dc solutions

8/8/2019 DC Solutions

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Page 1Foundation Test 2B

IIT-JEE

Test Prep

IIT-JEE : Foundation Test 2BAnswers and Explanations

1 b 11 b 1 a 11 c 1 a 11 b

2 c 12 c 2 c 12 b 2 d 12 c

3 c 13 a 3 c 13 d 3 b 13 a

4 a 14 c 4 b 14 c 4 c 14 a

5 d 15 d 5 a 15 d 5 b 15 c

6 c 16 c 6 c 16 b 6 a 16 b

7 d 17 d 7 c 17 d 7 d 17 d

8 b 18 b 8 b 18 c 8 b 18 c

9 a 19 c 9 b 19 a 9 a 19 d

10 a 20 b 10 d 20 b 10 b 20 b

MathematicsChemistry Physics

Batch : 1012

PartI : Chemistry

1. b Highest energy electron lies in 3d.

2. c Ionization energy of He+

=2

2

1 1I.E of Hydrogen z

1

= 2.17 10 18 4 = 8.68 10 18 J

3. c n 4

n 2

n 2 3

4. a

2 22

k

mV1 1 pE mV

2 2 m 2m

Wavelength,h

p or

hp

2

k 2

hE

2m

Kinetic energy of electron,2

k 2e e

hE

2m

Kinetic energy of proton,2

k 2p p

hE

2m

Since, k KE E ;2 2

pe2 2

p ee e p p

mh h

m2m 2m

A s mp

> me

or e p

i.e. the electron has greater wavelength than proton.

5. d

6. c

7. d

8. b

9. a The photoelectric effect gives the following equation

21h mv2

10. a

11. b Both have same number of electrons.

12. c

13. a

14. c

15. d

16. c After loosing one electron oxygen gains half filled

electronic configuration which is stable due to

symmetry.

17. d The increasing order of first ionization energy

= Bi < Te < Se < Cl < F

Therefore intermediate value in the seriescontains Se.

18. b

19. c The correct order of ionic radius would be

2 2Se I Br O F

20. b


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IIT-JEE

Test Prep

PartII : Mathematics

1. a Tn = a + (n 1) d

92 = 2 + (n 1) 3 90 = (n 1) 3 n = 31

16th term is middle term T16 = 2 + (16 1) 3 = 47

2. c Sp = qp

(2a (p 1)d) q2

p(2a (p 1)d) 2q (i)

Sq = pq

(2a (q 1)d) p2

q (2a + (q 1) d) = 2p (i i)On substacting equation (i) and (ii)

2a (p q) + d (p2 p q2 + q) = 2 (p q)

2a (p q) + d (p q) (p + q 1) = 2 (p q) 2a + d (p + q 1) = 2

Sp + q =p q

(2a (p q 1)d)2

p q

22

= p + q

3. c 21

n n5

n(n 1) 1 n(n 1)(2n 1)

2 5 6

2n + 1 = 15 n = 7.

4. b Let terms are2

2

a a, ,a,ar,arrr

(a = 4)

product of terms = a5 = 45

5. a 1 x x4 22log (2 1) 1 log (5.2 1)

1 x x

2 2 21

2. log (2 1) log 2 log (5.2 1)2

1 x x2 2log (2 1) log 2(5.2 1)

1 x x2 1 2(5.2 1)

x

x

21 2(5.2 1)

2

x 2 x10(2 ) 2 2 0

x x(5.2 2)(2.2 1) 0

x5.2 2 0

x2

25

x( 2.2 1 0)

22

x log5

6. c 2 + 5 + 8 + upto 2n terms = 57 + 59 + 61 + upto

n terms

2n n

2 2 (2n 1) 3 2 57 (n 1) 22 2

6n + 1 = n + 56 5n = 55 n = 111

7. c 2 2 2 2(a c) (c b) (b d) (d a)

2 22(b c ac bc bd ad) (i)

b c d[ a,b,c,dare in G.P.

a b c

2 2b ac,c bd and ad = bc]

= 0

8. b The given sum can be written as

2 2 3 3 4 13 14i i i i i i i i

= 2 3 13 2 3 4 14i i i i i i i i = i13 + i14

[ sum of four consecutive power of i is zero)

= i + i2 = i 1

9. b

4 2

2

i 3 i i 2 2 3 i 3 iz

1 3i4 2 2 3i4 1 i 3

Arg (z) = arg 3 i arg 1 3 i 6 3 6

10. d & are the roots of the equation x2 px + r = 0

p, r (A)

Again ,22

are the roots of the equation x2 qx+r=0

2 q, r2

(B)

p

4 2q

Subtracting1

3 p 2q (2q p)3

Again2 p 4p 2q

p p q3 3 3

2 1 2

r (2p q) (2q p) (2p q)(2q p)3 3 9


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IIT-JEE

Test Prep

11. c (1 p) is a root of the given equation

(1 p)2 + p (1 p) + (1 p) = 0

(1 p) {1 p + p + 1} = 0

p = 1.

The given equation can be written as

x2 + x = 0 x (x + 1) = 0

x = 0, 1.

12. b Let roots fo the equation x2 bx + c = 0 are and

1 .

1 b and ( 1) c

2 2b 4c (2 1) 4 ( 1)

2 2

4 4 1 4 4 1.

13. d P Q

tan and tan2 2

are the roots of the equation

ax2 + bx + c = 0

P Q b P Q c

tan tan ,tan tan2 2 a 2 2 a

(A)

Again P Q R P Q R

2 2 2 2

P Q

2 2 4

P Q

tan tan2 2 4

P Qtan tan

2 2 1P Q

1 tan tan2 2

b

a 1c

1a

[from A]

b = a c a + b = c.

14. c 100x iy (1 i 3) 100[2{cos( / 3) isin( / 3)} ]

100 100 1002 cos isin3 3

100 1 32 i

2 2

99 99x iy 2 i 2 3

15. d 2x 5 x 4 0

( x 1) ( x 4) 0

x 1, 4 (which is not possible)

so no real roots.

16. b

n3 i

13 i

n2

3 i1

3 1

n

3 1 2i 31

4

n

1 i 31

2

nn 21 or 1

17. d z2 + | z |2 = 0

2z z z 0 z z z 0

z = 0 or z z 0 z = 0 or Re (z) = 0

z = 0, z = ib

infinitely many solutions.

18. c 1 2z z 1

a i

1

1 bi

1 bi = a + i

a = 1, b = 1 a = 1, b = 1

19. ab

sin cos a

orc

sin cosa

2sin cos 1 2 sin cos

2

2

b 2c a 2c1

a aa

b2 = a2 + 2 ac

a2 b2 + 2ac = 0

20. b 3

3 3 3( ) 3 7

27 3 3 7

27 9 7 20

9

Equation whose roots are ,

2x ( ) x 0

2 20x 3x 09

9x2 27x + 20 = 0


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IIT-JEE

Test Prep

PartIII : Physics

1.a Both the angular displacement and time are the

same.So, angular speeds are same.

2. d 1 2 300rad /min

2 2 100rad /min

1 2 2 300 2 100

2 2

2 22 100rad/min 200 rad /min

3. b12 rad s

60 30

4. c If T is the tension in the rope, then the force exerted

by the boy on the rope is equal to the force exerted

by the rope on the boy. Let R be the normal reaction

between the boy and the frame.

T

R

40 g

T

360 g

R

Refer to the free body diagram of boy

T + R = 40 g (i)Refer to the free body diagram of frame

T = R + 360g (ii)

Adding, 2T = 400 g

or T = 200 g

5. b Comparing the given equation with y x tan

2

2 2

gx,

2v cos

we get

tan 3

6. a

2v sin2 2v sin200, 5

g g

Dividing,

2v 2sin cos g 20040

g 2v sin 5

or 1vcos 40ms

7. d V cos vcos

vcosV

cos

8. b If the relative initial velocity, relative acceleration & the

relative displacement of the coin with respect to the

floor of the lift be ur, a

r, and S

rthen

2r r r

1S u t a t

2

ur= u

c u

l= 10 10 = 0

ar= a

c a

l= 9.8 0 = 9.8

Sr= S

c S

l= 2.45

2.45 = 0.t + 1/2 (9.8)t2

1t sec

2

9. a Displacement = area under v-t graph with proper sign= 4 2 2 2 + 2 2 = 8m

distance = 4 2 + 2 2 + 2 2

10. b 2 3x at bt ct

2dxv a 2bt 2ctat

ordv

a 2b 6ctdt

11. b Let total length of road is 25

11

st

v or 2

2

st

v

average speed1 22s

t t

1 2

2sv

s s

v v

or1 2

1 2

2v vv

v v

12. c f iav

v vva

t t

45

5 2 m/s

v = 5 m/sf

v

v = 5 m/si

N

W E

S

v 5 2 m/ s

in north-west direction.

2av

5 2 1a m /s

10 2

(in north-west direction)

correct option is (c).


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IIT-JEE

Test Prep

13. a To cross the river in shortest time one has to swim

perpendicular to the river current.

14. a Acceleration at t = 8 s:

210 10a 5 m/s10 6

15. c Horizontal component of velocity of A is 10 cos 60or

5 m/s which is equal to the velocity of B in horizontaldirection. They will collide at C if time of flight of the

particles are equal ortA = tB

2u sin 2h

g g

2B

1h gt

2

or h =

2 22u sin

g

22 32 (10)

2

10

= 15 m

16. b Incase (A) acceleration4g 4

g4 5 9

Incase (B) acceleration5g 4g g

5 4 9

17. d 1T 15A , 2T 3A ,1

2

T 5

T 1

18. c2 2

1u sin

h2 g

2 2

2u cos

h2 g

21

2

htan

h

19. d Area under the curve ( 4 1)kg m/ s

3kgm/s

20. b

N

36 N 4kg

4g

N'

20g

20g RR

36 R = 4a ... (i)

R = 20a ... (ii)

R36 R 4

20

6R36

5

R = 30 N

dc solutions

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