IIT JAM Biology 2009
Previous Year Question Paper with Solution.

1. From among the following structures, the most acidic molecule is

    (a) CH3 = CH2–COOH

    (b) H2C=CHCOOH


    (d) CH3–CH2–CH2–OOH

Ans.    (c)

Sol.    HC=C–COOH

    Here, COOH group is attached to sp hybridised carbon. The sp hybridised carbon has maximum s-character i.e. 50% s character. Therefore, it is most electronegative carbon and hence, most acidic.

2. Which of the following compounds, upon mass spectral analysis, give a base peak at m/z 119?


    (a) P, Q and R

    (b) P and Q

    (c) Q and R

    (d) P and R

Ans.    (d)

Sol.    The base peak at m/z 119 occur in



3. The order of melting point of LiF, LiCl, LiBr and Lil is

    (a) LiF LiCl < LiBr > Lil    

    (b) LiF < LiCl < LiBr < Lil

    (c) LiF > LiCl LiBr < Lil    

    (d) LiF > LiCl > LiBr > Lil

Ans.    (d)

Sol.    The order of melting point is HF > HCl > HBr > HI

    As the sizee of anion increases, the covalent character increases. It means the ionic character decreases. So, the melting point also decreases.

4. Jahn-Teller distortion is a common phenomenon for octahedral complexes with

    (a) a high spin d5 configuration

    (b) a d9 configuration

    (c) a low spin d6 configuration

    (d) a d3 configuration

Ans.    (b)

Sol.    Jahn-Teller disortion is a common phenomenon for octahedral complexes with d9 configuration.

    When an odd electron is present in eg orbital i.e. in d9, d7, d4 than Jahn-Teller distortion is observed.

    Here, electrons present in eg orbitals causes distortion. Due to which change in shape occur.

5. The spin only magnetic moment of [CoF6]3– is

    (a) 0.0 BM

    (b) 1.73 Bm

    (c) 2.40 BM

    (d) 4.90 BM

Ans.    (d)

Sol.    The spin only magnetic moment of [CoF6]3– is

    Co = [Ar] 4s2 3d7

    Co2+ = [Ar] 4s0 3d6

    [Co Fe]3 =

    Number of unpaired electrons = 4

    Magnetic moment =

6. XeF5+ is isoelectronic and isostructural with

    (a) PF5

    (b) IF5

    (c) PtF5

    (d) CIF5

Ans.    (b)

Sol.    XeF5+ is isostructural and isoelectronic with IF5.

    In XeF5+,

    Total electrons = 54 + 9 × 5 – 1 = 98

    Hybrid orbital = 6 bond + LP = 5 + 1 = 6

    Hybridisation = sp3d2

    In IF5,

    Total electrons = 53 + 9 × 5 = 98

    Hybrid orbital = 6 bond + LP = 5 + 1 = 36

    Hybridisation = sp3d2

7.    Despite a large difference in their atomic number Zr(40) and Hf(72) have comparable atomicradii. This is because

    (a) both elements are in the same period of the periodic table

    (b) of the lanthanide contraction

    (c) of the actinide contraction

    (d) of the presence of half-filled f-orbitals in Hf

Ans.    (b)

Sol.    Za(40) and Hf(72) have comparable atomic radii. This is because of the lanthanide contraction. The filling of the 4f orbitals through the lanthanide elements cause a steady contraction, called the lanthanide contraction, in atomic and ionic sizes. Zr lies in second transition series and Hf lies in the third transition series.

    Thus, the expected size increases of elements of the third transition series, due to an increased number of electrons and higher principal quantum number of the outer ones, are almost exactly offset, and there is in general little difference in atomic and ionic sizes between the two heavy atoms of a group.

8. Let x and y be two numbers such that 0 < x < 1 and 0 < y < 1. Then the sum of the infinite series x(x + y) + x2(x2 + y2) + x3(x3 + y3) + ... + xn(xn + yn) + ... is





Ans.    (c)

Sol.    Sum of the series

    x (x + y) + x2 (x2 + y2) + x3 (x3 + y3) + ... xn (xn + yn) + .....


9. Let z = 2 + i represent the vertex of a square inscribed in the circle |z – 1| – 2. Then one of the adjacent vertices of the square is





Ans.    (b)

Sol.    Given, z = 2 +




    To find adjacent point of P

    multiply RHS of z – 1 = 2 cis


    Two adjacent points of P are



10. The area of the triangle formed by the lines joining the vertex of the parabola x2 8y to the ends of the latus rectum is

    (a) 8

    (b) 4

    (c) 16

    (d) 2

Ans.    (a)

Sol.    Equation of parabola

        x2 = 3y



    Compare with x2 = 4 ay

        a = 2

        Focus is (0, a) = (0, 2) and end points of latusrectum are (+ 2a, a) = (4, 2)


11. The co-ordinates of a point, where the line y = x + touches the circle x2 + y2 = 1, are





Ans.    (d)

Sol.    Given line y = x + touches the circle x2 + y2 = 1.

    Clearly, intersecting point is given by





12. Let y(x) be the solution of the differential equation x= x + y, x (0, ) satisfying the initial condition y(1) = 0. Then, as x 0

    (a) y(x)

    (b) y(x) 1

    (c) y(x) 0

    (d) y(x) does not have a limit

Ans.    (c)

Sol.    Given differential equation is




    Solution of the given differential equation is given by






13. The problem of minimizing the function 2x + 3y subject to the constraints x + y = 5, x< 2, y< 4, x> 0 and y> 0 has

    (a) multiple solutions

    (b) an optimal solution

    (c) no solution

    (d) an unbounded solution

Ans.    (b)

Sol.    Given, objective function

        Z = 2x + 3y

    and constraints x + y = 5, x < 2, y < 4, x > 0 and y >0.

    Graph of line x + y = 5, x = 2 and y = 0 is shown below.

    Clearly, feasible region in the line segment BC.

    Corner point    z = 2x + 3y

    B (2, 3)    2 × 2 + 3 × 3 = 13, minimum

    C (1, 4)    2 × 1 + 3 × 4 = 14

    Since, given objective function z = 2x + 3y minimise at point B (2, 3). So, it has an optimal solution.

14. The co-ordinates of the point, where the line joining the points A = (1, 2, 3) and B = (2, 3, 4) crosses the xy-plane, are

    (a) (–2, –1, 0)

    (b) (–2, 0, 0)

    (c) (–1, 2, 0)

    (d) (0, –1, 0)

Ans.    (a)

Sol.    Equation of the line passing through the points (x1, y1, z1) and (x2, y2, z2) is


    Equation of line joining (1, 2, 3) and (2, 3, 4) is


    and equation of xy-plane is z = 0

    Any point on the line is

    Since, line passes through xy-plane (z = 0)


    Required point of intersection of plane and line is


    =    (–3 + 1, –3 + 2, 0)

    =    (–2, –1, 0)

15. Electron affinity of nitrogen is

    (a) higher than that of phosphorus

    (b) lower than that of phosphorus

    (c) comparable to that of phosphorus

    (d) identical to that of phosphorus

Ans.    (b)

Sol.    Electron affinity of nitrogen is lower than that of phosphorus. As we move down the group, size of atom increases due to addition of new shell. Nitrogen is a smaller atom as compared to phosphorus. In 'N' atom electrons are near to the nucleas. It has more electron density as compare to P so addition of extra electron in N difficult as compare to P.

16. Kevlar (–[–NHC6H4–NHCO–C6H4–CO–]n–), polyethene (–[CH2–CH2–]n–), Dacron (–[–O–CH2–ch2–O–CO–C6H4–CO–]n–) and Lexan (–[–CO–O–C6H4–C(CH3)2–C6H4–O–]n–) represent

    (a) a polyamide, a polyolefin, a polyester and a polycarbonate respectively

    (b) a polyster, a polyolefin, a polyamide and a polycarbonate respectively

    (c) a polymide, a polyolefin, a polycarbonate and a polyester respectively

    (d) a polyamide, a polyolefin, a polyester and a polyester respectively

Ans.    (a)

Sol.    Kevlar is synthesised in solution from the monomers 1, 4-phenylene diamine and terephthaloyl chloride. It is a polyamide


    Polytene is formed by the condensation of ethylene. It is a polyolefin (CH–CH2)n

    Dacron is formed by the poly esterification process of ethylene glycol and terephthalic acid.



    Lexan is formed by condensation of diethyl carbonate and Bisphenol. It is a poly carbonate.


17.    Plasmid A and plasmid B were digested with BamHI and analyzed by agarose gel electrophoresis. If Plasmid A gave two fragments and plasmid B gave three fragments, then which of the following inferences are CORRECT?

    P.    Plasmid A has three sites and is circular

    Q.    Plasmid B has three sites and is circular

    R.    Plasmid A has two sites and is linear

    S.    Plasmid B has two sites and is linear

    (a) P and Q

    (b) Q and R

    (c) P and S

    (d) R and S

Ans.    (b, d)

Sol.    In the given question when BamHI will digest plasmid A and B, then two fragments of plasmid A are formed and three fragments of plasmid B are formed.

    This can only takes place in the following situations.

    (i)    For Plasmid A

        (a) If plasmid A has one restriction site and is linear

        (b) If plasmid A has two restriction sites and is circular

    (ii)    For Plasmid B

        (a) If plasmid B has three restriction sites and is circular

        (b) If plasmid B has two restriction sites and is linear

18. Cholera toxin manifests its action by

    P.    the ADP-ribolysation of Gi protein

    Q.    the transfer of ADP-ribose frmo NAD+ to the Gi protein

    R.    the inhibitino of phosphodiesterase

    S.    the activation of adenylate cyclase

    (a) P and R

    (b) P and S

    (c) Q and S

    (d) Q and R

Ans.    (c)

Sol.    Statements Q and S are true as 'A' subunit of cholera toxin has catalytic property that ADP-ribosylates G proteins. The A subunit ribosylates the GS alpha subunit of the heterotrimeric G protein. ADP ribose is provided by the intracellular NAD+.

    This ADP ribosylation alters the 'A' subunit so that it can longer hydrolyse its bound GTP, causing it to remain in an active state that stimulates adenylyl cyclase indefinitely. The resulting prolonged elevation in cyclic AMP levels within intenstinal epithelial cells causes a large efflux of Cl and water into the gut, thereby causing the severe diarrhoea that characterises cholera.

19. When cultured in vitro with a suitable combination of growth regulators, plant parenchyma and collenchyma cells becomes meristematic. This phenomenon is called

    (a) differnetiation

    (b) maturation

    (c) apoptosis

    (d) dedifferentiation

Ans.    (d)

Sol.    When cultured in vitro with a suitable combination of growth regulators, plant parenchyma and collenchyma cells become meristematic, this phenomenon is called dedifferentiation. In plants if the living differentiated cells, that have lost the capacity to divide have an ability to regain the capacity of division under certain conditions, this phenomenon is called dedifferentiated.

20. Which one of the following statements about sieve tube elements in plants is NOT CORRECT?

    (a) They are supported by companinon cells

    (b) They must die to become functional

    (c) They link end to end forming sieve tubes

    (d) They translocate organic nutrients

Ans.    (b)

Sol.    Sieve tube elements are living cells found in plants. Sieve tube elements are a specialised type of elongated cells in the phloem tissue of flowering plants. The ends of these cells are connected with other sieve tube members and together they constitute the sieve tube. The main function of the sieve tube is transport of organic nutrients from one part of the plant to another. They are well supported by companion cells.


    (a) –1

    (b) 1

    (c) 0

    (d) 2

Ans.    (b)









22. If a two digit numbers k is 4 times the sum of its digits and 2 times the product of its digits, then the number is

    (a) 36

    (b) 48

    (c) 20

    (d) 45

Ans.    (a)

Sol.    Let digit at unit place is y and at ten's place is x.

    Number is 10x + y

    According to question, we have

        10x + y = 4(x + y)    ...(i)

        and 10x + y = 2xy    ...(ii)

    From Eq. (i), we have

        10x + y = 4x + 4y

        6x = 3y

        2x = y    ...(iii)

    From Eqs. (ii) and (iii), we get

        10x + 2x = 2x (2x)

        12x = 4x2 3 = x x = 3    

    from Eq. (iii), we get

        2 × 3 = y y = 6

        Required two digit number is 36.

23. Two bicycles start off to a slow race with initial velocities 4 m/s and 2m/s and uniform accelerations 1 m/s2 and 2 m/s2, respectively. If both of them cover the same distance in the same time, then the distance covered is

    (a) 12 m

    (b) 16 m

    (c) 24 m

    (d) 36 m

Ans.    (c)

Sol.    Let the same distance covered be dm and time taken by both person be t second

    By Newton's law of motion


    For person 1, u = 4m/s and a = 1m/s2


    For person 2, u = 2m/s and a = 2m/s2


    From Eqs. (i) and (ii)


    As t cannot be zero, therefore, t = 4s

    From Eq. (i)


24. If the curve y = 5x3 + bx2 + cx + 5 touches the x-axis at the point (1, 0) then the pair (b, c) is

    (a) 5, 5

    (b) (–5, 5)

    (c) (5, –5)

    (d) (–5, –5)

Ans.    (d)

Sol.    Given the curve y = 5x3 + bx2 + cx + 5    ...(i)

    touches the X-axis at the point (1, 0)

    As the point (1, 0) lies on the curve, therefore





25. The value of is

    (a) 1

    (b) 0

    (c) –1

    (d) 2

Ans.    (b)





26. If 0 < y < 1, then the coefficient of yn in the expansion of is

    (a) 4n

    (b) 1

    (c) n – 1

    (d) 2n

Ans.    (a)


        = (1 + y)2 (1 – y)2

        = (1 + 2y + y2) [1 + 2y + 3y2 + ... + (n + 1)yn + ...]

    Coefficient of yn = (n + 1) + 2n + (n – 1) = 4n.

27.    If 0 <x</2 then the value of is





Ans.    (d)




28. Which one of the following is NOT part of a molluscan body plan?

    (a) Mantle

    (b) Radula

    (c) Visceral mass

    (d) Trachea

Ans.    (d)

Sol.    Trachea is not present in molluscan. Molluscans are soft bodied animals with body plan consisting of the head foot region containing the sensory and motor organs. The visceral mass containing the internal organs is the digestive, excretory and reproductive systems and the mantle (a specialised tissue surrounding the visceral mass).

    Radula is a characteristic feature of phyllum-Mollusca only.

29. A clade is

    (a) a type of phylogenetic tree

    (b) a group of evolutionary related species sharing a common ancestor

    (c) an extinct species

    (d) a tool for constructing a phylogenetic tree

Ans.    (b)

Sol.    A clade (Greek, Klados = branch) is a group of organisms that consist of a common ancestor and all its linear descendants and representing a single branch on the tree of life. Thus, it is a group of evolutionarily related species sharing a common ancestor. The study of clades is called cladistics.

30. Kupffer ceels are found in

    (a) stomach

    (b) liver

    (c) small intestine

    (d) large intestine

Ans.    (b)

Sol.    Kupffer cells are phagocytic cells which form the lining of the sinusolds of the liver and are involved in the breakdown of red blood cells.

31. Ecological succession refers to

    (a) changes in community composition after a disturbance

    (b) the process by which a species become abundant

    (c) the building of soil nutrients

    (d) changes in a forest as trees grow taller

Ans.    (a)

Sol.    Ecological succession refers to changes in community composition after a disturbance. The term ecological succession was first given by Hult (1885), as communities are never stable but keep on changing. This relatively definite sequence of communities over a period of time in the same area is called ecological succession.

32. During fertilization, the movement of the pollen tube towards the ovule is guided by a protein released from

    (a) Egg

    (b) Synergid

    (c) Antipodal cell

    (d) Polar nuclei

Ans.    (b)

Sol.    During fertilisation in plants the pollen tube enters the embryo sac only from the micropylar end irrespective of its mode of entry in the ovule. This growth of the pollen tube is guided by some chemotropic secretion or protein secreted by synergids. This allows the tip of pollen tube to enter into one of the synergid cell, which begins to degenerate after the pollen tube entry.

33. Choose the correct matches

    Disease    Glands/organs

    P.    Mumps    1. Pancreas

    Q.    Colitis    2. Stomach

    R.    Hepatitis    3. Salivary gland

    S.    Gastritis    4. Large intestine

            5. Liver

    (a) P–4, Q–1, R–2, S–5

    (b) P–3, Q–4, R–5, S–1

    (c) P–3, Q–4, R–5, S–2

    (d) P–2, Q–3, R–1, S–5

Ans.    (c)

Sol.    The correct match for the given question would be

    1. Mumps is epidermic parotitis in which a person has fever, swollen and tender parotid gland (i.e. salivary glands located beneath the ear) and headache.

    2. Colitis is inflammation of the lining of the colon, i.e. large intestine.

    3. Hepatitis is a common viral infecdtion in which the body makes antibodies against the liver tissue.

    4. Gastritis is inflammation of the lining of the stomach.

34. The correct sequence for sperm migration after its production in testis is

    (a) Seminiferous tubule epididymis vas deferens urethra

    (b) Urethra vas deferens epididymis seminiferous tubule

    (c) epididymis vas deferens urethra seminiferous tubule

    (d) Seminiferous tubule vas deferens epididymis urethra

Ans.    (a)

Sol.    The correct sequence of sperm migration after its production in testis is :

    Seminiferous tubule epididymis vas deferens urethra

    This takes place as seminiferous tubules are the site of spermatogenesis (i.e. sperm production). The semiinferous tubules open into rete testis. From rete testis sperm moves into a series of coiled efferent ducts called epididymis. Then, from epididymis sperms move through vasa deferentia (ducti deferentes) to urethra, from here they are released outside.

35. Statement 1: Isotopes are chemically identical

    Statement 2 : Chemical reactions do not depend on the number of neutrons in the atom. Which of the following is CORRECT?

    (a) Statement 1 and 2 are correct and 2 is the correct explanation for 1

    (b) Statement 1 and 2 are correct, but 2 is not the correct explanation for 1

    (c) Statement 1 is correct, but statement 2 is wrong

    (d) Statement 1 is wrong, but statement 2 is correct

Ans.    (b)

Sol.    Statement I : Isotopes are chemically identical.

    Statement II : Chemical reactions do not depend on the number of neurons in the atoms (as they are neutral particles).

    Isotopes are the atoms which have same atomic number but different mass number.

36. The wavelength of electromagnetic radiation emitted by the hydrogen atom in the first excited state is given as . The wavelength corresponding to the tansition from the third excited state to the ground state is





Ans.    (c)

Sol.    For H-spectrum, when electron from n2 comes to n1 shell, wavelength is given by


    Case I : Ist excited state mean n2 = 2 and n1 = 1.


    Case II : 3rd excited state mean n2 = 4 and n1 = 1.


37.    N0 atoms of a certain radioactive element undergo radioactive decay. After 100 days the number is reduced to . The mean life of the element, in days, is

    (a) 100 In(2)


    (c) 200

    (d) 100

Ans.    (b)

Sol.    Given, T1/2 = 100 days



38. A point particle of mass M moving with a constant speed u collides with another point particle of mass 2M, which is at rest. After the collision, if the second particle moves with a non-zero speed, the speed of the first particle is



    (c) 0


Ans.    (a)

Sol.    For eleastic collison, (when target body is at rest),




39. Two satellites of masses M and 2M are orbiting around the earth in circular orbits of radii R and 2R. respectively. The ratio of their speeds is



    (c) 1 : 4

    (d) 2 : 1

Ans.    (b)

Sol.    Orbital velocity of the satellite is inversely proportional to square root of radius of the orbit.


    [vorbit is independent of mass of revolving body]


40. A particle is projected vertically upwards from the surface of earth with an initial speed of 40 m/s. The acceleration of the particle when it reaches the maximum height is

    (a) 20 m/s2

    (b) 4.9 m/s2

    (c) 9.8 m/s2

    (d) 0

Ans.    (c)

Sol.    Since, projectile motion takes place under the constant acceleration of gravity (g = 9.8 m/s2). So, the value of acceleration at highest point is also 9.8 m/s2.

41. The phenomenon of expression of only one allele of an immunoglobulin gene in lymphocytes is known as

    (a) alletic exclusion

    (b) allelic inclusion

    (c) allelic variation

    (d) allelic heterogeneity

Ans.    (a)

Sol.    Allelic exclusion is the phenomenon when there is experssion of only one allele of an immunoglobulin gene in lymphocytes. In B-lymphocytes, successful heavy chain gene rearrangement on one chromosome results in the shutting down of rearrangement on the second chromosome. Basically, in allelic exclusion, only one allele of a gene gets expressed and the other gets silenced.

42. The antibody class that can pass from the mother to the fetus in human is

    (a) Ig A

    (b) Ig D

    (c) Ig G

    (d) Ig M

Ans.    (c)

Sol.    IgG is the only isotype that can pass through the human placenta, thereby providing protection to the fetus. Along with IgA secreted in the breast milk, the residual IgG absorbed through the placenta provides the neonate with humoral immunity much before its own immune system develops.

43. Which of the following statements is NOT CORRECT about MHC class II proteins?

    (a) They are recognized by CD4 co-receptors

    (b) They are composed by and chains

    (c) They are involved in presenting antigen to helper T cells

    (d) They are present in the T cell cytoplasm

Ans.    (d)

Sol.    MHC (major histocompatibility complex) class II molecules are family of molecules found only on antigen presenting cells such as dendritic cells, mononuclear phagocytes, thymic epithelial cell, some endothelial cells and B-cells. They are heterodimers composed of an and chain. The class II molecules interact mainly with immune cells like the T-helper cells (TCD4+).

44. Bilirubin is formed due to the degradation of

    (a) erythrocytes

    (b) leucocytes

    (c) hepatocytes

    (d) macrophages

Ans.    (a)

Sol.    Bilirubin is a yellowish pigment found in bile, a fluid made by the liver. It is a breakdown product of normal heme catabolism, caused by the body's clearance of aged erythrocytes (red blood cells) which contain hemoglobin.

    Bilirubin is excreted in bile and urine.

45. Match the diseases in Group-I with the corresponding hormones in Group-II.


    P.    Myxoedema

    Q.    Cushing's syndrome

    R.    Acromegaly

    S.    Grave's disease


    1. Excess secretion of T3 and T4

    2. Insufficient secretio of T3 and T4 in adults

    3. Growth hormone hypersecretion before complete ossification

    4. Glucocorticoid hypersecretion

    5. Growth hormone hypersecretion after complete ossification

    (a) P–3, Q–2, R–1, S–5

    (b) P–5, Q–3, R–2, S–4

    (c) P–2, Q–4, R–5, S–1

    (d) P–3, Q–1, R–4, S–2

Ans.    (c)

Sol.    The correct matching are as follows :

    1. Myxoedema is caused by insufficient production of thyroid hormone by thyroid gland which leads to decrease of T3 and T4 in adults.

    2. Cushing's syndrome is a metabolic disorder caused by overproduction of corticosteroid hormones by adrenal cortex.

    3. Acromegaly is overproduction of growth hormones after complete ossification (i.e. the natural process of bone formation).

    4. Graves disease is due to over activ thyroid gland which leads to excess secretion T3 and T4.

46. The function of a heterocyst in aerobic Cyanobacterim spp. is to facilitate

    (a) rapid cell division

    (b) DNA replication

    (c) nitrogen fixation

    (d) infection of host plants

Ans.    (c)

Sol.    Heterocyst is a differentiated cyanobacterial cell that carries out nitrogen fixation. An enzyme nitrogenase present in the heterocyst is responsible for the fixation of atmospheric ntirogen in the soil. Cyanobacteria are predominantly present in fresh water and Gram-negative photosynthesis prokarytoes.

47.    The number of distinct disaccharides that can be formed from two molecules of glucose is

    (a) 11

    (b) 6

    (c) 5

    (d) 1

Ans.    (a)

Sol.    Heterocyst is a differentiated cyanobacterial which is formed when two monosaccharide undergo a condensation reaction. There are 11 distinct disaccharides that can be formed from two molecules of glucose. They are maltose, trechalose, cellobiose, kojibiose, nigerose, isomaltose, -Trehalose, , -Trehalose, sophorose, laminaribiose and gentiobiose.

48. Which of the following covalent bond types are found in the structure of ATP?

    (a) N-glycoside, thioester, phosphomonoester

    (b) phosphoanhydride, phosphomonoester, N-glycoside

    (c) ester, ether, phosphoanhydride

    (d) ether, thioester, phosphomonoester

Ans.    (b)

Sol.    ATP is adenosine triphosphate, a major energy currency of the cell. The basic building block used to construct of the cell. The basic building block used to construct ATP are carbon, hydrogen, nitrogen, oxygen and phosphorus which are assembled in a complex. There is one phosphate ester bond and two phosphate anhydride bonds that, hold the three PO4 and the ribose together. The construction also contains a N-glycoside bond holding the ribose and adenine together.

49. Let = be the force acting at the point P = (1, 0, 0), where and are the unit vectors along the x-axis and y-axis, respectively. Then the moment of

    about the line through origin the direction of is



    (c) –1

    (d) 0

Ans.    (d)

Sol.    According to the question, F =


    The moment of force is given by f × F

    and x towards Y axis or j-direction is 0.

    So, moment of force about the line through origin in the direction of will be zero.

50. Let and be the area bounded by the curve y = f(x), then the x-axis and the ordinates at x = 0 and x = 4. Then the value of is

    (a) 16/3

    (b) 64/3

    (c) 59/12

    (d) 64

Ans.    (c)








51. An urn contains 3 red, 5 black and 7 yellow balls. If a ball is drawn at random, then the probability that the ball drawn is not yellow is

    (a) 7/15

    (b) 8/15

    (c) 7/8

    (d) 1/7

Ans.    (b)

Sol.    Given, red balls = 3

        black balls = 5

        and yellow balls = 7

    Total number of balls = 3 + 5 + 7 = 15

    Number of non-yellow balls = 8

    Required probability =

52. Let A be a non-singular square matrix of order 3. If B is the matrix obtained from A by adding 3-multiple of its first row to its second row then the value of det(2A–1B) is

    (a) 8

    (b) 6

    (c) 3

    (d) 2

Ans.    (a)

Sol.    Given, A be an non-singular square matrix of order 3.

    Matrix B is obtained from A by adding 3 multiple ofits first row to the second row, i.e.



53. The pressure difference between the inside and the outside of a liquid drop is

    (a) linearly proportional to the radius of the drop

    (b) inversely proportional to the radius of the drop

    (c) proportional to the square of the radius of the drop

    (d) zero

Ans.    (b)

Sol.    The pressure difference between the inside and outside of a liquid drop is given by


54. In a Young's double sit experiment using light of wavelength , the interference pattern of fringe width 2.5mm is observed. If the same set up is used with a light of wavelength , the fringe width would be

    (a) 1.25 mm

    (b) 2.5 mm

    (c) 5 mm

    (d) 10mm

Ans.    (c)

Sol.    Fringe width in Young's double slit experiment is given by



55. A ray of light travelling through a medium of refractive index 2 is incident on an interface with another medium of refractive index 1, at an angle of incidence of 30°. Which of the following statements is CORRECT?

    (a) The ray undergoes total internal reflection

    (b) The ray is fully transmitted

    (c) The ray just grazes the interface

    (d) The ray is partly reflected and partly transmitted

Ans.    (d)

Sol.    According to the question ray diagram can be drawn as shown in the figure.


    Now, from the concept of critical angle


    Since, angle of incidence is less than the critical angle, hence ray will not suffer total internal reflection. It partly shows reflection and partly refraction.

56. Choose the correct set of matches between the function and the corresponding cellular structure.

    Function    Structure

    P.    Protein synthesis    1. Lysosomes

    Q.    Intracellular digestion    2. Ribosomes

    R.    Protein secretion    3. Microtubules

    S.    Macromolecular traffic    4. Mitochondria

            5. Golgi apparatus

    (a) P–1, Q–2, R–5, S–3

    (b) P–2, Q–1, R–4, S–5

    (c) P–3, Q–5, R–1, S–4

    (d) P–2, Q–1, R–5, S–3

Ans.    (d)

Sol.    The correct matching for the following cellular structure with their functions are :

    1. Protein synthesis occurs in cellular structures called ribosomes, found outside the nucleus.

    2. Lysosomes are specialised for the intracellular digestion of macromolecules.

    3. Golgi apparatus functions as a factory in which protein received from the ER are further processed and transported to their destination in lysosomes, plasma membrane for secretion.

    4. Microtubules based motor proteins are able to perform intracellular transport reliably inspite of macromolecular traffic inside the cell.

57.    2, 4-Dinitrophenol inhibits mitochnodrial function by

    (a) inhibiting ATP synthesis

    (b) inhibitnig electron flow

    (c) dissipating the electrochemical gradient

    (d) decreasing oxygen permeability

Ans.    (c)

Sol.    In living cells, 2, 4 – Dinitrophenol acts as a proton ionophore (i.e. an agent that can shuttle protons across biological membrane).

    It dissipates the proton gradient across mitochondria and chloroplast membranes, collapsing the proton motive force that the cell uses to produce most of its ATPs chemical energy.

58. The premature termination of polypeptide synthesis due to a stop codon can be overcome by a compensatory mutation in tRNA. This genetic phenomenon is referred to as

    (a) intragenic suppression

    (b) extragenic suppression

    (c) codon bias

    (d) true reversion

Ans.    (b)

Sol.    Extragenic suppression is the one which relieves the effects of a mutation in one gene by a mutation somewhere else within the genome. For example, a mutation which disrupts complementary interaction between protein molecules may be compensated by a second mutation elsewhere in the genome that restores or provides a suitable alternative interaction between those molecules. Its application can be in prenature termination of polypeptide synthesis due to a stop codon by a compensatory mutation in tRNA.

59. Myoglobin shows a hyperbolic response, while hemoglobin shows a signoidal response for oxygen binding. Which of the following statemetns are TRUE with respect to this observation?

    P.    Hemoglobin blinds 2, 3-BPG while Myoglobin does not

    Q.    Hemoglobin exists in two different conformational states while Myoglobin does not

    R.    hemoglobin is a tetramer while Myioglobin is a monomer

    S.    Homoglobin is present in RBCs while Myoglobin is present in the muscle

    (a) R and S

    (b) S and P

    (c) P and Q

    (d) Q and R

Ans.    (d)

Sol.    Statement Q and R are true. The difference is the oxygen binding curves of myoglobin and hemoglobin is due to difference in shape i.e. the single subunit character of myoglobin and the multiple (i.e. tetramer of ) subunit character of hemoglobin which give each of these molecules their characteristic behaviour with respect to oxygen binding. Also, hemoglobin exists in two different conformational states, inrelaxed and tense while myoglobin does not. The relaxed state of hemoglobin has a higher affinity for O2.

60. An increase in enzyme activity in a cell is mechanistically due to transcription. This mechanism can be demonstrated by

    (a) measuring total enzyme activity in the cell free extract

    (b) ELISA

    (c) Northern blot

    (d) Western blot

Ans.    (c)

Sol.    Transcription is the process when a sequence of mRNA is formed from a particular segment of DNA by the enzyme RNA polymerase and northern blot is a technique used to study gene expression by detecting RNA (or isolated mRNA) in a sample.

    Thus, an increase in enzyme activity in a cell mechanistically due to transcription can be demonstrated by northern blot. Other techniques like western blot is used to identify specific amino acid sequence in protein, ELISA is used to measure concentration of antibodies or antigens in a solution and enzyme activity is the measure of quantity of active enzyme present in a sample.

61. Galactosemia is a recessive single gene genetic disorder, caused due to the mutation in any one of the three genes involved in galactose catabolism. A family consists of 10 normal children with both parents suffering from galactosemia. This is most likely because of

    (a) epistasis

    (b) reversion

    (c) suppression

    (d) complementation

Ans.    (d)

Sol.    Complementation occurs when two strains of an organisms with different homozygous recessive mutations that produce the same mutant phenotype produce offspring with the wild type phenotype when mated or crossed. Thus, in a situation when both parents are suffering from galactosemia (i.e. homozygous recessive disorder) they may produce normal children due to complementation.

62. If R is the resistance and C is not capacitance in an electric circuit, the dimensions of RC are the same as that of

    (a) Current

    (b) Voltage

    (c) Charge

    (d) Time

Ans.    (d)

Sol.    In the charging of a capacitor, the instantaneous charge stored is given by

    q = C(1 – e–t/CR)

    where, CR is called time constant and the dimension is same as that of time.

63. One mole of an ideal mono-atomic gas of initial volume V0 is compressed adiabatically to a volume . If the initial temperature is T0, then the final temperature is



    (c) T0(2)1.67

    (d) T0(2)0.67

Ans.    (d)

Sol.    Given, gas is monoatomic n = 1 and process is adiabatic.

        V1 = V0; T1 = T0





64. The internal energy of 3 moles of an ideal mono-atomic gas at absolute temperature T is given by





Ans.    (a)

Sol.    The internal energy of ideal monoatomic gas at absolute temperature = nRT.


65. A real gas behaves as an ideal gas at

    (a) constant volume

    (b) constant pressure

    (c) low density

    (d) high pressure

Ans.    (c)

Sol.    A real gas behaves ideally at low pressure and high temperature.

    Density (d) =

    So, density is directly proportional to pressure and hence, real gas behaves ideally at low density.

66. Treating the diol shown below with strong acid gives compound E.


    Compound E displays a prominent absorption band at 1710 cm–1 in its IR spectrum. The most likely structure of E is

    (a) *

    (b) *

    (c) *

    (d) *

Ans.    (b)

Sol.    Compound E displays absorption band at 1710 cm2 in its IR spectrum is due to the vinylic attachment with benzene ring.


67.    Okazaki fragments are formed during DNA synthesis because

    P.    DNA synthesis extends from 5' to 3' direction

    Q.    their synthesis are opposite to the direction of replication fork

    R.    DNA ahead of 1replication fork is positively supercoiled

    S.    DNA synthesis is semi-conservative

    (a) Q and R

    (b) P and Q

    (c) R and S

    (d) S and P

Ans.    (b)

Sol.    When DNA is being replicated from the 5' end to the 3' end. Each strand is anti parallel, meaning one goes 5' 3' and another in 3' 5' direction. As the strand unwinds, DNA polymerase can catalyse the attachment of bases on the non-lagging strand continuously, in a 5' 3' direction only. Thus, okazaki fragments are formed in 5' 3' direction, opposite the direction of unwinding (but continuing in the direction where the unwinding is occuring).

68. Which of the following elements must be present in a plasmid clonign vector?

    P.    Origin of replication

    Q.    Selectable marker

    R.    Unique restriction sites

    S.    Promoter element upstream of the unique restriction site

    (a) S and P

    (b) Q and R

    (c) 2, 1 and 2

    (d) 1, 2 and 1

Ans.    (d)

Sol.    A plasmid cloning vector has the following elements like an origin of replication, a selectable marker, a drug-resistance gene and a region where DNA can be inserted without interfering with plasmid replication or expression of the drug resistance gene.

69. The metabolite pair that is NOT formed directly from pyruvate is

    (a) Oxaloacetate and acetaldehyde

    (b) Alanine and ethanol

    (c) Acetyl CoA and alanine

    (d) Lactate and oxaloacetate

Ans.    (d)

Sol.    Pyruvate is a conjugate base which is a key intermediate in several metabolic pathways. It can both lactate and oxaloacetate directly as follows.


70. Which of the following pairs of protocols is used before ligating two DNA molecules incompatible 5' overhangs?

    P.    Filling in with a Klenow fragment

    Q.    S1 nuclease digestion

    R.    Dephosphorylation of 5' phosphate

    S.    Phosphorylation of 3' hydroxyl group

    (a) P and Q

    (b) Q and R

    (c) R and S

    (d) S and P

Ans.    (c)

Sol.    Ligation is the joining of two nucleic acid fragments through the action of an enzyme. It is an essential laboratory procedure in the molecular cloning of DNA to create rDNA, in which foreign DNA fragment is inserted into a plasmid. In this, the ends of DNA fragments are joined together by the formation of phosphodiester bonds between the 3'-hydroxyl of one DNA terminus with the 5'-phosphoryl of another RNA many be also ligated similarly. A cofactor is generally involved in the reaction and this is usually ATP or NAD+.

71. Which of the following amino acids is a major precursor of one carbon units?

    (a) Proline

    (b) Alanine

    (c) Serine

    (d) Methionine

Ans.    (d)

Sol.    All cells require the synthesis of macromolecules, such as proteins, lipids and nucleic acids for cellular renewal and proliferation. Amino acids, such as methionine, can be generated from one carbon metabolism and used to generate proteins, lipids and nucleic acids, i.e. DNA or RNA.

72. A mutation in the operator, which prevents the binding of the repressor resulting in the constitutive expression of the Iac operon, is referred to as

    (a) semi-dominant

    (b) trans-dominant

    (c) co-dominant

    (d) cis-dominant

Ans.    (d)

Sol.    In the absence of an inducer, inducible genes remains repressed in Lac operon system. But mutation can also occur here. Thus, when a mutation occurs in the operator site, making it impossible for any molecules to bind to the operator, this results in Lac operon becoming constitutive.

    This mutation can also result in either a homozygous or heterozygous mutant bacterial cell, in which case, it is referred to as cis-dominance mutation.

73. Two double stranded DNA samples that are identical with respect to the number of base pairs, but differ significantly in their GC content, can be separated by

    (a) Density gradient centrifugation

    (b) Agarose gel electrophoresis

    (c) Dialysis

    (d) Oligo-dT column chromatography

Ans.    (a)

Sol.    Under high centrifugal force, a solution of cesium chloride molecule will dissociate and the heavy Cs atoms will be forced towards the outer end of the tube, thus forming a shallow density gradient.

    DNA molecules placed in this gradient will migrate to the point where they have the same density as the gradient (isopyonic point). The gradient is sufficient to separate types of DNA with slight differences in density due to differing [G + C] content.

74. Match the test from Column 1 with the sample pairs in Column 2 so that positive identification is possible.


    (J)    Flame test

    (K)    Tollen's reagent test

    (M)    Water solubility test

    (N)    Iodoform test


    (1)    CH3–CO–CH3 and Ch3–CO–C6H5

    (2)    H–CO–H and CH3–CO–C2H5

    (3)    Glucose and Starch

    (4)    CH3–CO–C6H5 and C6H5–CO–C6H5

    (5)    NaCl and Na

    (6)    Ch3NH–CO–NHCH3 and CH3–CO–NH2

    (a) J–4, K–3, L–2 and M–1

    (b) J–4, K–4, L–5 and M–1

    (c) J–1, K–2, L–3 and M–4

    (d) J–3, K–5, L–6 and M–4

Ans.    (c)

Sol.    J 1, K 2, L 3 and M 4

    Flame test is used to differentiate between

    CH3–CO–CH3 and CH3–CO–C6H5.

    CH3–CO–C6H5, when comes in contact with flame gives sooty flame while CH3–CO–CH3 not. Tollen's reagent test is useful to distinguish aldehyde and alcohols.

    Water solubility test is used to distinguish glucose and starch.

    Iodoform test used to distinguish –CH3CO group and help in distinguishing CH3–CO–C6H5 and C6H5–CO–C6H5.

75. Under neutral conditions, O-glycosides, unlike the free sugar from which they are derived, do not exhibit mutarotation. This is because

    (a) the anomeric hydroxyl group exists as an acetal

    (b) the anormeric carbon atoms exists in the open chain form

    (c) the anomeric carbon atom exists as a hemi-acetal

    (d) O-gktcisudes exist as a racemic mixture and hence mutarotaion cannot be observed

Ans.    (a)

Sol.    Mutarotation is the change in specific rotation that accompanies the equilibrium of and anomers in aqueous solution.

    Mutarotation is common to all carbohydrates that exist in hemiacetal forms. On treatment of the hemlacetal with a molecule of alcohol yields an acetal. A cyclic acetal derived from a monosaccharide is called a glycoside. A carbohydrate in which the –OH group on its anomeric carbon is replaced with an –OR group and the bond from the anomeric carbon to the –OR group is called a glycosidic bond.

    Mutarotation is not possible in glycoside because an acetal unlike a glycosidic hemiacetal is no longer in equilibrium with the open chain carbonyl containing compound.

76. The reaction CaCO3(s) = CaO(s) + CO2(g) is not favoured at 298K. Given that for this reaction at 298K, = 200 kJ mol–1 and = 200 JK–1 mol–1, the lowest temerature at which this reaction will proceed in the forward direction is

    (a) 801 K

    (b) 901 K

    (c) 1001 K

    (d) 1101 K

Ans.    (c)


        = 200000 – 298 × 200

        = 140400 Jmol–1

    When T = 801 K,

        = 200000 – 801 × 200

        = 19800 Jmol–1

    When T = 901K,

        = 20000 – 901 × 200

        = 19800 Jmol–1

    When T = 1001K,

        = 200000 – 101 × 200

        = 200 Jmol–1

    When T = 1101K,

        = 20200 K

77.    The graphs P, Q and R show the variation of rate constant (k) with temperature. The reactions represented by P, Q and R, respectively, are


    (a) P-Arrhenius type,    Q-an enzyme catalysed and R-a chain reaction

    (b) P-an enzyme catalysed, Q-Arrhenius type, and R-a chain reaction

    (c) P-Arrhenius type, Q-a chain reaction and R-an enzyme catalysed reaction

    (d) P-a chain reaction, Q-an enzyme catalysed and R-Arrhenius type reaction

Ans.    (a)


    Graph (P) is an arrhenius type graph. It displays the logarithm of kinetic constants versus temperatures.


    Graph (Q) is an enzyme catalysed plot. The graph shows the relation between the substrate concentration and reaction rate.


    Graph (R) is a chain reaction plot. The graph shows the relation between the rate constant and temperature.

78. Which of the following pairs of amino acids have two chiral carbons in their structure?

    (a) Thr and Ile

    (b) Tyr and Trp

    (c) His and Met

    (d) Leu and Gly

Ans.    (a)

Sol.    Chiral carbon is an asymmetric carbon atom that is attached to four different types of atoms or four different groups of atoms.

    The structures of various amino acids given in the question are




    Thus, only Thr and Ile are having chiral carbons in their structures.

79. Base pairing between inosine and uridine occurs through

    (a) Hoogsteen base pairing

    (b) Waston-Crick base pairing

    (c) Wobble base pairing

    (d) Purine-purine base pairing

Ans.    (c)

Sol.    Wobble base pairing is a pairing between two nucleotides in RNA molecules that does not follow Watson-Crick base pair rules. The four main wobble base pairs are guanine uracil (G-U), inosine uracil (I-U), inosine-adenine (I-A) and inosine cytosine (I-C).

80. The net charge on a protein will be negative when the pH is

    (a) at its isoelectric pH

    (b) above its isoelctric pH

    (c) below its isoelectric pH

    (d) at neutral pH

Ans.    (b)

Sol.    The net charge on a protein will be negative when the pH is above its isoelectric pH. Generally, the net charge on the protein is affected by pH of their surrounding environment and can become more positively or negatively charged due to the loss or gain of protons (H+).

    The pI (isoelectric point) is the pH value at which the protein carries no electrical chage or the negative and positive charges are equal. Below the isoelectric point proteins carry a net positive charge, above it a net negative charge.

81. The molecule that functions as a natural thiol reducatant in a cell is

    (a) Glutathione

    (b) Methionine

    (c) Dithiothreitol

    (d) Cystine

Ans.    (a)

Sol.    Glutathione is a tripeptide. Glutathione, an antioxidant helps protect cells from reactive oxygen species such as free radicals and peroxides. Glutathione is converted into oxidised form, i.e. glutathione disulfide (GSSG).

82. The two pathways required for the net synthesis of glucose from triglycerides in germinating groudnut seeds are

    (a) Hexose monophosphat shunt and Gluconeogenesis

    (b) Calvin cycle and Glyoxalate cycle

    (c) Glycolysis and Cori cycle

    (d) Glyoxalate cycle and Gluconeogenesis

Ans.    (d)

Sol.    Acetyl CoA is generated by beta-oxidation of fatty acids. The glyoxylate cycle begins with the condensation of acetyl CoA and OAA to form citrate, which is then isomerised to isocitrate. Instead of being decarboxylated, isocitrate is cleaved by isocitrate lyase into succinate and glyoxylate.

    Acetyl CoA condenses with glyoxylate to form malate in a reaction catalysed by malate synthase, which resembles citrate synthase. Finally, malate is oxidised to oxaloacetate. In plants, these reactions take place in organelles called glyoxysomes. Succinate can be converted into carbohydrates by a combination of the citric acid and gluconeogenesis.

83. Mathc the antibiotic with its inhibitory mode of action.

    Antibiotic    Mode of action

    P.    Penicillin    1. Protein synthesis

    Q.    Rifamycin    2. Protein glycosylation

    R.    Tunicamycin    3. RNA polymerase

    S.    Sulfanilamide    4. Peptidoglycan synthesis

            5. Peptidoglycan synthesis

    (a) P–5, Q–3, R–1, S–2

    (b) P–3, Q–4, R–2, S–1

    (c) P–2, Q–1, R–3, S–4

    (d) P–5, Q–3, R–2, S–4

Ans.    (d)

Sol.    The correct matching for the following antibiotic with their inhibitory mode of action are :

    P.    Penicilin damages and penetrates the bacterial cell wall thus killing the bacteria. Bacterial cell wall has peptidoglycan as its major component.

    Q.    Rifamycin works on inhibiting the bacterial DNA dependent RNA synthesis. This is due to the high affinity of rifamycins for the prokaryotic RNA polymerase.

    R.    Tunicamycin is an antibiotic which inhibits the glycoprotein synthesis in the bacteria.

    S.    Sulfanilamde antibiotic functions by completely inhibiting enzymatic reactions involving para-aminobenzoic acid (PABA). PABA is needed to produce folic acid, which is coenzyme in the synthesis of purines and pyrimidines.

84. Which of the following statements are TRUE for a mutation that changes the codon from UAC to UAG?

    P.    It is a nonsense mutation

    Q.    It is a missense mutation

    R.    It is a point mutation

    S.    It is a tarnsversion

    (a) P, Q, R

    (b) P, Q, S

    (c) P, R, S

    (d) Q, R, S

Ans.    (c)

Sol.    The codon change from UAC to UAG can be a non-sense mutation (as UAG is a stop codon), point mutation (as its a single base modification) or a transversion (as in its orientation of purine and pyrimidine is reversed). It cannot be a missense mutation because in missense mutation a single nucleotide change results in a codon that codes for a different amino acid.

85. The product M obtained in the substitution reaction shown below is






Ans.    (b)

Sol.    By the trans effect,

    So,    Cl > NH3


86. The order of X-N-X bond angle in NH3, NF3 and NCl3 is

    (a) NH3 < NCl3 < NF3    

    (b) NCl3 > NH3 > NF3

    (c) NH3 > NCl3 > NF3    

    (d) NCl3 > NF3 > NH3

Ans.    (c)



    In N–F bond, 'F' is more electronegative and hence, attracts bp towards itself whereas in N–H bond. 'N' is more electromagentic then 'H' atom that's why bp's are more close to the N atom. As a result, bp-bp repulsion takes place which increases the bond angle.

87.    The dep pink colour of aqueous KMnO4 solution is due to

    (a) a very strong d-d transition

    (b) a very weak d-d transition

    (c) a strong ligand to metal charge transfer interaction

    (d) a strong metal to ligand charge transfer interaction

Ans.    (c)

Sol.    The deep pink colour of aqueous KMnO4 solution is due to a strong legand to metal charge transfer interaction.

    In MnO4 an electron is momentarily transferred from 'O' to the metal, thus momentarily charging O2– to O and reducing the oxidation state of the metal from Mn(VII) to Mn(VI). Charge transfer requires that the energy levels on the two different atoms are fairly close.

88. The concept of "Spontaneous generation of life" was disproved by experiments using swan neck flasks. This experiments was conducted by

    (a) Koch

    (b) Pasteur

    (c) Schwann

    (d) Lister

Ans.    (b)

Sol.    Louis Pasteur designed 'swan necked flasks' experiment. Before Louis Pasteur, it was widely believed that non-living things could produce living organisms. This was the idea of spontaneous generation, first proposed by Aristotle. From this swan necked flasks experiment, Pasteur proved that not only macroorganisms but the microorganisms also cannot arise spontaneously. All oragnisms arise from pre-existing organisms.

89. Consider the following three groups and choose the correct match.

    Vitamin    Cofactor

    P1. Riboflavin    Q1. TPP

    P2. Thiamine    Q2. CoA

    P3. Nicotinamide    Q3. Biocytin

    P4. Pantothenate    Q4. NADA

    P5. Cobalamine    Q5. FAD

    P6. Biotin


    R1. Pyruvate

    R2. Succinate dehydrogenase

    R3. Glucose 6-phosphate dehydrogenease

    R4. Pyruvate decarboxylase

    R5. Succinate thiokinase

    R6. Hexokinase

    (a) P1–Q5–R2, P2–Q1–R4, P3–Q4–R3, P4–Q2–R5, P6–Q3–R1

    (b) P1–Q5–R2, P4–Q1–R6, P3–Q4–R3, P4–Q2–R5, P6–Q3–R1

    (c) P1–Q5–R2, P2–Q1–R4, P1–Q4–R6, P4–Q2–R5, P6–Q3–R1

    (d) P1–Q5–R2, P2–Q1–R4, P3–Q4–R3, P5–Q2–R6, P6–Q3–R1

Ans.    (a)

Sol.    The correct match groups are as follows :


90. For a reversible reaction S P, the equilibrium concentration of P is 100 times that of S. The equilibrium constant for this reaction in the presence of an enzymes catalyst will be

    (a) 0.01

    (b) 1

    (c) 100

    (d) 1000

Ans.    (c)

Sol.    A reversible reaction is a chemical reaction that results in an equilibrium mixture of reactants and products.



91. Which of the following statements are TRUE with respect to membrane fluidity of a phospholipid bilayer?

    P.    Incresasing proportion of long chain fatty acids decrease membrane fluidity

    Q.    Increase in cholestreol content incresases membrane fluidity

    R.    Increasing proportion of cis unsaturated faty acids increase membrane fluidity

    S.    Increasing proportion of trans unsaturated fatty acids incresases membrane fluidity

    (a) P and Q

    (b) P and R

    (c) R and S

    (d) Q and S

Ans.    (b)

Sol.    Fluidity of lipid bilayer depends on chain length and the degree of saturation of its component fatty acid residues. Lipids with short or unsaturated fatty acyl chains undergoes the phase transition at lower temperatures than lipids with long or saturated chains. Short chains have less surface area with which they form van der Waal's interactions with each other.

    Hence, a shorter chain length reduces the tendency of the hydrocarbon tails to interact with one another. Similarly, fatty acyl chains with cis configuration form less van der Waal's interactions with each other and thus increases the fluidity.

92. Choose the correct set of words denoted by (P), (Q) and (R) for the following statement. Pheophytin is a (P) molecule in which the central atom (Q) has been replaced by two atom of (R).

    (a) P-plstocyanin, Q-copper, R-hydrogen

    (b) P-plastocyanin, Q-zinc, R-oxygenQR

    (c) P-chlorophyll, Q-magnesium, R-oxygen

    (d) P-chlorophyll, Q-magnesium, R-hydrogen

Ans.    (d)

Sol.    Pheophytin is a chlorophyll molecule lacking a central Mg2+ ion which serves as the first electron carrier intermediate in the electron transfer pathway of photosystem II in plants.

93. A wheel of radius R rolls on a horizontal surface without slipping. If the center of mass moves with a speed v, the instantaneous speed at the highest point on the wheel is

    (a) v/2

    (b) 0

    (c) v

    (d) 2v

Ans.    (d)

Sol.    In pure rolling, the velocity of the contact point is zero. The velocity of centre of mass is vcm = and that of the topmost point is vtop = = 2vcm.

94. A quantity Z = XY is to be estimated by measuring X and Y. If the absolute errors in the measurement of X and Y are and , respectively, then the absolute error in Z is





Ans.    (a)

Sol.    Since, the equation is in product form, i.e. Z = XY


    Absolute error in Z is given by


95. Which of the following in NOT CORRECT is electromagnetic waves propagating in vacuum?

    (a) Electromagnetic waves with different wavelength travel with the same speed

    (b) The electric and the magnetic fields are perpendicular to each other and perpendicular to the direction of propagation

    (c) The magnetic field is along the direction of propagation

    (d) Electromagnetic waves carry energy

Ans.    (c)

Sol.    For an electromagnetic wave, magnetic field vector, electric field vector and direction of propagation of electromagnetic wave are mutually perpendicular.

96. A charged particle, with an initial velocity in the xy plane, is subjected to a uniform magnetic field along the z-axis. Which of the following is the CORRECT statement?

    (a) The particle will experience a force along the z-axis

    (b) The speed of the particle remains constant

    (c) Acceleration of the particle is zero

    (d) The particle moves in a helical path

Ans.    (b)

Sol.    As the magentic force on a particle is perpendicular to the velocity, it doesn't do any work on the particle. Hence, the kinetic energy or the speed of the particle doesn't change due to magnetic force.


97.    A uniform electric field E0i^, where i^ is the unit vector along the x-axis, exists in a region. A cube of side 'a' is kept with one of its corners coinciding with the origin and three edges along the x, y , z axes. Which one of the following is the CORRECT statement?

    (a) The total charge inside the cube is non zero

    (b) The flux of electric field through each face is zero

    (c) The flux of electric field through all the faces are equal but non zero

    (d) The net flux of electric field through all the faces is zero

Ans.    (d)

Sol.    According to question = E =

    The net electric flux of electric field associated with a closed surface will always be zero.


98. The greater efficiency of detergents over soaps as cleaning agents in aqueous media is best described by


    (a) graph S

    (b) graph R

    (c) graph Q

    (d) graph P

Ans.    (d)


    When the concentration is less, 6 will be more but as the concentration increases, 6 drops. The concentration of 6 is much lower in detergent then in soap which describes the surface tension of water decreases with concentration.

99. The pairs of structures P, Q and R are shown as Fischer projection. P, Q and R, respectively, represent


    (a) the same molecules, disastereomers and enantiomers

    (b) the same molecules, enantiomers and disastereomers

    (c) the same molecules, the same molecules and enantiomers

    (d) enantiomers, diastereomers and enantiomers

Ans.    (a)



    They have same configuration around one chiral carbon and different configuration around the other chiral carbon. These represent a pair of diastereoisomers.

    They are non-superimpossible mirror images. They have different configuration around the both chiral carbon. These represent a pair of enantiomers.

100. The product obtained by the nitration, at low temperature, of meta xylene is





Ans.    (a)

Sol.    –CH3 group is an electron releasing group and hence, 'O' and 'P' directing.