Vidyamandir Classes SOLUTIONS - Amazon Web...

Vidyamandir Classes

VMC/Final Step [Part-A] 1 HPT/Class XIth/JEE-2015

SOLUTIONS

Home Practice Tests/JEE-2015 Final Step [Part-A]

HPT-1 Chemistry

1.(C)

2.(B) 2

2

anP V nb nRT

V

For 2

1 a b 1n , P V RT

2 2 24V

2

aP 2V b RT

4V

3.(B) 1

3RTv

M

2 2T 2T,M M / 2

2 1

3R 2Tv 2v

M / 2

4.(D) Both H2Se and H2S are polar but the former has higher dispersion forces due to its higher molecular weight and hence

the higher boiling point.

5.(B) 1

22(He )

EE

n Z

22

2

2 413.6 13.6 13.6 eV

4n

6.(A) Initially BF3(sp2, trigonal planar) and NH3(sp

3, pyramidal) will convert into BF3(sp

3, tetrahedral) and NH3 (sp

3,

tetrahedral)

7.(A) Basic strength of the oxides of alkali metals increases down the group.

8.(B) 0G G RTln Q ; where Q is reaction quotient; at equilibrium cG 0andQ K

0

cG RTln K

9.(A) Fe(OH)3 + NaOH no reaction. Other hydroxides make soluble mettalate in cases of Zn(OH)2 and Al(OH)3 or

mettalite in case of Sn(OH)2.

Vidyamandir Classes


10.(D) Meq. of HCl = 100 0.1 = 10

Meq.ofNaOH = 30 0.2 = 6 meq.

Meq.of HCl left = 4

4 = 0.25 V

4

V 4 4 16 ml0.25

11.(C) 3 3 aqaq

AgIO Ag IO S Solubility

2

spK S

or, 2 8 4S 1.0 10 or S=1.0 10 mol / lit

41.0 10 283 g / lit

41.0 10 283 100g /100ml

1000

428.3 10 g /100ml 32.83 10 g /100ml.

12.(D) 2H2SO4 4 2 3HSO H H O SO

13.(D)

H

H H

H

CH3

60º

CH3

14.(B)

CH3–CH–CH2–CH3 2BrCH3–CH–CH–CH3 + CH2–CH–CH2–CH3

CH3 CH3 CH3

Br Br

*

*

15.(A) (CH3)2CHBrCuI)ii(

Li)i( Li[(CH3)2CH]2Cu

BrCHCH)CH( 223(CH3)2CHCH2CH(CH3)2

(A) (B)

16.(B) Dipole moment of 3BF and 4CCl is zero. In 3NF , the dipole moment is diminished due to lone pair and it is

reinforced in the case of 3NH

17.(A) Let the mass of C-14 isotope be ‘x’ gm

Hence, the mass of C-12 isotope would be (12-x) gm

Number of C-14 atoms A

xN

14

and Number of C-12 atoms A

12 xN

12

A

A A

xN

14 100 2x 12 x

N N14 12

12x

100 212x 168 14x

1200x 336 4x

Vidyamandir Classes


336x gm

1204

Hence number of C-14 isotope 23336 16.023 10

1204 14

221.2 10

18.(C)

b

BpH 14 pK log

BOH

V = volume of acid required for the equivalence point.

(i) b1

10pH 9 14 pK log

V 10

(ii) b2

25pH 8 14 pK log

V 25

(i) – (ii)

25 10 25 V 101 log log log

V 25 V 10 10 V 25

V 30ml

19.(C) Spherical node = n – l – 1 = 3 – 1 – 1 = 1 Non-spherical node = l = 1

20.(D) Density Mass

Volume

For a sealed rigid container mass as well as volume remain a constant hence the density remains same even if the

temperature changes.

21.(D) 2

H 2 21 2

1 1 1R Z

n n

For Balmer Series 2H 2 2

1 1 1R 1

2 3

H 1R 5cm

36

22.(A) (g) (g) (g) (g)A B C D

Mole 1 1 1 1

Conc. 0.5 0.5 0.5 0.5

At 0.5 + x 0.5 + x 0.5–x 0.5 – x

2

C

[C] [D] (0.5 x) 1K

[A] [B] (0.5 x) 4

0.5 x 1

0.5 x 2

1 2x 0.5 x

1

x6

[C] at equilibrium 1 1

0.5 x2 6

3 1 1

M6 3

23.(A) S + O2 SO2; H 298.2kJ

2 2 3

1SO O SO

2 H 98.7kJ

3 2 2 4SO H O H SO H 130.2kJ

2 2 2

1H O H O

2 H 227.3kJ

Adding all the above equations we get :

S + 2O2 + H2 H2SO4 H = 754.4kJ

Vidyamandir Classes


24.(A) 4% of NaOHimplies 100 gm solution 4 gm NaOH

So 1000 gm solution 40 gm NaOH

1 mole NaOH

But 1000 ml solution weights 1200 gm (d = 1.2 gm/ml)

So 1000 ml solution 1.2 mole NaOH

Molarity = 1.2

25.(B) 2 3 2 2Li CO Li O CO

. Other carbonates are thermally stable.

26.(D) As the size of anion increases covalent character increases.

27.(C)

Cl alc.

KOH

NBS alc.

KOH

NBS

(A)

Br

(B) (C) (D) Br

alc.

KOH

(E)

28.(D) As XeF4 has square planar shape the dipole moments of Xe−F bonds compensate each other and hence it is

non-polar.

29.(D) 2NO2 = N2O4

Mavg = 2 × VD = 2 × 27.6

Let number of moles of 2NO and 2 4N O be m, n respectively

2 × 27.6 m 46 n 92

m n

m

0.8m n

30.(C) Experimental dipole moment = 1.03 D

Expected dipole moment assuming 100% ionic

Character e bond length

10 84.8 10 1.275 10 esu.cm

184.8 1.275 10 esucom

6.12D

% ionic character 1.03

100 16.67%6.12

17 %

Vidyamandir Classes


HPT-1 Physics

31.(C) 2 21

2mv ks

2

KV S

M

2

2 2

r t

t r

v dva ,a

R dt

a a a

32.(C) 0 w w k

0 d

w kdt

0 00

01

t

kt

ddt

w k

we

k

33.(B) Since every body has an acceleration of g sin on

the inclined plane force on nth block is zero.

34.(A) 1 1 1T m a

2 12T T

2 2 2 2 M g T M a

1 22a a

Solving above equations

2

21 24

M ga

M M

35.(A)

2

0180 37 180 37

Mg N

sin sin

2 N Mg

36.(A)

2

3

2

xW x dx , 3 6 18 yW ,

net x yW W W

2 23 18 x yW | x | W

4 9 xW

5 xW

So 13netW

37.(B) 2 21 2

2 2

l lt , t

g sin g sin cos

38.(D) P KE

39.(C)

21

2

2 0

0

VR =

VCos 45

R

= 2 2 R

40.(D) 2

01

4

330

Ve

V Cos

01 2 30V V Cos

41.(C) Initially there is no sliding between two blocks so

they move together with same acceleration but

when sliding starts B moves with constant

acceleration while of A increases with time.

42.(A) 2 22 12

2mR MR

t t

By using conservation of angular momentum.

43.(B)

andR m r Mm M m M

They rotate about their com,

2

2

GMm mv

r

On solving

2GMv

M m

44.(D) Cyclic process 5 AB BC CAW Q W W W

5 10 1 0 CAW

5CAW

45.(B) 21

3 2

GMm GMP.E, T .E. mv

R R h

Vidyamandir Classes


46.(A) 1 2

2At H H

a g

According to given data 1 2 0H h,H

2A

t ha g

--- (1) further

22

At' h

a g ----(2)

From (1) and (2) 2t ' t

47.(D) 2 2' . 2 4 [(3 ) ]W T A T r

2 264 8 8 8r T . r T W

48.(B) Tangential velocity of the rope, v r

Tangential acceleration, ( )dv d

a rdt dt

dr

dt

Here 2 n and dr

nddt

22a n d

221 1

a n dT w w

g g

49.(C) 2

5 40 8

5 20

realmdd .

m M

The distance of the dog from the shore is

10 – 3.2 = 6.8 m

50.(D) 2 2 2fw wi

20 2

0 24

……………..(1)

203

24

20

0 24

'

………………(2)

3'

,

3'

, So

3

n rotations

51.(B) 1A OP P gh

2B OP P gh

A BP P p.l.a

1 2

102

10

ah h l.

g = 2 cm

52.(B)

2

2

u sin y

g

; 2u gh

2

22

2

gh.sin y h sin

g

53.(A) Tension produced due to cooling yA

y.A. .

Net force mg yA

Stress mg

yA

Strainmg

Ay

Change in length mg

Ay

54.(B) As water is powered down,

water level increases first

than becomes constant. So

length of air column

decreases first and then

becomes constant.

55.(D) 0

s

v vn n

v v

;

0 5 1 53

0 5 0 5

v . v . vn n n n

v . v . v

Change in frequency 2n

56.(B) 0 05 20 50y . sin x t

1 (20 50 0)x

20 20 5 100x ;

2 [20 (5 25) 50 5]

600 250 350

Change in phase = 250

57.(A) Heat released if temp of water decreases from

30ºC to 0ºC

1 5 1 30 150cal Q

Heat required to increase the temperature of ice

Vidyamandir Classes


2 5 0.5 20 50cal Q

Heat required to melt the ice completely

3 5 80 400cal Q

Heat available would be less than heat required to

melt.

So final temperature 0ºC and mixture will be

heterogeneous.

58.(B)

22 2 2

2 1

3[1 ] [1 ]

2 4

l lCD l

2 22

2 1

3[1 2 ] [1 2 ]

4 4

l ll

2 22

2 1

3 3 12 ( )

4 4 4

l ll

1 24

59.(A)

2left rightn n

1 21 2

2 2

T T

1 24T T

Taking torque about com.

1 2. ( ) T x T l x , 2 24 . ( ) T x T l x ,

5

lx

60.(D) 2 Re ev g

3

2 e

kTgR

m,

2

3 emgR

Tk

HPT-1 Mathematics

61.(B) Put 2|x 2| t t t 2 0

t 1, 2

If t = 1 or if t 2 cannot be possible as | x 2| 2 not possible.

x 2 1 x 3, 1 Sum of roots be 4

62.(D) For P: 4

0 0 a , 4 5,5

For Q : a 4

0 0 a 4, 5a 5

For R: and20

D 0, 0, a a , 4 5,9

(1) TrueP Q (2) TrueR P

(3) P Q R' 4, 5 All are true.

Vidyamandir Classes


63.(D) Statement-1: Clearly y1 = | sin | x | | and y2 = x + | x |

has three pt. of intersection in 2 2 ,

Statement-2: True:

64.(B) Graph of 1 0 0 1| f |x| | x , , and Graph of 1 0 0 1| f x | x , ,

Graph of g (x) = | f |x|| | f x | is:

Graph of [ | g (x) |] is:

Range of is {0, 1}

65.(B) Case – I: 3 0 1log x x

2

3 32 0log x log x a

For Real solution : 2

1 4 2 0. .a 1

8a …(i)

Case-2: 3 0log x x < 1

2

3 32 0log x log x a

For real solution: (1)2 – 4.2.a > 0

1

8a …(ii)

3

1 1 8

4

alog x

Here log3 x < 0

1 1 89

04

1 1 8 0a

1 8 1a (+ve sign)

a > 0 …(iii)

Hence from (i), (ii) and (iii)

1

08

a

66.(A) 1 is a root of f (x), then

f (x) = (x + 1) (ax + b)

f (1) + f (2) = 0 2(a + b) + 3 (2a + b) = 0

8a + 5b = 0

8

5

b

a

= other root

2

O

2

-1 1

-1 1 O

-1 1

O 1 –1

2

O

2

Vidyamandir Classes


67.(D) 2 2 2 2 0x y gx fy c Put y = 0

2 2 0x gx c

1 2 2x x g

1 2x x C

Similarly put 0x

2 2 0 y fy c

1 2 1 22y y f y y c

Area of quadrilateral = 1 2 1 21

2x x y y

= 2 2

1 2 1 2 1 2 1 21

4 42

x x x x y y y y

= 2 214 4 4 4

2g c f c = 2 22 g c f c

68.(B) 2 2 1

8 2 2cos

Now other root is conjugate of this: 2 1

2 2

Sum of roots = 1b 1b

Product of root = 1

8c

11

8b, c ,

69.(C) Statement-1: 3 2 1tan x x 4

x

General solution is 4

n

But for this value, tan 2x = N.D, which does not satisfy the given equation as it reduces to Indeterminate form.

Statement is false!

Statement-2: sin 2x + cos 4 x = 2

sin 2x = 1, cos 4x = 1

cos 4x = 1 21 2 2 1sin x

1 2 1 not possible No solutions (False)

70.(C) We know that logax is defined when 0 0 1x , a , a

x

|x|log

x is defined if: 0 0 and 1

|x|, x x

x

Case-I: 0|x|

x x > 0 …(i)

Case-II: 0x 1x , …(ii)

Case-III: 1x 1 2x , …(iii)

From (i), (ii) and (iii), we get 2x ,

(0, y1)

(x, 0)

(0, y1)

(x2, 0)

Vidyamandir Classes


For 2x , we find that 1 0x x

|x|log log

x

1 0 for all 22

f x cos x ,

Domain 2 and range2

f x ,

71. (B) As 04

, , tan cot

Since 1 1tan and cot

1 and 1cot tan

tan cot

4 1t t which only holds in (B)

72.(B) Area of quadrilateral

= Area of (ABMD) + Area of BMC rectangle

= 1

2 22

r . x . r x

= 3 xr

But 3xr = 18 xr = 6 …(i)

OBE OBK

EOB KOB

OCN OCK

90NOC KOC

and In 2

90x r

ONC tanr

2x r

cotr

…(ii)

In x r

OEB tanr

…(iii)

From (ii) and (iii)

2

x r r

r x r

2 3 0x x r

30

2

rx x …(iv)

From (iii) and (iv) we get : r = 2

73.(C) Since is isosceles, hence centroid is desired point.

Coordinates are 4

33

,

74.(A) As 1 2 3 4 5 1 4 6 7 10 0a a a a a

x = 1 is root of a1x4 + a2x

3 + a3x

2 + a4x + a5 = 0

Maximum roots = 4

and complex root are in pair form

Hence the given equation has at least two real roots.

A

D C

B

x

x – r

2x – r

2x

O

r

r

r

M

E

90

N

K

r

P (3, 4)

O

(0, 0) (6, 0)

Q

R

Vidyamandir Classes


75.(B) Since AP PQ QB

The co-ordinate of P are (a, 0) and of Q are 2 0a, , equation of the circles an AP, PQ and QB as diameter are

2 0 x x a y , 22 0 x a x a y and 22 3 0 x a x a y

If h,k be any pt. on the locus, then

2 2 2 23 9 8 h k ah a b 2 2 2 23 9 8 0x y ax a b

76.(B) We have 2 21 x x x x

If f x is divisible by 2 21 0 0 x x , f , f

3 3 0 1 1 0 P Q P Q ………………… (1)

6 2 6 20 1 1 0 P Q P Q …………………. (2)

From (1) and (2) we get,

1 0 1 0 P , Q

3 P x and 3Q x are divisible by 3 1x , Hence by 1x

Since 3 3 f x P x xQ x , We get f x is divisible by 1x

77.(C)

1

131 2

13

S Choice 2 22 1 2 20 25 0 5 0 5 4

log S log /A . . .

Choice 5 5

25

3 1 2

3

10 008 0 2 8

5

log S log

logB . .

/

78.(D) 1

2 2 12 6 6

n

sin A B sin A B n

……………. (i)

Also A B C and 2 B A C 33

B B

From (i) 1n

5 5

26 4 12

A B A ,C

79.(A) 1 1 1

1 2 2 3 11 1 1

n n

d d dtan tan ............. tan

a a a a a a

1 1 13 2 12 1

1 2 2 3 11 1 1

n n

n n

a a a aa atan tan .......... tan

a a a a a a

1 1 1 1 1 1

2 1 3 2 1

n ntan a tan a tan a tan a ............ tan a tan a

1 1

1 ntan a tan a

1C

2C 3

C

P Q

A

B

3a a 2a

x-axis

0 0,

Vidyamandir Classes


80.(B)

R1

R2

R3

Case No. of Balls

R1 R2 R3

No. of different ways

I 4 1 1 26 2 1

4 1 1 4 2 C C C

II 3 2 1 26 3 1 4

3 2 1 3 3 2 C C C C

III 3 1 2 26 3 1 4

3 2 1 3 3 2 C C C C

IV 2 2 1 26 4 2 4

2 2 2 2 2 2 C C C C

Total = 26 6

81. (C) 2 3

1 1 1

3 3 39 3

.....

xtill

; 2 3

1 1 1

3 3 34

..........

ytill

= (4)1/4

4

1 1

12

ri /

r

r r

eZ i =

4

4

2

12

i /

i /

e

e= i

1 4

3 4 3 2 /

x y z i i

Argument = 1 2

3

tan

Choice: (C)

82. (A) 3 2 1 0 1 0 1 0 f x x bx cx , f f b cand

1 0 1 0 ' ' b / w i.e.lies to

1 1 1 11

tan tan tan cot = 2 2 / /

83. (C) 1 1 1 312

2

tan tan A tan cot A tan cot A

= 3

1 14

21

2 1

tan A cot A cot Atan tan cot A

cot A= 1 1

2 21 1

tan A cot Atan tan

tan A cot A

= 1 12 21 1

tan A tan Atan tan

tan A tan A= 14 4 1

4

. tan

Vidyamandir Classes


84. (A) 5 4 44

4 5

2 5 2 80 onr shoe outof pairseach pair

C

We need to select only one shoe out of any one pair. So we choose out of spairs & then choose one shoe out of each.

85.(A) 4 3 2 2 22

2 12 3 2 1 2 3

n n n n n n n

n n=

22 1 1

1 2n n nn n

=

22 1

1

n nn

2008 2008

1 1

1 1 11 1

1 1

n n

n n n n=

1 1 20082008 2008

1 2009 2009

86.(B) Let roots be: a, a d , a d then sum of roots = 3 1

B

aA

As a = 1/3 is a root so put in original equation 9 27

1 27 1 9 3 0 12 2

/ / /

For any 1 , x yline

Definitely passes then 9 27

2 2

, .

87.(B) 1 2 2 3 1 3

1 2 3 1 2 3 2

1

Required root at are 1 2 3

1 1 1

, ,

1 1 y x

x yor (Replace)

=

3 21 1

1 0

ay y

= 3 1 0 by ay = 3 1 0 bx ax

88.(D) This equation can be written as sin x . cos x [sin2 x + cos

2 x + sin x . cos x] = 1

1 1 sin x . cos x sin x . cos x 2 2 2 4 sin x sin x

= 2 4 16

2 1 52

sin x which is not possible.

89.(C)

90.(C) nP cos cos n

n

nP cos Re cos i sin 21

n

nP x Re x i x

2 1n

nZ cos i sin cos n i sin n x x

2 1

nn

Z cos i sin cos n i sin x x

2 22 1 1

n n

Z Z cos n x x x x = 2Pn(x)

Vidyamandir Classes


HPT-2 Chemistry

1.(B) 3CaCO will dissociate to give CaO and 2CO until 2P COK P 1.5atm

Use PV = nRTfor extra CO2 produced.Find ‘n’ which will equal to net of moles of CaO.

PV 1 1 1

nRT 0.0821 292 24

2.(A) 2 2 n

n n 2

N N XeF

XeF XeF N

r P M

r P M

2 2 n

n 2

n

N F XeF

XeN N

XeF

Molar volume

r t t 57

Molar volumer t 38

t

nXeFM57 0.8

38 1.6 28

n

2 2

XeF

57 1.6M 28

38 0.8

=252

Xe nF 252

131 n 19 252 = n = 6

Molecular formula = XeF6

3.(A) PV = nRT W

RTM

WP RT

V M

RT

M

A

3P RT

M

;

BP RT

2M

A

B

3RT

P 3 2MM

P MRT

2M

= 6

4.(C)

5.(C)

2H

HH

nr 0.529 A

Z

3

3

2

Be

BeBe

nr 0529

Z

3

3 3

22H H Be

2 2HBe Be

Zr n 2 4

r Z n 1 3

3

0

Be

a 16

r 9

30

Be

9ar

16

Vidyamandir Classes


6.(A) C(s) + CO2 (g) 2CO (g) ; …(i)

1

14pK 10 atm

2CO(g) + 2Cl2 (g) 2COCl2 (g) ; …(ii)

2

3 2 2pK (6 10 ) atm

Add (i) and (ii)

C(s) + CO2 (g) + 2Cl2(g) 2COCl2 (g) ;

Kp = 14 610 36 10

= 36 × 108

For given reaction ng = – 1

c pK K (RT)

= 36 × 108 × 0.0821 × 1120

11 1

cK 3.31 10 M

7.(A) Solubility of CuS and HgS (AB type) is spK ; solubility of 2 2Ag S (A B type) sp

3K

4

8.(D)

CH2CH2OH

H+

H2O CH2CH2

H

+

CHCH3 +

+

H

CH3 CH3

+ H2O

H+

OH

CH3

9.(B)

OH CH3CH=CH H+

OH CH3CH2CH

Br

(more stable)

OH CH3CH2CHBr

(major)

+

10.(B) The blue colouration is due to solvated electrons 11.(A)

12.(D) 3NH N g 3H g 1

1H x Kcal.mol

2 4N H 2N g 4H g 1

2H yKcal.mol

1 N HH 3 x …(1)

2 N H N NH 4 y …(2)

From (1) &(2) N N

xy 4.

3

1

N N

4x 3y 4xy Kcal.mol

3 3

13.(D) Factual

Vidyamandir Classes


14.(C) In the hydrides of group 15 and group 16 (except NH3 and H2O) the energy difference between 3s and 3p orbitals is

quite high. Hybridization increases the energy of 3s orbital so much that lone pair rather prefers to occupy

unhybridizeds orbital. For example in PH3, 600 kJ mol-1

of energy is required to hybridize the central atom. So to

avoid such energy demanding hybridization P forms bonds with unhybridizedp orbitals leaving the lone pair in the

spherical s orbital which leads to a bond angle close to 90°.

15.(D) [H+]1 = 10

–2

[H+]2 = 10

–6

Factor = 41

2

H10

H

16.(B) Peroxide effect is only shown by HBr 17.(A)

C=C H

Br

H

Br ; C=C

H

Br H

Br ; C=C

Br H

Br H

18.(B) Assuming mass of CH4 and H2 are 1 gram of each

4 2CH H

1 1Mole Mole

16 2

2Hp Mole fraction Ptotal

2H total

8p P

9

19.(B) Vacant d orbitals available only in sulphur.

20.(D) CH2=CH2 Cl2, H2O H2SO4

alc. KOH

CH2ClCH2OH ClCH2CH2OCH2CH2Cl

(B)

CH2=CHOCH=CH2

(C)

Heat

21.(D) Oxygen exist +ve oxidation state only with 'F' In F2O, 'F' in -1 oxidation state and 'O' has +2 oxidation state

22.(D) 4 2 4 2 2 4 2 4 4 2 22KMnO 3H SO 5H C O K SO 2MnsO 10CO 8H O

COOH

1 1 1 4 2 2 2 COOH

M V n KMnO M V n |

4 2

110 V 5 10 0.5 2 , 1V 20L

23.(B)

X 2Y, Z P Q

Initial mol 1 0 1 0 0

At equilibrium 1- 2 1-

1

2

12Y

PX

1

2P

P 1K

1PP

1

; 2

2 2P q

Pz

2

P PP P 1 1

K1P

P1

Vidyamandir Classes


1

21

P 2

4 PK

1

2

22

P 2

PK

1

Given is 1

2

P

P

K 1

K 9

Substituting values from equation (i) & (ii) into (iii) we get :

21

2 1 1

22 22

2

4 P

1 4P 1 P 11

9 P 9 P 36P

1

24.(D) For acidic buffer,

a

ApH pK log

HA

When the acid is 50% ionized, [A ] [HA] or apH pK log1 or apH pK

Given apK 4.5

pH 4.5

pOH 14 4.5 9.5

25.(C) maximum number of orbital’s in a sub-orbit = maximum number of electrons with parallel spin 2l 1

26.(D) 4HClO is a strong acid.

27.(B) Borazole is polar due to BN bond; benzene is nonpolar due to CC bonds.

28.(D)

Anhydrous

+ CH2Cl2 AlCl3

CH2Cl

Anhydrous AlCl3

CH2

29.(B) 3

vapVap

H 30 10S T 400K

T 75

30.(C) Three replaceable H so molarity 3 = Normality

HPT-2 Physics

31.(D) 0n

cos 0mg T

cosT mg

2sinT mrw

1

2Max

gP

h

Vidyamandir Classes


32.(D)

0 08 30 6 300 7

10

rod

sin sin. rad / s

33.(D) ( ) ( 1)T n mg m g a a n g

Max. possible retardation g

2

12( 1) .v n g h

2

22v gh

& 1 2 h h h

On solving 2 ( 1)gh n

vn

34.(A) N Fsin 60º 3g

Fcos60º [Fsin 60º 3g]

35.(B) Let masses on two sides be ' 'm and m m , where '2

mm

' ' 1m g T m a ' ' 2T m m g m m a

2 4 1

m'F T m' g mg

m

36.(B) 2

cos 1mv

T mgl

;

212

2 2

mglmgl mv

On solving T = 1.5 mg

37.(B) 1 cm cmy y

1 1

10 7 30 301

40 40

y y y

1 .y cm

38.(C) Applying conservation of momentum

0.50 5 0

2.5

v

v

Vidyamandir Classes


39.(A) 20

0

w= 10+0.50x 21

40.(C) cos cos2

v u

cos

cos / 2

uv

21 1cos

2 2gravityW K mv m u

2 2 21cos tan

2 2mu

.

41.(B) 2

5

R

T

K

K

5

7TK Mg R r

21 5

2 7mv mg R r

v

wR r

10

7( )

g

R r

42.(D) F

MR

F f Ma MR F

0f

43.(D) 21

2 2 2e

GMm GMmmv

l / l /

2

2e

GMv

l

44.(D) The buoyant force pushed body in upwards direction for which cube will rise

When 0.2 kg is removed, cube rises by 2 cm

So, ( ) mg Buoyancy force

2 2 30.2 ( 2 10 )10 g a g

2 210a

110a m or 10 cm

45.(A) According to perpendicular axis theorem

1 2 1 2and2

II I I I I

Vidyamandir Classes


Similarly 3 4 3 4and2

II I I I I

So, 1 3I I

46.(D) Using conservation of energy 21 1

2 2FX mv

2 11 120

2 2

YAXX mv v ms

L

47.(C)

0

1/201/2

0

2

w t

w

dw dwKw K dt w Kt

dt w

The body comes to stop in time 0

0

2 wt

K , further

2

02

Ktw w

0

20 00 0

0 2

0

2

12 2

t

t

wdtK w wKwK

w wKK

dt

0 00 0

3 3

w ww w

48.(A) 10 kx B g

10 kx kx g

10 5 5 10 1

2 500 10

g gx m

k k

10x cm

49.(D) Force by curved surface

= Fbottom – mg =

22 .

[ ] . .3

R RgR R g

33

3

g Rg R

32

3g R

50.(B) Force required to separate the plates 2TA

d

51.(B) 2 L a bt

Suppose 2ˆ ˆ L ai bt j

ˆ2t btj

Angle between L and t is 45º

So, 2a bt

a

tb

2a

t bb

Vidyamandir Classes


52.(D) So F > 15, then acceleration of two block together

15

15

Fa

15

15 515

Ff

Max. value of f = 25

15

25 153

F

30 15 F

45F

53.(D) Less/Gain in time 1

( )2

time 511.2 10 10 86400 7

2

36.28

54.(C) 3

rms

RTV

m

21 3

2 2rmsK mV KT

55.(A)

0

0

. . t

Q m s dt0

2

0

( ) t

m at bt c dt

3 20 0

03 2

mat bt

Q Ct

56.(A) 2 2AD (68) (53) 42.6

Speed of Car = 42.6

2130.2

57.(B) 0

1

vn

4( e)

; fundamental mode

0

2

3vn

4( e)

; first overtone

By above equations

1 23 3e e

2 13e 0.025m

2

58. 2 4x ay

Slope = tan1

.24 2

dy xx

dx a a

Restoring force sin tan2

mgxmg mg

a

2

2

g

a

Vidyamandir Classes


59.(A) 1 1 1

1 5[ ] . . . 100 125

2 2

RQ n cv T R

2

1 3[ ] . 100 37.5

4 2

RQ R

162.5 162.5 2 Q R

60.(B)

Area under P V curve with volume

Axis = work done

HPT-2 Mathematics

61.(B) are roots of 3 2 0 x px qx r p, q, r

2 2 2 2 2 2

2 2 2 2 2 2

1 1 1

2 2 2

2 2 2

2

2

2

2

q rp

r

62.(A) 2

2

2 1 9 40

8 32

ax a x af x

x x

2

2

2 1 9 40

4 16

ax a x a

x

2 2 1 9 4 0 ax a x a x R 2 22 1 9 4 0 a a a a and 0a

28 2 1 0 a a and 0a 28 4 2 1 0 a a a and 0a

4 2 1 1 2 1 0 a a a and 0a 2 1 4 1 0 a a and 0a

1 2 a / or 1 4 / and 0a

So, 1

2

a ,

63.(B) S1 : 2 0 ax bx c 0a

If 0 a b c then one root is 1 product of roots c

a

Vidyamandir Classes


So another root c

a

S2 : 2 f x ax bx c 0a

It has finite minimum value of graph is upward parabola roots are opposite sign

So 0 0f

S3 : is repeated root of 2 0 ax bx c

So 22 ax bx c a x

Sn : 2 0 ax bx c 0a

Irrational roots occur in conjugate pair only if a, b, c are rational.

64.(D)

2 4 6 200

1 3 5 199

100

a a a ..... a

a a a ...... a

...............................................

d

100

d

65.(C) 1 2 3 x x 3 4 12 x x

1 2 x x A 3 4 12 x x

x1 x2 x3 x4 G.P.

a ar ar2 ar

3 1r

1 3 a r

2 1 12 ar r

……………………………………..

2 2 r , Hence r =2

a = 1

2 2 A a.ar a r

2 5 52 32 B a r

66.(C) 2 2 2

2 1

1 2

r

rT

.....r

6 2 1 6

1 2 1 1

r

r r r r r

1 16

1

r r

1 1

61

n nS T

r r

1 6

6 11 1

n

nS

n n

6nn

S lim S

Vidyamandir Classes


67.(B) 3 3

32 182 2

cos x sin x sin x

19 56 9 7cos x cos x sin x

18 9sin x cos x sin x cos x cos x sin x 18 9 9cos x sin x sin x bcos x

27a b

68.(C) 2001 2001 2001

2001 20012 3 4 6

cos cot sec tan cos ec

667

667 1 0 1 1 1 22 4 2

cos cot sec tan cos ec

69.(A) 22 2

tan tan

2

2

2

2

123 5

13 5 2

5 31

25 3

12

tan

tancos

costan

tan

2 2

2 2

3 1 5 12 2

5 1 3 12 2

tan tan

tan tan

2 2 2

2 2 2

8 2 8 8 12 2 2

8 2 8 8 12 2 2

tan tan tan

cos

tan tan tan

70.(B) 2

2 12 2 2 4

2cos x cos x cos x

2 211 2 2 2 2 2 1

2cos x cos x cos x 2 2 1 3

1 2 2 2 2 2 22 2

cos x cos x cos x cos x

71.(B) 2 0x bx c

2

8cos

is the roots of equation

21

1 14 18 2 2 2

cos

cos

21 1 1 1

1 1 02 2 2 2

b c

1 1 1 1

1 2 1 04 2 2 2

b c

1 1

1 2 04 2 2 2 2

b bc

Since b, c Q so comparing rational and irrational parts with zero we get : 1

02 2 2 2

b and

30

2 8

bc

1b ; 1

8c

Vidyamandir Classes


72.(A) Domain of 1x x is 1 1x 1x x ….(1)

Case I For 1 0x , equation (i) holds.

Case II 0 1x ,

1x x or, 2 1x x

or 2 1 0x x or 1 5

02

x ,

1 5

12

x ,

73.(D) 3

2x iy

cos i sin

. . . .(i)

3

2x iy

cos i sin

. . . .(ii)

Multiplying (i) and (ii)

2 2 9

5 4x y

cos

Adding (i) and (ii)

3 2

5 4

cosx

cos

2 2 24 12 94

5 4

cosx x y

cos

2 2 5 5 4

45 4

cosx x y

cos

2 24 3x x y

74.(A) 5 5 3 1

4 46 6 2 2

z cos i sin z i

2 3 2z i

75.(A) 21 1 2 3x . x a a

2

1 1 1 4 1 3x x a a a

For solution

1 3 0a a At 1 3, ,

76.(A) 1 2

1 2

31

3

z z

z z

1 2 1 2

1 2 1 2

3 31

3 3

z z z z.

z z z z

2

1 1 1 2 1 2 23 9z z z z z z z 2 2

1 2 1 2 1 29 3 z z z z z z

2 2 2 2

1 2 1 29 9z z z z 2 21 29 1 0z z

But 2 1z 1 3z

Vidyamandir Classes


77.(A) 1 2 4x x 1 2 2y y

2 3 10x x 2 3 6y y

1 3 6x x 3 1 14y y

1 2 3 10x x x 1 2 3 11y y y

2 4x 2 3y

4 3B ,

Equation of medium 10 37 0x y

78.(B) 3 4 3 3 4 3 0a x y x y

1 2 0L L

3 4 3 0x y

3 4 3 0x y

Point is 1 0,

79.(A) 2 2 14 10 151 0x y x y . . . .(i) ;

Centre 7 5 15, , r

Let 2 7P ,

1 4 49 28 70 151 56 0S

maxD r PS 2 2

7 2 5 7 15 13 15 28 and minD PT ST PS r PS

15 13 2

80.(B) Equation of line (L) is

3

7 5 2 52

2 10 3 15y x

3 2 5x y

Point E 2 3 15 0x y 3 2 5 0x y

E (3, 2)

Point C 5

03

,

Area of circle CEB 1 1 10 46 230 5 3 3 0 0

120 0 2 4 02 2 3 6 3

/

81.(D) Let 1 1 2 2 3 3A r , r , B r , r , C r , r

1 2 32 2OA . OB . OC r r r

On putting point (r, r) in the curve

3 3 23 30 72 55 0r r r r

3 24 30 72 55 0r r r

It has roots 1 2 3r , r , r

Vidyamandir Classes


1 2 3

55

4r r r

55

2OA.OB.OC

4 24

55

OA.OB.OC

82.(C) Equation of line passing through A is 1 1 0px qy qx py

1 1 0px qy qx py

Passed (p, q) 2 2 1 2 1 0 p q pq

2 2 1 2 1 0p q pq

2 2 1

1 2

p q

pq

83.(D) Let the centre the (h, h)

3 6 10 h h 9 10h 10 9h /

line 2 3 x y is tangent to the circle 2 3

4 55

h h

3 3 20 h 20

13

h

20

13

h 23

3h

So centre is 23 23

3 3

,

Equation of circle is

2 223 23

803 3

x y

84.(A) Since given triangle is a right angled isosceles triangle

Angle between them is 1 2

1 2

45 451

m mtan

m m

. . . . .(i)

Where 1 2

1andm m

k

From (i)

1

2

12

11

1 1 11 1

m

mk

km

m

85.(A) 3 4 2 9 0y x , y bx are parallel 2

63 1

bb

4 3 10y x, ay x are perpendicular 1

3 1 3 3 6 18a . aba

86.(A) Let 1 1P x , y be a point in the locus.

Let OAB . Then 1 1y xsin , cos

b a

2 2

2 2 1 12 2

1 1x y

cos sina b

Locus is 2 2

2 21

x y

a b A

B

O

a

b

x1

x1

y1

P(x1, y1)

Vidyamandir Classes


87.(A) 3 3 3 3 3 30 3 0 3

a b c

b c a abc a b c a b c abc

c a b

G.E = 3 3 3

2 2 2 32 2 2 8a b c / abc

a / bc b / ca c / ab

88.(A)

10 10

10 10

f xe

x xe f x log

x x

222

2 2

2

20010

200 100 20 10100 2200 10100 100 2010

100

e

x

x x x xxf log log log f xx xx x x

x

2 2

200 1 200 12 0 5

2 2100 100

x xf f x f x f k .

x x

89.(B) If 0x . Then 2 3f x x x x . If x < 0. Then 2f x x x x

If 0x . Then 2 3 2 3 2 2f x f x f x x x x x x

If x < 0. Then 2 2 3 2 2f x f x f x x x x x x

90.(A) 1 2fofof

0 33fofof

1 2fofof

HPT-3 Chemistry

1.(C) In xyd both lobes is in xy plane so yz and zx are the nodal plane

2.(B) No of hybrid orbitals equal to number of atoms attached + lone pair = 5 + 1 = 6

So, hybridization is 3 2sp d

3.(D) meg of 2 4H SO 200 0.1 2 40

Meg of KOH 200 0.2 1 40 meq

Heat involved 40

57.3 2.292kJ1000

2.3k J

4.(C) 2 1Cl g 2Cl g ; H 242.3 kJ / mol

2 2I g 2I g ; H 151 kJ / mol ; 3ICl g I g Cl g ; H 211.3 kJ / mol

2 2 4I s I g ; H 62.8 kJ / mol

Required equation : 2 2

1 1I g Cl g ICl g ; H ?

2 2

62.8 151 242.3

H 211.32

=16.75 kJ/mol

Vidyamandir Classes


5.(B) 2 2 1s

C O CO x 1

2 2 22 g

1H O H O x 2

2

4 2 2 2 3CH 2O CO 2H O x 3

By (1) + 2 × (2) – (3) we get

(g) 2 4(g)C 2H (g) CH

1 2 3H x 2x x

6.(A) Change in 5 2

oxidation state (Mn) (C)

Coefficients 2 5

7.(C)

Fe

Cl2

Cl Cl Cl

Cl Cl

8.(D) 2H A H HA

51

2

H HAK 1 10

H A

2HA H A

2

102

2

H AK 5 10

H A

Overall dissociation constant = 1 2K=K K

5 10

15

1 10 5 10

5 10

9.(B) H+ + HOCl H2O + Cl

(Electrophile)

10.(D) The solubility of alkaline earth metal chlorides decreases down the group.

11.(B) Nitrogen is more electronegative than ‘H’.

12.(B)

CCl3CH=CH2

OH2

Cl2 CCl3CHCH2

Cl+

.. CCl3CHCH2

HO+H

Cl H+

CCl3CHCH2

OH

Cl

13.(C) The enol form is aromatic in nature.

Vidyamandir Classes


14.(D) 3.6 M solution means 3.6 moles of H2SO4 is present in 1000 ml of solution.

Mass of 3.6 moles of 2 4H SO 3.6 98g 352.8g

ass of 2 4H SO in 1000 ml of solution =352.8 g

Given 29 g of 2 4H SO in present in 100 g of solution

2 429% H SO by mass

352.8 g of 2 4H SO is present in

100352.8 1216g

29 of solution

Now, density Mass 1216

1.216 g/mlVolume 1000

=1.22 g/ml

15.(A) Milli eq. of Na2S2O3 = 24.35 1/10 = 2.435

This would be the milli eq. of I2 and therefore that of Cl2 (which liberates I2 from KI).

Millieq. of Cl2 in 500 ml. = 2.435 500/25 = 48.7

Meq.of Cl2 = meq of bleaching powder = Meq of available Cl2 in the bleaching powder.

% of chlorine = 48.7 35.5

100 30.33%1000 5.7

16.(B) 2)Y(32

)X(

NMgNMg3 ;

Mg3N2 + 6H2O 3Mg(OH)2 + 2NH3 (Y)

CuSO4 Blue colour

17.(A) 18.(B)

Note: chiral centre are marked.

.

.

19.(A) Because solution does not contain Cl+ ion and hence the cyclic chloronium ion is not formed.

20.(C) 2XeF and 3I

are linear in shape

21.(C) PV

ZnRT

9 V 0.082 273 0.9

0.9 V 2.24 L / mole1 0.082 273 9

For 1 millimole, the volume is 2.24 mL

22.(C) Factual

23.(C) Initial pair = 750 – 100 = 650mm

1

1 1 2 2 1 2

VP V P V 650 V P

3

Final Pair = 1950

Final Ptotal =1950 +100 = 2050 mm = 2050 torr

24.(A) Mortar is Ca(OH)2 mixed with silica and water.

25.(B)

148

5

Kh Kw 10h 10

C Kb.C 10 0.1

4h 10

2%h 10 0.01%

Vidyamandir Classes


26.(D) Solubility of salt of AB type is spK

27.(A) Factual

28.(C) h

x.m. v4

2 hx .m x v

4

2 h

x4 m

, h

x4 m

1

2

hx

m

29.(A) For 2A B and 2AB type 3

spK 4s and for 3AB type 3

spK 27s

30.(A) 2 10–4

= 0.5 2

= 2 10–2

[H+] = 0.5 2 10

–2

= 10

–2

Hence, pH = 2

HPT-3 Physics

31.(A) If the breadth of the lake is l and velocity of boat is vb. Time in going and coming back on a quite day

2

Q

b b b

l l lt

v v v .....(i)

Now if va is the velocity of air- current then time taken in going across the lake,

1

b a

lt

v v

[As current helps the motion]

and time taken in coming back 2

b a

lt

v v

[As current opposes the motion]

So 1 2Rt t t 2

2

[1 ( / ) ]b a b

l

v v v

.....(ii)

From equation (i) and (ii)

2

2 2

11 [ 1 1]

[1 ( / ) ]

aR

Q a b b

vtas

t v v v

i.e. R Qt t

i.e. time taken to complete the journey on quite day is lesser than that on rough day.

32.(D) dv dv dx

adt dx dt

2dvv x

dx (Given)

0

02

0

S

v

vdv x dx

0

02 3

02 3

S

v

v x

2 30

2 3

v S

12 303

2

vS

Vidyamandir Classes


33.(A) In this problem it is assumed that particle although moving in a vertical loop but its speed remain constant.

Tension at lowest point

2

max

mvT mg

r

;

Tension at highest point

2

min

mvT mg

r

2

max

2min

5

3

mvmg

T r

T mvmg

r

By solving we get, 4v gr 4 9.8 2.5 98 /m s

34.(B) As the mass of 10 kg has acceleration 12 m/s2 therefore it apply 120N force on mass 20kg in a backward direction.

Net forward force on 20 kg mass = 200 – 120 = 80N

Acceleration280

4 /20

m s .

35.(C) AB BGF f f

( )AB a BG A Bm g m m g

0.2 100 10 = 0.3(300) 10

200 900 1100 N

36.(C) The total time from A to C

Ac AB BCt t t

( / 4) BCT t

where T = time period of oscillation of spring mass system

BCt can be obtained from, sin(2 / ) BCBC AB T t

Putting 1

2

BC

AB we obtain

12BC

Tt

2

4 12 3AC

T T mt

k

.

37.(D)

2 2 2 2

2 22 2 2

u m u PS

g gm m g

38.(D)

If h is the common height when they are connected, by conservation of mass

1 1 2 2 1 2( )A h A h h A A

1 2( )/2h h h [as 1 2A A A given]

Vidyamandir Classes


As (h1/2) and (h2/2) are heights of initial centre of gravity of liquid in two vessels., the initial potential energy of the

system

2 21 2 1 2

1 2

( )( ) ( )

2 2 2i

h h h hU h A g h A gA

...(i)

When vessels are connected the height of centre of gravity of liquid in each vessel will be h/2,

i.e. ( 1 2( )

4

h h [as 1 2( )/2]h h h

Final potential energy of the system

1 2 1 2( )

2 4F

h h h hU A g

21 2( )

4

h hA g

…(ii)

Work done by gravity 2 2 21 2 1 2

1[2( ) ( ) ]

4i fW U U gA h h h h

2

1 2

1( )

4 gA h h

39.(A) Velocity of 50 kg. mass after 5 sec of projection v u gt 100 9.8 5 51 /m s

At this instant momentum of body is in upward direction

50 51 2550 /initialP kg m s

After breaking 20 kg piece travels upwards with 150 m/s let the speed of 30 kg mass is V

20 150 30finalP V

By the law of conservation of momentum

initial finalP P

2500 20 150 30 15 /V V m s

i.e. it moves in downward direction.

40.(A) Let mass A moves with velocity v and collides in elastically with mass B, which is at rest.

According to problem mass A moves in a perpendicular direction and let the mass B moves at angle with the

horizontal with velocity v.

Initial horizontal momentum of system

(before collision) = mv ....(i)

Final horizontal momentum of system

(after collision) = mV cos ....(ii)

From the conservation of horizontal linear momentum

mv = mV cos v = V cos ...(iii)

Initial vertical momentum of system (before collision) is zero.

Final vertical momentum of system sin3

mvmV

Vidyamandir Classes


From the conservation of vertical linear momentum

sin 03

mvmV

sin3

vV

...(iv)

By solving (iii) and (iv)

22 2 2 2(sin cos )

3

vv V

224

3

vV

2

3V v .

41.(B) Let 1 ,m m 2 2 ,m m 3 3 ,m m 4 4m m

1ˆ ˆ0 0r i j

2

3ˆ ˆ ˆcos60 sin602 2

a ar a i a j i j

3

3 3ˆ ˆ ˆ ˆ( cos60) sin602 2

ar a a i a j ai j

4ˆ ˆ0r ai j

by substituting above value in the following formula 1 1 2 2 3 3 4 4

1 2 3 4

3 ˆ0.954

m r m r m r m rr ai a J

m m m m

So the location of centre of mass 3

0.95 ,4

a a

42.(A) For translatory motion the force should be applied on the centre of

mass of the body.

Let mass of rod AB is m so the mass of rod CD will be 2m.

Let y1 is the centre of mass of rod AB and y2 is the centre of mass of

Rod CD. We can consider that whole mass of the rod is placed at

their respective centre of mass i.e., mass m is placed at 1y and mass

2m is placed at 2y .

Taking point ‘C’ at the origin position vector of point 1y and 2y can

be written as 1r = ˆ2l j , 2ˆr l j ,

and 1m m and 2 2m m

Position vector of centre of mass of the system 11 2 2

1 2

cmm r m r

rm m

=

ˆ ˆ2 2

2

m l j ml j

m m

=

ˆ4

3

ml j

mjl ˆ

3

4

Hence the distance of centre of mass from C =4

3l .

43.(A) Due to net force in downward direction and towards

left centre of mass will follow the path as shown in

figure

Vidyamandir Classes


44.(A) Mass of element .2dm xdx

MI of element 2.2 .dI xdx x

32

R

r

I x dx

4 4

2 22

4( )

M R r

R r

=

2 2( )

2

M R r

45.(A) Weight of the rod = W

Reaction of boy 4

B

WR Reaction of man

3

4M

WR

As the rod is in rotational equilibrium 0

02

B M

LR R x

30

4 2 4

W L Wx

6

Lx

Distance from other end, 2

Ly x

y

2

2 6 6 3

L L L L

46.(B) 2

2

2 2 4

1 311

2

gh ghv gh

K

R

47.(C) . .K E U

21 1 1

2eMV GM M

R R h

…(i)

Also 2

eGMg

R …(ii)

On solving (i) and (ii)

2

21

Rh

gR

V

48.(B) Initial length (circumference) of the ring = 2r

Final length (circumference) of the ring = 2R

Change in length = 2R – 2r.

lengthoriginal

length in changestrain

2 ( )

2

R r

r

R r

r

Now Young's modulus / /

/ ( )/

F A F AE

l L R r r

R r

F AEr

49.(B) Force acting on the base

F P A hdgA 30.4 900 10 2 10 7.2N

Vidyamandir Classes


50.(D) When the bob is immersed in water its effective weight = 1m

mg g mg

1effg g

'

eff

T g

T g 1 2

mT

K

51.(B) From Bernoulli's theorem,

2 21 1

2 2A A A B B BP dv dgh P dv dgh

Here, A Bh h

2 21 1

2 2A A B BP dv P dv

2 21[ ]

2A B B AP P d v v

Now, 0,A Bv v r and A BP P hdg

2 21

2hdg dr or

2 2

2

rh

g

52.(C) On heating the system; x, r, d all increases, since the expansion of isotropic solids is similar to true photographic

enlargement.

53.(C) It is given that the volume of air in the flask remains the same. This means that the expansion in volume of the vessel

is exactly equal to the volume expansion of mercury.

i.e., g LV V or g g L LV V

6

4

1000 (3 9 10 )150

1.8 10

g g

L

L

VV cc

54.(C) Since in the region AB temperature is constant therefore at this temperature phase of the material changes from solid

to liquid and (H2 – H1) heat will be absorb by the material. This heat is known as the heat of melting of the solid.

Similarly in the region CD temperature is constant therefore at this temperature phase of the material changes from

liquid to gas and (H4 – H3) heat will be absorb by the material. This heat as known as the heat of vaporisation of the

liquid.

55.(C) ( )f iQ U W U U W 30 ( 40) 10fU 60fU J

56.(B) ABW 0

BCW nR T 6R 1400 8400R

CDW 0

DAW 6R 600 3600R

W 4800R 4.8R kJ 4.8 8.31 kJ = 30 kJ

57.(D) Rate of cooling (here it is rate of loss of heat)

( ) ( )l l c c

dQ d dmc W m c m c

dt dt dt

60 55(0.5 2400 0.2 900)

60

dQ

dt

115

sec

J .

Vidyamandir Classes


58.(C) Frequency of vib. is stretched string 1

2(

Tn

Length) m

When the stone is completely immersed in water, length changes but frequency doesn’t ( unison reestablished)

Hence length T ( 1)

air

water

TL V g

l T V g

(Density of stone = and density of water =1)

1

L

l

2

2 2

L

L l

59.(D) Intensity 22a

Here, 2

1

A

B

a

a and

1

2

A

B

2 22 1 1

1 2 1

A

B

I

I

60.(B) Suppose N resonance occurred before tube coming out.

Hence by using (2 1)

4

N vl

n

(2 1) 3301.5

4 660

N

6N .

HPT-3 Mathematics

61.(D) Since, and are roots of 2 0 px qx r , 0p .

q r

,p p

as p, q and r are in AP. 2 q p r …..(i)

Also, 1 1

4

4

4 4

q r

p p 4 q r

2 4 r p r 9 p r [From (i)]

4 4 4

9 9

q r r

p p r and

1

9 9

r r

p r

2 2

4 16 4 16 36

81 9 81

52

81

213

3

62.(D) Let 32 3 f x x x k

On differentiating w.r.t. x, we get : 26 3 0 f x x , x R

Thus, f (x) is strictly increasing function.

Hence, f (x) = 0 has only one real root, so two roots are not possible.

Vidyamandir Classes


63.(A) Given equation are

2 2 3 0 x x .......(i)

and 2 0 ax bx c ..…(ii)

Since, equation (i) has imaginary roots.

So, equation (ii) will also have both roots same as equation (i).

Thus, 1 2 3

a b c

Hence, : :a b c is 1: 2 : 3.

64.(C) Let z x iy, given Re(z) = 1

x = 1 1 z iy

Since, the complex roots are conjugate of each other

1 z iy and 1 iy are two roots of 2 0 z z

Product of roots =

1 1 iy iy

21 1y 1,

65.(B) We have, 1 1 z z z i

Clearly, z is the circumcentre of the triangle formed by the

vertices (1, 0) and (0, 1) and 1 0 , which is unique.

Hence, the number of complex number z is one.

66.(A) 7

1 A B ,

7

2 A B [ 21 ]

14 A B

2 A B 14 12 2 2[ ]

1 A B

On comparing, we get : A = 1, B = 1

67.(D) Let and be the roots of equation Alternate Solution

2 2 1 0 x a x a Since, 2 a and

Then, 2 a and 1 a Let 22 2 2 f a

Now, 22 2 2

22 2 1 a a 2 2 6 a a

22 2 2 2 1a a 2 2 f a a

2 2 2 2 6 a a For maxima or minima, put 0 f a

22 2 1 5 a 2 2 0 a a = 1

Vidyamandir Classes


The value of 2 2 will be least, if 1 0 a Now, 1 2 0 f

a = 1 So, f (a) is minima at a = 1

68.(C) Given, that 0 z i w z i w

z = i w w i z

and arg(zw) = 2 arg i z

2 arg i arg z

22

arg z

2

arg i

3

4arg z

69.(C) 1 8 2 7 99 910 10 2 11 10 3 11 10 10 11 k. .....

2 911 11 11

1 2 3 1010 10 10

k .... ...(i)

2 9 1011 11 11 11 11

1 2 9 1010 10 10 10 10

k ..... …(ii)

On subtracting equation (ii) from equation (i), we get

2 9 1011 11 11 11 11

1 1 1010 10 10 10 10

k .....

10

10

111

1010 11 1110

1110 101

10

k

10 1011 11

10 10 10 1010 10

k k = 100

70.(A) Since x, y and z are in AP 2 y x z

Also, 1tan x, 1tan y and 1tan z are in AP

1 1 12 tan y tan x tan z

1 1

2

2

11

y x ztan tan

xzy

2 11

x z x z

xzy 2 y xz

Since x, y and z ar in AP as well as in GP. x y z

71.(C) Given 2 4 6 200 a a a ..... a ...(i)

and 1 3 5 199 a a a ....... a ...(ii)

On subtracting equation (ii) from equation (i), we get

2 1 4 3 6 5 200 199 a a a a a a ...... a a

100 d d d ..... times

100 d 100

d

Vidyamandir Classes


72.(D) Number of notes that the person counts in 10 min 10 150 1500

Since, a10, a11, a12,….. are in AP with common difference 2 .

Let n be the time taken to count remaining 3000 notes, then 2 148 1 2 30002

nn

2 149 3000 0 n n 24 125 0 n n

n = 24, 125

Then, the total time taken by the person to count all notes = 10 24 = 34 min

[neglecting n = 125 because for this value of n, a125 will be negative, which is not possible as currency notes cannot be

negative]

73.(C) Given that,

21 2

21 2

p

q

a a ..... a p

a a ...... a q

21

2

1

2 12

2 12

pa p d

p

q qa q d

[where, d is a common difference of an AP]

1

1

2

2

a d pd p

a d qd q 12 0 a d p q 1

2

da

Now, 6 1

21 1

5

20

a a d

a a d

5112

4120

2

dd

dd

74.(B)

2 3

11! 2! 3!

x x x xe . . . .

1 1 1

2! 3! 4! ... 11 1 1

1 12! 3! 4!

... e

75.(B) Since, 13 31 3 2 4 3 1x x, log , log . are in AP.

1 2

13 3 32 3 2 3 4 3 1x xlog log log . 1

3 33 2 3 4 3 1 x xlog log .

13 2 12 3 3x x.

3

2 12 3tt [Let 3x t ]

212 5 3 0t t 3 1 4 3 0t t

1 3

3 4t ,

33

4

x [Since, 3x cannot be negative]

33

4log x

31 4x log

76.(C) 2 3 4

12 3 4

x x xlog x x

2 31 11 1 1 1 12 3

x x x . . . log x log xe e e x

Vidyamandir Classes


77.(C) Let coordinate of the point be ,

Since, , lie on 4 2 0ax ay c and 5 2 0bx by d

4 2 02

ca a c

a

. . . .(i)

Also, 5 2 03

db b d

b

. . . .(ii)

From equations (i) and (ii), 2 3

c d

a b

3 2bc ad

78.(D) Let the coordinate of the centre of T be (0, k)

Distance between their centre

2

1 1 1k k 21 1 1 2k k k

21 2 2k k k 2 21 2 2 2k k k k

1

4k

So, the radius of circle T is k i.e., 1

4

79.(B) So, the sides of a triangle will be 2, 2 and 2 22 2 2 2

x-coordinate of incentre

2 0 2 2 0 2 2

2 2 2 2

. .

2 2 2

2 22 2 2 2

80.(A) 2 2 0x y ax and 2 2 2x y c touch each other.

(i) If circles touch internally,

2 2 2 2

a a a ac c

a c

(ii) If circles touch externally,

2 2 2 2

a a a ac c

c = 0, i.e., not possible as c > 0

Hence, the circles should touch internally and a c .

83.(C) Equation of AB is x, y

Y y dy

X x dx

. . . .(i)

x-intercept is 0dy

x y . ,dx

and y-intercept is 0

dy, y x

dx

As P is mid-point of AB,

Vidyamandir Classes


2dx

x x y .dy

dxy x

dy

0dx dy

x y

Integrating both sides, we get :

log x log y log c

xy c , as it passes through (2, 3).

6 6c xy

82.(C) As, x y a and 1ax y , intersect in I quadrant.

Therefore, x and y intercept are positive.

1

01

ax

a

and

10

1

a ay

a

1 0a and 1 0a a

1a and 1a a . . . .(i)

Case 1 If 1 0a

2 1a [Not possible]

Case 2 If 20 1a a

1a . . . .(ii)

1a or 1a ,

83.(A) 4 7 0sin sin sin

4 7 0sin sin sin 4 2 4 3 0sin sin . cos

4 1 2 3 0sin cos 1

4 0 32

sin , cos

Case I : 4 0sin

As, 0 0 4 4 4 2 3, , …..(i)

Case II : 1

32

cos

As, 0 0 3 3 2 4 5

33 3 3

, ,

…..(ii)

From (i) and (ii) we get : 3 2 4 8

4 2 4 9 9 9, , , , ,

84.(C) 3

2cos cos cos

2 3 0cos cos cos

2 2 2 2 2 22 0cos cos cos sin cos sin cos sin cos

2 2

0sin sin sin cos cos cos

It is possible when, 0sin sin sin and 0cos cos cos

Hence, both statements A and B are true.

Vidyamandir Classes


85.(B) Given that, 1 1

2

ycos x cos

2

1 21 12 4

xy ycos x

2

21 12 4

xy yx cos

2

22 1 1 24

yx cos xy

On comparing both sides, we get : 2 2

2 2 24 1 4

4 44

x ycos x y xycos

2 2 2 2 2 2 24 4 4 4x y x y cos x y xycos

2 2 24 4 4x xycos y sin

86.(A) Given that,

1 1cot cos tan cos x . . . .(i)

We know that,

1 1

2cot cos tan cos

. . . .(ii) 1 1

2cot x tan x

On adding equations (i) and (ii), we get :

122

cot cos x

4 2

xcos cot

12

12

xcot

cosx

cot

2 2

2 2

x xcos sin

cosx x

cos sin

On squaring both sides, we get :

2 2

2 2

22 2 2 2

22 2 2 2

x x x xcos tan sin cos

cosx x x x

cos tan sin cos

1

1

sin xcos

sin x

2

2

112

11

2

tansin x

sin xtan

Applying componendo and dividendo rule, we get : 2

2sin x tan

87.(C) Given, 2n A and 4n B

8n A B

The number of subsets of A B having 3 or more elements 8 8 83 4 8C C . . . C

8 8 8 80 1 22 C C C 256 1 8 28 219 0 12n n n n

nC C . . . C

88.(B) Given, 3n

nT C

11 3

nnT C

11 3 3 10n n

n nT T C C [Given]

Vidyamandir Classes


2 3 3 310 10n n n nC C C C

5n

89.(C) The number of ways in which 4 novels can be selected 64 15C

The number of ways in which 1 dictionary can be selected 31 3C

Now, we have 5 places in which middle place is fixed.

4 novels can be arranged in 4! ways.

The total number of ways = 15 4! 3 = 15 24 3 = 1080

90.(B) Total number of ways

10 10 10 101 2 3 4C C C C

10 45 120 210 385

HPT-4 Chemistry

1.(C) Let the initial pressure of HCN in the following reaction be Po atm.

2HCN(g) H2(g) + (CN)2(g)

Initial pressure Po

Pressure at equilib. Pox x/2 x/2

Given that 2

x = 0.5 x = 1

Kp = 2 2H (CN)

2HCN

(P ) (P

(P ) =

2o

0.5 0.5

(P 1)

= 4 × 10

6.

On solving, Po = 251 atm.

2.(D) For the reaction,

N2O4(g) 2NO2(g)

V).aa(

a4K

22

c

…(i)

where ‘a’ is initial mole of N2O4 present in ‘V’L and is its degree of dissociation.

Also, n

cp )RT(KK

On reducing the volume of container to V

L,2

initial concentration of N2O4 becomes 2a

.V

An increase in concentration

leads to more dissociation of N2O4 in order to have Kc constant, a characteristic constant for a given reaction at a

temperature.

3.(B) 2Na + O2 Na2O2 4.(A) Heating of AlCl3 results in Al(OH)3.

5.(B) NO3 NH4

+5

(n = 8)

Equivalent weight of =8

M3NO

= 8

62 = 7.75.

6.(A) Mg burns in air to react with N2 and gives Mg3 N2.

Vidyamandir Classes


7.(A) 2KOH + CO2 K2CO3 + H2O

m. mol 20 1 0

18 – 1

milli equivalent of acid = milli equivalent of K2CO3 + milli equivalent of KOH = 1 + 18 = 19

N = millequivalent 19

V(ml) 200 0.095 N.

8.(B) MnO2 + 4HCl MnCl2 + Cl2 + H2O

Cl2 + 2KI I2 + 2 KCl

Moles of MnO2 = 1.087

7.8

Moles of Cl2 = 0.1

Moles of I2 = 0.1

Mass of I2 = 0.1 254 = 25.4 gm.

9.(A) 10.(D) Both have same molecular mass.

11.(B)

Mg/ether CH3CN

Br

(A)

MgBr C=NMgBr +

CH3

H3O+

CH3C=O

12.(A) CH3 + D2 O

4

2

NaBD

Hg(OAc)

Actually oxymercuration demercuration reaction occur according to syn addition of D2O without any rearrangement

because of no carbocation formation.

13.(D) 14.(A) 2 4

4

CuCl H / Pd(BaSO )2 2 2

NH Cl2H CH CH C CH CH CH CH CH CH

15.(D)

16.(C) Down the group thermal stability of alkaline earth metal carbonates increases.

17.(B) CaH2 + 2H2O Ca(OH)2 + 2H2

Ca(OH)2 + CO2CaCO3 + H2O

18.(D)

CH3—CH2—CH2—CH2

1CHO

2 3 4 5

(Pentan-1-al)

CH3—CH2—CH—CH3

1CHO

2 3 4

(2-methyl butan-1-al)

19.(B)

CH2—CH==CH—C—H CH2==CH—CH==C—H

O O—H

H

Vidyamandir Classes


20.(A)

CHO

H

H

CH2OH

OH

HO

(I)

CHO

H OH

HOH2 C

(II)

OH

H

CHO

H First

interchange

OH

HO CH2OH

H

Second

interchange

CHO

H

H

CH2OH

OH

HO

(I)

Similarly

CHO HO

H

CH2OH

HO

(III)

CHO

H After two

interchanges at C

OH

H OH

(III)

H

*

*

CH2OH

(I) and (II) are identical; (I) and (III) are diastereomers.

21.(A) C—H is weakest bond so rate is maximum in C6H6.

22.(D) OH Strongly activating group because of +R effect.

Cl Weakly deactivating group because of I effect.

NO2 Strongly deactivating group because of R effect and I effect.

So, 4 > 1 > 3 > 2.

23.(B) N2 has maximum bond energy due NN bond.

24.(C) Out of N and P, N has higher IE, and out of O and S, O has higher IE and out of N and O, N has higher IE, due to

greater stability of the exactly half-filled 2p-subshell.

25.(D) Catalyst

2 2 2 2(steam)

CO H H O CO H

26.(A)

27.(B) Diborane is a dimer of 3BH having 3c-2e bonds, as well as 2c-2e bonds. There are two bridged bonds and four

terminal bonds.

28.(C) E W K.E

K.E E W

12400

4.2 2.0ev2000

19 19K.E 2.0 1.6 10 3.2 10 Joule

29.(D) 30.(A) AB5 has trigonal bypyramidal structure

Vidyamandir Classes


HPT-4 Physics

31.(A) By adjoining graph 0ABW and JWBC 24010]25[108 34

JWWW BCABAC 2402400

Now, JQQQ BCABAC 800200600

From FLOT ACACAC WUQ

240800 ACU .560 JUAC

32.(B) Volume of the gas is constant V = constant TP

i.e., pressure will be doubled if temperature is doubled

02PP

Now let F be the tension in the wire. Then equilibrium of any one piston gives APAPPAPPF 0000 )2()(

33.(A) The distance covered by the ball during the last t seconds of its upward motion = Distance covered by it in first t

seconds of its downward motion

From 2

2

1tguth

2

2

1tgh [As u = 0 for it downward motion]

34.(C) Both the cylinders are in parallel, for the heat flow from one end as shown.

Hence 1 1 2 2

1 2

eq

K A K AK

A A

; where A1 = Area of cross-section of inner cylinder = R

2 and 2A Area of cross-section

of cylindrical shell 2 2 2{(2 ) ( ) } 3R R R

2 21 2 1 2

2 2

( ) (3 ) 3

43eq

K R K R K KK

R R

35.(A) Since 2 22 1( )

2

Lt x x

k

2 2 ( )( )

( )2 2

L L x y x yt x y

k K

36.(B) For observer note of B will not change due to zero relative motion.

Observed frequency of sound produced by A = (330 30)

660 600330

Hz

No. of beats = 600 – 596 = 4

Vidyamandir Classes


37.(A) According to problem

1

2 4

T v

L m L …..(i) and

1 8 3

2 4

T v

L m L

..…(ii)

Dividing equation (i) and (ii), 1

18 3

TT N

T

38.(B) Time taken by particle to move from x=0 (mean position) to x =4 (extreme position)1.2

0.34 4

Ts

Let t be the time taken by the particle to move from x=0 to x=2 cm

2

sin 2 4siny a t tT

1 2sin

2 1.2t

20.1

6 1.2t t s

. Hence time to move from x = 2 to x = 4 will be equal to 0.3 – 0.1 = 0.2 s

Hence total time to move from x = 2 to x = 4 and back again sec4.02.02

39.(B) Loss of weight at 27ºC is

= 46 – 30 = 16 = V1 × 1.24 l × g …(i)

Loss of weight at 42ºC is

= 46 – 30.5 = 15.5 = V2× 1.2 l × g …(ii)

Now dividing (i) by (ii), we get 16

15.5 = 1

2

1.24

1.2

V

V

But 2

1

V

V = 1 + 3 (t2 – t1) =

15.5 1.24

16 1.2

= 1.001042

3 (42º – 27º) = 0.001042 = 2.316 × 10–5

/ºC.

40.(C) Since specific heat = 0.6 kcal/gm °C = 0.6 cal/gm°C

From graph it is clear that in a minute, the temperature is raised from 0°C to 50°C.

Heat required for a minute = 50 0.6 50 = 1500 cal.

Also from graph, Boiling point of wax is 200°C.

41.(B) Let specific gravities of concrete and saw dust are 1 and 2 respectively.

According to principle of floatation weight of whole sphere = up thrust on the sphere

3 3 3 3

1 2

4 4 4( ) 1

3 3 3R r g r g R g

3 3 3 3

1 1 2R r r R

3 3

1 1 2( 1) ( )R r

31 2

31 1

R

r

3 3

1 2 13

1

1

1

R r

r

3 3

1 2 13

1 22

( ) 1

1

R r

r

1 0.3 2.44

2.4 1 0.3

Mass of concrete

Mass of saw dust

Vidyamandir Classes


42.(B) mg N

So c

1.

2

If length decreases by cm/hr then candle moves down by 1 cm/hr.

43.(D)

2L dgl

2Y

2 32

6

(8) 1.5 10 109.6 10 m

2 5 10

44.(B) Let the width of each plate is b and due to surface tension liquid will rise upto

height h then upward force due to surface tension

= 2 cosTb …(i)

Weight of the liquid rises in between the plates

= ( )Vdg bxh dg …(ii)

Equating (i) and (ii) we get , 2 cosT bxhdg 2 cosT

hxdg

45.(B) Let velocities of these masses at r distance from each other be 1v and 2v respectively.

By conservation of momentum

1 1 2 2 0m v m v

1 1 2 2m v m v … (i)

By conservation of energy

Change in P.E.= change in K.E.

2 21 2

1 1 2 2

1 1

2 2

Gm mm v m v

r

2 2 2 21 1 2 2 1 2

1 2

2m v m v Gm m

m m r …(ii)

On solving equation (i) and (ii)

22

1

1 2

2

( )

Gmv

r m m

and

21

2

1 2

2

( )

Gmv

r m m

1 2 1 2

2| | | | ( )app

Gv v v m m

r

46.(B) Linear density of the rod varies with distance dm

dx (Given) dm dx

Position of centre of mass cm

dm xx

dm

3

0

3

0

( )dx x

dx

3

0

3

0

(2 )

(2 )

x xdx

x dx

33

2

0

33

0

3

22

xx

xx

9 9 36 12.

9 21 76

2

m

Vidyamandir Classes


47.(C) Conservation of angular momentum 1 1 2 2 1 2I I I I

Angular velocity of system 1 1 2 2

1 2

I I

I I

Rotational kinetic energy = 21 2

1

2I I

2 2

1 1 2 21 1 2 21 2

1 2 1 2

1

2 2

I II II I

I I I I

48.(A) By the conservation of energy

P.E. of rod = Rotational K.E.

21

sin2 2

lmg I 2

2

32

1sin

2

mllmg

3 sing

l

But in the problem length of the rod 2L is given

3 sin

2

g

L

49.(B) 2 20.3 0.7(1.4 )I x x

For minimum work moment of inertia of the system should be

minimum i.e. 0dI

dx

dI

dx 0.3 2 0.7 2 (1.4 ) 0x x 0.98x m

50.(B) Let two boys meet at point C after time 't' from the starting. Then AC vt , 1BC v t

2 2 2( ) ( ) ( )AC AB BC

2 2 2 2 21v t a v t

By solving we get

2

2 21

at

v v

51.(D) For upper half

2 2 2 / 2 2( sin ) / 2 sinv u al g l gl

For lower half

20 2 (sin cos )

2

lu g

sin (sin cos )gl gl

cos 2sin 2tan

52.(C) Maximum force by surface when friction works 2 2 2 2 2( ) 1F f R R R R

Minimum force R when there is no friction

Hence ranging from R to 2 1R

We get,

2 1Mg F Mg

Vidyamandir Classes


53.(C) Tension the string ( )m g a Breaking force 20( ) 25g a g 2/ 4 2.5 /a g m s

54.(A) v u at 2 2 2 2 cosv u a t u at

200 100 2 10 2 10 cos135v 10 /m s

sin 10sin135

tan 1cos 10 2 10cos135

at

u at

45

i.e. resultant velocity is 10 m/s towards East.

55.(B) 2 2 2(3 ) (2 ) 2 3 2 cosR P P P P …(i)

2 2 2(2 ) (6 ) (2 ) 2 6 2 cosR P P P P …(ii)

By solving (i) and (ii), cos 1/ 2 120

56.(C) ˆ ˆ5 2As v t i t j ˆ ˆx ya a i a j ˆ ˆ5 2i j

ˆ ˆ( )x yF ma i m g a j

2 2| | ( ) 26x yF m a g a N

57.(D) Force acting on plate, dp dm

F vdt dt

Mass of water reaching the plate per sec =dm

dt1 2( )Av A v v 1 2

2

( )V

v vv

( 1 2v v v velocity of water coming out of jet w.r.t. plate)

( A Area of cross section of jet

2

V

v ) 1 2 1 2

2

( ) ( )dm V

F v v v v vdt v

2

1 2

2

( )V

v vv

58.(D) Since the maximum tension BT in the string moving in the vertical circle is at the bottom and minimum tension TT is

at the top.

2B

B

mvT mg

L and

2T

T

mvT mg

L

2

2

4

1

B

B

T T

mvmg

T L

T mvmg

L

or

2

2

4

1

B

T

v gL

v gL

or 2 24 4B Tv gL v gL but

2 2 4B Tv v gL

Vidyamandir Classes


2 24 4 4T Tv gL gL v gL

23 9Tv gL

2 10

3 3 103

Tv g L or 10 / secTv m

59.(B) Angular momentum m(vcos )H

2 2sin 45º. .

22

v vm

g

3

4 2

mv

g

60.(D) At the highest point, vertical component of velocity becomes zero so there will be only horizontal velocity and it is

perpendicular to the acceleration due to gravity.

HPT-4 Mathematics

61.(C) As OA OB OC So O (0, 0) is circumcentre of ABC

1 1

3 3

cos cosG ,

So 1

3 3

cos

1 1cos , sin

Orthocenter 1 1cos , sin

62.(D) 1 1 0L y x i.e. Below the line

2 2 1 0L x y Above and on the line

63.(C) A, B, C lie on the circle of centre 2 3, and of radius 3 and they from equilateral triangle

incentre is 2 3,

64.(B) 1 1 2 21 1 1 8 1 2r , C , , C , , r Two circles are separated to each other.

2y x a, neither touch nor cut circle

1r length of perpendicular

2 1

14 1

a

5 1 5 1a , , . . . .(i)

Similarly 16 1

25

a 15 2 5 15 2 5a , , . . . .(ii)

Also line will be between the circles if their centres 1 2andC C are on the opposite sides of it

2 1 16 1 0a a 15 1a , . . . . (iii)

Now from (i), (ii) and (iii) 15 2 5 5 1a ,

Vidyamandir Classes


65.(C) Point of intersection of 3 0x y and 3 0x y are (3, 0)

Parametric form of lines 3 0 3 0

4 and 41 1 1 1

2 2 2 2

x y x y

Coordinates of point on lines and circle which are at a distance of 4 unit from (3, 0) are

2 2 3 2 2 and 2 2 3 2 2, , these are diametric and points

2 2 3 2 2 3 2 2 2 2 0x x y y

2 2 2 23 8 8 4 2 0 6 4 2 9 0 x y y x y x y

66.(A) Statement-2 is correct and 1 1 0 6 0 3S negative

67.(A) From given functional equation 2y x

f xy f x f y , x, y R

Putting 1y

2 1x

f x f x f

1010 10

10

1 1

3 3 1 33 3 1

3 1 2

r

r r

f r

68.(D) 4

xyf x y f

Putting ; 0y 0f x f . . . .(i)

Putting ; 4 4 0x , f f 0 4f [Using 4 4f ]

Putting ; 2011x in equation (i) 2011 0 4f f

69.(B) 1 2

3 3 13 3

f x x x x x sin x

1 1 2 2

3 3 13 3 3 3

x x x x x x x sin x

1 2

33 3

sin x x x x

3sin x is periodic with period with period 2

Period of 2 1 2

3 3 3f x LCM ,

70.(B) Graph is upward parabola so 0a

0c

x-co-ordinate pf vertex 02

b

a

0b

0b

2 4 2 0f a b c

2 4 2 0f a b c

Vidyamandir Classes


71.(D) are root of 3 25 2 0 x x x

Now 2

2

y

2 1

1

y

y

it is root of given equation so

3 2

3 2

8 1 1 2 15 4 2 0

11 1

y y y.

yy y

3 2 2 3

8 1 20 1 1 2 1 1 2 1 0 y y y y y y

3 23 2 9 8 0y y y

Roots of this equation are 2 2 2

2 2 2

, ,

so product of roots

2 2 2 8

2 2 2 3

72.(B) 1 2 60 80

2 10

. cos cos

sin

11 2 2 80

21

2 10

cos

sin

73.(C) 5 3 0 sin x sin x sin x

2 3 2 3 0 sin x cos x sin x 3 2 2 1 0 sin x cos x

3 0sin x or 1

22

cos x

3 0x , 2 3, , .... 2

03 3

x , , , ....

or 2 4 8 10

23 3 3 3

x , , , ,.....

2 4

3 3 3x , ,

But given 02

x ,

3

x

74.(B) (B)

5 5

4

1 1x x

x x

a af x f x f x

a a

Hence f x is neither even nor odd

(C) 2 1

1 1 1 12 2 21 1 1

x

x x x

x x x x ef x

e e e

2

x odd function

1

1

x

x

e

e

odd function

1 even function

f x odd function x odd function + even function = even function.

(D)

5

g x g xf x f x

f x is odd function

Vidyamandir Classes


75.(B) (i) 2 1f x 1f x

(ii)

10

2

f x

f x

1 2f x , ,

(iii)

1 1 7 77

f x, f x

(iv) 0cos sin f x f x R

Hence value of f x for which g x is defined 7 1 1 1 2 7, , ,

76.(C) Since for 0 1f x x ,

(A) 1 1

10 1 0 44 4

sin x sin xf , sin x

1

2 2sin x ,

102

sin x

0 1x ,

(B) 2

2 2

2 3

2

x xf log

x

2

2 2

2 30 1

2

x xlog

x

2

2

2 31 2

2

x x

x

1

2x ,

(C) 0 1f x x x

(D) f x f sgn x 0 1 0 1x sgn x

x R x x

77.(D) 1 2 11 1nnx x x x ..... x

Log 1 2 11 1nnx log x log x log x ..... log x

Differentiate w.r.t. x

1

1 2 1

1 1 1 1

11

n

nn

nx......

x x x xx

Put x = 2

11

1 2 1

2 2 11 1 1 21

2 2 2 2 1 2 1

nn

n nn

nn........

78.(B) 71 2 61 1x x x x ....... x

Put x = 3

71 2 63 1 2 3 3 3....... a

But 1 63 3 , 2 53 3 & 3 43 3

So 7

1 3 53 1

3 3 3 10932

Vidyamandir Classes


79.(A) 2 2 2 2 2 21 2 2 3 1

1 1 1

n n

......a a a a a a

2 2

12

1n

k kk a a

2 2 4 22

1

1

n

kk a r r

2 2 2 4

2

1 1

1

n

kka r r

12

2 2 2

1 11

1 1 1

nr

a r r

/

/

2 2 2

22 2 2 2

1

1

n

n

r r

a r r

80.(D)

2

2 2

4 4 2

10 6 5 33f

sin sinsin cos

So 21

14

f 1

12

f 1

12

max minf & f

81.(C) 1 1 1 1 1rT r n r

1r n r

21rT n r r

2

1 1 1

1 1 1 2 11

2 6

n n n

n r

n r r

n n n n n nS T n r r

1 2 1 1 21

2 3 6

n n n n n nn

82.(B) 1 1 1

12 3

nH ....n

50

1 1 12 1 2 2 2

2 3 50S ......

50

1 1 12 50 1

2 3 50S . .....

50 50100S H

83.(A)

2 3 1

4 7 10 3 21

5 5 5 5n

nS .......

…….(i)

2 3 1

1 1 4 7 3 5 3 2

5 5 5 5 5 5n n

n nS .......

…….(ii)

Now subtract (i) and (ii)

2

4 3 3 3 21 1

5 5 5 5n

nS ........ ...... n term

13 1

15 54 3 2

115 5

15

n

n

nS

1

4 3 1 3 21 1

5 4 5 5n n

nS

1 1

5 15 1 3 21

4 16 5 4 5n n

nS

.

84.(A) 2b a c & 2c ab

To prove that : c, a, b are in HP

2bc c a c

ab c b c

2ac c

b c

ac baa

b c

Vidyamandir Classes


Hence c, a, b are in HP

(2) Eliminating a from the two equation

2

2c

c bb 2 22c bc b 2 22 0b bc c

2 22 2 0b bc bc c 2 0b b c c b c

2c b or b = c which is not possible

2b a c

4b a

: : 4 : : 2a b c b b b

4 :1: 2

85.(B) 8 4

3 2

8 7 6 4 3336

6 2C C

86.(B) Total number of ways such that no row remains empty

= Total number of ways – those ways in which R2 is empty

– those ways in which R3 is empty

86 6! 6! 6!C .

866! 2 26 6!C .

87.(D) Total number of triangles that can be made by excluding R in the following ways.

(i) Take any two points lying on line 1, and are point lying on 2.

or

(ii) Take any one point on line 1, and two points on line 2.

Hence total number of ways = 1 22nC . nC 2 1n n

88.(B) Point of intersection of 8 unequal circles = 822 C =

8 72

2

= 56

Point of intersection of 4 unequal circles = 422 C = 12

Point of intersection of 4 non coincident = 421 C = 6

Point of intersection of 4 circles and 4 lines = 4 41 12 C C = 32

36

89.(A) x, 1, 2 are boys and y, 3, 4 are girls

No of arrangement 2! 2! 4

90.(B) Since, there are 14 stations in total.

Hence the total number of ways different tickets can be generated is 14

C2 = 91.

Since, these are 75 persons, hence total number way of distribution = 91

C75.

Vidyamandir Classes


HPT-5 Chemistry

1.(B) Acidic solution of Na2S2O3 is unstable due to different oxidation states of sulphur.

2.(A) Iron has more affinity towards oxygen therefore it releases more heat.

3.(B) In isoelectronic species anion is bigger than cation. So Cl

> K+.

4.(C) 2 2 4Be C 2H O 2BeO CH ; 2 2 2 2 2CaC 2H O Ca(OH) C H

5.(B) –CO group shows –R effect and –CH2 group shows +R effect

6.(D) In 13th

group, on moving down the stability of +1 oxidation state increases due to inert pair effect.

7.(A) This one is the Diel’s Alder reaction involving the addition of conjugated alkene with an alkene (dienophile) to give

cyclic product.

NBS/hv

CCl4

Br

A B 8.(A) Anti addition.

9.(B) Peroxide effect means reaction will occur according to free radical mechanism.

10.(A) 2 +6 0 +3

N2H4 + K2CrO4 N2 + Cr(OH)4

nfactor 4 3

Moles of K2CrO4 reacted = 24

194

Equivalents of K2CrO4 reacted = 24 3

194

=

72

194

Equivalents of N2H4oxidised = 72

194

Moles of N2H4oxidised = 72

194 4 =

18

194

Mass of N2H4oxidised = 18 32

194

= 2.97 g.

11.(A) Percentage of metal in its oxide = 60%.

Number of parts of metal that combines with 40 parts of oxygen = 60.

Number of parts of metal that combines with 8 parts of oxygen = 60 8

40

= 12

Equivalent weight of metal = 12.

12.(B) Due to presence of lone pair of electron on nitrogen atom 13.(C)

14.(B) H

ST

H S T (60.01 38.20)273

15.(D) When 2 2H O acts as oxidizing agent 2O is not evolved.

16.(C) In CO2, ‘C’ having it’s maximum oxidation state (4), so it cannot act as a reducing agent.

Vidyamandir Classes


17.(A) meq of acid = meq of Base i.e. M × V × n-factor = M × V × n-factor

(Acid) (Base)

100 0.01 2 = v 0.2 2

v = 5 ml

18.(A) sK Ag X ,

Where X Cl ,Br ,or I

sK

[Ag ][X ]

For iodide :

17 2168.5 10 M

8.5 10 M0.1M

For bromide:

13 2125.0 10 M

5.0 10 M0.1

19.(B) Initial temperature = 127°C = 400 K

Final temperature = 527°C = 800 K

Since KE T, when temperature is doubled, KE is doubled.

20.(B) For photoelectric effect 2

oh h 1/ 2mv

2

01/ 2mv h h

211/ 2mv = 2 hf0 – hf0 = hf0 …(1)

221/ 2mv = 5hf0 – hf0 = 4hf0 …(2)

2n1 1

n2 2

Eq (1) v 1 v 1

v 4 v 2Eq (2)

21.(B) Since entropy is a state function A B A C C D D BS S S S 50 30 20 60eu

B D D BS 20, S 20

22.(B) (s) (aq)NaCl Na (aq) Cl , H 1 kcal/mol

Heat of dissolution = Lattice energy Absorbed + hydration energy evolved

Let heat of hydration of Na and Cl the 6x and 5x respectively 1 kcal) mole 180 ( 6x) ( 5x)

1/ x 179

179

x11

6 179

6x 97.63kcal / mol11

23.(A) Bond length is inversely proportional to bond order

24.(B) 3 3 2 3 2 5CH COOH CH CH OH CH COOC H

Mass 60 gm 46 gm

(t = 0) moles 1 mole 1 mole

(At ) 0.5 mole 0.5 mole 44

gm 0.588

mole

C

0.5

5K 10

0.5 0.5

5 5

Vidyamandir Classes


3 2 5 3 2 5CH COOH C H OH CH COOC H

At t = 0 2moles 1 mole –

2 – x 1 – x x

C

x / 5K 10

(2 x) (1 x)

5 5

x

2(2 x)(1 x)

2x 4 6x 2x

22x 7x 4 0

x 0.72

Mass of 3 2 5CH COOHC H formed

0.72 88 63.3 gm

25.(D) Equation of 2 3Na CO eq. of NaOH eq. of HCl

20 0.1 20 0.15 25 N

N 0.2 and molarity = 0.2

26.(B) Mass of aluminium = 54 gm

Moles of aluminium = 54

27 = 2 moles i.e. A2 N atoms of Al

Mass of same number of Mg atoms = 24 × 2 = 48 gm

27.(C) 2BOH HCl BCl H O

2B H O BOH H

(t = 0) c – –

(At ) c(1 – h) chch

For titration: acid acid base baseN V N V

2 2

V 2.5 V volumeof HCl15 5

V 7.5 ml

In resulting Solution: 2 / 5 2.5

BCl 0.110

2w

b

ch K

1 h K

Or

14w

12b

K 10 1h

K C 1010 0.1

Now, H ch

21H 0.1 3.16 10 M

10

28.(B) 2 2 4

4 4 2

O O CH

CH CH O

r n M

r n M

3 16 16 3

2 32 32 4 2

Vidyamandir Classes


29.(D) H

He

1 1R

y 1 9 He1;

1 1 4y 5 H4R 1

4 9 9

When H x

He

9x

5

30. Error

HPT-5 Physics

31.(C)

23

2 3 2

FW t dt

M

24

3 43

F

W t dtM

2 22 3

2 23 4

3 2 5

74 3

W

W

32.(C) Stress

Strainy

Since stress = 0, so strain = 0

33.(B) 21

cos2

mgR mgR mv

2 2 (1 cosθ) v Rg … (i)

2

cosθ mv

mgR

… (ii)

34.(D) Work done = Change in KE 21

02

mgh F h mv

1

1 10 5 1 6 62

h h

15 18h

1.20h m

35.(C) Fundamental frequency

2

V

n

Beats2 2( )

V V

x

2 ( )

V x

x

22

Vx

36.(A) Stopping distance

2 2

2 2

v vS

a g

37.(C) ω2

v

R

2 2

1

1 1ω

2 2 K I mv

22 2

2

1 12

2 24

vmR mv

R

Vidyamandir Classes


2 22

1

3

4 2 4

mv mvK mv

Velocity of block ω V R3

2 2

V VV

2 2

2

1 9 9

2 4 8

v mvK m

21

22

3 4 3 8 2

4 9 39 8

K mv

K mv

38.(B) Linear momentum conservation

0 mv mv … (1)

Angular momentum conservation

2

0 ω12

mL

mv x … (2)

Point A is at rest

ω

2

LV

0 0 12

2

mv mv xL

M MLx

6

L

x

39.(B) P P C CK U K V

21

022

C

GMm GMmmv

aa,

2 2

2C

GM GMV

a a

40.(B) 2 3T r Kepler’s III

rd law

321

22

2 1

8 64

T R

RT

1 2: : :1:8T T

41.(C) Acc of earth = 2g, acc of mass = g, Relative acc = 3g

2

3

Ht

g

42.(C) Point of contact position will remain unchanged as friction is sufficient.

43.(C) Force of reaction at A is 2A

A A

dPF AV

dt

Force of reaction at B 2B

B B

dPF AV

dt

Net force 2 2( )net B A B AF F F A V V

Applying Bernouli’s equation at A & B, we get 2 2

2 2

1 1( )

2 2 a A a BP V h h g P V gh

2 2 2 2B AV V gh F A gh

Vidyamandir Classes


44.(A) Suppose V’ volume of shell of volume V is hollow, then 1

2B mg

1

ω ( ')5 ω2

V P g V V P g

3

'5

V V

45.(A) Given 5sinπ( 4) y t

5sin 2π 22

ty

Standard equation sin 2πλ

t xy A

y

So 5A

2T

46.(B) At point O wave A. Shows mean position while B shows positive extreme and C negative Extreme

47. Acceleration of system

P

m M

Force on block

MPMa

m M

48.(B) If 1 2&VV are volumes of ball in the upper and lower liquid respectively 1 2VV V

1 1 2 2g V gV g V

Fraction of volume in lower liquid is 2v V

49.(B) Fraction of the ice inside water = 0.9

Fraction of the ice outside water = 0.1

(by laws of flotation).

Minimum length of rope required = (0.1) 10 = 1m

50.(A) The lower plate will rise if spring force on it greater than its weight

2 2kx m g .

Here 2x maximum extension

For this it should be compressed by 1' 'x

2 21 2

1 1 2( )2 2

kx kxm g x x

Hence , 1 21

2m g m gx

k k

F force given to upper plate in downward direction

1 1F m g kx

1 2 F m g m g

Vidyamandir Classes


51.(A) sinR r l

2 2( sin )

tanm R r l

mg g

tan

sin

g

r l

52.(C) Δ

Δ /

PB

V V

Δ Δ

V P

V B

Δ Δ

100 100

P P

P B

9

Δ 1000.1

2 10

P

6Δ 2 10 P

53.(B) Tension at the midpoint of the lower wire = T1 = 1

10.2

2m g a

= 2.9 0.1 9.8 0.2 30N

54.(D) 2

3x t

x = 2 6 9t t

v = 2 6ds

tdt

When v = 0, 2t – 6 = 0, t = 3 sec.

Then v = 0, displacement x = (3 – 3 ) = 0

When t = 0, 1 2 1 0 6 6 /V t m s

When t = 6, 2 2 6 6 6V

L =

2 22 1

1 1

2 2mv mv =

2 216 6

2m

= 0

55.(B) 1 1 2 2

1 2cm

m x m xX

m m

2 2 22

2 2 20

b p.c. a b p.x

a b p

2

2 2 2

cbX

a b

56.(C) For the bead to reach point ‘B’ after it leaves point ‘A; the horizontal path which it travels should be 2R sin .

Then

2 22

u sinR sin

g

Where ‘u’ velocity at ‘A’

Let V velocity at “O”

According to law of conservation of energy 2 2 1 cosV u gR

12 2V gr cos

cos

Vidyamandir Classes


57.(A) This can be assumed as a pure rotation about point of contact, say O, with angular velocity = R

, where R is the

radius of hoop. Speed of P will be P = OP

2R Sin 2

or P= 2R Sin 2

or P= 2 Sin

2

58.(D) The momentum of third part will be equal and opposite to the

resultant of momentum of rest two equal parts

Let V is the velocity of third part.

By the conservation of linear momentum

3 12 2m V m 4 2 /V m s

59.(C) Tension at distance 3

4

L from lower end

1

3

4T

Stress 1

3/

4S

60.(A) Slope is irrelevant hence

1/2

22

MT

K

HPT-5 Mathematics

61.(D) 1

2 2sin cosec sinsin

2 2 1 0sin sin 1sin cosec

Given expression 2 2 4 .

62.(B) log x k y z , log y k z x , log z k x y

x y z x y zlog x y z log x log y log z . . .(i)

= xlog x ylog y zlog z 0k x y z y z x z x y

0 1x y z x y zlog x y z x y z

63.(C) 1sin cos

x xcos sin sin cos

and

2 21 1 cos siny cos

cos cos cos

2sin ycos

Now 2

2

2 2 3

1 1sinx y .

cossin cos cos

and

32

3

sinxy

cos

2 3 2 32 2 1x y xy

O

)

R

= v/R

Vidyamandir Classes


64.(A)

1 2 3

2 1 3 25 0

5 5 4

D System of equations have exactly one solution.

65.(C) 0x x I and 2 4 0x x

2 4 0x x x I

0 4x ,

66. (B) 1

1

1

2 11

11 2

r

r r

r rtanr r

.r r

1 1

1

1

2 1r

r rtan tan

r r

1 1 1 1 1 12 1 3 2 1

3 2 4 3 2 1n

n nlim tan tan tan tan . . . tan tan

n n

1 1

4 2tan

1 1 2 1

1 1 2 3

tan

67.(A) 21 12 1 1sin sin x sin sin x

LHS is defined only for 1

2x But 1 1 1

12

sin sin sin sin

68.(B) 1 11 1S cosec x sec y

x y

0xy let x < 0 and y > 0

11 1

2 06

x cosec x ,x x

and

12y

y 1 1

3 2sec y ,

y

1 11 1

6 2cosec x sec y ,

x y

If x > 0, y < 0

Then 1 1

06

cosec x ,x

and 1 1 2

2 3sec y ,

y

1 11 1 5

2 6cosec x sec y ,

x y

Range = 5

6 2 2 6, ,

which contains the integers 1 and 2.

69.(B) (A) 1 1 2sin x

1cos cos x x when 0 x

(B) 1 2 2 1 2sin sin cos x sin x sin sin cos x

2 1 2 1cos x, cos x

1

2 2sin sin x x ; x

Vidyamandir Classes


(C) 21 2 2x x

1

2 2tan tan x x ; x

Option Not possible

(D) 1 1sin x

3 3 3sin x

1 3 3tan tan sin x sin x (Not possible)

70-71. 70.(C) 71.(A)

70. 10 10 10 7 7 710 75 5P P

7 3 7 3 7 35 5 5a 3 3 35 5 5, ,

8 8 88P

71. 6

1r

n n n

6 2

3

1

6 7 6 75 6 5

2 2r

. .n n .

441 21 30 390

72.(A) 3 7 4P , , AB

4 2AC

Let equation of AC

3 7

2 245 45

x y

cos sin

3 2 7 2 5 9 1 5x, y , , , ,

Similarly equation of BD

3 7

2 2135 135

x y

cos sin

3 2 7 2 1 9 5 5x, y , , , ,

73.(B) Equation of a line through (1, 1) and parallel to 4x y and 2x y

Now area of 1

22

PAB . AB

2 2

3 2 1 1t t 4 2t ,

4 2B , or 2 4,

74-75. 74.(B) 75.(D)

74. Point from which length of tangents to these circle is same as radical centre :

1 2 0 4 4 4 1 0S S x y x y

2 3 0 6 14 10 0S S x y

3 7 5 0x y

3 3 3 0

4 8 0 2 3

x y

y y , x

75. If circle he drawn taking radical centre as centre and length of tangents from radical centre to any circle s radius will

cut all the three circles orthogonally.

Vidyamandir Classes


Length of tangent 9 4 9 4 1 1 27S

Equation of circle 2 2

3 2 27x y

2 24 : 6 4 14 0S x y x y

76.(B) 6 3 9f x f x f x f x . . . .(i)

In equation (i) replace x by (x + 3)

3 9 6 12f x f x f x f x . . . .(ii)

(i) and (ii) 12f x f x

f x is a periodic function having period 12

12 12

0 0

13 1

3f x dx f x dx

12

0

12 4f x dx

77.(C) π 3π 5π 7π

1 cos 1 cos 1 cos 1 cos8 8 8 8

2 2π 3π

1 cos 1 cos8 8

2 2

1 1π 3π1 11 cos 1 cos

π 3π 14 4 2 2sin sin

8 8 2 4 8

78.(C) α,β are roots of 2 0x ax a b

2 2α α and β βa a b a a b

2 2

1 1 2 1 1 20

α α β β a b a b a b a ba a

79.(C) We have 1 1 1 1 1

12 2 3 3 4

S . . . .

S = 1

Given 1 2 1 3 25 2 4 3 3cos x cos x x x

1 2 5cos x and 1 3 22 4 3 2cos x x x

2 5 1x and 3 2 4 3 1x x x 2 4x and 3 2 4 4 0x x x

2 2x , and 1 2 2 0x x x

Finally 2 2x ,

Hence sum 2 2 0

80.(A) 22f x f x x x . . . .(i)

Replace x x

Vidyamandir Classes


22f x f x x x . . . .(ii)

Solving (i) and (ii) we get : 2 3

3

x xf x

23

03

f x f xf f x

0f x or 3f x

2 3

03

x x or

2 33

3

x x 0 3x ; or 2 3 9 0x x (No real roots)

Sum = –3

81.(D) RHS =

20072 4 21 1 1 1 2 1 1

1

x x x x . . . .

x

RHS = 20072 2 21 1 1 1x x . . . . x

=

2 2 20072 2 21 1 1 1x x . . . . x

Hence

2008

20082

2 11 1

11

x

g x . xx

2008

20082 122 2 1 2 1 1g .

2008

2008

2

22 2

2 2 2 22

g . x g

82.(B)

2

21 1

5 6 3 2 2 22

1 4 45 4

n n

n nn n

k k

k k k k nS lim S lim

k k nk k

83.(A) As 1, z , 1z , z cos i sin , 0, .

1 1

1 1

z cos i sin

2 2

1 1

2 11

cos i sin cos i sin

coscos sin

1 1

2 2 2 2 2 2 2

i icot tan

This shows that lies on the line x = 1/2 and 2 2 / arg / , 0Arg . The maximum value of

Arg is never attained.

84.(A) We can choose one denomination in 131C ways, then 3 cards of this denomination can be chosen in 4

3C ways and one

remaining card can be chosen in 481C ways. Thus, the total number of choices is

13 4 481 3 1C C C 13 4 48 2496 .

85.(D) We can permute M, I, I, I, I, P, P in 7!

4! 2!ways. Corresponding to each arrangement of these seven letters, we have 8

places where S can be arranged as shown below with X.

X X X X X X X

We can choose 4 places out of 8 in 84C ways.

Thus, the required number of ways 8 8 64 4 4

7!7

4! 2!

C C C

Vidyamandir Classes


86.(A) 2

: 4 : 5

abab

a b

2 4

5

ab

a b

5

2

a b

ab

5

2

a b

b a

12

2

a,

b

: 4 :1a b or 1: 4

87.(B) From the given equation we have 1 cos a x

sin y

and

1 cos a x

sin y

Multiplying we get : 2 2 2

2 2 2

2 2

1 cos a xx y a

sin y

88.(A) Let A (1, 1) and B(2, 4)

If P(x, y) divides the line segment AB in the ratio 3 : 2, then

3 2 2 1 3 4 2 1

5 5x , y

8 14

5 5P ,

As the line 2x y k passes through P

8 14

2 65 5

k k

89.(D) Since 2

1 5 5 1f

2

2 20 5 2f

2

3 45 5 3f 25f x x for 1 2 3x , ,

25 0f x x has 1, 2, 3 as roots 25 1 2 3f x x a x x x

21 2 3 5f x a x x x x

Now, 2 23 2 2f x x f x x f x x f x x

Product of roots of 2 23 2 0f x x f x x

= (Product of roots, 0f x x ) (Product of roots of 2 0f x x )

= 1 2 3 1 2 3 36

90.(D) Two lines are to a common line if these two lines are parallel.

22

2

2

11

1

pp p

p

21 1 0 1p p p

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