IIT JAM Biology 2011
Previous Year Question Paper with Solution.

1. The Iac repressor binds its operator with a Kd = 10–10 M. In a strain of E. coli, the intracellular repressor concentration is 10–12 M. Induction of Iac operon in this strain

(a) does not require lactose

(b) requires lactose

(c) requires glucose

(d) requires both glucose and lactose

Ans. (b)

Sol. The given case is an example of inducible lac operon. The various components of operon are :

(i) Regulatory gene or i — for repressor of lac operon

(ii) Structural genes z — for beta-galactosidase (-gal) that catalyses the hydrolysis of lactose into galactose and glucose.

y — for permease, inceases the permeability of the cell

a — for transcetylase.

(iii) Promoter gene (P) sequence of DNA where, RNA polymerase binds and initiates transcription.

(iv) Operator sequence of DNA adjacent to promoter, here specific repressor protein binds.

(v) Inducer prevents the repressor from binding to the operator

In case of inducible lac operon, one of the regulatory element is the CRP or cyclic — AMP Receptor Protein binding site.

When E. coli doesn't have glucose, it makes cyclic AMP (cAMP) as in intracellular signal that it requries sugar. One of the proteins that receives this signal is the cAMP Receptor Protein (CRP). The CRP site is right next to the promoter site. When the cAMP—CRP complex is formed, it helps push RNA polymerase can't transcribe genus until it the closed form. This is the positive (inducible) control, because the regulatory protein, CRP, stimulates transcription.

2. The DNA of an organism is 0.68 meters. What is the length of the DNA in kilobases?

(a) 2 × 108

(b) 2 × 106

(c) 2 × 105

(d) 2 × 103

Ans. (b)

Sol. The length of one base pair of DNA = 0.34 nm

The 0.68 m is the length of DNa with x base pair

x =

x =

x = 2 × 109 base pairs

x = 2 × 106 kilobase pairs

3. A segment of DNA (underlined) in a gene sequence 5'-GGAGTATCATTGCA-3'


undergoes an inversion. The new gene sequence is









Ans. (a)

Sol. An inversion is a chromosome rearrangement in which a segment of chromosome is reversed end to end. Thus, inversion of the DNA sequence, i.e.



would result in



4. The product of an enzyme reaction is estimated in a colorimeter. For a 1cm path length cuvette, 20% of the incident light intensity is absorbed. If the path length is increased to 2cm, the percentage of absorption is

(a) 36

(b) 40

(c) 64

(d) 90

Ans. (c)

Sol. Absorbance (A) =

l0 = Incident light intensity

lt = Transmitted light intensity

Case I A1 =

A1 = 0.0969

Since, 20% of the incident light intensity aborbed. Hence, 80% of light intensity will be transmitted.

Again A1 = × C × l1

C = Concentration of solution

= Molar absorption coefficient

l1 = Path length

A1 = × C × 1

A1 = C

Case II When path length is increased to 2 cm

A2 = × C × 2

A2 = 2C ∈

= = 2

A2 = 2A1

= 2(0.0969) = 0.1938

Now, percentage of absorption


= 66.66 64%

5. Neisseria meningitides is a

(a) Gram + ve bacillus

(b) Gram +ve coccus

(c) Gram –ve bacillus

(d) Gram –ve coccus

Ans. (d)

Sol. Neisseria meningitidis is often referred to as meningococcus is a gram negative bacterium that can cause meningitis and other forms of meningococcal disease such as meningococemia, a life threatening species.

6. An enzyme solution is sterilized by

(a) use of an autoclave

(b) ethylene oxide

(c) memebrane filtration

(d) -radiation-radiation

Ans. (c)

Sol. An enzymes solution is sterilised by microfiltration using membrane filters. The procedure is well suited for beverages and phamaceuticals industries. As sterilisation by heat denatues of destroys the enzyme. Microfiltration are employed in these industries as a method to remove bacteria and other undrived suspensions from liquids, a procedure termed as cold sterilisation, which negate the use of heat.

7. RNA interference is induced by

(a) double strand DNA

(b) double strand RNA

(c) both (A) and (B)

(d) specific proteins

Ans. (b)

Sol. RNA interference (RNA), is a biological process in which RNA molecules inhibit gene expression, typically by causing the destruction of specific mRNA molecule. It is basically an envoutionary conserved mechanism of gene regulation that is induced by silencening small dsRNA in a sequence specific manner.

8. The detection and estimation of DNA products by Real Time PCR is achieved with

(a) SYBR green

(b) Ethidium bromide

(c) Acridine orange

(d) Green fluorescent protein

Ans. (a)

Sol. SYBR green is an asymmetrical cyanine dye used as a nucleic acid stain in molecular biology. SYBR green binds to the minor groove of the dsDNA only. In the solution, the unbound dye exhibits very little fluorescence. This fluorescence is substantially enhanced, when the dye is bound to dsDNA. SYBR green has, however, a limitation that includes preferential binding to G=C rich sequences.

9. A peritrichous arrangement of flagella in bacilli is a

(a) single falgellum at one pole

(b) single flagellum at each pole

(c) cluster of flagella at one place

(d) uniform distribution of flagella around the cell

Ans. (d)

Sol. Bacterial flagella are long, thin (about 20 nm), whip like appendages that move the bacteria towards nutrients and other attractants. The number and location of flagella are distinctivew for each genus. There are four types of flagellar arrangement.

(i) Monotrichous : Single polar flagellum (e.g. Vibrio cholerae)

(ii) Amphitrichous : Single flagellum at both ends (e.g. Alcaligenes faecalis)

(iii) Lophotrichous : Tuft of flagella at one or both ends (e.g. Spirilla)

(iv) Peritrichous : Flagella surrounding the cell (e.g. Typhoid bacilli)

10. Fibroblast cells in culture survive limited number of passages due to

(a) exhaustion of nutrients

(b) contact inhibition

(c) shorting of telomeres

(d) loss of cell adherence

Ans. (c)

Sol. Fibroblast cells in culture survive limited number of passages due to shortening of telomeres. Passaging or subculture or splitting of cell involves transferring a small number of cells into a new vessel.

Shortening of telomeres causes the fibroblast cells in culture to survive limited number of passages. As during the process of DNA replication, small segments of DNA at each end of the DNA strand (telomeres) are unable to be copied and are lost after each time DNA is duplicated. The telomere region of DNA does not code for any protein, it is simply a repeated code on the end region of DNA that is lost. After many divisions the telomeres become depleted and the cell begins apoptosis.

11. Fetal bovine serum is a component of cell culture medium primarily to provide

(a) vitamins

(b) growth factors

(c) amino acids

(d) trace elements

Ans. (b)

Sol. Fatal bovine serum is the blood fraction remaining after the natural coagulation of blood, followed by centrifugation to remove any remaining blood cells. it is the most widely used serum-supplement for the in vitro cell culture of eukaryotic cells. This is due to it having a very low level of antibodies and containing more growth factors, allowing for versatilty in many different cell culture applications.

12. Identify the correct combination between the groups

Group-I Group-II

P. Mast cells 1. Cytotoxic effect on tumour cells

Q. Macrophage 2. Release of histamine

R. Natural Killer cells 3. Production of immunoglobulins

S. B cells 4. Ingestion of particulate antigen


(a) P–2, Q–4, R–1, S–3

(b) P–4, Q–1, R–3, S–2

(c) P–1, Q–3, R–2, S–4

(d) P–4, Q–3, R–2, S–1

Ans. (a)

Sol. The correct match for the following are

(i) Mast cells are derived from the myeloid stem cell. It is a part of the immune system and contains many granules rich in histamin and heparin.

(ii) Macrophage are large specialised cells of immune system that recognise, engulf and destory cells.

(iii) Natural killer cells are a type of cytotoxix lymphocyte critical to the innate immune system.

(iv) B-cells functions to make antibodies in response to antigens, to perform the role of antigen-presenting cells (APCs)

13. Mad cow disease is caused by a

(a) bacterium

(b) virus

(c) viroid

(d) prion

Ans. (d)

Sol. The infectious agent that causes mad cow disease in an abnormal version of a protein normally found on cell surfaces called as prion. Mad cow disease is also known as bovine spongiform encephalopathy.

14. The phenomenon in which one gene inhibits the expression of another gene is called

(a) dominance

(b) epistasis

(c) penetrance

(d) expressivity

Ans. (b)

Sol. Epistasis is a phenomenon occuring when one gene locus masks or modifies the phenotypes of a second gene locus. It is a type of gene interaction that are not alleles, in particular the suppression of the effect of one such gene by another.

15. The major protease involved in apoptosis has at its active site

(a) serine

(b) aspartate

(c) cysteine

(d) histidine

Ans. (c)

Sol. Caspases have proteolytic activity. These are cys dependent asparatate specific protease. They contain a key cysteine residue in the catalytic site and selectively cleavage proteins at sites just C-terminal to asparatate residues.

16. The techniques to show that a protein is a homotetramer with a subunit moleculear weight of 25kDa site

(a) gel filteration and native PAGE

(b) affinity and ion exchange hromatography

(c) SDS-PAGE and gel filteration

(d) isoelectric focusing and SDS-PAGE

Ans. (c)

Sol. The techniques to show that a protein in a homotetramer with a subunit molecular weight of 25 kDa are SDS-PAGE and isoelectric focusing these both techniques combinly used in 2-D electrophoresis. The advantages of adding SDS-PAGE is, it allows protein to be separate on the basis of approximate mass and isoelectric focusing separate the protein on basis on charge.

17. A phage infects bacteria at a multipicity of infection (moi) of 0.1. This means that

(a) every bacterium is infected by the phage

(b) one out of 10 bacteria is infected by the phage

(c) ten phage infect one bacterium

(d) only 1/10 of the phage population is infectious

Ans. (b)

Sol. When phage and their bacterial host are mixed with each other then bacteria and phage collide randomly, leading to phage infection. The percentage of the bacterial cells that will be infected depends upon the concentration of the phage and bacteria. Higher concentration of the phage and bacteria leads to more infection.

The efficiency of infection is affected not only by the concentration of the phage and bacteria, the multiplicity of infection.

18. Which one of the following fatty acids has the highest melting temperature?

(a) Myristate (14 : 0)

(b) Stearate (18 : 0)

(c) Oleate (18 : 1)

(d) Linoleate (18 : 2)

Ans. (b)

Sol. Melting point of fatty acids increases with molecular weight. Unsatruated fatty acids have lower melting points than satrauted fatty acids of the same length. Stearic acid has the highest melting point of 69.6°C.

19. Which one of the following lipids will not form a biological membrane?

(a) phosphatidyl ethanolamine

(b) cerebroside

(c) triacylglycerol

(d) sphingomyelin

Ans. (c)

Sol. Triacylglycerol is an ester derived from glycerol and three fatty acid. It is a lipid found in blood. Other three phosphatidyl ethanolamine, cerebroside and sphingomyelin are biological membrane lipids.

20. Transketolases and tarnsladolases are involved in

(a) oxidative phase of pentose phosphate

(b) non-oxidative phase of pentose phosphate pathway

(c) Embden-Meyerhof pathway

(d) glyoxylate pathway

Ans. (b)

Sol. Transketolase and transadolase are enzymes of non-oxidative phase of pentose phosphate pathway which catalyse transfer of 2-carbon and 3-carbon molecular fragments respectively in each case from a ketose donor to an aldose acceptor.

(i) Transketolase transfer 2-C fragments from xylulose-5-phosphate to either ribose-5-phosphate or erythrose-4-phosphate.

(ii) Transaldolase catalyses transfer of a 3-C dihydroxyacetone moiety from sedoheptulose-7-phosphate to glyceraldehyde-3-phosphate

21. In which of the following conversion is ATP synthesized by substrate level phosphorylation?

(a) isocitrate to -ketoglutarate

(b) -ketoglutarate to succinyl CoA

(c) succinyl CoA to succinate

(d) succinate to fumarate

Ans. (c)


22. Which one of the following in not the function of photosystem II?

(a) ATP synthesis

(b) light collection

(c) NADPH synthesis

(d) charge collection

Ans. (c)

Sol. During phostosynthesis it appeared from non-cyclic photophosphorylation that photosystem I was responsible for NADPH production while in cyclic photophosphorylation. It is important for ATP production.

23. Which one is the incorrect statement?

(a) cellulose has , 4 linkages

(b) amylose has , 6 linkages

(c) glycogen has , 4 and , 6 linkages

(d) chitin has , 4 linkages

Ans. (b)

Sol. Amylose is a helical polymer mode of α-glucose units, bound to each other through α (1 → 4) glycosidic bonds.

24. O-glycosidic bonds are present in

(a) only polysaccharides but not glycoproteins

(b) both glycoproteins and polysaccharides

(c) DNA between base and sugar

(d) RNA between base and sugar

Ans. (b)

Sol. O-glycosidic bonds are present in glycoproteins, polysaccharides or lipids in living organisms. In these bonds the glycosidic oxygen links the glycoside to the aglycone or reducting end sugar.

25. Under physiological conditions when [S] « KM, the catalytic efficiency is estimated by

(a) kcat

(b) KM

(c) kcat/kM

(d) Vmax

Ans. (c)

Sol. Under physiological condition when [S] << KM the catalytic efficiency is estimated by kcat/KM. This measure of efficiency is helpful in determining whether the rate is limited by the creation of product or the amount of substrate in the environment.

26. Diacylglycerol is known to activate

(a) phospholipase A

(b) phospholipase C

(c) protein kinase A

(d) protein kinase C

Ans. (d)

Sol. Diacylgcerol is known to activate protein kinase C. In biochemical signaling, diacylglycerol functions is a second messenger signaling lipid, and is a product of the hydrolysis of the phosphoolipid phosphattidylinositol 4,5-biphosphate (PIP2) by the enzyme phospholipase C(PLC) (a membrane-found enzyme) that, through the same reaction, produces inositol trisphosphate (IP3). Although inositol trisphosphate diffuses into the cytosol, diacylglycerol remains within the plasma membrane, due to its hydrophobic properties. IP3 stimulates the release of calcium ions from the smooth endoplasmic reticulum, whereas DAG is a physiological activator of protein kinase C (PKC). The production of DAG in the membrane facilitates translocation of PKC from the cytosol to the plasma membrane.

27. The reagent required to cleave the carboxyl side of methionine is

(a) cyanogen bromide

(b) tryspin

(c) chymotryspin

(d) performic acid

Ans. (a)

Sol. Cynogen bromide hydrolyses peptide bonds at the C-terminus of methionine residues. This reaction is used to reduce the size of polypeptide segments for identification and sequencing.

28. Hormones that act on cells near the point of their synthesis and not transmported the rough blood circulation are

(a) prostaglandins and thromboxane

(b) estradiol and cortisol

(c) prednisolone and prednisone

(d) thyroxine and glucagon

Ans. (a)

Sol. Prostaglandins have two derivatives that are prostacyclins and thromboxanes. These are powerful locally acting vasodilators and inhibit the aggregation of blood platelets. They are snthesised in the wallls of blood vessels and serve the physiological function of preventing needless clot formation.

29. Match the coenzymes in Group I with the corresponding units of Group II

Group I Group II

P. Tetrahydofolate 1. Acyl

Q. Biotin 2. Electrons

R. FMNH2 3. One carbon unit

S. Coenzyme A 4. CO2

Codes :

(a) P–3, Q–4, R–2, S–1

(b) P–1, Q–4, R–3, S–2

(c) P–2, Q–3, R–4, S–1

(d) P–4, Q–2, R–1, S–3

Ans. (a)

Sol. The correct match for the following are

(i) Tetrahydrofolate is a cofactor in many reactions, especially in the metabolism of amino acid and nucleic acid. It acts as a donor of a group with one carbon unit.

(ii) Biotin is also known as vitamin H assists in various metabolic reactions involving the transfer of carbondioxide.

(iii) FMNH2 (Flavin mononueletide) is an strong oxidising agent than NAD and is particularly useful because it can take part in both one and two electron transfer.

(iv) A molecule of coenzyme A carrying an acetyl group is also referred to as acetyl-CoA.

30. Bacteria, auxotrophic for fatty acids were grown at 40°C in the presence of a mixture of fatty acids. Which of the fatty acid combinations will be incorporated into the membrane?

(a) saturated and long chain fatty acids

(b) saturated and short chain fatty acids

(c) unsaturated and short chain fatty acids

(d) unsaturated and long chain fatty acids

Ans. (a)

Sol. Bacterium auxotrophic for fatty acids when grown at 40°C in the presence of mixture of fatty acid, they will incorporate only saturated and long chain fatty acids into the membrane. As above 37°C unsaturated fatty with long chains survive above 40°C because they have high melting points.

31. Which one of the following statements is incorrect?

(a) Mer B is a bacterial protein similar to eukaryotic actin

(b) FtsZ protein found in most of the bacteria is a tubulin hormologues

(c) AmiC hydrolyzes peptidoglycan to separate daughter cells

(d) AmiC hydrolyzes peptidoglycan to separate daughter cells

Ans. (d)

Sol. Fts Z is protein encoded by fts Z gene that assembles into a ring at the future site of the septum of bacterial cell divison. Fts Z has the ability to bind to GTP and also exhibits a GTPase domain that allows it to hydrolyse GTP to GDP and a phosphate group.

In vivo Fts Z, forms filaments with a repeating arrangements of subunits all arranged head to tail. These filaments form a ring around the lingitudinal midpoint or septum of the cell. This ing is called the Z-ring.

32. Massive doses of methylend blud are sometimes given for cyanide poisoning. Which one of the following statements is incorrect?

(a) reduction potential of methylene blue is similar to oxygen

(b) cyanide blocks transfer of electrons from cytochrome oxidase to oxygen

(c) in cyanide poisoning, all the respiratory chain components become reduced and electron transport stops

(d) methylene blue can reduce the various components of the respiratory chain to restore ATP synthesis.

Ans. (b)

Sol. methylene blue is given during cyanide poisoning as it can reduce the various components of the respiratory chain to restore ATP synthesis, since its reduction potential is similar to that of oxygen and can be reduced by the components of the electron transport chain.

33. A disease manifests only in the homozygous recessive condition. A couple has two children. If both parents are heterozygous for the disease causing gene, what is the probability that both the children are normal? Assume that the disease causing gene is not sex linked.

(a) 1/16

(b) 3/16

(c) 9/16

(d) 12/16

Ans. (c)

Sol. Since, the parents are homozygous for the diseased condition for disease causing gene,

Probability that a child is normal


Similarly, probability that second child is normal =

Since, both the chances are independent

34. Match the entries in Group-I with those in Group-II.

Group-I Group-II

P. Glucagon 1. Eicosanoid

Q. Prednisolone 2. Peptide

R. Prostaglandin E1 3. Catecholamine

S. Epinephrine 4. Steroid

Codes :

(a) P–3, Q–2, R–4, S–1

(b) P–2, Q–4, R–1, S–3

(c) P–2, Q–3, R–4, S–1

(d) P–4, Q–1, R–3, S–2

Ans. (b)

Sol. The correct match are as follows :

(i) Glucagon is peptide hormone produced by alpha cells of pancreas

(ii) Prednisolone is a synthetic glucocorticoid, a derivative of cortisol and is used to treat variety of inflammatory and auto immune conditions.

(iii) Prostaglandin E1 are subclass of eicosanoids and form the prostanoid class of fatty acid derivatives.

(iv) Epinephrine is also known as adrenlaline. The adrenal medulla is a minor contributor to total circulating catecholamines, through it contributes over 90% of circulating epinephrine.

35. An amino acid has a non-ionizable R group. The pKa for the NH2 group is 9.4 and for the COOH group is 2.8. Consider the following statements :

P. At a pH of 6.1, 50% of the amino acid molecules which migrate towards the cathode when placed between two electrodes

Q. At a pH of 2.8, 50% of the amino acid molecules in solution are of the form

R. The pl of the amino acid is 6.1

S. On titration of the amino acid solution with NaOH, the amino group is deprotonated before the carboxylic group

Which pair of the above statement is correct?

(a) Q, R

(b) P, S

(c) P, R

(d) Q, S

Ans. (c)

Sol. pKa for the —NH2 group = 9.4

Ka = Antilog(–9.4)

Ka = 3.98 × 10–10

[H+] =

= 1.99 × 10–5

pHNH2 = –loog [H+]

= + 4.7 ...(i)

PKa for the — COOH group = 2.8

Ka = Antilog (–2.8)

= 1.548 × 10–5

[H+] = = 0.0398

pHCOOH = –log[H+]

= + 1.4 ...(ii)

Thus, from Eqs. (i) and (ii)

pH of amino acid is 4.7 + 1.4 = 6.1

At pH of 6.1, 50% of amino acid molecules will migrates towards cathode as pH is slighly acidic. In acidic medium protonation will takes palace (show below)

The isoelectronic point of this amino acid is

Pl =

= PI = 6.1

For its pH value

Ka = Antilog (–6.1)

= 7.94 × 10–7

H+ = = 8.9 × 10–4 = 3.05

Thus, statement Q is incorrect.

36. A wild type (W) strain of a bacterium is red due to the conversion of a colourless precursor X to R. Two colourless mutants M1 and M2 (with mutations in Gene 1 and Gene 2, respectively) were obtained. The following chart shows the observed colour of the bacterium when grown in a medium supplemented with either X, Y or Z.

Which one of the following describes the correct pathway for red pigment synthesis?





Ans. (c)

Sol. In the given condition, wild type strain of a bacterium in the presence of 'X' supplement undergoes mutation (M1) in gene 1, and the colour changes into white. Further, when it supplemented with 'Z', the colour remains same but mutation (M2) occurs in gene 2 and again when supplemented with 'Y' the colour changes into red.

37. Which one of the following is not an autoimmune disorder?

(a) AIDS

(b) Systemic Lupus erythematosus

(c) Rheumatoid Arthritis

(d) Myestenia Gravis

Ans. (a )

Sol. AIDS is not a autoimmune disease. An autoimmune disease develops when your immune system, which defends your body against disease, decides your healthy cells are foreign but in AIDS, i.e., acquired immuno deficiency syndorme the person's immune system becomes too weak to fight off infections.

38. Which one of the following is not common to both chloroplast and mitochondria?

(a) electron transport

(b) electron donor-acceptor pair

(c) proton pump

(d) independent genomes

Ans. (b)

Sol. Electron donor-acceptor pair is nor common to both mitochondria and chloroplast. Mitochondria performs cellular respiration and chloroplast is the site for photosynthesis. They both share similarities like both are cell organelles, have double membrane, have independent genomes in them, have electron transport chains in their inner membrane, have proton pump, etc.

39. Match the entries in Group I with those in Group II.

Group I Group II

P. LH 1. Corpus Luteum for production of estrogen and progesterone

Q. hCG 2. Leydig cells for production of testosterone

R. GnRH 3. Sertoli cells for maintencance of sprmatogenesis

S. Testosterone 4. Pituitary gonadotrophs for production of LH and FSH

Codes :

(a) P–3, Q–2, R–1, S–4

(b) P–2, Q–4, R–1, S–3

(c) P–2, Q–1, R–4, S–3

(d) P–3, Q–2, R–4, S–1

Ans. (c)

Sol. The correct match are as follows :

(i) The leyding cells in males produce testosterone under the control of Luteinising hormones (LH).

(ii) Human chorionic gonadotropin (hCG) hormone stimulates the corpus luteum to produce estrogen and progesterone in the first 10 weeks after conception, until the placental cells can do so themselves.

(iii) GnRH is gonadotropin releasing hormone is synthesised and released by hypothalamus and acts upon the anterior pituitory to release follicle stimulating hormone and luteinising hormone.

(iv) Testosterone is necessary for normal sperm development. It activates genes in sertoli cells, which promote differentiation of spermatogonia.

40. A competitive reversible enzyme inhibitor

(a) increase KM, decrease Vmax

(b) increases Vmax, decreases KM

(c) increases KM, does not change Vmax

(d) increase Vmax, does not change KM

Ans. (c)

Sol. In competitive inhibition, Vmax stays same and KM increases, but the inhibitor does not affect the turnover number of the enzyme.

41. A 7kb ciruclar plamid was completely digested with either EcoRI or BamHI or both. The digestion pattern is shown below.

Which one of the following is the correct restriction map of the plasmid?





Ans. (d)

Sol. The correct restriction map of the plasmid is

As Eco RI has one restriction fragment of 7 kb alone and Bam HI produces 2 fragments of 5 kb + 2 kb = 7 kb alone of the 7 kb circular plasmid.

42. Consider the following statements

Statement 1 : A sudden 100 meter spring by an athlete will increase the pH of blood.

Statement 2 : Fate of pyruvate under anaerobic condition in skeletal muscle is lactate.

Which one of the following statement is true?

(a) Statement 1 and 2 are correct and 1 is the outcome of 2

(b) Statement 2 is incorrect and statement 1 is correct

(c) Statement 1 and 2 are correct and 1 is not the outcome of 2

(d) Statement 2 is correct and statement 1 is incorrect

Ans. (d)

Sol. Statement II is correct and statement I is correct. As lactate production increases during physical exercise and a result leads to increase production of hydrogen ions. Hydrogen ions accumulation decreases the pH of blood.

43. Mitochondria were isolated and permeabilized with a detergent. They were incubated with succinate, ADP and Pi in the presence of O2. ATP synthesis and O2 consumption were measured. Which one of the following statements is true?

(a) ATP synthesis will occur but O2 consumption will not occur

(b) O2 consumption will occur but ATP synthesis will not occur

(c) both O2 consumption and ATP synthesis will occur

(d) both O2 synthesis and ATP synthesis will not occur

Ans. (c)

Sol. If mitochondria are incubated in an oxygraph apparatus (oxygen containing) often in an isotonic medium containing substrate (succinate) and inorganic phosphate (Pi), then addition of ADP causes a sudden burst of oxygen (O2) uptake as the ADP is converted in ATP.

44. The frequency of occurrence of a disease is one in a million individuals. The disease results from the homozygosity in a recessive allele. The population satisfies al the assumptions of the Hardy-Weinberg equilibrium. The frequency of the dominant allele and frequency of carriers of the disease respectively are

(a) 0.9 and 0.19

(b) 0.09 and 0.019

(c) 0.999 and 0.0019

(d) 0.009 and 0.00019

Ans. (c)

Sol. According to the question to the question the homozyogous recessive percentage which is equal to q2 is 1/1,000,000 = 0.000001

So, q = 0.001 = frequency of allele A

and p + q = 1 thus, p = 0.999

Therefore, the frequency of allele A = p = 0.999

and carriers are heterozygous and are equal to 2 pq

So, 2pq = 2(0.999) (0.001)

= 0.001998

Thus, the answer is 0.999 and 0.0019 as the frequency of the dominant allele and frequency of carriers of the disease respectively.

45. Identify the major product of the following reaction





Ans. (a)



AlCl3 is a Lewis acid. It will attack on the lone pair of oxygen.

Step I

The activated epoxide will react with benzene and opened up from highly hindered site.

46. Which one of the following compounds is optically active?





Ans. (a)


This compound is optically active. In this molecule, symmetry is not present. We get non-superimosable mirror images.

47. Which one of the following compounds shows a fragmentation peak corresponding to an enolic species (m/z = 58) in the mass spectrum?





Ans. (c)

Sol. The ketone readily undergoes Mc Lafferty rearrangement

The pentanone contain -H and hence, undergoes Mc Lafferty rearrangement.

48. Correlate the reactions in Group I with the reagent (s) and conditions (s) in Group II :

(a) P–4, Q–3, R–2

(b) P–3, Q–1, R–4

(c) P–4, Q–3, R–1

(d) P–3, Q–2, R–1

Ans. (d)


49. Synthetic Rubber Buna-N is a copolymer of

(a) 1, 3-butadiene and acrylonitrile

(b) 2-chloro-1, 3-butadiene and acetonitrile

(c) 2-chloro-1, 3-butadiene and acrylonitrile

(d) 1, 3-butadiend and acetonitrile

Ans. (a)

Sol. Synthetic rubber Buna-N is a compolymer of 1, 3-butadiene and acrylonitrile.

50. Arrange the following compounds in the increasing order of their reactivity towards hydrolysis

(a) P–2, Q–4, R–1

(b) P–4, Q–1, R–2

(c) P–1, Q–3, R–4

(d) P–4, Q–3, R–1

Ans. (d)

Sol. The reactivity of hydrolysis affected by the substituent

—OCH3, —Cl, —NH — CH3 and —OCOCH3

Chlorine is more electronegative among all and hence, increase the reactivity of hydrolysis and then

It takes the electrons from —CH3CO group.

So, the order is —Cl > —OCOCH3 > —OCH3 > —NHCH

51. Match the reactants and reagents in Column I with the product in Column II

(a) P-2, Q-4, R-1

(b) P-4, Q-1, R-2

(c) P-1, Q-3, R-4

(d) P-4, Q-3, R-1

Ans. (d)

Sol. Mechanism of Q reaction

52. The total number of isomers exhibited by the complexes, [Co(en)2Br]+ and [Co(en)3]3+, respectively, are {'en' is ethylene diamine}

(a) 3 and 2

(b) 2 and 2

(c) 2 and 3

(d) 3 and 3

Ans. (a)

Sol. The total anumber of isomers exhibited by the complexes, [Co(en)2Br2]+ and [Co(en)3]3+ are 3 and 2.

53. The order in which increases for the homoleptic octahedral complexes of Fe3+ is

(a) F < Br < CN < NCS

(b) BR < F < NCS < CN

(c) F < CN < Br < NCS

(d) Br < NCS < F < CN

Ans. (b)

Sol. The order of increases for the homoleptic octahedral complexes of Fe3+ is

Br < F < NCS CN

The crystal field splitting produced by the strong field CN ligand is about double that of weak field ligands like the halide ions. This is attributed to -bonding in which the metal donates electrons from a filled t2g orbital into a vacant orbital on the ligand.

54. The spin-only magnetic moments of [Fe(CN)6]4– and [FeCl6]2– in Bohr magnetons respectively are

(a) 0 and 1.73

(b) 4.73 and 5.73

(c) 0 and 5.73

(d) 4.73 and 1.73

Ans. (c)

Sol. The spin only magnetic moment of

[Fe(CN)6]4– =

Fe2+ = 3d6 45°

[Fe(CN)6]4– =

No unpaired electron present hence, spin only magnetic moment is 0.

For [Fe(Cl)6]2–, Fe2+ = 3d6 4s°

[Fe(Cl)6]2– = 3d6 =

Magnetic moment =

55. A dilute solution of MnCl2 is almost colourless. The reason for this is

(a) only spin forbidden transition

(b) only Lapporte forbidden transition

(c) odd number of unpaired spins

(d) both spin and Lapporte forbidden transition

Ans. (b)

Sol. A dilute solution of MnCl2 is almost colourless. The reason for this is only Lapporte forbidden transition.

Transitions which involve a change in the subsidiary quantum number = ± 1 are 'Lapporte allowed'

Mn2+ = 3d5 4s0 =

Since, the change in 'l' is 0, so, this a Lapporte forbidden transition.

56. Nitrobacter oxidizes nitrite to nitrate. The number of electrons involved in this oxidation process is

(a) 1

(b) 2

(c) 3

(d) 4

Ans. (b)

Sol. Plants require nitrates for their survival, bacteria in the soil will readily convert the nitrogeneous compounds into nitrates.

57. The bond order in carbon monoxide (CO) decreases from three, when bonded to transition metals in their low oxidation states, because

(a) -electrons of CO are donated to metal center

(b) -electrons of CO are donated to metal center

(c) p-electrons of metal center are transferred to empty orbitals of CO

(d) d-electrons of metal center are transferred to empty * orbitals of CO

Ans. (b)

Sol. As bond order is equal to

i.e. nb = bonding electrons

na = anti-bonding electrons

When carbon monoxide (CO) bonded to transition metals in their low oxidation states, then CO donates its -electrons to metal centre. As a result (nb–na) decreases and hence bond order decreases and hence, bond order decreases from three to lower value.

58. For [ICl4]–, the number of lone pairs present on the central atom, and the shape of the ion respectively, are

(a) 2, octahedral

(b) 1, trigonal bipyramidal

(c) 1, square pyramidal

(d) 0, tetrahedral

Ans. (a)

Sol. Number of valence electrons of l = 7

Number of bond pairs present = 4

Number of lone pairs present on central atom =

Hybridisation =

Structure of [Cl4]

The shape of [lCl4] is octahedral. Its hybridisation is sp3d2.

59. The correct orders of the reactions deduced from the graphs given below, are

(a) P-First order, Q-Zero order, R-Half order

(b) P-Zero order, Q-First order, R-Second order

(c) P-Pseudo-first order, Q-Second order, R-Third order

(d) P-Second order, Q-First order, R-Zero order

Ans. (b)

Sol. In graph P, the rate isdirectly proportional to concentration, which exist in zero order reactions.

Rate (concentration)0

In graph Q, the rate increases with increases in concentration, so, the rate is first order reaction.

Rate (concentration)1

In graph R, the rate first remains constant and then increases stepply. So, this is second order reaction in which

Rate (concentration)2

60. A large hot water bath at a constant temperature of 360 K is kept in the laboratory. The water bath loses 60 Joules of heat energy to the surroundings which is at 300K. Assuming that the heat transfer is a reversible process, the total entropy change (in JK–1) would be

(a) +0.33

(b) –0.033

(c) +0.20

(d) +0.033

Ans. (c)


So, total entropy change = + 0.20 JK–1

61. The equilibrium constant for the reaction, N2(g) + 3H2(g) 2NH3(g) at 298K, is Kp. The equilibrium constant for the reaction, 1/2N2(g) + 3/2H2(g) NH3(g), at the same temperature is

(a) Kp1/2

(b) Kp

(c) Kp2

(d) Kp/2

Ans. (a)


The equilibrium constant drop by .

62. The value of the equilibrium constant of an electrochemical cell reaction is 10 and its standard e.m.f. is 0.0148V at 298K. The number of electrons transferred in the overall cell reaction is

(a) 2

(b) 1

(c) 4

(d) 3

Ans. (c)

Sol. Given, k = 10

cell = 0.0148 V

Number of electrons transferred in cell reaction,

–nFE°cell = –2.303 RT log K

63. The difference in the number of nodes for the quantum numbers n = 4 and n = 1, for a 1 – D box is

(a) 1

(b) 2

(c) 3

(d) 4

Ans. (c)

Sol. For 1-D box, Number of nodes = n – 1

To find the difference in the number of nodes for the

quantum numbers n = 4 and n = 1 are

For n = 1, number of nodes = 1 – 1 = 0

For n = 4, number of nodes = 4 – 1 = 3

Difference in number of nodes for quantum number 4 and 1 = 3 – 3 = 0

n = Energy level

64. An aqueous solution contains 0.01 mol of formic acid (pKa = 3.8) and 0.1 mol of sodium formate. The pH of the solution is

(a) 3.8

(b) 4.8

(c) 2.8

(d) 1.8

Ans. (b)

Sol. Formic acid and sodium formate together forms acidic buffer.

pH for acidic buffer.

pH = pKa + log = 3.8 + log

= 3.8 + log 10 = 4.8

65. An electron is released with a speed of 1600 ms–1 in the x–y plane. There is a uniform magnetic field of 1 × 10–3 T along the z-direction. The radius of circular path that the electron will traverse will be close to (Mass of the electron = 9 × 10–31kg, Charge of the electron = 1.6 × 10–19C)

(a) 10 nm

(b) 10

(c) 10mm

(d) 10 m

Ans. (b)

Sol. As we know that force experienced by a change in magnetic field is given by

q(v × B) = F

And this force provides centripetal force for circular motion of the given charge particle

= 9 × 10–31 × 103 × 103 × 1019

= 9 × 10–31 × 1025

= 9 × 10–6 m

66. A ball of mass 100g is thrown vertically upwards with a velocity of 10 ms–1 from the top of a building 10m high. The kinetic energy of the ball as it hits the ground is (Acceleration due to gravity = 10 ms–2)

(a) 1 J

(b) 15 J

(c) 150 J

(d) 1500 J

Ans. (b)

Sol. By conservation of energy

Total energy at top of building = Energy at the ground

= 100 × 10–3(100 + 50)

= 150 × 10–1

= 15J

67. A slab of material is kept within a region of a constant uniform electrostatic field (Eout). For the materials given in Group I, choose the correct option from Group II that describes the fiedl inside (Ein) the slab

Group-I Group-II

P. Conductor 1. Ein = 0

Q. Dielectric 2. Ein < Eout

3. Ein > Eout

4. Ein = Eout

(a) P–1 and Q–2

(b) P–2 and Q–4

(c) P–4 and Q–3

(d) P–3 and Q–1

Ans. (a)

Sol. As we know in case of a conductor, charge accumulates at the surface. So, inside the conductor electric field is zero.

When dielectric is inserted in the region of constant uniform electric field, the material gets polarised and an opposite electric field is devloped inside the material which is less than the value the external uniform electric field in case of dielectric material.

68. In reverse bias, the current in p-n junction diode is negligible because the

(a) currents due to the electron and the holes cancel each other

(b) current is only due to the holes and the current due to the electron is shut off

(c) curent due to the majority carriers is negligible and the current is only due to the minority carriers

(d) current due to the majority carriers reverses its direction

Ans. (c)

Sol. In case of reverse bias, the p-junction is connected with negative terminal and n-junction is connected with positive terminal of the battery. Due to which potential barrier increases and current due to majority carriers become negligible, as p-n junction in reverse bias provides high resistance to the flow of current.

69. Exposure to X-rays is considered harmful due to their ability to ionizee molecules in tissue. The following property of X-ray is directly responsible for this aspect

(a) X-rays have large amplitude

(b) X-rays have high frequency

(c) X-rays travel with a velocity of 3 × 108 ms–1

(d) the magnetic field associated with X-rays in large

Ans. (b)

Sol. X-rays having high frequency can ionise molecules in the tissue, so they are considered harmful.

70. The electrical resistance of a metallic wire decreases with

(a) increasing temperature and increasing radius of the wire

(b) decreasing temperature and decreasing radius of the wire

(c) increasing temperature and decreasing radius of the wire

(d) decreasing temperature and increasing radius of the wire

Ans. (d)

Sol. Dependence of electrical resistance of a metallic wire is as follows. For temperature

R =

R = Resistance at any temperature

R0 = Resistance at given temperature

= Temperature coefficient

= Change in temperature

So, on increasing temperature, resistance of metallic wire increases and vice-versa.

In terms of length and area

R =

Resistance of metallic wire is inversely proportional to area of cross-section of the wire. So, on increasing radius of cross-section, resistance of metallic wire decreases.

71. A stone of mas m is tied to a string of length l and is rotated in the horizontal plane at a constant angular velocity . When the length of the string is decreased to l/2, the angular velocity becomes





Ans. (d)

Sol. Applying conservation of angular momentum in the two cases

72. A stick partially immersed in a half-filled glass of water appears broken at the air-water interface because of

(a) refraction of light

(b) total internal reflection

(c) diffraction of light

(d) dispersion of light

Ans. (a)

Sol. Due to refraction, light coming from partially immersed stick bends away from the normal and hence it appears broken at the air water interface.

73. Two independent and identical circular conducting loops have radius r. They are symmetrically placed parallel to the x-y plane about the z-axis with their centers at

(0, 0 + z0. The currents in the loops are equal and counter-propagating. The magnetic field along the z-direction (Bz) at the origin is

(a) zero, as the Bz fields produced by the individual loops cancel each other

(b) non-zero and two times the Bz field produced by the individual loops

(c) non-zero and half the Bz field produced by the individual loop

(d) non-zero and four times the Bz field produced by the individual loop

Ans. (a)

Sol. According to question,

Since, current is in opposite direction in the circular loops, so according to right hand thumb rule magnetic field produced in Z-axis will be equal and in opposite direction, so magnetic field will be zero.

74. A satellite has an elliptical orbit around the earth. Its maximum distance from earth is d1 and the minimum distance is d2 and the corresponding tangential velocities are V1 and V2 respectively. The ratio of the velocities at these distances is





Ans. (c)

Sol. By conserving the angular momentum of the satellite at different positions, we have md1v1 = md2v2

75. The dimensions of the ratio are

(a) ML–4T–2

(b) ML–1T–2

(c) ML–2T–1

(d) ML–2T–2

Ans. (b)

Sol. Dimensions of is given by ML–1 T–2

76. A large aircraft of mass 105 Kg travels at a speed of 600 km per hour. Assuming the surface area to be 600 m2, the percentage difference between the speed of air on the upper and the lower surface of the aircraft is (acceleration due to gravity = 10 ms–2, air density = 1.2 Kg m–3)

(a) 5%

(b) 25%

(c) 50%

(d) 95%

Ans. (a)

Sol. Applying Bernoulli's therorem to two different situations

And we know that

From Eqs. (i) and (ii), we get

77. The surface temperature of the Sun in around 6000K and its peak wavelength of emission is 5000Å. Given that the temperature of the Moon surface is 200K, the peak wavelength of the radiation form the Moon is

(a) 3000 Å

(b) 1200 Å



Ans. (c)

Sol. Applying Wien's displacement law

= constant

For two different given situations

78. Two waves on a string have a displacement given by y1 = y2sin(kx – ) and y2 = y0cos(kx – + ). If superposition of these waves results in a null displacement, then what should be the choice of?




(d) Such a is not possible

Ans. (b)

Sol. As given

79. An optical communication system operating at 1.5 is used to transmit a number of audio channels of bandwidth 8 KHz. Supposing that 1% of the optical source frequency is not available channel bandwidth, the number of audio channels that can be accomodated are

(a) 10

(b) 104

(c) 108

(d) 1012

Ans. (c)

Sol. The frequency of optical communication system is given by v =

v = = 2 × 1014 Hz

Now, 1% of 2 × 1014 Hz = 2 × 1012 Hz

So, number of audio channels that can be accommodated so =

80. Consider the following statements

Statement 1 : The large energy released in a nuclear fission reaction is due to the extraction of the binding energies of the nucleons.

Statement 2 : The coulomb binding energies are comparable to the nuclear binding energy at an inter-nucleon distance of 1 femtometer.

(a) statement 1 and 2 are true, and 2 is the correct explanation for 1

(b) statement 1 and 2 are true, and 2 is not the correct explanation for 1

(c) statement 1 is false and the statement 2 is true

(d) statement 1 is true and the statement 2 is false

Ans. (d)

Sol. Option (d) is correct because statement I is true (the energy released during nuclear reaction is extracted from binding energy of nucleous (neutron and proton) and statement II is false (At such a small distance colombic binding energy is not comparable with nuclear binding energy).

81. Match the elements in Group II that are used widely in the devices listed in Group I

Group I Group II

P. Detector 1. Soft-iron core

Q. Rectifier 2. Polymer

R. Transformer 3. Photodiode

S. Amplifier 4. Diode

5. Transistor

(a) P–4, Q–1, R–2, S–3

(b) P–2, Q–5, R–1, S–4

(c) P–1, Q–2, R–5, S–3

(d) P–3, Q–4, R–1, S–5

Ans. (d)

Sol. Photodiode is used as a detector. Diode can be used as a half-wave or full wave rectifier. Soft iron core is used is transformer.

Transistor is used as an amplifier to amplify signals.

82. The wattage rating of Heater I and Heater II are marked as 500W and 1000W, respectively and are specified for operation at 200V. Both these heaters are connected in series to a 200 V dc supply. Which one of them will produce more heat and what is the total heat generated?

(a) heater I produces more heat, and the total heat generated is 333W

(b) heater I produces more heat, and the total heat generated is 1500W

(c) heater II produces more heat, and the total heat generated is 333W

(d) heater II produces more heat, and the total heat generated is 1500W

Ans. (a)

Sol. Resistance of I heater, R1 =

Resistance of II heater, R2 =

So , heater I produces more heat.

Equivalent resistance = R1 + R2 =

Power by equivalent resistance =

= = 333 W

83. For a real number x, let [x] denote the greatest integer less than or equal to x. Let K be a real number and function f is defined by

If exists, then the value of f(2) is equal to

(a) –2

(b) –1

(c) 1

(d) 2

Ans. (b)

Sol. Given function

f(2) = k(2) + 1 [f(x) = kx + 1, x > 1]

= 2k + 1 = 2(–1) + 1 = –2 + 1

f(2) = –1

84. The area of a circle is increasing at the rate of 10 cm2/sec. If the initial area is 1 cm2, then the time at which the perimeter of the circle equals is





Ans. (b)

Sol. Given, initial area of a circle, = 1

85. The value of the integral is

(a) 0




Ans. (d)


86. Three balls were chosen randomly from a bag containing 5 white balls are 6 red balls. The probability that exactly 2 white balls and 1 red ball were chosen is equal to

(a) 1/12

(b) 1/10

(c) 3/10

(d) 2/3

Ans. (c)

Sol. In a bag there are 4 white balls and 6 red balls.

If three balls are chosen randomly. Then, the probability that exactly two white balls and one red ball were chosen is

87. Consider the following statement

"All peacocks dance and some elephants sing."

Which one of the following statement is the negation of the above?

(a) some peacocks sing and all elephants dance

(b) all peacocks do not dance and some elephants do not sing

(c) all peacocks sing and some elephants do not dance

(d) some peacocks do not dance for all elephants do not sing

Ans. (d)

Sol. Given statement :

p: "All peacocks dance and some elephants sing."

Negation of the following statement

~p: Some peacocks do not dance or all elephants do not sing.

88. The total number of relations on the set {a, b, c} is equal to

(a) 23

(b) 26

(c) 29

(d) 212

Ans. (c)

Sol. Given, se = {a, b, c}

Cardinality of the set = n = 3

Total number of relation =

89. Given that 1 + 2i and 2 are roots of the cubic equation x3 – 4x2 + 9x + K = 0 where K is a real number, then the value of K is equal to

(a) –10

(b) –6

(c) 6

(d) 10

Ans. (a)

Sol. Given, roots of equation x3 – 4x2 + 9x + k = 0 are

1 + 1i and 2.


{x – (1 – 2i)}{x –(1 + 2i)}(x – 2) = 0

(x – 1 + 2i)(x – 1 –2i)(x – 2) = 0

{(x – 1)2 – (2i)2} (x – 2) = 0

(x2 – 2x + 1 – 4i2)(x – 2) = 0

{x2 – 2x + 1 – 4 (–1)}(x – 2) = 0

(x2 – 2x + 1 + 4)(x – 2) = 0

(x2 – 2x + 5)(x – 2) = 0

x3 – 2x2 – 2x2 + 4x + 5x – 10 = 0

x3 – 4x2 + 9x – 10 = 0 ...(i)

and given equation is x3 – 4x2 + 9x + k = 0 ...(ii)

Since, Eqs. (i) and (ii) are having equal roots. So, their corresponding terms are same.

Comparing constant terms we get, k = –10

90. A frog is moving along a straight line by jumping. It always jumps the same distance which is a natural number. If the frog touches the ground exactly 12 times in a stretch of 90 units length, the size of each step is

(a) 6 units

(b) 7 units

(c) 8 units

(d) 9 units

Ans. (c)

Sol. Let size of each jump = m units

Frog touches the ground exactly 12 times means 12 jumps

According to question

91. The value of the determinant is

(a) xyz(x – y)(x – z)(z – y)

(b) xyz(y – x)(x – z)(z – y)

(c) 6xyz(x – y)(x – z)(z – x)

(d) 6xyz(y – x)(x – z)(z – y)

Ans. (d)

Sol. Given, determinant

x × 2y × (–3z)

[Taking common out factor x, 2y and – 3z from C1, C2 and C3]

[Taking out common factor (x – y), (y – z) from C1 and C2 respectively]

= –6xyz(x – y)(y – z) {(y + z) – (x + y)}

(Explanding along C3)

= –6xyz(x – y)(y – z) (z – x)

= 6xyz(y – x)(z – y)(x – z)

92. Let I, k be real numbers such that the following non-homogeneous system of linear equations is consistent.

5x + 2y + 5z = 7

2x + z = I

2y + 2zx = k

Then I and k satisfy

(a) k + 3I = 7

(b) k = 0 and I = 1

(c) k = 1 and I = 0

(d) 2I – k = 7

Ans. (a)

Sol. Given non-homogeneous system of linear equation

5x + 2y + 5z = 7

2x + z = l

2y + 2z – x = k

On writing the above system of linear equation as matrix form

AX = B,


A =

B =

Now, |A| = 5(0 – 2) –2(4 + 1) +5(4 – 0)

|A| = –10 – 10 + 20 |A| = 0

Since, system of linear equation is consistant and |A| = 0, so

(adj A)B = 0

For, adj A, cofactors of A

A11 = = –2

A12 = = – (4 + 1) = –5

A13 = = (4 – 0) = 4

A21 = = –(4 – 10) = 6

A22 = = (10 + 5) = 15

A23 = = –(10 + 2) = – 12

A31 = = (2 – 0) = 2

A32 = = –(5 – 10) = 5

A33 = = (2 – 0) = 2

adj A =

adj A =

adj A =

(adj A) B = 0

= 0

–14 + 6l + 2k = 0, –32 + 15l + 5k = 0

and 28 – 12l – 4k = 0

Above three equations are same,

So, –14 + 6l + 2k = 0

k + 3l – 7 = 0

k + 3l = 7

93. If the straight lines are perpendicular to each other, then the value of k is equal to

(a) –16/3

(b) –4/3

(c) –8/9

(d) –1/9

Ans. (d)

Sol. Given straight lines

and are perpendicular or We can write above equations as

Directions ratios of these lines are 2, –3, k, –5 respectively.

Two lines with direction ratios a1, b1, c1 and a2, b2, c2 are perpendicular to each other if a1a2 + b1b2 + c1c2 = 0

94. The value of is equal to





Ans. (d)


95. Consider the following data 8, 3, 4, 1, 6, 8. The mean deviation about the mean for the above data is equal to

(a) 13/6

(b) 7/3

(c) 15/6

(d) 8/3

Ans. (b)

Sol. Given data 8, 3, 4, 1, 6, 8

96. The area of the triangle whose vertices have the coordinates (2, 6), (0, 0) and (3, 1) is

(a) 1/2 sq. units

(b) 3/2 sq. units

(c) 8 sq. units

(d) 10 sq. units

Ans. (c)

Sol. Given coorinates are (2, 6), (0, 0) and (3, 1).

Area of triangle with coordinates (x1, y1), (x2, y2) and (x3, y3) is

= 8 sq units

97. The coordinates of a point that divides the line segment joining the points (–1, 2) and (1, –4) internally in the ratio 2 : 3 are





Ans. (a)

Sol. Let coordinates of point C(x, y) which divides the line segment joining the two points (–1, 2) and (1 –4) in the ratio 2 : 3.

So, m = 2, n = 3

By section formula, coordinate of point C(x, y)

98. For x > 0, define f(x) = x2 – 3x – 4. Invoking the chain rule, the derivative of the inverse function f–1 at x = 0 is



(c) 3

(d) –5

Ans. (b)

Sol. Given f(x) i x2 – 3x – 4

Ley y be the f-image of x, then

99. Let K be a real number and let the function F be defined by F(x) = . If the derivatie of F at x = 1 is , then the value of K is equal to





Ans. (c)


100. The maximum value of the function f(x) = sin() – + 2 on the interval [–1, 1] is equal to



(c) 2

(d) 4

Ans. (b)


So, f(x) is monotonically decreasing function on [–1, 1].

Hence, maximum of f(x) will be obtain at x = –1.