NARAYANA IIT ACADEMY

Transcription

1

2 IIT-JEE8-CODE--Question and Solutions- CODE NARAYANA IIT ACADEMY presents IIT-JEE 8 SOLUTIONS PAPER - I Mathematics + x [{x cos (cot x) + sin (cot x)} } / =. If < x <, then x (A) (B) x + x x + x (D) / x Sol.: + x x x = + x : / + x = x + x.. Consider the two curves C : y = 4x C : x + y 6x + = Then, (A) C and C touch each other only at one point (B) C and C touch each other exactly at two points C and C intersect (but do not touch) at exactly two points (D) C and C neither intersect nor touch each other + x Sol.: (x ) + (y ) = ( ) NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

3 IIT-JEE8-CODE--Question and Solutions- (, ) y = 4x (,) (,) (, ) (, -) x + 4x 6x + = (x ) = x =, Ans. touch each other exactly two points. (B). The edges of a parallelopiped are of unit length and are parallel to non coplanar unit vectors a,b,c ˆ ˆ ˆ such that Sol.: ˆ ˆ a.b ˆ = b.cˆ= c.a ˆ ˆ= Then, the volume of the parallelopiped is (A) aa. ab. ac. V = ba. bb. bc. = ca. cb. cc. / / / / / / (B) (D) V = V = (A) 4. Let a and b be non zero real numbers. Then, the equation (ax + by + c) (x 5xy + 6y ) = represents (A) four straight lines, when c = and a, b are of the same sign (B) two straight lines and a circle, when a = b, and c is of sign opposite to that of a two straight lines and a hyperbola, when a and b are of the same sign and c is of sign opposite to that of a (D) a circle and an ellipse, when a and b are of the same sign and c is of sign opposite to that of a. Sol.: x 5xy + 6y = represent a pair of lines passing through origin ax + ay = c x + y = c a > ax + ay + c = represent a circle (B) 5. The total number of local maxima and local minima of the function f(x) = Sol.: (A) (B) (D) f(x) = ( + x), < x = x /, < x < + < < < ( x), x / x, x is NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

4 IIT-JEE8-CODE--Question and Solutions-4 O x The total number of local maximum or minimum =. 6. Let g(x) = n (x ) ; < x <, m and n are integers, m, n >, and let p be the left hand derivative of g(x) = p, then m log cos (x ) x at x =. If lim x + (A) n =, m = (B) n =, m = n =, m = (D) n >, m = n Sol.: According to question we get, lim + g(x) = x ( h) lim log cos ( h = ( h) lim ) + h log(cosh) + h Alternate: From graph : p = = h = lim h tanh g(x) = g(x) = x = n ( x ) < x <, m, n N m log(cos ( x ) n ( x ) lim = m log cos ( x ) + x n ( h) g(+) = lim h log cos n h m h, h > = lim h m(ln cosh) n h = lim m h (ln cosh) n nh = lim cosh m h sinh n n h = lim (cosh) m h sinh h n =, m = 7. Let P (x, y ) and Q(x, y ), y <, y <, be the end points of the latus rectum of the ellipse x + 4y = 4. The equations of parabolas with latus rectum PQ are (A) x + y = + (B) x y = + x + y = (D) x y = NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

5 IIT-JEE8-CODE--Question and Solutions-5 Sol.: Given ellipse is x y + = e = 4 = 4 (-ae,) Q, P (ae, b /a) = (ae,) P,, Q ( ae, b /a) = (, ) length of PQ = Q V S / P V, V V +, VS = SV = PQ 4 = Equations of parabolas are x = y x + y = and x = + y + x y = + (B,C) 8. A straight line through the vertex P of a triangle PQR intersects the side QR at the point S and the circumcircle of the triangle PQR at the point T. If S is not the centre of the circumcircle, then (A) + < (B) + > PS ST QS SR PS ST QS SR 4 + < (D) PS ST QR P 4 + > PS ST QR Sol.: Q S T H.M. < G.M. < (PS. ST) / + PS ST R NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

6 IIT-JEE8-CODE--Question and Solutions-6 + > PS ST PS.ST As PS ST = QS SR + >... (i) Option (B) is correct PS ST QS SR A.M. > G.M QS + SR > (QS.SR) QS SR < QR... (ii) From (i) and (ii) 4 + > PS ST QR (D) 9. Let f(x) be a non constant twice differentiable function defined on (, ) such that f(x) = f( x) and f 4 =. Then, (A) f (x) vanishes at least twice on [, ] (B) f = / / f x+ / sinx dx = (D) sin t f(t)e π dt = f( t) e sinπt dt / Sol.: (A) f (x) = f ( x) f (/4) = f (/4) = f (/4) = f (/4) = a point (/4, /4) in which f (x) = f (/) = (B) f(x) = f( x) f (x) = f ( x) f (/) = f (/) f (/) =. f(x + /) = f( x /) f(x + /) = f(/ x) i.e., f(x + /) is even i.e., / f ( x + /)sin x = / / (D) f () t esinπt = Put ( t) = u f ( t ) esinπ t dt / R.H.S. = f (4) esinπ( u) du = / K ey (A,B,C,D) / f ( u ) esinπu dx. Let S n and T n = n n = k= n + kn+ k π (A) S n < π T n < n n for n =,,,... then, k= n + kn+ k π (B) S n > π (D) T n > NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

7 IIT-JEE8-CODE--Question and Solutions-7 Sol.: Consider: lim n n n = k + kn+ n k f(x) = x + x+ = x y = dx x + x + π = -/ x As f(x) is decreasing for x >, S n < dx < T n x + x+ Hence A & D are correct. : (A), (D) SECTION III Assertion Reason Type This section contains 4 reasoning type questions. Each question has 4 choices (A), (B), and (D) out of which ONLY ONE is correct.. Consider the system of equations x y + z = x + y z = k x y + 4z = STATEMENT : The system of equations has no solutions for k and STATEMENT : The determinant k, for k 4 (A) STATEMENT is True, STATEMENT is True; STATEMENT is a correct explanation for STATEMENT (B) STATEMENT is True, STATEMENT is True; STATEMENT is NOT a correct explanation for STATEMENT STATEMENT is True, STATEMENT is False (D) STATEMENT is False, STATEMENT is True Sol.: = = 4 z = k = k, k (A). Consider the system of equations ax + by =, cx + dy =, where a, b, c, d {, } STATEMENT : The probability that the system of equations has a unique solution is /8 and STATEMENT : The probability that the system of equations has a solution is. (A) STATEMENT is True, STATEMENT is True; STATEMENT is a correct explanation for STATEMENT (B) STATEMENT is True, STATEMENT is True; STATEMENT is NOT a correct explanation for STATEMENT NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

8 IIT-JEE8-CODE--Question and Solutions-8 STATEMENT is True, STATEMENT is False (D) STATEMENT is False, STATEMENT is True Sol.: a, b, c, d {, } D = a b c d n(s) = 4 = 6 (I) To have unique solution.,,,,, Total = 6 cases, The probability that system of equations has unique solution = 6/6 = /8 II. Homogenous system is always consistent (B). Let f and g be real valued functions defined on interval (, ) such that g (x) is continuous g(), g () = g (), and f(x) = g(x) sinx. STATEMENT : lim [g(x) cotx g(x) cosec x] = f (). and STATEMENT : x f () = g() (A) STATEMENT is True, STATEMENT is True; STATEMENT is a correct explanation for STATEMENT (B) STATEMENT is True, STATEMENT is True; STATEMENT is NOT a correct explanation for STATEMENT STATEMENT is True, STATEMENT is False (D) STATEMENT is False, STATEMENT is True Sol.: f(x) = g(x) sinx f (x) = g (x) sinx + g (x) cosx + g(x) sinx f () = g () = f (x) = g (x) sinx + g(x) cosx f () = g() For statement : g(x)cos x g() g (x) cos x lim = lim g(x)sin x = g () = = f () x sin x x cos x For statement f () = g() : (B) 4. Consider three planes P : x y + z = P : x + y z = P : x y + z = Let L, L, L be the lines of intersection of the planes P and P, P and P, and P and P, respectively STATEMENT : At least two of the lines L, L and L are non-parallel and STATEMENT : The three planes do not have a common point (A) STATEMENT is True, STATEMENT is True; STATEMENT is a correct explanation for STATEMENT (B) STATEMENT is True, STATEMENT is True; STATEMENT is NOT a correct explanation for STATEMENT STATEMENT is True, STATEMENT is False (D) STATEMENT is False, STATEMENT is True Sol.: L, L, L are parallel to each other statement () is not true NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

9 D = = The three planes do not have a common point. statement () is correct. : (D) IIT-JEE8-CODE--Question and Solutions-9 SECTION IV Linked Comprehension Type This section contains paragraphs. Based upon each paragraph, multiple choice questions have to be answered. Each question has 4 choices (A), (B), and (D) out of which ONLY ONE is correct. Paragraph for Questions Nos. 5 to 7 Let A, B, C be three sets of complex numbers as defined below A = {z : Imz } B = {z : z i = } C = {z : Re (( i)z ) = } 5. The number of element in the set A B C is (A) (B) (D) 6. Let z be any point in A B C. Then, z + i + z 5 i lies between (A) 5 and 9 (B) and 4 5 and 9 (D) 4 and Let z be any point in A B C and let w be any point satisfying w i <. Then, z w + lies between (A) 6 and (B) and 6 6 and 6 (D) and 9 Sol (5 7) P(z) A B C R (-, ) y O (, ) 5. From graph only one point in A B C (B) 6. z + i + z 5 i = PR + PQ = RQ = 6 = 6 7. (D) As z ω < z ω < 6 z ω < 6 6 < z ω < 6 or 6 + < z ω + < 9 < z ω + < 9 (D). B x +y -4x-y-4 = Q (5,) A y x C x + y = Paragraph for Questions Nos. 8 to NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

10 IIT-JEE8-CODE--Question and Solutions- A circle C of radius is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ, QR, RP are D,E, F, respectively. The line PQ is given by the equation x + y 6 = and the point D is,. Further, it is given that the origin and the centre of C are on the same side of the line PQ. 8. The equation of circle C is (A) (x ) + (y ) = (B) + + = (x ) + (y + ) = (D) (x ) + (y ) = 9. Points E and F are given by (A),,(,) (B),,(,),,, (D),,,. Equations of the sides QR, RP are (A) y = x+, y = x (B) y = x, y = (x ) (y ) y = x+, y= x (D) y = x, y = Sol.: 8 P, = (, ) Q(, ) E, R (, ) π/6 I π/6 π/6 D, F (, ) P(,) x + y = 6 I = (, ) 8. (x ) + (y ) = : (D) 9. E = π π cos, sin, F (, ) : (A). Equation of PR, y = Equation of QR, y = (D) x Paragraph for Questions Nos. to Consider the functions defined implicitly by the equation y y + x = on various intervals in the real line. If x (, ) (, ), the equation implicitly defines a unique real valued differentiable function y = f(x). NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

11 IIT-JEE8-CODE--Question and Solutions- If x (, ), the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g () =.. If f( ) =, then f ( ) = 4 4 (A) (B) (D) 7 7. The area of the region bounded by the curve y = f(x), the x axis, and the lines x = a and x = b, where < a < b <, is b b x x (A) dx + bf (b) af (a) (B) dx bf (b) af (a) a [((f (x)) ] + a ((f (x)) ) b b x x dx bf (b) + af (a) (D) dx bf (b) af (a) ((f (x)) ) + ((f (x)) ). a g(x)dx = (A) g( ) (B) g () (D) g() a Sol.: y y + x = y y y + = y = ( y ) 6 yy ( ) y. y = = = ( y ) ( y ).9 ( 8).9 (B). x = y + y as x (, ) x < y + y < y y > (y + ) (y ) > y > x (, ) (As y = x = ) f(x) is positive x (, ) b b Hence required area = f(x)dx = ydx = yx = a b xdx bf (b) a ((f (x)) ) + (A) af(a) a 4 = 7 b a b a dy xdx dx. Consider: (g(x)) g(x) + x = and (g( x)) g( x) x = (g(x)) + (g( x)) (g(x) + g( x)) = [g(x) + g( x)] [(g(x)) + (g( x)) g(x) g( x) ] = Let (g(x)) + (g( x)) g(x) g( x) = g() = g() = + or which is not the case as g() =, given g(x) + g( x) = g(x) is an odd function for x (, ) (D) g ( x ) dx = g() g( ) = g(). NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

12 IIT-JEE8-CODE--Question and Solutions- Physics Useful Data : Planck s constant Velocity of light h = 4. 4 ev.s c = 8 m/s. SECTION I Straight Objective Type This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), and (D), out of which ONLY ONE is correct. 4. Figure shows three resistor configurations R, R and R connected to V battery. If the power dissipated by the configuration R, R and R is P, P and P, respectively, then Figure : Ω Ω Ω Ω Ω V Ω Ω Ω Ω Ω V Ω Ω Ω Ω Ω V R R (A) P > P > P (B) P > P > P (B) P > P > P (D) P > P > P. R Sol. The resistor configurations reduce to Ω V Ω V Ω V R R R V Now P = R eq P > P > P is correct. 5. Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table. Least count for length =. cm NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

13 Sol. IIT-JEE8-CODE--Question and Solutions- Least count for time =. s. Student Length of the pendulum (cm) Number of oscillations (n) Total time for (n) oscillations (s) Time period (s) I II III g If E I, E II and E III are the percentage errors in g, i.e., for students I, II and III, respectively, g (A) E I = (B) E I is minimum E I = E II (D) E II is maximum. g = 4π T g T E = = + g T Student Length of the pendulum, No. of Time Period, error error (cm),, Oscillation T, T I 64.± ± 8 II 64.± ± 4 III.± ± E I =, E II =, EIII = E is the minimum correct choice is (B) 6. Which one of the following statement is WRONG in the context of X rays generated from X ray tube? (A) wavelength of characteristic X rays decreases when the atomic number of the target increases (B) cut off wavelength of the continuous X rays depends on the atomic number of the target intensity of the characteristic X rays depends on the electrical power given to the X ray tube (D) cut off wavelength of the continuous X rays depends on the energy of the electrons in the X ray tube. Sol. Cut off wavelength depends upon accelerating potential. (B) is the correct choice. 7. Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 6º). In the position of minimum deviation, the angle of refraction will be (A) º for both the colours (B) greater for the violet colour greater for the red colour (D) equal but not º for both the colours. Sol. In the position of minimum deviation A r = irrespective of colour (A) is correct. 8. A spherically symmetric gravitational system of particles has a mass density ρ for r R ρ= forr> R where ρ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed V as a function of distance r ( < r < ) from the centre of the system is represented by NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

14 IIT-JEE8-CODE--Question and Solutions-4 V V (A) V R r (B) V R r R r (D) R r Sol. For points r R, 4 G ρ πr m mv = r r v r for points r > R 4 G ρ πr m mv = r r v r is correct. 9. An ideal gas is expanding such that PT = constant. The coefficient of volume expansion of the gas is (A) (B) T T (D) 4 T T. dv Sol. Coefficient of volume expansion γ= VdT Now, PT = constant T = constant V dv T.V T dt = V γ =. T is correct. SECTION II Multiple Correct Answers Type This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), and (D), out of which ONE OR MORE is/are correct.. Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this plot to choose the correct choice (s) given below. Figure NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

15 B/A IIT-JEE8-CODE--Question and Solutions A (A) fusion of two nuclei with mass numbers lying in the range of < A < 5 will release energy (B) fusion of two nuclei with mass numbers lying in the range of 5 < A < will release energy fission of a nucleus lying in the mass range of < A < will release energy when broken into two equal fragments (D) fission of a nucleus lying in the mass range of < A < 6 will release energy when broken into two equal fragments Sol. (Energy released) = Total B.E. of products Total B.E. of reactants. (B) and (D) are correct.. Two balls, having linear momenta ˆ p ˆ = pi and p = pi, undergo a collision in free space. There is not ' external force acting on the balls. Let p ' and p be their final momenta. The following option (s) is (are) NOT ALLOWED for any non zero value of p, a, a, b, b, c and c. ' ˆ ˆ ˆ ' p = ai + bj + ck p = ckˆ (A) (B) ' p ˆ ˆ ' = ai+ bj p ˆ = ck ' ˆ ˆ ˆ ' p = ai + bj + ck p = ai ˆ + bj ˆ (D) ' p = ai ˆ+ bj ˆ ckˆ. ' p = ai ˆ+ bj ˆ Sol. Initial momentum of the system =. In the absence of external forces, final momentum of the system must also be zero. i.e., P + P = (A) and (D) are correct.. In a Young s double slit experiment, the separation between the two slits is d and the wavelength of the light is λ. The intensity of light falling on slit is four times the intensity of light falling on slit. Choose the correct choice(s), (A) if d = λ, the screen will contain only one maximum (B) if λ< d < λ, at least one more maximum (besides the central maximum) will be observed on the screen if the intensity of light falling on slit is reduced so that it becomes equal to that of slit, the intensities of the observed dark and bright fringes will increase (D) if the intensity of light falling on slit is reduced so that it becomes equal to that of slit, the intensities of the observed dark and bright fringes will increase. Sol. Condition for maxima is d sin θ = nλ If d = λ, sin θ = n Possible value is only one maxima will be obtained (A) is correct. IF λ < d < λ nλ λ < < λ sin θ n < < sin θ n =, ±. (B) is correct. NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

16 IIT-JEE8-CODE--Question and Solutions-6 ( ) ( ) I = I + I & I = I I max min with I = 4I Imax = 9I, Imin = I but when I = I I max = 4I and I min = and (D) are wrong.. A particle of mass m and charge q, moving with velocity V enters Region II normal to the boundary as shown in the figure. Region II has a uniform magnetic field B perpendicular to the plane of the paper. The length of the Region II is. Choose the correct choice(s) Figure : Region I Region II Region III V qb (A) the particle enters Region III only if its velocity V > m qb (B) the particle enters Region III only if its velocity V < m qb path length of the particle in Region II is maximum when velocity V = m (D) time spent in Region II is same for any velocity V as long as the particle returns to Region I. Sol. mv The radius of the circular path in region II r = qb If r >, the particle enters the region III qb V > m (A) is correct. The path length will be maximum if it is able to describe a semi circle qb i.e., = r V = m is correct. If it is to return to region I, it must trace a semi circle, the time for which is dependent of (D) is correct. SECTION III πm V T = qb Assertion Reason Type This section contains 4 reasoning type questions. Each question has 4 choices (A), (B), and (D) out of which ONLY ONE is correct. 4. STATEMENT Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first. because STATEMENT By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline. (A) Statement is True, Statement is True; Statement is a correct explanation for Statement. NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

17 IIT-JEE8-CODE--Question and Solutions-7 (B) Statement is True, Statement is True; Statement is NOT a correct explanation for Statement. Statement is True, Statement is False. (D) Statement is False, Statement is True. Sol. The acceleration of an object down an incline of angle θ is gsinθ a = I + mr Now, I hollow > I solid for same mass and dimensions a hollow < a solid solid cylinder will reach the bottom first. Statement is false. (D) is correct. 5. STATEMENT The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down. because STATEMENT In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant. (A) Statement is True, Statement is True; Statement is a correct explanation for Statement. (B) Statement is True, Statement is True; Statement is NOT a correct explanation for Statement. Statement is True, Statement is False. (D) Statement is False, Statement is True. Sol. As the water stream moves up, its speed decreases (due to gravity) and since flow rate (= Av) remains constant, the area increases making it spread like a fountain. The reverse is true when it moves down. Both Statements are correct and Statement is a correct explanation of Statement. (A) is correct. 6. STATEMENT In a Meter Bridge experiment, null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance. because STATEMENT Resistance of a metal increases with increase in temperature. (A) Statement is True, Statement is True; Statement is a correct explanation for Statement. (B) Statement is True, Statement is True; Statement is NOT a correct explanation for Statement. Statement is True, Statement is False. (D) Statement is False, Statement is True. Sol. In a meter bridge R X X = at null point RS X When unknown resistance is put in an enclosure maintained at a higher temperature, R X will increase. To keep the null point same, R S has to be increased. Statement is false. (D) is correct. 7. STATEMENT An astronaut in an orbiting space station above the Earth experiences weightlessness. because STATEMENT An object moving around the Earth under the influence of Earth s gravitational force is in a state of free fall. (A) Statement is True, Statement is True; Statement is a correct explanation for Statement. (B) Statement is True, Statement is True; Statement is NOT a correct explanation for Statement. Statement is True, Statement is False. (D) Statement is False, Statement is True. NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

18 IIT-JEE8-CODE--Question and Solutions-8 Sol. Statement is correct and so is Statement. Both the astronaut and space station are in a state of free fall. (A) is correct. NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

19 Section IV IIT-JEE8-CODE--Question and Solutions-9 Linked Comprehension Type This section contains paragraphs. Based upon each paragraph, multiple choice questions have to be answered. Each question has 4 choices (A), (B), and (D) out of which ONLY ONE is correct. Paragraph for Question Nos. 8 to 4 A small block of mass M moves on a frictionless surface of an inclined plane, as show in figure. The angle of the incline suddenly changes from 6º to º at point B. The block is initially at rest at A. Assume that collision between the block and the incline are totally inelastic (g = m/s ). Figure : A M v B º C m m 8. The speed of the block at point B immediately after it strikes the second incline is (A) 6 m / s (B) 45 m / s m / s (D) 5 m / s. Sol. At B before collision v = 6m/s at an angle of º with downward vertical. Hence component of v along BC = v cos º impact force acts along the normal to the plane, hence v cos º will not change and component of v perpendicular to the plane becomes zero as collision is completely inelastic. (B) is correct. 9. The speed of the block at point C immediately before it leaves the second incline is (A) m / s (B) 5 m / s 9 m / s (D) 75 m / s Sol. Potential Energy + Kinetic Energy = constant m(vcos ) + mg m = mve vc = 5 m / s. (B) is the correct choice. 4. If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point B, immediately after it strikes the second incline is (A) m / s (B) 5 m / s (D) 5 m / s. Sol. Velocity after collision 5 º 45 6º 5 5 º Uy = 5cos 45cos6 º 6º 45 NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

20 IIT-JEE8-CODE--Question and Solutions = =. is correct. Paragraph for Question Nos. 4 to 4 5 A small spherical monoatomic ideal gas bubble γ= is trapped inside a liquid of density ρ (see figure). Assume that the bubble does not exchange any heat with eth liquid. The bubble contains n moles of gas. The temperature of the gas when the bubble is at the bottom is T, the height of the liquid is H and the atmospheric pressure is P (Neglect surface tension). Figure : P Liquid H y 4. As the bubble moves upwards, besides the buoyancy force the following forces are acting on it (A) only the force of gravity (B) the force due to gravity and the force due to the pressure of the liquid the force due to gravity, the force due to the pressure of the liquid and the force due to viscosity of the liquid (D) the force due to gravity and the force due to viscosity of the liquid. Sol. Besides the buoyancy force, the other forces are (i) force of gravity, and (ii) force due to viscosity The buoyant force is due to the pressure of the liquid. (D) is correct. 4. When the gas bubble is at a height y from the bottom, its temperature is Sol. 5 P +ρ gh P +ρ g(h y) (A) T (B) T P +ρ gy P +ρ gh P 5 +ρ gh P 5 +ρ g(h y) T (D) T. P +ρ gy P +ρ gh For the gas inside the bubble, the adiabatic equation T Y P Y = constant applies 5 5 { } T (P +ρ gh) = T P +ρ g(h y) { +ρ } 5 T P g(h y) T = (B) is correct. (P +ρ gh) The buoyancy force acting on the gas bubble is (Assume R is the universal gas constant) (A) ρ nrgt 5 (P +ρ gh) 7 5 (P +ρ gy) (B) ρ nrgt 5 +ρ +ρ 5 (P gh) ]P g(h y)] NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

21 IIT-JEE8-CODE--Question and Solutions- ρ nrgt 5 (P +ρ gh) 8 5 (P +ρ gy) (D) ρ nrgt 5 +ρ +ρ 5 (P gh) [P g(h y)]. Sol. Buoyancy force = Vρ g where V is the volume of the bubble (at location y) nrt Now V = (P +ρ gh) and PV Y = constant Y nrt (P +ρ gh) = { P +ρ (H y) } V P +ρ gh V = nrt γ γ (P +ρ H) { P +ρ g(h y) } γ ρ g Buoyant force = V (B) is correct. γ Paragraph for Question Nos. 44 to 46 In a mixture of H He + gas (He + is singly ionized He atom), H atoms and He + ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He + ions (by collisions). Assume that the Bohr model of atom is exactly valid. 44. The quantum number n of the state finally populated in He + ions is (A) (B) 4 (D) 5. Sol. The excitation energy of the H atoms =.4 (.6) =. ev Energy of the He + ions in their first excited state =.4 4 =.6 ev Energy after transferrence of energy =.6 +. =.4 ev. Now let n be the quantum number of the final state of He + ions, then ().4 =.6 n n = 4. is correct The wavelength of light emitted in the visible region by He + ions after collisions with H atoms is (A) m (B) m m (D) 4. 7 m. Sol. E = E E4 hc 7 = E 4 λ evm 6 7 λ= 4.8 m 7.6 ev is correct. 46. The ratio of the kinetic energy of the n = electron for the H atom to that of He + ion is (A) 4 (B) NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

22 IIT-JEE8-CODE--Question and Solutions- (D). Sol. Kinetic energy of n = electron for H atom =.4 ev Kinetic energy of n = electron for He + ion =.4 4 ev ratio = 4 (A) is correct. Chemistry SECTION I Straight Objective Type This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), and (D), out of which ONLY ONE is correct. 47. The major product of the following reaction is Me Br F PhS Na dim ethylformamide (A) NO Me SPh (B) Me SPh F F Me NO Br (D) Me NO SPh SPh SPh NO NO Sol. In DMF solvent S N being not possible, S N takes place resulting into inversion of configuration. (A) 48. Hperconjugation involves overlap of the following orbitals (A) σ - σ (B) σ - p p p (D) π - π Sol Hyperconjuation is σ - π, σ - odd e and σ - cationic carbon conjugation and it results into π bond formation by p p overlap. 49. Aqueous solution of Na S O on reaction with Cl gives (A) Na S 4 O 6 (B) NaHSO 4 NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

23 NaCl (D) NaOH IIT-JEE8-CODE--Question and Solutions- Sol Na SO + 4Cl + 5HO NaHSO4 + 8HCl (B) 5. Native silver metal forms a water soluble complex with a dilute aqueous solution of NaCN in the presence of (A) nitrogen (B) oxygen carbon dioxide (D) argon Sol Ag + NaCN + O + HO Na[Ag(CN) ] + NaOH (B) 5..5 ml of 5 M weak monoacidic base (K b = at 5 C) is titrated with 5 concentration of H + at equivalence point is (K w = 4 at 5 C) (A).7 M (B). 7 M. M (D).7 M M HCl in water at 5 C. the Sol.4M BOH (K b = ` ) + MHCl 5 ph = [pk w pkb log c] = [4 log.] =.5, [H + ] =.5 =.5 =. M 5. Under the same reaction conditions, initial concentration of.86 mol dm of a substance becomes half in 4 k seconds and seconds through first order and zero order kinetics, respectively. Ratio k of the rate constants for first order (k ) and zero order (k ) of the reaction is (A).5 mol dm (B). mol dm.5 mol dm (D). mol dm.69 Sol K = K = = min 4 K.69 = = K.86 (A) SECTION II Multiple Correct Answer Type This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), and (D), out of which ONE OR MORE is/are correct. 5. The correct statement(s) concerning the structures E, F and G is (are) NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

24 IIT-JEE8-CODE--Question and Solutions-4 H C O H C OH C H OH H C CH H C CH C H OH (E) (F) (G) (A) E, F and G are resonance structures F and G are geometrical isomers (B) E, F and E, G are tautomers (D) F and G are diastereomeers Sol F is the enol form of E and so is also G for E. F and G are also geometrical isomers. (B),, (D) 54. The correct statement(s) about the compound given below is (are) Cl H H C Cl (A) The compound is optically active The compound possesses plane of symmetry CH CH (B) The compound possesses centre of symmetry (D) The compound possesses axis of symmetry Sol CH Two chiral centres are of same configurations and hence optically active. H Cl Cl H (A) CH 55. A gas described by van der Waals equation (A) behaves similar to an ideal gas in the limit of large molar volumes (B) behaves similar to an ideal gas in the limit of large pressure is characterized by van der Waals coefficients that are dependent on the identity of the gas but are independent of the temperature (D) has the pressure that is lower than the pressure exerted by the same gas behaving ideally a Sol P + (V b) = RT V a When V is very large V b V and P+ P, so PV = RT (for mole of an ideal gas). Due to inward pull V acting on molecule striking the wall, the pressure decreases. The van der Waal s constants a and b are characteristics of a gas and as per van der Waal s they are temperature independent. (A),, (D) 56. A solution of colourless salt H on boiling with excess NaOH produces a non-flammable gas. The gas evolution ceases after sometime. Upon addition of Zn dust to the same solution, the gas evolution restarts. The colourless salt(s) H is (are) (A) NH 4 NO (B) NH 4 NO NH 4 Cl (D) (NH 4 ) SO 4 Sol NH4X + NaOH NaX + HO + NH (X = NO, NO, Cl, SO 4 ) NO + 4Zn + 7OH 4ZnO + HO + NH NO + Zn + 5OH ZnO + H O + NH (A), (B) SECTION III NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

25 IIT-JEE8-CODE--Question and Solutions-5 Assertion Reason Type This section contains 4 reasoning type questions. Each question has 4 choices (A), (B), and (D) out of which ONLY ONE is correct. 57. STATEMENT-: The plot of atomic number (y-axis versus number of neutrons (x-axis) for stable nuclei shows a curvature towards x-axis from theline of 45 slope as the atomic numbr is increased. and STATEMENT-: Proton-proton electrostatic repulsions begin to overcome attractive forces involving protons and neutrons in heavier nuclides. (A) STATEMENT- is True, STATEMENT- is True; STATEMENT- is a correct explanation for STATEMENT- (B) STATEMENT- is True, STATEMENT- is True; STATEMENT- is NOT a correct explanation for STATEMENT- STATEMENT- is True, STATEMENT- is False (D) STATEMENT- is False, STATEMENT- is True Sol (A) (A) When Z >, the number of neutron must increase above the number of protons so as to overcome protonproton repulsion i.e., n p >. 58. STATEMENT-: For every chemical reaction at equilibrium, standard Gibbs energy of reaction is zero. and STATEMENT-: At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy. (A) STATEMENT- is True, STATEMENT- is True; STATEMENT- is a correct explanation for STATEMENT- (B) STATEMENT- is True, STATEMENT- is True; STATEMENT- is NOT a correct explanation for STATEMENT- STATEMENT- is True, STATEMENT- is False (D) STATEMENT- is False, STATEMENT- is True Sol G is minimum while G = at equilibrium (D) 59. STATEMENT-: Bromobenzene upon reaction with Br /Fe gives, 4-dibromobenzeen as the major product. and STATEMENT-: In bromobenzene, the inductive effect of the bromo group is more dominant than the mesomeric effect in directing the incoming electrophile.. (A) STATEMENT- is True, STATEMENT- is True; STATEMENT- is a correct explanation for STATEMENT- (B) STATEMENT- is True, STATEMENT- is True; STATEMENT- is NOT a correct explanation for STATEMENT- STATEMENT- is True, STATEMENT- is False (D) STATEMENT- is False, STATEMENT- is True Sol In bromobenzene, inductive effect is responsible for deactivating the benzene nucleus and has no effect on directive influence. The directive influence is governed solely by mesomeeric effect. 6. STATEMENT-: Pb 4+ compounds are stronger oxidizing agents than Sn 4+ compounds. and NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

26 IIT-JEE8-CODE--Question and Solutions-6 STATEMENT-: The higher oxidation states for the group 4 elements are more stable for the heavier members of the group due to inert pair effect. (A) STATEMENT- is True, STATEMENT- is True; STATEMENT- is a correct explanation for STATEMENT- (B) STATEMENT- is True, STATEMENT- is True; STATEMENT- is NOT a correct explanation for STATEMENT- STATEMENT- is True, STATEMENT- is False (D) STATEMENT- is False, STATEMENT- is True Sol Sn and Pb both have ns np configurations of their valence shells. Moving down a group, the ns electron pair becomes more and more inert towards bonding called inter pair effect. This is maximum in Pb. So Pb 4+ tends to get reduced to Pb + i.e., Pb 4+ is stronger oxidizing agent then Sn 4+ in which the inert pair effect is relatively less. SECTION IV Linked Comprehension Type This section contains paragraphs. Based upon each paragraph, multiple choice questions have to be answered. Each question has 4 choices (A), (B), and (D) out of which ONLY ONE is correct. Paragraph for Questions Nos. 6 to 6 Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixtures as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is.9 Given: Freezing point depression constant of water Freezing point depression constant of ethanol Boiling point elevation constant of water Boiling point elevation constant of ethanol Standard freezing point of water = 7K Standard freezing point of ethanol = 55.7 K Standard boiling point of water = 7 K Standard boiling point of ethanol = 5.5 K Vapour pressure of pure water =.8 mm Hg Vapour pressure of pure ethanol = 4 mm Hg Molecular weight of water = 8 g mol Molecular weight of ethanol = 46 g mol water (K f ) =.86 K kg mol ethanol (K f ) =. K kg mol water (K b ) =.5 K kg mol ethanol (K b ) =. K kg mol In answering the following questions, consider the solutions to be ideal dilute solutions and slutes to be non-volatile and non-dissociateive. 6. The freezing point of the solution M is (A) 68.7 K 4. K (B) 68.5 K (D) 5.9 K Sol. mole water in.9 mol i.e..9 46g ethanol. molality of water in the solution =.9 46 =.5 4 = T = k C =.5= 5 f f m NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4

27 T f of solution = = 5.7K (D) IIT-JEE8-CODE--Question and Solutions-7 6. The vapour pressure of the solution M is (A) 9. mm Hg 9.5 mm Hg Sol p = xp + xp mix =., = = 9.8 mm (A) (B) 6. mm Hg (D) 8.8 mm Hg 6. Water is added to the solution M such that the mole fraction of water in the solution becomes.9 The boiling point of this solution is (A) 8.4 K (B) 76. K 7.5 K (D) 54.7 K Sol x =., x =.9 CHOH 5 HO 8.9g of H O has. mol of C H 5 OH. molality of C H 5 OH = = Tb =.5 = = boiling point = 7 +. = 76.K (B) mole KS Paragraph for Questions Nos. 64 to 66 In the following reaction sequence product I, J and L are formed. K represents a reagent.. Mg / ether. nabh4. CO K H H. PBr + C. H Pd / BaO O Cl 4 quinoline Hex ynal I J L 64. The structure of the product I is: (A) Me Me Br Br O (B) Me Br (D) Me Br Sol C H O NaBH 4 C H OH (D) PBr H C Br 65. The structures of compound J and K, respectively, are (A) Me COOH and SOCl (B) Me OH and SO Cl Me COOH O and SOCl (D) Me COOH and CH SO Cl Sol I.Mg / ether H C Br II. CO III.HO + C H NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4 OH O

28 IIT-JEE8-CODE--Question and Solutions-8 (A) 66. The structure of product L is (A) Me CHO (B) Me CHO Me CHO (D) Me CHO Sol C H OH O SOCl C H Cl O H / Pd BaSO4 quinoline H C H H O Paragraph for Questions Nos. 67 to 69 There are some deposits of nitrates and phosphates in earth s crust. Nitrates are more soluble in water. Nitrates re difficult to reduce under the laboratory conditions but microbes do it easily. Ammonia forms large number of complexes with transition metal ions. Hybridization easily explains the ease of sigma donation capability of NH and PH. Phosphine is a flammable gas and is prepared from white phosphorous. 67. Among the following, the correct statement is (A) Phosphates have no biological significance in humans (B) Between nitrates and phosphates, phosphates are less abundant in earth s crust Between nitrates and phosphates, nitrates are less abundant in earth s crust (D) Oxidation of nitrates is possible in soil. Sol Nitrates being water soluble and as they get reduced by microbes so obviously its abundance will decrease. 68. Among the following, the correct statement is (A) Between NH and PH, NH is a better electron donor because the lone pair of electrons occupies spherical s orbital and is less directional (B) Between NH and PH, PH si a better electron donor because the lone pair of electrons occupies sp orbital and is more directional Between NH and PH, NH is a better electron donor because the lone pair of electrons occupies sp orbital and is more directional (D) Between NH and PH, PH is a better electron donor because the lone pair of electrons occupies spherical s orbital and is less directional. Sol NH is a stronger Lewis base than PH. 69. White phosphours on reaction with NaOH gives ph as one of the products. This is a (A) dimerization reaction (B) disproportionation reaction (D) condensation reaction (D) precipitation reaction Sol + P4+ NaOH + HO NaH P O + P H Here P 4 is oxidized to NaH PO and it is reduced to PH. (B) NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS: 47-B, KALU SARAI, NEW DELHI 6. PH: 4686//4