NARAYANA I I T / P M T A C A D E M Y

CPT–12

XIS-IIT-IC-SPARK (JEE ADVANCE MODEL 2013-P1)(24.12.2018)

KEY PHYSICS 1. (C) 2. (D) 3. (B) 4. (D) 5 (C) 6. (D) 7. (A) 8. (D) 9. (C) 10. (B) 11. (A,C) 12. (B,D) 13. (A) 14. (A,B,C,D) 15. (A,B,D) 16. (6) 17. (1) 18. (2) 19. (2) 20. (2)

CHMISTRY 21. (C) 22. (A) 23. (B) 24. (D) 25. (B) 26. (C) 27. (C) 28. (C) 29. ()A 30. (A) 31. (A,B,C) 32. (B,C) 33. (A,B,C,D) 34. (B,C,D) 35. (A,B) 36. (5) 37. (5) 38. (4) 39. (4) 40. (9)

MATHEMATICS 41. (A) 42. (C) 43. (B) 44. (A) 45. (A) 46. (A) 47. (B) 48. (A) 49. (A) 50. (A) 51. (A,B,C) 52. (A,B,C) 53. (A,B) 54. (A,B,D) 55. (A,B) 56. (4) 57. (8) 58. (4) 59. (3) 60. (2)

SOLUTION Physics

1. (C) As weight of liquid in capillary is balanced by surface tension, then ghrrT ρππ 1

22 =× (for

uniform r radius tube) gr

Thρ2

1 = ;

but weight of liquid in tapered tube is more than uniform tube of radius r, then in order to balance 1hh <

; gr

Thρ2

<

2. (D) ghagalaT ρρ 220120cos4. =+ ( )hlgaaT −= ρ22.

( )ga

Thlρ2

=−

3. B; m

Kf eff

π21

=

4. (D) ; By kxx =+ 21 and 2211 xkxk = 5. Let P0 be atmospheric pressure and P1 the pressure of air within the sealed tube. Therefore equilibrium pressure just below the meniscus should be equal to atmospheric pressure

because levels of water inside and outside the tube is same.

i.e., 1 02TP pr

− =

or 1 02TP pr

= +

If L=0.11m is the length of tube and x the

Length of immersed part, then from Boyle's law

1 1 2 2p V p V ;= 0 02TP La Pr

= +

( )L x a−

Where a is the cross-sectional area of tube,

i.e., ( )0 02TP L P L xr

× = + −

( ) ( )0 02TP L P L x L xr

× = − + −

o

4TLxP d 4T

=+

6. Energy released 2 2n 4 a 4 b × π − π σ

Now, 3 34 4n ra b3 3

= π or 3

3

bna

=

Therefore, energy released is

3

2 23

b 4 a 4 ba

× π − π

2 b4 b 1a

σ = π − σ

Now, 21 4 b2 3 π

2 2v 4 bρ = πb 1a − σ

or1/2

6 1 1va b

σ = − ρ .

7. Total upward force due to surface tension = ( )1 22 r 2 rσ π + π .

This supports the weight of the liquid column of height h. Weight of liquid column 2 2

2 1h r r g = = π − π ρ Equating,

We get ( )( )2 1 2 1h r r r rπ + − ( )1 2g 2 r rρ = πσ +

or ( )2 1h r r g 2− ρ = σ or ( )2 1

2hr r g

σ=

− ρ.

8. (D) ; At equilibrium, weight of the given block is balanced by force due to surface tension, i.e.,

2L.T = W or 12

025.03.02

105.12

−−

=××

== Nmm

NL

WT

9. (C) ; Work done = Change in surface energy

( )2

12242 RRTW −×=⇒ π ; ( ) ( )[ ] mJJ ππ 4.01035403.02 422 =×−××= −

10. (B) ; 2

21

144rTpand

rTp =∆=∆ ; 21 rr <

21 pp ∆>∆∴ ∴ Air will flow from 1 to 2 and volume of bubble at end 1 will decrease. Therefore, correct option is (B)

13. (A) ; glTl λθ =cos2 ; ( )ayay ≈<< θcos ;

yaT

2λ

=

14. (A, B, C, D) i) Excess pressure inside a spherical liquid drop 2T/R because there in only one free surface here. ii) In case of spherical meniscus of radius of curvature R, also excess pressure = 2T/R because

again there is only one free surface. iii) Excess pressure inside a cylindrical drop of radius R = T/R, hence, for a cylindrical bubble in

air, excess pressure is 2T/R, because there exist two free surface = T/R, in this case iv) For a spherical bubble in water, excess pressure = T/R, as there is only one free surface. Hence, all the four optional are correct.

16. Although not given in the question, but we will have to assume that temperatures of A and B are same.

B B B B B

A A A A A

n p V / RT p Vn p V / RT p V

= =

( )( ) ( )

3A A

3B B

p 4s / r 4 / 3 rp 4s / r 4 / 3 r+ × π

=+ × π

(s=surface tension)

Substituting the values, we get B

A

n 6n

= .

17. 1; Pressure at point A and B are dTPPA

20 += are

ndTPPB /

20 += .

From continuity equation at point A and B. 2

122

2

22

1 nVVndVdV =⇒=

Applying Bernoulli’s equation at point A and B. Hence n = 1 18. From equation of continuity (Av=constant)

( ) ( ) ( ) ( )2 28 0.25 2 v4 4π π

= …... (i)

Here, v is the velocity of water with which water comes out of the syringe (Horizontally). Solving Eq. (i), we get

V=4 m/s The path of water after leaving the syringe will be a parabola. Substituting proper values in equation of trajectory.

2

2 2

gxy x tan2u cos

= θ−θ

According to question, we have,

( )( )

( )( )

2o

2 2 o

10 R1.25 R tan 0

2 4 cos 0− = −

(R=horizontal range) Solving this equation, we get R=2m. 19. Downward force = Buoyant force.

gxaTaMg ρ24 =+ ; gaTaMgx

ρ2

4+=

9

2084.090084.2.0

101010910310704101020

34

233

=+

=×××

××××+××= −

−−− cmm 315.202315. == ; cm

gaaTmgx 3.24

2 =

+=

ρ

20. The bubble will separate from the tube when thrust force due to striking air at B is equal to the force due to excess pressure.

ArTAv

=∴

42ρ (A = area of bubble at B where air strikes)

= 2

4vTr

ρ

CHEMISTRY

argch e separation2aromaticring stabilisedby resonance

⊕Θ

argch e separation2

more stable duetoextra resonanceand has aromatic rings

⊕Θ

argch e separation ⊕Θ arg

one aromatic and oneanti aromatic ring soch e separationis less

c1)

22) (a) With the increase in – I Group, acidic strength increases. 23) (b)

2

3 2 3 2 2 2

5 2H H

CH CH C H CH CH CH CH C Hstabilises by and I of double

hyperconjugation soless stableα α

⊕ ⊕

− − − > = − −−

24) (d) Stability of alkene α No. of Hα or more hyper conjugation

25) (b) acidic strength 1II

α α−+

( )c6)

NHH

MeMeN

H Me

NH2 NH2

Me

> > >

03 02 01 SIP

Due to I+ effect of methyl basic strength increases in 03 amine. ( )c7)

C C

CH3

CH3

CH3

CH3 Has highest C-C bond length because it has maximum hyper conjugation. More single bond

character by hyper conjugation. ( )c8)

N

NN N< <

minleast basicduetoa einversion

' ;I of one Non another− I of alkyl group+

( )minno a einversion ( )a9)

C⊕

Most stable carbocation due to ‘σ ’ resonance. 30) (a) In resonance we cannot change position of atom. 31) ( ) 3,, CHcba − is I+ group. 32) ( ),b c → less no. of Hα so are less stable and high value of heat of hydrogenation.

33) ( ), , ,a b c d 34) (b, c, d)

2

2

2

2

35) ( ),a b

Aromatic

Aromatic

⊕

⊕Θ

Θ

O

36.

O

O+

Θ

O+

Θ

O+Θ O

+

Θ

37.

NH

O

⊕

O

⊕

, , , ,

38. compound contain 3 Hα

10 Hα 6 Hα 6 Hα 4 Hα 39.

Cl C-

Cl

Cl

PH3+

Cl O Cl

, , , P

F FF

40. No. of hyperconjugating structure = no of Hα .

CH3 C+

CH3

CH3

α

α

α

9Hα =

MATHEMATICS 41. (A) ( ) ( )r 2 or r 3 and r 1 or r 5⇒ < − > < > r 2 or r 5⇒ < − >

Also 2 2 11r r 6 r 6r 5 r5

− − ≠ − + ⇒ ≠

42. (C) Given ellipse is ( )2 2x y 1 ... i

16 4+ =

Equation of a circle centered at (1,0) can be written as ( ) ( )2 2 2x 1 y r ... ii− + = The abscissa of the intersection points of the circle and the ellipse is given by the equation

( )2

2 216 xx 1 r4−

− + =

i.e. ( )2 2 24 x 2x 1 16 x 4r− + + − = , i.e. 2 23x 3y 6x 8 0+ − − =

43. (B)

2x 2y 3 2x y 1

5 5 14 1/ 4

− + − + + =

PA PB 2a 2 2 4+ = = × = 44. (A) 46. (A)

47. (B)

48. (A)

50. (A) The given ratio is b 1a 2=

Now 2

22

b 1 3 3e 1 1 ea 4 4 2

= − = − = ⇒ =

51. (a) Equation of the variable circles

(x – h)(x – 2) + (y – k)(y – 1) = 0 x2 + y2 – (2 + h) x – (k + 1) y + k + 2h = 0 As x –intercept = 0

⇒g2 = C ∴( ) hkh 2

42 2

+=+

⇒ (h + 2)2 =4k + 8h ⇒ (h – 2)2 = 4k ∴Locus is (x – 2)2 = 4y ⇒L.R. = 4

(b) N = (2, -1) (c) Figure is a square

∴Area ( )( ) 82

44== sq. Unit

(d) 51. Equation of the variable circles (x – h)(x – 2) + (y – k)(y – 1) = 0 x2 + y2 – (2 + h) x – (k + 1) y + k + 2h = 0 As x –intercept = 0

⇒g2 = C ∴( ) hkh 2

42 2

+=+

⇒ (h + 2)2 =4k + 8h ⇒ (h – 2)2 = 4k ∴Locus is (x – 2)2 = 4y ⇒L.R. = 4

(e) N = (2, -1) (f) Figure is a square

∴Area ( )( ) 82

44== sq. Unit

52. Equation of AB y = 2x – 1 solving it y2 = 4x y2 = 2(y + 1) ⇒ y2 – 2y – 2 = 0 but y1 + y2 + y3 = 0 but y1 + y2 = 2 ∴y3 = - 2 putting in y2 = 4x; x3 = 1 Hence coordinates c are (1, -2) ∴sum of the x and y coordinates of C are –1 ⇒ (a) is correct Obviously normal at C passes through the lower end of the latus rectum⇒ (b) is correct

Again centroid of ΔABC3

321 xxx ++=

Now solving y = 2x – 1 with y2 = 4x (2x – 1)2 = 4x 4x2 – 8x + 1 = 0⇒x1 + x2 = 2; also x3 = 1

∴centroid of the ΔABC ( )0,13

12=

+=

Again equation of the normal at C y + 2 = -(-2/2)(x – 1) y + 2 = x – 1⇒x – y – 3 = 0 hence gradient of chord at C is 1⇒ (d) is incorrect

53. mmmm 3, 21 == and eliminate m. ; Thus, the locus is xy 163 2 = , which is a parabola 54. The equation of normal to xy 42 = is 32 mmmxy −−= . As it passes through (9, 6) 3296 mmm −−=∴ ; ( ) ( ) 061067 23 =−+−⇒=+−⇒ mmmmm ( ) ( ) ( ) 3,2,10231 −=⇒=−+−⇒ mmmm ∴ Normal is 3331223 +−=−=−= xyorxyorxy ; ∴ a, b, d are the correct option. 55. If cmxy += is tangent to 2xy = , then 02 =−− cmxx has equal roots

4

042

2 mccm −=⇒=+⇒ ; 4

2mmxy −=∴ is tangent to 2xy = ;

∴ This is also tangent to ( )22−−= xy ; 444

22

−+−=−⇒ xxmmx

( ) 04

442

2 =

−+−+⇒

mxmx has equal roots.

4,016168 22 =⇒+−=+−⇒ mmmm ; 440 −==⇒ xyory are the tangents. 56.

T : ty = x + t2,t1tan =θ

( )( ) ( )( )ttPNANA 2221

21 2==

A = 2t3 = 2(t2)3/2 i.e. t2 [1, 4] & Amax occurs when t2 = 4⇒Amax = 16 Thus 4X = 16 ⇒X = 4

57. SS1 = T2 (y2 – 4x)(y1

2 – 4x1) = (yy1 – 2 (x + x1))2 (y2 – 4x)(4 + 4) = [2y – 2(x – 1)]2 = 4 (y – x + 1)2 2(y2 – 4x) = (y – x + 1)2 Solving with the line x = 2 we get 2(y2 – 8) = (y – 1)2 or 2(y2 – 8) = y2 – 2y + 1 or y2 + 2y – 17 = 0 Where y1 + y2 = –2 and y1y2 = - 17

Now ( ) 212

212

21 4 yyyyyy −+=+ or ( ) 721744221 =−−=− yy

∴ ( ) 267221 ==− yy Thus, A = 6 B = 2 ⇒A + B = 6 + 2 = 8

59. Conceptual