aiats iitjee 2012 test-2 p-ii solution
Post on 07-Jul-2018
221 Views
Preview:
TRANSCRIPT
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
1/18
Test - 2 (Paper - II) (Answers & Hints) All India Aakash Test Series for IIT-JEE 2012
1/18
TEST - 2 (Paper - II)
A N S W E R S
1. (B)
2. (A)
3. (B)
4. (A)
5. (A)
6. (A)
7. (D)
8. (C)
9. (A, B, D)
10. (A, C)
11. (A)
12. (A, B, C)
13. (3)
14. (6)
15. (9)
16. (8)
17. (0)
18. (9)
19. A → (p, q, s)
B → (q, s)
C → (q, t)
D → (q, r)
20. A → (p, q)
B → (q)
C → (q, s, t)
D → (p, r)
CHEMISTRY MATHEMATICS PHYSICS
21. (A)
22. (B)
23. (B)
24. (D)
25. (D)
26. (D)
27. (C)
28. (B)
29. (A, B, C, D)
30. (A, B, C, D)
31. (A, B)
32. (B, D)
33. (1)
34. (9)
35. (4)
36. (2)
37. (3)
38. (3)
39. A → (p, r)
B → (t)
C → (p, q, s)
D → (p, q, r, s)
40. A → (s)
B → (q)
C → (p)
D → (q)
41. (C)
42. (B)
43. (C)
44. (C)
45. (D)
46. (B)
47. (B)
48. (B)
49. (A, B, C)
50. (A, C)
51. (B, C, D)
52. (B, C)
53. (7)
54. (2)
55. (2)
56. (2)
57. (6)
58. (4)
59. A → (t)
B → (r)
C → (s)
D → (q)
60. A → (t)
B → (t)
C → (r)
D → (p)
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
2/18
All India Aakash Test Series for IIT-JEE 2012 Test - 2 (Paper - II) (Answers & Hints)
2/18
1. Answer (B)
Decreases in denser medium.
2. Answer (A)
Wave in same phase add-up together.
3. Answer (B)
π∗p y has two nodal planes
4. Answer (A)
7 22 2
1 1 11.09677 10 (1)
(4) (7)
= × −
λ
⇒ λ = 2.17 × 10 –9 m/s
5. Answer (A)
1 atomst : 5 4 3 2 1 4 lines (different)
2 atomnd
3 atomrd
4 atomth
5 atomth
6 atomth
7 atomth
:
::
:
:
:
5
55
5
5
5
4
43
3
2
1
11
1
1
1
1 line
1 line1 line
1 line
1 line
1 line
10 lines
4
2
5 → 3 → 2 → 1 has repeated lines
6. Answer (A)
Factual.7. Answer (D)
3 21
2 32
T (4) (2) 4T 9(3) (4)
= × =
8. Answer (C)
0 01
h h KE (KE)h
ν = ν + ∴ ν = ν + .
9. Answer (A, B, D)
O N O
O
–
ANSWERS & HINTS
PART - I (CHEMISTRY)
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
3/18
Test - 2 (Paper - II) (Answers & Hints) All India Aakash Test Series for IIT-JEE 2012
3/18
10. Answer (A, C)
Bond order of O 2 – = 1.5
Bond order of O 22– = 1
Bond order of O 2 = 2
Bond order of O 3 = 1.5
11. Answer (A)
Cp – C v = R ⇒ (1.25 M – 0.75 M) = 2
0.5 M = 2 ∴ Molar mass = 4 g
12. Answer (A, B, C)
Factual
13. Answer (3)
5 mole steam absorbs 660 kJ.
To maintain temperature constant CO(g) must evolve.
660 kJ ∴ n CO = 6 ⇒ 2On = 3
14. Answer (6)
Wrev =3.99V
2.303 nRT logV
−
= –2.303 × 1 × 8.314 × 522.27 log(3.99)
= –6000 J = –6 kJ
15. Answer (9)
∆G = ∆H – T ∆S for minimum ∆H, ∆G = 0
∆H = T ∆S = 298 × 30.2 = 9000 = 9 kJ
16. Answer (8)
Equation (i) × 2 + Equation (ii) × 2 – Equation (iii)
= –65 × 2 + (–97) × 2 – (–340) = 340 – 194 – 130 = 16 kcal
= (2 kcal) × 8
17. Answer (0)
H2O( ) H2O(g)
Molar heat capacity = ∞ =10
18. Answer (9)
∆G° = –2.303 × 8.314 × 298 log K = 55.6 × 10 3
∴
355.6 10log K 9.744 ( x 0.744)
2.303 8.314 298×= − = − = − −
× ×
∴ x = 9
19. Answer → A(p, q, s), B(q, s), C(q, t), D(q, r)
20. Answer → A(p, q), B(q), C(q, s, t), D(p, r)
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
4/18
All India Aakash Test Series for IIT-JEE 2012 Test - 2 (Paper - II) (Answers & Hints)
4/18
PART - II (MATHEMATICS)
21. Answer (A)
Let,1 2 2 3 1 3
1 3 5(say)
| | | | | |
= = =
− − −
k
z z z z z z
∴ 1 2 2 3 1 31 3 5| | , | | and | |− = − = − =z z z z z z k k k
or, 1 2 1 2 2 3 2 3 1 3 1 32 2 21 9 25
( )( ) , ( )( ) and ( )( )− − = − − = − − =z z z z z z z z z z z z k k k
So, 2 1 21 2
1( )= −
−k z z
z z
22 3
2 3
9( )= −
−k z z
z z
23 1
3 1
25 ( )= −−
k z z z z
∴1 2 2 3 3 1
1 9 250+ + =
− − −z z z z z z
22. Answer (B)
From question
2a 2α 2 + b α + c = 0
∴ b α + c = –2 a 2α 2 …(i)
also, 3 a 2β2 – 2 b β – 2 c = 0
∴2 23
2β+ = βb c a …(ii)
Now, let f (x ) = 4 a 2x 2 + 3 bx + 3 c
Then f (α ) = 4 a 2α 2 + 3( b α + c )
= 4 a 2α 2 – 6 a 2α 2 …[from (i)]
= –2 a 2α 2 < 0
f (β) = 4 a 2β2 + 3 b β + 3 c
= 4 a 2β2 + 3( b β + c )
= 2 2 2 29
42
β + βa a …[from (ii)]α
βx -axisγ
= 2 217
02
β >a
f (α ) f (β) < 0, then one root of the equation f (x ) = 0 must lie between α and β.
23. Answer (B)
Let, the seating arrangements of the persons are as shown below,
S N S N N S N S N N
where, S → persons may be selected
N → persons may not be selected
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
5/18
Test - 2 (Paper - II) (Answers & Hints) All India Aakash Test Series for IIT-JEE 2012
5/18
As there are 6N’s and 4S’s.
So, the problem is similar to that of selecting any 4 places out of 7 gaps (shown below)
X N X N X N X N X N X N X
which can be done is 7C 4 ways
= 35 ways
24. Answer (D)
6 = 1 + 5 Required number of ways
= 2 + 43
161 61 61
2!51 2! 4! 3! 3! 2!
C
= + + ×
= 3 + 3
25. Answer (D)
Let, f (x ) = ax 2 + bx + c ; a , b , c ∈ R ; a ≠ 0
Given, α and β are the roots of the equation f (x ) = 0
For (1) :
2
2 21 − + − = − + =
a b a bx cx f c
x x x x
∴ cx 2 – bx + a = 0 has roots1 1
and− −α β
For (2) :
Equation with roots ( α + β) and αβ is
Given by x 2
– ( α + β + αβ )x + (α + β)αβ = 0
or, 2 2 0
− − + + − = b c bc
x x a a a
or, a 2x 2 + (b – c )ax – bc = 0
For (3) :
22 2 4 ( ) (say)
2 4 2
= + + = + + = x ax bx f c ax bx c g x
Now, g (x + 1) = a (x + 1) 2 + 2 b (x + 1) + 4 c
= a (x 2 + 1 + 2 x ) + 2 bx + 2 b + 4 c
= ax 2 + 2( a + b )x + a + 2 b + 4 c
∴ Roots of the equation ax 2 + 2( a + b )x + a + 2 b + 4 c = 0 are 2 α – 1 and 2 β – 1.
26. Answer (D)
5 5| | = − +z z
z z
5 5| |
≤ − +z z z
or, 5| | 4 | |≤ +z z
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
6/18
All India Aakash Test Series for IIT-JEE 2012 Test - 2 (Paper - II) (Answers & Hints)
6/18
or,5
| | 4| |
− ≤z z
or, | z |2 – 4| z | – 5 ≤ 0
or, | z |2 – 4| z | + 4 – 9 ≤ 0
or, ( | z | – 2) 2 ≤ 9
∴ | z | – 2 ≤ 3
or | z | ≤ 5
Obviously | z | max = 5
27. Answer (C)
Let, n 1, n 2, n 3 and n 4 be the number of white, red, yellow and brown balls that are selected.
∴ n 1 + n 2 + n 3 + n 4 = 8
So, the required number of ways
= coefficient of x 8 in (1 + x + x 2 + ........) 4
= coefficient of x 8 in (1 – x ) –4
= coefficient of x 8 in (1 + 4C 1x +5C 2x
2 + 6C 3x 3 + ........)
= 11C 8
= 11C 3
28. Answer (B)
210 = 2 × 3 × 5 × 7
x 1 x 2 x 3 x 4 = 210∴ Each of x i may be equal to 2 or 3 or 5 or 7
So, the possible number of solutions
= 4 × 4 × 4 × 4
= 256
29. Answer (A, B, C, D)
Let, z 1 = e i α
and z 2 = re i θ
∴( ) – ( )
1 2α−θ θ−α= =i i z z r e r e
OP = | z 1 | = 1 = OA
Also, 1 2 1 2 2= = = =OR z z z z z OQ
α –( – )αθ
θ
P z ( )1
A x O
R z z ( )1 2
y
(1, 0)
Q z ( )2
and ∠QOP = ( θ – α ) = ∠ROA
Obviously ∆POQ and ∆AOR are congruent
∴ PQ = AR
or, 2 1 1 2 1− = −z z z z
Also, Area ∆POQ = Area ∆AOR
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
7/18
Test - 2 (Paper - II) (Answers & Hints) All India Aakash Test Series for IIT-JEE 2012
7/18
Now, 1 1 1 21 2 1 2 2 1
1arg arg arg arg
= = =
z z z z
z z z z z z
∆POQ and ∆AOR are congruent
∴ ∠PQO = ∠ARO
⇒2 1 1 2
2 1 2arg arg
1
−=
− z z z z
z z z
or, 12 1 2
1arg 1 arg 1
1
− = +
− z z z z
or, ( )1 11 2 1 2
1 1arg 1 arg 1 11
z z z z z z
− = − =
− ∵
So, all the options are correct.
30. Answer (A, B, C, D)
Number of straight lines = 7C 2 – 3C 2 + 1 = 19Number of triangles = 7C 3 –
3C 3 = 34
For quadrilateral ; it can be formed in following cases :Selected collinear
points (3)Selected non-collinear
points (4)Number of
quadrilaterals formed
(i) 1 3 = × = 123 4C C 1 3
(ii)
(iii)
2
0
2
4
= × = 183 4C C 2 2
= × = 13 4C C 0 4
So, number of quadrilaterals = 12 + 18 + 1 = 31
Similarly, number of pentagons = 3C 1 × 4C 4 + 3C 2 × 4C 3= 3 + 12 = 15
31. Answer (A, B)
21 1= + + −z t i t
⇒ Re( z ) = x = 1 + t
2Im( ) 1= = −z y t
So, ( x – 1) 2 + y 2 = t 2 + 1 – t 2O
y
P z ( )
c Q (2, 0)
x (0, 0) (1, 0) 1
1
or, ( x – 1) 2 + y 2 = 1
i.e. , | z – 1| = 1
⇒ z lies on a circle with radius 1 and centered at (1, 0). Obviously the circle passes throgh origin.
Also, | z |max = 2
⇒ | z | ≤ 2 ∀ t ∈ R
From figure, its clear that ∠CPQ ≠ 2π
2π
∠ = ∵ OPQ
⇒1
arg2 2
− π ≠ −
z z
∆CPQ is an isosceles ∆, so its quite possible that if PQ also becomes 1 (for some P on the circle),
Then ∆CPQ may become equilateralSo, it is possible that | z – 1| = | z – 2| for some z or for some values of t .
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
8/18
All India Aakash Test Series for IIT-JEE 2012 Test - 2 (Paper - II) (Answers & Hints)
8/18
32. Answer (B, D)
The given equation is x 2 + 2( m – 1) x + (m + 5) = 0
Here, D r = (2( m – 1)) 2 – 4( m + 5)
= 4[m 2 + 1 – 2 m – m – 5]
= 4[ m 2 – 3 m – 4]
= 4( m + 1) ( m – 4)
For, the equation to have at least one negative root
(i) Either both the roots are real and negative.
OR
(ii) Exactly one root is negative.
Case I : D ≥ 0 and α + β < 0 and αβ > 0
⇒ m ≤ –1 or m ≥ 4 …(i)
and m > 1 …(ii)and m > –5 …(iii)
(i) ∩ (ii) ∩ (iii) : –5 –1 1 4
⇒ m ≥ 4 …(a)
Case II : αβ < 0
m < –5 …(b)
From (a) and (b)
m ∈ (– ∞, –5) ∪ [4, ∞]33. Answer (1)
This is possible when
k 2 – 4 = 0 ⇒ k = ±2
k 2 – 5 k + 6 = 0 ⇒ k = 3, 2
and k 2 – 3 k + 2 = 0 ⇒ k = 1, 2
The common value of k for which all the conditions are satisfied is k = 2
So, there is only one value of such ‘ k ’ is possible.
34. Answer (9)Since, ω and ω2 are the roots of the equation
1 1 1 3+ + =+ + +a x b x c x x
or [x 2 + (b + c )x + bc + x 2 + ( a + c )x + ac + x 2 + ( a + b )x + ab ]x = 3( a + x )(b + x )(c + x )
or 3 x 3 + 2( a + b + c )x 2 + x (ab + bc + ca ) = 3[ x 3 + ( a + b + c )x 2 + ( ab + bc + ca )x + abc ]
or (a + b + c )x 2 + 2( ab + bc + ca )x + 3 abc = 0
Obviously, 22( ) 1+ +ω+ω = − = −
+ +ab bc ca
a b c …(i)
and 2 3· 1ω ω = =+ +abc
a b c …(ii)
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
9/18
Test - 2 (Paper - II) (Answers & Hints) All India Aakash Test Series for IIT-JEE 2012
9/18
From (i) and (ii), we have
2(ab + bc + ca ) = 3 abc
or,1 1 1 3
2+ + =
c a b
or, 1 1 1 3 932 2
+ + ++ + = + =c a b c a b
or,1 1 1
2 9+ + +
+ + = a b c
a b c 35. Answer (4)
Let, 2 n – 1 and 2 n + 1 be two consecutive odd integers and roots of the equation ax 2 + bx + c = 0.
Then, 2 n – 1 + 2 n + 1 = −b a
…(i)
and (2 n – 1)(2 n + 1) =c a
…(ii)
Therefore,2
4
− b c a a
= (4 n )2 – 4 × (4 n 2 – 1)
= 16 n 2 – 16 n 2 + 4
= 4
∴
2
24
4− =b c
a a
or,2
24 4− =b ac
a
36. Answer (2)1 Row
st
2 Rownd
3 Rowrd
As there are 1, 3 and 4 boxes in the 1 st, 2 nd and 3 rd row respectively.
Then the required number of ways of selecting the boxes, so that none of the row may remain empty is
= Coefficient of x 6 in the expansion of ( 1C 1x )(3C 1x +
3C 2x 2 + 3C 3x
3)(4C 1x +4C 2x
2 + 4C 3x 3 + 4C 4x
4)
= Coefficient of x 6 in x (3x + 3 x 2 + x 3)(4x + 6 x 2 + 4 x 3 + x 4)
= Coefficient of x 6 in x 3(3 + 3 x + x 2)(4 + 6 x + 4 x 2 + x 3)= Coefficient of x 3 in (3 + 3 x + x 2)(4 + 6 x + 4 x 2 + x 3)
= 21
Now,
The word LOKPAL may be arranged in6!2
ways in the boxes. Therefore, total number of required ways
=6!
212
×
=2(5 2 1) (2 2 2)!
2
× + × × +
∴ n = 2
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
10/18
All India Aakash Test Series for IIT-JEE 2012 Test - 2 (Paper - II) (Answers & Hints)
10/18
37. Answer (3)
9x 2 – 6 x + y 2 = 24
or, (3 x – 1) 2 + y 2 = 25
The possible solutions are
3 – 1x
y
±5
0
0
±5
±3
±4
±4
±3
(2, 0), (–1, 3) and (–1, –3) only
∴ Number of ordered pair = 3
38. Answer (3)
N = 5880 = 2 3 × 3 × 5 × 7 2
= 2 × (2 2 × 3 × 5 × 7 2)
Number of even divisors = (2 + 1)(1 + 1)(1 + 1)(2 + 1)
= 3 × 2 × 2 × 3
= 36
= 4 × (3) 2
∴ n = 3
39. Answer → A(p, r), B(t), C(p, q, s), D(p, q, r, s)
(A) If we select 4 vertices, two diagonals can be formed and we get one point of intersection.
So, total number of points of intersection of diagonals = 6C 4 × 1 =6C 4 = 15
(B) One line divides the plane in 2 regions
Two lines divides the plane in 4 regions
Three lines divides the plane in 7 regions
Four lines divides the plane in 11 regions
So, there is a series like
2, 4, 7, 11, ........
and we have to find the eighth term of the series
S T n n = 2 + 4 + 7 + 11 + ........ +S T T n n n = 2 + 4 + 7 + ........ + + – 1or, (–) (–)
0 = 2 + 2 + 3 + 4 + ........ + – n T n
Let,
∴ T n = 2 + 2 + 3 + 4 + ........ + n
=( 1)
12+
+ n n
∴ T 8 = 1 + 8 ×92
= 1 + 36
= 37
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
11/18
Test - 2 (Paper - II) (Answers & Hints) All India Aakash Test Series for IIT-JEE 2012
11/18
(C) f : {1, 2, 3, 4} → {p , q , r , s }
1 may be related to any of p , q , r , s i.e. , 4 ways
f has to be one-one onto
∴ 2 may be related in 3 ways
Now, 3 may be related in 2 ways
and 4 may be related in only 1 way
So, the number of one-one onto function = 4 × 3 × 2 × 1
= 24
∴ The possible number of invertible function is = 24
(D) Let number consider the side A1A2.
Now to form a ∆ whose one side is common with that of then -gon, available vertices are,
A4, A
5, ........, A
n – 1.
i.e. , (n – 4) vertices are available.
A1 A2
A3An
Now, as there is n -gon. So, we may take any of the n sides of n -gon to be the common side.
Therefore, total number of such triangles = ( n – 4) × n
As we have n = 10
∴ Number of such triangles = (10 – 4 ) × 10
= 6 × 10
= 60
40. Answer → A(s), B(q), C(p), D(q)
(A) Obviously,
| z 1 – z 2 |min = 0
and | z 1 – z 2 |max = 8
∴ sum = 8
(B) (1 + αω 5)n = (α + ω4)n
or, (1 + αω 2)n = (α + ω)n
or, ω2n (ω + α )n = (α + ω)n
B
A z z ( ) ( )2 1≡
2
2
x
y
( )z 1 | | = 4z 1
O
| – 1 + 3| = 2z i 2
(1, – 3 )
4
or, ( ω + α )n [ω2n – 1] = 0
⇒ ω 2n = 1
∴ Least positive value of n = 3
(C)
1
1 2 2
11 2
2
3 31 13 4 4 4333 4 1144
+ ++= =
− −−
z ik z z z
z z z ik z
314 1314
+= =
− +
ik
ik
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
12/18
All India Aakash Test Series for IIT-JEE 2012 Test - 2 (Paper - II) (Answers & Hints)
12/18
Since, 1 2 1 2 0+ =z z z z
or, 1 12 2
= −z z z z
or, 1 12 2
= −
z z z z
⇒1
2=
z ik
z (i.e. , 1
2
z z
is purely imaginary)
(D) Let, z 1 = cos α + i sin α , z 2 = cos β + i sin β and z 3 = cos γ + i sin γ
∴ z 1 + z 2 + z 3 = (cos α + cos β + cos γ ) + i (sin α + sin β + sin γ )
= 0 + i 0 (from question)
= 0
Now, (z 1 + z 2 + z 3)2 = z 12 + z 22 + z 32 + 2 ( z 1z 2 + z 2z 3 + z 1z 3)
or, 0 = z 12 + z 2
2 + z 32 + 2 ∑z 1z 2 …(i)
Now, 1 1 11 2 31 2 3
1 1 1 − − −+ + = + +z z z z z z
= (cos α + cos β + cos γ ) – i (sin α + sin β + sin γ )
= 0
i.e. , 2 3 1 3 1 21 2 3
0+ +
=z z z z z z
z z z
⇒ z 1z 2 + z 2z 3 + z 1z 3 = 0 …(ii)
From (i) and (ii), we get
z 12 + z 2
2 + z 32 = 0
⇒ cos2 α + cos2 β + cos2 γ = 0 = sin2 α + sin2 β + sin2 γ
Now, cos2 α + cos2 β + cos2 γ = 0
⇒ 2(cos 2α + cos 2β + cos 2γ ) – 3 = 0 ( cos2 θ = cos 2θ – sin 2θ)
or, 2(cos 2α + cos 2β + cos 2γ ) = 3
PART - III (PHYSICS)41. Answer (C)
2
(cos sin ) mv T mg − θ+ θ =
21 (1 cos ) sin2
mv mg F = − − θ + θ
3 2sin 24
T mg π = θ+ −
max 3 2 2T mg = −
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
13/18
Test - 2 (Paper - II) (Answers & Hints) All India Aakash Test Series for IIT-JEE 2012
13/18
42. Answer (B)
2
2cos mv mg
R β =
2
1 2
1cos cos2mgR mv mgR β = + β
v 2 = 2 gR (cos β1 – cos β2)
Rg cos β2 = 2 Rg (cos β1 – cos β2)
12
2coscos
3β
β =
43. Answer (C)
0du dx
=
50,2
x =
x = 0 is local maxima
at52
x = ± u = – 6.25 J
K.E. = 42.25 J
v max = 6.5 m/s
44. Answer (C)
Applying energy conservation
2 21 12 (1 cos ) 2 (1 sin ) 22 2
mgr mgr mgr mv mv = − θ + − θ + +
[ ]2 2 cos 2sin 13
v gr = θ+ θ−
for maximum v 2
tan θ = 2
45. Answer (D)
By work energy theorem2
gravity1 02 F
mv W W − = +
212 2
r mv Fr mg = −
21 10 (200 50) 102
v × = − ×
v = 300
= 10 3
= 17.3 m/s
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
14/18
All India Aakash Test Series for IIT-JEE 2012 Test - 2 (Paper - II) (Answers & Hints)
14/18
46. Answer (B)
2 cos2
T mr f ω = θ+
sin ;2T
f f mg θ = = µ
2 2 sin(45 )g r µω = + θ
max2 g r µ
ω =
47. Answer (B)
v = 4 t + 5
a = 4
m = 2
∆P = 2[8 – 0]
= 16 N-s
48. Answer (B)
a 1 = a (1 + cos θ)
ma 1 = mg sin α + ma cos α – T
N = mg cos α – ma sin α
For horizontal motion of wedge
ma = T + T cos θ – T cos α + N sin θa 1 a
2 2
(1 cos )sin
(1 cos cos ) sin
mg a M m
+ θ α= + + θ− α + α
here m = M
θ = 60°
α = 30°
49. Answer (A, B, C)
Maximum acceleration of 7 kg block = 2 215
m/s 2.1 m/s7
=
Maximum acceleration of 7 kg and 3 kg block as one system = 0.4 m/s 2
Acceleration of whole system if it moves together = 210 5
m/s12 6
=
Then 2 kg block will slip on 3 kg block and there will be no slipping between 3 kg and 7 kg block.
Acceleration of 2 kg = 210 4 3 m/s
2− =
Acceleration of 3 kg and 7 kg block = 0.4 m/s 2
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
15/18
Test - 2 (Paper - II) (Answers & Hints) All India Aakash Test Series for IIT-JEE 2012
15/18
50. Answer (A, C)
25 kg 15 kg F
Wall
25 kg
0.4 F 0.4 F
F F
2T
250 N
⇒ 2T = 0.8 F + 250 …(i)
T + 0.4 F = 150 …(ii)
0.8 F + 125 = 150
0.4 N
F
T
15 kg
150 N
F
25 250 31.25 N0.8 8
F = = =
T = 150 – 0.4 × 31.25
T = 137.5 N
51. Answer (B, C, D)
Work done by friction on plank in plank frame = 0
Work done by friction on block in plank frame = 212
mv −
Total work done = heat produced = 212
mv
Work done on block in ground frame = 21
K.E.2
mv = ∆
Work done on plank in ground frame = 2 2 21 12 2
mv mv mv − − =
52. Answer (B, C)
2
a g g
−=
2 1
2a a
g −
=
g
g a
a 1
a 23g
a 2a
1
g 2
13
2g a
a +
=
Solving the three equations
a = 3 g
a 2 = 7 g
a 1 = 5 g
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
16/18
All India Aakash Test Series for IIT-JEE 2012 Test - 2 (Paper - II) (Answers & Hints)
16/18
53. Answer (7)
2 20
1 12
2 2mv mg R mv + = …(i)
20
B
mv mg N R − = …(ii)
2
Amv
N mg R
− = …(iii)
54. Answer (2)
45
T Ma m a a
= + +
95
R M m a
= +
930 4·5 5
ma a m Ma ma − − − =
37°
M a a
a m
T ma
1730
5ma Ma = +
= (18 + 42) a
55. Answer (2)
T sin θx 1 – Tx 2 = 0
1 1 2sin cos 0d v x v
dt
θθ+ θ − =
as x 1 is very small
v 1sin θ = v 2
1 1 2sin cos d a v v
dt θθ+ θ =
m
m
30°x 1
T
T m
x 2
T 0
m
m T 2
T
mg
as v 1 = 0
a 1sinθ = a 2
a 1 = 2 a 2
⇒ 1 222T ma ma = =
mg – T = ma 2
mg = 5 ma 2
22 2 m/s5
g a = =
56. Answer (2)
Applying conservation of energy
21 1 1500 0.4 0.4 15 500 (0.2)(0.2)2 2 2
v × × × = × + × ×
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
17/18
Test - 2 (Paper - II) (Answers & Hints) All India Aakash Test Series for IIT-JEE 2012
17/18
57. Answer (6)
·d F dr ω=
ˆ ˆ ˆ ˆ( )·( )yi xj dxi dyj = + +
ydx xdy = +∫( )d xy = ∫4,3
2,1/2
2 312 9 J
2xy
× = = − = ∫Work done equals change in kinetic energy
219 82
v = × ×
3 m/s2v =
6 m/s4
v =
58. Answer (4)
26dy F r dx
= − = −
2mv F r
=
mv 2 = 6 r 3
Total energy = P.E. + K.E.
= 2 r 3 + 3 r 3
= 5 r 3
r = 5
⇒ 54 J
59. Answer → A(t), B(r), C(s), D(q)
For reacting point C C v gR ≥
52R
h ≥
For reacting point D 16D gR
v ≥
6932
h R ≥
h should be greater of the two
2
16
C C
mv N mg
R − =
N C = 17 mg
-
8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution
18/18
All India Aakash Test Series for IIT-JEE 2012 Test - 2 (Paper - II) (Answers & Hints)
18/18
BP = v D × t
21 5 22 2 8D
R R mv mg mg R
= − +
175 4P v gR gR = −
34gR =
2 178
t g
=
5116
BP R =
60. Answer → A(t), B(t), C(r), D(p)2a 1 + a 2 = a 3
48 – 2 T = 4 a 1
24 – T = 2 a 2
T – 12 = a 3
a 1 = a 2 =212 m/s
5
a 1
a 2a 3
a 3 = 236
m/s5
[here a 1, a 2 and a 3 are acceleration w.r.t. elevator]
top related