Answer Key and Solution for Wb Jee 2012

45
Q.No. μ β γ δ Q.No. μ β γ δ Q.No. μ β γ δ 01 D C A C 01 C C D C 01 C D D C 02 A D C D 02 D D A A 02 B A D B 03 C B A B 03 B B C B 03 A B C A 04 A A C C 04 A A A C 04 B D B D 05 D B A D 05 D* C A C 05 B A C B 06 D D B A 06 A D A B 06 B B B D 07 D B B D 07 B C D B 07 D C B B 08 D C A B 08 D C C A 08 C C C C 09 D B D B 09 C A D D 09 B C C B 10 C D B C 10 D B B A 10 B D C B 11 D B D B 11 B C A C 11 A A C D 12 A D D D 12 A C D* A 12 A A B B 13 D D B B 13 C B A A 13 C D A C 14 B B A B 14 D B B A 14 C A C B 15 C C A C 15 C A D D 15 A A A C 16 C C D A 16 D C C C 16 D C C D 17 D D A C 17 A D A D 17 A B B D 18 B B C A 18 C B B B 18 B A A C 19 A C A C 19 A A C A 19 D D B B 20 B D D A 20 A D* C C 20 A B B C 21 D A D B 21 A A B D 21 B D B B 22 B D D B 22 D B B C 22 C B D B 23 C B D A 23 C D A C 23 C C C C 24 B B D D 24 A D C D 24 C B B C 25 D C C B 25 B A D B 25 D B B C 26 B B D D 26 C C B A 26 A D A C 27 D D A D 27 C A A D* 27 A B A B 28 D B D B 28 B A C A 28 D C C A 29 B B B A 29 B A D B 29 A B C C 30 C C C A 30 A D C D 30 A C A A 31 A D C C 31 B C B D 31 D C C D 32 C A D D 32 B D D B 32 D B B A 33 A C B B 33 C C D B 33 C A A B 34 C A C A 34 D D B D 34 B B D D 35 A D D B 35 C B B D 35 C B B A 36 B D A D 36 D B C B 36 B B D B 37 B D D B 37 B D D B 37 B D B C 38 A D B C 38 B D C C 38 C C C C 39 D D B B 39 D B D D 39 C B B C 40 B C C D 40 D B B C 40 C B B D 41 D D B B 41 A D A D 41 C A D A 42 D A D D 42 B A D B 42 B A B A 43 B D B D 43 A D A C 43 A C C D 44 A B B B 44 C B C A 44 C C B A 45 A C C C 45 D B C A 45 A A C A 46 C A C D 46 C A D C 46 C D D C 47 D C D A 47 B D C D 47 B D A B 48 B A B C 48 A D A C 48 A C B A 49 C C A A 49 D B B A 49 D B D B 50 D A B D 50 A C A D 50 B C A B 51 A B D D 51 D A C A 51 D B B B 52 D B B D 52 B A D C 52 B B C D 53 B A C D 53 B C C C 53 C C C C 54 B D B D 54 A D B D 54 B C C B 55 C B D C 55 D C A C 55 B C D B 56 B D B D 56 A A D D 56 D C A A 57 D D D A 57 D B B A 57 B B A A 58 B B D D 58 A A C D 58 C A D C 59 B A B B 59 C C A B 59 B C A C 60 C A C C 60 C D A B 60 C A A A 61 B B C A 61 D C C A 61 C A C C 62 C C B C 62 C B D D 62 B C B A 63 B C A D 63 D A C A 63 B A D B 64 D B B B 64 B A D B 64 A A D D 65 D A C A 65 C D A A 65 B A A A 66 B A B C 66 A A D C 66 A C C C 67 C C C B 67 A C B D 67 C A B B 68 C D B A 68 C C B C 68 A B B D 69 B B D B 69 D D A B 69 A D A D 70 A A D C 70 C C D A 70 A A B A 71 A C B B 71 D B B C 71 C C A C 72 C B C C 72 A A A C 72 A B C B 73 D A C B 73 B D B B 73 B D A B 74 B B B D 74 A C D A 74 D D A A 75 A C A D 75 D C A B 75 A A A B 76 C B A B 76 C B B D 76 C C C A 77 B C C C 77 C A A A 77 B B A C 78 A B D C 78 B B D B 78 D B B A 79 B D B B 79 A D C A 79 D A D A 80 C B A A 80 B A C D 80 A B A A TOP RANKERS ALWAYS FROM AAKASH WB-JEE - 2012 Answer Keys by Aakash Institute, Kolkata Centre MATHEMATICS P H Y & C H E M B I O L O G Y * Assuming screening effect zero.

Transcript of Answer Key and Solution for Wb Jee 2012

Page 1: Answer Key and Solution for Wb Jee 2012

Q.No. μ β γ δ Q.No. μ β γ δ Q.No. μ β γ δ01 D C A C 01 C C D C 01 C D D C02 A D C D 02 D D A A 02 B A D B03 C B A B 03 B B C B 03 A B C A04 A A C C 04 A A A C 04 B D B D05 D B A D 05 D* C A C 05 B A C B06 D D B A 06 A D A B 06 B B B D07 D B B D 07 B C D B 07 D C B B08 D C A B 08 D C C A 08 C C C C09 D B D B 09 C A D D 09 B C C B10 C D B C 10 D B B A 10 B D C B11 D B D B 11 B C A C 11 A A C D12 A D D D 12 A C D* A 12 A A B B13 D D B B 13 C B A A 13 C D A C14 B B A B 14 D B B A 14 C A C B15 C C A C 15 C A D D 15 A A A C16 C C D A 16 D C C C 16 D C C D17 D D A C 17 A D A D 17 A B B D18 B B C A 18 C B B B 18 B A A C19 A C A C 19 A A C A 19 D D B B20 B D D A 20 A D* C C 20 A B B C21 D A D B 21 A A B D 21 B D B B22 B D D B 22 D B B C 22 C B D B23 C B D A 23 C D A C 23 C C C C24 B B D D 24 A D C D 24 C B B C25 D C C B 25 B A D B 25 D B B C26 B B D D 26 C C B A 26 A D A C27 D D A D 27 C A A D* 27 A B A B28 D B D B 28 B A C A 28 D C C A29 B B B A 29 B A D B 29 A B C C30 C C C A 30 A D C D 30 A C A A31 A D C C 31 B C B D 31 D C C D32 C A D D 32 B D D B 32 D B B A33 A C B B 33 C C D B 33 C A A B34 C A C A 34 D D B D 34 B B D D35 A D D B 35 C B B D 35 C B B A36 B D A D 36 D B C B 36 B B D B37 B D D B 37 B D D B 37 B D B C38 A D B C 38 B D C C 38 C C C C39 D D B B 39 D B D D 39 C B B C40 B C C D 40 D B B C 40 C B B D41 D D B B 41 A D A D 41 C A D A42 D A D D 42 B A D B 42 B A B A43 B D B D 43 A D A C 43 A C C D44 A B B B 44 C B C A 44 C C B A45 A C C C 45 D B C A 45 A A C A46 C A C D 46 C A D C 46 C D D C47 D C D A 47 B D C D 47 B D A B48 B A B C 48 A D A C 48 A C B A49 C C A A 49 D B B A 49 D B D B50 D A B D 50 A C A D 50 B C A B51 A B D D 51 D A C A 51 D B B B52 D B B D 52 B A D C 52 B B C D53 B A C D 53 B C C C 53 C C C C54 B D B D 54 A D B D 54 B C C B55 C B D C 55 D C A C 55 B C D B56 B D B D 56 A A D D 56 D C A A57 D D D A 57 D B B A 57 B B A A58 B B D D 58 A A C D 58 C A D C59 B A B B 59 C C A B 59 B C A C60 C A C C 60 C D A B 60 C A A A61 B B C A 61 D C C A 61 C A C C62 C C B C 62 C B D D 62 B C B A63 B C A D 63 D A C A 63 B A D B64 D B B B 64 B A D B 64 A A D D65 D A C A 65 C D A A 65 B A A A66 B A B C 66 A A D C 66 A C C C67 C C C B 67 A C B D 67 C A B B68 C D B A 68 C C B C 68 A B B D69 B B D B 69 D D A B 69 A D A D70 A A D C 70 C C D A 70 A A B A71 A C B B 71 D B B C 71 C C A C72 C B C C 72 A A A C 72 A B C B73 D A C B 73 B D B B 73 B D A B74 B B B D 74 A C D A 74 D D A A75 A C A D 75 D C A B 75 A A A B76 C B A B 76 C B B D 76 C C C A77 B C C C 77 C A A A 77 B B A C78 A B D C 78 B B D B 78 D B B A79 B D B B 79 A D C A 79 D A D A80 C B A A 80 B A C D 80 A B A A

TOP RANKERS ALWAYS FROM AAKASH

WB-JEE - 2012Answer Keys by

Aakash Institute, Kolkata Centre MATHEMATICS P H Y & C H E M B I O L O G Y

* Assuming screening effect zero.

Page 2: Answer Key and Solution for Wb Jee 2012

Q. 1 – Q. 60 carry one mark each

1. The maximum value of |z| when the complex number z satisfies the condition 2z 2z

+ = is

(A) 3 (B) 3 2+ (C) 3 1+ (D) 3 1−Ans : (C)

Hints : 2 2 2z | z | | z | 2z | z | | z |

+ ≥ − ⇒ − ≤ 2| z | 2 | z | 2 0 | z | 3 1⇒ − − ≤ ⇒ ≤ +

2. If ( )50

253 3i 3 x iy2 2

⎛ ⎞+ = +⎜ ⎟⎜ ⎟

⎝ ⎠, where x and y are real, then the ordered pair (x, y) is

(A) (–3, 0) (B) (0, 3) (C) (0, –3) (D)1 3,2 2

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

Ans : (D)

Hints : ( )50 50

503 3 3 ii 32 2 2 2

⎛ ⎞ ⎛ ⎞+ = +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠( )50253 Cos iSin6 6

π π= +25 50 503 Cos iSin

6 6π π⎛ ⎞= +⎜ ⎟

⎝ ⎠

25 1 33 i

2 2⎛ ⎞

= +⎜ ⎟⎜ ⎟⎝ ⎠

3. If z 1z 1−+

is purely imaginary, then

(A)1| z |2

= (B) | z | 1= (C) |z| =2 (D) |z|=3

Ans : (B)

Hints : Let z 1 iz 1−

= λ+

, such that Rλ∈ .

On taking Componendo and Dividendo

1 iz1 i+ λ

=− λ

; 1 iz 11 i+ λ

= =− λ

Code-δδδδδWBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(1)

ANSWERS & HINTSfor

WBJEE - 2012SUB : MATHEMATICS

Page 3: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(2)

4. There are 100 students in a class. In an examination, 50 of them failed in Mathematics, 45 failed in Physics, 40 failedin Biology and 32 failed in exactly two of the three subjects. Only one student passed in all the subjects. Then thenumber of students failing in all the three subjects

(A) is 12 (B) is 4

(C) is 2 (D) cannot be determined from given information

Ans : (C)

Hints : n(M P B) n(M) n(P) n(B) n(M P) n(PMB) n(B P) n(M P B)∪ ∪ = + + − ∩ − − ∩ + ∩ ∩

given n(M) = 50 , n (P) = 45, n(B) = 40; n(M P) n(P B) n(B M) 3(M P B) 32⇒ ∩ + ∩ + ∩ − ∩ ∩ =

99 = 50 + 45 + 40 – (32 + 3 n (M P B)∩ ∩ )+ n(M P B)∩ ∩ ; 2 n(M P B)∩ ∩ = 36 – 32; n(M P B)∩ ∩ = 2

5. A vehicle registration number consists of 2 letters of English alphabets followed by 4 digits, where the first digit is notzero. Then the total number of vehicles with distinct registration numbers is

(A) 2 426 10× (B) 26 102 4P P× (C) 26 10

2 3P 9 P× × (D) 2 326 9 10× ×

Ans : (D)

Hints : 26 26 9 1010 10

. Total No. of ways = 262 × 9 × 103

6. The number of words that can be written using all the letters of the word ‘IRRATIONAL’ is

(A) 310!

(2!) (B) 210!

(2!) (C) 10!2!

(D) 10!

Ans : (A)

Hints : IRRATIONAL. There are 2 I, 2 R, 2 A, one T, one O, One N, One L. Total Number of words = 310!

(2!)

7. Four speakers will address a meeting where speaker Q will always speak after speaker P. Then the number of waysin which the order of speakers can be prepared is

(A) 256 (B) 128 (C) 24 (D) 12

Ans : (D)

Hints : The number of ways in which speaker P and Q can deliver their speach is 4C2.

Total number of ways = 4C2 × 2! = 12

8. The number of diagonals in a regular polygon of 100 sides is

(A) 4950 (B) 4850 (C) 4750 (D) 4650

Ans : (B)

Hints : Number of Diagonals in n sided Polygon is equal to nC2 – n. Total number of Diagonals = 100C2 – 100=4850

9. Let the coefficients of powers of x in the 2nd, 3rd and 4th terms in the expansion of (1+x)n, where n is a positive integer,be in arithmetic progression. Then the sum of the coefficients of odd powers of x in the expansion is

(A) 32 (B) 64 (C) 128 (D) 256

Ans : (B)

Hints : Co-efficient of 2nd, 3rd and 4th terms are nC1, nC2,

nC3. It is given 2.nC2 = nC2 + nC3

⇒ n = 7.

Sum of Co-efficient of odd power of x = 26

Page 4: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(3)

10. Let f(x) = ax2 + bx + c, g(x) = px2 + qx + r, such that f(1) = g(1), f(2) = g(2) and f(3) – g(3) = 2. Then f(4) – g(4) is(A) 4 (B) 5 (C) 6 (D) 7Ans : (C)Hints : Let a – p = w, b – q = y and c – r = z⇒ f(1) = g(1) ⇒ w + y + z = 0; f(2) = g(2) ⇒ 4w + 2y + z = 0; f(3) = g(3) + 2 ⇒ 9w + 2y + z = 2 ⇒ w = 1, y = –3, z = 2f(4) – g(4) = 16 w + 4y + z = 6

11. The sum 1 × 1! + 2 × 2! + .... + 50 × 50! equals(A) 51! (B) 51! – 1 (C) 51! + 1 (D) 2 × 51!

Ans : (B)

Hints : Let 50

r 1S r.r !

== ∑ ;

50

r 1(r 1 1)r!

== + −∑ ;

50

r 1(r 1)! r !

== + −∑ ; 51! 1= −

12. Six numbers are in A.P. such that their sum is 3. The first term is 4 times the third term. Then the fifth term is(A) –15 (B) –3 (C) 9 (D) –4Ans : (D)Hints : Let α–5d, α–3d, α–d, α + d, α + 3d, α + 5d are six numbers in A.P.

⇒ 6 α = 3 ⇒α = 12

; (α – 5d) = 4 (α – d) ⇒d = 32

− . Hence 1 33d 3 42 2

−α + = + × = −

13. The sum of the infinite series 1 1.3 1.3.5 1.3.5.71 ....3 3.6 3.6.9 3.6.9.12

+ + + + + is equal to

(A) 2 (B) 3 (C) 32

(D) 13

Ans : (B)

Hints : Let ( ) n 1 1.3 1.3.5 1.3.5.71 x 1 ....3 3.6 3.6.9 3.6.9.12

−− = + + + + +

21 n(n 1) 1.3 2 1nx , x x ,n3 2 3.6 3 2

+= = ⇒ = =

14. The equations x2 +x +a=0 and x2 +ax +1=0 have a common real root(A) for no value of a (B) for exactly one value of a(C) for exactly two values of a (D) for exactly three values of aAns : (B)Hints : x2 + x + a = 0 ..............(A); x2 + ax +1 = 0 ..............(B)⇒ (A) – (B);(1–a)x + (a–1) = 0; (1–a) (x–1) = 0 if a ≠ 1, x = 1, so, only solution is a = –2

15. If 64, 27, 36 are the Pth , Qth and Rth terms of a G.P., then P + 2Q is equal to(A) R (B) 2R (C) 3 R (D) 4 RAns : (C)Hints : tP = 64, a . rP–1 = 64.............(1); tQ = 27, a . rQ–1 = 27.............(2) tR = 36, a . rR–1 = 36.............(3)(2)2 × (1) ÷ (3)3 ; 2 Q + P = 3R

Page 5: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(4)

16. If ( )α + β and ( )α − β are the roots of the equation x2 + px + q = 0 where , ,α β p and q are real, then the roots of

the equation (p2– 4q) (p2x2 + 4 px) – 16 q = 0 are

(A) 1 1 1 1and

⎛ ⎞ ⎛ ⎞+ −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟α αβ β⎝ ⎠ ⎝ ⎠

(B)1 1 1 1and⎛ ⎞ ⎛ ⎞

+ −⎜ ⎟ ⎜ ⎟β βα α⎝ ⎠ ⎝ ⎠

(C) 1 1 1 1and

⎛ ⎞ ⎛ ⎞+ −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟α β α β⎝ ⎠ ⎝ ⎠

(D) ( ) ( )andα + β α − β

Ans : (A)Hints : 2 α = – P, α2 − β = q; so the given equation is 4 β(4α2x2 – 8 α x) – 16 α2 + 16β =0

( )2 2 2x (2 )x ( ) 0α β − αβ + β − α = ; 1 1x = ±α β

17. The number of solutions of the equation log2(x2 + 2x –1) = 1 is

(A) 0 (B) 1 (C) 2 (D) 3Ans : (C)Hints : x2 + 2x –1 = 2 ⇒ x2 + 2x – 3=0 ⇒ (x + 3) (x – 1) = 0; x = 1, –3

18. The sum of the series n n n1 2 n

1 1 11 C C ..... C2 3 n 1

+ + + ++

is equal to

(A) n 12 1n 1

+ −+

(B) ( )n3 2 1

2n

−(C)

n2 1n 1++

(D) n2 12n+

Ans : (A)

Hints : (1+x)n = C0 + C1x + C2x2 + ...............+ Cnx

n; Integrating 1 1n 2 n

0 1 2 n0 0(1 x) dx (C C x C x ..... C x )dx+ = + + + +∫ ∫ ;

n 131 2 n

0CC C C2 1 C ..........

n 1 n 1 2 2 3 n 1

+− = + + + + +

+ + +

19. The value of r 2

1 2 .... (r 1)r !

=

+ + + −∑ is equal to

(A) e (B) 2e (C)e2

(D) 3e2

Ans : (C)

Hints : r 2 r 2

r(r 1) 1 1 1 1 1 ........2 r 2 r 2 2 0 1 2

∞ ∞

= =

⎛ ⎞−= = + + + +⎜ ⎟− ⎝ ⎠

∑ ∑ e2=

20. If 1 2 1

P1 3 1⎡ ⎤

= ⎢ ⎥⎣ ⎦

, Q = PPT, then the value of the determinant of Q is equal to

(A) 2 (B) –2 (C) 1 (D) 0Ans : (A)

Hints : T

1 11 2 1 6 8

PP 2 31 3 1 8 11

1 1

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

. So det (PPT) = 66–64=2.

Page 6: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(5)

21. The remainder obtained when 1!+2!+...+95! is divided by 15 is(A) 14 (B) 3(C) 1 (D) 0Ans : (B)Hints : Starting from 5! all the other numbers are divisible by 15 since 15 will be a factorSo 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33So the remainder on dividing 33 by 15 is 3.

22. If P,Q,R are angles of triangle PQR, then the value of

1 cosR cosQcosR 1 cosPcosQ cosP 1

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−⎣ ⎦

is equal to

(A) -1 (B) 0 (C)12

(D) 1

Ans : (B)Hints : On expanding the determinant,Δ = – 1 + cos2p + cos2Q + cos2R + 2cosP cosQ cosR= – 1 cos2 P + cos2Q + cos2Q + cos2R + (cos(P+Q) + cos(P – Q)) cosRnow P + Q = π – RSo Δ = – 1+ cos2P + cos2Q + cos2R – cos2R – cos(P – Q) cos(P + Q)= – 1 + cos2P + cos2Q – cos2P + sin2Q= – 1 + 1 = 0

23. The number of real values of α for which the system of equations x + 3y+5z = α x 5x+y+3z = α y 3x+5y+z = α z

has infinite number of solutions is(A) 1 (B) 2 (C) 4 (D) 6Ans : (A)

Hints :

x(1 ) 3y 5z 05x (1 )y 3z 03x 5y (1 )z 0

− α + + =+ − α + =+ + − α =

So for infinite solution,

1 3 55 1 3 03 5 1

− αΔ = − α =

− α 1 1 2 3

9 3 59 1 3 0 C C C C9 5 1

− αΔ = − α − α = → + +

− α − α

making two zeroes,

0 2 20 4 2 0

9 5 1

+ α− α− + α =

− α − α

⇒ (9 – α) (α2 + 4α + 4 + 2α + 8) = 0⇒ (9 – α) (α2 + 6α + 12) = 0⇒ α = 9

24. The total number of injections (one-one into mappings) from { }1 2 3 4a ,a ,a ,a to { }1 2 3 4 5 6 7b ,b ,b ,b ,b ,b ,b is

(A) 400 (B) 420 (C) 800 (D) 840Ans : (D)Hints : So total = 7 × 6 × 5 × 4 = 840 one-one into functions

Page 7: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(6)

25. Let 10

10 rr

r 0

(1 x) c x=

+ = ∑ and 7

7 rr

r 0

(1 x) d x=

+ =∑ If 5 3

2r 2r 1r 0 r 0

P c and Q d += =

= =∑ ∑ , then PQ is equal to

(A) 4 (B) 8 (C) 16 (D) 32Ans : (B)

Hints : 10 10 10 10 10 100 2 4 6 8 10P C C C C C C= + + + + +

= 210–1 = 29

Q = 7 7 7 7 7 1 61 3 5 7C C C C 2 2−+ + + = =

So 9

6

2P 8Q 2= =

26. Two decks of playing cards are well schuffled and 26 cards are randomly distributed to a player. Then the probabilitythat the player gets all distinct cards is(A) 52C26 /

104C26 (B) 2 X 52C26 / 104C26

(C) 213 x 52C26 / 104C26 (D) 226 x 52C26 /

104C26

Ans : (D)

Hints : Total no. of way of distribution = 10426C

Favourable no. of ways of ditribution

= 52 2626C 2× (Any one card can be from any one deck)

So probability = 26 52

26104

26

2 CC

×

27. An urn contains 8 red and 5 white balls. Three balls are drawn at random. Then the probability that balls of bothcolours are drawn is

(A)40

143 (B)70

143 (C) 3

13 (D) 1013

Ans : (D)

Hints : Total no. of selection = 133C

So probability = 8 5 8 5

1 2 2 113

3

C C C C 1013C

× + ×=

28. Two coins are avilable, one fair and the other two-headed. Choose a coin and toss it once; assume that the unbiased

coin is chosen with probalility 34

. Given that the outcome is head, the probability that the two-headed coin was

chosen is

(A)35 (B)

25 (C)

15 (D)

27

Ans : (B)Hints : B is the event of selecting a biased coinUB is the event of selecting an unbiased coin

3 1P(UB) , P(B)4 4

= =

B P(B H)PH P(H)⎛ ⎞ =⎜ ⎟⎝ ⎠

( )( ) ( )

HP(B).P BH HP(B).P P(UB).PB UB

=+ =

1 1 241 3 1 514 4 2

×=

× + ×

Page 8: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(7)

29. Let be the set of real numbers and the funtions f: → and g: → be defined by f(x)=x2 + 2x-3 and g(x) =x+1. Then the value of x for which f(g(x)) = g(f(x)) is(A) -1 (B) 0 (C) 1 (D) 2Ans : (A)Hints : fog(x) = (x + 1)2 + 2(x + 1) – 3 = x2 + 4x= x2 + 4xgof(x) = x2 + 2x – 3 + 1 = x2 + 2x – 2fog(x) = gof(x)⇒ x2 + 4x = x2 + 2x – 2⇒ x = – 1

30. If a,b,c are in arithmetic progression, then the roots of the equation ax2 - 2bx + c = 0 are

(A) 1 and ca (B) -

1a and -c (C) -1 and -

ca (D) -2 and -

c2a

Ans : (A)Hints : a – 2b + c = 0So x = 1 is a root

Now product of roots = ca

So the other root is = ca

31. If sin-1x + sin-1y + sin-1z = 32π

, then the value of x9 + y9 + z9 - 9 9 91

x y z is equal to

(A) 0 (B) 1 (C) 2 (D) 3Ans : (C)

Hints : 1 1 1sin x sin y sin z2

− − − π= = =∵

⇒ x = y = z = 132. Let p,q,r be the sides opposite to the angles P,Q,R respectively in a triangle PQR, if r2sin P sin Q = pq, then the

triangle is(A) equilateral (B) acute angled but not equilateral(C) obtuse angled (D) right angledAns : (D)

Hints : 1p q r 2R

sinP sinQ sinR= = =∵ (R1 : Circum radius)

2r sin PsinQ pq=∵

21 1 1(2R sinR) sinPsinQ (2R sinP)(2R sinQ)⇒ =

2sin R 1⇒ = 1 1[ P Q 0, R 0]≠ ≠∵

sin R 1 [sinR 1]⇒ = ≠ −

0R 90⇒ =

Page 9: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(8)

33. Let p,q,r be the sides opposite to the angles P,Q,R respectively in a triangle PQR. Then 2pr sin P Q R

2− +⎛ ⎞

⎜ ⎟⎝ ⎠

equals

(A) p2+q2+r2 (B) p2+r2-q2 (C) q2+r2-p2 (D) p2+q2-r2

Ans : (B)

Hints : P Q R2pr sin

2− +⎛ ⎞

⎜ ⎟⎝ ⎠

∵ = 2pr sin Q2π⎛ ⎞−⎜ ⎟

⎝ ⎠ [ ]P R Q+ = π −∵

= 2pr cosQ = p2 + r2 – q2

34. Let P ( )2, 3− , Q ( )2,1− be the vertices of the triangle PQR. If the centroid of ΔPQR lies on the line 2x+3y = 1, then thelocus of R is(A) 2x + 3y=9 (B) 2x - 3y =7 (C) 3x + 2y=5 (D) 3x - 2y=5Ans : (A)Hints : Let R(h, k)

∴ Co-ordinate of centroid G ≡2 2 h 3 1 k h k 2, ,

3 3 3 3− + − + + −⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

∵ G lies on the line

h k 22 3 13 3

−⎛ ⎞ ⎛ ⎞⇒ + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⇒ 2h + 3k – 6 = 3 ⇒ 2h + 3k = 9

35.x

x 0

1lim1 x 1→

π −

+ −

(A) does not exist (B) equals loge(π2) (C) equals 1 (D) Lies between 10 and 11

Ans : (B)

Hints : x

x 0

1Lt1 x 1→

π −

+ −

= ( )x

x 0 x 0

1Lt Lt 1 x 1x→ →

π −× + +

= ( )elog 2π × = 2logeπ = loge(π2)

36. If f is a real-valued differentiable function such that f(x) f’(x) < 0 for all real x, then(A) f(x) must be an increasing function (B) f(x) must be a decreasing function

(C) f(x) must be an increasing function (D) f(x) must be a drecreasing function

Ans : (D)Hints : f(x) f′(x) < 0⇒ f(x) < 0 ; f′(x) > 0 or f(x) > 0 ; f′(x) < 0 ............... (ii)

So possible graphs of f(x) are

|f(x)| is decreasing function in both cases.37. Rolle’s theorem is applicable in the interval [-2,2] for the function.

(A) (A) f(x) = x3 (B) f(x) = 4x4 (C) f(x) = 2x3 +3 (D) f(x) = xπ

Ans : (B)Hints : For f(x) = 4x4 ⇒ f(–2) = f(2) & f′(0) exist

Page 10: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(9)

38. The solution of 2

2d y dy25 10 y 0,

dxdx− + = y(0) = 1, y(1) = 2e1/5 is

(A) y = e5x + e–5x (B) y = (1 + x)e5x (C)

x5y (1 x)e= + (D)

x5y (1 x)e

−= +

Ans : (C)Hints : Let y = emx is a trial solution

dy mdx

⇒ =

225m 10m 1 0⇒ − + =

1m (equal roots)5

⇒ =

⇒ The differential solution x5y (A Bx)e= + (A & B are two arbitrary constant)

Initially y (0) = 1 0(A 0) e 1 A 1⇒ + = ⇒ =

y (1) = 2 e1/5 1 15 5(A B)e 2e⇒ + =

⇒ B = 2 – A = 2 – 1 = 1∴ Required solution y = (1 + x)ex/5

39. Let P be the midpoint of a chord joining the vertex of the parabola y2 = 8x to another point on it. Then the locus of Pis

(A) y2 = 2x (B) y2 = 4x (C) 2

2x y 14+ = (D)

22 yx 1

4+ =

Ans : (B)

Hints : Let P(h, k) be the midpoint of chord AB through vertex (0, 0) of parabola y2 = 8x

∴ Co-ordinate of B (2h 0, 2k 0) (2h, 2k)≡ − − =

∵ B (2h, 2k) lies on y2 = 8x⇒ 4k2 = 8 × 2h⇒ k2 = 4h

40. The line x = 2y intersects the ellipse 2

2x y 14+ = at the points P and Q. The equation of the circle with PQ as

diameter is

(A) 2 2 1x y2

+ = (B) x2 + y2 = 1 (C) x2 + y2 = 2 (D) 2 2 5x y2

+ =

Ans : (D)

Hints : Solving the given equations then common points are P 12,2

⎛ ⎞⎜ ⎟⎝ ⎠

and Q 12,2

⎛ ⎞− −⎜ ⎟⎝ ⎠

∴ Eqn. of Circle PQ as

diameter 1 1(x 2)(x 2) (y )(y )2 2

− + + − + = 0 ⇒ x2 - 2 + y2 - 12

= 0 ⇒ x2 + y2 = 52

Page 11: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(10)

41. The eccentric angle in the first quadrant of a point on the ellipse 2 2x y 1

10 8+ = at a distance 3 units from the centre of

the ellipse is

(A) 6π

(B)4π

(C) 3π

(D)2π

Ans : (B)

Hints : ( )P 10 cos , 2 2 sin ,θ θ ∵centre (0, 0)

OP 3⇒ = (given)⇒ 10cos2θ + 8sin2θ = 9 [apply distance formula]

∴θ = as angle in 1st quadrant

42. The transverse axis of a hyperbola is along the x-axis and its length is 2a. The vertex of the hyperbola bisects the linesegment joining the centre and the focus. The equation of the hyperbola is(A) 6x2 – y2 = 3a2 (B) x2 – 3y2 = 3a2 (C) x2 – 6y2 = 3a2 (D) 3x2 – y2 = 3a2

Ans : (D)

Hints : aeVertex at , 02

⎛ ⎞⎜ ⎟⎝ ⎠

aea e 22

= ⇒ =∵

2 22 2

2 2

b be 1 2 1a a

= + ⇒ = +∵

⇒ b2 = 3a2

∴ Equation of hyperbola 3x2 – y2 = 3a2

43. A point moves in such a way that the difference of its distance from two points (8, 0) and (–8, 0) always remains 4.Then the locus of the point is(A) a circle (B) a parabola (C) an ellipse (D) a hyperbolaAns : (D)Hints : Let P(h, k) be the moving point

( )2 2 2 2h 8 k (h 8) k 4∴ + + − − + = ........................ (i)

Conjugate equation of (i)

2 2 2 2 32h(h 8) k (h 8) k 8h4

+ + + − + = = ................ (ii)

[ using the form A2 – B2 = 32h 32h 32hA B 8h

A B 4⇒ + = = =

− ]

(i) + (ii) and squaring (h + 8)2 + k2 = (2 + 4h)2

or h2 + 16h + 64 + k2 = 4 + 16h2 + 16h⇒ 15h2 – k2 = 60

44. The number of integer values of m, for which the x-coordinate of the point of intersection of the lines 3x + 4y = 9 andy = mx + 1 is also an integer, is(A) 0 (B) 2 (C) 4 (D) 1Ans : (B)

Hints : From given equation 5x

3 4m=

+ x∵ is integer

⇒ 3 + 4m = ± 5 or ± 1⇒ m = – 2 or m = – 1

Page 12: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(11)

45. If a straight line passes through the point (α, β) and the portion of the line intercepted between the axes is divided

equally at that point, then x y+

α β is

(A) 0 (B) 1 (C) 2 (D) 4Ans : (C)

Hints : Let, the equation of line be x y 1a b+ = , ∵ passes through P(α, β)

∵ P(α, β) is the mid point of the portion intercepted between the axes

a 02+

⇒ = α ; b 0

2+

⇒ = β

⇒ a = 2α ; b = 2β x y 1

2 2∴ + =

α β x y 2⇒ + =α β

46. The equation y2 + 4x + 4y + k = 0 represents a parabola whose latus rectum is(A) 1 (B) 2 (C) 3 (D) 4Ans : (D)Hints : y2 + 4x + 4y + k = 0⇒ (y + 2)2 = 4 (–x + 1 – k/4)Latus rectum = 4

47. If the circles x2 + y2 + 2x + 2ky + 6 = 0 and x2 + y2 + 2ky + k = 0 intersect orthogonally, then k is equal to

(A) 32 or2

− (B) 32 or2

− − (C) 32 or2

(D) 32 or2

Ans : (A)Hints :2g1g2 + 2f1f2 = c1 + c2

⇒ 2k2 – k – 6 = 0⇒ k = 2, –3/2

48. If four distinct points (2k, 3k), (2, 0), (0, 3) (0, 0) lie on a circle, then(A) k < 0 (B) o < k < 1 (C) k = 1 (D) k > 1Ans : (C)Hints : Equation of circle x(x – 2) + y (y – 3) = 0(2k, 3k) will satisfySo, K = 1

49. The line joining A (b cos α, b sin α) and B (a cos β, a sin β), where a b≠ , is produced to the point M(x, y) so that

AM : MB = b : a. Then xcos y sin2 2

α + β α + β+ is equal to

(A) 0 (B) 1 (C) –1 (D) a2 + b2

Ans : (A)

Hints : abcos abcos basin absinx , y

b a b aβ − α β − α

= =− −

ab abx 2sin .sin , y 2cos .sinb a 2 2 b a 2 2

α +β α −β α + β α −β⎛ ⎞⎛ ⎞= = −⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠

⇒⇒⇒⇒⇒ x cos ysin 02 2

α + β α +β+ =

Page 13: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(12)

50. Let the foci of the ellipse 2

2x y 19+ = subtend a right angle at a point P. Then the locus of P is

(A) x2 + y2 = 1 (B) x2 + y2 = 2 (C) x2 + y2 = 4 (D) x2 + y2 = 8Ans : (D)

Hints : P(h, k) be a point on the ellipse, 2 2e

3= .

S1 ( )2 2, 0 S2 ( )2 2, 0− , m1m2 = –1 ⇒ k k 1

h 2 2 h 2 2× = −

− +⇒ h2 + k2 = 8

51. The general solution of the differential equation dy x y 1dx 2x 2y 1

+ +=

+ + is

(A) loge|3x + 3y + 2| + 3x + 6y = c (B) loge|3x + 3y + 2| – 3x + 6y = c(C) loge|3x + 3y + 2| – 3x – 6y = c (D) loge|3x + 3y + 2| + 3x – 6y = cAns : (D)Hints : Put x + y = θ

52. The value of the integral /2

/6

1 sin2x cos2x dxsinx cos x

π

π

+ +⎛ ⎞⎜ ⎟+⎝ ⎠∫ is equal to

(A) 16 (B) 8 (C) 4 (D) 1Ans : (D)

Hints : / 2

/6

1 sin2x cos2x dxsin x cos x

π

π

+ ++∫

/ 2 2 2 2

/6

(sinx cos x) cos x sin x dxsinx cos x

π

π

+ + −+∫ =

/2

/ 6

(sin x cosx cos x sin x)dxπ

π

+ + −∫12 1 12

⎛ ⎞= − =⎜ ⎟⎝ ⎠

53. the value of the integral 21010

1 dx1 (tan x)

π

+∫ is equal to

(A) 1 (B) 6π

(C) 8π

(D)4π

Ans : (D)

Hints : ( )101/2

101 1010

cos / 2 xI dx

sin( / 2 x) cos ( / 2 x)

π π −=

π − + π −∫

I4π

=

54. The integrating factor of the differential equation e edy3x log x y 2log xdx

+ = is given by

(A) 3e(log x) (B) e elog (log x) (C) elog x (D)

13

e(log x)Ans : (D)

Hints : dy3x logx y 2logxdx

+ =

dy 1 2ydx 3x logx 3x

⇒ + =

I.F. =1 dx

3x logxe∫ = 13

1 loglogx3e (logx)=

Page 14: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(13)

55. Number of solutions of the equation tan x + sec x = 2 cos x, x∈[0, π] is(A) 0 (B) 1 (C) 2 (D) 3Ans : (C)Hints : tanx + secx = 2cosx2sin2x + sinx – 1 = 0

sinx = 12

56. The value of the integral 4

0

sin x cosx dx3 sin2x

π

++∫ is equal to

(A) loge2 (B) loge3 (C)14

loge2 (D)14

loge3

Ans : (D)

Hints : /4

0

sin x cos xdx3 sin2x

π ++∫

= / 4

20

sin x cos x(sin x cos x) 4

π +−

− −∫ dx

Let sinx – cosx = t

(cox + sinx)dx = dt 0

21

dt 1 log344 t−

= =−∫

57. Let x

ex

3 1y sin x log (1 x), x 13 1

⎛ ⎞−= + + > −⎜ ⎟+⎝ ⎠

. Then at x = 0, dydx equals

(A) 1 (B) 0 (C) –1 (D) –2Ans : (A)

Hints : x

ex

3 1y sinx log (1 x) x 13 1

−= + + > −

+

x 0

dy 1dx =

=

58. Maximum value of function x 2f(x)8 x

= + on the interval [1, 6] is

(A) 1 (B) 98 (C)

1312

(D)178

Ans : (D)

Hints : 2

dy 1 2 [1, 6]dx 8 x

= − f(x) is decreasing

Max. value will be at x = 1

max1 17f 28 8

= + =

Page 15: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(14)

59. For 3x ,

2 2π π

− < < the value of 1d costan

dx 1 sinx−⎧ ⎫

⎨ ⎬+⎩ ⎭

is equal to

(A) 12

(B) 12

− (C) 1 (D) 2

sin(1 sinx)+

Ans : (B)

Hints : 1d cos xtan

dx 1 sinx−⎛ ⎞⎛ ⎞

⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠

1sin x

d 2tandx 1 cos x

2

⎛ ⎞⎛ ⎞π⎛ ⎞−⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟⎜ ⎟=

⎜ ⎟π⎜ ⎟⎛ ⎞+ −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

d x 1dx 4 2 2

π⎛ ⎞= − = −⎜ ⎟⎝ ⎠

60. The value of the integral 2

|x|

2

(1 2sin x)e dx−

+∫ is equal to

(A) 0 (B) e2 – 1 (C) 2(e2 – 1) (D) 1Ans : (C)

Hints : 2 2 2

|x| |x| |x|

2 2 2

(1 2sin x)e dx e dx 2sinx e dx− − −

+ = +∫ ∫ ∫

= 2x

02.e 0+ = 2(e2 – 1)

Q. 61 – Q. 80 carry two marks each

61. The points representing the complex number z for which z 2argz 2 3− π⎛ ⎞ =⎜ ⎟+⎝ ⎠

lie on

(A) a circle (B) a straight line (C) an ellipse (D) a parabolaAns : (A)

Hints : z 2argz 2 3− π⎛ ⎞ =⎜ ⎟+⎝ ⎠

⇒⇒⇒⇒⇒ locus of z is circle3π

2–2

Z

62. Let a, b, c, p,q, r be positive real numbers such that a, b, c are in G.P. and ap = bq = cr. Then(A) p, q, r are in G.P. (B) p, q, r are in A.P. (C) p, q, r are in H.P. (D) p2, q2, r2 are in A.P.Ans : (C)Hints : b2 = ac ; 2logb = loga + log c

ap = bq ; loga qlogb p

= . . . . . . (ii)

Page 16: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(15)

and bq = cr ; logc qlogb r

= . . . . . . (iii)

Hence

2 1 1q r p= + ; p, q, r are in H.P.

63. Let Sk be the sum of an infinite G.P. series whose first term is k and common ratio is k (k 0)

k 1>

+. Then the value of

k

kk 1

( 1)S

=

−∑ is equal to

(A) loge4 (B) loge2 – 1 (C) 1 – loge2 (D) 1 – loge4Ans : (D)

Hints : kSk k(k 1)

k1k 1

= = +−

+

; k

kk 1

( 1)S

=

−∑

= 1 1 1 1 .......2 2.3 3.4 4.5

− + − + = 1 1 11 2[1 .....]2 3 4

− − + − + = e e1 2log 2 1 log 4− = −

64. The quadratic equation 2x2 – (a3+ 8a – 1)x + a2 – 4a = 0 possesses roots of opposite sign. Then(A) a ≤ 0 (B) 0 < a < 4 (C) 4 ≤ a < 8 (D) a ≥ 8Ans : (B)Hints : 0

a(a 4) 0

0 a 4

αβ <− <

< <

65. If loge(x2 – 16) ≤ loge(4x – 11), then

(A) 4 < x ≤ 5 (B) x < – 4 or x > 4 (C) – 1 ≤ x ≤ 5 (D) x < – 1 or x > 5Ans : (A)Hints : x2 – 16 > 0

x∈ (– ∞ , – 4) ∪ (4 , ∞) .......(i)

4x – 11 > 0 ; 11x ,4

⎛ ⎞∈ ∞⎜ ⎟⎝ ⎠

.......(ii)

loge(x2 – 16) ≤ loge(4x – 11)

(x – 5) (x + 1) ≤ 0 ; x∈[ –1, 5] ....... (iii)x∈(4, 5] ⇒ 4 < x ≤ 5

66. The coefficient of x10 in the expansion of 1 + (1 + x) + ......... + (1 + x)20 is(A) 19C9 (B) 20C10 (C) 21C11 (D) 22C12

Ans : (C)Hints : 1 + (1 + x) + ......... + (1 + x)20

= 21(1 x) 1

x+ −

coefficient of x10 in 21(1 x) 1

x x⎡ ⎤+

−⎢ ⎥⎢ ⎥⎣ ⎦

= 21C11

Page 17: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(16)

67. The system of linear equations

λx + y + z = 3

x – y – 2z = 6

– x + y + z = μ

has

(A) infinite number of solutions for λ ≠ – 1 and all μ (B) infinite number of solutions of λ = – 1 and μ = 3

(C) no solution for λ ≠ – 1 (D) unique solution for λ = –1 and μ = 3

Ans : (B)

Hints : 1 1

1 1 21 1 1

λΔ = − −

= λ + 1

if λ = – 1 then Δ = 0

so equation has infinite number of solution for λ = – 1 and μ = 3

68. Let A and B be two events with P(AC) = 0.3, P(B) = 0.4 and P(A ∩ BC) = 0.5. Then P(B/A ∪ BC) is equal to

(A) 14

(B) 13 (C)

12

(D) 23

Ans : (A)Hints : P(A∩ BC) = P(A) – P(A∩ B) ; P (A ∩ B) = 0.2

( )CC

C

P B (A B )P(B / A B )

P(A B )

∩ ∪∪ =

∪ = C C

P(B A)P(A) P(B ) P(A B )

+ − ∩ = 0.2 0.2 1

0.7 0.6 0.5 0.8 4= =

+ −

69. Let p, q, r be the altitudes of a triangle with area S and perimeter 2t. Then the value of 1 1 1p q r+ + is

(A) st (B)

ts (C)

s2t (D)

2st

Ans : (B)

Hints : Area of 1ABC base height2

Δ = × × A

B C

cp

b

a

1S a p2

= × × ; 1 ap 2s= ........ (i)

similarly 1 bq 2s= ...... (ii)

1 cr 2s= ........ (iii)

(i) + (ii) + (iii)

1 1 1 b c a 2t tp q r 2s 2s s

+ ++ + = = =

Page 18: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(17)

70. Let C1 and C2 denote the centres of the circles x2 + y2 = 4 and (x – 2)2 + y2 = 1 respectively and let P and Q be theirpoints of intersection. Then the areas of triangles C1PQ and C2PQ are in the ratio(A) 3 : 1 (B) 5 : 1 (C) 7 : 1 (D) 9 : 1Ans : (C)Hints : Equation of chord S2 – S1 = 0 ; x2 + y2 – 4x + 3 = 0

7x4

=

C1T = Perpendicular distance of point (0, 0) from 7x4

= is 47

P

C

Q

1 C2

T

7x4

=

C2T = Perpendicular distance of point (2, 0) from 7x4

= is 14

area of ΔC1 PQ = 11 C T PQ2

× = 1 7 PQ2 4× ×

area of ΔC2 PQ = 21C T PQ2

× = 1 1 PQ2 4× ×

1

2

area of C PQarea of C PQ

ΔΔ =

71

= 7 : 1

71. A straight line through the point of intersection of the lines x + 2y = 4 and 2x + y = 4 meets the coordinate axes at Aand B. The locus of the midpoint of AB is(A) 3(x + y) = 2xy (B) 2(x + y) = 3xy (C) 2(x + y) = xy (D) x + y = 3xyAns : (B)

Hints : Point of intersection of given lines 4 4x , y3 3

= = B(0, b)

4 4,3 3

⎛ ⎞⎜ ⎟⎝ ⎠

(h, k)

A(a, o)

Equation of lines AB is x y 1a b+ =

It is passes through 4 4,3 3

⎛ ⎞⎜ ⎟⎝ ⎠

, so, 4 4 1

3a 3b+ = .... (1)

(h1, k) be the mid point of ABa = 2h , b = 2k

from (1) 2 2 13h 3k

+ =

locus of (h, k) is 2(x y) 3xy+ =

72. Let P and Q be the points on the parabola y2 = 4x so that the line segment PQ subtends right angle at the vertex.If PQ intersects the axis of the parabola at R, then the distance of the vertex from R is(A) 1 (B) 2 (C) 4 (D) 6Ans : (C)Hints : PQ subtants right angle at vertex.

90°O

P

R

Q22 2t ,2t

21 1t ,2t

so, t1 t2 = – 4equation of PQ is

( )211 12

1

2ty 2t x t

t 4− = −

it intersect the x axis so, put y = 0we get x = 4

Page 19: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(18)

73. The incentre of an equilateral triangle is (1, 1) and the equation of one side is 3x + 4y + 3 = 0. Then the equation of thecircumcircle of the triangle is(A) x2 + y2 – 2x – 2y – 2 = 0 (B) x2 + y2 – 2x – 2y – 14 = 0(C) x2 + y2 – 2x – 2y + 2 = 0 (D) x2 + y2 – 2x – 2y + 14 = 0Ans : (B)

Hints : 2 2

3 1 4 1 3r3 4

× + × +=

+ = 2

30°

(1, 1)

R

r

3x+4y+3=0

For equalateral triangle R = 2r = 2 × 2 = 4equation of circumcircle(x – 1)2 + (y – 1)2 = 42 ; x2 + y2 – 2x – 2y – 14 = 0

74. The value of 1n

n

(n!)limn→∞

is

(A) 1 (B) 21

e(C)

12e (D)

1e

Ans : (D)

Hints : 1n

nn

nA limn→∞

⎛ ⎞= ⎜ ⎟⎝ ⎠

log A = nn

1 nlim logn n→∞

⎡ ⎛ ⎞⎜ ⎟⎢ ⎝ ⎠⎣

= n

1 1.2.........nlim logn n...........n→∞

⎡ ⎤⎛⎜⎢ ⎥⎝⎣ ⎦

= 1

0logxdx∫ = – 1

A = e–1 ; 1Ae

=

75. The area of the region bounded by the curves y = x3, y = 1x

, x = 2 is

(A) 4 – loge2 (B) e1 log 24+ (C) 3 – loge2 (D) e

15 log 24−

Ans : (D)

Hints : Area of region 2 3

1

1x dxx

⎛ ⎞−⎜ ⎟⎝ ⎠∫ = e

15 log 24−

76. Let y be the solution of the differential equation 2dy yx

dx 1 y logx=

− satisfying y(1) = 1. Then y satisfies

(A) y = xy – 1 (B) y = xy (C) y = xy + 1 (D) y = xy + 2

Ans : (B)

Hints : . 21 dx 1 1logxx dy y y

+ = ; logx = θ

2d 1 1.dy y yθ+ θ = ;

1dylogyyIF e e y= = =

Ans. 21y y.dyy

θ = ∫y logy cθ = +

y.logx = logy + c x = 1, y = 1. ∴ c = 0

log (x)y = logy ; y = (x)y

Page 20: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Mathematics

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(19)

77. The area of the region, bounded by the curves y = sin–1x +x (1 – x) and y = sin –1x – x(1 – x) in the first quadrant, is

(A) 1 (B) 12

(C) 13 (D)

14

Ans : (C)

Hints : 1 1 2

0 0A 2x(1 x) (2x 2x )dx= − = −∫ ∫ =

13

78. The value of the integral

5

1

x 3 1 x dx⎡ ⎤− + −⎣ ⎦∫ is equal to

(A) 4 (B) 8 (C) 12 (D) 16Ans : (C)

Hints : 5 5

1 1| x 3 | dx |1 x | dx− + −∫ ∫ =

3 5 5

1 3 1(3 x)dx (x 3)dx (x 1)dx− + − + −∫ ∫ ∫ = 12

79. If f(x) and g(x) are twice differentiable functions on (0, 3) satisfying f″(x) = g″ (x), f′(1) = 4, g′(1) = 6, f(2) = 3, g(2) =9,then f(1) – g(1) is(A) 4 (B) – 4 (C) 0 (D) – 2Ans : (B)Hints : f″(x) = g″(x) f(x) = g(x) – 2x + D

f′(x) = g′(x) + c f(2) = g(2) – 4 + Df′(1) = g′(1) +c 3 = 9 – 4 + D4 = 6 + c – 2 = Dc = – 2f′(x) = g′(x) – 2 f(x) – g(x) = – 2x – 2

f(1) – g(1) = – 2 – 2 = – 4

80. Let [x] denote the greatest integer less than or equal to x, then the value of the integral

1

1

(| x | 2[x])dx−

−∫ is equal to

(A) 3 (B) 2 (C) – 2 (D) – 3Ans : (A)

Hints : 1 1

0 1I 2 | x | dx 2 [x]dx

−= −∫ ∫

= 1 0 1

0 1 02 x dx 2 ( 1)dx 0dx

⎡ ⎤− − +⎢ ⎥⎣ ⎦∫ ∫ ∫ = 2 1 00 1(x ) 2(x) 0−+ +

= 1 + 2 (0– (– 1) = 1 + 2 = 3

Page 21: Answer Key and Solution for Wb Jee 2012

Q. 1 – Q. 30 carry one mark each1. In a mercury thermometer the ice point (lower fixed point) is marked as 10o and the steam point (upper fixed point) is

marked as 130o. At 40oC temperature, what will this thermometer read?(A) 78o (B) 66o (C) 62o (D) 58o

Ans : (D)

Hints : 40 0 T 10

100 0 130 10− −

= ∴− −

T = 58o

2. The magnetic flux linked with a coil satisfies the relation φ = 4t2 + 6t + 9 Wb, where t is the time in second. The e.m.f.induced in the coil at t = 2 second is(A) 22 V (B) 18 V (C) 16 V (D) 40 VAns : (A)

Hints : d| | 8t 6dtφ

ε = = + ∴ at t = 2 s. | |ε = 22 V

3. Water is flowing through a very narrow tube. The velocity of water below which the flow remains a streamline flow isknown as(A) Relative velocity (B) Terminal velocity (C) Critical velocity (D) Particle velocityAns : (C)

4. If the velocity of light in vacuum is 3 × 108 ms–1, the time taken (in nanosecond) to travel through a glass of thickness10 cm and refractive index 1.5 is(A) 0.5 (B) 1.0 (C) 2.0 (D) 3.0Ans : (A)

Hints : 8

8oC 3 10C 2 10 m / s1.5×= = = ×

μ

2

8d 10 10tc 2 10

−×∴ = =

× = 0.5 ns

5. A charge +q is placed at the orgin O of X – Y axes as shown in the figure. The work done in taking a charge Q fromA to B along the straight line AB is

O X

Y

B(0,b)

A(a,0)

(A) 0

qQ a b4 ab

−⎛ ⎞⎜ ⎟πε ⎝ ⎠ (B)

0

qQ b a4 ab

−⎛ ⎞⎜ ⎟πε ⎝ ⎠ (C) 2

0

qQ b 14 ba

⎛ ⎞−⎜ ⎟πε ⎝ ⎠ (D) 20

qQ a 14 bb

⎛ ⎞−⎜ ⎟πε ⎝ ⎠

Ans : (A)

Hints : ( )B A0

Qq 1 1W Q V V4 b a

⎡ ⎤= − = −⎢ ⎥πε ⎣ ⎦=

0

qQ a b4 ab

−⎛ ⎞⎜ ⎟πε ⎝ ⎠

Code -γγγγγWBJEE - 2012 (Answers & Hints) Physics & Chemistry

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(1)

ANSWERS & HINTSfor

WBJEE - 2012SUB : PHYSICS & CHEMISTRY

Page 22: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Physics & Chemistry

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(2)

6. What current will flow through 2 kΩ resistor in the circuit shown in the figure?6kΩ 4kΩ

3kΩ2kΩ

72V

(A) 3 mA (B) 6 mA (C) 12 mA (D) 36 mAAns : (A)

Hints : Req

6 3 6 8k6 3×

= + = Ω+

, 33

V 72I 9 10 AR 8 10

−= = = ××

, 31

1I I 3 10 A3

−= × = × , I1= current through 2 kΩ

7. In a region, the intensity of an electric field is given by ˆ ˆ ˆE 2i 3 j k= + + in NC–1. The electric flux through a surface

ˆS 10i= m2 in the region is

(A) 5 Nm2C–1 (B) 10 Nm2C–1 (C) 15 Nm2C–1 (D) 20 Nm2C–1

Ans : (D)Hints : e E.Sφ = = 20 Nm2C–1

8. A train approaching a railway platform with a speed of 20 ms–1 starts blowing the whistle. Speed of sound in air is 340ms–1. If the frequency of the emitted sound from the whistle is 640 Hz, the frequency of sound to a person standing onthe platform will appear to be(A) 600 Hz (B) 640 Hz (C) 680 Hz (D) 720 HzAns : (C)

Hints : a 0s

V 340n .n 640 680HzV V 320

= = × =−

9. A straight wire of length 2 m carries a current of 10 A. If this wire is placed in a uniform magnetic field of 0.15 T makingan angle of 45o with the magnetic field, the applied force on the wire will be

(A) 1.5 N (B) 3 N (C) 3 2 N (D) 32 N

Ans : (D)

Hints : o 3F i l B i lBsin 10 2 0.15 sin45 .N2

= × = θ= × × × =

10. What is the phase difference between two simple harmonic motions represented by 1x A sin t6π⎛ ⎞= ω +⎜ ⎟

⎝ ⎠ and x2 = A

cos (ω t)?

(A)6π

(B)3π

(C)2π

(D)23π

Ans : (B)

Hints : 1 2X A sin t ,X A sin t6 2π π⎛ ⎞ ⎛ ⎞= ω + = ω +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠2 1 t t

2 6π π⎛ ⎞ ⎛ ⎞∴Δφ = φ − φ = ω + − ω +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠=

11. Heat is produced at a rate given by H in a resistor when it is connected across a supply of voltage V. If now theresistance of the resistor is doubled and the supply voltage is made V/3 then the rate of production of heat in theresistor will be(A) H/18 (B) H/9 (C) 6H (D) 18HAns : (A)

Hints : ( )22 2VV 1 V H3H ,H .R 2R 18 R 18

′= = = =

Page 23: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Physics & Chemistry

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(3)

12. Two elements A and B with atomic numbers ZA and ZB are used to produce characteristic x–rays with frequencies Aν

and Bν respectively. If ZA : ZB = 1 : 2, then Aν : Bν will be

(A) 1 : 2 (B) 1 : 8 (C) 4 : 1 (D) 1 : 4Ans : (D)

Hints : a(z b)ν = − , Ignoring screening effect (i.e. b=0), 2

A A

A B

Z 1Z 4

⎛ ⎞ν∴ = =⎜ ⎟ν ⎝ ⎠

13. The de Broglie wavelength of an electron moving with a velocity c/2 (c = velocity of light in vacuum) is equal to thewavelength of a photon. The ratio of the kinetic energies of electron and photon is(A) 1 : 4 (B) 1 : 2 (C) 1 : 1 (D) 2 : 1Ans : (A)

Hints : 2

2e

1 c 1K m mc2 2 8

⎛ ⎞= =⎜ ⎟⎝ ⎠

, ee

h h 2hcp mcm 2

λ = = =ne pλ = λ∵ ∴ kp=K.E. of photon

2hc mcp 2

= =λ

e

p

k 1k 4

∴ =

14. Two infinite parallel metal planes, contain electric charges with charge densities +σ and – σ respectively and they areseparated by a small distance in air. If the permittivity of air is 0ε then the magnitude of the field between the twoplanes with its direction will be

(A) σ / 0ε towards the positively charged plane (B) σ / 0ε towards the negatively charged plane

(C) σ /(2 0ε ) towards the positively charged plane (D) 0 and towards any directionAns : (B)

15. A box of mass 2 kg is placed on the roof of a car. The box would remain stationary until the car attains a maximumacceleration. Coefficient of static friction between the box and the roof of the car is 0.2 and g = 10 ms–2.This maximum acceleration of the car, for the box to remain stationary, is(A) 8 ms–2 (B) 6 ms–2 (C) 4 ms–2 (D) 2 ms–2

Ans : (D)Hints : amax

= μ g =0.2 × 10 = 2 m/s2

16. The dimension of angular momentum is(A) M0L1T–1 (B) M1L2T–2 (C) M1L2T–1 (D) M2L1T–2

Ans : (C)Hints : [L]= [m r v] = [M1L2T–1]

17. if A B C= + and A,B,C have scalar magnitudes of 5, 4 , 3 units respectivley then the angle between A and C is(A) cos–1 (3/5)(B) cos–1 (4/5)(C) π/2(D) sin–1 (3/4)

Ans : (A)

Hints : ( )13 3Cos ; Cos 55−α = α =

α

α

3

4

5A

C

B

18. A particle is travelling along a straight line OX. The distance x (in meters) of the particle from O at a time t is given byx = 37 + 27 t – t3 where t is time in seconds. The distance of the particle from O when it comes to rest is(A) 81 m (B) 91 m (C) 101 m (D) 111 mAns : (B)

Hints : 2dxv 27 3t 0dt

= = − = , t = 3 sec. ∴x = 37 + 27 × 3 – 33, = 37 + 81 – 27 = 91 m

Page 24: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Physics & Chemistry

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(4)

19. A particle is projected from the ground with kinetic energy E at an angle of 600 with the horizontal. Its kinetic energyat the highest point of its motion will be

(A) E / 2 (B) E / 2 (C) E / 4 (D) E / 8Ans : (C)

Hints : 2 o 2 2i f

1 1 1 1 EK E mu ,K E m(ucos60 ) mu2 2 2 4 4

⎛ ⎞′= = = = = × =⎜ ⎟⎝ ⎠

20. A bullet on penetrating 30 cm into its target loses its velocity by 50%. What additional distance will it penetrate intothe target before it comes to rest?(A) 30 cm (B) 20 cm (C) 10 cm (D) 5 cmAns : (C)

Hints : 2 2 2

2 2V V VV 2a(0.3)..........(1), 0 2as a ..............(2)2 4 8s

⎛ ⎞ = + = + ∴ = −⎜ ⎟⎝ ⎠

From (1) and (2) s = 10 m

21. When a spring is stretched by 10 cm, the potential energy stored is E. When the spring is stretched by 10 cm more,the potential energy stored in the spring becomes(A) 2 E (B) 4 E (C) 6 E (D) 10 EAns : (B)

Hints : 20 0

1E kx ,x 10cm2

= =

21E kx2

′ = , x = 10 + 10= 20 cm = 2 x0

2

01 k(2x ) 4E2

= =

22. Average distance of the earth from the Sun is L1. If one year of the Earth = D days, one year of another planet whoseaverage distance from the Sun is L2 will be

(A)

122

1

LDL

⎛ ⎞⎜ ⎟⎝ ⎠

days (B)

322

1

LDL

⎛ ⎞⎜ ⎟⎝ ⎠

days (C)

232

1

LDL

⎛ ⎞⎜ ⎟⎝ ⎠

days (D) 2

1

LDL

⎛ ⎞⎜ ⎟⎝ ⎠

days

Ans : (B)

Hints :

3 32 22 3 2 2 2

21 1 1

T L LT L D DT L L

⎛ ⎞ ⎛ ⎞∝ ∴ = ∴ =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

23. A spherical ball A of mass 4 kg, moving along a straight line strikes another spherical ball B of mass 1 kg at rest.After the collision, A and B move with velocities v1 ms–1 and v2 ms–1 respectively making angles of 30o and 60o with

respect to the original direction of motion of A. The ratio 1

2

vv will be

(A) 3

4(B)

43 (C)

13 (D) 3

Ans : (A)Hints : Apply conservation of momentum along Normal

m1V1 sin 30o = m2V2 sin 60o ∴ 1

2

V 3V 4

=

24. The decimal number equivalent ot a binary number 1011001 is(A) 13 (B) 17 (C) 89 (D) 178Ans : (C)Hints : (1011001)2= 1× 26 + 0 × 25 +1× 24 + 1× 23 + 0× 22 + 0× 21 + 1× 20 = 64 + 16 + 8 + 1 =89

Page 25: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Physics & Chemistry

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(5)

25. The frequency of the first overtone of a closed pipe of length l1 is equal to that of the first overtone of an open pipe oflength l2. The ratio of their lenghts (l1 : l2) is(A) 2 : 3 (B) 4 : 5 (C) 3 : 5 (D) 3 : 4Ans : (D)

Hints : 1

1 2 2

l3 2v 3.v4l 2l l 4

= ∴ =

26. The I – V characteristics of a metal wire at two different temperatures (T1 and T2) are given in the adjoining figure. Here,we can conclude that

I

V

T

T

2

1

(A) T1 > T2 (B) T1 < T2 (C) T1 = T2 (D) T1 = 2T2

Ans : (B)

Hints : I 1V R= = slope of curve, R T∝ ∴ greater the slope smaller will be temperature. T1 < T2.

27. In a slide calipers, (m + 1) number of vernier divisions is equal to m number of smallest main scale divisions. If d unitis the magnitude of the smallest main scale division, then the magnitude of the vernier constant is(A) d/(m+1) unit (B) d/m unit (C) md/(m+1) unit (D) (m+1)d/m unitAns : (A)

Hints : m dL.C 1 d

m 1 m 1⎛ ⎞= − =⎜ ⎟+ +⎝ ⎠

28. From the top of a tower, 80 m high from the ground, a stone is thrown in the horizontal direction with a velocity of 8ms–1. The stone reaches the ground after a time ‘t’ and falls at a distance of ‘d’ from the foot of the tower. Assumingg = 10 ms–2, the time t and distance d are given respectively by(A) 6 s, 64 m (B) 6 s, 48 m (C) 4 s, 32 m (D) 4 s, 16 mAns : (C)

Hints : time of flight (t) = 2h 2 80 4s x 8 4 32mg 10

×= = ∴ = × =

29. A Wheatstone bridge has the resistances 10 Ω, 10Ω, 10Ω and 30 Ω in its four arms. What resistance joined in parallelto the 30 Ω resistance will bring it to the balanced condition?(A) 2 Ω (B) 5 Ω (C) 10 Ω (D) 15 ΩAns : (D)

Hints : for a balanced bridge. R = 10 Ω, ∴30 x 10 30x 300 10x 20x 300, x 1530 x

×= ⇒ = + ⇒ = = Ω

+30. An electric bulb marked as 50 W–200 V is connected across a 100 V supply. The present power of the bulb is

(A) 37.5 W (B) 25 W (C) 12.5 W (D) 10 WAns : (C)

Hints : Pactual consume =2 2

sr

r

V 100 50P 50 12.5 WV 200 4

⎛ ⎞ ⎛ ⎞= × = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

Page 26: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Physics & Chemistry

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(6)

Q. 31 - Q. 40 carry two marks each31. A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an

angle of 600 and the work done is W. The torque on the magnetic needle at this poition is

(A) 2 3 W (B) 3 W (C)3

2W (D)

34

W

Ans : (B)

Hints : o MBW MB(1 cos ) MB(1 cos60 )2

= − θ = − = , o MBMBsin MBsin60 3 3W2

τ = θ = = =

32. In the adjoining figure the potential difference between X and Y is 60 V. The potential difference between the points Mand N will be

2C

2C

C C

X

Y

M

N

(A) 10 V (B) 15 V (C) 20 V (D) 30 VAns : (D)

Hints : eq net MN

190CC C 3C C 3 , q 90C V 30V2 2 C

×= + = = = =

33. A body when fully immersed in a liquid of specific gravity 1.2 weighs 44 gwt. The same body when fully immersed inwater weighs 50 gwt. The mass of the body is(A) 36 g (B) 48 g (C) 64 g (D) 80 gAns : (D)Hints : mg – σ vg = 44,σ = density of liquid, mg –1.2 ρ vg = 44 ..............(1), ρ = density of watermg – ρ vg = 50.............(2), (1) – 1.2 × (2) ∴ M = 80 g

34. When a certain metal surface is illuminated with light of frequency ν , the stopping potential for photoelectric current

is V0. When the same surface is illuminated by light of frequency 2ν

, the stopping potential is 0V4

. The threshold

frequency for photoelectric emission is

(A) 6ν

(B)3ν

(C)23ν

(D)43ν

Ans : (B)

Hints : e V0 = h ν – φ0.........(1), 0eV4

= h2ν

– φ0.........(2), solving equation (1) and (2) , 0 3ν

ν =

35. Three blocks of mass 4 kg, 2 kg, 1 kg respectively are in contact on a frictionless table as shown in the figure. If aforce of 14 N is applied on the 4 kg block, the contact force between the 4 kg block, the contact force between the 4kg and the 2 kg block will be

14 N 4 kg 2 kg 1kg

(A) 2 N (B) 6 N (C) 8 N (D) 14 NAns : (B)

Hints : 2net

net

F 14a 2m / s R 3 2 6Nm 7

= = = = × =

Page 27: Answer Key and Solution for Wb Jee 2012

WBJEE - 2012 (Answers & Hints) Physics & Chemistry

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(7)

36. Let L be the length and d be the diameter of cross section of a wire. Wires of the same material with different L andd are subjected to the same tension along the length of the wire. In which of the following cases, the extension of wirewill be the maximum?(A) L = 200 cm, d = 0.5 mm (B) L = 300 cm, d = 1.0 mm(C) L = 50 cm, d = 0.05 mm (D) L = 100 cm, d = 0.2 mmAns : (C)

Hints : TL LL LAY A

Δ = ⇒ Δ ∝

37. An object placed in front of a concave mirror at a distance of x cm from the pole gives a 3 times magnified real image.If it is moved to a distance of (x + 5) cm, the magnification of the image becomes 2. The focal length of the mirror is(A) 15 cm (B) 20 cm (C) 25 cm (D) 30 cmAns : (D)

Hints : 1uf ( x)( f) 4v ,v 3x x f.........(1)

u f x f 3− −

= = = − ∴ =− − +

, 2(x 5)( f )v , 2x 10 3f

x 5 f+ −

= − + =− + +

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

so f = 30 cm38. 22320 cal of heat is supplied to 100 g of ice at 0oC. If the latent heat of fusion of ice is 80 cal g–1 and latent heat of

vaporization of water is 540 cal g–1, the final amount of water thus obtained and its temperature respectively are(A) 8g, 100oC (B) 100 g, 90oC (C) 92g, 100oC (D) 82g, 100oCAns : (C)Hints : total energy required to melt and boil at 100oC is

(Q)min = 8000 + 10000 = 18000 cal, (Qrequired)min < Qgiven for amount of vapour 22320=18000+ m′ × 540

⇒ m′ = 8 gm. ∴ temperature = 100oC and water remaining = 92 gm.

39. A progressive wave moving along x–axis is represented by ( )2y A sin t xπ⎡ ⎤= ν −⎢ ⎥λ⎣ ⎦. The wavelength ( λ ) at which the

maximum particle velocity is 3 times the wave velocity is(A) A / 3 (B) 2A / (3 π) (C) (3/4)π A (D) (2/3)π AAns : (D)

Hints : 2 2A 3 A

3πν = ν ∴λ = π

λ

40. Two radioactive substances A and B have decay constant 5 λ and λ respectively. At t = 0, they have the same numberof nuclei the ratio of number of nuclei of A to that of B will be (1/e)2 after a time interval of

(A) 1λ

(B) 12λ

(C) 1

3λ(D)

14λ

Ans : (B)

Hints : 5 t

1t 2

2

N e 1 1, tN 2e e

− λ

−λ= = =λ

Q. 41 - Q. 70 carry one mark each41. Li occupies higher position in the electrochemical series of metals as compared to Cu since

(A) the standard reduction potential of Li+/Li is lower than that of Cu2+/Cu(B) the standard reduction potential of Cu2+/Cu is lower than that of Li+/Li(C) the standard oxidation potential of Li/Li+ is lower than that of Cu/Cu2+

(D) Li is smaller in size as compared to CuAns : (A)Hints : E° Li+/Li = –3.04V ; E°Cu2+/Cu = 0.34V

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WBJEE - 2012 (Answers & Hints) Physics & Chemistry

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42. 11Na24 is radioactive and it decays to(A) 9F20 and α-particles (B) 13Al24 and positron (C) 11Na23 and neutron (D) 12Mg24 and β-particlesAns : (D)

Hints : 2424 01211 1Na Mg e−⎯⎯→ +

(n/p = 1.18) (n/p=1)43. The paramagnetic behavior of B2 is due to the presence of

(A) 2 unpaired electrons in Πb MO (B) 2 unpaired electrons in Π∗ MO(C) 2 unpaired electrons in σ∗ MO (D) 2 unpaired electrons in σb MOAns : (A)

Hints : _

2 *2 2 *22 1s 1s 2 s 2 sB (10 e ) → σ σ σ σ {π2px

1 = π2py1 }

Two unpaired e–

44. A 100 ml 0.1 (M) solution of ammonium acetate is diluted by adding 100 ml of water. The pH of the resulting solutionwill be (pKa of acetic acid is nearly equal to pKb of NH4OH)(A) 4.9 (B) 5.0 (C) 7.0 (D) 10.0Ans : (C)Hints : Hydrolysis of a salt of weak acid and weak base

pH = 7 + ½ (pKa– pKb) = 7 a b[ pK pK ]=∵

45. In 2-butene, which one of the following statements is true ?(A) C1-C2 bond is a sp3 - sp3 σ-bond (B) C2-C3 bond is a sp3 - sp2 σ-bond(C) C1-C2 bond is a sp3 - sp2 σ-bond (D) C1-C2 bond is a sp2 - sp2 σ-bondAns : (C)

Hints : CH –CH = CH – CH3 31 2 3 4

↓ ↓sp3 sp2

But-2-ene46. The well known compounds, (+)- lactic acid and (–) lactic acid have the same molecular formula, C3H6O3. The correct

relationship between them is(A) constitutional isomerism (B) geometrical isomerism (C) identicalness (D) optical isomerismAns : (D)

Hints :

COOH

CH3

OHH

COOH

CH3

HHO

They are enantiomers and correct relationship between them is optical isomerism.47. The stability of Me2C=CH2 is more than that of MeCH2CH = CH2 due to

(A) inductive effect of the Me group (B) resonance effect of the Me group(C) hyperconjugative effect of the Me group (D) resonance as well as inductive effect of the Me groupAns : (C)Hints : Me2 C = CH2 is more stable than MeCH2CH = CH2 due to hyperconjugative effect.

(6 number of α – H) (2 number of α – H)

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WBJEE - 2012 (Answers & Hints) Physics & Chemistry

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48. Which one of the following characteristics belongs to an electrophile ?(A) It is any species having electron deficiency, which reacts at an electron rich C-centre(B) It is any species having electron enrichment that reacts at an electron deficient C-centre(C) It is cationic in nature(D) It is anionic in natureAns : (A)Hints : Fact

49. Which one of the following methods is used to prepare Me3COEt with a good yield?(A) Mixing EtONa with Me3CCl(B) Mixing Me3CONa with EtCl(C) Heating a mixture of (1:1) EtOH and Me3COH in presence of conc. H2SO4

(D) Treatment of Me3COH with EtMglAns : (B)

Hints : CH – C – O Na3– +

CH3

CH3

+ CH3CH2Cl NS 2⎯⎯⎯→ CH – C – O – CH CH + NaCl3 2 3

CH3

CH3

⇒ Williamson’s synthesis (SN2) given by 1° alkyl halide.50. 58.5 gm of NaCl and 180 gm of glucose were separately dissolved in 1000 ml of water. Identify the correct statement

regarding the elevation of boiling point (b.p.) of the resulting solutions.(A) NaCl solution will show higher elevation of b.p.(B) Glucose solution will show higher elevation of b.p.(C) Both the solutions will show equal elevation of b.p.(D) The b.p. elevation will be shown by neither of the solutionsAns : (A)Hints : For NaCl (i = 2) for glucose (i = 1),

Hence NaCl solution will show higher elevation of bp (Colligative property) as both are one Molar Solution.51. Equal weights of CH4 and H2 are mixed in an empty container at 25oC. The fraction of the total pressure exerted by H2

is(A) 1/9 (B) 1/2 (C) 8/9 (D) 16/17Ans : (C)Hints : Let equal weights be w g.

mole fraction of 2w / 2H

w / 2 w / 16 98

= =+

Partial pressure = mole fraction × Total pressure

2H8p9

= × Total Pressure

52. Which of the following will show a negative deviation from Raoult’s law?(A) Acetone-benzene (B) Acetone-ethanol(C) Benzene-methanol (D) Acetone-chloroformAns : (D)Hints : Acetone – chloroform due to formation of intermolecular Hydrogen bonding

C = O H — C — Cl Cl

Cl

CH3

CH3

δ– δ+

H-bonding

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WBJEE - 2012 (Answers & Hints) Physics & Chemistry

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53. In a reversible chemical reaction at equilibrium, if the concentration of any one of the reactants is doubled, then theequilibrium constant will(A) also be doubled (B) be halved(C) remain the same (D) become one-fourthAns : (C)Hints : Equilibrium constant does not depend on conc. It is only a function of temperature

54. Identify the correct statement from the following in a chemical reaction.(A) The entropy always increases(B) The change in entropy along with suitable change in enthalpy decides the fate of a reaction(C) The enthalpy always decreases(D) Both the enthalpy and the entropy remain constantAns : (B)Hints : Spontaneity of a reaction is decided by a combined effect of suitable change in values of enthalpy andentropy. [ΔG = ΔH – TΔS]

55. Which one of the following is wrong about molecularity of a reaction?(A) It may be whole number or fractional(B) It is calculated from reaction mechanism(C) It is the number of molecules of the reactants taking part in a single step chemical reaction(D) It is always equal to the order of elementary reactionAns : (A)Hints : Molecularity can never be a fraction.

56. Which of the following does not represent the mathematical expression for the Heisenberg uncertainty principle?(A) Δx.Δp ≥ h/(4Π) (B) Δx.Δv ≥ h/(4Πm) (C) ΔE.Δt ≥ h/(4Π) (D) ΔE.Δx ≥ h/(4Π)Ans : (D)Hints : Fact.

57. The stable bivalency of Pb and trivalency of Bi is(A) due to d contraction in Pb and Bi(B) due to relativistic contraction of the 6s orbitals of Pb and Bi, leading to inert pair effect(C) due to screening effect(D) due to attainment of noble liquid configurationAns : (B)Hints : Lower oxidation state becomes more stable for group 14 and group 15 elements as we move down the groupdue to inert pair effect. Hence Pb+2 and Bi+3 are more stable.

58. The equivalent weight of K2Cr2O7 in acidic medium is expressed in terms of its molecular weight (M) as(A) M/3 (B) M/4 (C) M/6 (D) M/7Ans : (C)Hints : Cr2O7

–2 + 14H+ + 6e– → 2Cr+3 + 7H2ONumber of electrons gained by one ion = 6

Eq.wt M6

=

59. Which of the following is correct ?(A) radius of Ca2+ < Cl- < S2- (B) radius of Cl- < S2- < Ca2+

(C) radius of S2- = Cl- = Ca2+ (D) radius of S2- < Cl- < Ca2+

Ans : (A)

Hints : Ca+2, S–2, Cl– are isoelectronic (consist of 18e–) and for isoelectronic species, ionic radii ez

⎛ ⎞∝ ⎜ ⎟⎝ ⎠

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WBJEE - 2012 (Answers & Hints) Physics & Chemistry

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(11)

60. CO is practically non-polar since(A) the σ-electron drift from C to O is almost nullified by the Π-electron drift from O to C(B) the σ-electron drift from O to C is almost nullified by the Π-electron drift from C to O(C) the bond moment is low(D) there is a triple bond between C and OAns : (A)Hints : Factual

61. The number of acidic protons in H3PO3 are(A) 0 (B) 1 (C) 2 (D) 3Ans : (C)

Hints :

P

O

HO OHH

⇒ Number of OH group is 2 hence dibasic in nature.

62. When H2O2 is shaken with an acidified solution of K2Cr2O7 in presence of ether, the ethereal layer turns blue due tothe formation of(A) Cr2O3 (B) CrO4

2- (C) Cr2(SO4)3 (D) CrO5

Ans : (D)

Hints : K2Cr2O7 + H2SO4 + 4H2O2 ether⎯⎯⎯→ 2CrOs + K2SO4 + 5H2O

(Blue)63. The state of hybridization of the central atom and the number of lone pairs over the central atom in POCl3 are

(A) sp, 0 (B) sp2, 0 (C) sp3, 0 (D) dsp2, 1Ans : (C)

Hints : P

O sp3

Cl Cl Cl

Number of lone pairs on P atom = 0

64. Upon treatment with l2 and aqueous NaOH, which of the following compounds will form iodoform?(A) CH3CH2CH2CH2CHO (B) CH3CH2COCH2CH3

(C) CH3CH2CH2CH2CH2OH ( D) CH3CH2CH2CH(OH)CH3

Ans : (D)

Hints : Iodoform reaction is given by all alcohols having CH – CH – group3

OH.

65. Upon treatment with Al(OEt)3 followed by usual reactions (work up), CH3CHO will produce(A) only CH3COOCH2CH3 (B) a mixture of CH3COOH and EtOH(C) only CH3COOH (D) only EtOHAns : (A)

Hints : 2CH –CH3

O 3Al(OEt)⎯⎯⎯⎯→ CH –COCH CH3 2 3

O

This reaction is known as Tischenko reaction

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66. Friedel-Craft’s reaction using MeCl and anhydrous AlCl3 will take place most efficiently with

(A) Benzene (B) Nitrobenzene (C) Acetophenone (D) TolueneAns : (D)Hints : As – CH3 group in toluene activates the benzene ring for electrophilic substitution reaction.

67. Which one of the following properties is exhibited by phenol?(A) It is soluble in aq. NaOH and evolves CO2 with aq. NaHCO3

(B) It is soluble in aq. NaOH and does not evolve CO2 with aq. NaHCO3

(C) It is not soluble in aq. NaOH but evolves CO2 with aq. NaHCO3

(D) It is insoluble in aq. NaOH and does not evolve CO2 with aq. NaHCO3

Ans : (B)

Hints :

OH ONa– +

NaOH+ –

+ H O2 (Soluble in aqueous NaOH)

Phenol is less acidic than H2CO3. So it cannot liberate CO2 from NaHCO3 (aq).68. The basicity of aniline is weaker in comparison to that of methyl amine due to

(A) hyperconjugative effect of Me-group in MeNH2

(B) resonance effect of phenyl group in aniline(C) lower molecular weight of methyl amine as compared to that of aniline(D) resonance effect of – NH2 group in MeNH2

Ans : (B)

Hints :

H N2H N2

+

Here delocalisation of lone pair of nitrogen.

CH – NH23 → No delocalisation of lone pair

69. Under identical conditions, the SN1 reaction will occur most efficiently with(A) tert-butyl chloride (B) 1-chlorobutane(C) 2-methyl-1-chloropropane (D) 2-chlorobutaneAns : (A)Hints : Reactivity order for SN1 reaction in case of alkyl halide is 3°> 2°> 1°

70. Identify the method by which Me3CCO2H can be prepared(A) Treating 1 mol of MeCOMe with 2 moles of MeMgl (B) Treating 1 mol of MeCO2Me with 3 moles of MeMgl(C) Treating 1 mol of MeCHO with 3 moles of MeMgl (D) Treating 1 mol of dry ice with 1 mol of Me3CMglAns : (D)

Hints : Reaction Me C Mg I + O = C = O Me C – C – OMg I3 3→+

O

H O3+

Me C – C – OH3

O

– – +

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WBJEE - 2012 (Answers & Hints) Physics & Chemistry

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(13)

Q. 71 to Q. 80 carry two marks each.

71. 20 ml 0.1 (N) acetic acid is mixed with 10 ml 0.1 (N) solution of NaOH. The pH of the resulting solution is (pKa of aceticacid is 4.74)

(A) 3.74 (B) 4.74 (C) 5.74 (D) 6.74

Ans : (B)

Hints : CH3COOH + NaOH → CH3COONa + H2O

Initial 20 ml 0.1N 10ml 0.1N 0 meq

= 2 meq = 1 meq

Final 1 meq 0 meq 1 meq

Mixed solution behaves as acidic buffer

a[Salt]pH pK log[Acid]

= + = 4.74 + log 11

= 4.74

72. In the brown ring complex [Fe(H2O)5(NO)]SO4, nitric oxide behaves as

(A) NO+ (B) neutral NO molecule (C) NO– (D) NO2–

Ans : (A)

Hints : ‘NO’ ligand when attached to Fe behaves as positively charged ligand.

73. The most contributing tautomeric enol form of MeCOCH2CO2Et is

(A) CH2 = C(OH)CH2CO2Et (B) MeC(OH) = CHCO2Et

(C) MeCOCH = C(OH)OEt (D) CH2 = C(OH)CH = C(OH) OEt

Ans : (B)

Hints : CH – C – CH – COOEt3CH –C = CH – COOEt3

OH

Enol form

O H

74. By passing excess Cl2(g) in boiling toluene, which one of the following compounds is exclusively formed?

(A)

Me

ClCl

Cl

(B)

Me

Cl

ClCl

(C)

CCl

ClCl

Cl

3

(D)

CCl3

Ans : (D)

Hints :

CH3

+ Cl (g)2

CCl3

Free Radical benzylic substitution

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WBJEE - 2012 (Answers & Hints) Physics & Chemistry

Aakash Institute - Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472(14)

75. An equimolar mixture of toluene and chlorobenzene is treated with a mixture of conc. H2SO4 and conc. HNO3.Indicate the correct statement from the following(A) p-nitrotoluene is formed in excess(B) equimolar amounts of p-nitrotoluene and p-nitrochlorobenzene are formed(C) p-nitrochlorobenzene is formed in excess(D) m-nitrochlorobenzene is formed in excessAns : (A)

Hints : +

CH3 Cl

+

CH3 Cl

NO2 NO2

(excess)

conc.H SO2 4

+ conc.HNO3

– CH3 is activating group due to +I effect and hyperconjugation.But for – Cl, – I > + R effect.∴ – CH3 is weakly activating and – Cl is weakly deactivating

76. Among the following carbocations : Ph2C+CH2Me(I), PhCH2CH2CH+Ph (II), Ph2CHCH+Me (III) and Ph2C(Me)CH2

+ (IV),the order of stability is(A) IV > II > I > III (B) I > II > III > IV (C) II > I > IV > III (D) I > IV > III > IIAns : (B)

Hints : Ph CCH Me PhCH –CH –CH–Ph Ph CHCH–CH Ph C(Me)–CH 2 2 2 2 2 3 2 2

+ + + +

(I) (II) (III) (IV)(2°) (1°)

Number of resonating structures decide stability in I and II.77. Which of the following is correct?

(A) Evaporation of water causes an increase in disorder of the system(B) Melting of ice causes a decrease in randomness of the system(C) Condensation of steam causes an increase in disorder of the system(D) There is practically no change in the randomness of the system when water is evaporatedAns : (A)Hints : Evaparation increases randomness as liquid water is converted to water vapours.

78. On passing ‘C’ Ampere of current for time ‘t’ sec through 1 litre of 2 (M) CuSO4 solution (atomic weight of Cu = 63.5),the amount ‘m’ of Cu (in gm) deposited on cathode will be(A) m = Ct / (63.5 × 96500) (B) m = Ct/ (31.25 × 96500)(C) m = (C × 96500) / (31.25 × t) (D) m = (31.25 × C × t) / 96500Ans : (D)Hints : 1000 ml 2(M) CuSO4 ≡ 2(M) CuSO4 solution contains 2 moles Cu+2

Cu2 + 2e → Cu 2F 63.5g q = c × t coulomb or, 2 × 96500C

63.5 31.25 Ctm Ct g2 96500 96500

×= × =

×

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79. If the 1st ionization energy of H atom is 13.6 eV, then the 2nd ionization energy of He atom is(A) 27.2 eV (B) 40.8 eV (C) 54.4 eV (D) 108.8 eVAns : (C)

Hints : H 213.6En

= − ( )1 HE 13.6eV= −

( )1 HeE + 2

2z13.6n

= − × 2213.61

= − × = – 54.4eV

The 2nd IE = E∞ – ( )1 HeE + = 0 – (–54.4) eV = 54.4 eV

80. The weight of oxalic acid that will be required to prepare a 1000 ml (N/20) solution is(A) 126/100 gm (B) 63 / 40 gm (C) 63 / 20 gm (D) 126 / 20 gmAns : (C)

Hints : M.wt. of H2C2O4. 2H2O is 126 (n.factor = 2) , E.wt = 126

2= 63

no. of gm equivalentNvolume of solution in L

−=

∴ 1 w / 63 63w gm

20 1 20= =∵

Page 36: Answer Key and Solution for Wb Jee 2012

Q.1 – Q. 60 carry one mark each1. In higher plants, continuity of cytoplasm from one cell to its neighbouring cells is established through

(A) Apoplast (B) Chloroplast(C) Leucoplast (D) SymplastAns : (D)Hints : A number of plasmodesmata are present in pit through which the cytoplasm of one cell is in contact withother. These form the symplastic system between two cells

2. In plant tissue culture, the callus tissues can be regenerated into complete plantlets primarily by altering the concen-tration of(A) Sugars (B) Vitamins(C) Amino acids (D) HormonesAns : (D)Hints : Differential concentration of auxin and cytokinin regulates morphogenesis

3. If you want to develop hybrid seeds within a bisexual flower, the following part needs to be removed from the sameflower(A) Stigma (B) Ovary(C) Anther (D) OviductAns : (C)Hints : Removal of anther (i.e., emasculation) is required for the development of hybrid seeds

4. Xenogamy is essentially a type of(A) Cleistogamy (B) Allogamy(C) Autogamy (D) HomogamyAns : (B)Hints : Xenogamy is true cross pollination i.e. allogamy

5. Basic principle of developing transgenic plants and animals is to introduce the gene of interest into the nucleus of

(A) Somatic Cell (B) Vegetative Cell(C) Germ Cell (D) Body cellAns : (C)Hints : The genes are always introduced in the genome.

6. Who is regarded as the ‘Father of Taxonomy’ ?(A) John Ray (B) Carolus Linnaeus(C) A.P. de Candolle (D) Charles DarwinAns : (B)Hints : Carolus Linnaeus

Code-γγγγγWBJEE - 2012 (Answers & Hints) Biology

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ANSWERS & HINTSfor

WBJEE - 2012SUB : BIOLOGY

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WBJEE - 2012 (Answers & Hints) Biology

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7. Which pair of the following diseases are caused by two genes located on human X chromosome ?(A) Colour blindness and Phenylketonuria(B) Colour blindness and Haemophilia(C) Colour blindness and Albinism(D) Colour blindness and HypertrichosisAns : (B)Hints : Both colour blindness and haemophilia are X-linked recessive disorders.

8. Which one of the following triplet codon is known as initiation codon ?(A) UUU (B) UAA(C) AUG (D) UGAAns : (C)Hints : AUG codes for methionine

9. Muga silk worm feeds on(A) Shorea (B) Terminalia(C) Machilus (D) MorusAns : (C)

10. Mammals have originated from which of the following ?(A) Pisces (B) Amphibia(C) Reptilia (D) AvesAns : (C)

11. Which one of the following animal phyla possesses spicule ?(A) Annelida (B) Mollusca(C) Porifera (D) PlatyhelminthesAns : (C)Hints : Spicules are characteristic feature of porifera.

12. Fin rot of fish is caused by(A) Aeromonas (B) Pseudomonas(C) Branchiomyces (D) XenopsyllaAns : (B)

13. Which one of the following animal possesses giant chromosome ?(A) Drosophila (B) Mouse(C) Pigeon (D) ElephantAns : (A)

14. The vein which is formed from the capillaries of an organ and terminates into capillaries in some other organ beforeentering the heart is called(A) Pulmonary vein (B) Coronary vein(C) Portal vein (D) Systemic veinAns : (C)

15. Which one of the following genus of insects prefer to breed in clean water and their larvae lie parallel to the surface ofwater ?(A) Anopheles (B) Culex(C) Aedes (D) PhlebotomusAns : (A)

16. After forceful inspiration, the amount of air that can be breathed out by maximum forced expiration is equal to(A) Inspiratory Reserve Volume (IRV) + Expiratory Reserve Volume (ERV) + Tidal Volume (TV) + Residual Volume(RV)(B) IRV + RV + ERV(C) IRV + TV + ERV(D) TV + RV + ERVAns : (C)Hints : It is also called as vital capacity.

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WBJEE - 2012 (Answers & Hints) Biology

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17. Amount of oxygen suplied by 100 ml arterial blood while passing through the tissues is(A) 0.4-0.6 ml (B) 4-6 ml(C) 14-15 ml (D) 19-20 mlAns : (B)Hints : 100 ml blood carries 19–20 ml oxygen but it delivers 4–6 ml of oxygen at tissue level derving normal condition

18. Cholecystokinin stimulates the secretion of(A) Bile (B) Gastric juice(C) Pancreatic juice (D) Succus entericusAns : (A)

19. In a normal adult human, the average cardiac output (stroke volume) is(A) 47 ml (B) 70 ml(C) 5 litre (D) 3.3 litreAns : (B)Hints : Storke volume is volume of blood ejected by left ventricle in one systole/stroke.

20. Humoral immunity is mediated by(A) Cytotoxic T-cell (B) Plasma cell(C) Eosinophil (D) NeutrophilAns : (B)Hints : Plasma cell secretes antibodies to perform humoral immunity.

21. The expanded name of MRI is(A) Medical Research Instrument (B) Magnetic Resonance Imaging(C) Magnetic Research Institute (D) Medical Resonance ImagingAns : (B)

22. The volume of ‘Anatomical Dead Space’ air is normally(A) 230 ml (B) 210 ml(C) 190 ml (D) 150 mlAns : (D )Hints : It is volume of air from nostril to terminal bronchiole which is not taking part in exchange of gases.

23. Bone is mainly composed of(A) Iron and Phosphorus (B) Sulphur and Calcium(C) Calcium and Phosphorus (D) Calcium and MagnesiumAns : (C)

24. Which one of the following is incorrect for ‘Atherosclerosis’ ?(A) Constriction of arterial lumen reduces the blood flow(B) Loss of dilatation ability of the arterial wall and its rupture(C) Cholesterol deposition at the inner wall of the artery(D) Proliferation of the vascular musclesAns : (B)Hints : Atherosclerosis in due to depositon of chloesterol and proliferation of vascular muscle in presence of PlateletDerived Growth Factor (PDGF)

25. Which one of the following statement is incorrect ?(A) Glucagon is secreted by pancreas (B) Androgen is produced by ovary(C) Thyroxine is secreted by thyroid (D) Oxytocin is secreted by pituitaryAns : (B)

26. Juxta-glomerular apparatus is made-up of(A) Juxta-glomerular cell, Macula densa and Lacis cell(B) Juxta-glomerular cell, Purkinje cell and Chief cell(C) Juxta-glomerular cell, Lacis cell and Myoepithelial cell(D) Juxta-glomerular cell, Macula densa and Argentaffin cellAns : (A)

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27. Which vitamin helps in blood coagulation ?(A) Vitamin K (B) Vitamin C(C) Vitamin A (D) Vitamin DAns : (A)

28. Presence of which of the following hormone in the urine confirms pregnancy ?(A) Progesterone (B) Estrogen(C) Human choriogenic gonadotropin (D) ProlactinAns : (C)Hints : hCG secreted by syncytiotrophoblast of placenta confirms pregnancy, when detected in urine

29. Which one of the following is mainly responsible for the second heart sound ?(A) Closure of atrioventricular valves(B) Opening of atrioventricular valves(C) Closure of semilunar valves(D) Thrust of blood on ventricular wall during atrial contractionAns : (C)Hints : Second heart sound DUP is produced by closure of semilunar valves during onset of ventricular diastole

30. Brunner’s gland is present in(A) Duodenum (B) Jejunum(C) Ileum (D) RectumAns : (A)

31. Numerical aperture of microscope lens is expressed by(A) Angular aperture only(B) Refractive index only(C) Both angular aperture and refractive index(D) Wave length of the light usedAns : (C)Hints : Numerical (NA) = nsin θ

aperturewhere, n = Refractive index of the mediumθ = Half of angular aperture

32. Which one of the following is not true for enzymes ?(A) They act on a specific substrate(B) They are made up of fat and sugar(C) They act at a specific temperature(D) They act at a specific pHAns : (B)Hints : Enzymes are exclusively made up of protein

33. Living cells placed in isotonic solution (0.9% saline) retain their size and shape. This is based on the concept of(A) Osmosis (B) Diffusion(C) Facilitated diffusion (D) TranspirationAns : (A)Hints : Osmosis

34. Identify the polysaccharide with β-glycosidic bond(A) Starch (B) Glycogen(C) Sucrose (D) CelluloseAns : (D)

35. Simple storage protein that coagulates upon heating but remains soluble in dilute salt solution is correctly exempli-fied by(A) Globulin (B) Albumin(C) Histone (D) CollagenAns : (B)

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36. Melting of DNA at an elevated temperature (70oC) is primarily due to the breakdown of(A) Phosphodiester bonds (B) Glycosidic bonds(C) Disulfide bonds (D) Hydrogen bondsAns : (D)Hints : At high temperature, 700C or above denaturation of ds DNA occurs i.e., separation of two strands due tobreakdown of H-bonds.

37. N-acetyl muramic acid is found in(A) Cell wall component of plant(B) Cell wall component of Gram positive bacteria(C) Cell wall component of fungi(D) Viral coat materialAns : (B)Hints : N-acetyl muramic acid is found in cell wal of Gram positive bacteria.

38. Identify the bacterium that appears violet after Gram staining(A) Salmonella enterica (B) Escherichia coli(C) Mycobacterium tuberculosis (D) Rhizobium melilotiAns : (C)Hints : Mycobacterium tuberculosis

39. The usual source of restriction endonucleases used in gene cloning is from(A) Fungi (B) Bacteria(C) Plants (D) VirusesAns : (B)

40. Portion of apical meristem that gives rise to xylem tissue is called(A) Protoxylem (B) Procambium(C) Metaxylem (D) TracheidAns : (B)Hints : Procambium gives rise to xylem.

41. For a clean environment, which one of the following is not essential ?(A) Producer (B) Consumer(C) Decomposer (D) PolluterAns : (D)Hints : Polluter

42. Inulin is a type of(A) Protein (B) Carbohydrate(C) Lipid (D) Nucleic acidAns : (B)Hints : Inulin is polymer of fructose.

43. Higher animals cannot synthesize few fatty acids which are very essential for their growth and development. Thesefatty acids are typically(A) Saturated (B) Cyclic(C) Unsaturated (D) BranchedAns : (C)

44. Identify the membrane across which the proton (H+) gradient facilitates ATP synthesis in a typical eukaryotic cell.(A) Plasma membrane (B) Mitochondrial inner membrane(C) Mitochondrial outer membrane (D) Nuclear membraneAns : (B)Hints : Mitochondrial inner membrane.

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45. What is the substrate for lipase enzyme ?(A) Protein (B) Carbohydrate(C) Lipid (D) Nucleic acidAns : (C)

46. Which one of the following life cycle stages of Wuchereria bancrofti is infective to man ?(A) Microfilaria (B) 1st stage larva(C) 2nd stage larva (D) 3rd stage larvaAns : (D)

47. Which one of the following life cycle stages of malarial parasite is responsible for relapse of malarial symptoms ?(A) Merozoite (B) Sporozoite(C) Hypnozoite (D) GametocyteAns : (A)

48. The correct sequence of different phases of cell cycle is(A) S - G1 - G2 - M (B) G1 - S - G2 - M(C) G2 - S - G1 - M (D) G1 - G2 - S - MAns : (B)Hints : G1 → S → G2 → M

49. Which one of the following poultry birds is not an English breed ?(A) Sussex (B) Australorp(C) Orpington (D) MinorcaAns : (D)Hints : Minorca is Mediterranean breed

50. The important gas which was absent during the formation of earth is(A) Oxygen gas (B) Hydrogen gas(C) Nitrogen gas (D) Carbon dioxide gasAns : (A)

51. In spermatogenesis, reduction division of chromosome occurs during conversion of(A) Spermatogonia to primary spermatocytes(B) Primary spermatocytes to secondary spermatocytes(C) Secondary spermatocytes to spermatids(D) Spermatids to spermsAns : (B)

52. The carcinoma, a type of cancer, originates from(A) Blood (B) Connective tissue(C) Epithelial tissue (D) Lymph glandAns : (C)

53. Homo erectus evolved during(A) Oligocene (B) Pliocene(C) Pleistocene (D) MioceneAns : (C)

54. Which one of the following is not included under in situ conservation ?(A) National park (B) Wild life sanctuary(C) Zoological garden (D) Biosphere reserveAns : (C)Hints : Zoological Garden

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WBJEE - 2012 (Answers & Hints) Biology

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55. Animals are classified into hierarchical groups. In which one of the following, the largest number of species is found?(A) Genus (B) Order(C) Family (D) CohortAns : (D)Hints : Cohort is a category placed in between Infra-class and super order.

CLASSSUBCLASSINFRACLASSCOHORTSUPER ORDERORDER

56. The blood vessel which suply oxygenated blood to cardiac tissue is(A) Coronary artery (B) Coronary vein(C) Coronary sinus (D) Pulmonary veinAns : (A)

57. Vagus nerve is a(A) Mixed Xth cranial nerve (B) Mixed XIth cranial nerve(C) Mixed Xth thoracic nerve (D) Sensory Xth cranial nerveAns : (A)

58. The number of spinal nerves in human is(A) 10 pairs (B) 12 pairs(C) 43 pairs (D) 31 pairsAns : (D)

59. Third and fourth ventricles of the brain are connected by(A) Aqueduct of Sylvius (B) Foramen of Monro(C) Foramen of Magnum (D) Corpus CollosumAns : (A)

60. All or none law is not applicable for(A) Whole skeletal muscle (B) Single skeletal muscle fibre(C) Whole cardiac muscle (D) Single smooth muscle fibreAns. (A)Hints : All or none law is applicable to single muscle fibre and not for entire muscle.

Q. 61 to Q. 80 carry two marks each61. How many ATP molecules will be generated in a plant system during complete oxidation of 40 moles of glucose ?

(A) 190 (B) 380 (C) 1520 (D) 3040Ans : (C)Hints : 40 × 38 ATP = 1520 ATP Molecules

62. A – DNA is(A) Left-handed helix with 12 nucleotide pairs per turn(B) Right-handed helix with 11 nucleotide pairs per turn(C) Right-handed helix with 12 nucleotide pairs per turn(D) Left-handed helix with 11 nucleotide pairs per turnAns : (B)Hints : Right handed helix with 11 nucleotide pairs per turn.

63. Bursa fabricius is an important organ of birds. This organ is associated with(A) Generation of basophils (B) Production of uric acid(C) Metabolism of fatty acid (D) Generation of B-cellAns : (D)Hints : Bursa of fabricius is the site of B-cell maturation because bone marrow is absent in birds.

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64. Which one of the following is a population ?(A) A spider and some trapped flies in its web(B) Earthworm that lives in a grassland along with other arthropods(C) All the plants in a forest(D) All the oak trees in a forestAns : (D)Hints : All oak trees in a forest constitute a population

65. In most simple type of canal system of porifera, water flows through which one of the following ways ?(A) Ostia → Spongocoel → Osculum → Exterior(B) Spongocoel → Ostia → Osculum → Exterior(C) Osculum → Spongocoel → Ostia → Exterior(D) Osculum → Ostia → Spongocoel → ExteriorAns : (A)Hints : It is the Ascon type of canal system, which is the simplest one.

66. What would be the percentage of thymine in a double stranded DNA sample which contains 20% cytosine of the totalbases ?(A) 10% (B) 20% (C) 30% (D) 40%Ans : (C)Hints : C = 20% ∴ G = 20%

So, A or T = 100 (20 20)

2− +

A or T = 30%67. Which one of the following is a correct match ?

(A) Filariasis – Taenia solium (B) Encephalitis – Cules vishnui(C) Malaria – Phlebotomus sp. (D) Kala-azar – Anopheles stephensiAns : (B)

68. One micrometer is equal to(A) 0.0001 mm (B) 0.001 mm (C) 0.01 mm (D) 0.00001 mmAns : (B)Hints : 1 micrometer = 0.001 mm

69. Identify the origin of sympathetic nerve fibres and the location of their ganglia.(A) They arise from thoraco-lumber region of spinal cord and form ganglia just beside the vertebral column(B) They arise from thoraco-cervical region of spinal cord and form ganglia just beside the vertebral column(C) They arise from cranio-sacral region of spinal cord and form ganglia very close to effector organ(D) They arise from thoraco-lumber region of spinal cord and form ganglia very close to effector organAns : (A)

70. Among the following characteristics, indicate the correct combiantion applicable to conditional reflex.P : Acquired by practice or learningQ : Non acquired by birthR : Does not abolish by lack of practiceS : Participation of cerebral cortexT : Originates spontaneously(A) P, Q, R (B) P, Q, S (C) P, R, T (D) Q, R, TAns : (B)

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71. During synaptic transmission of nerve impulse, neurotransmitter (P) is released from synaptic vesicles by the actionof ions (Q). Choose the correct P and Q.(A) P = Acetylcholine, Q = Ca++ (B) P = Acetylcholine, Q=Na+

(C) P = GABA, Q=Na+ (D) P = Cholinesterase, Q=Ca++

Ans : (A)Hints : Ca++ is responsible for release of neurotransmitters Acetylcholine from axon terminal by rupturing synapticalvesicle.

72. Choose the right sequential phenomena among the following during the delivery of O2 from blood to tissueP : Absorption of CO2 by the bloodQ : Reaction of absorbed CO2 with H2O to form H2CO3 within RBC and its convesion into H+ and HCO3

– ionsR : Reaction of absorbed CO2 with H2O in plasma to form H2CO3 and its conversion into H+ and HCO3

– ionsS : Combination of H+ with heme portion of HbO2 to release O2

T : Combination of HCO3– with heme portion of HbO2 to form reduced hemoglobin and release of O2

(A) P, Q, T (B) P, R, S (C) P, Q, S (D) P, R, TAns : (C)

73. Oral contraceptive pill is composed of(A) Estrogen and Progesterone (B) Estrogen and Testosterone(C) Progesterone and Testosterone (D) Estrogen and Growth hormoneAns : (A)

74. Match the items in column I with those in column II and choose the correct answer.

Column I Column II

P. Electroencephalography i. A graphic recording of the electrical activity of heart

Q. Electrocardiography ii. A graphic recording of the electrical activity of brain

R. Endoscopy iii. A technique that gives image automatically in multiple planes

S. MRI Iv. To view within the body without cutting through the overlaying tissue

(A) P-ii, Q-i, R-iv, S-iii (B) P-iv, Q-ii, R-iii, S-i (C) P-i, Q-iii, R-ii, S-iv (D) P-iii, Q-iv, R-i, S-iiAns : (A)

75. What are ketone bodies ?(A) Acetoacetic acid, Acetone and Beta-hydroxybutyric acid(B) Nicotinic acid, Folic acid and Ascorbic acid(C) Acetone, Beta-hydroxybutyryl CoA, and Acetoacetic acid(D) Acetic acid, acetone and Beta-hydroxybutyric acidAns : (A)

76. Identify the correct sequence of events in the viral replication process.1. Eclipse2. Maturation3. Adsorption4. Assembly5. Penetration6. Lysis(A) I→ II→ III→ IV→ V→ VI (B) II→ I→ III→ IV→ V→ VI(C) III→ V→ I→ II→ IV→ VI (D) III→ V→ I→ IV→ II→ VIAns : (C)Hints : Adsorption → Penetration → Eclipse → Maturation→ Assembly→ Lysis[Ref. A Text Book of Biology – Mitra, Chattopadhyay & Chanda Pg. 21, 22, 5th Edition – 2011, Vol-II]

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77. Match the items in column I with those in column II and choose the correct answer.

Column I Column II

P. Blue green algae as biofertilizer i. Ectomycorrhiza

Q. Fungi as biofertilizer ii. Thiobacillus sp.R. Free living nitrogen fixing bacteria

iii. Anabena sp.

S. Phosphate solubilizing bacteria iv. Clostridium sp.v. Azozpirillum sp.

(A) P-iii, Q-i, R-v, S-ii (B) P-v, Q-i, R-ii, S-iv(C) P-v, Q-iv, R-i, S-ii (D) P-iv, Q-ii, R-v, S-iiAns : (A)

78. A man having the genotype EEFfGgHH can produce P number of genetically different sperms, and a woman of geno-type IiLLMmNn can generate Q number of genetically different eggs. Determine the values of P and Q.(A) P=4, q=4 (B) P=4, Q=8(C) P=8; Q=4 (D) P=8, Q=8Ans : (B)Hints : Given genotype of man = EEFfGgHH∴ P = 2n = 22 = 4Given genotype of woman = IiLLMmNn∴ Q = 2n = 23 = 8When n = Number of heterozygosity.

79. In an organism, tall phenotype is dominant over recessive dwarf phenotype, and the alleles are designated as T andt, respectively. Upon crossing two different individuals, total 250 offsprings were obtained, out of which 124 displayedtall phenotype and rest were dwarf. Thus, the genotype of the parents were(A) TT x TT (B) TT x tt(C) Tt x Tt (D) Tt x ttAns : (D)Hints : Test Cross.

80. Identify the incorrect statements from the following.(P) Cymose inflorescence found in Hibiscus sp.(Q) Hypenthodium is found in Ficus benghalensis(R) Synandrous stamen is found in Calotropis(D) Hesperidium type of fruit is Mango(A) R, S (B) P, Q(C) Q, R (D) P, SAns : (A)