JEE CHEMISTRY PHYSICS & Mathematics 2013

1. An unknown alcohol is treated with the "Lucas reagent" to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism:
(1) tertiary alcohol by S_N1
(2) secondary alcohol by S_N2
(3) tertiary alcohol by S_N2
(4) secondary alcohol by S_N1

Sol. (1)
Reaction proceeds through carbocation formation as 3° carbocation is highly stable, hence reaction proceeds through S_N1 with 3° alcohol.

2. The first ionization potential of Na is 5.1 eV. The value of electron gain enthalpy of Na⁺ will be:
(1) –5.1 eV
(2) –10.2 eV
(3) +2.55 eV
(4) –2.55 eV

Sol. (1)
Na → Na⁺ + e⁻, ΔH = –5.1 eV; here the backward reaction releases same amount of energy and is known as electron gain enthalpy.

3. Stability of the species Li₂, Li₂⁻ and Li₂⁺ increases in the order of:
(1) Li₂⁺ < Li₂⁺ < Li₂⁻
(2) Li₂⁺ < Li₂⁺ < Li₂⁻
(3) Li₂⁺ < Li₂⁺ < Li₂⁻
(4) Li₂⁺ < Li₂⁺ < Li₂⁻

Sol. (1)
Li₂(6) = σ1s² σ*1s² σ2s², B.O. = (4 – 2)/2 = 1
Li₂⁺(5) = σ1s² σ*1s² σ2s¹, B.O. = (3 – 2)/2 = 0.5
Li₂⁻(7) = σ1s² σ*1s² σ2s² σ*2s¹, B.O. = (4 – 3)/2 = 0.5
Li₂⁺ is more stable than Li₂⁻ because Li₂⁺ has fewer antibonding electrons.

4. The molarity of a solution obtained by mixing 750 mL of 0.5 M HCl with 250 mL of 2 M HCl will be:
(1) 1.00 M
(2) 1.75 M
(3) 0.975 M
(4) 0.875 M

Sol. (4)
M₁V₁ + M₂V₂ = MV
M = (M₁V₁ + M₂V₂)/V = (0.5×750 + 2×250)/1000
M = 0.875

5. Which of the following is the wrong statement?
(1) O₃ molecule is bent
(2) Ozone is violet-black in solid state
(3) Ozone is diamagnetic gas
(4) ONCl and ONO⁻ are not isoelectronic

Sol. (All the options are correct statements)
(1) Correct, as O₃ is bent.
(2) Correct, as ozone is violet-black solid.
(3) Correct, as ozone is diamagnetic.
(4) Correct, as ONCl = 32 electrons and ONO⁻ = 24 electrons hence are not isoelectronic.
All options are correct statements.

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JEE (MAIN)-2013-CMP-3

6. Four successive members of the first row transition elements are listed below with atomic numbers. Which one of them is expected to have the highest E°(M³⁺/M²⁺) value?
(1) Mn(Z = 25)
(2) Fe(Z = 26)
(3) Co(Z = 27)
(4) Cr(Z = 24)

Sol. (3)
E°(Mn³⁺/Mn²⁺) = 1.57 V
E°(Fe³⁺/Fe²⁺) = 0.77 V
E°(Co³⁺/Co²⁺) = 1.97 V
E°(Cr³⁺/Cr²⁺) = –0.41 V

7. A solution of (–)-1-chloro-1-phenylethane in toluene racemises slowly in the presence of a small amount of SbCl₅ due to the formation of:
(1) carbene
(2) carbocation
(3) free radical
(4) carbanion

Sol. (2)
C₆H₅–CHCl–CH₃ —SbCl₅→ [Ph–CH–CH₃]⁺ [SbCl₆]⁻
Phenyl carbocation

8. The coagulating power of electrolytes having ions Na⁺, Al³⁺ and Ba²⁺ for arsenic sulphide sol increases in the order:
(1) Na⁺ < Ba²⁺ < Al³⁺
(2) Ba²⁺ < Na⁺ < Al³⁺
(3) Al³⁺ < Na⁺ < Ba²⁺
(4) Al³⁺ < Ba²⁺ < Na⁺

Sol. (1)
As₂S₃ is an anionic sol (negative sol) hence coagulation will depend upon coagulating power of cation, which is directly proportional to the valency of cation (Hardy-Schulze rule).

9. How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?
(1) 0.9 L
(2) 2.0 L
(3) 9.0 L
(4) 0.1 L

Sol. (3)
Initial pH = 1, i.e., [H⁺] = 0.1 mole/litre
New pH = 2, i.e., [H⁺] = 0.01 mole/litre
In case of dilution: M₁V₁ = M₂V₂
0.1 × 1 = 0.01 × V₂
V₂ = 10 litre.
Volume of water added = 9 litre.

10. Which one of the following molecules is expected to exhibit diamagnetic behaviour?
(1) N₂
(2) O₂
(3) S₂
(4) C₂

Sol. (1) & (4) both are correct answers.
N₂ → Diamagnetic
O₂ → Paramagnetic
S₂ → Paramagnetic
C₂ → Diamagnetic

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11. Which of the following arrangements does not represent the correct order of the property stated against it?
(1) Ni²⁺ < Co²⁺ < Fe²⁺ < Mn²⁺ : ionic size
(2) Co³⁺ < Fe³⁺ < Cr³⁺ < Sc³⁺ : stability in aqueous solution
(3) Sc < Ti < Cr < Mn : number of oxidation states
(4) V²⁺ < Cr²⁺ < Mn²⁺ < Fe²⁺ : paramagnetic behaviour

Sol. (2) & (4) both are correct answers
The exothermic hydration enthalpies of the given trivalent cations are:
Sc³⁺ = 3960 kJ/mole
Fe³⁺ = 4429 kJ/mole
Co³⁺ = 4653 kJ/mole
Cr³⁺ = 4563 kJ/mole
Hence Sc³⁺ is least hydrated; so least stable (not most stable)
Fe²⁺ contains 4 unpaired electrons whereas Mn²⁺ contains 5 unpaired electrons hence (4) is incorrect.

12. Experimentally it was found that a metal oxide has formula M₀.₉₈O. Metal M is present as M²⁺ and M³⁺ in its oxide. Fraction of the metal which exists as M³⁺ would be:
(1) 4.08%
(2) 6.05%
(3) 5.08%
(4) 7.01%

Sol.
Metal oxide = M₀.₉₈O
If x⁺ ions of M are in +3 state, then
3x + (0.98 – x)×2 = 2
x = 0.04
So the percentage of metal in +3 state would be 0.04/0.98 × 100 = 4.08%

13. A compound with molecular mass 180 is acylated with CH₃COCl to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is:
(1) 5
(2) 4
(3) 6
(4) 2

Sol.
R–NH₂ + CH₃COCl → R–NH–COCH₃ + HCl
Each CH₃CO addition increases the molecular wt. by 42.
Total increase in m.wt. = 390 – 180 = 210
Then number of NH₂ groups = 210/42 = 5

14. Given
E°(Cr³⁺/Cr) = –0.74 V; E°(MnO₂/Mn) = 1.51 V
E°(Cr₂O₇²⁻/Cr³⁺) = 1.33 V; E°(Cl₂/Cl⁻) = 1.36 V
Based on the data given above, strongest oxidising agent will be:
(1) Cr³⁺
(2) Mn²⁺
(3) MnO₄⁻
(4) Cl⁻

Sol.
As per data mentioned MnO₄⁻ is strongest oxidising agent as it has maximum SRP value.

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15. Arrange the following compounds in order of decreasing acidity:
OH OH OH OH OH
Cl CH₃ NO₂ OCH₃
(I) (II) (III) (IV)
(1) I>II>III>IV
(2) III>I>II>IV
(3) IV>III>I>II
(4) II>IV>I>III

Sol. (2)
Correct order of acidic strength is III>I>II>IV

16. The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be:
(R = 8.314 JK⁻¹ mol⁻¹ and log 2 = 0.301)
(1) 48.6 kJ mol⁻¹
(2) 58.5 kJ mol⁻¹
(3) 60.5 kJ mol⁻¹
(4) 53.6 kJ mol⁻¹

Sol. (4)
As per Arrhenius equation:
ln(k₂/k₁) = –Eₐ/R (1/T₁ – 1/T₂)
2.303 log 2 = –Eₐ/8.314 (1/300 – 1/310)
⇒ Eₐ = 53.6 kJ/mole

17. Synthesis of each molecule of glucose in photosynthesis involves:
(1) 10 molecules of ATP
(2) 8 molecules of ATP
(3) 6 molecules of ATP
(4) 18 molecules of ATP

Sol. (4)
12H₂O + 12NADP + 18ADP —light reaction→ 6O₂ + 18ATP + 12NADPH
6CO₂ + 12NADPH + 18ATP —Dark reaction→ C₆H₁₂O₆ + 12NADP + 18ADP + 6H₂O
Net reaction: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

18. Which of the following complex species is not expected to exhibit optical isomerism?
(1) [Co(en)₂Cl₂]⁺
(2) [Co(NH₃)₃Cl₃]
(3) [Co(en)(NH₃)₂Cl₂]⁺
(4) [Co(en)₃]³⁺

Sol. (2)
[Co(NH₃)₃Cl₃] exists in two forms (facial and meridional). Both of these forms are achiral. Hence, [Co(NH₃)₃Cl₃] does not show optical isomerism.

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19. A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0°C. As it does so, it absorbs 208J of heat. The values of q and w for the process will be:
(R = 8.314 J/molK) (ln 7.5 = 2.01)
(1) q = –208 J, w = –208 J
(2) q = –208 J, w = +208 J
(3) q = +208 J, w = +208 J
(4) q = +208 J, w = –208 J

Sol. (4)
Process is isothermal reversible expansion, hence ΔU = 0
∴ q = –W
As q = +208 J
Hence W = –208 J

20. A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO₂. The empirical formula of the hydrocarbon is:
(1) C₃H₄
(2) C₆H₅
(3) C₇H₈
(4) C₂H₄

Sol. (3)
CₓHᵧ + (x + y/4)O₂ → xCO₂ + (y/2)H₂O
Weight(g) 3.08 g 0.72 g
moles 0.07 0.04
x/(y/2) = 0.07/0.04
x/y = 7/8
Empirical formula = C₇H₈

21. The order of stability of the following carbocations:
I: H₂C=CH–CH₂⁺
II: H₃C–CH₂–CH₂⁺
III: C₆H₅–CH₂⁺
is:
(1) II > III > I
(2) III > I > II
(3) III > I > II
(4) III > I > II

Sol.
Order of stability is III > I > II

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22. Which of the following represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se and Ar?
(1) S < Se < Ca < Ba < Ar
(2) Ba < Ca < Se < S < Ar
(3) Ca < Ba < S < Se < Ar
(4) Ca < S < Ba < Se < Ar

Sol. (2)
Increasing order of first ionization enthalpy is Ba < Ca < Se < S < Ar

23. For gaseous state, if most probable speed is denoted by C*, average speed by C̄ and mean square speed by C, then for a large number of molecules the ratios of these speeds are:
(1) C* : C̄ : C = 1.128 : 1.225 : 1
(2) C* : C̄ : C = 1 : 1.128 : 1.225
(3) C* : C̄ : C = 1 : 1.125 : 1.128
(4) C* : C̄ : C = 1.225 : 1.128 : 1

Sol. (2)
C* = √(2RT/M)
C̄ = √(8RT/πM)
C = √(3RT/M)

24. The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was:
(1) Methylamine
(2) Ammonia
(3) Phosgene
(4) Methylisocyanate

Sol. (4)
It was methyl isocyanate (CH₃NCO)

25. Consider the following reaction:
xMnO₄⁻ + yC₂O₄²⁻ + zH⁺ → xMn²⁺ + 2yCO₂ + (z/2)H₂O
The values of x, y and z in the reaction are, respectively:
(1) 2,5 and 8
(2) 2,5 and 16
(3) 5,2 and 8
(4) 5,2 and 16

Sol. (2)
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
x = 2, y = 5, z = 16

26. Which of the following exists as covalent crystals in the solid state?
(1) Silicon
(2) Sulphur
(3) Phosphorous
(4) Iodine

Sol. (1)
Silicon (Si) – covalent solid
Sulphur (S₈) – molecular solid
Phosphorous (P₄) – molecular solid
Iodine (I₂) – molecular solid

27. Compound (A), C₈H₈Br, gives a white precipitate when warmed with alcoholic AgNO₃. Oxidation of (A) gives an acid (B), C₈H₆O₄. (B) easily forms anhydride on heating. Identify the compound (A).

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(OCR खराब)

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30. In which of the following pairs of molecules/ions, both the species are not likely to exist?
(1) H₂, He₂²⁺
(2) H₂²⁺, He₂
(3) H₂, He₂²⁺
(4) H₂²⁺, He₂²⁺

Sol.

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31. The circle passing through (1, –2) and touching the axis of x at (3, 0) also passes through the point
(1) (2, –5)
(2) (5, –2)
(3) (–2, 5)
(4) (–5, 2)

Sol. (2)
(x – 3)² + y² + λy = 0
The circle passes through (1, –2)
⇒ 4 + 4 – 2λ = 0 ⇒ λ = 4
(x – 3)² + y² + 4y = 0 ⇒ Clearly (5, –2) satisfies.

32. ABCD is a trapezium such that AB and CD are parallel and BC⊥CD. If ∠ADB = θ, BC = p and CD = q, then AB is equal to
(1) (p² + q² cosθ)/(p cosθ + q sinθ)
(2) (p² + q²)/(p² cosθ + q² sinθ)
(3) ((p² + q²) sinθ)/(p cosθ + q sinθ)²
(4) ((p² + q²) sinθ)/(p cosθ + q sinθ)

Sol. (4)
Using sine rule in triangle ABD
AB/sinθ = BD/sin(θ + α)
⇒ AB = √(p² + q²) sinθ / (sinθ cosα + cosθ sinα)
= ((p² + q²) sinθ)/(p cosθ + q sinθ)

33. Given: A circle, 2x² + 2y² = 5 and a parabola, y² = 4√5 x
Statement-I: An equation of a common tangent to these curves is y = x + √5
Statement-II: If the line, y = mx + √5/m (m ≠ 0) is their common tangent, then m satisfies m⁴ – 3m² + 2 = 0
(1) Statement-I is True; Statement-II is true; Statement-II is not a correct explanation for Statement-I
(2) Statement-I is True; Statement-II is False.
(3) Statement-I is False; Statement-II is True
(4) Statement-I is True; Statement-II is True; Statement-II is a correct explanation for Statement-I

Sol. (1)
Let the tangent to the parabola be y = mx + √5/m (m ≠ 0)
Now, its distance from the centre of the circle must be equal to the radius of the circle.
So, |√5/m| = √5/√2 √(1 + m²) ⇒ (1 + m²)m² = 2 ⇒ m⁴ + m² – 2 = 0
⇒ (m² – 1)(m² + 2) = 0 ⇒ m = ±1
So, the common tangents are y = x + √5 and y = –x – √5

34. A ray of light along x + √3 y = √3 gets reflected upon reaching x-axis, the equation of the reflected rays is

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(1) √3 y = x – √3
(2) y = √3 x – √3
(3) √3 y = x – 1
(4) y = x + √3

Sol. (1)
Slope of the incident ray is 1/√3.
So, the slope of the reflected ray must be 1/√3.
The point of incidence is (√3, 0). So, the equation of reflected ray is y = (1/√3)(x – √3).

35. All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given?
(1) median
(2) mode
(3) variance
(4) mean

Sol. (3)
Variance is not changed by the change of origin.

36. If x, y, z are in A.P. and tan⁻¹ x, tan⁻¹ y and tan⁻¹ z are also in A.P., then
(1) 2x = 3y = 6z
(2) 6x = 3y = 2z
(3) 6x = 4y = 3z
(4) x = y = z

Sol. (4)
If x, y, z are in A.P.
2y = x + z
and tan⁻¹ x, tan⁻¹ y, tan⁻¹ z are in A.P.
2 tan⁻¹ y = tan⁻¹ x + tan⁻¹ z ⇒ x = y = z.
Note: If y = 0, then none of the options is appropriate.

37. If ∫ f(x) dx = Ψ(x), then ∫ x⁵ f(x³) dx is equal to
(1) (1/3)x³ Ψ(x³) – 3∫ x³ Ψ(x³) dx + C
(2) (1/3)[x³ Ψ(x³) – ∫ x³ Ψ(x³) dx] + C

Sol. (2)
∫ f(x) dx = Ψ(x)
Let x³ = t
3x² dx = dt
then ∫ x⁵ f(x³) dx = (1/3)∫ t f(t) dt
= (1/3)[t ∫ f(t) dt – ∫ {t·∫ f(t)dt} dt] = (1/3)x³ Ψ(x³) – ∫ x² Ψ(x³) dx + C.

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38. The equation of the circle passing through the foci of the ellipse x²/16 + y²/9 = 1, and having centre at (0,3) is
(1) x² + y² – 6y + 7 = 0
(2) x² + y² – 6y – 5 = 0
(3) x² + y² – 6y + 5 = 0
(4) x² + y² – 6y – 7 = 0

Sol. (4)
foci ≡ (±ae, 0)
We have a²e² = a² – b² = 7
Equation of circle (x – 0)² + (y – 3)² = (√7 – 0)² + (0 – 3)²
⇒ x² + y² – 6y – 7 = 0

39. The x-coordinate of the centre of the triangle that has the coordinates of mid points of its sides as (0,1), (1,1) and (1,0) is
(1) 2 – √2
(2) 1 + √2
(3) 1 – √2
(4) 2 + √2

Sol. (1)
x-coordinate = (a x₁ + b x₂ + c x₃)/(a + b + c)
= (2×2 + 2√2×0 + 2×0)/(2 + 2 + 2√2)
= 4/(4 + 2√2) = 2/(2 + √2) = 2 – √2.

40. The intercepts on x-axis made by tangents to the curve, y = ∫₀ˣ |t| dt, x ∈ R, which are parallel to the line y = 2x, are equal to
(1) ±2
(2) ±3
(3) ±4
(4) ±1

Sol. (4)
dy/dx = |x| = 2 ⇒ x = ±2 ⇒ y = ∫₀² |t| dt = 2 for x = 2
and y = ∫₀⁻² |t| dt = –2 for x = –2
∴ tangents are y – 2 = 2(x – 2) ⇒ y = 2x – 2
and y + 2 = 2(x + 2) ⇒ y = 2x + 2
Putting y = 0, we get x = 1 and –1

41. The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ...., is
(1) (7/9)(99 – 10⁻²⁰)
(2) (7/81)(179 + 10⁻²⁰)
(3) (7/9)(99 + 10⁻²⁰)
(4) (7/81)(179 – 10⁻²⁰)

Sol. (2)
tᵣ = 0.777…… r times

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= 7/9(1 – 10⁻ʳ)
S₂₀ = ∑_{r=1}^{20} tᵣ = 7/9(20 – ∑_{r=1}^{20} 10⁻ʳ) = 7/9(20 – 1/9(1 – 10⁻²⁰)) = 7/81(179 + 10⁻²⁰)

42. Consider:
Statement-I: (p ∧ ~q) ∧ (~p ∧ q) is a fallacy.
Statement-II: (p → q) ↔ (~q → ~p) is a tautology.
(1) Statement-I is True; Statement-II is true; Statement-II is not a correct explanation for Statement-I
(2) Statement-I is True; Statement-II is False.
(3) Statement-I is False; Statement-II is True
(4) Statement-I is True; Statement-II is True; Statement-II is a correct explanation for Statement-I

Sol. (1)
S1:
p q ~p ~q p∧~q ~p∧~q (p∧~q)∧(~p∧~q)
T T F F F F F
T F F T T F F
F T T F F T F
F F T T F F F

S2:
p q ~p ~q p→q ~q→~p (p→q)↔(~q→~p)
T T F F T T T
T F F T F F T
F T T F T T T
F F T T T T T

S2 is not an explanation of S1.

43. The area (in square units) bounded by the curves y = √x, 2y – x + 3 = 0, x-axis, and lying in the first quadrant is
(1) 36
(2) 18
(3) 27/4
(4) 9

Sol. (4)
2√x = x – 3
4x = x² – 6x + 9
x² – 10x + 9 = 0
x = 9, x = 1
∫₀³ (2y + 3) – y² dy
= [y² + 3y – y³/3]₀³ = 9 + 9 – 9 = 9

44. The expression tan A/(1 – cot A) + cot A/(1 – tan A) can be written as
(1) sec A cosec A + 1
(2) tan A + cot A
(3) sec A + cosec A
(4) sin A cos A + 1

Sol. (1)

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1/(1 – cot A) + cot A/(1 – tan A)
= 1 + sec A cosec A

45. The real number k for which the equation, 2x³ + 3x + k = 0 has two distinct real roots in [0, 1]
(1) lies between 2 and 3
(2) lies between –1 and 0
(3) does not exist
(4) lies between 1 and 2

Sol. (3)
If 2x³ + 3x + k = 0 has 2 distinct real roots in [0, 1], then f'(x) will change sign but f'(x) = 6x² + 3 > 0
So no value of k exists.

46. lim_{x→0} ((1 – cos 2x)(3 + cos x))/(tan 4x) is equal to
(1) 1/2
(2) 1
(3) 2
(4) –1/4

Sol. (3)
lim_{x→0} (1 – cos 2x)/(x tan 4x)(3 + cos x)
= lim 2(sin x/x)² · 1/4(4x/tan 4x)(3 + cos x) = 2×1×1/4×1×(3+1) = 2.

47. Let Tₙ be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If T_{n+1} – Tₙ = 10, then the value of n is
(1) 5
(2) 10
(3) 8
(4) 7

Sol. (1)
n+1C₃ – nC₃ = 10 ⇒ nC₂ = 10 ⇒ n = 5.

48. At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production P w.r.t. additional number of workers x is given by dP/dx = 100 – 12√x. If the firm employs 25 more workers, then the new level of production of items is
(1) 3000
(2) 3500
(3) 4500
(4) 2500

Sol.
∫₂₀₀₀ᴾ dP = ∫₀²⁵ (100 – 12√x) dx
(P – 2000) = 25×100 – (12×2/3)(25)^{3/2}
P = 3500.

49. Statement-I: The value of the integral ∫_{π/6}^{π/3} dx/√tan x is equal to π/6.
Statement-II: ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b – x) dx.
(1) Statement-I is True; Statement-II is true; Statement-II is not a correct explanation for Statement-I
(2) Statement-I is True; Statement-II is False.
(3) Statement-I is False; Statement-II is True
(4) Statement-I is True; Statement-II is True; Statement-II is a correct explanation for Statement-I

Sol.

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50. If P = [[1, α, 3], [1, 3, 3], [2, 4, 4]] is the adjoint of a 3×3 matrix A and |A| = 4, then α is equal to
(1) 11
(2) 5
(3) 0
(4) 4

Sol.
|Adj A| = |A|²
|Adj A| = 16
(12 – 12) – α(4 – 6) + 3(4 – 6) = 16
2α – 6 = 16
2α = 22
α = 11

51. The number of values of k, for which the system of equations
(k + 1)x + 8y = 4k
kx + (k + 3)y = 3k – 1
has no solution, is
(1) 1
(2) 2
(3) 3
(4) infinite

Sol.
For no solution (k + 1)/k = 8/(k + 3) ≠ 4k/(3k – 1) … (1)
⇒ (k + 1)(k + 3) – 8k = 0 or k² – 4k + 3 = 0 ⇒ k = 1, 3
But for k = 1, equation (1) is not satisfied
Hence k = 3

52. If y = sec(tan⁻¹ x), then dy/dx at x = 1 is equal to
(1) 1/2
(2) 1
(3) √2
(4) 1/√2

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53. If the lines (x – 2)/1 = (y – 3)/1 = (z – 4)/–k and (x – 1)/k = (y – 4)/2 = (z – 5)/1 are coplanar, then k can have
(1) exactly one value
(2) exactly two values
(3) exactly three values
(4) any value

Sol.

54. Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A × B having 3 or more elements is
(1) 220
(2) 219
(3) 211
(4) 256

Sol.
A × B will have 8 elements.
2⁸ – ⁸C₀ – ⁸C₁ – ⁸C₂ = 256 – 1 – 8 – 28 = 219.

55. If the vectors AB = 3i + 4k and AC = 5i – 2j + 4k are the sides of a triangle ABC, then the length of the median through A is
(1) √72
(2) √33
(3) √45
(4) √18

Sol.

56. A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is
(1) 13/3⁵
(2) 11/3⁵
(3) 10/3⁵
(4) 17/3⁵

Sol. (2)
P(correct answer) = 1/3

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JEE (MAIN)-2013-CMP-17

5C₄(1/3)⁴(2/3) + 5C₅(1/3)⁵
= (5×2 + 1)/3⁵ = 11/3⁵.

57. If z is a complex number of unit modulus and argument θ, then arg((1 + z)/(1 + z̄)) equals
(1) π/2 – θ
(2) θ
(3) π – θ
(4) –θ

Sol. (2)
|z| = 1 ⇒ z z̄ = 1
(1 + z)/(1 + z̄) = (1 + z)/(1 + 1/z) = z.
⇒ arg = θ

58. If the equations x² + 2x + 3 = 0 and ax² + bx + c = 0, a, b, c ∈ R, have a common root, then a : b : c is
(1) 3 : 2 : 1
(2) 1 : 3 : 2
(3) 3 : 1 : 2
(4) 1 : 2 : 3

Sol. (4)
For equation x² + 2x + 3 = 0
both roots are imaginary.
Since a, b, c ∈ R.
If one root is common then both roots are common
Hence, a/1 = b/2 = c/3
a : b : c = 1 : 2 : 3.

59. Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is
(1) 5/2
(2) 7/2
(3) 9/2
(4) 3/2

Sol. (2)
4x + 2y + 4z = 16
4x + 2y + 4z = –5
d = |21|/√36 = 21/6 = 7/2.

60. The term independent of x in expansion of ((x + 1)/(x^{2/3} – x^{1/3} + 1) – (x – 1)/(x – x^{1/2}))¹⁰ is
(1) 120
(2) 210
(3) 310
(4) 4

Sol. (2)
((x^{1/3} + 1)(x^{2/3} – x^{1/3} + 1)/(x^{2/3} – x^{1/3} + 1) – 1/√x · (√x + 1)(√x – 1)/(√x(√x – 1)))¹⁰
= (x^{1/3} – x^{−1/2})¹⁰
T_{r+1} = (−1)ʳ ¹⁰Cᵣ x^{(20−5r)/6} ⇒ r = 4
¹⁰C₄ = 210.

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61. In an LCR circuit as shown below both switches are open initially. Now switch S₁ is closed, S₂ kept open. (q is charge on the capacitor and τ = RC is capacitive time constant). Which of the following statement is correct?
(1) At t = τ, q = CV/2
(2) At t = 2τ, q = CV(1 – e⁻²)
(3) At t = τ/2, q = CV(1 – e⁻¹)
(4) Work done by the battery is half of the energy dissipated in the resistor.

Sol.
Charge on the capacitor at any time 't' is q = CV(1 – e^{−t/τ})
at t = 2τ, q = CV(1 – e⁻²)

62. A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance 100 kilo ohm. Find the maximum modulated frequency which could be detected by it.
(1) 10.62 kHz
(2) 5.31 MHz
(3) 5.31 kHz
(4) 10.62 MHz

Sol. (3)
f_c = 1/(2πRC) = 1/(2×3.14×100×10³×250×10⁻¹²) = 6.37 kHz
f_c = cut off frequency
As we know that f_m ≤ f_c
(3) is correct
Note: The maximum frequency of modulation must be less than f_m, where f_m = f_c√(1 – m²)/m
m ⇒ modulation index

63. The supply voltage to a room is 120 V. The resistance of the lead wires is 6 Ω. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?
(1) 2.9 Volt
(2) 13.3 Volt
(3) 10.04 Volt
(4) zero volt

Sol. (3)
Resistance of bulb = 120×120/60 = 240 Ω
Resistance of Heater = 120×120/240 = 60 Ω

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JEE (MAIN)-2013-CMP-19

Voltage across bulb before heater is switched on, V₁ = 120/246×240
Voltage across bulb after heater is switched on, V₂ = 120/54×48
Decrease in the voltage = V₁ – V₂ = 10.04 (approximately)
Note: Here supply voltage is taken as rated voltage.

64. A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density σ at equilibrium position. The extension x₀ of the spring when it is in equilibrium is:
(1) Mg/k (1 – LAσ/M)
(2) Mg/k (1 – LAσ/2M)
(3) Mg/k (1 + LAσ/M)
(4) Mg/k
(Here k is spring constant)

Sol. (2)
At equilibrium ΣF = 0
kx₀ + (AL/2)σg – Mg = 0
x₀ = Mg/k (1 – LAσ/2M)

65. Two charges, each equal to q, are kept at x = –a and x = a on the x-axis. A particle of mass m and charge q₀ = q/2 is placed at the origin. If charge q₀ is given a small displacement (y << a) along the y-axis, the net force acting on the particle is proportional to:
(1) –y
(2) 1/y
(3) –1/y
(4) y

Sol. (4)
F_net = 2F cosθ
= 2 · k q q/2 / (√(a² + y²))² · y/√(a² + y²)
= kq² y/a³ (y << a)

66. A beam of unpolarised light of intensity I₀ is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A. The intensity of the emergent light is:
(1) I₀/2
(2) I₀/4
(3) I₀/8
(4) I₀

Sol. (2)

67. The anode voltage of a photocell is kept fixed. The wavelength λ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows:

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68. Two coherent point sources S₁ and S₂ are separated by a small distance 'd' as shown. The fringes obtained on the screen will be:
(1) straight lines
(2) semi-circles
(3) concentric circles
(4) points

Sol.

69. (OCR खराब)

70. (OCR खराब)

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71. Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is ρ and L is its latent heat of vaporization.
(1) √(T/ρL)
(2) T/ρL
(3) 2T/ρL
(4) ρL/T

Sol. (3)
ρ 4πR² ΔR L = T 4π[R² – (R – ΔR)²]
ρ R² ΔR L = T[R² – R² + 2R ΔR – ΔR²]
ρ R² ΔR L = T 2R ΔR (ΔR is very small)
R = 2T/ρL

72. The graph between angle of deviation (δ) and angle of incidence (i) for a triangular prism is represented by:

Sol. (2)

73. Let [ε₀] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then:
(1) [ε₀] = [M⁻¹ L⁻³ T⁴ A²]
(2) [ε₀] = [M⁻¹ L⁻³ T⁴ A²]
(3) [ε₀] = [M⁻¹ L⁻³ T⁴ A]
(4) [ε₀] = [M⁻¹ L⁻³ T⁴ A²]

Sol. (1)
1/(4πε₀) q²/r² = F
ε₀ = [A² T²]/[M L² L⁻²] = [M⁻¹ L⁻³ A² T⁴]

74. The above p-V diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat extracted from the source in a single cycle is
(1) (13/2)p₀v₀
(2) (11/2)p₀v₀
(3) 4p₀v₀
(4) p₀v₀

Sol. (2)
Heat is extracted from the source in path DA and AB is

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75. A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are 7.7×10³ kg/m³ and 2.2×10¹¹ N/m² respectively?
(1) 178.2 Hz
(2) 200.5 Hz
(3) 770 Hz
(4) 188.5 Hz

Sol. (1)
Fundamental frequency f = (1/2ℓ)√(T/μ)
= (1/2ℓ)√(T/Aρ)
= (1/2ℓ)√(stress/ρ) = 1/(2×1.5)√(2.2×10¹¹×10⁻² / 7.7×10³).

76. This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement-I: Higher the range, greater is the resistance of ammeter.
Statement-II: To increase the range of ammeter, additional shunt needs to be used across it.
(1) Statement-I is true, Statement-II is true, Statement-II is not the correct explanation of Statement-I.
(2) Statement-I is true, statement-II is false.
(3) Statement-I is false, Statement-II is true.
(4) Statement-I is true, Statement-II is true, Statement-II is the correct explanation of statement-I.

Sol. (3)
For Ammeter, S = I_g G/(I – I_g)
So for I to increase, S should decrease, so additional S can be connected across it.

77. What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?
(1) 2GmM/3R
(2) GmM/2R
(3) GmM/3R
(4) 5GmM/6R

Sol. (4)
T.E._f = –GmM/6R
T.E._i = –GmM/R
ΔW = T.E._f – T.E._i = 5GmM/6R

78. A projectile is given an initial velocity of (i + 2j) m/s where i is along the ground and j is along the vertical. If g = 10 m/s², the equation of its trajectory is:
(1) y = 2x – 5x²
(2) 4y = 2x – 5x²
(3) 4y = 2x – 25x²
(4) y = x – 5x²

Sol. (1)

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79. Two capacitors C₁ and C₂ are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then:
(1) 3C₁ = 5C₂
(2) 3C₁ + 5C₂ = 0
(3) 9C₁ = 4C₂
(4) 5C₁ = 3C₂

Sol. (1)
120C₁ = 200C₂
6C₁ = 10C₂
3C₁ = 5C₂

80. A hoop of radius r and mass m rotating with an angular velocity ω₁ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?
(1) rω₁/3
(2) rω₁/2
(3) rω₁
(4) rω₁/4

Sol.
From conservation of angular momentum about any fix point on the surface
mr² ω₁ = 2mr² ω
ω = ω₁/2
V_CM = ω₁ r/2

81. An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is V₀ and its pressure is P₀. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency:
(1) (1/2π)√(V₀ M P₀/A² γ)
(2) (1/2π)√(A² γ P₀/V₀ M)
(3) (1/2π)√(A γ P₀/V₀ M)
(4) (1/2π)√(V₀ M/A γ P₀)

Sol.

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82. A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at a distance L from the end A is:
(1) Q/(ε₀ L)
(2) Q/(4πε₀ L ln 2)
(3) Q ln 2/(4πε₀ L)
(4) Q/(8πε₀ L)

Sol. (3)
V = ∫_{x=L}^{x=2L} k(Q/L)/x dx = Q ln 2/(4πε₀ L)

83. A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is
(1) 6 × 10⁻¹¹ Weber
(2) 3.3 × 10⁻¹¹ Weber
(3) 6.6 × 10⁻⁹ Weber
(4) 9.1 × 10⁻¹¹ Weber

Sol. (4)
Let M₁₂ be the coefficient of mutual induction between loops
φ₁ = M₁₂ i₂
⇒ μ₀ i₂ R²/(2(d² + R²)^{3/2}) πr² = M₁₂ i₂
⇒ M₁₂ = μ₀ R² πr²/(2(d² + R²)^{3/2})
φ₂ = M₁₂ i₁ ⇒ φ₂ = 9.1×10⁻¹¹ weber

84. If a piece of metal is heated to temperature θ and then allowed to cool in a room which is at temperature θ₀, the graph between the temperature T of the metal and time t will be closest to:

(1) [graph]
(2) [graph]
(3) [graph]
(4) [graph]

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(OCR खराब)

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87. The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10s it will decrease to α times its original magnitude, where α equals.
(1) 0.81
(2) 0.729
(3) 0.6
(4) 0.7

Sol. (2)
A = A₀ e^{−kt}
⇒ 0.9A₀ = A₀ e^{−5k}
and αA₀ = A₀ e^{−15k}
solving ⇒ α = 0.729

88. Diameter of plano-convex lens is 6 cm and thickness at the centre is 3 mm. If speed of light in material of lens is 2×10⁸ m/s, the focal length of the lens is:
(1) 20 cm
(2) 30 cm
(3) 10 cm
(4) 15 cm

Sol. (2)
R² = d² + (R – t)²
R² – d² = R²{1 – t/R}²
1 – d²/R² = 1 – 2t/R
R = (3)²/(2×0.3) = 90/6 = 15 cm
1/f = (μ – 1)(1/R₁ – 1/R₂)
1/f = (3/2 – 1)(1/15)
f = 30 cm

89. The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is:
(1) 6 V/m
(2) 9 V/m
(3) 12 V/m
(4) 3 V/m

Sol. (1)
E₀ = cB₀
= 3×10⁸ × 20×10⁻⁹
= 6 V/m

90. Two short bar magnets of length 1 cm each have magnetic moments 1.20 Am² and 1.00 Am² respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centres is close to (Horizontal component of earth's magnetic induction is 3.6×10⁻⁵ Wb/m²)

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