = 0 These equations give no solution for b = S is singleton set *. The following statement (p q) [(~pq)q] is: () a tautology () equivalent to ~p q () equivalent to p~q a fallacy Sol. () (p q) ((~p q) q) = (p q) ((p q) q) = (p q) ((~p ~q) q) = (p q) ((~p q) (~q q)) = (p q) (~p q) = (p q) (p q) = T *. If 5(tan x cos x) = cos x + 9, then the value of cos 4x is: () () 5 () (sec x cos x) = (cos x ) t t 9 t 5( t t ) = 4t + 7t 9t + t 5 = 0 5 t, cos x FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

3 JEE-MAIN-MPC-07- cos x 7 cos 4x 9 4. For three events A, B and C, P(Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P (Exactly one of C or A occurs) = 4 and P(All the three events occur simultaneously) =. Then the 6 probability that at least one of the events occurs, is: 7 7 () () 6 7 () 64 6 Sol. () P (A) + P (B) P (A B) = 4 P (B) + P (C) P (B C) = 4 P (A) + P (C) P (A C) = 4 PA PA B 8 P A P A B P A B C P (A B C) = = *5. Let be a complex number such that + = z where z. If k Sol. () 7, then k is equal to: () z () z () Determinant simplifies to k = 0 0 = z k = z FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

4 JEE-MAIN-MPC-07-4 *6. Let k be an integer such that the triangle with vertices (k, k), (5, k) and ( k, ) has area 8 sq. units. Then the orthocentre of this triangle is at the point: (), (), 4 (),, 4 k k 5 k 8 k k = (since k I) rthocentre is, *7. Twenty meters of wire is available for fencing off a flowerbed in the form of a circular sector. Then the maximum area (in sq. m) of the flowerbed, is: ().5 () 0 () 5 0 Sol. () Length of wire = r ( + ) = 0 m Area A = r A (r) = 0r r Area is maximum if r = 5. Maximum area A = 5 sq. m r r r 8. The area (in sq. units) of the region {(x, y) : x 0, x + y, x 4y and y + x } is: () 59 () () 7 5 Required area x x dx dx 4 = 0 0 y (, ) x = 4y y x = 5 sq. units (, ) 0 x x + y = FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

5 JEE-MAIN-MPC If the image of the point P(,, ) in the plane, x + y 4z + = 0 measured parallel to the line, x y z is Q, then PQ is equal to: 4 5 () 5 () 4 () Sol. () x y z 4 5 Let midpoint of PQ be M which lines on the plane M(x, y, z) = ( +, 4, 5 + ) ( + ) + (4 ) 4 (5 + ) + = = 0 = M (,, 8), P (,, ) PM = PQ = If for () 9 9x () x 9x x 0, 4, the derivative of 6x x tan 9x is x. g(x), then g(x) equals: () x x 9x 9x Sol. () Here, y tan x dy 9x dx 9x 9 g x 9x xg x. If ( + sinx) dy dx + (y + )cosx = 0 and y(0) =, then y is equal to: () () () 4 Sol. () ( + sin x)dy + cos x(y + )dx = 0 (y + )( + sin x) = C ( + )( + 0) = C = 4 (y + ) ( + sin x) = 4 Put x y FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

6 JEE-MAIN-MPC-07-6 *. Let a vertical tower AB have its end A on the level ground. Let C be the midpoint of AB and P be a point on the ground such that AP = AB. If BPC =, then tan is equal to: () 6 () 7 4 () Sol. () tan B and tan = 4 x tan 4 tan 4 9 tan = P 4x C x A tan = 9. If A 4, then adj (A + A) is equal to: 7 84 () () 6 7 Sol. () A = (A A) = 84 5 Adj (A A) = () *4. For any three positive real numbers a, b and c, 9(5a + b ) + 5(c ac) = 5b(a + c). Then: () b, c and a are in G.P. () b, c and a are in A.P. () a, b and c are in A.P. a, b and c are in G.P. Sol. () (5a b) + (5a 5c) + (b 5c) = 0 Let, 5a = b = 5c = 45 a = ; b = 5; c = 9 c = a + b b, c, a are in A.P. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

7 JEE-MAIN-MPC The distance of the point (,, 7) from the plane passing through the point (,, ), having normal perpendicular to both the lines x y z 4 and x y z 7, is: () 0 0 () 74 8 () Sol. () Equation of plane is x y z = 0 5x + 7y + z + 5 = Distance from (,, 7) = 8 8 n 6. Let I tan x dx, n. If I 4 + I 6 = a tan 5 x + bx 5 + C, where C is a constant of integration, then the Sol. () n ordered pair (a, b) is equal to: (), 5 (), 5 In n tan xdx, n > n = tan x sec x dx = n tan x I n n C n tan x In In C n tan x I6 I4 C 5 Given, I I a tan x bx C 4 6 a, b = (), 0 5, 0 5 *7. The eccentricity of an ellipse whose centre is at the origin is. If one of its directrices is x = 4, then the equation of the normal to it at, is: () y x = () 4x y = () 4x + y = 7 x + y = 4 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

8 JEE-MAIN-MPC-07-8 Sol. () Eccentricity, e Let a be the length of major axis and b be the length of minor axis a 4 e a = Also, b, as e x y Equation of ellipse is 4 Equation of normal at, is 4x y = *8. A hyperbola passes through the point hyperbola at P also passes through the point: P, and has foci at, 0 (), (), (),,. Then the tangent to this Sol. () x a y b x y a 4 a a = 8,, (a 8) x y. Hence equation of tangent at P 6x y, is x y 9. The function f : R, defined as x f x, is: x () invertible. () injective but not surjective. () surjective but not injective. neither injective nor surjective. Sol. () For, f x x the curve has graph as shown x Y / / (, ½) Which is onto but not one-one for, f : R, (, ½) X FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

9 JEE-MAIN-MPC cot x cos x lim x x () 4 () 8 equals: () 6 4 Sol. () cos x cot x lim x 8 x Put x t ; x t sin t tan t lim t0 8 t sin t cos t lim t0 8 t cos t t = 8 6. Let a i ˆ ˆj kˆ and b ˆi ˆj. Let c be a vector such that c a between c and a b be 0º. Then a c is equal to: () 5 8 () () 5 8 a b c, and the angle Sol. () c a a c 9 c a c 0 and a b c sin 0º c c a c. The normal to the curve y(x )(x ) = x + 6 at the point where the curve intersects the yaxis passes through the point: (), (), (),, FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

10 JEE-MAIN-MPC-07-0 Sol. () dy at x = 0 is dx Slope of normal at (0, ) is Equation of normal is x + y =. If two different numbers are taken from the set {0,,,,, 0}; then the probability that their sum as well as absolute difference are both multiple of 4, is: () 6 () () Sol. () Consider two sequences : 0, 4, 8 and, 6, 0 Take both numbers from either of these sequences. C C 6 Hence, probability =. C 55 *4. A man X has 7 friends, 4 of them are ladies and are men. His wife Y also has 7 friends, of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting ladies and men, so that friends of each of X and Y are in this party, is: () 485 () 468 () Sol. () X : 4L, M; Y : L, 4M Possible combinations () () () X L L, M L, M M Y M L, M L, M L Number of ways = 4 C 4 C 4 C C C 4 C 4 C C C 4 C C C = *5. The value of C C C C C C C C... C 0 C 4 4 () () 0 () Let S = C C C C... C0 C0 S = C C... C0 C0 C... C0 S = 0 0. is: A box contains 5 green and 0 yellow balls. If 0 balls are randomly drawn, onebyone, with replacement, then the variance of the number of green balls drawn is: () () () 4 5 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

11 JEE-MAIN-MPC-07- Sol. () p = 5 5, q = 0 5, n = 0 = npq = Let a, b, c R. If f(x) = ax + bx + c is such that a + b + c = and f(x + y) = f(x) + f(y) + xy, x, y R, 0 then f n is equal to: n () 0 () 65 () Sol. () Partially differentiating, we get f(x) x = constant = x f (x) = x k f (0) = 0 k = 0 = f n a n b n n n n n n n 5 n n = 6 = *8. The radius of a circle, having minimum area, which touches the curve y = 4 x and the lines, y = x is: () () () 4 4 Sol. () t t Let P, 4 be point where circle touches the 4 parabola y = 4 x t 7t Normal at P: ty x 0 to the parabola 4 passes through centre (c) of the circle (0, ). t 4t + 4t = 0... () Also, radius r t 4 + 4t + 8t t = 6r t 4 + (8 8)t = 0... () From equation () and (), we get Either 8 4 for t = 0 7 or for t = 4 4 A(0, 4) P r C (0, ) r r /4 0 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

12 JEE-MAIN-MPC-07- As, r r = 4 4, Minimum possible radius, r 7 7 [But of the given options r = 4 is minimum] 9. If, for a positive integer n, the quadratic equation, x(x + ) + (x + )(x + ) + + has two consecutive integral solutions, then n is equal to: () () 9 () 0 x + nx + n 0 Let I and I + be the roots of the equation I + = n () n I (I + ) = () Eliminating I from () and (), we get n n 4 n = n =. 0. The integral Sol. () 4 dx is equal to: cos x 4 () () () I cosec x cosec x cot x dx = x n (x + n) = 0n FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

13 JEE-MAIN-MPC-07- PART B - PHYSICS ALL THE GRAPHS/DIAGRAMS GIVEN ARE SCHEMATIC AND NT DRAWN T SCALE.. A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.. Then, t is given by : T T log () t () t log(.) log. () log. t T t T log(.) log Sol. () N0 N 0. N N N = 0. N = N 0 e t t e. n(.) n(.) t = T n() n. T A B t = 0 N 0 0 t = t N N 0 N. The following observations were taken for determining surface tension T of water by capillary method: diameter of capillary, D = m rise of water, h = m Using g = 9.80 m/s rhg and the simplified relation T = 0 N / m, the possible error in surface tension is closest to : () 0 % () 0.5 % ().5 %.4 % Sol. () rhg T = 0 N / m T r h T r h.5.45 T % error = % T An electron beam is accelerated by a potential difference V to hit a metallic target to produce X rays. It produces continuous as well as characteristic X-rays. If min is the smallest possible wavelength of X-ray in the spectrum, the variation of log min with log V is correctly represented in: () () log min log min log V log V FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

14 JEE-MAIN-MPC-07-4 () log min log min log V log V Sol. () hc min ev hc log log log V min e log k log V min log min log V *4. The moment of inertia of a uniform cylinder of length and radius R about its perpendicular bisector is I. What is the ratio /R such that the moment of inertia is minimum? () () () Sol. () m I = R m m = di m m d For minima R m m 0 = R R *5. A slender uniform rod of mass M and length is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle with the vertical is: z x () g cos () g sin () g sin g cos Sol. () FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

15 JEE-MAIN-MPC-07-5 = I M Mg sin = g sin z Mg x *6. C p and C v are specific heats at constant pressure and constant volume respectively. It is observed that C P C v = a for hydrogen gas C P C v = b for nitrogen gas The correct relation between a and b is () a = 8 b () a = b 4 () a = b a = 4 b For ideal gas C P C v = R/M If C P and C v are specific heats (J/kg 0 C) M = molar mass of gas a = R/ and b = R/8 a = 4b *7. A copper ball of mass 00 gm is at a temperature T. It is dropped in a copper calorimeter of mass 00 gm, filled with 70 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75 0 C. T is given by: (Given : room temperature = 0 0 C, specific heat of copper = 0. cal/gm 0 C) () 85 0 C () C () C 50 0 C Sol. () Final temperature of calorimeter and its contents is given, T 0 = 75 0 C (75 T) (75 0) + 70 (75 0) = 0 75 T = 0 T = C 8. In amplitude modulation, sinusoidal carrier frequency used is denoted by c and the signal frequency is denoted by m. The bandwidth ( m ) of the signal is such that m < < c. Which of the following frequencies is not contained in the modulated wave? () c - m () m () c m + c Sol. () Modulated signal can be written as C m (t) = (A c + A m sin m t) sin c t Ac Ac C m (t) = A c sin c t + cos( c m)t cos( c m )t Am where = A c *9. The temperature of an open room of volume 0 m increases from 7 0 C to 7 0 C due to the sunshine. The atmospheric pressure in the room remains 0 5 Pa. If n i and n f are the number of molecules in the room before and after heating, then n f n i will be : () ().6 0 () FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

16 JEE-MAIN-MPC-07-6 Sol. () PV Using, n = RT PV n f n i = moles R Tf Ti = = molecules molecules 40. In a Young s double slit experiment, slits are separated by 0.5 mm, and the screen is placed 50 cm away. A beam of light consisting of two wavelengths, 650 nm and 50 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is () 5.6 mm ().56 mm () 7.8 mm 9.75 mm Sol. () y = 9 9 m 6500 D n 500 D d d m 4 minimum values of m and n will be 4 and 5 respectively. n y = 50 = 7.8 mm 4 meter 4. A particle A of mass m and initial velocity v collides with a particle B of mass m which is at rest. The collision is head on, and elastic. The ratio of the de-broglie wavelengths A to B after the collision is: A () A () () B A B B A B Sol. () m mv mva vb (conservation of linear momentum) vb v va vb va (elastic collision) vb 4 v A A mbv B m v B A A A m v B m Before collision A v A B After collision V B 4. A magnetic needle of magnetic moment Am and moment of inertia kg m is performing simple harmonic oscillations in a magnetic field of 0.0 T. Time taken for 0 complete oscillations is: () 8.76 s () 6.65 s () 8.89 s 6.98 s FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

17 JEE-MAIN-MPC-07-7 Sol. () T I MB = sec So, time of 0 oscillations = 6.65 sec 4. An electric dipole has a fixed dipole moment p, which makes angle with respect to x-axis. When subjected to an electric field E ˆ Ei, it experiences a torque T ˆ k. When subjected to another electric field E Eˆ j it experiences a torque T T. The angle is: () 90 () 0 () From the given information y E y E E x x P Case - I PE sin = PE sin = 60 P Case - II 44. In a coil of resistance 00, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is: () 75 Wb () 00 Wb () 5 Wb 50 Wb 0 Current (amp.) Time 0.5 sec Change in flux = R idt = 50 Wb *45. A time dependent force F = 6t acts on a particle of mass kg. If the particle starts from rest, the work done by the force during the first sec. will be: () 8 J () 4.5 J () J 9 J Sol. () From impulse momentum theorem 6tdt mv 0 v = m/s FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

18 JEE-MAIN-MPC-07-8 So, work done by the force = K.E. ()() 4.5J 46. Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = /, is given by: 4 () r () r () r r 4 E 4 E E E Sol. () E E E 47. In the given circuit, the current in each resistance is: () 0 A () A () 0.5 A 0.5 A v v v v v v Sol. () Potential difference across each resistor is zero. *48. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time? () () v v t () v v t t t v = u gt FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

19 JEE-MAIN-MPC A capacitance of F is required in an electrical circuit across a potential difference of.0 kv. A large number of F capacitors are available which can withstand a potential difference of not more than 00 V. The minimum number of capacitors required to achieve this is: () () () 6 4 Sol. () C 4 C = 8 F Which requires eight F capacitors in parallel. Minimum number of capacitors required is. C eq = C 50. In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be: r () CE () CE (r r) r r () CE CE (r r) (r r ) E C r r r q = CV CEr r r 5. In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be: () 80 () 45 () 90 5 Sol. () In common emitter amplifier circuit the output voltage is out of phase w.r.t. input voltage. 5. Which of the following statements is false? () Kirchhoff s second law represents energy conservation. () Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude. () In a balanced wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed. A rheostat can be used as a potential divider. Sol. () The balanced condition is given by P R ; When battery and Galvanometer Q S are exchanged, it become P Q ; which is same as previous R S P R G Q S FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

20 JEE-MAIN-MPC-07-0 *5. A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energy time graph of the particle will look like: () KE () KE () 0 KE T 4 T T t 0 T T KE T t 0 T t 0 T T t Sol. () For given SHM x = A sint dx v Acos t dt cos t KE mv ma cos t KEmax 54. An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 0 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 0 8 ms ) () 5. GHz () 0. GHz (). GHz 7. GHz v f ' f, where c So, f = 7. GHz *55. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of: () 8 () 9 () 9 8 Sol. () F mg Ag Stress g A A A Stressf So, 9 Stress i 56. When a current of 5 ma is passed through a galvanometer having a coil of resistance 5, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0 0V is: () () () FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

21 JEE-MAIN-MPC-07- Sol. () g i R R V g g V R R i g 0 R i g G R *57. The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth s radius): () g () g d d () g R g d d R R Sol. () The variation of magnitude of acceleration due to gravity is given by GM g d, where 0 d R R GM, where d R d *58. An external pressure P is applied on a cube at 0 C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by: P () PK () K () P K Sol. () BV By applying pressure, P V V P P given B K V B K By increasing temperature, fractional increase in volume V V P K P K PK FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

22 JEE-MAIN-MPC A diverging lens with magnitude of focal length 5 cm is placed at a distance of 5cm from a converging lens of magnitude of focal length 0cm. A beam of parallel light falls on the diverging lens. The final image formed is: () real and at a distance of 6 cm from the convergent lens. () real and at a distance of 40 cm from convergent lens. () virtual and at a distance of 40 cm from convergent lens. real and at a distance of 40 cm from the divergent lens. Sol. () For diverging lens, v u f v 5 v 5 cm First image is formed at a distance 5cm left to the diverging lens. For the converging lens. v u f v 40 0 v 0 40 v 40 cm f = 5 cm d = 5 cm f = 0 cm *60. A body of mass m = 0 kg is moving in a medium and experiences a frictional force F = kv. Its initial speed is v 0 = 0 ms. If, after 0 s, its energy is mv0, the value of k will be: 8 () 0 kg m s () 0 kg m () 0 kg s v0 mvf mv0 vf 8 mdv Now, kv dt v0 dv 0 m k dt v 0 v 0 m v v0 0 v 0 k t m 0 k v0 v0 m 0k v 0 m 0 k 0 kg m 0v kg m FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

23 JEE-MAIN-MPC-07- PART C - CHEMISTRY *6. gram of a carbonate (M C ) on treatment with excess HCl produces mole of C. The molar mass of M C in g mol is: () 84. () 8.6 () Sol. () M C HCl C MCl H Moles of M C = Moles of C produced. w moles of MC molar mass Molar mass = 84. g mol - So, option () is correct. *6. Given C g C g ; graphite H 0 r 9.5kJ mol H g g H ; kj mol r H C g H CH g g ; 4 H 0 r 890.kJ mol 0 Based on the above thermochemical equations, the value of at 98 K for the reaction C H g CH g will be: graphite 4 () kj mol () 74.8 kj mol () 44.0 kj mol kj mol Sol. () o C g C g ; H 9.5 kj / mol () graphite r o H g g H ; rh 85.8 kj / mol () C g H CH g g ; H 890. kj / mol () o 4 r C H g CH g ; H? graphite 4 [Eq. () + Eq. ()] + [ Eq. ()] = Eq. H H H H kj / mol = 74.8 kj/mol - 6. The freezing point of benzene decreases by C when 0. g of acetic acid is added to 0g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be: (K f for benzene = 5. K kg mol ) () 80.4 % () 74.6 % () 94.6 % 64.6 % Sol. () T i K m f f i r H FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

24 JEE-MAIN-MPC-07-4 i 0.57 CH CH CH CH i % dissociation is 94.6%. *64. The most abundant elements by mass in the body of a healthy human adult are: xygen (6.4%); Carbon (.9%), Hydrogen (0.0%); and Nitrogen (.6%). The weight which a 75 kg person would gain if all H atoms are replaced by H atoms is: () 7.5 kg () 7.5 kg () 0 kg 5 kg Sol. () Total hydrogen ( H ) = 0 75 = 7.5 kg 00 If it is replaced by H then mass will be doubled so now hydrogen mass = 5 kg So, mass of person will be increased by 7.5 kg. *65. U is equal to () Isobaric work () Adiabatic work () Isothermal work Isochoric work Sol. () U q w q 0 in adiabatic process. So, U w 66. The formation of which of the following polymers involves hydrolysis reaction? () Bakelite () Nylon 6, 6 () Terylene Nylon 6 NH Caprolactam 554 K H C CH Nylo n 6 H N 5 n 67. Given 0 0 E.6V, E 0.74 V Cl /Cl Cr /Cr E. V,E.5V. 0 0 Cr 7 /Cr Mn 4 /Mn Among the following, the strongest reducing agent is () Mn + () Cr + () Cl Cr FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

25 JEE-MAIN-MPC-07-5 Reduction potential of o E 0.74 V Cr /Cr So, E 0.74 V Cr/ Cr Cr would be strongest reducing agent. 68. The Tyndall effect is observed only when following conditions are satisfied: (a) The diameter of the dispersed particles is much smaller than the wavelength of the light used. (b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used. (c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude. (d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude. () (b) and (d) () (a) and (c) () (b) and (c) (a) and (d) Sol. () Tyndall effect is observed only when (i) The diameter of the dispersed particle is not much smaller than the wavelength of the light used. (ii) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude. So, (b) and (d) are correct. *69. In the following reactions, Zn is respectively acting as a/an: Zn Na Na Zn (a) (b) Zn C ZnC () base and base () acid and acid () acid and base base and acid Sol. () Zn Na Na Zn ; Zn behaving as an acid. Zn C ZnC ; Zn behaving as a base. 70. Which of the following compounds will behave as a reducing sugar in an aqueous KH solution? HH C CH H HH C CH H () H () H CH HH C H H () H H CH CH HH C CH H H CCH H H FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

26 JEE-MAIN-MPC-07-6 HH C H CH H CCH KH HH C H H CH H H Reducing sugar H H H H Hemi-acetal 7. The major product obtained in the following reaction is: Br H t BuK C 6 H 5 C 6 H 5 () C6H5CH CHC6H5 () t t () C6H5CH BuCH C6H5 t C H CH Bu CH C H C H CH Bu CH C H Sol. () Br H C 6 H 5 C 6 H 5 t BuK C 6 H 5 C 6 H 5 E lim ination major product *7. Which of the following species is not paramagnetic? () C () () B N Sol. () C diamagnetic (6) paramagnetic (0) B paramagnetic (5) N paramagnetic 7. n treatment of 00 ml of 0. M solution of CoCl. 6H with excess AgN ;. 0 ions are precipitated. The complex is: () CoH Cl.H () CoH Cl 6 () CoH Cl Cl.H 5 Moles of CoCl.6H 00 ml 0.M 0 0 moles Ions = ions Precipitated ions =. 0 Ag + ion and Cl ion. So CoH Cl 5 Cl.H is correct. Co H Cl 4 Cl.H FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

27 JEE-MAIN-MPC-07-7 *74. pk a of a weak acid (HA) and pk b of a weak base (BH) are. and.4, respectively. The ph of their salt (AB) solution is: () 6.9 () 7.0 ().0 7. Sol. () pka pkb ph = The increasing order of the reactivity of the following halides for the S N reaction is: CH CHCH CH CHCHCHCl p HC C6H4 CHCl Cl I II III () (II) < (I) < (III) () (I) < (III) < (II) () (II) < (III) < (I) (III) < (II) < (I) Sol. () Rate of S N carbocation stability CH CHCH CH (I) II < I < III CH CH CH (II) p CH C 6 H 5 CH (III) *76. Both lithium and magnesium display several similar properties due to the diagonal relationship; however, the one which is incorrect, is: () both form soluble bicarbonates () both from nitrides () nitrates of both Li and Mg yield N and on heating both form basic carbonates Sol. () 77. The correct sequence of reagents for the following conversion will be: H CH H CH CH CH () CH MgBr, H / CH H, Ag NH () H CH MgBr, Ag NH H,H / CH H () Ag NH Ag NH H,CH MgBr,,H / CH H H,H / CH H,CH MgBr FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

28 JEE-MAIN-MPC-07-8 H CH Ag NH H H /CH H CHMgBr CH CH CCH H C H 78. The products obtained when chlorine gas reacts with cold and dilute aqueous NaH are: () Cl and Cl () Cl and Cl () Cl and Cl Cl and Cl Sol. () Cl NaH NaCl NaCl H cold & dilute 79. Which of the following compounds will form significant amount of meta product during mono-nitration reaction? NH CCH CH () () NHCCH H () Sol. () NH NH NH NH N conc. HN conc. HS4 47% N % NH NH + N 5% NH + H N N FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

29 JEE-MAIN-MPC-07-9 *80. -Methyl-pent--ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is : () Zero () Two () Four Six Sol. () H H C CH C C H 5 HBr, RR H C CH C C H 5 CH Stereoisomers = n = = 4 Br CH 8. Two reactions R and R have identical pre-exponential factors. Activation energy of R exceeds that of R by 0 kj mol. If k and k are rate constants for reactions R and R respectively at 00 K, then ln k / k is equal to : Sol. () (R = 8.4 J mol K ) () () 6 () 4 8 k k Ae Ae Ea k e k k k E / RT a E /RT a E a RT e k k ln 4 *8. Which of the following molecules is least resonance stabilized? () () N () Sol. () Except, all are aromatic in nature, so it has least resonance energy. C H CH FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

30 JEE-MAIN-MPC-07-0 *8. The group having isoelectronic species is: (), F, Na, Mg (), F, Na, Mg, F, Na + and Mg +, all have 0 electrons each. (), F, Na, Mg, F, Na,Mg *84. The radius of the second Bohr orbit for hydrogen atom is: (Planck s Const. h = Js; mass of electron = kg; charge of electron e = C; permittivity of vacuum 0 = kg - m - A ) () 4.76 o A () 0.59 o A (). o A.65 o A Sol. () 0.5n rn Z n = Z = A r 0.5 4A. A 85. The major product obtained in the following reaction is: DIBAL H CH H () CH () CH CH H CH () CH CH CH CH H DIBAL H CH CH CH FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

31 JEE-MAIN-MPC-07- *86. Which of the following reactions is an example of a redox reaction? 5 6 () XeF PF XeF PF () XeF6 H XeF4 HF () XeF6 H XeF 4HF XeF4 F XeF X ef F Xe F 4 6 Xenon oxidises and oxygen gets reduced. 87. A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is a, the closest approach between two atoms in metallic crystal will be: () a () a () a a Sol. () In FCC structure 4r = a r a = closest approach between two atoms. 88. Sodium salt of an organic acid X produces effervescence with conc. H S 4. X reacts with the acidified aqueous CaCl solution to give a white precipitate which decolourises acidic solution of KMn 4. X is: () HCNa () CH CNa () Na C 4 C 6 H 5 CNa Sol. () conc. X Na C C C Na S HS4 4 4 CaCl CaC C acidic 4 white ppt. KMn4 *89. A water sample has ppm level concentration of following anions F 0; S 00; N 50 4 The anion/anions that make/makes the water sample unsuitable for drinking is/are: () both S 4 and N () only F () only S 4 only N Sol. () Permissible limit for S 4 = 500 ppm Permissible limit for N = 50 ppm Permissible limit for F = ppm FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

32 JEE-MAIN-MPC Which of the following, upon treatment with tert-buna followed by addition of bromine water, fails to decolourize the colour of bromine? C 6 H 5 () Br () Br () Br Br Br tert BuNa No elimination FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

33 JEE-MAIN-MPC-07- Read the following instructions carefully: (Instructions as printed on question paper). The candidates should fill in the required particulars on the Test Booklet and Answer Sheet (Side) with Black Ball Point Pen.. For writing/marking particulars on Side of the Answer Sheet, use Blue/Black Ball Point Pen only.. The candidates should not write their Roll Numbers anywhere else (except in the specified space) on the Test Booklet/Answer Sheet. 4. ut of the four options given for each question, only one option is the correct answer. 5. For each incorrect response, (¼) (one-fourth) marks of the total marks allotted to the question (i.e. marks) will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Sheet. 6. Handle the Test Booklet and Answer Sheet with care, as under no circumstances (except for discrepancy in Test Booklet Code and Answer Sheet Code), another set will be provided. 7. The candidates are not allowed to do any rough work or writing work on the Answer Sheet. All calculations/writing work are to be done in the space provided for this purpose in the Test Booklet itself, marked Space for Rough Work. This space is given at the bottom of each page and in four pages (Pages 0 ) at the end of the booklet. 8. n completion of the test, the candidates must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the candidates are allowed to take away this Test Booklet with them. 9. Each candidate must show on demand his/her Admit Card to the Invigilator. 0. No candidate, without special permission of the Superintendent or Invigilator, should leave his/her seat.. The candidates should not leave the Examination Hall without handing over their Answer Sheet to the Invigilator on duty and sign the Attendance Sheet again. Cases where a candidate has not signed the Attendance Sheet a second time will be deemed not to have handed over the Answer Sheet and dealt with as an unfair means case. The candidates are also required to put their left hand THUMB impression in the space provided in the Attendance Sheet.. Use of Electronic/Manual Calculator and any Electronic device like mobile phone, pager etc. is prohibited.. The candidates are governed by all Rules and Regulations of the JAB/Board with regard to their conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and Regulations of the JAB/Board. 4. No part of the Test Booklet and Answer Sheet shall be detached under any circumstance. 5. Candidates are not allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, electronic device or any other material except the Admit Card inside the examination room/hall. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594