CRASH COURSE BITSAT-2017 MOCK TEST-1( ) ANSWER KEY

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1 CRAS CURSE BISA-07 MCK ES-(6.0.07) ANSWER KEY PAR-I_(PYSICS) Q. C Q. C Q. A Q. C Q.5 C Q.6 B Q.7 A Q.8 A Q.9 C Q.0 D Q. C Q. D Q. C Q. C Q.5 B Q.6 D Q.7 A Q.8 C Q.9 D Q.0 A Q. C Q. C Q. B Q. A Q.5 C Q.6 A Q.7 C Q.8 D Q.9 C Q.0 B Q. B Q. B Q. A Q. C Q.5 A Q.6 D Q.7 C Q.8 A Q.9 C Q.0 C PAR-II (CEMISRY) Q. C Q. C Q. C Q. D Q.5 B Q.6 D Q.7 D Q.8 D Q.9 B Q.50 B Q.5 C Q.5 A Q.5 C Q.5 B Q.55 B Q.56 A Q.57 C Q.58 B Q.59 A Q.60 D Q.6 A Q.6 D Q.6 C Q.6 A Q.65 A Q.66 B Q.67 C Q.68 B Q.69 B Q.70 D Q.7 B Q.7 B Q.7 B Q.7 B Q.75 A Q.76 C Q.77 C Q.78 A Q.79 A Q.80 D PAR-III (A)ENGLIS PRICIENCY Q.8 B Q.8 A Q.8 C Q.8 C Q.85 C Q.86 C Q.87 A Q.88 C Q.89 D Q.90 D Q.9 B Q.9 C Q.9 B Q.9 C Q.95 A (B) LGICAL REASNNG Q.96 A Q.97 C Q.98 B Q.99 C Q.00 D Q.0 C Q.0 B Q.0 A Q.0 D Q.05 A PAR-IV (MAEMAICS) Q.06 A Q.07 D Q.08 B Q.09 C Q.0 C Q. D Q. C Q. D Q. A Q.5 C Q.6 B Q.7 D Q.8 A Q.9 B Q.0 B Q. B Q. A Q. C Q. C Q.5 A Q.6 C Q.7 B Q.8 C Q.9 D Q.0 A Q. C Q. A Q. A Q. C Q.5 A Q.6 A Q.7 D Q.8 A Q.9 C Q.0 B Q. B Q. D Q. A Q. B Q.5 A Q.6 C Q.7 B Q.8 D Q.9 B Q.50 A Page #

2 IN & SLUIN PYSICS Q. 0 (0 + R) R 0 Q. I R I (R) as R and R in parallel Q. V I R Q. C is correct. irst set the force equal to mass times acceleration Eq ma. hen use the uniform accelerated motion equation x at. Q.5 electric field is perpendicular to equipotential surface, so field lines connect one charge with another in one sense. ne of the charge should be positive and other negative. Q.6 I V I V V I 5A, V 00 I 50 ma Q.7 L 0.07, C.5 0, R, 00 Z L R 5 C E 0 50, I 0 Z E 0A Q.8 P gh Q v, B 0.65 vg ; m' m 0.65 V.7 Page #

3 .8 Q.0 B.0 g.8 W g B W Q. g Gm m /r, e kq q /r Q. Q mc, gradient / Q /mc Q. Net work done is area under loop of PV graph Q. Absorb energy per cycle 55 cal, total absorb (500 g)(80 cal/g) 55 t 60/0 Q.5 Low pressure and high temperature Q.6 Isobaric I and III, adiabatic only II is possible. Q C C Q.9 D is correct. he pressure at the bottom of the column is given by gh. Sitting this equal to maximum pressure we get h. Q.0 t h g (000).7 08 (approx) sec..8 min. Q. Acceleration is slope of velocity time graph. Q. Velocity tangential, acceleration towards centre. Q. du d Page #

4 Q. B Idl ( ĵ) B ( kˆ ) IdlB (î) Q.5 Coulomb : Charge strong, Strong interaction between nucleons. Q.6 Coulomb interaction is exhibited by particles that possess charge only while nuclear interaction is independent of charge. Q.7 h mev V V, Q.8 Emission and absorption spectra provide evidence for the existence of atomic energy levels. Q.9 mg r r mg cos Q.0 Uniform speed means zero tangential acceleration means zero net horizontal force along direction of motion centripetal acceleration mean net force normal to direction of motion is non-zero. Q. a g sin independent of mass a Q. Intensity after polarizer is I 0, it transmits completely by polarizer. Q. K I (I) L L : conserved k k Page #

5 Q. otal K.E. translational K.E. + Rotational K.E. Q.5 ma m( w x) mw x Q.6 V A x v 0 x A V + A upper half of ellipse Q.7 00 gm N Q.8 Pdt mcd k d dt P mc 0 Q cm 5.5 mm v v Q.0 f, f ' f L L Page # 5

6 SLUIN Q. BaC + K Cr BaCr + K + + (A) (B) Yellow CEMISRY C BaC + S BaS + C + (C) white BaC + Cl BaCl + C + Clear sol. (D) Q. hyd. stability (A) (B) 6 (C) 0 (D) hyd. A > D > B > C Minimum C Q.5 ydrolysis followed by condensation polymerization of RSiCl produces -D cross-linked silicones which are hard. Q.6 (A) K + 6 e aromatic (B) (C) Br I 6 e aromatic 6 e aromatic (D) Br e antiaromatic Q.8 As Pt 6 is a powerful oxidizing agent hence. Na + Pt 6 Na + [Pt 6 ] N + Pt 6 N + [Pt 6 ] Xe + Pt 6 Xe + [Pt 6 ] Q.9 Structures of given compound are Me C Et and Et C Me so metamers. Q.5 + [] Q.5 ence, total stereoisomer 8 Page # 6

7 A l( ) Q.5 Na : K : Ba() : Mg() : Al() If we take equal mass of each then for Al(), we need maximum Cl + Cl AlCl V( l) 6.5 V(l) 6 litre Q.5 In Bessemer converter slag of esi is also formed. Blister appearance is due to escape of S, gas from molten copper. Q.55 Ph / Ph CC Me Br Anti addition Ph Me Br Br (B) Br Br Ph Me Ph Me Anti addition Br CCl Br Ph Me Br Br Ph Me Br (C) (B) and (C) are diastereomers Q.56 A(g) B(g) + C(s) t t t (00 - P') p' 0 Since the reaction is of first order.0 00 k log t (00 p') log 8 (00 p') n solving p' 00 After 8 minimum the pressure of the reaction vessel is (00 + p') 700 mm g. Q.57 A e : by carbon reduction. B Al : by Electrolytic reduction. Page # 7

8 Q.58 / + C + C + CC keto acid + C Q.59 C CAg is a salt of weak acid strong base. he solubility of any salt of weak acid strong base is highest in acidic buffer, less in pure water and least in basic buffer. C CAg(s) l Ag + (aq) + C C (aq) C C (aq) + + (aq) l C C(aq) (from buffer) Q.60 Anthracite - is purest form of coal. Q.6 P P P Q.6 K K sp sp [CaC] x.0 [CaC ] y CaC l Ca + + C x + y x CaC l Ca + + C y + x K sp (x + y) x y K sp (x + y) y (x + y) So, K sp Q.6 Among d-block elements max. M.P. of first transition series Cr min. M.P. of second transition series Cd Q.66 Due to deliquescent nature of MgCl.6, ecl.6, ZnCl.6 they get hydrolysed by their own water of crystallization and hence they are made anhydrous by heating in presence of dry Cl gas. CuS.5 Not deliquescent. Q.67 undsdicker reaction Page # 8

9 Q.69 hence, hybridization : d sp Q.70 D is amine Q.7 No. of cis isomers 6 hey are : No. of rans isomers Q.7 C N Example of offmann Bromamide reaction Q.75 he hybrid orbitals used for forming C bonds contain more s character than hybrid orbitals used for forming C bonds. As more the s character in hybrid orbital larger will be bond angle. Q.76 DBC Me I (excess) + MeI + EtI DBC Et Q.77 Cr e Cr Q.78 + C -C Ethanal will give iodoform test but Methyl propanal will fail in iodoform test Ans is A Page # 9

10 (A) LGICAL REASNING SLUINS ENGLIS & REASNING Q.8 At 0 o'clock the two hands are 0min. spaces apart. o be in opposite directions. he minute hand will have to gain (0 0) 0 minute spaces. 60 Now, 5 min. spaces can be gained in 0 min min. min. he two hands will be in opposite directions at 9 min. post 0. Q.8 he year 00 is a leap year. It has odd days. he day on 8th eb, 00 is days before the day on 8th eb 005. ence, this day is Sunday. Q and 000 are leap years. herefore, alternatives (C) and (D) are ruled out. If falls on Monday, then falls on Wednesday. owever, falls on Monday again. Calendars for 990 and 00 are exactly the same. Q.8 00 years contain 5 odd days. Last day of st Century is riday. 00 years contain (5 ) odd days. Last day of nd Century is Wednesday. 00 years contain (5 ) 5 odd days odd day. Last day of rd Century is Monday. 00 years contain 0 odd day. Last day of th Century is Sunday. his cycle is repeated. Last day of a century cannot be uesday or hursday or Saturday. Page # 0

11 Q.85 We shall find the day on st April 00. Ist April, 00 (000 years + Period from..00 to..00) dd days in 600 years 0 dd days in 00 years 0. Jan. eb. March April ( ) 7 days 0 odd days. otal number of odd days 0 odd day on st April, 00 it was Sunday. In April, 00 Wednesday falls on th, th, 8th and 5th. Q.86 No. of cubes Volume of larger cube Volume of a smaller cube 5cmcm9cm cmcmcm 60 Q.87 Number of cubes with one face painted ( ) Q.88 Number of cubes with two faces painted ( ) + ( ) + 0 Q.89 Number of cubes with two faces painted with same colour Q.90 Number of cubes with two faces painted with different colour Page #

12 SLUINS MAEMAICS dy Q.06 x + y (ln y) 0 dx dx dy + x y ( ln y) C ; ln (x ln y) C. If x then y e ln(ln e) C C 0 (A)] Q.07 (x) f (x) f (x) g(x) f (x) g(x) f '(x) f (x) f(x) g'(x) '(x) f (x) g(x) f (x) g'(x) g(x) f '(x) f (x) g(x) () (5) () (6) '(C) ( ) Ans. Q.08 he denial of the given statement "he is not rich or not happy." or not happy (As, ~(p q) (~p) (~q) Aliter: p q p q ~ (p q) ~ p ~ q (~ p) (~ q) ~ (p q) (~ p) (~ q) Q.09 a b a c a (b c) 0 Since a 0, therefore b c a b c a sin x Q.0 l n ln (sec x + tan x) cos x d dx l n (secx tan x) [tan x sec x] sec x sec x tan x sec x sec x dx sin x l n. Ans. cos x Page #

13 Q. Let (a, a a ) be the point of tangency. By using derivatives, the slope of the line through this point is a. By using the definition of slope, the slope of the line through this point is a a 5. Setting these equal. a We must have a a a 5 a a + 8a 8 a a 5 a a + (a ) (a ) 0 a or a, and the slopes would be or, So the product of these slopes is ( ). Ans. Q. y' x + 6x + 9 y" 6x x ence maximum slope y'(x ) Q. f(x) (x 5a) + 5 a. 5 a a. Q. In ABC, a + b + c ab + bc + ca (a b) + (b c) + (c a) 0 a b c ABC is equilateral So, A B C (each). Ans. y Q.5 A ( x ) dx + (x ) dx 0 x ] Q.6! 5C 0 0. Ans. Q.7 P î ĵ kˆ, Q î ĵ 5kˆ P + Q î ĵ kˆ [otal force applied] Page #

14 Distance moved AB B A 5î 7ĵ kˆ Work done d î ĵ kˆ 5î 7ĵ kˆ units Q.8 Equation of AB : A(,,) x y z 0 B(, +, ) (say) C x y0 6 Mid point of AB, C,, lies on x y 0 9, B is,, So, all choices are wrong. Ans. B 9 Q.9 x 0x 9 dx 6 (x 5) dx. 5 5 Put x 5 t dx dt 9 I 6 t dt Put t sin 0 6 cos d 6. Ans. 0 Q.0 m 5. Ans. (, ) B A (, ) Q. Since () + () + 5( ) 0 he line is perpendicular to the normal to the plane. ence line is parallel to plane in (B). Page #

15 Q. D 0] x ( 0) ( x) 0 x + ( + x) 0 x x. Ans. Q. Since, line is tangent to the circle. ence, r r r. Ans. Q. r î ĵ (î ĵ kˆ ) (î ĵ kˆ ) is a plane passing through î ĵ and parallel to î ĵ kˆ and î ĵ kˆ. So, it is perpendicular to ( î ĵ kˆ ) (î ĵ kˆ ) 5î ĵ kˆ Equation of plane can be written as : r (î ĵ) (5î ĵ kˆ) 0 x y z or 5 Sum of intercepts 5 0 Q.5 It is a concept. Q.6 Equation of tangent with slope, is y x + C Now, C (Using condition of tangency) y x 6 x + y It meets the coordinate axes in A and B. So A(8, 0) and B(0, 6). ence, required area of AB (8) (6). Ans. Page # 5

16 Q.7 A G I N V A G! A! A I! 08 A N! A V G! 6 A V! 6 A V I G N Number of words that appear before the word AVIGN is 08 Aliter : A 5! 0 0 G! 96! I! 0 0 N! 0 0 V 08 Ans. Q.8 It is a known fact. Q.9 he given differential equation is not a polynomial equation in terms of derivatives. Ans. Q.0 Writing the data in ascending order, we get 5, 6, 8, 9, 0,,,, 6, we find median is 0. Q. Distance of origin (0, 0) from given three straight line are equal. Q. Standrad property Q. Area of shaded region 9 Area of sector 9 9 y (,) 9 9. Ans. x Q. Use R R R & R R R and expand to get f(x) + sin x Page # 6

17 Q.5 Using power of point x (x + 5) 6 (6 + ) x + 5x 60 0 x 5 5 ()( 60) () x 5 65 ( 5 65) (J K ) J 5 and K 65 K 65 J 5 5. Ans. dx Q.6 I cosx dx cosx 0 / sec x dx tan x0 0 Q.7 csc Lim x0 x (cos x ) tan x Lim x sec x x 0 cos x sin x Lim x sin x x0 x (cos x ) sin x Lim Lim x0 x x 0 (cos x ). Ans. Q.8 n C 0, n C, n C, n C,..., n C are binomial coefficients which are in odd numbers n (because n is even) and middle binomial coefficient is n C n, which is required median. Q.9 Given b a and ab 8. Substituting, we get a and b. he length of the latus rectum is Ans. b a 8. Page # 7

18 Q.0 Let ABC be the equilateral triangle with vertex A(h, k) and let D () be the point on BC. hen h 0 k k 0. Also 0 and h 0 ( ). Q. P(E) P() + P() + P(5) + P(7) P() P() + P() P() 0.5 P (E ) P() + P() 0.5 P(E ) Ans. Q. Since n divides n, n N, R is reflexive. R is not symmetric since for, 6 N, here R 6 is defined but 6 R is not defined. R is transitive since for n, m, r whenever m and n r m n r, i.e. n divides m and m divides r, then n will divide r. Q. cot x + cot y [tan x + tan y]. Ans. 5 5 Q. P(W/W) 5 ; P(L/W) 5 E: Mr. A wins the nd game A: Mr. A wins the first game, P() A B: Mr. B wins the first game P() E B P(E) P(E A) + P(E B) P() P(E/A) + P() P(E/B) Q.5 Let a î ĵ kˆ and b î ĵ kˆ Since cos a b a b () (î ĵ kˆ) (î ĵ kˆ) () () () ( ) ( ) ( 9 ) ( ) 0 Page # 8

19 Q.6 he probability of getting a defective bulb from the box, is 0. ence using binomial distribution, the required probability, is (0.9) 5. Q.7 ( 6y ) dy (sin x e ) dx graph. x y + y cos x + e x + C and since 0, is on the cos 0 + e 0 + C C 8 herefore, y + y cos x + e x cos x y + e x y 0. Ans. Q.8 Standard deviation 8 j j 8 (x 8) 8 j (x j 8) Ans.] Q.9 a ij ( ix) Lim x0 x / j e Lim x0 ln ( ix) j x ln ( ix) j Lim x 0 x j i A / / / / A / / / / / / / / 6 9 / 9 / / / / / A. Ans. Page # 9

20 0 Q.50 he initial radius is, so we are looking for the rate of increase of area when the 0 radius is, which occurs when + t ( + t) t t 0.5 da dr he rate of increase of the area is r (A r ) dt dt t 0. 5 ( 0.5) ( 0.5) ( 0.5) 6. Ans. Page # 0