VISION40 IIT-JEE PRE-MED ACADEMY

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1 VISIN0 VISIN0 IIT-JEE PE-MED ACADEMY JEE(MAIN) 07 TEST PAPE WITH SLUTIN (HELD N SUNDAY 0 nd APIL, 07) PAT A PHYSICS. 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 K.E. = ma cos t CDE-B Hence graph of K.E. v/s time is square cos function KE () 0 T/ T t KE 0 T () Ans. () KE KE 0 T/ T/ T t () 0 T/ T T t KE () 0 T t V = 0 V max V = 0 Equilibrium position Extreme position Time taken to reach the extreme position from equilibrium position is T. Velocity is maximum at equilibrium position and zero at extreme position. V = A cos t K.E. = mv (m is the mass of particle and v is the velocity of particle. The temperature of an open room of volume 0 m increases from 7 C to 7 C due to 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 ().8 0 Ans. () Using ideal gas equation PV = NT (N is number of moles) P 0 V 0 = N i () After heating P 0 V 0 = N f () from equation () and () N f N i = [T i = = 90 K] [T f = = 00 K] P0V0 P0V P0V0 0 difference in number of moles 9000 Hence n f n i is = P0 V putting P 0 = 0 5 P A and V 0 = 0 m Number of molecules n f n i = VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

2 JEE(MAIN)-07. Which of the following statements is false? () A rheostat can be used as a potential divider () 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. Ans. () () n interchanging Cell & Galvanometer.. The following observations were taken for Ans. () determining surface tensiton T of water by capillary method : Diameter of capilary, D =.5 0 m rise of water, h =.5 0 m Using g = 9.80 m/s and the simplified relation T = rhg 0 N/m, the possible error in surface tension is closest to : ().% () 0% () 0.5% ().5% rhg T 0 VISIN0 T r h 0 T r h () G T T = ( ) = (.89) T % T n balancing condition G n balancing condition...()...().5% 5. 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? () m + c () c m () m () c efer NCET Page No. 56 Three frequencies are contained As we see both equation () & () are same. So th statement is false. m + c, c m & c VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

3 VISIN0 6. A diverging lens with magnitude of focal length 5 cm is placed at a distance of 5 cm from a converging lens of magnitude of focal length 0 cm. A beam of parallel light falls on the diverging lens. The final image formed is : () real and at a distance of 0 cm from the divergent lens () real and at a distance of 6 cm from the convergent lens () real and at a distance of 0 cm from convergent lens () virtual and at a distance of 0 cm from convergent lens. As parallel beam incident on diverging lens if forms virtual image at v = 5 cm from the diverging lense which works as a object for the converging lense (f = 0 cm) f = 5 cm 5 cm f = 0 cm So for converging lens u = 0 cm, f = 0 cm Final image m m I For maxima & minima di m m 0 d m or or CDE-B 8. 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 V V 0 0 V = 0 cm from converging lenses. 7. The moment of inertia of a uniform cylinder of length and radius about its perpendicular bisector is I. What is the ratio / such that the moment of inertia is minimum? () () m m I or Also m I m = () ()...()] log min () log min () log min () hc min ev log V log V log V Put in equation () m ev hc min VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

4 JEE(MAIN)-07 e n nv n min hc e n( min ) = nv + n hc e n min nv n hc It is a straight line with ve slope. 9. 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 = T log log. Ans. () At time t N N B A. N B =. N A T log(.) () t = T log. log also let initially there are total N 0 number of nuclei N A + N B = N 0 N N. 0 A Also as we know N A = N 0 e t N0 N0e. e. t t n(.) = t or n. n. t T n n T n. t 0. 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 torque T T. The angle is : () 60 () 90 () 0 () 5 Ans. () So from p E p x axis x y ˆ y ˆ x y kˆ kˆ p ˆi p ˆj Ei ˆ Ej ˆ p E k p E k x 0 Ekˆ p p p p y x tan = = 60. In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be : () 5 () 80 () 5 () 90 Ans. () In common emitter amplifier circuit input and out put voltage are out of phase. When input voltage is increased then i b is increased, i c also increases so voltage drop across c is increased. However increase in voltage across C is in opposite sense.. 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 = b () a = 8 b () a = b () a = b Ans. () C P C V = where C P and C V are molar specific heat capacities As per the question a = b a = b = 8 VISIN0 VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

5 VISIN0. 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 C. T is given by : (Given : room temperature = 0 C, specific heat of copper = 0. cal/gm C () 50 C () 85 C () 800 C () 885 C Ans. () Heat given = Heat taken (00) (0.)(T 75) = (00)(0.)(5) + (70)()(5) 0(T 75) = = 800 T 75 = 80 T = 885 C. A body of mass m = 0 kg is moving in a medium and experiences a frictional force F = kv. Its intial speed is v 0 = 0 ms. If, after 0 s, its energy is mv0, the value of k will be:- 8 () 0 kg m () 0 kg m s () 0 kg m () 0 kg s Ans. () mvf mv0 8 v 0 v f = = 5 m/s dv (0 ) dt kv 5 dv = 00 0 v k 0 0 dt = 00 k (0) 5 0 k = 0 5. 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 to voltmeter of range 0 0 V is: ().55 0 () () () = (5 0 ) (5 + ) CDE-B r = 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 g () cos g () sin mg mgsin x g () cos g () sin Taking torque about pivot = I m mgsin = = g sin 7. Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = /, is given by : () r () r E E E E () r () r 5 VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

6 JEE(MAIN)-07 VISIN0 Ans. () using E = hc Current 0A for E ( E) = hc 0.5 sec time 6 = hc E for E E hc = = hc E = r = 8. 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 () 8 () 9 () 9 Force mg volume density g Stress area A Area Stress = Stress L L g L 9. In a coil of resistance 00, a current is induced Ans. () by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is : 0 Current (amp.) Time 0.5 sec () 50 Wb () 75 Wb () 00 Wb () 5 Wb q = = change in flux q = Idt = Area of current-time graph = =.5 coloumb q = =.5 00 = 50 wb 0. In a Young's double slit experiment, slits are Ans. () 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 : () 9.75 mm () 5.6 mm ().56 mm () 7.8 mm For common maxima n = n n 650 = n 50 n n 5 yd n D y 0.50 y = 7.8 mm VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

7 VISIN0. 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 : () 6.98 s () 8.76 s () 6.65 s () 8.89 s I T = MB I = kg m M = Am By substituting value in the formula T =.665 sec for 0 oscillation, time taken will be Time = 0 T = 6.65 sec Answer option. The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by ( = Earth's radius): () () Ans. () g g GMx g = GM g = r d d () () g g inside the Earth (straight line) outside the Earth where M is Mass of Earth g option () d d d CDE-B. In the above circuit the current in each resistance is : V V V V V V () 0.5 A () 0 A () A () 0.5 A Ans. () 6V 6V V V V V 0V A 0V Taking voltage of point A as = 0 Then voltage at other points can be written as shown in figure Hence voltage across all resistance is zero. Hence current = 0. A particle A of mass m and initial velocity v collides Ans. () with a particle B of mass m which is at rest. The collision is head on, and elastic. The ratio of the de oglie wavelengths A to B after the collision is : () A B () A B () A B () A After collision m V M m V M V rest By conservation of linear momentum m mv = mv + v v = v + v...() by law of collision v v e u u...() u = v v B 7 VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

8 JEE(MAIN)-07 8 By equation () and () v v v ; v h p ; h o ption () 5. 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 : () () PK PK () P K Due to thermal expansion v T v Due to External pressure v P v K Equating both w get T P K () P K P T K 6. 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 : () 9 J () 8 J ().5 J () J F = 6t = ma a = 6t dv 6t dt v dv 0 0 6t dt v = (t ) 0 = m/s From work energy theorem WF K.E mv u 9 0 =.5 J 7. 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 ) () 7. GHz () 5. GHz () 0. GHz (). GHz Ans. () Doppler effect in light (speed of observer is not very small compare to speed of light) f V / C fsource / 0 GHz V / C / = 7. GHz 8. In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be : E r () C r r r CE () r r () CE () r CE r r r CE r r Ans. () It steady state, current through AB = 0 E r B D V AB = V CD C r r V r V r r AB CD Q C = CV AB r CE r r A C VISIN0 VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

9 VISIN0 9. 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. CDE-B 0. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time? () v t The minimum number of capacitors required to achieve this is : () () () v t () () 6 Ans. () To hold KV potential difference () v t minimum four capacitors are required in series v C for one series. () t So for Ceq to be F, 8 parallel combinations are required. 8 Parallel Ans. () Velocity at any time t is given by V 0 g v = u + at KV Minimum no. of capacitors = 8 = v = v 0 + ( g)t v = v 0 gt straight line with negative slope v t 9 VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

10 JEE(MAIN)-07. Let k be an integer such that triangle with 0 vertices (k, k), (5, k) and ( k, ) has area 8 sq. units. Then the orthocentre of this triangle is at the point : () (),, Ans. () We have k k 5 k 8 k () (),, 5k + k 6 = 0 or 5k + k + 66 = 0 (no real solution exist) k = 5 or k = As k is an integer, so k = B (5, ) H ( ) A(, 6) D orthocentre is, E C(, ). If, for a positive integer n, the quadratic equation, x(x + ) + (x + ) (x + ) (x + n ) (x + n) = 0n has two consecutive integral solutions, then n is equal to : () () () 9 () 0 Ans. () PAT B MATHEMATICS We have n r x r (x r) 0n n r (x (r )x (r r)) 0n n solving, we get n x + nx + 0 (n ) ( + ) = n...() and n ( )...() n = (using () in ()) or n =. The function f : x f(x) = x, is :, () neither injective nor surjective. () invertible. () injective but not surjective. () surjective but not injective Ans. () f :,, f(x) = x x x defined as ( x ) xx (x )(x ) f (x) ( x ) ( x ) + x = x =, y, (0,0) x = sign of f(x) From above diagram of f(x), f(x) is surjective but not injective. VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD- x VISIN0

11 VISIN0 CDE-B. The following statement (p q ) [(~p q) q] is : () a fallacy () a tautology () equivalent to ~ p q () equivalent to p ~q Ans. () (p q) [(~p q) q] p q ~ p pq ~ p q (~ pq) q (pq) ((~ pq) q) T F F F T F T T T F T T T T F F T T F T T F T T T T T T x + y = (,) (0,) (,) (0,0) (,0) (,0) y = + x y = x It is tautology 5. If S is the set of distinct values of 'b' for which the following system of linear equations Ans. () x + y + z = x + ay + z = ax + by + z = 0 has no solution, then S is : () a singleton () an empty set () an infinite set () a finite set containing two or more elements D a 0 a = a b and at a = D = D = D =0 but at a = and b = First two equationsare x+y+z= There is nosolution. and third equation is x+y+z = 0 b = {} it is a singleton set 6. The area (in sq. units) of the region {(x, y} : x 0, x + y, x y and y + x } is : () 5 Ans. () () 59 () () 7 x Area = ( x) dx ( x) dx dx = For any three positive real numbers a, b and c, 9(5a + b ) + 5(c ac) = 5b(a + c). Then : () a, b and c are in G.P. () b, c and a are in G.P. () b, c and a are in A.P. () a, b and c are in A.P. (5a) + (b) + (5c) (5a) (5c) (5a) (b) (b) (5c) = 0 [(5a b) + (b 5c) + (5c 5a) ]=0 it is possible when 5a = b = 5c b = 5c, a = c a + b = c b, c, a in A.P. VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

12 JEE(MAIN) A man X has 7 friends, of them are ladies and are men. His wife Y also has 7 friends, of them are ladies and 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 : () 8 () 85 () 68 () 69 Ans. () Ladies Ladies x y Men Men Total number of ways C 0 C C C 0 + C C C C + C C C C + C C 0 C 0 C = The normal to the curve y(x )(x ) = x + 6 at the point where the curve intersects the y-axis passes through the point : (), (), () y =, x 6 (x )(x) (), Point of intersection with y axis (0, ) (x 5x 6)() (x 6)(x 5) y (x 5x 6) y= at point (0, ) Slope of normal is Hence equation of normal is x + y =, satisfy it. 0. A hyperbola passes through the point P, and has foci at (±, 0). Then the tangent to this hyperbola at P also passes through the point : (), () (),, (), x y Equation of hyperbola is a b foci is (±, 0) hence ae =, a e = b = a (e ) a + b =...() Hyperbola passes through,...() a b n solving () and () a = 8 (is rejected) and a = and b = x y x y Equation of tangent is Hence, satisfy it.. Let a, b, c. If f(x) = ax + bx + c is such that a + b + c = and f(x + y) = f(x) + f(y) + xy, x, y, then 0 n f(n) is equal to : () 55 () 0 () 65 () 90 Ans. () f(x) = ax + bx + c f() = a + b + c = Now f(x + y) = f(x) + f(y) + xy put y = f(x + ) = f(x) + f() + x f(x + ) = f(x) + x + Now f() = 7 f() = Now S n = t n...() S n = t n + t n...() n subtracting () from () t n = upto n terms t n = (n 5n) S n = t n = S n (n 5n) n(n )(n ) 5n(n ) 6 S 0 = 0 VISIN0 VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

13 VISIN0 CDE-B. Let a i ˆ ˆj kˆ and b ˆi ˆj. Let c be a vector such that c a =, (a b) c = and AB AP Let APC = the angle between c and a b is equal to : be 0º. Then a c AC AB tan = = = AP AP AC AB () 8 () 5 8 () () 5 a i ˆ ˆj kˆ, ˆi ˆ b j and a B C A Now P a b i ˆ ˆj kˆ a b Now : ( a b ) c = a b c sin 0 ˆn ( a b ) c = c = c c = Now : c a = c + a c a a c 9 a c tan tan tan( ) tan tan tan tan tan tan on solving tan = 9 AB tan( ) AP tan( ). Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower -bed, is :- Ans. () () 0 ().5 () 0 () 5 Total length = r + r + r = 0 0 r r. Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = AB. Area = A = 0r r 0 r r r r r If BPC =, then tanis equal to :- () 9 () 6 7 () () 9 da 0 dr 0 r = 0, r = 5 Ans. () A = 50 5 = 5 VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

14 JEE(MAIN) The integral dx cosx is equal to :- () () () () dx I cosx...() dx I cosx...() Adding () and () I dx sin x I cosec x dx / I cot x / dy 6. If ( sin x) (y )cos x 0 dx then y is equal to :- () Ans. () () () dy ( + sinx) (y )cosx 0 dx d ( sin x)(y ) 0 dx ( + sinx) (y + ) = c x = 0, y = c = y sin x y and y(0) =, () n 7. Let I tan x dx,(n ). I + I 6 = a tan 5 x + bx 5 n + C, where C is a constant of integration, then the ordered pair (a, b) is equal to :- (),0 (), (),0 (), I + I 6 = (tan x tan x) dx = = 5 tan5 x + c a = 5, b = 0 tan x sec x dx 8. Let be a complex number such that + = z where z =. If k, 7 then k is equal to :- () () z () z () Ans. () Here is complex cube root of unity + + = 0 0 = ( ) = z k = z 9. The value of ( C 0 C ) + ( C 0 C ) + ( C 0 C ) + ( C 0 C ) ( C 0 0 C ) is :- 0 () 0 0 () () 0 () 0 9 Ans. () ( C + C...+ C 0 ) ( 0 C + 0 C... 0 C 0 ) = [( C C 0 ) + ( C +... C 0 )] ( 0 ) = [ ] ( 0 ) = ( 0 ) ( 0 ) = 0 0 VISIN0 VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

15 VISIN0 CDE-B cot x cosx lim 50. ( x) equals :- x () () () 6 () 8 Line PQ ; x y = = z 5 Let F( +,, 5 + ) P(,,) lim x cot x( sin x) 8 x F tan x lim 8 x = x = 8 = 6 cos x x 5. If 5(tan x cos x) = cos x + 9, then the value of cosx is :- () Ans. () 7 9 () 5 () () 9 5 t t t = (t ) + 9 {Let cos x = t} 5( t t ) = t(t + 7) 9t + t 5 = 0 9t + 5t t 5 = 0 (t ) (t + 5) = 0 t = as t 5. cos x = = cos x = = If the image of the point P(,, ) in the plane, x + y z + = 0 measured parallel to line, x y z is Q, then PQ is equal to :- 5 () 6 5 () 5 () () Q F lies on the plane ( + ) + ( ) (5 + ) + = = 0 = F (,, 8) PQ = PF =. 5. The distantce of the point (,, 7) from the plane passing through the point (,, ), having normal perpendicular to both the lines x y z and () () Normal vector ˆ ˆ ˆ i j k x y z 7, is :- () () = 5i ˆ 7ˆj kˆ So plane is 5(x ) + 7(y + ) + (z + ) = 0 5x + 7y + z + 5 = 0 Distance 5 5 = VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

16 JEE(MAIN) If for x 0,, 6x x tan is 9x () Ans. () 9x () 9 9x the derivative of x g(x), then g(x) equals :- x x () 9x () x 9x 6x x Let y = tan 9x where x 0, = tan /.(x ) / (x ) = tan (x / ) As x / 0, 8 dy dx = 9x 9 = 9x 9 g(x) = 9x x x/ 55. The radius of a circle, having minimum area, which touches the curve y = x and the lines, y = x is :- () () () () Ans. (Bonus or ) x y y y y y r which is not in options therefore it must be bonus. But according to history of JEE-Mains it seems they had following line of thinking. Given curves are y = x and y = x (0, r) (0, ) VISIN0 x + (y ) = r x y = 0 0 r r There are two circles satisfying the given conditions. The circle shown is of least area. Let radius of circle is 'r' co-ordinates of centre = (0, r) circle touches the line y = x in first quadrant 0 ( r) = r r = ± r r = = ( ) which is given in option. 6 VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

17 VISIN0 56. A box contains 5 green and 0 yellow balls. If 0 balls are randomly drawn, one by one, with replacement, then the variance of the number of green balls drawn is :- 6 () 5 () 5 () 6 () Ans. () We can apply binomial probability distribution Variance = npq = = The eccentricity of an ellipse whose centre is at the origin is. If one of its directices is x=, then the equation of the normal to it at, is :- () x + y = () y x = () x y = () x + y = 7 Eccentricity of ellipse = CDE-B 58. 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, is :- () Ans. () 7 55 () 55 () 6 55 () 5 Let A {0,,,,..., 0 } n(s) = C (where 'S' denotes sample space) Let E be the given event E {(0, ), (0, 8), (, 6), (, 0), (, 8), (6, 0)} n(e) = 6 P(E) = For three events A, B and C, P(Exactly one of A or B occurs) = P(Exactly one of B or C occurs) x = (0, 0) x = Now, a e = a = = b = a ( e ) = a = Equation of ellipse x + y x + y y' (, / ) y' = 0 y' = x y Equation of normal at, y = (x ) y = x x y = = P(Exactly one of C or A occurs) = and P(All the three events occur simultaneously) = 6. Then the probability that at least one of the events occurs, is :- () 6 () 7 () 7 6 P(exactly one of A or B occurs) = P(A) + P(B) P(A B) = P(Exactly one of B or C occurs) = P(B) + P(C) P(B C) = P(Exactly one of C or A occurs) () VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

18 JEE(MAIN)-07 VISIN0 = P(C) + P(A) P(C A) = Adding all, we get Given A = P(A) P(A B) = A = 6 9 P(A) P(A B) = 8 Now, P(A B C) = P(A B C) 6 (Given) = P(A) P(A B) + P(A B C) = = A = A + A = adj (A + A) = If A, then adj (A + A) is equal to :- () () () () VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

19 VISIN0 6. Which of the following compounds will significant amont of meta product during mono-nitration reaction? () () H NH () () CCH NHCCH (i) Nitration is carried out in presence of concentrated HN + concentrated H S. (ii) Aniline acts as base. In presence of H S its protonation takes place and anilinium ion is formed NH H + H S + NH HS (iii) Anilinium ion is strongly deactivating group and meta directing in nature so it give meta nitration product in significant amount. NH + NH HS conc.hs + conc. HN N 6. U is equal to () Isochoric work () Isobaric work () Adiabatic work () Isothermal work From st law : U = q + w For adiabatic process : q = 0 U = w Work involve in adiabatic process is at the expense of change in internal energy of the system. PAT C CHEMISTY CDE-B 6. The increasing order of the reactivity of the following halides for the S N reaction is CH CHCH CH CH CH CH Cl Cl (I) (II) p-h C C 6 H CHCl (III) () (III) < (II) < (I) () (II) < (I) < (III) () (I) < (III) < (II) () (II) < (III) < (I) Ans. () For any S N reaction reactivity is decided by ease of dissociation of alkyl halide X X Higher the stability of (carbocation) higher would be reactivity of S N reaction. Since stability of cation follows order. CH CH CH CH CH CH CH p H C C H CH Hence correct order is II I III 6 6. The radius of the second Bohr orbit for hydrogen atom is : Ans. () (Plank's const. h = Js ; mass of electron = kg ; charge of electron e = C ; permittivity of vaccum = kg m A ) ().65Å ().76Å () 0.59Å () Å adius of n th Bohr orbit in H-atom = 0.5 n Å adius of II Bohr orbit = 0.5 () =. Å VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD- 9

20 JEE(MAIN) pk a of a weak acid (HA) and pk b of a weak base (BH) are. and., respectively. The ph of their salt (AB) solution is () 7. () 6.9 () 7.0 ().0 Ans. () Given 0 pk a (HA) =. pk b (BH) =. As given salt is of weak acid and weak bas ph = 7 + pk a pkb = 7 + (.) (.) = The formation of which of the following polymers involves hydrolysis reaction? () Nylon 6 () Bakelite () Nylon 6, 6 () Terylene Ans. () Formation of Nylon-6 involves hydrolysis of its monomer (caprolactum) in initial state. NH Caprolactum Hydrolysis + HN CH CH CH CH CH C HN ( CH ) 5 C Nylon-6 /Polymerization 67. The most abundant elements by mass in the body of a healthy human adult are : xygen (6.%) ; 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 () 5 kg () 7.5 kg () 7.5 kg () 0 kg Mass in the body of a healthy human adult has :- xygen = 6.%, Carbon =.9%, Hydrogen = 0.0% and Nitrogen =.6% n Total weight of person = 75 kg Mass due to H is = kg 00 H atoms are replaced by H atoms. So mass gain by person =7.5 kg 68. 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 () () Ans. () 68. Ans.() tert-buna () () tert-buna -tbu (fails to decolorise the colour of bromine) C 6 H 5 C H tert-buna tert-buna VISIN0 6 5 (it decolorises bromine solution) (it decolorises bromine solution due to unsaturation) (it decolorises bromine solution due to unsaturation). VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

21 VISIN0 69. In the following reactions, Zn is respectively acting as a/an : (a) Zn + Na Na Zn (b) Zn + C ZnC () base and acid () base and base () acid and acid () acid and base Ans. () Although Zn is an amphoteric oxide but in given reaction. (A) Zn + Na Na Zn acid base salt (B) Zn + C ZnC base acid salt 70. Both lithium and magnesium display several similar properties due to the diagonal relationship ; however, the one which is incorrect is : () Both form basic carbonates () Both form soluble bicarbonates () Both form nitrides () Nitrates of both Li and Mg yield N and on heating Ans. () Mg can form basic carbonate like 5Mg + + 6C + 7H MgC. Mg(H). 5H + HC While Li can form only carbonate (Li C ) not basic carbonate. 7. -Methyl-pent--ene on reaction with H in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is :- Ans. () () Six () Zero () Two () Four -methyl pent--ene 5 H +H Anti markownikov product ( stereo isomers possible due to chiral centre as molecule is nonsymmetric) H H H CH Et (I) CH Et (III) H CH CH H C H H C CH Et (II) CH Et CDE-B H H H (IV) 7. 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 :- Ans. () () a () a () a () In FCC unit cell atoms are in contant along face diagonal So, a a a closest distance () = 7. Two reactions and have identical preexponential factors. Activation energy of exceeds that of by 0 kj mol. If k and k are rate constants for reactions and respectively at 00 K, then ln(k /k ) is equal to :- Ans. () ( = 8. J mol K ) () 8 () () 6 () From arrhenius equation, Ea K A.e T E a / T so, K A.e...() K E a / T A.e...() a VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

22 JEE(MAIN)-07 so, equation ()/() K K (Ea E a ) T e (As pre-exponential factors of both reactions is same) K E a E a 0,000 ln K T The correct sequence of reagents for the following conversion will be :- Ans. () () [Ag(NH ) ] + H, H + /CH H, CH Mg () CH Mg, H + /CH H, [Ag(NH ) ] + H () CH Mg, [Ag(NH ) ] + H, H + /CH H () [Ag(NH ) ] + H, CH Mg, H + /CH H 75. 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 ligh 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. () (a) and (d) () (b) and (d) () (a) and (c) () (b) and (c) Ans. () As per NCET book (fact) 76. Which of the following compounds will behave as a reducing sugar in an aqueous KH solution? () VISIN0 [Ag(NH ) ]H Tollens reagent CH C H + H /CHH (esterification) () C CH () CH Mg CH () HC C CH H Ans. () VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

23 VISIN0 () Ester in presence of Aqueous KH solution give SN AE reaction so following reaction takes place CDE-B 78. Which of the following reactions is an example of a redox reaction? () XeF + F XeF 6 + HCH H CH H C Me Aq. KH SN AE HCH H CH H + Me C () XeF + PF 5 [XeF] + PF 6 () XeF 6 + H XeF + HF () XeF 6 + H Xe F + HF Hemiketal Ans. () ing opening In the reaction ve silver mirror test Tollen's eagent HCH H CH H () In above compound in presence of Aq. KH (SN AE ) reaction takes place & -Hydroxy carbonyl compound is formed which give ve Tollen's test So this compound behave as reducing sugar in an aqueous KH solution. 77. Given C (grahite) + (g) C (g) ; r H = 9.5 kj mol H (g) (g) H(l); r H = 85.8 kj mol C (g) + H (l) CH (g) + (g); r H = kj mol Based on the above thermochemical equations, the value of r H at 98 K for the reaction C (grahite) + H (g) CH (g) will be :- () +7.8 kj mol () +.0 kj mol () 7.8 kj mol ().0 kj mol C (g)+h () CH (g)+ (g); r H = 890. f H ? 0 r H = ( fh ) products ( fh ) e actan ts 890. ( fh ) CH 0 ( 9.5) ( 85.8) ( f H ) CH kj / mol 6 0 X ef F XeF6 Xenon undergoes oxidation while oxygen undergoes reduction. 79. 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 80. The major product obtained in the following reaction is :- Ans. () H C 6 H 5 BuK C 6 H 5 () t C H CH Bu CH CH H () C 6 H 5 CH=CHC 6 H 5 () (+)C 6 H 5 CH( t Bu)CH H 5 () ( )C 6 H 5 CH( t Bu)CH C 6 H 5 VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

24 JEE(MAIN)-07 Elimination reaction is highly favoured if (a) Bulkier base is used (b) Higher temperature is used Hence in given reaction biomolecular ellimination reaction provides major product. H C 6 H 5 C 6 H 5 H H t Bu C H 6 5 C 6 H 5 t + BuH + 8. Sodium salt of an organic acid 'X' produces effervescence with conc. H S. 'X' reacts with the acidified aqueous CaCl solution to give a white precipitate which decolourises acidic solution of KMn. 'X' is :- Ans. () () C 6 H 5 CNa () HCNa () CHCNa () Na C N ne unpaired electron is present in * molecular orbital. C No unpaired electron is present Two unpaired electrons are present in * molecular orbitals. B Two unpaired electrons are present in bonding molecular orbitals. 8. The freezing point of benzene decreases by 0.5 C when 0. g of acetic acid is added to 0 g 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 ) () 6.6% () 80.% () 7.6% () 9.6% Ans. () In benzene CH CH (CH CH) i = i = Here is degree of association T f ik m f VISIN = Which of the following species is not paramagnetic :- () N () C 0.57 = 0.95 % degree of association = 9.5% 8. Which of the following molecules is least resonance stabilized? () () () () B Ans. () () N () Ans. () VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

25 VISIN0 () is nonaromatic and hence least reasonance stabilized CDE-B Number of Cl present in ionization sphere = Mole of ion precipitated with exess AgN 0.0 mole of complex 0.0 It means Cl ions present in ionization sphere complex is [Co(H ) 5 Cl]Cl.H 86. The major product obtained in the following reaction is :- () Benzene is aromatic DIBAL H () furan is aromatic CH H () N pyridine is aromatic () CH CH 85. n treatment of 00 ml of 0. M solution of CoCl. 6H with excess AgN ;. 0 ions are precipitated. The complex is :- () [Co(H ) Cl ]Cl.H () [Co(H ) Cl ].H () H CH CH () [Co(H ) 6 ]Cl Ans. () () [Co(H ) 5 Cl]Cl.H Moles of complex = = Molarityvolume(ml) 000 = 0.0 mole () () CH CH CH CH Moles of ions precipitated with excess of AgN = = 0.0 moles Ans. () DIBAL H is electrophilic reducing agent reduces cynide, esters, lactone, amide, carboxylic acid into corresponding Aldehyde (partial reduction) 5 VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-

26 JEE(MAIN) A water sample has ppm level concentration of following anions F 0; S 00; N 50 the anion/anions that make / makes the water sample unsuitable for drinking is / are : () only N () both S and N () only F () only S N : The maximum limit of nitrate in drinking water is 50 ppm. Excess nitrate in drinking water can cause disease. Such as methemoglobinemia. S : above 500 ppm of S ion in drinking water causes laxative effect otherwise at moderate levels it is harmless F : Above ppm concentration of F drinking water cause brown mottling of teeth. The concentration given in question of S & N in water is suitable for drinking but the concentration of F (i.e 0 ppm) make water unsuitable for drinking purpose : 88. 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 :- () 86 () 8. () 8.6 ().86 Ans. () Given chemical eq n M C + HCl MCl + H + C gm 0.086mol from the balanced chemical eq n. M in 89. Given o Ans. () E.6 V, E 0.7 V Cl / Cl Cr / Cr o o o Cr / Cr Mn / Mn E. V, E.5V. 7 Among the following, the strongest reducing agent is () Cr () Mn + () Cr + () Cl o Mn / Mn E.5V..(i) E E o Cl / Cl o Cr.6 V..(ii). V..(iii) 7 / Cr o Cr / Cr E 0.7..(iv) Since Cr + is having least reducing potential, so Cr is the best educing agent. 90. The group having isoelectronic species is :- Ans. () (), F, Na +, Mg + (), F, Na, Mg + (), F, Na, Mg + (), F, Na +, Mg + VISIN0 ions F Na + Mg + Atomic number = 8 9 No. of e = therefore, F, Na +, Mg + are isoelectronic M 8. gm/mol 6 VISIN0 IIT-JEE PE-MED ACADEMY AMBEPET, 6 N X AD, HYDEABAD-