# Structures Of The Molecular Formula

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| By Catherine Halcomb
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Catherine Halcomb
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Quizzes Created: 1518 | Total Attempts: 5,459,314
Questions: 90 | Attempts: 717  Settings  • 1.

### The variation of concentration of products with respect to time for a first order reaction A→ B is given by which ofthe following ?

• A.

P

• B.

Q

• C.

R

• D.

S

B. Q
Explanation
The correct answer is Q. In a first order reaction, the rate of reaction is directly proportional to the concentration of the reactant. Therefore, as the concentration of reactant A decreases with time, the concentration of product B increases. This is represented by the variation of concentration of products with respect to time, which is given by answer Q.

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• 2.

### The concentration of Zn2+  and Cu2+  in the Daniel cell is increased ten times. The emf of the cell

• A.

Increases 10 times

• B.

Decreases by 10 times

• C.

Increases by 1/10 times

• D.

Remains constant

C. Increases by 1/10 times
Explanation
When the concentration of Zn2+ and Cu2+ in the Daniel cell is increased ten times, the emf of the cell increases by 1/10 times. This is because the emf of a Daniel cell is directly proportional to the logarithm of the ratio of the concentrations of the two ions. Increasing the concentration of the ions by ten times would result in an increase in the emf by a factor of 1/10, as the logarithm of 10 is 1. Therefore, the emf of the cell increases by 1/10 times.

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• 3.

### In an aqueous solution the ionisation constants for carbonic acid are K1 = 4.2 x 10-7 and K2 = 4.8 x 10-11  The correct statement for a saturated 0.034 M solution of carbonic acid is

• A.

The concentration of H+ is double than that of CO32-

• B.

The concentration of CO32- is 4.8 x 10-11 M

• C.

The concentration of CO32- is greater than that of HCO3-

• D.

The concentration of H+ and HCO3- are approximately equal.

B. The concentration of CO32- is 4.8 x 10-11 M
Explanation
The concentration of CO32- is 4.8 x 10-11 M because K2 is the ionization constant for the reaction CO32- + H2O ⇌ HCO3- + OH-, and the concentration of CO32- can be determined using the expression [CO32-] = √(K2/[HCO3-]).

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• 4.

• A.

217.5

• B.

390.7

• C.

417.5

• D.

582.7

A. 217.5
• 5.

### Which of the following complexes will show geometrical as well as optical isomerism ?

• A.

[Pt(NH3)2Cl2]

• B.

[Pt(NH3)Cl4]

• C.

L Pt(en)3 l4+

• D.

L Pt(en)2 cl2 l

C. L Pt(en)3 l4+
Explanation
The complex [Pt(en)3]4+ will show both geometrical and optical isomerism. This is because it contains a chelating ligand, ethylenediamine (en), which can form two five-membered rings with the platinum ion. The two rings can be arranged in different orientations in space, resulting in geometrical isomers. Additionally, the complex contains three identical en ligands, which can give rise to optical isomers due to their different spatial arrangements around the central platinum ion.

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• 6.

### Out of the possible structures of the molecular formula C5H11Br, how many are optically active ?

• A.

1

• B.

4

• C.

5

• D.

3

B. 4
Explanation
There are four possible structures of the molecular formula C5H11Br that are optically active. Optically active compounds are those that have chiral centers, which are carbon atoms bonded to four different groups. In C5H11Br, there are four carbon atoms, each of which can potentially have four different groups bonded to them. Therefore, all four carbon atoms can form chiral centers, resulting in four optically active structures.

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• 7.

### In the following benzyl/alkyl system        (R is alkyl group) the decreasing order of inductive effect is

• A.

(CH3)3C - > (CH3)2CH- > CH3CH2 -

• B.

(CH3)3CH2 - > (CH3)2CH - > (CH3)3C -

• C.

(CH3)2CH - > CH3CH2 - > (CH3)3C -

• D.

(CH3)2C - > CH3CH2- > (CH3)3CH -

B. (CH3)3CH2 - > (CH3)2CH - > (CH3)3C -
Explanation
The given answer indicates that the decreasing order of inductive effect in the benzyl/alkyl system is (CH3)3CH2- > (CH3)2CH- > (CH3)3C-. This means that (CH3)3CH2- has the strongest inductive effect, followed by (CH3)2CH-, and then (CH3)3C-. The inductive effect refers to the ability of an atom or group to withdraw or donate electron density through sigma bonds. In this case, the alkyl groups are donating electron density to the benzyl group, and the order of decreasing inductive effect suggests that (CH3)3CH2- is the best donor, followed by (CH3)2CH-, and then (CH3)3C-.

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• 8.

• A.

KC=

• B.

KC=

• C.

KC=

• D.

KC=

D. KC=
• 9.

### Bottles containing C6H5I and C6H5CH2I lost their original labels. They were labelled A and B for testing. A and B wereseparately taken in test tubes and boiled with NaOH solution. The end solution in each tube was made acidic withdilute HNO3 and then some AgNO3 solution was added. Substance B gave a yellow precipitate. Which one of thefollowing statemens is true for this experiment ?

• A.

A was C6H5CH2I

• B.

B was C6H5I

• C.

• D.

A was C6H5I

D. A was C6H5I
Explanation
The given question describes an experiment where two bottles containing unknown substances, labeled A and B, are tested. Both substances are boiled with NaOH solution and then made acidic with dilute HNO3. After adding AgNO3 solution, it is observed that substance B gives a yellow precipitate. From this observation, we can conclude that substance B contains chloride ions (Cl-), as the yellow precipitate formed is AgCl. Therefore, it can be inferred that substance B is C6H5I, which contains a phenyl group (C6H5) and an iodine atom (I). Hence, the correct answer is A was C6H5I.

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• 10.

### The separation of lanthanides by ion-exchange method is based on :

• A.

Size of the ions

• B.

Oxidation state of the ions

• C.

The solubilities of their nitrates

• D.

Basicity of hydroxides of Lanthanides

B. Oxidation state of the ions
Explanation
The separation of lanthanides by ion-exchange method is based on the oxidation state of the ions. This is because lanthanides have similar sizes and similar solubilities of their nitrates, making it difficult to separate them based on these factors. However, lanthanides can have different oxidation states, which can be exploited to separate them using ion-exchange methods. By manipulating the pH and the complexing agents used in the ion-exchange process, different lanthanides with different oxidation states can be selectively bound to the ion-exchange resin and then eluted to achieve separation.

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• 11.

### Which of the following will not give butyl acetate when treated with I-butanol ?

C.
Explanation
When I-butanol is treated with acetic acid in the presence of a catalyst such as sulfuric acid, it undergoes an esterification reaction to form butyl acetate. Therefore, the correct answer would be a compound that does not have an -OH group to react with acetic acid. One such compound is butylamine, which has an -NH2 group instead of an -OH group. Therefore, it will not give butyl acetate when treated with I-butanol.

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• 12.

### Rate of the reaction is fastest when Z is

• A.

Cl

• B.

NH2

• C.

OC2H5

• D.

OCOCH3

B. NH2
Explanation
The rate of a reaction is determined by the reactivity of the reactants. In this case, NH2 is the fastest reacting group because it contains a lone pair of electrons on the nitrogen atom, which can readily participate in chemical reactions. The lone pair can act as a nucleophile, attacking other molecules and facilitating the reaction. On the other hand, Cl, OC2H5, and OCOCH3 do not have a lone pair of electrons readily available for reaction, making them less reactive.

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• 13.

### Select the rate law that corresponds to the data shown for the following reaction : A + B → C

• A.

K I B I4

• B.

K [A][B]2

• C.

K[A]2 [B]2

• D.

Rate = R[B]3

C. K[A]2 [B]2
Explanation
The given rate law, K[A]2 [B]2, is the correct answer because it shows that the rate of the reaction is directly proportional to the square of the concentrations of both reactants, A and B. This suggests that the reaction is second order with respect to both A and B. The rate constant, K, represents the proportionality constant between the rate and the concentrations of the reactants.

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• 14.

### Consider the following species (1) RCH+CH3 (2) RCH2CH2+ (3) RCH2CH2+CHO In dehydration of 1 alcohols, the correct sequence of formation of species involved is

• A.

2, 1

• B.

1, 2

• C.

3, 2

• D.

2, 3

C. 3, 2
Explanation
In the dehydration of 1 alcohols, the correct sequence of formation of species involved is 3, 2. This means that the species RCH2CH2+CHO (3) is formed first, followed by the formation of RCH2CH2+ (2). This sequence makes sense because the dehydration of an alcohol typically involves the loss of a water molecule, which can result in the formation of an aldehyde (CHO) group. Therefore, the species with the aldehyde group (3) is formed first, and then it loses a proton to form the species without the aldehyde group (2).

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• 15.

### Amount of oxalic acid present in a solution can be determined by its titration with KMnO4 solution in the presence of H2SO4. The titration gives unsatisfactory result when carried out in the presence of HCl, because HCl

• A.

Gets oxidised by oxalic acid to chlorine

• B.

Furnishes H+ ions in addition to those from oxalic acid

• C.

Reduces permanganate to Mn2+

• D.

Oxidises oxalic acid to carbon dioxide and water

C. Reduces permanganate to Mn2+
Explanation
HCl reduces permanganate to Mn2+ during the titration with oxalic acid. This means that HCl interferes with the reaction between oxalic acid and permanganate, leading to inaccurate results. Therefore, the titration gives unsatisfactory results when carried out in the presence of HCl.

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• 16.

### The reaction is described as

• A.

SE2

• B.

SN1

• C.

SN0

• D.

SN2

B. SN1
Explanation
The reaction is described as SN1 because it involves a unimolecular nucleophilic substitution. In SN1 reactions, the rate-determining step is the formation of a carbocation intermediate, followed by the attack of a nucleophile. This reaction typically occurs with tertiary substrates, where the carbocation stability is high.

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• 17.

### A photon of hard gamma radiation knocks a proton out of   nucleus to form

• A.

The isobar of

• B.

The nuclide

• C.

The isobar of parent nucleus

• D.

The isotope of parent nucleus

B. The nuclide
Explanation
When a photon of hard gamma radiation knocks a proton out of the nucleus, it forms a nuclide. A nuclide refers to a specific atomic nucleus with a specific number of protons and neutrons. In this case, the photon has enough energy to dislodge a proton from the nucleus, resulting in a different nuclide. The other options, such as "the isobar of the parent nucleus" or "the isotope of the parent nucleus," do not accurately describe the outcome of the interaction between the photon and the nucleus.

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• 18.

### 25 ml. of a solution of barium hydroxide on titration with 0.1 molar solution of hydrochloric acid gave a titre value of 35 ml. The molarity of barium hydroxide solution was

• A.

0.07

• B.

0.14

• C.

0.28

• D.

0.35

A. 0.07
Explanation
The molarity of a solution can be calculated using the formula:

M1V1 = M2V2

Where M1 is the molarity of the first solution, V1 is the volume of the first solution, M2 is the molarity of the second solution, and V2 is the volume of the second solution.

In this case, the molarity of the hydrochloric acid solution is 0.1 M and the volume used is 35 ml. The volume of the barium hydroxide solution used is 25 ml. Plugging these values into the formula:

M1 * 25 ml = 0.1 M * 35 ml

M1 = (0.1 M * 35 ml) / 25 ml

M1 = 0.14 M

Therefore, the molarity of the barium hydroxide solution is 0.14 M.

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• 19.

### Which of the following is not a surfactant ?

• A.

CH2(CH2)15N+(CH3)Br-

• B.

CH3 - (CH2)14 - CH2NH2

• C.

CH3(CH2)16 - CH2 OSO2- Na+

• D.

OHC - (CH2)14 - CH2 - COO-Na+

B. CH3 - (CH2)14 - CH2NH2
Explanation
CH3 - (CH2)14 - CH2NH2 is not a surfactant because it does not have both hydrophilic and hydrophobic regions. Surfactants are compounds that have both a hydrophilic (water-loving) and a hydrophobic (water-repelling) part in their structure, which allows them to lower the surface tension between two immiscible substances like oil and water. CH3 - (CH2)14 - CH2NH2 is a simple amine compound and lacks the necessary hydrophilic or hydrophobic groups to function as a surfactant.

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• 20.

### The magnetic moment of a transition metal ion is 3.87 BM. The number of unpaired electrons present in it is

• A.

2

• B.

3

• C.

4

• D.

5

B. 3
Explanation
The magnetic moment of a transition metal ion is directly related to the number of unpaired electrons present in it. In this case, the magnetic moment is given as 3.87 BM. According to Hund's rule, electrons fill up orbitals singly before pairing up. Each unpaired electron contributes 1 BM to the magnetic moment. Therefore, if the magnetic moment is 3.87 BM, it means that there are 3 unpaired electrons present in the transition metal ion.

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• 21.

### When a certain amount of a non-volatile substance B is dissolved in a solvent A having a vapour pressure of 0.80 atm in the pure state, the vapour pressure is lowered to 0.60 atm. The mole fraction of B in the soluton is

• A.

0.75

• B.

0.48

• C.

0.25

• D.

0.2

C. 0.25
Explanation
When a non-volatile substance is dissolved in a solvent, it lowers the vapor pressure of the solvent. In this case, the vapor pressure of solvent A is lowered from 0.80 atm to 0.60 atm when substance B is dissolved in it. The mole fraction of substance B in the solution can be calculated using Raoult's law, which states that the vapor pressure of a solvent in a solution is equal to the mole fraction of the solvent multiplied by its vapor pressure in the pure state. Rearranging the equation, we can find the mole fraction of B by dividing the change in vapor pressure (0.80 atm - 0.60 atm = 0.20 atm) by the vapor pressure of the pure solvent (0.80 atm). This gives us a mole fraction of 0.25.

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• 22.

### A mole of Argon at 506.5 kPa expands from 10 dm3 against a constant opposing pressure of 253.25 kPa at a constant temperature till pressure of the gas equalizes the opposing pressure. What is the work done ?

• A.

-2533 J

• B.

2000 J

• C.

854 J

• D.

-533 J

A. -2533 J
Explanation
The work done by a gas during expansion can be calculated using the equation: work = -PΔV, where P is the pressure and ΔV is the change in volume. In this case, the gas expands from 10 dm3 to a volume where the pressure of the gas equals the opposing pressure. Since the pressure of the gas is initially 506.5 kPa and the opposing pressure is 253.25 kPa, the change in pressure is 506.5 kPa - 253.25 kPa = 253.25 kPa. The change in volume is the final volume minus the initial volume, which is equal to 0 dm3 - 10 dm3 = -10 dm3. Plugging these values into the equation, we get work = -(253.25 kPa)(-10 dm3) = -2533 J.

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• 23.

### The equilibrium constant for the reaction HONO + CN- = HCN + ONO- is 1.1 x 10-4 The magnitude of this equilibrium constant indicates that

• A.

CN- is weaker base than ONO-

• B.

HCN is weaker acid than HONO

• C.

NO- is the conjugate base of HONO

• D.

The conjugate acid of CN- is HCN.

A. CN- is weaker base than ONO-
Explanation
The magnitude of the equilibrium constant indicates the extent to which the reaction reaches equilibrium. A smaller magnitude indicates a lower concentration of the products compared to the reactants at equilibrium. Since the equilibrium constant for the reaction HONO + CN- = HCN + ONO- is very small (1.1 x 10-4), it suggests that the concentration of the products (HCN and ONO-) is much lower than the concentration of the reactants (HONO and CN-) at equilibrium. Therefore, CN- is a weaker base than ONO- because it is less likely to accept a proton and form the product HCN.

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• 24.

### Consider the following compounds 1. CH3CH2CH2Br  2. CH3CHBrCH3  3. (CH3)3CBr   The compounds are dehydrogenated by treatment with strong base under identical conditions. The correctsequence of reactivity of these compounds in the given reaction is

• A.

1 > 2 > 3

• B.

2 > 1 > 3

• C.

3 > 1 > 2

• D.

3 > 2 > 1

D. 3 > 2 > 1
Explanation
In the given question, the compounds are dehydrogenated by treatment with a strong base under identical conditions. Dehydrogenation is a process in which hydrogen atoms are removed from a compound. In this case, the strongest base is likely to abstract the hydrogen atom most easily.
Compound 3, (CH3)3CBr, has a tertiary carbon atom, which means it has the most substituted carbon atom among the three compounds. Tertiary carbons are more stable and less reactive compared to primary (compound 1) and secondary (compound 2) carbons. Therefore, compound 3 is the least reactive and will be dehydrogenated last.
Compound 2, CH3CHBrCH3, has a secondary carbon atom and will be dehydrogenated before compound 3.
Compound 1, CH3CH2CH2Br, has a primary carbon atom and will be the most reactive and dehydrogenated first.
Hence, the correct sequence of reactivity is 3 > 2 > 1.

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• 25.

### List the hydrogen halide acid in decreasing order of reactivity in the following reaction ROH + HX → RX + H2O

• A.

HBr > HI > HCl > HF

• B.

HI > HBr > HCl > HF

• C.

HI > HF > HBr > HCl

• D.

HI > HCl > HBr > HF

B. HI > HBr > HCl > HF
Explanation
In the given reaction, ROH (an alcohol) reacts with HX (a hydrogen halide acid) to form RX (an alkyl halide) and H2O (water). The reactivity of the hydrogen halide acids in this reaction is determined by their ability to donate a proton (H+). HI (hydrogen iodide) is the most reactive because iodine is the largest halogen and has the weakest bond with hydrogen, making it easier to donate the proton. HBr (hydrogen bromide) is the next most reactive due to bromine's larger size compared to chlorine and fluorine. HCl (hydrogen chloride) is less reactive than HBr because chlorine is smaller and has a stronger bond with hydrogen. HF (hydrogen fluoride) is the least reactive due to fluorine's small size and strong bond with hydrogen. Therefore, the correct order of decreasing reactivity is HI > HBr > HCl > HF.

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• 26.
• A.

0.2

• B.

0.4

• C.

0.8

• D.

1

B. 0.4
• 27.

### A 2° alcohol (A), having the molecular formula C5H12O, on oxidation gives a compound (B) which gives phenyl hydrazone derivative, but does not give negative Tollen’s test. The structure of the compound (A) is

• A.

CH3 COCH2 CH2 CH3

• B.

CH3 CHOHCH2 CH2 CH3

• C.

CH3 CH2 COCH2 CH3

• D.

CH3 CH2 CHOHCH2 CH3

A. CH3 COCH2 CH2 CH3
Explanation
The correct answer is CH3 COCH2 CH2 CH3. This is because the given compound is a 2° alcohol, which means that the hydroxyl group is attached to a carbon atom that is bonded to two other carbon atoms. The molecular formula C5H12O fits this description. On oxidation, this compound would form a ketone, which is represented by the structure CH3 COCH2 CH2 CH3. This compound does not give a negative Tollen's test, indicating that it does not have an aldehyde group. Therefore, the correct structure is CH3 COCH2 CH2 CH3.

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• 28.

### Consider the following amines   1) n-butyl amine   2) ethyl dimethyl amine   3) diethylamine               The correct sequence of boiling point is

• A.

1 > 3 > 2

• B.

1 > 2 > 3

• C.

2 > 3 > 1

• D.

2 > 1 > 3

A. 1 > 3 > 2
Explanation
The boiling point of amines is influenced by the strength of intermolecular forces, which in turn depends on the molecular size and shape. In this case, n-butyl amine (1) has the highest boiling point because it has the longest carbon chain, resulting in more surface area for intermolecular interactions. Diethylamine (3) has a lower boiling point than n-butyl amine because it has a shorter carbon chain. Ethyl dimethyl amine (2) has the lowest boiling point because it is a smaller molecule with no branching, resulting in weaker intermolecular forces. Therefore, the correct sequence of boiling points is 1 > 3 > 2.

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• 29.

### Polystyrene, dacron and orlon are classified respectively as

• A.

Chain growth, step growth, step growth

• B.

Chain growth, chain growth, step growth

• C.

Chain growth, step growth, chain growth

• D.

Step growth, step growth, chain growth

D. Step growth, step growth, chain growth
Explanation
Polystyrene is classified as a step growth polymer because it is formed through a stepwise reaction where two monomers combine and form a dimer, which then reacts with another monomer to form a trimer, and so on. Dacron is also classified as a step growth polymer because it is formed through the condensation reaction between ethylene glycol and terephthalic acid. Orlon, on the other hand, is classified as a chain growth polymer because it is formed through the addition polymerization of vinyl chloride monomers.

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• 30.

### The oxidation number of sulphur in S8, S2F2  and  H2S  respectively are

• A.

0, +1, -2

• B.

+2, +1 and -2

• C.

0, +1 and +2

• D.

-2, +1, -2

A. 0, +1, -2
Explanation
The oxidation number of an element is the charge it would have in a compound if all the bonding electrons were assigned to the more electronegative atom. In S8, sulphur exists as an element and does not gain or lose any electrons, so its oxidation number is 0. In S2F2, each sulphur atom gains one electron from fluorine, resulting in an oxidation number of +1. In H2S, hydrogen is assigned an oxidation number of +1, so the remaining charge must be balanced by sulphur, resulting in an oxidation number of -2. Therefore, the oxidation numbers of sulphur in S8, S2F2, and H2S are 0, +1, and -2 respectively.

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• 31.

• A.

2

• B.

1

• C.

√2

• D.

√2-1

D. √2-1
• 32.

### The equation of the common tangent to the curves y2 = 8x and xy = -1 is

• A.

3y = 9x + 2

• B.

Y = 2x + 1

• C.

2y = x + 8

• D.

Y = x + 2

C. 2y = x + 8
Explanation
The equation of the common tangent to the curves y^2 = 8x and xy = -1 is given by 2y = x + 8. This can be derived by finding the point of intersection between the two curves and then finding the slope of the tangent at that point. The slope of the tangent is equal to the product of the slopes of the curves at the point of intersection. By substituting the coordinates of the point of intersection into the equation of the tangent, we can find the equation of the common tangent.

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• 33.

### The condition that the equation      has real roots that are equal in magnitude but opposite in sign is

• A.

B2 = m2

• B.

B2 = 2m2

• C.

2b2 = m2

• D.

None of these

D. None of these
• 34.

### If the function f(x) = x2 + a/x  has a local minimum at x = 2, then the value of a is

• A.

8

• B.

16

• C.

18

• D.

None of these

D. None of these
Explanation
If the function has a local minimum at x = 2, then the derivative of the function at x = 2 must be equal to zero. Taking the derivative of f(x) with respect to x, we get f'(x) = 2x - a/x^2. Setting this equal to zero and solving for x, we get 2x - a/x^2 = 0. Multiplying through by x^2, we get 2x^3 - a = 0. Solving for x, we find that x = (a/2)^(1/3). Since x = 2, we can substitute this into the equation to find that (a/2)^(1/3) = 2. Solving for a, we get a = 16. Therefore, the value of a is 16.

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• 35.

• A.

0

• B.

2

• C.

4

• D.

Infinite

B. 2
• 36.

### Suppose a, b, c are in A.P. and a2, b2, c2 are in G.P. where a < b < c and a + b + c = 3/2, thenthe value of a is

D.
Explanation
Since a, b, c are in arithmetic progression (A.P.), the common difference between any two consecutive terms is the same. Let's assume the common difference is 'd'. Therefore, we can write b = a + d and c = a + 2d.

Given that a^2, b^2, c^2 are in geometric progression (G.P.), the common ratio between any two consecutive terms is the same. Let's assume the common ratio is 'r'. Therefore, we can write b^2 = a^2 * r and c^2 = b^2 * r.

Using the given equation a + b + c = 3/2, we can substitute the values of b and c to get a + (a + d) + (a + 2d) = 3/2. Simplifying this equation, we get 3a + 3d = 3/2.

Substituting the value of b^2 = a^2 * r, we can rewrite it as (a + d)^2 = a^2 * r. Expanding and simplifying this equation, we get a^2 + 2ad + d^2 = a^2 * r.

Substituting the value of c^2 = b^2 * r, we can rewrite it as (a + 2d)^2 = (a + d)^2 * r. Expanding and simplifying this equation, we get a^2 + 4ad + 4d^2 = (a^2 + 2ad + d^2) * r.

Now we have two equations:
1) 3a + 3d = 3/2
2) a^2 + 4ad + 4d^2 = (a^2 + 2ad + d^2) * r

By solving these equations simultaneously, we can find the value of 'a'.

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• 37.

### Equation of the normal at a point on the parabola y2 = 36x, whose ordinate is three times itsabscissa is

• A.

2x+3y+44 = 0

• B.

2x - 3y + 44 = 0

• C.

2x +3y -44 = 0

• D.

2x = 3y

C. 2x +3y -44 = 0
Explanation
The equation of the normal at a point on the parabola y^2 = 36x, whose ordinate is three times its abscissa, can be found using the formula: y = mx - 2am - am^3, where m is the slope of the tangent at that point and a is the distance from the focus to the directrix. In this case, since the ordinate is three times the abscissa, we can substitute y = 3x into the formula. Simplifying the equation, we get 2x + 3y - 44 = 0.

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• 38.

### Sum to n terms of the series 1 . 3 . 5 + 3 . 5 . 7 + 5 . 7 . 9 + ................. Is

• A.

8n2 + 12n2 - 2n - 3

• B.

N(8n3 + 11n2 - n - 3)

• C.

N(2n3 + 8n2 + 7n - 2)

• D.

None of these

C. N(2n3 + 8n2 + 7n - 2)
Explanation
The given series is a product of three consecutive odd numbers starting from 1. The general term of the series can be written as (2n - 1)(2n + 1)(2n + 3). By simplifying this expression, we get 8n^3 + 24n^2 + 18n - 2n^2 - 6n - 3. Combining like terms, we get 8n^3 + 22n^2 + 12n - 3. Factoring out n from this expression, we get n(8n^3 + 22n^2 + 12n - 3). Therefore, the correct answer is n(8n^3 + 22n^2 + 12n - 3).

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• 39.

### The solution of (1+y+x2y) dx + (x + x3) dy = 0 is

• A.

Y + tan-1 x = C

• B.

Xy + tan-1 x = C

• C.

Y2- tan-1 x = C

• D.

X2 + tan-1 y / x = C

B. Xy + tan-1 x = C
Explanation
The given equation is a first-order linear differential equation. To solve it, we can use the method of integrating factors. After finding the integrating factor, we multiply the entire equation by it to make it exact. Upon simplification and integration, we obtain the solution in the form of xy + tan-1 x = C.

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• 40.

### The number of words that can be formed by using the letters of the word MATHEMATICS that start as well as ends with T is

• A.

80720

• B.

90720

• C.

20860

• D.

37528

B. 90720
Explanation
The word "MATHEMATICS" has 11 letters. To form a word that starts and ends with "T", we need to fix "T" at the first and last position. The remaining 9 letters can be arranged in 9! ways. However, the letter "M" appears twice, so we need to divide by 2!. Similarly, the letter "A" appears twice, so we need to divide by 2! again. Therefore, the total number of words that can be formed is 9! / (2! * 2!) = 90720.

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• 41.

### The solution of the differential equation (x+y) dy - (x - y) dx = 0 is

• A.

Y2 + 2xy + x2 = k

• B.

Y2 + 2xy + x2 = k

• C.

Y2 +2xy + x2 = 0

• D.

Y2 - 2xy + x2 = k

C. Y2 +2xy + x2 = 0
Explanation
The given differential equation is a homogeneous equation, which means it can be written in the form of M(x,y)dx + N(x,y)dy = 0, where M(x,y) = (x - y) and N(x,y) = (x + y). To solve this type of equation, we can make a substitution y = vx, where v is a function of x. Substituting this into the equation and simplifying, we get (1 + v)dv = -dx/x. Integrating both sides, we obtain v + ln|x| = C, where C is the constant of integration. Substituting y = vx back in, we have y + ln|x| = C, which can be rearranged as y^2 + 2xy + x^2 = k, where k is another constant. Therefore, the correct answer is y^2 + 2xy + x^2 = k.

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• 42.

### If in the expansion of   ,  n ϵ N , sum of the coefficients of  x5 is 0,  then value of n is

• A.

5

• B.

10

• C.

15

• D.

None of these

C. 15
Explanation
In the expansion of (1+x)^n, the coefficient of x^k represents the number of ways to choose k items from a set of n items. If the sum of the coefficients of x^5 is 0, it means that there are no ways to choose 5 items from a set of n items. This can only happen if n is less than 5, so the value of n cannot be 5 or 10. Therefore, the only possible value for n is 15.

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• 43.

### Let A and B be two 2 X 2 matrices. Consider the statements :  i) AB= O A = O  or  B = O  ii)  AB =I2   A = B-1  iii) (A + B)2 = A2 + 2AB + B2        In this case

• A.

(i) is false but (ii) and (iii) are true

• B.

(i) and (iii) are false but (ii) is true

• C.

(i) and (ii) are false but (iii) is true

• D.

(ii) and (iii) are false but (i) is true

B. (i) and (iii) are false but (ii) is true
Explanation
The statement (i) states that if AB = O (zero matrix), then either A = O or B = O. This statement is false because it is possible for AB to equal O without either A or B being O.

The statement (ii) states that if AB = I2 (identity matrix), then A = B-1 (inverse of B). This statement is true because for AB to equal the identity matrix, A must be the inverse of B.

The statement (iii) states that (A + B)2 = A2 + 2AB + B2. This statement is false because in general, (A + B)2 is not equal to A2 + 2AB + B2.

Therefore, the correct answer is that (i) and (iii) are false but (ii) is true.

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• 44.

### If A= , then A100 is equal to

• A.
• B.
• C.
• D.

None of these

A.
• 45.

• A.

{1, 2}

• B.

{2, 3}

• C.

{1, p, 2}

• D.

{1, 2, -p}

A. {1, 2}
• 46.

### If  I = , then I equals

• A.
• B.
• C.

0

• D.

None of these

C. 0
Explanation
If the equation is "I = ", it means that the value of I is not given or not specified. In this case, if I equals none of these options, the only remaining possibility is that I equals 0.

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• 47.

B.
• 48.

### If  , then the numerical value of K is equal to

• A.

1/2

• B.

1/4

• C.

1/8

• D.

1/18

A. 1/2
Explanation
The given question is incomplete as it does not provide any condition or equation to determine the value of K. Therefore, an explanation for the correct answer cannot be generated.

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• 49.

### The greatest value of f(x) = 2 sin x + sin2x in    is

• A.

9/2

• B.

5/2

• C.
• D.

3/2

C.
Explanation
The greatest value of the function f(x) = 2 sin x + sin^2x is 5/2. To find the maximum value, we can take the derivative of f(x) and set it equal to zero. The derivative of f(x) is 2 cos x + 2 sin x cos x. Setting this equal to zero, we get cos x + sin x cos x = 0. Simplifying further, we have cos x (1 + sin x) = 0. This equation is satisfied when cos x = 0 or sin x = -1. The maximum value of f(x) occurs when sin x = -1, which happens at x = 3π/2. Plugging this value into f(x), we get f(3π/2) = 2(-1) + (-1)^2 = 5/2.

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• 50.

### The number of solutions of the pair of equations 2sin2Ө - cos 2Ө=0  and  2cos2Ө - 2sinӨ=0 in the interval  is

• A.

0

• B.

1

• C.

2

• D.

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