Page # 1

CRASH COURSE BITSAT-2017MOCK TEST-1(16.04.2017)

ANSWER KEY

PART-I_(PHYSICS)Q.1 C Q.2 C Q.3 A Q.4 C Q.5 CQ.6 B Q.7 A Q.8 A Q.9 C Q.10 DQ.11 C Q.12 D Q.13 C Q.14 C Q.15 BQ.16 D Q.17 A Q.18 C Q.19 D Q.20 AQ.21 C Q.22 C Q.23 B Q.24 A Q.25 CQ.26 A Q.27 C Q.28 D Q.29 C Q.30 BQ.31 B Q.32 B Q.33 A Q.34 C Q.35 AQ.36 D Q.37 C Q.38 A Q.39 C Q.40 C

PART-II (CHEMISTRY)Q.41 C Q.42 C Q.43 C Q.44 D Q.45 B

Q.46 D Q.47 D Q.48 D Q.49 B Q.50 B

Q.51 C Q.52 A Q.53 C Q.54 B Q.55 B

Q.56 A Q.57 C Q.58 B Q.59 A Q.60 D

Q.61 A Q.62 D Q.63 C Q.64 A Q.65 A

Q.66 B Q.67 C Q.68 B Q.69 B Q.70 D

Q.71 B Q.72 B Q.73 B Q.74 B Q.75 A

Q.76 C Q.77 C Q.78 A Q.79 A Q.80 D

PART-III (A)ENGLISH PROFICIENCYQ.81 B Q.82 A Q.83 C Q.84 C Q.85 CQ.86 C Q.87 A Q.88 C Q.89 D Q.90 DQ.91 B Q.92 C Q.93 B Q.94 C Q.95 A

(B) LOGICAL REASONNGQ.96 A Q.97 C Q.98 B Q.99 C Q.100 DQ.101 C Q.102 B Q.103 A Q.104 D Q.105 A

PART-IV (MATHEMATICS)Q.106 A Q.107 D Q.108 B Q.109 C Q.110 CQ.111 D Q.112 C Q.113 D Q.114 A Q.115 CQ.116 B Q.117 D Q.118 A Q.119 B Q.120 BQ.121 B Q.122 A Q.123 C Q.124 C Q.125 AQ.126 C Q.127 B Q.128 C Q.129 D Q.130 AQ.131 C Q.132 A Q.133 A Q.134 C Q.135 AQ.136 A Q.137 D Q.138 A Q.139 C Q.140 BQ.141 B Q.142 D Q.143 A Q.144 B Q.145 AQ.146 C Q.147 B Q.148 D Q.149 B Q.150 A

Page # 2

HINT & SOLUTION

PHYSICS

Q.1 20 = 1 (10 + R)

R = 10

Q.2 I2R = I3(2R) as R and 2R in parallel

Q.3 V1 = I1 R

Q.4 C is correct. First set the force equal to mass times acceleration Eq = ma. Then use the

uniform accelerated motion equation x =2

1 at2.

Q.5 electric field is perpendicular to equipotential surface, so field lines connect one charge with

another in one sense. One of the charge should be positive and other negative.

Q.6 I1V1 = I2V2

I1 = 5A,1

2

V

V = 100

I2 = 50 mA

Q.7 L = 0.07, C = 2.5 × 10–3, R = 4, = 100

Z =2

2

RC

1L

= 5

E0 = 50, I

0 =

Z

E = 10A

Q.8 P = gh

Q.97.2

775 = v, B = 0.65 vg ; m' = m – 0.65 VV

Page # 3

Q.10 B = 1.0 ×3

4 g

2

8.33

W = 0.0840 ×3

4

3

2

8.3

g

F = B – W

Q.11 Fg = Gm1m2/r2, Fe = kq1q2/r2

Q.12 Q = mc T, gradient = T / Q = 1/mc

Q.13 Net work done is area under loop of PV graph

Q.14 Absorb energy per cycle = 55 cal, total absorb = (500 g)(80 cal/g) = 55 × t × 60/10

Q.15 Low pressure and high temperature

Q.16 Isobaric I and III, adiabatic only II is possible.

Q.17 T = 90°F = 50°C

T =

=5010

015.0

= 3 × 10–5°C–1

Q.19 D is correct. The pressure at the bottom of the column is given by gh. Sitting this equal to

maximum pressure we get 12 × 105 – 4000 × 10 × h.

Q.20 t =g

h2 =

72.3

)22000(2 = 108 (approx) sec.

= 1.8 min.

Q.21 Acceleration is slope of velocity time graph.

Q.22 Velocity tangential, acceleration towards centre.

Q.23 =d

dU

Page # 4

Q.24B

Idl )j( × B )k( = IdlB )i(

Q.25 Coulomb : Charge strong, Strong interaction between nucleons.

Q.26 Coulomb interaction is exhibited by particles that possess charge only while nuclear interaction is

independent of charge.

Q.27 =meV2

hV 2V,

2

Q.28 Emission and absorption spectra provide evidence for the existence of atomic energy levels.

Q.29

mg

FrFr = mg cos

Q.30 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.31 a = g sin independent of massa

Q.32 Intensity after polarizer 1 is2

I0 , it transmits completely by polarizer 2.

Q.33 K =2

1I2 =

2

1(I) =

2

1L

L : conserved

1

2

k

k =

1

2

=5.2

6= 2.4

Page # 5

Q.34 Total K.E. = translational K.E. + Rotational K.E.

Q.35 F = ma = m(–w2x) = –mw2x

Q.36 V = 22 xA v 0

2

2

A

x + 22

2

A

V

= 1 upper half of ellipse

Q.37 100 gm 1 N

Q.38mcP

dtdT

kmcdTPdt

Q.39 = 31060

330

= 5.5 × 10–3 cm = 5.5 mm

Q.40 f =L4

v, f ' =

L2

v = 2f

Page # 6

SOLUTION

CHEMISTRY

Q.42 BaCO3 + K2CrO4 BaCrO4+ 2K+ + 23CO

(A) (B) Yellow

BaCO3 + H2SO4 BaSO4 + CO2+ H2O (C) white

BaCO3 + 2HCl BaCl2 + CO2 + H2O Clear sol.

(D)

Q.43 Hhyd. stability

1

(A) 3 H (B) 6 H (C) 10 H (D) 4 HHhyd. A > D > B > C Minimum HOH = C

Q.45 Hydrolysis followed by condensation polymerization of RSiCl3 produces 3-D cross-linked siliconeswhich are hard.

Q.46 (A)

H H

K H

+ H2 6 e¯ aromatic

(B)

O OH

H–Br

6 e¯ aromatic

(C)HI

6 e¯ aromatic (D)

O

H Br

O–H

4 e¯ antiaromatic

Q.48 As PtF6 is a powerful oxidizing agent hence.Na + PtF6 Na+[PtF6]–

NO + PtF6 NO+[PtF6]–

Xe + PtF6 Xe+ [PtF6]–

Q.49 Structures of given compound are Me–O–CH2–Et and Et–O–CH2Me so metamers.

Q.51 O3 O2 + [O]

Q.52 Hence, total stereoisomer = 23 = 8

Page # 7

Q.53 NaOH : KOH : Ba(OH)2 : Mg(OH)2 : Al(OH)3

40 56 171 58 78

If we take equal mass of each

then for Al(OH)3 , we need maximum HCl

A l ( O H ) 3 + 3HCl AlCl3 + 3H2O

2 6 = )(V5.36

1065.3l

V(l) = 6 litre

Q.54 In Bessemer converter slag of FeSiO3 is also formed. Blister appearance is due to escape of SO2, gas from molten copper.

Q.55 PhOH

/H

Ph

C=C

H

H

MeadditionAnti

2BrH HBr Br

Br BrH H

Me Me

Ph Ph

(B)

Ph

H H

Me

4

2

CCl

Br

additionAnti

H HBr BrBr BrH H

Me Me

Ph Ph

(C)

(B) and (C) are diastereomers

Q.56 A(g) 2B(g) + C(s)t = 0 400 0 0t = t (400 - P') 2p' 0

Since the reaction is of first order

)'p400(

400log

t

303.2k

)'p400(

400log

48

303.2

24

6932.0

On solving p'= 300After 48 minimum the pressure of the reaction vessel is (400 + p') = 700 mm Hg.

Q.57 A Fe : by carbon reduction.B Al : by Electrolytic reduction.

Page # 8

Q.58 OH/O 23

COOH + CH COOH3

O + CO + H O2 2

O + CO2

– keto acid

Q.59 CH3COOAg is a salt of weak acid strong base. The solubility of any salt of weak acid strong base ishighest in acidic buffer, less in pure water and least in basic buffer.

CH3COOAg(s) l Ag+(aq) + CH3COO¯ (aq)CH3COO¯ (aq) + H+(aq) l CH3COOH(aq)

(from buffer)

Q.60 Anthracite - is purest form of coal.

Q.61

O

P1 P2

O

O

P3

OHOH

Q.62y

x087.2

]OCaC[K

]CaCO[K

42sp

3sp

2

1 2.0 = y

x

CaCO3 l Ca+2 + CO32–

x + y xCaC2O4 l Ca+2 + C2O4

2–

y + x y

1spK = (x + y) x

2spK = (x + y) y

(x + y) = 9.0 × 10–5

So, 1spK = 5.4 × 10–9.

Q.63 Among d-block elementsmax. M.P. of first transition series = Crmin. M.P. of second transition series = Cd

Q.66 Due to deliquescent nature of MgCl2.6H2O , FeCl3.6H2O, ZnCl2.6H2O they get hydrolysed by their

own water of crystallization and hence they are made anhydrous by heating in presence of dry HCl gas.

CuSO4.5H2O Not deliquescent.

Q.67 Hundsdicker reaction

Page # 9

Q.69

hence, hybridization : d2sp3

Q.70 D is 2° amine

Q.72

No. of cis isomers = 6They are :

No. of Trans isomers = 3

Q.73CH –NH2 2

Example of Hoffmann Bromamide reaction

Q.75 The hybrid orbitals used for forming C–H bonds contain more s–character than hybridorbitals used for forming C–F bonds. As more the s–character in hybrid orbital largerwill be bond angle.

Q.76

OMe

OEt

HI (excess)DBC

DBC

OH

OH

+ MeI + EtI

Q.77 Cr2O72– + 14 H+ + 6 e–2 Cr3+ + 7 H2O

Q.78 OH3 +

CH3-CHO

Ethanal will give iodoform test but 2–Methyl propanal will fail in iodoform test Ans is A

Page # 10

SOLUTIONS

ENGLISH & REASONING

(A) LOGICAL REASONING

Q.81 At 10 o'clock the two hands are 10min. spaces apart. To be in opposite directions. The

minute hand will have to gain (30 – 10) = 20 minute spaces.

Now, 5 min. spaces can be gained in55

60 × 20 min.

=11

240min. = 21

11

9 min.

The two hands will be in opposite directions at 2111

9 min. post 10.

Q.82 The year 2004 is a leap year. It has 2 odd days.

The day on 8th Feb, 2004 is 2 days before the day on 8th Feb 2005.

Hence, this day is Sunday.

Q.83 1996 and 2000 are leap years.

Therefore, alternatives (C) and (D) are ruled out. If 1 - 1 - 1990 falls on Monday, then

1 - 1 - 1997 falls on Wednesday.

However, 1 - 1 - 2001 falls on Monday again.

Calendars for 1990 and 2001 are exactly the same.

Q.84 100 years contain 5 odd days.

Last day of 1st Century is Friday.

200 years contain (5 × 2) = 3 odd days.

Last day of 2nd Century is Wednesday.

300 years contain (5 × 3) = 15 odd days 1 odd day.

Last day of 3rd Century is Monday.

400 years contain 0 odd day.

Last day of 4th Century is Sunday.

This cycle is repeated.

Last day of a century cannot be Tuesday or Thursday or Saturday.

Page # 11

Q.85 We shall find the day on 1st April 2001.

Ist April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)

Odd days in 1600 years = 0

Odd days in 400 years = 0.

Jan. Feb. March April

(3 + 0 + 3 + 1 ) = 7 days 0 odd days.

Total number of odd days = 0 odd day

on 1st April, 2001 it was Sunday.

In April, 2001 Wednesday falls on 4th, 11th, 18th and 25th.

Q.86 No. of cubes = cubesmalleraofVolume

cubeerarglofVolume

=cm3cm3cm3

cm9cm12cm15

= 60

Q.87 Number of cubes with one face painted

= (3 + 2 + 3 + 2 + 6 + 6) = 22

Q.88 Number of cubes with two faces painted

= (1 + 1 + 1 + 1) + (6 + 6 + 4 + 4) = 4 + 20 = 24

Q.89 Number of cubes with two faces painted with same colour

= 3 + 3 + 5 + 1 = 12

Q.90 Number of cubes with two faces painted with different colour

= 3 + 1 + 2 + 2 + 3 + 1 = 12

Page # 12

SOLUTIONS

MATHEMATICS

Q.106 xdx

dy + y (ln y) = 0 x

dx + )yn(y

dy

l = C ;

ln (x ln y) = C. If x = 1 then y = e

ln(ln e) = C C = 0 (A)]

Q.107 F(x) =)x(g)x(f

)x(f

F'(x) =

2)x(g)x(f

)x('g)x(f)x(f)x('f)x(g)x(f

= 2)x(g)x(f

)x('f)x(g)x('g)x(f

F'(C) = 2)24(

)6()2()5()4(

= 5 + 3 = 8. Ans.

Q.108 The denial of the given statement "he is not rich or not happy." or not happy

(As, ~(p q) = (~p) (~q)

Aliter:

TTTF

TFTF

TTFF

TTTF

FFFT

FTFT

FFTT

)q(~)p(~q~p~)qp(~qpqp

)qp(~ = )q(~)p(~

Q.109 ca2ba 0)c2b(a

Since 0a

, therefore a||c2b

ac2b

Q.110

x2cos

x2sin1nl = ln (sec 2x + tan 2x)

)x2tanx2(secndx

dl =

x2tanx2sec

x2sec]x2secx2[tan2

= 2sec 2x

2 dxx2sec =

x2cos

x2sin1nl . Ans.

Page # 13

Q.111 Let (a, 4a – a2) be the point of tangency. By using derivatives, the slope of the line through

this point is 4 – 2a. By using the definition of slope, the slope of the line through this point is

2a

5aa4 2

. Setting these equal.

We must have 4 – 2a =2a

5aa4 2

–2a2 + 8a – 8 = 4a – a2 – 5 a2 – 4a + 3 = (a – 1) (a –3) = 0 a = 1 or a = 3, and the slopes would be 4 – 2·1 = 2 or 4 – 2·3 = –2,So the product of these slopes is 2(– 2) = – 4. Ans.

Q.112 y' = –3x2 + 6x + 9

y" = –6x + 6 = 0 x = 1

Hence maximum slope = y'(x = 1) = 12

Q.113 f(x) = (x – 5a)2 + 5 – a. 5 – a = 1 a = 4.

Q.114 In ABC, a2 + b2 + c2 = ab + bc + ca

(a – b)2 + (b – c)2 + (c – a)2 = 0

a = b = c

ABC is equilateral

So, A =B =C =3

(each). Ans.

Q.115 A = 1

0

2 dx)1x( + 2

1

2 dx)1x(

y

xO 21

]

Q.116 4! · 5C3 · 1 = 24 × 10 = 240. Ans.

Q.117 P = kj2i3 , Q = k5j3i

P + Q = k4ji4 [Total force applied]

Page # 14

Distance moved = k3j7i5ABAB

Work done = k3j7i5·k4ji4d·F = 20 – 7 + 12 = 25 units

Q.118 Equation of AB :

1

1x =

2

3y

=0

4z = (say) C x–2y=0

A(–1,3,4)

B B( – 1, – 2 + 3, 4)

Mid point of AB, C

4,2

62,

2

2 lies on x – 2y = 0

=5

14, B is

4,5

13,

5

9

So, all choices are wrong. Ans.

Q.119 9

5

2 dx9x10x = 9

5

2 dx)5x(16 .

Put x – 5 = t dx = dt

I = dtt164

0

2 Put t = 4 sin

= 16

0

2 dcos = 16 ×4

= 4. Ans.

Q.120 m =4

4

= – 1

O

(– 1, 3)

(3, – 1)

A

B

= 135°. Ans.

Q.121 Since 3(2) + 4(1) + 5(–2) = 0 The line is perpendicular to the normal to the plane. Hence line is parallel to plane in (B).

Page # 15

Q.122 D = 0]

x (2 – 0) – 2 (– 2 – x) = 02x + 2 (2 + x) = 0

4x = – 4x = – 1. Ans.

Q.123 Since, line is tangent to the circle.

Hence,2

r = r r = 2. Ans.

Q.124 )k3j2i()kji(jir is a plane passing through ji and parallel to kji

and k3j2i . So, it is perpendicular to

)k3j2i()kji( = k3j2i5

Equation of plane can be written as :

0)k3j2i5(·)ji(r

or 1

33

z

23

y

53x

Sum of intercepts =3

3

2

3

5

=

10

31

Q.125 It is a concept.

Q.126 Equation of tangent with slope =4

3, is y =

4

3x + C

Now, C = 184

332

2

= 1818 = 6 (Using condition of tangency)

y = 6x4

3

3x + 4y = 24

It meets the coordinate axes in A and B. So A(8, 0) and B(0, 6).

Hence, required area of AOB = )6()8(2

1 = 24. Ans.

Page # 16

Q.127 A G H I N V

A G __ __ __ __ = 4 ! = 24

A H __ __ __ __ = 4 ! = 24

A I __ __ __ __ = 4 ! = 24

A N __ __ __ __ = 4 ! = 24108

A V G __ __ __ = 3 ! = 6

A V H __ __ __ = 3 ! = 6

A V I G H N

Number of words that appear before the word AVIGHN is = 108 Ans.

Aliter :

A 5! × 0 = 0

G 4! × 4 = 96

H 3! × 2 = 12

I 2! × 0 = 0

N 1! × 0 = 0

V

= 108

Q.128 It is a known fact.

Q.129 The given differential equation is not a polynomial equation in terms of derivatives. Ans.

Q.130 Writing the data in ascending order, we get 5, 6, 8, 9, 10, 11, 12, 14, 16, we find median is

10.

Q.131 Distance of origin (0, 0) from given three straight line are equal.

Q.132 Standrad property

Q.133 Area of shaded region = 9 – Area of sector

= 9 – 9··4

1

(3,3)

y

xO=

4

99 . Ans.

Q.134 Use R1 R1 R2 & R2 R2 R3 and expand to get f(x) = 2 + 4 sin 2x

Page # 17

Q.135 Using power of point

x (x + 5) = 6 (6 + 4)

x2 + 5x – 60 = 0

x =)1(2

)60()1(455 2

x =2

2655 = )2655(

2

1 )KJ(

2

1

J = – 5 and K = 265

J

K =

5

265

= –53. Ans.

Q.136 I =

4

4

x2cos1

dx = 2

4

0x2cos1

dx =

4

0

2 dxxsec = 4/0xtan = 1

Q.137xsecx

xtan)1x(cosxcscLim

2

0x

=xsinx

1xcosLim

0x

=)1x(cosx

xsinLim

0x

=

x

xsinLim

0x· )1x(cos

1Lim

0x

=2

1. Ans.

Q.138 0n2 C , 1

n2 C , 2n2 C , 3

n2 C , ......., nn2 C are binomial coefficients which are in odd numbers

(because n is even) and middle binomial coefficient is2n

n2 C , which is required median.

Q.139 Givenb

a = 3 and ab = 48.

Substituting, we get a = 12 and b = 4. The length of the latus rectum is =a

b2 2

=12

4·2 2

=3

8.

Ans.

Page # 18

Q.140 Let ABC be the equilateral triangle with vertex A(h, k) and let D () be the point on BC.

Then3

h2 = 0 =

3

k2 . Also – 2 = 0 and

0h

0k × (–1) = –1.

Q.141 P(E) = P(2) + P(3) + P(5) + P(7)

= 0.23 + 0.12 + 0.20 + 0.07 = 0.62

P(F) = P(1) + P(2) P(3) = 0.5

P (E F) = P(2) + P(3) = 0.35

P(E F) = 0.62 + 0.5 – 0.35 = 0.77. Ans.

Q.142 Since n divides n, n N, R is reflexive. R is not symmetric since for 3, 6 N, here 3 R 6

is defined but 6 R 3 is not defined. R is transitive since for n, m, r whenevern

m and

m

r n

r, i.e. n divides m and m divides r, then n will divide r..

Q.143 cot–1x + cot–1y = – [tan–1x + tan–1y] = –5

4 =

5

. Ans.

Q.144 P(W/W) =5

4 ; P(L/W) =

5

1

E: Mr. A wins the 2nd game

A: Mr. A wins the first game, P(1) =4

3

B: Mr. B wins the first game P(2) =4

1 EB

A

P(E) = P(E A) + P(E B)

= P(1) · P(E/A) + P(2) · P(E/B)

=20

13

5

1·

4

1

5

4·

4

3

Q.145 Let a

= kj3i2 and b

= kji2

Since cos =|b|a|

b·a

=

222222 )1()1()2()1()3()2(

)kji2(·)kj3i2(

=

)114()194(

134

= 0

Page # 19

=2

Q.146 The probability of getting a defective bulb from the box, is10

1. Hence using binomial

distribution, the required probability, is (0.9)5.

Q.147 dy)2y6( = dx)ex(sin x 3y2 + 2y = – cos x + ex + C and since

3

4,0 is on the

graph.

3

42

3

43

2

= – cos 0 + e0 + C C = 8

Therefore, 3y2 + 2y = – cos x + ex + 8 8 – cos x – 2y + ex – 3y2 = 0. Ans.

Q.148 Standard deviation =

218

1jj

18

1j

2j

18

)8x(

18

)8x(

=2

18

9

18

45

=

4

1

18

45 =

36

81 =

6

9 =

2

3. Ans.]

Q.149 aij = x

1)ix1(Lim

j/1

0x

=x

1eLim

)ix1(nj1

0x

l

=x

)ix1(nj1

Lim0x

l =

j

i

A =

12/333/2123/12/11

A2 =

12/333/2123/12/11

12/333/2123/12/11

=

32/9923612/33

= 3

12/333/2123/12/11

= 3A. Ans.

Page # 20

Q.150 The initial radius is01

0·41

= 1, so we are looking for the rate of increase of area when the

radius is 2, which occurs when

1 + 4t = 2(1 + t) 2t = 1 t = 0.5

The rate of increase of the area isdt

dA =

5.0tdt

drr2

(A = r2)

= 2· 2 · 2)5.01(

)5.0·41()5.01(4

=3

4·4 =

3

16. Ans.