unit test-6 · 2017-06-21 · mentors all india test series 2015 jee advanced pattern (paper-i...

MENTORS ALL INDIA TEST SERIES 2015

JEE ADVANCED Pattern

(Paper-I & Paper-II)

Unit Test-6

Test Date: 30-11-2014

Mentors Eduserv: Plot No.-136/137, Parus Lok Complex, Boring Road Crossing, Patna-1, Ph. No. : 0612-3223680 / 81, 7781005550 / 51

[ 2 ] MAITS 2015_Unit Test-6 (Advanced-Paper-I & Paper-II)_30-11-14

Mentors Eduserv: Plot No.-136/137, Parus Lok Complex, Boring Road Crossing,Patna-1, Ph. No. : 0612-3223680 / 81, 7781005550 / 51

PAPER-1PHYSICS

1. (A, C)

02kx cos mg …(i)

2RF 2kcos x

22kcosm

2. (B, C)2 2v a bx

differentiating both side w.r.t. time

dv dx dv2v 2bx Adt dt dt

(acceleration)

A bx

Hence motion will be S.H.M. 2 b , 2 24 f b ,bf

2

T.E. 21m2

(Amplitude)2 1 a 1m b ma2 b 2

3. (A, C, D)

dBE d l Adt

×

× ×

× × × × × ×

×

R

O

d

P

E

2 2 2E2 x d R k 2

2 2

R kE2 x d

2

ext0

q RW qE dx k4

MAITS 2015_Unit Test-6 (Advanced-Paper-I & Paper-II)_30-11-14 [ 3 ]


4. (A, D)

2d dBE rdt dt

( E = induced emf)

or 2E r × ×

×

×

××

×××

×

×× ×

×× ×

×

×

××r

Hence, choice (A) is correct and choice (B) is wrong.Let i = current in the ring

R = Resistance of loopConsider a small element dl of the ring.

Emf induced in elementEde dl

2 r

and resistance of elementRdR dl

2 r

Potential difference across the element dV de i dR

E E Rdl dl2 r R 2 r

i

dR de

dV zeroor V is constant.So, all points on the ring are at same potential.Hence, choice (C) is wrong and choice (D) is correct.

5. (B, C, D)

t/0qdq edt

rate is different

6. (B, C)

Applying KVL at any instant of time t, 1 2

1 2

q q dILC C dt

when current in the circuit is maximum 1 2

1 2

q qdI 0dt C C

q1 + + + +

– – – –

+ + + +

– – – –

K

q2

L

Potential diff. is same across the capacitors, and since dI 0,dt

induced emf in the inductor

is zero



7. (A, B, C)

rmsR

V 20IR 200

0.2 AA

rmsC

C3 6

V 20I 1X2 5 10 0.2 10

= 0.126 AA

2 2R CI I I 0.24A

8. (A, B, C, D)

LXtan4 R XL = R = 100

tan4

= CX

R

XC= R = 100

2 2X 100 100 100 100

Because XL = XC condition is of resonance.So (D) is also correct

9. (A, C)Work done by force equal to the elastic potential energy stored in the wire.

10. (C)

Net elongation = FL F(L / 2) FLAY AY 2AY

11. (2)Let elongation of spring be x0 in equilibrium. Then,

02T mgsin 2kx …(i)

and T mg …(ii)

Let Block B is displaced by x down the inclinationF.B.D. of B

2T

2k(x0 + 2x) aB

mg sin

B 0ma 2k(x 2x) 2T mgsin …(iii)



F.B.D. of A

T

aA

mg

mg – T = maA

Also, A Ba 2a

BT mg 2ma

B 0 Bma 2kx 4kx 2mg 4ma mgsin

B Bma 4kx 4ma

B4ka x5m

This gives,

5mT 24k

T 2 12. (1)

cm

Rm 2mR52x R

m 2m 6

R

R/2

A y

x

A 1 2 ,

2 2 2

1R R mRm m2 2 2

2 2 22 2mR 2mR 4mR

2A

9I mR2

for compound pendulum T 2Mgd

here M = 3m, 5d R6

9RT 25g



13. (4)

T 2g a

, a is the downward acceleration of box

03gT 2 a

g 4

MgMg R Ma R4

, Mgv4k

14. (3)

Voltage across rod = 20 0 0

1B l2

Charge on capacitor = CV0

1 2

20 R R

1v q CV H H2

2 2

2 2 20 0 0 0 0 R R

1

R1 1CV B l CV H H2 2 R

2 20 0 0 0 0

1 1CV B l CV2 2

= 2R 2 1

1

H [R R ]R

2

2 2 21 1R 0 0 0 0 0 0 0 0 0 0

1 2 1 2

R R1 1 1H CV B l CV CV [B l V ]R R 2 2 R R 2

= 2

01 CV2

15. (9)16. (2)

30 V

I L = 2H R = 20+10

At t = 2 sec. circuit is like as shown in figureBy Kirchoff’s voltage law

dI30 2dtI 2

30

17. (1)Conserving flux in the inductor of inductance L

L . L i 'R

New current after the switch is shifted to 2 is R

which is the steady current in the circuit. So,

current will not change.



18. (4)

2 2c

220 21R X

2 2CR X 220 220 2

222cX 2 220 220 CX 220

Xc 220tan 1R 220

045

n = 419. (1)

1 22 R / L (f f ) 2

20. (4)

1 2W W W

CHEMISTRY21. (A, B, C)

(A) is wrong because zeolites are not used as such but are first heated in vaccum so that the water ofhydration is lost.(B) is wrong, because enzyme have maximum activity at pH of 7.4(C) is wrong because enzyme has maximum activity at 37oC.

22. (A, B, C)Particle size of an adsorbent affects the amount of adsorption.

23. (A, D) 24. (A, B, C)25. (A, B, C, D)

Case (i)

OH

+ KOH

O

2H O K

Case (ii)

O+ : CCl2

OCCl2OO

Case (iii)

OO

CCl2 IMPT

OCHCl2H O



26. (A, C)

OHCH – NH + NO2 2

OH

CH2

OHOH

CH2

27. (C)

2 2 4 2 2YZ X

H C O H O(s) CO(s) CO (s)

28. (A, C, D)29. (B, C, D)

4 6 2 3 3 3 4 3Disproportionation

P O 6H O H PO 3H PO PH

30. (B, C, D)31. (4)

Haemoglobin is positively charged sol. Hence the sol with negative charge can coagulate haemoglobic,i.e., starch, clay, As2S3, CdS.

32. (4)

Mass of HCl acid adsorbed by 10 g charcoal 3548 10 (0.5 0.4) 36.5 2

(Mw of HCl = 36.5 g mol-1)

The amount of adsorptionx 2 4m 0.5

33. (6)

x 1log log K log Pm n

plot of log xm

versus log P is linear with slope = 1n

and intercept = log K

Thus o1 tan tan45 1n or n = 1

log K = 0.301 or K = antilog 0.301 = 2At P = 3 atm

1/n 1x KP 2 (3) 6m



34. (3) 35. (4) 36. (3)37. (4)

5 1Cu

3 2(V.dil)

HNO N O

Changge in O./N. = |5.1| = 4

3 3 2Zn 2HNO Zn(NO ) 2[H] 4

3 2 22HNO 8[H] N O 5H O

3 2 3 2 2 24Zn 10HNO (NO ) 4Zn(NO ) N O 5H O

38. (7)

Crystal of 2 4 7 2Na B O .10H O exist as 2 4 5 4 2Na [B O (OH) ]SH O

B

B

O

O

O

O

B OHHO

B

O

OH

OH

B-O-B bonds = 5(X)B-B bonds = 0 (Y)Sp2 = 2 (2)X + Y + Z = 5 + 0 + 2 = 7.

39. (1)40. (1)

MATHEMATICS41. (A,B)

2 22f x f ' x f x f ' x

2f x f ' x 0 f x f ' x x 1xf x Ae e 2f 1 e .

42. (A, B)

2(y x) (y x) 4xydydx 2y

dy dy x1 ordx dx y

2 2x y 1 0 and x y 25



43. (B, C)Since AP : PB = 4 : 1

5yA(5x,0) and B 0,4

B

AO

P(x, y)

y=f(x)

dy y dy dxat P(x,y), 4dx 4x y x

4 cy

x

Also f(1) = 1 and y = f(x) c = 1 4 1yx

44. (A, C, D)

Area of OABC 1 (OB)(AC)2

, 1 32

z z 3 3OE z2 4 4

E is mid point of AC, so AC = 2AE 9 2 72 1

16 4

1 2 7 7Area of OABC (1)2 4 4

O

C(z )3

B(z )2

A(z )1

3/4

Eand i1

2

z 3 7 ie cos isinz 4

2 31

z zAD z2

2 2 1

1 2 1

3z z z 12AB z | 5z 6z |2 4

2 2

11 1

5z 5z1 1| z | 6 6 14 z 4 z

.

45. (A, B)

i1 2 1 2

3 2 3 2

z z | z z |e

z z | z z |

1 i1

2 2

ww ew | w |

B(z )2A(z )1

C(z )3

2 2i2 2

1 1 1 1

w ww we (cos isin ) Re 1

w w w w

2 2i11 2 2 1 2 1 1 2

2

ww w w w w e ( w i w w sin )

w

21 2 1Re (w w ) w



46. (B, C)

3 2z (1 i)z (1 i)z i 0 2z (z i) z(z i) (z i) 0

2 2(z i)(z z 1) 0 (z i)(z )(z ) 0 2z i, ,

For 994 995 994 995z i, z z 1 ( i) ( i) 1 i 0

994 995 994 995z , z z 1 1 0

2 994 995z , z z 1 0

and 2 are common roots.

47. (A, B, C, D)A necessary and sufficient condition for a triangle having vertices z1, z2 and z3 to form an equilateral

triangle is 2 2 21 2 3 1 2 1 3 2 3z z z z z z z z z

Which follows (A), (B) and (C).

To prove (D) note that z1 + z2 + z3 = 0 can be changed to 1 2 3

1 1 1 0z z z

1 2 3( | z | | z | | z |)

48. (A, B, C, D)Let y = f (x) , then

2 2dyx x y ydx

2

2 2 2xdy ydx xor dx

x x y

22

xdy ydx dxyx 1x

2

ydx dxy1x

Integrating both sides gives 1 ytan x cx

y x tan x f (x) x tan x ( f( ) 0) 49. (C, D)

x

0

1 1f(x)dx x f(x)2 2

31f(x) (x f '(x) f(x)) f(x) cx4

f(2) 2 31 xc f(x)

4 4 .



50. (A, B)1 4 1 4 1 4x sin (a 1) cos (a 1) tan (a 1) is

defined at a = 0 only.

x2 4 4

y = x

A

/4

/4

21Area A2 4 4 32

.

51. (3)Given equation can be written as

2 2 2 22x y x dy 2xy dx y dy x dx 0

2 2 2 2or 2x yd x y y dy x dx 0

Integrating, we get, 22 3 33 x y y x c

52. (0)

Let the roots of 3 2z az bz c 0 be , , , then

1 1 1

= a a b

53. (2)(2 x)dy y dx 0 y m(x 2) Fixed point is P(– 2, 0)

PL 2

P(– 2, 0)

x + y = 0

L

2 2side length

3

23 2 2 2Area

4 3 3

54. (1)

dy = x tan y (sin y – 1) dx dy x dx

tany(siny 1)

cot ycosecy dy x dxcosecy 1

2x2 c2xlog | cosecy 1| log | c | cosecy 1 e

2 2xc

2

1sin y k 1

1 e

.

55. (6)

2 3 1x cx x 0 or xc

Also, graph of y = x2 lies above that of y = cx3 in 10,c

1/ c

2 33

0

2 1 2(x cx )dx or3 312c

1c2

2

1 1 2 4 6c c

.



56. (3)

x x 2 (x – 2)2 = x x2 – 5x + 4 = 0 x = 1 or 4

O

A(4, 2)

B(4, -2)

x = 4 ( | y | x 2 0 x 2) Also | y | = x – 2 y = ± 2

Area bounded by curves 2

2

0

202 ((y 2) y )dy3

57. (4)

Solving the two functions we get x 2 or 3 1

So, required area 2

3 1

32 | x 1| dx(x 1)

1 2

13 1

3 32 (x 1) dx 2 (1 x) dx(x 1) (x 1)

3A 2 ln3 sq. units2

2A 4 3ln3 2A + 3 ln 3 = 4.58. (1)

Let C be the centre of the arc, then C(z )c

Q B(– 4 + 2i)

A(3i)

4

2

i / 2c

c

z 3iBCA e

2 z 2i 4

c

1z (9i 5)2

k 1. 59. (2)

3 7 3 7 7 15 35z | z | | | | | | z | | | ...(i)

Again 5 11 5 11 15 33z 1 | z | | | 1 | z | | | 1 ...(ii)

From (i) & (ii) | z | | | 1

Again 35 2 2

331( ) ( ) 1 i

i or i

The number of ordered pairs is 2.60. (8)

O M

PC(4, 3)

rX

Y

r = 3

Maximum |z| = OP = 5 + 3 = 8.



PAPER-2PHYSICS

1. (B)E = BlVAs the rod is in SHM, velocity of rod varies with time as

V d cos t where is the angular frequency..

Also polarity of ends depends on the direction of magnetic field.2. (D)

2Q 0R

2

1

Q 0Q

3. (B)

Required emf =2 dB( R )

dt 2

2 dBR2 dt

4. (B)Current through the inductor before closing the switch = 1 A

Current through the inductor after closing the switch (in steady state) 20 4A5

L 1.5 Wb 5. (B)

1Z R j L ; 2iZ R

j C

Zeq = 1 2

1 2

1(R j L) (R )Z Z j C

1Z Z 2R j( L )C

2 LRC

Z R

6. (D)

C R3 P

Q

'Q

Q

net2 2

2B (2R) B R 3 B R2 2 2

= 6V



7. The induced emf between two ends of a segment dx = dE = Bvdx

dx

x

v

i

where B = magnetic field due to straight current carrying wire

at the segment dx = 0i2 x

dE = 0ivdx2 x

The induced emf between the ends of the rod

= E = a b

0a

iv dxdE2 x

E = 0iv bln 12 a

(B)8. Let’s assume that speed of the puck at the time of maximum elongation is v then using conservation of

mechanical energy

2 2 20

1 1 1mv mv kx2 2 2

...... (i)

More over using conservation of angular momentum

00 0 0

lmv l mv l10

0vv 11

10 ..... (ii)

Solving (i) and (ii)20

20

mv2100 2100k 121 2100 N/m121 l 121

(A)

9.2

2 20V1 1 1KA 2m KA '2 2 4 2

(by energy and momentum considerations)

2 2 20

1KA mV KA '2

A ' 2A

A ' 2A 54

x A 'sin ' t

K'2m

Kx 2A sin t2m 4

(A)



10. 2mg (0.10 kg) (10 m/s ) 1 N

21/20 0

1 N 1F mg 4x x4 16

mg

F

O x0 x

eXX

kdxdF

0

0

1/2e

0x

dF 1 2k 4 x 2 4 8 N/mdx 2 x

e

n

8k4 50.1mf

2 2 2

(B)

11. The voltage across the resistor and the inductor are 900 out of phase

Hence, 2 2out R LV V V

L C1X L, XC

Hence, 2

2 1Z R LC

sVIZ

and outV2 2 2

2 2L s 2

2

R LI R X V1R LC

2 2 2out

2s 2

V R LV 1R L

C

(B)

12. As gets small, 2

2 1R LC

2 2

1C

2

out2 2

s

V R R CV 1/ C

(A)

13. At terminal velocity 1 2mgv P P v = 1 m/s

(D)14. = Bvl = 0.6 volt

2

1R 0.76 w R1 = 0.47

= Bvl = 0.6 volt2

2R 1.2 w R2 = 0.3

(C)



Q. No. 15 & 16

Ky2T 2mg2

eq

mg 4mgyK K

eqKK4

eq

mg 4mg mgK K 2K

eq2KK9

eq

mg mg mgK 16K 4K

eq16K K5

15. (C) 16. (A) 17. (A)18. (C)

When the switch is connected with ‘a’ long time, current in the circuit would be ER

. So in (a) and (C)

combination, current and equal 0

ER then in (b) and (d) in which current is

0

F2R . Next comparison is

made on the basis of time constant. Shorter time constant means faster decay of the current in (a) and(c), (c) has greater time constant means slower decay of current, corresponding to graph IV.

19. (B) 20. (D)

CHEMISTRY21. (D)22. (B)

The nuclear reaction ism m 12 4z z 6 2A B 3 He

Given, 1/2 0t 10day, t 20day N m g 1g atm 1/2

Tn 2 nt

Amount left in 2 halves = 21 1 g atom

2 4

Amount of A decayed in 2 halves 1 31 g atom4 4

1 g-atom of A gives 3mole of He at STP

3 g atom4

of give 334

mole of He at STP = 94

mole of He at STP

Volume of He at STP = 9 22.4L4 50.4 Litre at STP

23. (D)Enthalpy of adsorption regarding physisorption is not positive. Rather it is negative.



24. (C)

Volume of 2N at STP required to cover the iron surface with monolayer i.e.,

1 18.15mv ml g

Area occupied by singe 2N molecule i.e. 20 216 10S m

20 4 2 16 216 10 10 16 10cm cm

Now, 22400 ml of 2N at NTP contains NA molecules

8.15ml of 2N at NTP contains 23

206.022 10 8.15 2.19 1022400

molecules

Area occupied by single molecule 16 216 10 cm

Area (A) occupied by 202.19 10 molecules at

NTP 20 16 2 12.19 10 16 10 cm g 4 2 135.06 10 cm g

Surface area of the iron adsorbent 2 135.06 m g In short, use the relationship

25. (A)Reactivity order of alcohols towards HCl/ZnCl2 is 3o R–OH > 2o R–OH > 1o R–OH > Me – OH. There fore3o R–OH reacts immediately.

26. (D)

5PCl3R OH R Cl POCl HCl

3 3 33R OH PCl 3R Cl H PO

2 2R OH SOCl R Cl SO HCl

27. (B)Ph

O+ H

Ph

OH

Ph

OH Ph

OH

II

28. (B)

2 6 11 2 2Ca B O 2SO 11H O

(colemanite) 3 2 3 32Ca(HSO ) 6H BO



29. (D)30. (A)

Inert pair effect.31. (B)

1. Adsorption decreases with increases in temperature but

G ve at high temperature

H ve and S ve for, thus T S ve

G H T S ve( v) ve

2. Gases having high critical temperature are easily liquefiable due to higher forces ofattractions among molecules and thus, also show more adsorption.

3. Adsorption extent of an adsorbent is more if its surface area is more.4. Chemisorption requires energy of activation.5. H2O is absorbed in CaCl2 and not adsorbed.

32. (A)2MnO (s)

3 22KClO (s) 2KCl 3O Reset all are facts.

33. (D)

CH — C = CH3 2

CH3

Hg(OAC)2

PhOH CH — C — CH3 2

OPh

CH3

NaBH4

Hg

OAc

CH — C — CH3 3

OPhHI

CH — C—I + Ph — OH3

CH3CH3

CH3

34. (A)

CH — C = CH3 2 CH — C — CH3 2

CH3CH3

O

LiAlH43RCO H CH — C — CH3 3

CH3

OH35. (C)

3 23Zn(NO )3 2 24Zn 10HNO (dil.) 5H O N O



36. (D)

HCl

3 4(from AgNa regia)

Au 3[Cl] AuCl H[AuCl ]

2HCl4 2 6

(from AgNa regia)Pt 4[Cl] PtCl H [PtCl ]

37. (A)(P4) Peptization is a process of conversion of a freshly precipitated substances into colloidal sol byshaking with suitable electrolyte.(Q3) Flocculation is minimum numbers of millimoles of the electrolyte required for complete coagulationof one liter or a colloidal sol.(R 1) Blood is a natural colloid.(S 2) fact

38. (D)39. (A)40. (B)

MATHEMATICS41. (A)

1 20

A 4 (1 x )dx

(0,1)

(1,0)(–1,0)

(0,–1)

83

42. (C)

2 2xdy ydx dxx y

1 yd tan dxx

1 ytan x cx

y x tan(c x)

43. (B)

Required area 3 3 9

22 2

(3,3)

(3,0)

44. (C)

f (x) 2f(x) 3f(x)

( 2 3)xf(x) ke



45. (B)|z – 1| = 1 represent a circle with centre at 1 radius equal to 1.

OPA2

2 z z 2 AParg i0 z 2 z OP

1 2

Ao

P(z)

Now in OAP , APtanOP

z 2 i tan

z

46. (B)

1A 2 12 2 2

(1,0)

(1, )

(0, )

(–1,0) x

y

0

47. (C)

1 1(x 2) i(y 2) | z | cos sin ,2 2

where = argz

2 2 1(x 2) (y 1) | x y |,2

It is Parabola48. (C)

xy 4x 6y 0 (x 6)(y 4) 24 is a rectangular hyperbola and x = – 6 and y = 4 are it'sasymptotes.Area of the region enclosed by a tangent to the rectangular hyperbola xy = c2 and it's asymptotes is = 2c2

Here, c2 = 24Hence, Area = 2 × 24 = 48 sq units.

49. (A)

y xlogx a bx

2 2

dyx y a bx a bx bxdxxx (a bx)

22

32

d y dyx x ydxdx



50. (B)

25 5 5 5 10 10i j 1 2 1 2

1 i j 1002 .... ...

= 0 – 0 = 0

because r r r1 2 100

100 if r 100k.....

0 if r 100k

51. (C)Clearly x = a sin3 t,

y = a cos3 t , (0 t 2 )

23 2 2

0

3S a sin t a 3cos t( sin t)dt a8

52. (B)

22

0S a(1 cos t)a(1 cos t)dt 3 a

53. (A)

21 (y / x)yd dxx x

1sin y / xcx e

54. (B)xy4 dx + ydx – x dy = 0

23 x xx dx d 0

y y

34x 1 x c

4 3 y

55. (D)Here |z – z1| + |z – z2| = |z1 – z2|, so the equation represents a line segment.

56. (D)Given equation represents a hyperbola, if |z1 –z2| > t and pair of rays, if |z1 – z2| = t.

57. (B)58. (D)59. (C)60. (B)

unit test-6 · 2017-06-21 · mentors all india test series 2015 jee advanced pattern (paper-i...

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