JEE Advanced Prahaar

Code: RTJP26-JAP2-200425-02

Answer Key & Solutions

1 2 3 4 5 6 7 8 9 10
D A D D A,B,C B,C D 0.5 0.5 3
11 12 13 14 15 16 17 18 19 20
4 3 135 B C D C A B A
21 22 23 24 25 26 27 28 29 30
A A,B,C B,D A,B,C,D 1 10 3 5 2 18
31 32 33 34 35 36 37 38 39 40
A D D B A C A C A,B,D B,C,D
41 42 43 44 45 46 47 48 49 50
B,D 3 5 3 2017 4 3 B B A
51
A

Page 1
1.

σ 1 σ 2 3σ
+ =.
2ε 0 2ε 0 2ε 0

σ 1 + σ 2 = 3σ
σ1 σ 2 σ
− =
2ε 0 2ε 0 20

σ1 − σ 2 = σ
∴ σ 1 = 2σ

σ2 = σ

3 
2. ma = (mg ) −  mg 
4 
g
a=
4
v0 4v0
t
= =
a g

4R σπ R 2
3. τ= × 2E ×
3π 2
4ER 3σ
= .
3

4. this combination is equivalent to a wire of

λ λ0 a 2k λ0 a
λ ⇒ λ0 a ∴=
E = =
2πε 0 r 2πε 0 r r

1
5. FR = 2 × 2
q
4π ∈ × cos 60°
a2
Kq 2 Kq 2
⇒ FR =2 × 2 × cos 60 = 2
°

a a

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 θ T qB
θ
qA
6. q
mg
T cos θ + qE sin θ =mg
T sin θ = qE cos θ
E cos θ A,
= = E sin θ B
qA cos θ
+ qB = mg
sin θ
mg
q=
A cot θ + B
mg A
T=
A cos θ + Bsin θ

7. Time period = 4T and is independent of d.

1 th 4T T
So, if CB = 2d coverage of AO required of time period
= = and if OB= 2d , t =
12 12 3
4T T Qq
= and ω 2 ∝ 3 (for small d)
8 2 R

KQ 3KQy
8. 2
= 3
y +2
(
y2 + 2 2 )
1
∴y =
2

λ sin 45 λ ′ sin 45 λ 1
9. − =0 ∴ ′ =
2πε 0 R 2πε 0 (2k ) λ 2
⇒ 0.5

10. NLM on system

T − 2mg =0
T = 2mg
9 ×109 q 2
=T + mg
R2
9 ×109 × q 2 ×10−12
=2mg + mg
9
mg= q 2 ×10−3
0.9 ×10−3 ×10 = q 2 ×10−3
q2 = 9
q=3

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 λ
11.
= E [sin 45 + sin 45]
l
4πε 0
2
λ 1
= × ×2
2πε 0l 2
λ 4 2πε 0
= = = 4
2πε 0l 2πε 0

12. Net Electric field at P

  
E= E1 + E2 due to vertical wire
 λ1  λ2
E1 = in +y direction, E2 = due to vertical wire in x direction.
2πε 0 y 2πε 0 x
α angle of electric field with x direction
E λ / 2πε 0 y 1 λx λ x λ
tan α = 1 = 1 ⇒ = 1 ⇒ 1 = 3; 1 = 3
E2 λ2 / 2πε 0 x 3 λ2 y λ2 y λ2

13.

180 − 45 45
= 90 −
2 2
= 67.5
angle with x axis = 45 + 22.5 + 67
=135°

15.
For small ' θ ' separation of balls = lθ and hence, T = mg
θ θ 1 q2
T sin = T = ⇒ q 2 ∝ θ 3 , q ∝ , T ∝  −2
2 2 4π 0 (1θ ) 2

In satellite
1 q2
T= ( g eff = 0 ) ⇒ T ∝  −2 , θ = π
4π o (21) 2

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18. for Zero Order
[A0 ] − [A t ] =
kt
0.2 − 0.1 =k × 6
1
k= M / hr
60
1
and 0.5 − 0.2 = ×t
60
t = 18hrs.

−d[A] d[B]
19. log= log + 0.3010
dt dt
−d[A] d[B]
= 2×
dt dt
1 −d[A] d[B]
× =
2 dt dt
2 A 
→B
2C2 H 4 
→ C4 H8

1  1 1 
22. kt
 n −1 − n −1  =
n − 1  ct c0 
1
n=
2

d [O2 ]
23. − ⇒ 32gL−1sec −1 =
1molL−1sec −1
dt
d [O2 ] d [SO3 ]
d [SO 2 ] − dt +
dt
R=
− = =
dt 1/ 2 1

c0 c
25. t1/2 = = 0
2nk 4k

1 P∞ − P0 1 1 ln 2
26.
= k ln
= = = ln 8 1/2
t P∞ − Pt 30 30 t
P∞
= 4=
P∞ 400
P0

27. x=1
y=2
z=0
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 4 − x2 4 − x2
35. | x|+ x+
=
x x

 4 − x2 
⇒ x  ≥ 0, x ≠ 0
 x 
⇒ x 2 − 4 ≤ 0 ⇒ x ∈ [−2, 2] − {0}

36. Let M be the set of students who study Mathematics, P be the set of students who study
Physics and C be the set of students who study Chemistry.
Then,
= n(U ) 220,= n( M ) 120,
= n( P ) 90,
n(=
C ) 70, n( M ∩ =
P) 40, n( P ∩ =
C ) 30,
n(C ∩= (
M ) 50, n M ′ ∩ P′ ∩
= C′ 20 )
n(U ) − n( M ∪ P ∪ C ) =20
n( M ∪ P ∪ C )= 220 − 20= 200
 n( M ∪ P ∪ C=
) n( M ) + n( P) + n(C ) − n( M ∩ P) − n( P ∩ C ) − n(C ∩ M ) + n( M ∩ P ∩ C )
⇒ 200 = 120 + 90 + 70 − 40 − 30 − 50 + n( M ∩ P ∩ C )
⇒ 200= 160 + n( M ∩ P ∩ C )
⇒ n( M ∩ P ∩ C )= 200 − 160= 40

37. We must have 9 − x 2 ≥ 0, x 2 − 1 > 0 and x 2 − 4 < 0

⇒ 1 < x 2 ≤ 9 and x 2 < 4
⇒ 1 < x2 < 4
⇒ x ∈ (−2, −1) ∪ (1, 2)

1
cos x −
38. f ( x) = 2 is defined if
6 + 35 x − 6 x 2
1
cos x ≥ and 6 + 35 x − 6 x 2 > 0
2
Now, 6 x 2 − 35 x − 6 < 0
⇒ (6 x + 1)( x − 6) < 0
⇒ − 1/ 6 < x < 6 ….(1)
1
cos x ≥
2
 π π   5π 7π 
⇒ x ∈  − ,  ∪  ,  (keeping (1) in mind)
 3 3  3 3 
 1 π   5π 
Therefore, x ∈  − ,  ∪  , 6 
 6 3  3 
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 43. Given that, A = {1, 2,3}
Now, number of equivalence relations are as follows:
R1 = {(1,1), (2, 2), (3,3)}
R2 = {(1,1), (2, 2), (3,3), (1, 2), (2,1)}
R3 = {(1,1), (2, 2), (3,3), (1,3), (3,1)}
R4 = {(1,1), (2, 2), (3,3), (2,3), (3, 2)}

R5
= {(1, 2,3) ⇔ A=
×A A2 }
∴ Maximum number of equivalence relation = 5

44.
= f (π ) 2=
and f (22 / 7) 1
So, f (π ) + f (22 / 7) =
3

46. We have (| x − 1| −3)(| x + 2 | −5) < 0

Case I: x ≥ 1
∴ ( x − 4)( x − 3) < 0
∴ x ∈ (3, 4) ….(i)
Case II: −2 ≤ x < 1
∴ ( x − 3)( x + 2) > 0
∴ x ∈ (−∞, −2) ∪ (3, ∞)
∴ x ∈φ ( as − 2 ≤ x < 1)
Case III: x < −2
∴ ( x + 2)( x + 7) < 0
∴ x ∈ (−7, −2) ….(ii)
From (i) and (ii), x ∈ (−7, −2) ∪ (3, 4)

47. We have relation, R = {( p, p ), (q, r ), ( p, q)}

To make R reflexive, we must add (q, q ) and (r , r ) to R .
Also, to make R transitive, we must add ( p, r ) to R .
So, ordered pairs to be added are (q, q ), (r , r ), ( p, r ) .

( ) ( )
′ ′
48. (1)  A′ ∪ B′ − A =  A′ ∪ B′ ∩ A′  = ( A ∩ B )′ ∩ A′ 

( )
′ ′
= ( A ∩ B)′  ∪ A′ = ( A ∩ B) ∪ A = A

( ( )
(2)  B′ ∪ B′ − A  =  B′ ∪ B′ ∩ A′  ( )
′ ′



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 ( ) ( )
′
=  B′ ∪ B′ ∩ B′ ∪ A′ 

( ) ( ) ( )
′ ′ ′
=  B′ ∩ B′ ∪ A′  = B′ ∪ B′ ∪ A′

(( ) ( ) ) =B ∪ ( B ∩ A) =B
′ ′
=B ∪ B′ ∩ A′

(3) ( A − B) − ( B − C ) = ( A ∩ B′ ) − ( B ∩ C ′ )

( ) ( )
′
= A ∩ B′ ∩ B ∩ C ′

=( A ∩ B ) ∩ ( B ∪ C )
′ ′

=  A ∩ ( B ∪ C )  ∩  B ∩ ( B ∪ C ) 
′ ′ ′

=  A ∩ ( B ∪ C )  ∩ B
′ ′

= ( A ∩ B ) ∩ ( B ∪ C ) ∩ B 
′ ′ ′

=( A ∩ B ) ∩ B = A ∩ B = A − B
′ ′ ′

(4) ( A − B) ∩ (C − B) = ( A ∩ B ) ∩ ( C ∩ B ) ′ ′
 A − B = A ∩ B′ 

= ( A ∩ C ) ∩ B′ = ( A ∩ C ) − B

51. Obvious (according diffenations of Reflexive, Symmectric and Transitive relation)

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