Solutions to Major Test - 2 (JEE)

Test Id: 778265

1. n2 =1 from n1 = 2, 3, 4 .... ⇒ Lyman series (UV)

n2 = 2 from n1 = 3, 4, 5 ... ⇒ Balmer series (Visible)

n2 = 3 from n1 = 4, 5, 6 ... ⇒ Paschen series (IR) ⇒
n NaOH 1
=
n H3 PO2 1
n2 = 4 from n1 = 5, 6, 7 ... ⇒ Brackett series (IR) Number of m.moles of H3PO2 = 2 × 100 = 200
Number of m.moles of NaOH = 1 × V = V
n2 = 5 from n1 = 6, 7, 8 ... ⇒ Pfund series (IR)
where V = Volume of NaOH used in ml for
neutralisation.
2. c 3 × 10 8 ms -1
⇒
λ= = ≅2.90 m V 1
= ⇒ V NaOH = 200 ml
ν 1.034 × 10 8 s -1 200 1
3. The balanced equation is given below: ⇒ Option (C) is CORRECT.
2SO2 + O2 → 2SO3 6. The chemical formula of methane is CH4
From the balanced equation, In one mole of CH4, one mole of carbon atoms and 4
1 mole of O2 reacts with 2 moles of SO2 moles of hydrogen atoms are present. (1C + 4H atoms)
The amount of O2 taken = 1 mole ⇒ Mole % of C =
1
× 100
But, the amount of SO2 taken = 1 mole
1+4
⇒ Mole % of C = 20%
⇒ SO2 is the limiting reagent and it will be fully ⇒ Option (C) is CORRECT.
consumed.
7. According to the Law of Conservation of Mass:
The stoichiometry of the product can be related to the
The total masses of reactants undergoing a chemical
stoichiometry of the limiting reagent.
reaction is equal to the total masses of the products
2 moles of SO3 is formed by 2 moles of SO2 (limiting
formed.
reagent) mass of C used = 6 g
⇒ Amount of SO3 formed by 1 mole of SO2 = 1 mole. mass of O2 used = 8 g
⇒ Option (B) is CORRECT. mass of product formed = 6+8 = 14 g.
Hence, it follows the Law of Conservation of Mass.
4.
m % of urea =
6g
× 100 = 12
(6 + 9 + 35)g 8. Since neutrons are electrically neutral, they do not
experience any force while moving through electric or
m % of glucose =
9g
50g
× 100 = 18 magnetic fields.
Hence, neutrons are not deflected by electric or
5. We know H3PO3 is dibasic acid and there is 2 acidic magnetic fields.
hydrogen.
⇒ 1 mole of H3PO3 reacts with 2 moles of NaOH in the 9. 2.4
2.4 g C = = 0.2 mol
balanced reaction as shown.
12
1.2 × 10 23
1.2 × 10 23 atoms of H = = 0.2 mol
6 × 10 23
0.2 mole of ’O’ atoms
⇒ Simplest ratio = C : H : O :: 0.2 : 0.2 : 0.2
⇒ ⇒ Simplest ratio = 1 : 1 : 1
n NaOH 2
=
n H3 PO3 1
Number of m.moles of H3PO3 = Molarity × Volume (ml) ⇒ Empirical Formula = CHO
= 1 × 50 = 50 10.
CH 4 + 2 O 2 → CO 2 + 2 H 2 O - Combination reaction
Δ
Number of m.moles of NaOH = Molarity × Volume (ml)
−
=1×V=V
2H 2 O → 2 H 2 + O 2 - Decomposition reaction
Δ
where V = Volume of NaOH used in ml for
−
neutralisation. Cl2 + 2Br- → 2Cl- + Br2 - Displacement reaction
⇒
V 2
50
= ⇒ V NaOH = 100 ml
1 2H2O2 → 2H2O + O2 - Disproportionation reaction
We know H3PO2 is monobasic acid and there is 1 acidic 11. Triads consist of three elements with similar properties,
hydrogen. where the atomic mass of the middle element is roughly
⇒ 1 mole of H3PO2 reacts with 1 moles of NaOH in the the average of the atomic masses of the first and third
balanced reaction as shown. elements.
K, Ca are metals; Cl is a non-metal, so they are not traid.

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 Solutions to Major Test - 2 (JEE)

12. The rules for the nomenclature of Elements of Atomic ⇒ The molecular mass of gas = 8 × 2 = 16
Numbers greater than 100: ⇒ Option (B) is CORRECT.
(1) The name is derived directly from the atomic 19. Ground state energy of the electron in H-atom = –13.6
number of the element using the following numerical eV
roots: ⇒ Ionization energy of H-atom = 13.6 eV
0 = nil
1 = un 20. Bohr's theory accounts for the stability and spectrum of
2 = bi single–electron species like H, He+, Li2+ etc.
3 = tri It doesn't explain the spectrum and stability of the
4 = quad multielectron system.
5 = pent
Li+ has 2 electrons.
6 = hex
7 = sept ⇒ Bohr's theory does not account for the stability and
8 = oct line spectrum of Li+ ion
9 = enn ⇒ Statement-I is false.
(2) The roots are put together in the order of the digits
One of the shortcomings of Bohr's model is that it is
which make up the atomic number and terminated by
unable to explain the splitting of spectral lines in the
'ium' to spell out the name. The final 'n' of 'enn' is elided
presence of a magnetic field.
when it occurs before 'nil', and the final 'i' of 'bi' and of
'tri' when it occurs before 'ium'. ⇒ Statement II is true.
1 0 9
un enn+ium ⇒ Option (B) is CORRECT.
nil
Hence the atomic number of the element unnilennium is
21. – (wave no.) = 1 = 1
109. v = 17241
λ 5800 × 10 −8 cm
13. NH4NO3 1724.1 × 10 cm −1
⇒ x = 1724
The nitrogen in the ammonium ion (NH4+) has an
oxidation number of -3, while the nitrogen in the nitrate 22. Given Solution:
ion (NO3–) has an oxidation number of +5 5% by weight of a NaCl solution.
⇒ 5 g NaCl is present in 100 g of solution.
14. Alkali metals (group 1) = ns1 ⇒ 0.3 g NaCl is present in
100
× 0.3 g of solution.
Alkaline earth metals (group 2) = ns2
5
⇒ Mass of solution,m solution = 6 g
Halogens (group 17) = ns2np5 ⇒ Answer = 6
Noble gases (group 18) = ns2np6
23. By absorbing energy the electron in a sample of Li2+
15. mass(g) ions goes from the 1st Bohr orbit to the 3rd Bohr orbit.
⇒ The electron will make the transition from the 3rd
Number of Moles =
gram atomic mass
46 orbit ( n = 3)
We know, if an electron is in any higher state n = n and
Moles of Na = = 2.
23
makes a transition to the ground state, then maximum
16. No. of Protons = 16
no. of different photons emitted =
n(n − 1)
No. of Neutrons = 32 –16 = 16 2
No. of electrons = 16 + 2 = 18 ⇒The total number of spectral lines are observed in the
emission spectrum =
3(3 − 1)
17. Law of Multiple proportions states that : When two =3
elements combine to form more than one compound,
2
These transitions are: 3 → 2, 2 → 1, 3 → 1
then the mass of one element which combines with a
⇒ Answer = 3
fixed mass of the other element in the various
compounds bean a simple whole number ratio with one 24. In the combustion reaction of hydrocarbon, the
another. products are CO2 and H2O
Given that the ratio of masses of Oxygen which The balanced reaction of combustion of butane (C4H10)
combines with a fixed mass of Hydrogen (2 g) in the
is written below:
two compounds is in the ratio 1:2.
Thus, it is in accordance with the Law of Multiple
Proportions.
18. We know, the vapour density of a gas is defined as,
The molar mass of butane = 58 g mol–1
Vapour density = The molar mass of water = 18 g mol–1
Molecular mass of gas
Molecular mass of H 2
In the balanced reaction, we can see,
⇒ The molecular mass of gas = Vapour density ×
10 moles of water is obtained from the combustion of 2
Molecular mass of H2
moles of butane.
⇒ 10 × 18 g of water is obtained from the combustion of

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 Solutions to Major Test - 2 (JEE)
2 × 58 g of butane. 31. Given: n(A) = 10, n(B) = 15 & n(A ∪ B) = x
⇒ 72 g of water is obtained from the combustion of
Now n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
× 72 g of butane.
2 × 58
10 × 18 ⇒ x = 10 + 15 − n(A ∩ B)
⇒ Amount of butane utilized =
2 × 58
× 72 = 46.4 g ⇒ x = 25 − n(A ∩ B)
We know 0 ≤ n(A ∩ B) ≤ min{n(A), n(B)}
10 × 18
⇒ Amount of butane utilized = 464 × 10–1
i.e. 0 ≤ n(A ∩ B) ≤ 10
⇒ Answer = 464 Thus,
25. Number of electrons in single N atom = 7 xmax = 25 − n(A ∩ B)∣∣ & xmin = 25 − n(A ∩ B)∣∣ = 15
min max
⇒ Number of electrons in single N3– ion = (7 + 3) = 10 ∴ xmax = 25 − 0 = 25 & xmin = 25 − 10 = 15
⇒ Number of moles of electrons in 1 mole of N3– ion = ∴ 15 ≤ x ≤ 25
10 moles
⇒ Number of moles electrons in 0.5 mole of N3– ion
= 0.5 x 10 moles = 5 moles
32. 1 1
⇒ Answer = 5 cos θ =
2
(x + )
x
26. Let y = 27 cos 2x 81 sin 2x = 3 3 cos 2x+4 sin 2x x+
1
≥ 0 or x +
1
≤ −2 ⇒ cos θ ≥ 1 or cos θ ≤ −1
Now −√3 2 + 4 2 ≤ 3 cos 2x + 4 sin 2x ≤ √3 2 + 4 2
x x
⇒ cos θ = x = ±1 ⇒ x 2 = 1 and sin θ = 0 ⇒
or −5 ≤ 3 cos 2x + 4 sin 2x ≤ 5 1 1
∴ 3 −5 ≤ 3 3 cos 2x+4 sin 2x ≤ 3 5 (x 2 + 2 ) = 1
2 x
27. We have, sin 6 θ + cos 6 θ + k cos 2 2θ = 1 cos θ = 1 − 2 sin 2 θ = 1 = sec 2θ
⇒ (sin2 θ + cos2 θ)(sin4 θ − sin2 θ ⋅ cos2 θ + cos4 θ) + k cos2 2θ = 1 sin 2θ = tan 2θ = 0
⇒ sin 4 θ + cos 4 θ − sin 2 θ ⋅ cos 2 θ + k cos 2 2θ = 1
⇒ (sin 2 θ + cos 2 θ) 2 − 3 sin 2 θ ⋅ cos 2 θ + k cos 2 2θ = 1 33. Let A denote the set of Americans who like cheese and
let B denote the set of Americans who like apples.
⇒ 3 sin 2 θ ⋅ cos 2 θ = k cos 2 2θ Let total Population of America be100.
⇒ sin 2 2θ = k cos 2 2θ
3 Then n(A) = 63, n(B) = 76
4 Now, n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
⇒ k = tan 2 2θ
3 = 63 − 76 − n(A ∩ B)
4 ⇒ n(A ∩ B) = 139 − n(A ∪ B)
28. tan 15 ∘ + tan 15 ∘ − tan 15 ∘ + tan 15 ∘ But n(A ∪ B) ≤ 100
⇒ n(A ∩ B)min = 139 − n(A ∪ B)max
⇒ n(A ∩ B)
= 2 tan 15 ∘
= 2(2 − √3) = 2a ⇒ a = 2 − √3 min = 139 − 100 = 39
+ a ⇒ (2 + √3) + (2 − √3) = 4
1 ∴ n(A ∩ B) ≥ 39 . . . (1)
∴
a Again, A ∩ B ⊆ A, A ∩ B ⊆ B
29. 3A = 2A + A n(A ∩ B) max = min{n(A), n(B)} = 63
⇒ tan 3A = tan(2A + A) ∴ n(A ∩ B) ≤ 63 . . . (2)

⇒ tan 3A = tan 2A+tan A From (1) and (2) we get

1−tan 2A.tan A
39 ≤ n(A ∩ B) ≤ 63
⇒ tan 3A − tan 3A tan 2A tan A = tan 2A + tan A ⇒39 ≤ x ≤ 63.
⇒ tan 3A − tan 2A − tan A = tan 3A tan 2A tan A
34.
1 1 1 1 1 1 1 1
(a 8 + a − 8 )(a 8 − a − 8 )(a 4 + a − 4 )(a 2 + a − 2 )
30. Let = [(a + a − )(a − a − )](a + a − )(a + a − 2 )
1 1 1 1 1 1 1 1
8 8 8 8 4 4 2

Use the identity (a + b)(a − b) = a 2 − b 2

∴ sin θ + sin 2θ + sin 4θ + sin 5θ
= sin θ + sin 5θ + sin 2θ + sin 4θ
= [(a 4 − a − 4 )(a 4 + a − 4 )](a 2 + a − 2 )
1 1 1 1 1 1
= 2 sin 3θ cos 2θ + 2 sin 3θ cos θ
= (a 4 − a − 4 )(a 2 + a − 2 )
= 2 sin 3θ(cos 2θ + cos θ) 2 2 1 1

= (a 2 − a − 2 )(a 2 + a − 2 )
3π 2π π 1 1 1 1
= 2 sin (cos + cos )
= (a 2 − a − 2 )
18 18 18 2 2
π π
= (a − a −1 )
= cos + cos
9 18
π π π π
= sin ( − ) + sin ( − )
2 9 2 18
7π 8π
= sin + sin
18 18
7π 4π
= sin + sin
18 9
35. Using A. M. ≥ G. M.

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 Solutions to Major Test - 2 (JEE)

By sine rule
⇒
1
(3 tan 2 θ + 12 cot 2 θ) ≥ 6
We know that
2 sin A sin B sin C
= = = k(let)
⇒ 3 tan θ + 12 cot θ has minimum value 12.
2 2 a b c
⇒ sin A = ak, sin B = bk, sin C = ck,
36. 2π 4π 6π
Put this in (iii), so we get
cos + cos + cos
7 7 7 2(ck) 2 = (ak) 2 + (bk) 2
⇒ 2c 2 = a 2 + b 2
π 2π 6π
sin (3 × ) ⎛ + ⎞
⇒ b 2 , c 2 and a 2 are in A.P.
7 7 7
= π × cos
sin 2
⎝ ⎠
39.

37.
=

=−
2 sin (

2
sin

sin (

− sin

2 sin
1
7π

Given that
7
7
3π

π
7

7π

π
7
7
7
)
× cos (

) + sin (

2 sin
π
7
−π
7

A ∪ B = A ∩ B . . . . . (i)
Let x ∈ A
⇒x ∈ A ∪ B . . . . . (ii)
[as A ⊆ (A ∪ B)]
Also,
4π
7

)
⎜⎟
)

40.

41.

42.
8 sin
π
5π
2 sin (
12

= cos (
5π
12

8
) sin

−
π
12

= cos − cos =
3
π
2

cos
π
12
Use the formula 2 sin A sin B = cos(A − B) − cos(A + B)

) − cos (

1
2

2
cos
5π
12
+

4
π
12
)

The given expression is a continued product in cos. We

will simplify it by taking two terms at a time, and
repeatedly applying the formula 2 sin cos = sin x

So,
x x x
cos
x
8
x
2

sin θ + cosecθ ≥ 2 and sin θ + cosecθ ≤ −2

For sin θ + cosecθ = 2; sin θ = 1 = cosecθ
⇒ sin 8 θ + cosec 8 θ = 1 8 + 1 8 = 2

Let the angles be α − d, α, α + d

∴ 3α = 180 ∘ ⇒ α = 60 ∘
x
2

= 8 sin
x
8
cos
x
8
cos
x
4
cos
x
2
= 4 (2 sin
x
8
cos
x
8
) cos
x
4
cos
x
2
= 4 sin
x
4
cos
x
4
cos
x
2
= 2 (2 sin
x
4
cos
x
4
) cos
x
2
= 2 sin
x
2
cos
x
2
= sin x

⇒x ∈ A ∩ B . . . . . (iii)
α−d 180 60
× =
[as from equation (i)(A ∪ B) = (A ∩ B)] α+d π π
Now from the equation (iii), we can see that x is an
(60 − d) × 3 = 60 + d
element of both sets A & set B ⇒ d = 30 ∘
⇒x ∈ A ⇒ x ∈ B ⇒A ⊆ B . . . . . (iv) Hence the angles are 30 ∘ , 60 ∘ , 90 ∘

Similarly, 43. A number is divisible by 99, iff it is divisble by 9 and 11.

⇒x ∈ B ⇒ x ∈ A ⇒B ⊆ A . . . . . (v) 3572403 and 1357329 are not divisble by 99 as they
Now from equations (iv) & (v) are not divisble by 9.
A⊆B&B⊆A 9134640 is not divisble by 99 as it is not divisble by 11.
⇒A = B For 1143549,
Sum of digits = 27 ⇒ divisible by 9
38. Sum of digits at even places = 4 + 3 + 1 = 8
Given:
sin A sin(A − C)
Sum of digits at odd places = 9 + 5 + 4 + 1 = 19
=
sin B sin(C − B)
As A, B, C are angles of triangle Difference of 'sum of digits at odd places' and 'sum of
⇒A+B+C =π odd digits at even places' = 11 ⇒ divisible by 11
Hence, 1143549 is divisible by 99.
⇒ A = π − (B + C)
Take sin both sides 44. Given:
⇒ sin A = sin(π − (B + C)) 1 + r + r 2 + . . . + r n = (1 + r)(1 + r 2 )(1 + r 4 )(1 + r 8 )
⇒ sin A = sin(B + C) . . (i) As the two polynomials on the LHS and RHS of the
Similarly sin B = sin(A + C) equation are equal, their degrees (highest power of r)
will also be the same.
. . (ii)
Divide (i) and (ii)
sin A sin(B + C) Clearly, degree of LHS = n ...(1)
The highest power of r in RHS will be obtained when we
=
sin B sin(A + C)
multiply r, r 2 , r 4 and r 8 .
⇒
sin(B + C) sin(A − C)
= Hence, (r)(r 2 )(r 4 )(r 8 ) = r 1+2+4+8 = r 15
⇒ Degree of RHS = 15 ...(2)
sin(A + C) sin(C − B)
⇒ sin(C + B) ⋅ sin(C − B) = sin(A − C) sin(A + C)
⇒ sin2 C − sin2 B = sin2 A − sin2 C {∵ sin(x + y) sin(x − y) = sin2 x − sin2 y} Now,
Degree of LHS = Degree of RHS
⇒ 2 sin 2 C = sin 2 A + sin 2 B
⇒ n = 15 (From (1) and (2))
. . (iii)

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 Solutions to Major Test - 2 (JEE)

45. LHS = (1 + tan 1 ∘ )(1 + tan 44 ∘ )(1 + tan 2 ∘ )(1 + tan 43 ∘ ). . . . (1 + tan 45 ∘ ) 53.

As if A + B = , then
π
4
π tan A + tan B
(1 + tan A)(1 + tan B) = 2 {tan = }
4 1 − tan A ⋅ tan B
⇒ (1 + tan 1 ∘
+ tan 44 ∘ + tan 1 ∘ tan 44 ∘ ) + (1 + tan 2 ∘ + tan 43 ∘ + tan 2 ∘ + tan 43 ∘ ). . . . (1 + tan 45 ∘ )

⇒ (1 + 1)(1 + 1). . . . . (1 + 1)23 times = 2 23

⇒ n = 23

46.
Given log 10 ( Slope is tan θ, where θ is the angle made by the tangent
sin 2x
) = −1
2 to the curve with the positive x-axis.
sin 2x 1 1
⇒ = ⇒ sin 2x =
54. According to definition of definite integration,
2 10 5
n t=b
log 10 ( )
Also log 10 (sin x + cos x) = 10 ∫ f(t)dt = [F (t)] t=b
t=a = F (b) − F (a)
2 t=a
n 2 2 2
⇒ log 10 (sin x + cos x) = log 10 ( )
10 ∫ 2t dt = 2 ∫ t dt = [t 2 ] 20 = 4
n
⇒ 1 + sin 2x = 0 0
10
1 n 6 n 55. ds d 2
⇒1+ = ⇒ = = (t + 5t + 3)
5 10 5 10 dt dt
n = 2t + 5
⇒ =4
56.
3 → → → → →
Here A and B sets having 2 elements in common, so
R = (2Q + 2P) + (2Q − 2P)
47. →
A × B and B × A have 2 2 i.e., 4 elements in common.
→
R = 4Q
Hence, n[(A × B) ∩ (B × A)] = 4
The angle between Q and R is zero.
→ →

48. Sum of interior angles of the 5 triangles = 5(180°) =

57.
s π cm π
900° θ= = = rad = 30 ∘
Sum of interior angles of the pentagon = 540°
r 6 cm 6

Total = 900° + 540° = 1440° = 16(90°) = 16 right angles 58. (x – 2)2 + 1 = 0

49.
x2 – 4x + 5 = 0
1 − sin 2 x − sin x + a = 0
The highest power of the variable is equal to 2.
⇒ sin 2 x + sin x − (a + 1) = 0
59.
From Eq. (i), we get sin 2 x + sin x = (a + 1)
For x ∈ (0, ), the range of sin 2 x + sin x is (0, 2)
π
2
⇒ 0 < (a + 1) < 2 ⇒ a ∈ (−1, 1)
i.e. a = 0 is the only integer value in this interval.
50.
We have abc + 1 =
bc
5
Similarly, from other expressions,
Tension always acts away from the body.
bc −ac ab
abc + 1 = = =
5 15 3
⇒ = −3 and = −5
a c 60. Both lines have positive slope.
According to newton's second law, when external force
b b
c 61.
−1 acts on a body, it produces an acceleration of the body
Now
c−b −5 − 1
in the direction of the force.
b
= c a = −5 − (−3) = 3
c−a −
62.
b b
v = 20 + 0.1t
51. Statement II does not explain I. a=
dv
= 0.1 m/s 2
dt
52. The slope is steeper at 1 than 2 which in turn is steeper Acceleration is uniform.
than the slope at 3.
63. We are given the parabola y = x 2 . We have to find the
shaded area.

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 Solutions to Major Test - 2 (JEE)

69. →
v = v cos θˆi + v sin θˆj
= 10 cos 60ˆi + 10 sin 60ˆj
1 √3
= 10 × ˆi + 10 × ˆj = 5ˆi + 5√3ˆj
2 2
70. Factual. The magnitude of distance is always greater
than the magnitude of displacement.
71. If a vector makes angles
π π
, and
π
with theX-axis,
3 3 n
the Y-axis and the Z-axis respectively, then we have to
find the value of n.

If a vector makes angles α, β and γ with the X-axis, the

Y-axis and the Z-axis respectively, cos α, cos β and cos γ
are called the direction cosines of the vector.
Now, cos 2 α + cos 2 β + cos 2 γ = 1.
⇒ cos 2 + cos 2 + cos 2
π π π
=1
3 3 n
⇒ cos 2
π 1
=
n 2
⇒ cos = ±
π 1
n √2
⇒ (Taking only the value in the first quadrant)
π π
=
n 4
⇒n=4

72.
Given that a force F = 20ˆi + 10ˆj is being applied on a
→
2kg body at rest.
On applying the newtons second law on the body we get
→
20ˆi + 10ˆj
acceleration as, a =
→ F
=
m 2
⇒ a = 10ˆi + 5ˆj
→

⇒ Displacement of the block is,

The infinitesimal shaded area is ydx. → 1→ 2 1
s = at = (10ˆt + 5ˆj) × (10) 2
The equation is y = x 2 . 2 2
∴ The infinitesimal shaded area is x 2 dx. ⇒ s = 50(10ˆi + 5ˆj) m.
→
The total shaded area is obtained by adding these The displacement along x-axis = 50 × 10 = 500 m
infinitesimally small areas (integrating).
1 73. x = −3t 3 + 18t 2 + 16t ⇒ v = −9t 2 + 36t + 16
So, the total shaded area is ∫ x 2 dx. a = −18t + 36 = 0 ⇒ t = 2 sec
0 v t=2 = −9(2) 2 + 36(2) + 16 = 52 m/s
The limits have been placed as x varies from 0 to 1 for
the shaded portion in the question statement. 74. ∣ î ĵ k̂ ∣
→ → ∣ ∣
The cross product P × Q = ∣ 3 2 ∣
1
x3
So, the total shaded area is, = . √3
1
∣ ∣
A = [ ]
3 0 3
∣4 √3 2.5 ∣
64. Vector product of two parallel vectors.
⇒ P × Q= ˆi (
→ → √3
magnitude of AXB = AB sin θ; θ = 0 is 0
→ → 1
) − ˆj (− ) + ˆ
k(−√3)
2 2
65. According to Galileo, uniform motion is possible when
⇒ P × Q=
→ → √3 ˆ
no frictional force oppose. ˆi + j − √3ˆ
k
2 2
66. Vector division is not allowed. ⇒ |P × Q| = √ + + 3 = 2
→ → 3 1
4 4
67. Distance s = (2.5)t 2 Unit vector along
Speed v =
ds → →
= (2.5)(2t) → → P×Q 1
dt P×Q= = (√3ˆi + ˆj − 2√3ˆ
k).
v = 5t → → 4
At t = 5, v = 5 × 5 = 25 m/s
|P × Q|
∴ x = 4.
68. d
(8) = 0
dx
d(c)
{ = 0; where c is constant}
dx

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 Solutions to Major Test - 2 (JEE)

75.
→ A ˆ A ˆ
R = (A + )i + (A − )j
√2 √2
2 2
→ A A
|R| = √(A + ) + (A − ) = √3A
√2 √2

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