NARAYANA. Co m m o n Pr act ice T e st 3 Date: XII STD BATCHES [CF] (Hint & Solution) PART A : PHYSICS

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1 NARAYANA I I T / N E E T A C A D E M Y Co m m o n Pr act ice T e st Date: XII STD BATCHES [CF] ANSWER PHYSICS CHEMISTRY MATHEMATICS (Hint & Solution) PART A : PHYSICS In non uniform electric field. Intensity is more, where the lines are more denser. Electric field due to a hollow spherical conductor is governed by following euation E = 0, for r < R (i) And E for r R.(ii) 4 0 r i.e., inside the conductor field will be zero and outside the conductor will vary according to E Gauss s law is valid for any closed surface, no matter what its shape or size. Total A B C 0 B and A C ' (assumed) ' 7. r According to Gausss theorem ()

2 Electric flux coming out through a closed surface is. To apply Gauss s theorem it is essential that charge should be placed inside a closed surface. So imagine another similar cylindrical vessel above it as shown in fig (dotted) O. E.ds = [ + (a + c) ]ab - [ + (a )]ab = abc (a + c) Force = electric field = mass acceleration At any point over the spherical Gaussian surface, net electric field is the vector sum of electric fields due to, and According to Gauss s theorem, the total outward normal flux over a closed pipe is eual to / 0 times the total charge enclosed within the surface tan 0 mg 8. Dividing, tan Force = electric field = mass acceleration. (C ). 5 x in dv 0 4 x dx 4 R 0 0 E 4 r in r r ()

3 , E L 0 L Independent of shape Electric flux, E E.dS E EdS cos EdS cos Electric field near the conductor surface is given by and it is perpendicular to surface conceptual Net flux = = 4 0 From the geometry of the figure. OA = OO and OA = OO. Thus, OAO is euilateral triangle. Hence AOO 60º or AOB 0º. The are AOB of the ring subtends an angle 0º at the centre O. Thus, third of the ring is inside the sphere. The charge enclosed by the sphere. From Gauss s law, the flux of the electric field through the surface of the sphere is. 0 ( B) E. A PART B : CHEMISTRY SN reaction proceeds through the formation of transition state. The configuration in the product is inverted. The sign of rotation amy and may not be changed. Rearrangement during the substitution is a characteristic of SN reaction 8. In CH = CH Cl, the C Cl bond has double bond character due to resonance. It is difficult to break this bond. 9. For isomeric halides, n-isomer containing straight chain has highest b.p During NBS reaction free radical as intermediate is obtained. So, reaction will depend upon the stability of free radical ()

4 Conjugative system is more stable Fridel craft Acylation Benzene ( ev.) / Anh. AlCl Air / h CHCl COCl C6H5COCl PART C : MATHEMATICS 6. cos (sinx) is defined for all real x logx {x} 0 Case I, 0 < x < {x} ; x (0, ) Case II, x > and {x} > but 0 < {x} > not possible 6. Least value of x + x + = ¾ (at x = - ½_ least value of Also x x x x x x sin x x sin sin x x sin Range of f, Thus 6. f x x x 4 y x x 4 x x + 4 = yx + xy + 4y or (y )x + x(y + ) + 4u 4 0 Since x is real (y + ) 4(y ) (4y 4) 0,7 7 y 7 7 y 0 y 64. Since f(x) is an odd degree polynomial, f(x) is not. Since f (x) = x 6x + 6 > 0 for all x, f is one-one (4)

5 65. Let f(x) = x + sin x, where is any real number then f(x) is continuous. f, f f x 0 for some x x + sin x = for some x. f is onto. Again f (x) = + cos x > 0 for all x f is one-one 66. Since f ' x x 0 for all x f is one-one but f(x) cannot become for any x R f is not onto {x} < 68. Since lies between 9 and 0, [ ] = 9, [- ] = - 0 f(x) = cos 9x + cos 0x 69. If y x 4 6 x It is clear that domain of y = [4, 6] Now dy dx x 4 6 x 0 x=5 Checking the function at 4, 5, 6 we easily get the minimum and maximum value of y as and. Whence the range is, which is an interval of length. 70. Factual 7. Factual 7. Since, f(x) = f( x), hence many-one also range f(x) is +ve real numbers which is subset of codomain, thus into. Hence, many-one 7. We have, f x x 5 It can be easily checked that f x is one-one and onto. f x k k 5 k x k 5 x 5, x R Put cos x = t, - t f x 74. f f t y t t, t t 0 t dy 4t 0 dt t t 4 (5)

6 f and f f 0 4 Hence is the correct answer. f x x x x AM GM x x x f x, f x x x B 0, sin x x 4 cos cos x. x sin x The period of cos x is and that of cos is 4 sin x period of is 4. x sin sin x sin 5 x sin 4 x period of, and sin x / sin x / 5 sin x / 4 For n =, we have Cannot be 4. Since f(x) is odd periodic function with period = f(-x) = - f(x) and f(x + ) = f(x) f() = f(0) and f(-) = f(- + ) = f (0) Now f(0) = f(-) = - f() = - f(0) f(0) = 0 f(0) = 0 f(4) = f() = f(0) = 0. Thus f(4) = 0. Solve the euations f(x) = k f(x) Expand and it is f(x) Put n = 7 nx x n 7 (6)