ANSWERS, HINTS & SOLUTIONS HALF COURSE TEST VII (Paper - 1) Q. No. PHYSICS CHEMISTRY MATHEMATICS. 1. p); (D q, r) p) (D s) 2.

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1 1 AIITS-HCT-VII (Paper-1)-PCM (Sol)-JEE(Advanced)/16 FIITJEE Students From All Programs have bagged in Top 100, 77 in Top 00 and 05 in Top 500 All India Ranks. FIITJEE Performance in JEE (Advanced), 015: 551 FIITJEE Students from All Programs have qualified in JEE (Advanced), 015. FIITJEE ALL INDIA INTEGRATED TEST SERIES JEE(Advanced)-016 ANSWERS, HINTS & SOLUTIONS HALF COURSE TEST VII (Paper - 1) Q. No. PHYSICS CHEMISTRY MATHEMATICS 1. B A C. B A C. A A A. C A B 5. B A A 6. A D C 7. A D A 8. A, C A, B, D A, B, C 9. A, C A, B, C, D A, C 10. B, C, D A, B, C, D A, B, C, D 11. A, B, C, D A, B, C, D A, B 1. B C C 1. A A A 1. C A C 15. D B B 16. A A D 1. (A p); (B q); (C (A t); (B s); (C (A q); (B r); (C s); (D p) p); (D q, r) p) (D s). (A s); (B p); (C (A q, r); (B p); (C (A s) (B p); (C q); (D r) s); (D t) q); (D r) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi , Ph , 65699, Fax 6519

2 AIITS-HCT-VII (Paper-1)-PCM (Sol)-JEE(Advanced)/16 Physics PART I Hints & Solution SECTION A 1. Pressure in the air inside the column of mercury is equal to the weight of mercury over the air divided by the internal cross sectional area of the tube. When the temperature increases, the weight of the upper part of the mercury column does not change. That is why the pressure in the air is also constant. For the isobaric process, the change in volume is proportional to the change in temperature. The same is true for the lengths of the air column. l T l 0 T l = l T 0 = 11 cm. 0 T 0 v w u. Frequency received by car B is f 1 f v w u Now the car B will be treated as a source of frequency f 1 wavelength of reflected sound received by the driver of car A is. v w u v w uv w u ' f1 v w u f m Hence (B) is correct v max F0 Kx F0 Kx a e v dv e dx m m 0 0 v F F K mk K max 0 0 max. Let initial location of center of mass is 0 then find location of cm (90 cm)(5 cm) (10 cm)(5) X cm = = 9 cm (90) (10) 5. As escape velocity is times of orbital velocity. 6. The power developed by the man = power required in the reference frame of the man = Force velocity = m(g + a).(h/t) Hence, (A) is correct. 7. x + y = 100 (1) Heat lost = Heat gained y = x y x = 1.6 6y = 1x 100 () Using (1) & () y = = 86.1 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi , Ph , 65699, Fax 6519

3 AIITS-HCT-VII (Paper-1)-PCM (Sol)-JEE(Advanced)/16 8. For a plane wave intensity (energy crossing per unit area per unit time) is constant at all point For spherical wave p 1 I = r r 9. Acceleration of block is = 10 m/s Displacement s = 1 at = = m Tension in the string is 0 N 10. N = 50 0 = 0 N Limiting friction force = µn = 1 N and applied force in horizontal direction is less than the limiting friction force, therefore the block will not slide. For equilibrium in horizontal direction, friction force must be equal to 5 N. N 0 N 5 N mg = 50 N From the top view, it is clear that = 7 i.e. 17 from the x-axis that is the direction of the friction force. It is opposite to the applied force. Contact force = N f = 95 N f 5 o 11. P = V, for ideal gas PV = nrt (A) relation between V and T V (V ) = nrt V nr = T (B) Relation between P and T PV = nrt 1. 1/ V T P P = nrt P / T (C) For expansion V increases work-done : positive internal energy : increases Hence, heat will have to supplied to the gas. (D) As T V with increase in temperature, volume increases Hence work done is positive. So = 0.8 and effective length of air column = = 1m l = 5, so five half loops will be formed i ˆ ˆj = 5 so second overtone FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi , Ph , 65699, Fax 6519

4 AIITS-HCT-VII (Paper-1)-PCM (Sol)-JEE(Advanced)/16 1. Pex = B d dx = (5 105) ( ) sin (y + 1cm) cos (00) t = (15 N/m) sin (y + 1cm) cos (00t) m..v m m 0 v 1 8 v0 v0 v1 v0 v0 v1 mg 1 1 cos I SECTION C 1. P = = 7 watt.. h h T N > 0 N T > 0 T min = Mg a = g/ = 5 m/s. mg T = m(g/) Mg g mg = m Mg = mg m M = 1. The acceleration of center of mass of the system m1a 1 ma ma a cm = m m m 1 The net force acting on the system m a m a m a = 1 1 F net = (m 1 a 1 + m a m a ) 1 = (1)(1) ()() () N = N. a 1 a a A B C m 1 m m FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi , Ph , 65699, Fax 6519

5 5 AIITS-HCT-VII (Paper-1)-PCM (Sol)-JEE(Advanced)/16. R max u g So, maximum area covered = u R. m g max 5. Velocity of block w.r.t. ground V = 0i ˆ 15ˆj 1G V = 15j ˆ 1P Velocity of block w.r.t. ground after 10 sec. is zero (since block will stop w.r.t. board after 5 sec.) V ' 1P = V ˆ 1P gt j 0 (t = 5 sec) V ' 1G = ˆ 0 i 6. Area under graph = 1 dx t v FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi , Ph , 65699, Fax 6519

6 AIITS-HCT-VII (Paper-1)-PCM (Sol)-JEE(Advanced)/16 6 Chemistry PART II 1.. Molecular weight E = 98/1 = 98 Replaced H B : He C : He Sr +, Ba + are relatively easily formed in order to achieve completely filled valence shell orbitals... K 1 c sp Ag NH Ag NH K Ag Cl 1 K NH sp c F K Ag NH Cl XeOF : Xe O Hence, or F 5. h mv = m 6. CH H C CH C CH CH HBr /HO or HO _ Br H HO Br CH H C CH C CH CH Br HBr CH H C CH C CH CH Br COg O g CO g H Br = G G CO G CO G O Spontaneous (ve sign) But G H T S H 0 98 K = = 57. kj FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi , Ph , 65699, Fax 6519

7 7 AIITS-HCT-VII (Paper-1)-PCM (Sol)-JEE(Advanced)/16 0 H 85. kj exothermic ( ve sign) 8. Allyl carbocation is less stable than benzyl carbocation due to less number of resonating structures. 9. In (A) Cl and O In (B) S and O In (C) only O In (D) P and O 10. (A) Br (B) (C) (D) CHI (E) CH COO (E) CH CO 11. The balanced equation is K Cr O 1HCl KCl CrCl 7H O Cl 7 9 g 16.5 g 71g Mass of K Cr O 7 = g 100 Mass of HCl = g 100 Hence, K Cr O 7 is limiting reagent. Mass of Cl produced =.6 g Also equivalent of K Cr O 7 = M 6 Equivalent of HCl = 7M 16. All the given molecules are cyclic isomers with molecular formula = C 5 H 10 Number of pairs of diastereomers = total number of pairs of stereomerms number of pairs of enantiomers = C 1 = C H CH is th missing cyclic isomer. 1. SECTION C HO HO HO HO x = 1 so x/ = 7 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi , Ph , 65699, Fax 6519

8 AIITS-HCT-VII (Paper-1)-PCM (Sol)-JEE(Advanced)/16 8. O Pure O Cr Cr O O O O O As these two does not participate in resonance.. It will produce most stable free radical.. meq of K = 0 meq = meq of acid n factor n-factor for acid = = number of replaceable proton. 5. [H CO ] in blood = M [NaHCO ] = 10 M Volume of blood = 10 ml Let the volume of NaHCO used = x ml 10 HCO in mixture x 10 NaHCO in mixture salt ph pka log acid 10 x x xx log log x x 10x log 1 so x = 0 ml Hence x/10 = N g H g NH g Ratio of initial and final volumes = / = 10x log 0 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi , Ph , 65699, Fax 6519

9 9 AIITS-HCT-VII (Paper-1)-PCM (Sol)-JEE(Advanced)/16 Mathematics PART III SECTION A C 1. Since O(0, 0) does not lie on 11x + 7y = 9 hence it is the equation of AC, 7 solving OC with AC we get C, 7x+y=0 5 solving OA with AC we get A, A (0, 0) 1 1 hence M = mid point of C & A, x+5y=0 hence equation of OB y = x.. The curve remains unchanged when y & x are interchanged.. Let the circle be x + y 1 + gx + fy + c = 0, if m i, lies on it then m i m + gm + fm + cm + 1 = 0 and m 1 m, m, m are its roots r 1. Coefficient of x r is r r! 5. r r r r r 1 r! r =. r r! r! r!r! M A B 11x+7y=9 x1 1 a 6. 1 a1 a 1 a 1 a x1 8 i.e. 1 a 1 x (multiply by (1 a) on both the sides) x = ABC can speak in 10 C 1 ways other 7 in 7! ways so totally 10 10! C. 7! ways Putting x = in the equation, 0 = a 0 + a 1 +a +a (1) Putting x = in the equation, 0 = a 0 + a 1 +a + a () Putting x = 1 in the equation, n = a 0 + a 1 +a + a () adding (1), () and (), n = (a 0 +a +a )... (a) a 0 + a +a = n-1 (option C) subtracting () from (1), 0 = ( - ) (a 1 a + a a ) Since - 0, a 1 +a + a = a + a 5 +a () FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi , Ph , 65699, Fax 6519

10 AIITS-HCT-VII (Paper-1)-PCM (Sol)-JEE(Advanced)/16 10 Also from () (a), a 1 + a +a +a = n n-1 =. n-1....(5) From () and (5), a 1 +a + a = a + a 5 + a = n-1 = a 0 + a + a Clearly the inscried triangle is equilateral. z z i z0 z 1 0 e, z z i z 0 z 1 0 e z = -1 +i( + ) and z = -1 + i( - ) z z z 0 (i) z 1( +i) 10. b is the H.M. of a and c and their G.M. = ac G.M > H.M ac > b And A.M. of a n and c n > G.M. of a n and c n a n n c n a c n n n n a c ac b n Put n = 100,, 5, 11. A A 1 sin A = sin cos =sin A cos A if sin A cos A ie A 5 A can be written in the following ways using 1 s and s only. Six 1 s in 1 way Four 1 s and one s in 5!! = 5 ways! Two 1 s and two s in!! = 6 ways Three s in 1 way. So f(6) = = We have f(6) = 1, so f{f(6)} = f(1). Now 1 can be expressed in the following ways: Thirteen 1 s in 1 way Eleven 1 s and one s in 1! = 1 ways 11! Nine 1 s and two s in 11! 9!! Seven 1 s and three s in 10! 7!! Five 1 s and four s in Three 1 s and five s in 9! 5!! 8!! 5! = 55 ways = 10 ways = 16 ways = 56 ways One 1 s and six s in 7! 6! = 7 ways. f(1) = x y + (1 + ) = 0 (x + ) (y ) = 0 pass through (, ) x y + (1 ) = 0 (x ) (y ) = 0 pass through (, ). FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi , Ph , 65699, Fax 6519

11 11 AIITS-HCT-VII (Paper-1)-PCM (Sol)-JEE(Advanced)/16 Clearly these represent the foci of ellipse, so ae =. The circle x + y y 5 = 0 x + (y ) = 9 represents auxiliary circle thus a = 9 e = and b = 5. SECTION B 1 (A). a + ib z = z (a 1) + ib 1 a 1 b and a + b = 1 1 a = 0 a = 1 and b = 1 b = 1 a b = 0. (B) We must have either 6a > a. b or 6b > a. b. If b = 0, then 6a > a a = 1,,,. If a = 0, then 6b > b b = 1, suppose a > 0 and b > 0 if 6a > a. b thus 6 a > a. so a > a (not possible) similarly if 6b > a. b than b > b not possible only solutions are,, 8, 76,, 9. z z (C). Re 0, 1, Im 0, 1 means that if z = a + ib than a, b (0, ) a bi Now a ib a b a b a b 0 < a, b < (a ) + b > and a + (b ) > so we want area inside the square and outside the two circles area = 16 + ( ) 1. (D). x + y + z = 5 let a i ˆ j ˆ kˆ b xi ˆ yj ˆ zkˆ a b a b x y z 6x y z 5 6 (A) sincos x = sin(cosx) 0 n cosx (n + 1) 0 cosx 1 x n, n 1 (B) cossin x cos(sinx) > 0 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi , Ph , 65699, Fax 6519

12 AIITS-HCT-VII (Paper-1)-PCM (Sol)-JEE(Advanced)/16 1 sin x x R. (C) tan( sinx) sinx 1 sinx x n 6 x R n 6. (D) ln(tanx) tanx > 0 0 < x < n < x < n. 1. 1/ 1 x log 5 log5 1 5 SECTION C. 7, c i.e. i.e. x c = 6.. The given expression can be written as k = k sin A cosb C sin A sin B sinc = k sin A sinbsin A cosb cos A sinb = k sin A sinbsinc k sin A sinbsinc abc. 5. Line L passes through the centre of second circle. a b 1 0 a(b 1)(c 1) (b 1)(1 a) (c 1)(1 a) = c a a 1 b 1 c FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi , Ph , 65699, Fax 6519