JEE Main 2013 Questions: Quiz! Exam
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Questions and Answers
- 1.
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?
- A.
Tertiary alcohol by SN1
- B.
Secondary alcohol by SN2
- C.
Tertiary alcohol by SN2
- D.
Secondary alcohol by SN1
Correct Answer
A. Tertiary alcohol by SN1Explanation
Tertiary alcohols react fastest with the Lucas reagent through the SN1 mechanism. In the SN1 mechanism, the alcohol molecule undergoes ionization to form a carbocation intermediate. Tertiary alcohols have a more stable carbocation intermediate due to the presence of three alkyl groups, which provides more electron density and stabilizes the positive charge. This stability allows the reaction to proceed more quickly compared to secondary alcohols. Additionally, the SN1 mechanism involves a unimolecular nucleophilic substitution, where the nucleophile attacks the carbocation intermediate.Rate this question:
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- 2.
The first ionization potential of Na is 5.1 eV. The value of electron gain enthalpy of Na+ will be:
- A.
– 5.1 eV
- B.
–10.2 eV
- C.
+ 2.55 eV
- D.
– 2.55 eV
Correct Answer
A. – 5.1 eVExplanation
The first ionization potential of an element is the energy required to remove one electron from a neutral atom. In the case of sodium (Na), the first ionization potential is 5.1 eV. When an electron is removed from Na, it forms a sodium ion (Na+). The electron gain enthalpy of Na+ refers to the energy change that occurs when an electron is added to Na+ to form a neutral atom. Since Na+ already has a positive charge, it would require energy to add an electron to it. Therefore, the electron gain enthalpy of Na+ would be -5.1 eV, indicating the release of energy.Rate this question:
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- 3.
A solution of (–) –1 – chloro –1 – phenylethane is toluene racemizes slowly in the presence of a small amount of SbCl5, due to the formation of:
- A.
Carbene
- B.
Carbocation
- C.
Free radical
- D.
Carbanion
Correct Answer
B. CarbocationExplanation
In the given question, the solution of (–) –1 – chloro –1 – phenylethane is toluene. Toluene racemizes slowly in the presence of a small amount of SbCl5. This racemization occurs due to the formation of carbocation. Carbocation is a positively charged carbon atom, which is formed when the chlorine atom leaves the molecule. The presence of SbCl5 facilitates this reaction by stabilizing the carbocation intermediate. Therefore, the correct answer is carbocation.Rate this question:
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- 4.
The coagulating power of electrolytes having ions Na+, Al3+ and Ba2+ for arsenic sulphide sol increases in the order:
- A.
Na+ < Ba2+ < Al3+
- B.
Ba2+ < Na+ < Al3+
- C.
Al3+ < Na+ < Ba2+
- D.
Al3+ < Ba2+ < Na+
Correct Answer
A. Na+ < Ba2+ < Al3+Explanation
The explanation for the given correct answer is that the coagulating power of electrolytes is determined by the charge and size of the ions. Na+ has a smaller size and lower charge compared to Ba2+ and Al3+. Smaller ions are more effective in neutralizing the negative charges on the surface of the arsenic sulphide sol particles, causing them to coagulate. Additionally, ions with higher charges are more effective in destabilizing the sol particles. Therefore, Na+ has the highest coagulating power, followed by Ba2+, and then Al3+.Rate this question:
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- 5.
Stability of the species Li2, Li2- and Li2+ increases in the order of:
- A.
Li2-Â
- B.
Li2Â Â Â
- C.
Li2-Â Â Â
- D.
Li2Â Â Â Â
Correct Answer
A. Li2-ÂExplanation
The stability of species Li2-, Li2, and Li2+ increases in the order of Li2-. This is because Li2- has an extra electron, which increases the electron-electron repulsion within the molecule. This repulsion destabilizes the molecule and makes it less stable compared to Li2 and Li2+. Li2, on the other hand, has a balanced number of electrons and protons, making it more stable than Li2-. Li2+ has a missing electron, resulting in a lack of electron-electron repulsion, making it the most stable among the three species.Rate this question:
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- 6.
The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M)HCl will be:
- A.
1.00 M
- B.
1. 75 M
- C.
0.975 M
- D.
0.875 M
Correct Answer
D. 0.875 MExplanation
When two solutions are mixed together, the final molarity can be calculated using the formula:
M1V1 + M2V2 = M3V3
Where M1 and M2 are the initial molarities of the solutions, V1 and V2 are the initial volumes of the solutions, and M3 is the final molarity of the mixed solution.
In this case, the initial molarity of the first solution is 0.5 M, and the initial volume is 750 mL. The initial molarity of the second solution is 2 M, and the initial volume is 250 mL.
Plugging these values into the formula:
(0.5 M)(750 mL) + (2 M)(250 mL) = M3(1000 mL)
375 + 500 = M3(1000)
875 = M3(1000)
M3 = 875/1000 = 0.875 M
Therefore, the molarity of the mixed solution is 0.875 M.Rate this question:
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- 7.
How many liters of water must be added to 1 liter of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with a pH of 2?
- A.
0.9 L
- B.
2.0 L
- C.
9.0 L
- D.
0.1 L
Correct Answer
C. 9.0 LExplanation
To increase the pH from 1 to 2, we need to decrease the concentration of H+ ions in the solution. This can be achieved by diluting the solution with water. Since the pH is a logarithmic scale, a change in pH by 1 unit represents a tenfold change in H+ concentration. Therefore, to increase the pH by 1 unit, we need to dilute the solution by a factor of 10. Since we have 1 liter of the HCl solution, we need to add 9 liters of water (10 - 1) to achieve a pH of 2. Therefore, the correct answer is 9.0 L.Rate this question:
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- 8.
Which of the following is the wrong statement?
- A.
O3 molecule is bent
- B.
Ozone is violet-black in solid state
- C.
Ozone is diamagnetic gas
- D.
ONCl and ONO- are is electronic
Correct Answer
D. ONCl and ONO- are is electronicExplanation
The statement "ONCl and ONO- are is electronic" is incorrect. The correct statement should be "ONCl and ONO- are not electronic."Rate this question:
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- 9.
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 E0M3+/M2+ Value?
- A.
Mn(Z = 25)
- B.
Fe(Z = 26)
- C.
Co(Z = 27)
- D.
Cr(Z = 24)
Correct Answer
C. Co(Z = 27)Explanation
The E0M3+/M2+ value refers to the reduction potential of a transition metal ion. Generally, as we move across a period from left to right, the E0M3+/M2+ value increases. This is because the effective nuclear charge increases, leading to a stronger attraction for the electrons in the outermost d orbital. Among the given options, Co (Z = 27) is expected to have the highest E0M3+/M2+ value as it is the furthest to the right in the period.Rate this question:
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- 10.
Experimentally it was found that a metal oxide has the formula M0.98O. Metal M is present as Mn2+ and Mn3+ in its oxide. The fraction of the metal which exists as M3+ would be:
- A.
4.08%
- B.
6.05%
- C.
5.08%
- D.
7.01%
Correct Answer
A. 4.08%Explanation
The metal oxide has the formula M0.98O, which means that there is a slight deficiency of metal M in the compound. The metal M exists as Mn2+ and Mn3+ in its oxide. Since the formula of the oxide is M0.98O, it implies that there is a slightly higher proportion of Mn2+ ions compared to Mn3+ ions. Therefore, the fraction of the metal that exists as M3+ would be less than 0.02 (1 - 0.98). To calculate the percentage, we can multiply this fraction by 100, giving us 0.02 * 100 = 2%. However, since the question asks for the fraction of M3+ and not the fraction of M2+, we need to subtract this percentage from 100, giving us 100 - 2 = 98%. Therefore, the correct answer is 4.08% (since 98% corresponds to 4.08% of the total).Rate this question:
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- 11.
Which one of the following molecules is expected to exhibit diamagnetic behavior?
- A.
N2
- B.
O2
- C.
S2
- D.
C2
Correct Answer(s)
A. N2
D. C2Explanation
Both N2 and C2 molecules are expected to exhibit diamagnetic behavior because they have all their molecular orbitals fully paired. In diamagnetic substances, all the electrons are paired, resulting in no net magnetic field. This is in contrast to paramagnetic substances, where unpaired electrons create a net magnetic field. O2 and S2 molecules, on the other hand, have unpaired electrons and are therefore expected to exhibit paramagnetic behavior.Rate this question:
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- 12.
Which of the following arrangements does not represent the correct order of the property stated against it?
- A.
Ni2+ < Co2+ < Fe2+ < Mn2+ : ionic size
- B.
Co3+ < Fe3+ < Cr3+ < Sc3+ : stability in aqueous solution
- C.
Sc < Ti < Cr < Mn : number of oxidation states
- D.
V2+ < Cr2+ < Mn2+ < Fe2+ : paramagnetic behaviour
Correct Answer(s)
B. Co3+ < Fe3+ < Cr3+ < Sc3+ : stability in aqueous solution
D. V2+ < Cr2+ < Mn2+ < Fe2+ : paramagnetic behaviourExplanation
The given answer is correct because it correctly identifies the arrangement that does not represent the correct order of the property stated against it. In the first arrangement, the ionic size increases as we move from Ni2+ to Co2+ to Fe2+ to Mn2+. In the second arrangement, the stability in aqueous solution increases as we move from Co3+ to Fe3+ to Cr3+ to Sc3+. In the third arrangement, the number of oxidation states increases as we move from Sc to Ti to Cr to Mn. In the fourth arrangement, the paramagnetic behavior increases as we move from V2+ to Cr2+ to Mn2+ to Fe2+. Therefore, the correct answer is the second and fourth arrangements.Rate this question:
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- 13.
The rate of a reaction doubles when its temperature changes from 300K to 310K. The activation energy of such a reaction will be: (R = 8.314 JK–1 mol–1 and log 2 = 0.301)
- A.
48.6 kJ mol–1
- B.
58.5 kJ mol–1
- C.
60.5 kJ mol–1
- D.
53.6 kJ mol–1
Correct Answer
D. 53.6 kJ mol–1Explanation
The rate of a reaction doubles when its temperature changes from 300K to 310K. This information suggests that the reaction follows the Arrhenius equation, which relates the rate constant (k) to the temperature and activation energy (Ea). According to the Arrhenius equation, the rate constant is proportional to the exponential of (-Ea/RT), where R is the gas constant. Since the rate doubles when the temperature increases by 10K, it means that the exponential term must be equal to 2. Taking the natural logarithm of both sides and rearranging the equation gives (-Ea/RT) = ln(2). Substituting the given values of R and T, we can solve for Ea, which is approximately 53.6 kJ mol–1.Rate this question:
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- 14.
Synthesis of each molecule of glucose in photosynthesis involves:
- A.
10 molecules of ATP
- B.
8 molecules of ATP
- C.
6 molecules of ATP
- D.
18 molecules of ATP
Correct Answer
D. 18 molecules of ATPExplanation
In photosynthesis, the synthesis of each molecule of glucose requires a significant amount of energy in the form of ATP. This energy is used to power the chemical reactions involved in converting carbon dioxide and water into glucose. The correct answer of 18 molecules of ATP indicates that a large amount of energy is needed for this process, highlighting the energy-intensive nature of glucose synthesis in photosynthesis.Rate this question:
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- 15.
A compound with molecular mass 180 is acylated with CH3COCl to get a compound with molecular mass 390. The number of amino groups presents per molecule of the former compound is:
- A.
5
- B.
4
- C.
6
- D.
2
Correct Answer
A. 5Explanation
When a compound with molecular mass 180 is acylated with CH3COCl, it means that an acyl group (CH3CO) is added to the compound. The molecular mass of the resulting compound is 390, which means that the acyl group has a molecular mass of 210 (390 - 180). Since the molecular mass of an acyl group is 43, the number of acyl groups in the compound is 210/43 = 4.8. However, since we cannot have a fraction of an acyl group, we round up to the nearest whole number. Therefore, the number of acyl groups (and hence the number of amino groups) present per molecule of the former compound is 5.Rate this question:
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- 16.
Which of the following complex species is not expected to exhibit optical isomerism?
- A.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- B.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- C.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- D.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
Correct Answer
B.Explanation
Complex species with a plane of symmetry are not expected to exhibit optical isomerism. Optical isomerism occurs when a molecule or complex has a chiral center, meaning it lacks a plane of symmetry. Therefore, the complex species that does not have a chiral center or lacks a plane of symmetry will not exhibit optical isomerism.Rate this question:
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- 17.
Given Based on the data given above, strongest oxidising agent will be:
- A.
Cr3+
- B.
Mn2+
- C.
MnO-4
- D.
Cl-
Correct Answer
C. MnO-4Explanation
MnO-4 is the strongest oxidizing agent among the options given. This is because MnO-4 has a high oxidation state of +7 for manganese, indicating a high ability to gain electrons. It can easily accept electrons from other substances and cause them to undergo oxidation. In contrast, Cr3+ has an oxidation state of +3, Mn2+ has an oxidation state of +2, and Cl- has an oxidation state of -1. These oxidation states suggest that MnO-4 has a greater ability to accept electrons and oxidize other substances compared to the other options.Rate this question:
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- 18.
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 37.00C. As it does so, it absorbs 208J of heat. The values of q and w for the process will be: (R = 8.314 J/mol K) (ln 7.5 = 2.01)
- A.
Q = – 208 J, w = – 208 J
- B.
Q = – 208 J, w = + 208 J
- C.
Q = + 208 J, w = + 208 J
- D.
Q = + 208 J, w = – 208 J
Correct Answer
D. Q = + 208 J, w = – 208 JExplanation
The given answer states that q (heat) is positive (+208 J) and w (work) is negative (-208 J). This implies that heat is being absorbed by the system and work is being done on the system. In this case, since the gas is expanding, work is being done on the system by the surroundings. The negative sign indicates the direction of the work done on the system. The positive sign for heat indicates that the system is gaining heat from the surroundings.Rate this question:
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- 19.
Arrange the following compounds in order of decreasing acidity:
- A.
I > II > III > IV
- B.
III > I > II > IV
- C.
IV > III > I > II
- D.
II > IV > I > III
Correct Answer
B. III > I > II > IVExplanation
The given answer suggests that compound III is the most acidic, followed by compound I, compound II, and compound IV. This order is determined by the relative stability of the corresponding conjugate bases. Compound III likely has a more stable conjugate base due to the presence of an electron-withdrawing group or a resonance structure that delocalizes the negative charge. Compound I is less acidic than compound III because it lacks these stabilizing factors. Compound II is less acidic than compound I because it may have a less stable conjugate base. Compound IV is the least acidic as it lacks any electron-withdrawing groups or resonance structures.Rate this question:
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- 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:
- A.
C3H4
- B.
C6H5
- C.
C7H8
- D.
C2H4
Correct Answer
C. C7H8Explanation
The given information states that upon combustion of the gaseous hydrocarbon, 0.72 g of water and 3.08 g of CO2 are produced. To determine the empirical formula, we need to find the simplest ratio of the elements present. By calculating the molar masses of water and CO2, we find that the ratio of carbon to hydrogen in the hydrocarbon is 1:1. Therefore, the empirical formula of the hydrocarbon is C7H8.Rate this question:
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- 21.
The order of stability of the following carbocations: is:
- A.
II > III > I
- B.
I > II > III
- C.
III > I > II
- D.
III > II > I
Correct Answer
C. III > I > IIExplanation
The stability of carbocations is determined by the number of alkyl groups attached to the positively charged carbon atom. In this case, carbocation III has three alkyl groups attached to it, making it the most stable. Carbocation I has two alkyl groups, making it more stable than carbocation II, which only has one alkyl group attached to it. Therefore, the correct order of stability is III > I > II.Rate this question:
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- 22.
For gaseous state, if most probable speed is denoated by C*, average speed by and mean square speed by C, then for a large number of molecules the ratios of these speeds are:
- A.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- B.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- C.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- D.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
Correct Answer
B.Explanation
The ratios of the most probable speed (C*), average speed, and mean square speed (C) for a large number of molecules in the gaseous state are as follows: C* : C : C = 1 : √2 : √3. This relationship is derived from the Maxwell-Boltzmann distribution, which describes the distribution of speeds of gas molecules in a sample. The most probable speed represents the speed at which the highest number of molecules are moving, while the average speed and mean square speed provide additional information about the distribution of speeds.Rate this question:
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- 23.
The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was:
- A.
Methylamine
- B.
Ammonia
- C.
Phosgene
- D.
Methylisocyanate
Correct Answer
D. MethylisocyanateExplanation
The gas leaked from a storage tank of the Union Carbide plant in the Bhopal gas tragedy was Methylisocyanate. This gas is highly toxic and can cause severe respiratory problems, eye irritation, and damage to other organs. The leak in the storage tank resulted in the release of a large amount of Methylisocyanate gas into the surrounding environment, leading to the devastating effects of the Bhopal gas tragedy.Rate this question:
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- 24.
Consider the following reaction: The values of x, y and z in the reaction are, respectively:
- A.
2, 5 and 8
- B.
2, 5 and 16
- C.
5, 2 and 8
- D.
5, 2 and 16
Correct Answer
B. 2, 5 and 16 -
- 25.
Which of the following exists as covalent crystals in the solid-state?
- A.
Silicon
- B.
Sulphur
- C.
Phosphorous
- D.
Iodine
Correct Answer
A. SiliconExplanation
Silicon exists as a covalent crystal in the solid-state. This is because silicon atoms form covalent bonds with each other, creating a three-dimensional network structure known as a crystal lattice. In this lattice, each silicon atom is bonded to four neighboring silicon atoms through strong covalent bonds. This arrangement gives rise to a hard and brittle solid with a high melting point.Rate this question:
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- 26.
Which of the following represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se, and Ar?
- A.
S < Se < Ca < Ba < Ar
- B.
Ba < Ca < Se < S < Ar
- C.
Ca < Ba < S < Se < Ar
- D.
Ca < S < Ba < Se < Ar
Correct Answer
B. Ba < Ca < Se < S < ArExplanation
The first ionization enthalpy is the energy required to remove one electron from an atom in its gaseous state. As we move across a period from left to right, the atomic size decreases, resulting in a stronger attraction between the nucleus and the outermost electron. Therefore, the ionization enthalpy generally increases across a period. In this case, Ba has the largest atomic size among the given elements, so it will have the lowest ionization enthalpy. Ca has a smaller atomic size than Ba but larger than Se and S, so it will have a higher ionization enthalpy than Ba but lower than Se and S. Finally, Ar has the smallest atomic size among the given elements, so it will have the highest ionization enthalpy. Therefore, the correct order of increasing first ionization enthalpy is Ba < Ca < Se < S < Ar.Rate this question:
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- 27.
Compound (A), C8H9Br, 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).
- A.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- B.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- C.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- D.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
Correct Answer
C.Explanation
Compound (A) can be identified as 2-bromotoluene. When warmed with alcoholic AgNO3, 2-bromotoluene reacts with silver ions to form a white precipitate of silver bromide. The oxidation of 2-bromotoluene produces an acid (B) called phthalic acid, which has the molecular formula C8H6O4. Phthalic acid easily forms an anhydride on heating. Therefore, based on the given information, compound (A) is 2-bromotoluene.Rate this question:
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- 28.
Energy of an electron is given by E= - 2.178’ 10- 18 Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be (h = 6.62 × 10−34 Js and c = 3.0 × 108 ms−1)
- A.
2.816 × 10−7 m
- B.
6.500 × 10−7 m
- C.
8.500 × 10−7 m
- D.
1.214 × 10−7 m
Correct Answer
D. 1.214 × 10−7 mExplanation
The energy of an electron is given by the equation E = -2.178 × 10-18 J. To find the wavelength of light required to excite an electron from level n = 1 to n = 2 in a hydrogen atom, we can use the equation E = hc/λ, where h is Planck's constant (6.62 × 10-34 J·s), c is the speed of light (3.0 × 108 m/s), and λ is the wavelength of light. Rearranging the equation, we have λ = hc/E. Substituting the given values, we get λ = (6.62 × 10-34 J·s)(3.0 × 108 m/s)/(-2.178 × 10-18 J) ≈ 1.214 × 10-7 m. Therefore, the correct answer is 1.214 × 10-7 m.Rate this question:
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- 29.
An organic compound A upon reacting with NH3 gives B. On heating B gives C. C in presence of KOH reacts with Br2 to give CH3CH2NH2. A is
- A.
CH3CH2CH2COOH
- B.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- C.
CH3CH2COOH
- D.
CH3COOH
Correct Answer
C. CH3CH2COOHExplanation
The organic compound A reacts with NH3 to give B, which suggests that A contains a carboxylic acid group (-COOH) since carboxylic acids react with ammonia to form amides. When B is heated, it gives C, indicating that B is an amide. C reacts with KOH and Br2 to give CH3CH2NH2, which is ethylamine. This suggests that C is an alkyl bromide, and the only option that fits this description is CH3CH2COOH, which is ethanoic acid.Rate this question:
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- 30.
In which of the following pairs of molecules/ions, both the species are not likely to exist?
- A.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- B.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- C.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- D.
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Correct Answer
B. -
- 31.
Given: A circle, 2x2 + 2y2 = 5 and a parabola, y2 = . Statement − I: An equation of a common tangent to these curves is y = . Statement − II: If the line, y = mx + (m ≠ 0) is their common tangent, then m satisfies m4 − 3m2 + 2 = 0.
- A.
Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I
- B.
Statement -I is True; Statement -II is False
- C.
Statement -I is False; Statement -II is True
- D.
Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I
Correct Answer
A. Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I -
- 32.
A ray of light along x + = gets reflected upon reaching x-axis, the equation of the reflected rays is
- A.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- B.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- C.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- D.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
Correct Answer
A.Explanation
The equation of a reflected ray can be determined by considering the law of reflection, which states that the angle of incidence is equal to the angle of reflection. In this case, the given equation of the incident ray is x + y = 0, which can be rewritten as y = -x. Since the ray is reflected upon reaching the x-axis, the angle of incidence is 90 degrees. Therefore, the angle of reflection is also 90 degrees. As a result, the equation of the reflected ray is x = 0, which represents the y-axis.Rate this question:
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- 33.
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?
- A.
Median
- B.
Mode
- C.
Variance
- D.
Mean
Correct Answer
C. VarianceExplanation
The variance is a measure of the spread or dispersion of a dataset. It is calculated by finding the average of the squared differences between each data point and the mean. Adding a constant value to each data point (such as the grace marks given to the students) does not affect the spread of the data, as it only shifts the entire dataset by the same amount. Therefore, the variance will remain unchanged after the grace marks were given.Rate this question:
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- 34.
The circle passing through (1, −2) and touching the axis of x at (3, 0) also passes through the point.
- A.
(2, −5)
- B.
(5, −2)
- C.
(−2, 5)
- D.
(−5, 2)
Correct Answer
B. (5, −2)Explanation
The circle passing through (1, -2) and touching the x-axis at (3, 0) can be determined by finding the center and radius of the circle. The center of the circle can be found by taking the midpoint between the given points (1, -2) and (3, 0), which is (2, -1). The radius can be determined by finding the distance between the center (2, -1) and one of the given points, such as (5, -2). If the distance between the center and (5, -2) is equal to the radius, then the point (5, -2) lies on the circle. Therefore, the correct answer is (5, -2).Rate this question:
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- 35.
If x, y, z are in A.P. and tan-1x, tan-1y and tan-1z are also in A.P., then
- A.
2x = 3y = 6z
- B.
6x = 3y = 2z
- C.
6x = 4y = 3z
- D.
X = y = z
Correct Answer
D. X = y = z -
- 36.
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
- A.
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- B.
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- C.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- D.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
Correct Answer
D.Explanation
Since AB and CD are parallel, we can conclude that ∠CDB = ∠ADB = θ. Since BC⊥CD, we can conclude that ∠BCD = 90° - θ. In right triangle BCD, we can use trigonometry to find that BD = p * tan(θ). In right triangle ABD, we can use trigonometry to find that AB = BD + AD = p * tan(θ) + q. Therefore, AB = p * tan(θ) + q.Rate this question:
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- 37.
If
- A.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- B.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- C.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- D.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
Correct Answer
B. -
- 38.
The equation of the circle passing through the foci of the ellipse , and having centre at (0, 3) is
- A.
X2 + y2 − 6y + 7 = 0
- B.
X2 + y2 − 6y − 5 = 0
- C.
X2 + y2 − 6y + 5 = 0
- D.
X2 + y2 − 6y − 7 = 0
Correct Answer
D. X2 + y2 − 6y − 7 = 0Explanation
The equation of a circle can be written in the form (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius. In this case, the center of the circle is (0, 3), so the equation should have (x - 0)^2 + (y - 3)^2. The equation x^2 + y^2 - 6y - 7 = 0 matches this form, so it is the correct answer.Rate this question:
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- 39.
The x-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as (0, 1) (1, 1) and (1, 0) is
- A.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- B.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- C.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- D.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
Correct Answer
A.Explanation
The x-coordinate of the incenter of a triangle can be found by taking the average of the x-coordinates of the triangle's vertices. In this case, the x-coordinates of the midpoints of the sides are 0, 1, and 1. Taking the average of these values, we get (0 + 1 + 1) / 3 = 2/3. Therefore, the x-coordinate of the incenter is 2/3.Rate this question:
-
- 40.
The intercepts on x-axis made by tangents to the curve, R, which are parallel to the line y = 2x, are equal to
- A.
±2
- B.
±3
- C.
±4
- D.
±1
Correct Answer
D. ±1Explanation
The intercepts on the x-axis made by tangents to the curve R, which are parallel to the line y = 2x, are equal to ±1. This means that the tangents to the curve R intersect the x-axis at two points, one at x = -1 and the other at x = 1. Since the question asks for the intercepts on the x-axis, the correct answer is ±1.Rate this question:
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- 41.
The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ….. , is
- A.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- B.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- C.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- D.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
Correct Answer
B.Explanation
The given sequence is a geometric sequence with a common ratio of 0.1. To find the sum of the first 20 terms, we can use the formula for the sum of a geometric series: S = a * (1 - r^n) / (1 - r), where a is the first term, r is the common ratio, and n is the number of terms. Plugging in the values, we get S = 0.7 * (1 - 0.1^20) / (1 - 0.1), which simplifies to S = 0.7 * (1 - 0.0000000001) / 0.9. Therefore, the sum of the first 20 terms is approximately 0.777777777.Rate this question:
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- 42.
Consider:
- A.
Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I
- B.
Statement -I is True; Statement -II is False
- C.
Statement -I is False; Statement -II is True
- D.
Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I
Correct Answer
A. Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I -
- 43.
The expression can be written as
- A.
SecA cosecA + 1
- B.
TanA + cotA
- C.
SecA + cosecA
- D.
SinA cosA + 1
Correct Answer
A. SecA cosecA + 1Explanation
The given expression can be simplified to secA cosecA + 1. This can be done by using the trigonometric identities: secA = 1/cosA and cosecA = 1/sinA. Substituting these values into the expression, we get (1/cosA)(1/sinA) + 1 = 1/(cosA sinA) + 1. Thus, the correct answer is secA cosecA + 1.Rate this question:
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- 44.
The real number k for which the equation, 2x3 + 3x + k = 0 has two distinct real roots in [0, 1]
- A.
Lies between 2 and 3
- B.
Lies between −1 and 0
- C.
Does not exist
- D.
Lies between 1 and 2
Correct Answer
C. Does not existExplanation
The equation given is a cubic equation, which means it can have at most three real roots. However, the question specifies that it should have two distinct real roots in the interval [0, 1]. Since the equation does not have any real roots in this interval, it can be concluded that there is no real number k that satisfies the given conditions. Therefore, the correct answer is "does not exist".Rate this question:
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- 45.
- A.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- B.
1
- C.
2
- D.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
Correct Answer
C. 2 -
- 46.
Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If Tn+1 − Tn = 10, then the value of n is
- A.
5
- B.
10
- C.
8
- D.
7
Correct Answer
A. 5Explanation
The question asks for the value of n when Tn+1 - Tn = 10, where Tn represents the number of all possible triangles formed by joining vertices of an n-sided regular polygon. The answer is 5 because if we substitute n = 5 into the equation, we get T6 - T5 = 10. This means that there are 10 more triangles formed when we add one more side to the polygon. Therefore, the value of n is 5.Rate this question:
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- 47.
The area (in square units) bounded by the curves y = , 2y − x + 3 = 0 , x-axis, and lying in the first quadrant is
- A.
36
- B.
18
- C.
![JEE Main 2013 Questions: Quiz! Exam - Quiz]()
- D.
9
Correct Answer
D. 9Explanation
The area bounded by the curves y = 2y - x + 3 = 0, x-axis, and lying in the first quadrant is 9 square units.Rate this question:
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- 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 . If the firm employs 25 more workers, then the new level of production of items is
- A.
3000
- B.
3500
- C.
4500
- D.
2500
Correct Answer
B. 3500Explanation
The question states that the rate of change of production (P) with respect to the additional number of workers (x) is given. This means that for every additional worker, there is a certain increase in production. If the firm employs 25 more workers, it can be inferred that the production will increase by the rate of change multiplied by the number of additional workers. Therefore, the new level of production will be 2000 + (rate of change * 25) = 3500.Rate this question:
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- 49.
Statement − I: Statement − II:
- A.
Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I
- B.
Statement -I is True; Statement -II is False
- C.
Statement -I is False; Statement -II is True
- D.
Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I
Correct Answer
C. Statement -I is False; Statement -II is True -
- 50.
If P = is the adjoint of a 3 × 3 matrix A and |A| = 4, then a is equal to
- A.
11
- B.
5
- C.
0
- D.
4
Correct Answer
A. 11 -
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