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HINT SHEET

1.2

A B

= A2+ B2 A B

2. Particle will execute SHM in tunnel along

diameter so time period will T = 2 Rg

So time taken from surface to centre

t1=

T

4=

2

4

Rg

If gravity (g) remains constant then time taken by

using equation of motion

s = ut +1

2at2 R = 0 +

1

2g t2

2t2=2R

g

so1

2

t

t=

2 2

3. B =1 2 506 500 32 2

Now

2 2

m ax 1 2

min 1 2

I A A 4 2A

I A A 4 2

= 9

So A = 9 B =3

5. R = 2 2A B 2ABcos

As increases cosdecreases

so R also decrease

At = 90 R2

= A2

+ B2

> 90 R2< A2+ B2

ANSWER KEY

DISTANCE LEARNING PROGRAMME(Academic Session : 2015 - 2016)

LEADER TEST SERIES / JOINT PACKAGE COURSE

TARGET : PRE-MEDICAL 2016Test Type : ALL INDIA OPEN TEST (MAJOR) Test Pattern :AIIMS

TEST DATE :17 - 04 - 2016

HS - 1/70999DM310315028

Que 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Ans. 4 1 2 1 2 1 2 1 4 1 3 1 3 1 1 3 1 3 1 4

Que 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

Ans. 1 2 3 2 2 3 1 1 3 3 2 2 2 1 2 3 2 1 1 1

Que 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60Ans. 2 1 4 4 1 1 2 2 3 2 3 4 3 1 3 2 4 3 3 2

Que 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

Ans. 4 1 3 1 2 1 3 3 1 4 1 4 2 3 4 3 3 4 3 4

Que 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

Ans. 4 1 3 3 2 3 1 4 3 2 4 4 3 3 4 2 3 1 3 3

Que 10 1 102 103 1 04 105 106 1 07 108 109 110 111 11 2 113 1 14 115 116 117 118 11 9 120

Ans. 3 3 1 2 3 3 4 3 1 2 1 2 4 4 1 4 1 4 1 2

Que 12 1 122 123 1 24 125 126 1 27 128 129 130 131 13 2 133 1 34 135 136 137 138 13 9 140

Ans. 4 2 3 1 4 1 3 4 3 4 3 4 3 4 3 4 2 1 1 1

Que 14 1 142 143 1 44 145 146 1 47 148 149 150 151 15 2 153 1 54 155 156 157 158 15 9 160

Ans. 3 2 4 1 2 4 1 1 3 1 4 3 4 4 1 3 3 3 2 3

Que 16 1 162 163 1 64 165 166 1 67 168 169 170 171 17 2 173 1 74 175 176 177 178 17 9 180

Ans. 1 1 1 1 3 2 1 2 3 1 1 3 2 1 1 2 2 3 1 2

Que 18 1 182 183 1 84 185 186 1 87 188 189 190 191 19 2 193 1 94 195 196 197 198 19 9 200

Ans. 4 4 3 3 3 3 4 1 4 3 3 4 3 3 1 4 3 2 3 2

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6.21

mgr mv2

2gr

7. When source is moving towards observer

n' = ns

v

v v

When source is moving away

s

vn" n

v v

s

s

v vn '

n" v v 8

= 8v + 8vs= 9v 9v

s

17vs= v

vs=

340

17= 20m/s

8. CD = 2R sin 45

= 2 mV

qB

1

2

= 0.14 m

10. According to theorem of parallel axes,

I =

2 2

22 R 2 RM M(2R) M5 2 5 2

=2 2 21 214MR MR MR

5 5

11. y

H

E

D

z

x

C

B

F

G

A

net

= 6y (area of EFGH) 4x (Area of BCFG)

= 6 1 12 4 1 12

0

q2

= 2

0

12. When seen from air through nearest surface,

1 2

5 u

=

1 2

20

2

u=

1 1

20 5

=

1 4

20

=

1

4

O C

n = 2

40 cm

u = 8 cm.

for second case,

u = (40 8) = 32 cm

1 2

v 32

=

1 2

20

1

v=

1 1

16 20 =

5 4

80

v = 80 cm.

13. According to work energy theorem

K = W(KE)

f (KE)

i= Fx

(KE)i= 0 & F = constant

f(KE) x

KE

x

14. The height h to which the liquid rises in a capillary

tube is given by:

2Tcosh

r g

Since, T cos , and g are constants.Hence, hr = constant.

16.

I

O =

v

u

If O and Iare on same sides of PA .I

Owill be

positive which implies v and u will be of opposite

signs.

Similarly if O and Iare on opp. sides,I

Owill be

-ve which implies v and u will have same sign.

If O is on PA, I=V

u

(O) = 0

I will also be on. P.A.

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17. W = K =1

2mv

22

1

2mv

12

W =1

2(2) ( 20)2

1

2(2) (10)2= 400 100

= 300 J18. Let d be the inner diameter of hemispherical bowl.

In just floating condition mg = FB

3 3 3

3 42 1 2 1 d(1.2 10 ) (2 10 )

3 2 3 2 2

Solving, we get; d = 0.98 m.

19. When G then potential difference across Ris 20 V.

then potential difference across 1= 4V

I =4

1 = 4A

I =24

1 R=4

R = 5

20. Lets look at the screen.

as we know that 75% intensity will correspond

to a point where intensity is 3 0.

{ Imax= 4 I0}

I= I0+ I

0+ 0 02 I I cos

3I0= 2I

0(1 + cos )

cos () =1

2

=3

, 2

3

, 2+

3

,...........

p =6

,

6

, +

6

,...........

p =yd

D

D

y

y

3I0

3I0

l

yd

D=

6

y =

D

d

6

,........

y =6

,

6

, +

6

2y =2

6

d

D=

10

3

6000 10 1

310= 0.2 mm

21.

1

F F

2

Mg Mg

a a

T T

For block (1)

Mg + F T = Ma .....(1)

For block (2)

F + T Mg = Ma .....(2)

From equation (2) (1)

2T 2Mg = 0

T Mg

22. Take x-axis along the flow and y-axis

perpendicular to it.

initial final i; j

j i

2 2( ) 2

Force exerted on the pipe =P m

t t

= 2S ( 2) 2S .

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24. E1= 1.23 eV

E 1

E2= E

11

2

= 1.2310,000

5000

E2 = 2.46 eV

20V = 1.36 V (Given)

e20

V = E2

1.36 = 2.46

= 1.10 eV

25. avg x avg y avg V (V ) i (V ) j

(Vx)avg

= u cos = constant

(Vy)avg

=Vertical disp

0time

So Ans. Vavg

= ucos26. I r2 I = 2r (r)

or2

I 2r( r) 2( r)

I r r

butr

( )( t)r

I

I

100=2()(t)100= 2(11 106)(10)(100)

= 0.022.

27. Potential difference of 3= 2 3 = 6voltand i = 3A

12V r BA

3A

VB VA= 6VApply KVL from A to B

VA+ 12 3r = V

B

r = 2

28. N = N0

nt / T1

2

=

0N

30

10

0

1

2

N

30

1512

=

1

2 = 0.5

29.N1

(1)g

f2

f1

2g

F

N2

N1

N1= (1)g .....(1)

N2= N

1+ 2g .....(2)

From (1) & (2)

N2= 3g ....(3)

f1=

1N

1= 0.1 1 g = 1 N ...(4)

f2=

2N

2= 0.2 (3g) = 6 N .... (5)

Now force required to pull the 2kg block

F = f1+ f2 = 1 + 6 = 7 N30. V = VT

=V

V T

Now PT2= constant

3T

V= constant

3 T V0

T V

3 V 0T V T

3T

31. R =1

c= X

c

Z = 2 R

I0= 0

V

2R.......(1)

when becomes1

3 times, X

cwill becomes

3 times Xcwill be 3 R

Z1= 2R

I0'=

0V

2R.......(2)

I0'= 0

I

2

32. m = (1 0.993) gmm = 0.007 gm = m C2

= 0.007 103 9 1016

= 63 1010 J

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33.4gR = VC

TCmg

TB

TA

mg

VA 8gR

VB

6gR

mg

By second Eqaution of motion

VC

2= V

A

2 2gh

VC

2= 8gR 2g(2R)

VC

2= 4gR

CV 4gR

At point (A)

TA

mg =

2

Amv

R

TA

= mg + 8mg = 9mg

At point C

TC+ mg =

2

CmvR

TC= 4mg mg = 3mg

C

A

T 3mg 1

T 9mg 3

34. msT =21 1 m

2 2

2 2 4(200) 4 10T 80 C.4s 4 125 4 125

36. ni= 2.5 1013cm3

neN

D= 0.5 1017cm3

nnN

A= ?

ni2= n

e n

n

ni2= N

D N

A

NA

=

2i

D

n

N

37. Before collision

2Kgu = 4 m/s1

3Kgu = 4 m/s2

After collision

2Kgv = ?1

3Kg

Rest

v = 02

By conservation of linear momentum

2(4) + 3( 4) = 2v1

8 12 = 2v1

4 = 2v1

1v 2 m / s

38. 1st case:As it is a series combination,

Ks=

1 2

1 2

2K K

K K

2nd case:As it is a parallel combination,

Kp= (K

1+ K

2)/2

s 1 2

2

p 1 2

K 4K K

K (K K )

40. =C

B

i

i

iC= iB= 100 5 106

Vout

= iC R

0= 5 104 10 103= 5V

46. Mg (OH)2

Mg+2+ 2OH

S' 2S'

Mg(NO3)

2 Mg+2+ 2NO

31

C 2C

Ksp of Mg (OH)2

= [Mg+2] [OH]2

1.8 1011 = (S' + C) (2S')2

1.8 1011 = C4S'2

(S') =

1/ 211

1.8 10

4 0.02

mol

L

S' =

1/ 2111.8 10

584 0.02

g/L

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50. Kp1= 8 102 = PCO

2

Kp2=

2

2

CO

CO

P

P

54. 5H2O () 5 H2O(g)H = E + ng RT

540 90= E + 5 (2) 373

55. (H2SO

4)N

1V

1= N

2V

2= N

2V

2(dilute acid)

N2= (10 36)/1000 = 0.36 N

58. NH

most stable

Complete octel

Resonating structure

NH

All other carbocations have incomplete octel

resonating structure.

59. N =6 1000

40 100

= 1.5 N

It is show highest normality than others

62. Heat of Hydrogenation Number of bond

1

Stability(if number of bond same)

stability Resonance No of H

* min bond* min. heat of hydrogenation

Resonance stablised

less heat of Hydrogenation than (III)

* max. bondmax. Heat of Hydrogenation

63. When cation shifts from lattice to interstitial site,

the defect is called Frenkel defect.

65. Valency of metal (x) =2VD

EW 35.5

x = 80

4.5 35.5=2

EW =Atwt

x

4.5 =Atwt

2

66. CH CH C3 2 CCH3

OH Cl

HOCl

CHCH C CCH3 2 3

CH CH CCCH3 2 3 CH CH CCCH3 2 3

OH OHCl Cl

Cl Cl

HOCl HOCl

more stable carbocation

CH CH CCCH3 2 3 CH CH CCCH3 2 3

less stable

OH

O OHCl Cl

Cl OHCl

(major)

67. Order of reaction is sum of the power raisedconcentration terms to express rate expression

69.2

2 2

1 2

1 1 1Z R

n n

2

2 2

1 1 1(1) R

(2) (3)

70. CH3

F

CH3

Alc.KOH

CH3CH2

(major)

HBr

R O2 2

CH3CH Br2

(major)

(anti maukovnikov addition)

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134. Statement 1 is false because constructive

interference can be obtained if phase difference

of sources is 2, 4, 6, etc.

155. I and II are structural isomers because connectivity

is different

157. It is not necessary that a good base is always a

good nucleophile

For example :

Basic strength OH > SH

Nucleophilicity OH < SH

159. Ethene is more reactive than ethyne towards

electrophilic addition reaction becauseintermediate carbocation in ethene is more stable

than ethyne

CH =CH2 2 CH CH

CH CH2 2 CH=CH

Ethene Ethyne

E

E

E Esp

2sp

more stable less stable

162. Module, Page : 188

171. NCERT Pg.# 248

176. NCERT XII, Pg.# 196,197(E), 213, 214 (H)

178. NCERT XII, Pg.# 89(E), 97,98 (H)

180. NCERT XII, Pg.# 213(E), 232 (H)

71. The concentration of reactant does not change

time for zero order reaction (unit of K suggests

order) since rectant is in excess

73. Bond energy of CH bond = 4004

= 100 kCal/mol

Bond energy of C-C + bond energy of 6C-H bonds

= 670

Bond energy of C-C = 670 6 100 = 70 kCal

74.

NH2

Cl

OH3

2

16

54

OH6 1

4

Cl

5

NH232 Indentical

75. It connect two solution and complete the circuit.

79. H2 undergoes oxidation and AgCl(Ag+)

undergoes reduction.

86. NCERT Pg.# 197,198

87. Module, Page : 179

91.

NCERT, Page : 12694. NCERT Pg.# 231,232

101. NCERT -I Pg.# 56 & 57

103. NCERT Pg # 176

104. NCERT XII, Pg.#89 (E), 97 (H)

108. NCERT XII, Pg.#288, 289 (E), 314,315 (H)

109. NCERT XI Pg.# 142 Para 2

112. NCERT XII, Pg.#81, 82 (E), 89,90,91(H)

116. NCERT XII, Pg.#187 (E), 204(H)

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