100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS Page 1906 InterviewSolution

95251.	Element with electronic configuration as `[Ar]^(18)3d^(5)4s^(2)` is placed in `:`A. `1^(st)`group, `s-` blockB. `2^(nd) ` group, `s-` blockC. `5^(th) ` group, `d-` blockD. `7^(th) ` group, `d-` block
Answer» Correct Answer - 4 Last electrons enters in `d-` orbital. So it belong to `d-` block. For `d-` block elements the group number is equal to the number of valence electrons `+` number of electrons in `(n-1)d` subshell `i.e., 2+5=7^(th)` group.

Discussion

95252.	A metal has electronic configuration `[Ar]^(18)3d^7 4s^2`.On the basis of this electronic configuration find out the group number.
Answer» Correct Answer - 9 Last electron enter in d-subshell so it belongs to d-block elements. For d-block elements group number =no of electrons in (n-1)d subshell+number of electrons in valence shell (i.e. ns) =7+2=9

Discussion

95253.	A short circuited stub is shunt connected to a transmission line as shown in figure. If z0 = 50Ω, the admittance Y seen at the junction of the stub and transmission line is(a) (0.01 – j 0.02) mho (b) (0.02 – j 0.01) mho (c) (0.04 + j 0.02) mho (d) (0.02 + j 0) mho
Answer» Correct option (a) (0.01 – j 0.02) mho Explanation: For both Transmission line and stub, z₀ = 50Ω For λ/2 line input impedance z_il= z_L z_il = zL = 100Ω y_il = 0.01mho For short circuited stub input impedance z_i2 = jz₀tan(βl) = jz0tan(2πλ/λ8) = jz₀tan(π/4) = jz₀ = j50 Y_i2 = 1/j50 = - j 0.02 Y = Y_il+ Y_i2 = (0.01 - j 0.02) mho

95253.

A short circuited stub is shunt connected to a transmission line as shown in figure. If z0 = 50Ω, the admittance Y seen at the junction of the stub and transmission line is(a) (0.01 – j 0.02) mho (b) (0.02 – j 0.01) mho (c) (0.04 + j 0.02) mho (d) (0.02 + j 0) mho

Answer»

Correct option (a) (0.01 – j 0.02) mho

Explanation:

For both Transmission line and stub, z₀ = 50Ω

For λ/2 line input impedance z_il= z_L

z_il = zL = 100Ω

y_il = 0.01mho

For short circuited stub input impedance

z_i2 = jz₀tan(βl)

= jz0tan(2πλ/λ8)

= jz₀tan(π/4)

= jz₀ = j50

Y_i2 = 1/j50 = - j 0.02

Y = Y_il+ Y_i2

= (0.01 - j 0.02) mho

Discussion

95254.	A transmission line of characteristic impedance 50Ω is terminated by a 50Ω load. When excited by a sinusoidal voltage source at 10 GHz, the phase difference between two points spaced 2 mm apart on the line is found to be π/4 radians. The phase velocity of the wave along the line is (a) 0.8 × 108m/s (b) 1.2 × 108m/s (c) 1.6 × 108 m/s (d) 3 × 108 m/s
Answer» Correct option (c) 1.6 × 10⁸ m/s Explanation: Phase difference βl = 2π/λ path difference π/4 = 2π/λ(2 x 10^-3) λ = 8 x 2 x 10^-3 = 16 x 10^-3m Given, f = 10GHz The phase velocity of the wave: Vp = fλ =10 x 10⁹ x 16 x 10^-3 = 160 x 10⁶m/sec = 1.6 x 10⁸m/sec

95254.

A transmission line of characteristic impedance 50Ω is terminated by a 50Ω load. When excited by a sinusoidal voltage source at 10 GHz, the phase difference between two points spaced 2 mm apart on the line is found to be π/4 radians. The phase velocity of the wave along the line is (a) 0.8 × 108m/s (b) 1.2 × 108m/s (c) 1.6 × 108 m/s (d) 3 × 108 m/s

Answer»

Correct option (c) 1.6 × 10⁸ m/s

Explanation:

Phase difference βl = 2π/λ path difference

π/4 = 2π/λ(2 x 10^-3)

λ = 8 x 2 x 10^-3

= 16 x 10^-3m

Given, f = 10GHz

The phase velocity of the wave:

Vp = fλ

=10 x 10⁹ x 16 x 10^-3

= 160 x 10⁶m/sec

= 1.6 x 10⁸m/sec

Discussion

95255.	A charge 'q' move in region where electric field 'E' and magnetic field 'B' both exist, then force on it is :-
Answer» Correct answer is (C)

Discussion

95256.	Coefficient of friction formulae
Answer» coefficient of friction, ratio of the frictional force resisting the motion of two surfaces in contact to the normal force pressing the two surfaces together. It is usually symbolized by the Greek letter mu (μ). Mathematically, μ = F/N, where F is the frictional force and N is the normal force.

Discussion

95257.	An instrumentation amplifier has a high a. Output impedance b. Power gain c. CMRR d. Supply voltage
Answer» The correct answer is: (c) CMRR

Discussion

95258.	Compared to the ac resistance of the emitter diode, the feedback resistance of a swamped amplifier should be a. Small b. Equal c. Large d. Zero
Answer» The correct answer is: (c) Large

Discussion

95259.	The emitter of a swamped amplifier a. Is grounded b. Has no de voltage c. Has an ac voltage d. Has no ac voltage
Answer» (c) Has an ac voltage

Discussion

95260.	Compared to a CE stage, a swamped amplifier has an input impedance that is a. Smaller b. Equal c. Larger d. Zero
Answer» The correct answer is: (c) Larger

Discussion

95261.	In a swamped amplifier, the effects of the emitter diode become a. Important to voltage gain b. Critical to input impedance c. Significant to the analysis d. Unimportant
Answer» (d) Unimportant

Discussion

95262.	Limits of poissons ratio
Answer» Generally, the value of Poisson's ratio ranges from -1 to +0.5 . Normally, the value of σ is between 0.2 to 0.4 [For Steel, σ=0.19, Copper σ=0.32 and for Brass it is σ=0.26.]

Discussion

95263.	Why HF acid is stored in wax coated glass bottle ?
Answer» This is because HF does not attack wax but reacts with glass.It dissolves SiO₂ present in glass forming hydrofluorosilicic acid. SiO₂ + 6HF → H₂SiF₆ + 2H₂O [Hint : HF is corrosive, hence HF attacks glass surface.]

95263.

Why HF acid is stored in wax coated glass bottle ?

Answer»

This is because HF does not attack wax but reacts with glass.It dissolves SiO₂ present in glass forming hydrofluorosilicic acid.

SiO₂ + 6HF → H₂SiF₆ + 2H₂O

[Hint : HF is corrosive, hence HF attacks glass surface.]

Discussion

95264.	Bond dissociation enthalpy of F2 is less than that of Cl2. Explain why ?
Answer» [Hint : F₂ is having higher electron-electron repulsion due to its smaller size, as compared to Cl₂.]

Discussion

95265.	The ionic charges of manganate and permanganate ion are respectivelyA. `-2,-2`B. `-1,-2`C. `-2,-1`D. `-1,-1`
Answer» Correct Answer - C Chemical formula for manganate and permanganate ion respectively are `Mn_(4)^(2-)` and `mnO_(4)^(-)`. Thus, the ionic charge of manganate and permangante ion are respectively -2 and -1.

Discussion

95266.	(i). State Kohlraush law of independent migration of ions (ii). Write Nernst’s equation for the following cell :Zn (s) I Zn2+ (aq) II Cu2+(aq) I Cu (s) (iii).Calculate ∆G0 for the following reaction:Mg (s) + Cu2+(aq) → Mg2+(aq) + Cu (s)[Given E0cell = +2.71 V, 1F = 96500 C/Mol]
Answer» (i) Kohlrausch's law of independent migration of ions states that the limiting molar conductivity of an electrolyte can be represented as the sum individual contributions of its cations & anions. (ii) E_cell= E⁰_cell^-\(\frac{0.0591}{2}log\frac{[zn^{2+}]}{[cu^{2+}]}\) (iii) ∆G = -nFE⁰ = - 2× 96500 × 2.71 = 523030 J

95266.

(i). State Kohlraush law of independent migration of ions (ii). Write Nernst’s equation for the following cell :Zn (s) I Zn2+ (aq) II Cu2+(aq) I Cu (s) (iii).Calculate ∆G0 for the following reaction:Mg (s) + Cu2+(aq) → Mg2+(aq) + Cu (s)[Given E0cell = +2.71 V, 1F = 96500 C/Mol]

Answer»

(i) Kohlrausch's law of independent migration of ions states that the limiting molar conductivity of an electrolyte can be represented as the sum individual contributions of its cations & anions.

(ii) E_cell= E⁰_cell^-\(\frac{0.0591}{2}log\frac{[zn^{2+}]}{[cu^{2+}]}\)

(iii) ∆G = -nFE⁰

= - 2× 96500 × 2.71

= 523030 J

Discussion

95267.	Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory: (i). `[Fe(CN)_6]^(4-)` (ii). `[FeF_6]^(3-)` (iii). `[Co(C_2O_4)_3]^(3-)` (iv). `[CoF_6]^(3-)`
Answer» (i). `[Fe(CN)_6]^(4-)` `-d^(2)sp^(3)` octahedral, diamagnetic (ii). `[FeF_6]^(3-)` `-sp^(3)d^(2)`, octahedral, paramagnetic (iii). `[Co(C_2O_4)_3]^(3-)` `-d^(2)sp^(3)` octahedral diamagnetic `(C_2O_4^(2-)` is a stong ligand) (iv). `[CoF_6]^(3-)` `-sp^(3)d^(2)`, octahedral, paramagnetic.

Discussion

95268.	Assertion: A solution of `[Ni(H_(2)O)_(6)]^(2+)` is green but a solution of `[Ni(CN)_(4)]^(2-)` is colourless Reason: `[Ni(CN)_(4)]^(2-)` is square planar complex .
Answer» In `[Ni(H_2O)_6]^(2+)`, Ni is in `+2` state with the configuration `3d^(8)`, i.e., it has two unpaired electrons which do not pair up in the presence of the weak `H_2O` ligand. Hence, it is coloured. The d-d transition, absorbs red light and the complementary light emitted is green. In case of `[Ni(CN)_4]^(2-)`, Ni is again in `+2` state with the configuration `3d^(8)` but in presence of the strong `CN^(ɵ)` ligand, the two unpaired electrons in the `3d` orbitals pair up. Thus, there is no unpaired electron present. Hence, it is colourless.

Discussion

95269.	What are `t_(2g)` and `e_(g)` orbitals?
Answer» In a free stransition metal ion, the five d-orbita,ls are degenerate. When it forms a complex, the degeneracy is split. Three d-orbitals having lower energy are called `t_(2g)` orbitals and the remaining two d-orbitals of higher energy are called `e_(g)` orbitals.

Discussion

95270.	(i). What is crystal field splitting energy? How does the magnitude of ∆0 decide the actual configuration of d-orbital in a coordination entity? (ii). Co (II) is very stable in aqueous solutions but get easily oxidised in presence of strong ligands. Why?
Answer» (i) When ligands approach a transition metal ion, the d-orbitals split into two sets, one with lower energy and the other with higher energy. The difference of energy between the two sets of orbitals is called crystal field splitting energy (△₀) for octahedral field. If △₀<P (pairing energy) the 4^th electron enters one of the egeg orbitals giving the configuration t³_2ge_g¹ Thus forming high spin complexes. Such ligands for which △₀<P are called weak field ligands. If △₀<P the 4^th electron pairs up in one of the t_2g orbitals giving the configuration t_2g4e_g0 thereby forming low spin complexes. Such ligands for which △₀>P are called strong field ligands. (ii) Water is weak ligand and cannot paired up the electrons, but strong ligand have a tendency to pair up the electrons, so CO²⁺ is stable in aqueous solution but in the presence of strong ligands it is easily oxidised.

95270.

(i). What is crystal field splitting energy? How does the magnitude of ∆0 decide the actual configuration of d-orbital in a coordination entity? (ii). Co (II) is very stable in aqueous solutions but get easily oxidised in presence of strong ligands. Why?

Answer»

(i) When ligands approach a transition metal ion, the d-orbitals split into two sets, one with lower energy and the other with higher energy. The difference of energy between the two sets of orbitals is called crystal field splitting energy (△₀) for octahedral field.

If △₀<P (pairing energy) the 4^th electron enters one of the egeg orbitals giving the configuration t³_2ge_g¹ Thus forming high spin complexes. Such ligands for which △₀<P are called weak field ligands.

If △₀<P the 4^th electron pairs up in one of the t_2g orbitals giving the configuration t_2g4e_g0 thereby forming low spin complexes. Such ligands for which △₀>P are called strong field ligands.

(ii) Water is weak ligand and cannot paired up the electrons, but strong ligand have a tendency to pair up the electrons, so CO²⁺ is stable in aqueous solution but in the presence of strong ligands it is easily oxidised.

Discussion

95271.	From where dose the electrical energy come in a generator ?
Answer» A generator is a device that transforms mechanical energy into electrical energy. A generator do not create electricity instead it uses the mechanical supplied to it to force the movement of electric charge present in the wire of its windings though an external electric circuit. This Flow of electrons constitutes the output electric current supplied by the generator.

Discussion

95272.	If \(f(n) =\begin{cases}\frac{x^2 - 4}{x - 2},\,x \neq 2 \\k,x=2\end{cases}\) is continuous at x = 2, then k =f(n) = {(x2 - 4)/(x - 2), x ≠ 2, k, x = 2 is continuous at x = 2, then k =(a) 2 (b) 4 (c) 6 (d) 3
Answer» Correct answer is (b) 4

95272.

If \(f(n) =\begin{cases}\frac{x^2 - 4}{x - 2},\,x \neq 2 \\k,x=2\end{cases}\) is continuous at x = 2, then k =f(n) = {(x2 - 4)/(x - 2), x ≠ 2, k, x = 2 is continuous at x = 2, then k =(a) 2 (b) 4 (c) 6 (d) 3

Answer»

Correct answer is

(b) 4

Discussion

95273.	the power house of the cell is
Answer» The power house of the cell is Mitochondria.

Discussion

95274.	what is the general formula for monosaccharides
Answer» Monosaccharides are the simplest carbohydrates; they conform to the general chemical formula (CH₂O)_x and are termed simple sugars.

Discussion

95275.	the general formula of monosaccharides is
Answer» Monosaccharides are the simplest unit of carbohydrates. They're composed of carbon, hydrogen, and oxygen atoms, and they cannot be broken down further since they are already in their simplest form. Their general formula is (CH₂O)_n, where n is any number equal or greater than 3

Discussion

95276.	`0.01` mol of a gaseous compound `C_(2)H_(2)O_(x)` was treated with `224 mL` of `O_(2)` at `STP`. After combustion the total volume of the gases is `560 mL` at `STP`. On treatment with `KOH` solution the volume decreases to `112 mL`. The volue of `x` is:A. `4`B. `2`C. `3`D. None of these
Answer» Correct Answer - A vol.of `CO_(2) +` vol. of remaining oxygen `= 560 ml`. or vol. of remaining oxygen `= 112 mL` `C_(2)H_(2)O_(x)(g) + ((5-x)/(2))O_(2)(g)overset(Delta)rarr2CO_(2)(g) + H_(2)O(l)` `{:(224 mL,224 mL,0,-),(0,112 mL,448 mL,-):}` for used mole of a`O_(2)`: `((5-x)/(2)) xx 0.1 = (0.01)/(2)` or `x = 4`

Discussion

95277.	`M(OH)_(X)` has `K_(SP) 4xx10^(-12)` and solubility `10^(-4) M`. The value of `x` is:A. `1`B. `2`C. `3`D. `4`
Answer» Correct Answer - B `{:(M(OH)_(x)hArr,M^(+x)+,xOH^(-)),(,s,xs):}` `K_(sp) = S.(XS)^(x) = X^(x).S^((x+1)) = 4 xx 10^(-12)` given `() S = 10^(-4)` `:. x = 2`

Discussion

95278.	If the vectors transmit the infection mechanically they are called a. Biological vectors b. Mechanical vectors c. Biological reservoir d. Both a and c
Answer» b. Mechanical vectors

Discussion

95279.	Small pox vaccine was first discovered by a. Robert Koch b. Louis Pasteur c. Lister d. Edward Jenner
Answer» Small pox vaccine was first discovered by Edward Jenner.

Discussion

95280.	Which is the following enzyme acts as a spreading factor? a. Hyaluronidase b. Coagulase c. Catalase d. DNase
Answer» a. Hyaluronidase

Discussion

95281.	B.anthracis was isolated by a. Louis Pasteur b. Robert Koch c. Antonyvon Leewenhok d. None of these
Answer» b. Robert Koch

Discussion

95282.	The principle light- trapping pigment molecule in plants, Algae, and cyanobacteria is a. Chlorophyll a b. Chlorophyll b c. Porphyrin d. Rhodapsin
Answer» a. Chlorophyll a

Discussion

95283.	Reverse isolation would be appropriate for a. a patient with tuberculosis b. a patient who has had minor surgery c. a patient with glaucoma d. a patient with leukemia
Answer» a. a patient with tuberculosis

Discussion

95284.	The capacity of a given strain of microbial species to produce disease is known as a. Pathogen b. Virulence c. Infection d. None of these
Answer» The capacity of a given strain of microbial species to produce disease is known as Virulence.

Discussion

95285.	Disease that effects many people at different countries is termed as a. Sporadic b. Pandemic c. Epidemic d. Endemic
Answer» Disease that effects many people at different countries is termed as Sporadic.

Discussion

95286.	The ability of a pathogen to spread in this host tissues after establishing the infection is known as a. Adhesion b. Invasiveness c. Toxigenicity d. None of these
Answer» b. Invasiveness

Discussion

95287.	Tuberculosis is a a. Water borne disease b. Air borne disease c. Food borne disease d. Atthropod borne disease
Answer» Tuberculosis is a Airborne disease.

Discussion

95288.	The ability of Microscope to distinguish two objects into two separate objects, is called. a. Resolving power b. Wave length c. N.A. d. None of these
Answer» a. Resolving power

Discussion

95289.	The resolution power of the compound microscope is a. 0.2 micron b. 0.2 millimeter c. 0.2 Angstrom units d. 0.2 centimeter
Answer» The resolution power of the compound micro scope is 0.2 micron.

Discussion

95290.	Light gathering capacity of Microscope is called a. Numerical aperture b. Angular aperture c. Both a and b d. None of these
Answer» Light gathering capacity of Microscope is called Numerical aperture.

Discussion

95291.	Streptococcus pneumoniae was isolated by a. Robert Koch b. Edward Jenner c. Antony von Leewenhock d. Louis Pasteur
Answer» Streptococcus pneumoniae was isolated by Louis Pasteur.

Discussion

95292.	L – forms are discovered by a. Klein Berger b. Louis Pasteur c. Robert Koch d. Antony von Leeuwenhock
Answer» L – forms are discovered by Klein Berger.

Discussion

95293.	Staphylococcus aureus was isolated by a. Rosenbach b. Louis Pasteur c. Passet d. Sir Alexander Ogston
Answer» Staphylococcus aureus was isolated by Louis Pasteur.

Discussion

95294.	Phagocytic phenomenon was discovered by a. Louis Pasteur b. Alexander Fleming c. Metchnikoff d. Robert Koch
Answer» Phagocytic phenomenon was discovered by Metchnikoff.

Discussion

95295.	Limit of resolution of compound microscope is a. 0.018 Åb. 0.1 mm c. 5 ìm d. 1 mm
Answer» Limit of resolution of compound microscope is 0.1 mm.

Discussion

95296.	Who perfected a magnetic lens in 1927 a. Gabor b. Broglie c. Busch d. None of these
Answer» Who perfected a magnetic lens in 1927 Gabor.

Discussion

95297.	Pseudomonas aeruginosa was first named a. Schroeter and Gessard b. Robert Koch c. Louis Pasteur d. Edward Jenner
Answer» Pseudomonas aeruginosa was first named Schroeter and Gossard.

Discussion

95298.	Mycobacterium lepree was discovered by a. Robert Koch b. Hansen c. Edward Jenner d. Louis Pasteur
Answer» Mycobacterium leper was discovered by Hansen.

Discussion

95299.	Source of light in fluorescence microscopy is from a. Mercury lamp b. Sunlight c. Both a and b d. None of these
Answer» Source of light in fluorescence microscopy is from Mercury lamp.

Discussion

95300.	Rh factor of the blood was discovered by scientist a. Louis Pasteur b. Landsteiner and Weiner c. Janskey d. Moss e. None of these
Answer» b. Landsteiner and Weiner

Discussion

Explore topic-wise InterviewSolutions in .

Element with electronic configuration as `[Ar]^(18)3d^(5)4s^(2)` is placed in `:`A. `1^(st)`group, `s-` blockB. `2^(nd) ` group, `s-` blockC. `5^(th) ` group, `d-` blockD. `7^(th) ` group, `d-` block

A metal has electronic configuration `[Ar]^(18)3d^7 4s^2`.On the basis of this electronic configuration find out the group number.

A short circuited stub is shunt connected to a transmission line as shown in figure. If z0 = 50Ω, the admittance Y seen at the junction of the stub and transmission line is(a) (0.01 – j 0.02) mho (b) (0.02 – j 0.01) mho (c) (0.04 + j 0.02) mho (d) (0.02 + j 0) mho

A charge 'q' move in region where electric field 'E' and magnetic field 'B' both exist, then force on it is :-

Coefficient of friction formulae

An instrumentation amplifier has a high a. Output impedance b. Power gain c. CMRR d. Supply voltage

Compared to the ac resistance of the emitter diode, the feedback resistance of a swamped amplifier should be a. Small b. Equal c. Large d. Zero

The emitter of a swamped amplifier a. Is grounded b. Has no de voltage c. Has an ac voltage d. Has no ac voltage

Compared to a CE stage, a swamped amplifier has an input impedance that is a. Smaller b. Equal c. Larger d. Zero

In a swamped amplifier, the effects of the emitter diode become a. Important to voltage gain b. Critical to input impedance c. Significant to the analysis d. Unimportant

Limits of poissons ratio

Why HF acid is stored in wax coated glass bottle ?

Bond dissociation enthalpy of F2 is less than that of Cl2. Explain why ?

The ionic charges of manganate and permanganate ion are respectivelyA. `-2,-2`B. `-1,-2`C. `-2,-1`D. `-1,-1`

(i). State Kohlraush law of independent migration of ions (ii). Write Nernst’s equation for the following cell :Zn (s) I Zn2+ (aq) II Cu2+(aq) I Cu (s) (iii).Calculate ∆G0 for the following reaction:Mg (s) + Cu2+(aq) → Mg2+(aq) + Cu (s)[Given E0cell = +2.71 V, 1F = 96500 C/Mol]

Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory: (i). `[Fe(CN)_6]^(4-)` (ii). `[FeF_6]^(3-)` (iii). `[Co(C_2O_4)_3]^(3-)` (iv). `[CoF_6]^(3-)`

Assertion: A solution of `[Ni(H_(2)O)_(6)]^(2+)` is green but a solution of `[Ni(CN)_(4)]^(2-)` is colourless Reason: `[Ni(CN)_(4)]^(2-)` is square planar complex .

What are `t_(2g)` and `e_(g)` orbitals?

(i). What is crystal field splitting energy? How does the magnitude of ∆0 decide the actual configuration of d-orbital in a coordination entity? (ii). Co (II) is very stable in aqueous solutions but get easily oxidised in presence of strong ligands. Why?

From where dose the electrical energy come in a generator ?

If \(f(n) =\begin{cases}\frac{x^2 - 4}{x - 2},\,x \neq 2 \\k,x=2\end{cases}\) is continuous at x = 2, then k =f(n) = {(x2 - 4)/(x - 2), x ≠ 2, k, x = 2 is continuous at x = 2, then k =(a) 2 (b) 4 (c) 6 (d) 3

the power house of the cell is

what is the general formula for monosaccharides

the general formula of monosaccharides is

`0.01` mol of a gaseous compound `C_(2)H_(2)O_(x)` was treated with `224 mL` of `O_(2)` at `STP`. After combustion the total volume of the gases is `560 mL` at `STP`. On treatment with `KOH` solution the volume decreases to `112 mL`. The volue of `x` is:A. `4`B. `2`C. `3`D. None of these

`M(OH)_(X)` has `K_(SP) 4xx10^(-12)` and solubility `10^(-4) M`. The value of `x` is:A. `1`B. `2`C. `3`D. `4`

If the vectors transmit the infection mechanically they are called a. Biological vectors b. Mechanical vectors c. Biological reservoir d. Both a and c

Small pox vaccine was first discovered by a. Robert Koch b. Louis Pasteur c. Lister d. Edward Jenner

Which is the following enzyme acts as a spreading factor? a. Hyaluronidase b. Coagulase c. Catalase d. DNase

B.anthracis was isolated by a. Louis Pasteur b. Robert Koch c. Antonyvon Leewenhok d. None of these

The principle light- trapping pigment molecule in plants, Algae, and cyanobacteria is a. Chlorophyll a b. Chlorophyll b c. Porphyrin d. Rhodapsin

Reverse isolation would be appropriate for a. a patient with tuberculosis b. a patient who has had minor surgery c. a patient with glaucoma d. a patient with leukemia

The capacity of a given strain of microbial species to produce disease is known as a. Pathogen b. Virulence c. Infection d. None of these

Disease that effects many people at different countries is termed as a. Sporadic b. Pandemic c. Epidemic d. Endemic

The ability of a pathogen to spread in this host tissues after establishing the infection is known as a. Adhesion b. Invasiveness c. Toxigenicity d. None of these

Tuberculosis is a a. Water borne disease b. Air borne disease c. Food borne disease d. Atthropod borne disease

The ability of Microscope to distinguish two objects into two separate objects, is called. a. Resolving power b. Wave length c. N.A. d. None of these

The resolution power of the compound microscope is a. 0.2 micron b. 0.2 millimeter c. 0.2 Angstrom units d. 0.2 centimeter

Light gathering capacity of Microscope is called a. Numerical aperture b. Angular aperture c. Both a and b d. None of these

Streptococcus pneumoniae was isolated by a. Robert Koch b. Edward Jenner c. Antony von Leewenhock d. Louis Pasteur

L – forms are discovered by a. Klein Berger b. Louis Pasteur c. Robert Koch d. Antony von Leeuwenhock

Staphylococcus aureus was isolated by a. Rosenbach b. Louis Pasteur c. Passet d. Sir Alexander Ogston

Phagocytic phenomenon was discovered by a. Louis Pasteur b. Alexander Fleming c. Metchnikoff d. Robert Koch

Limit of resolution of compound microscope is a. 0.018 Åb. 0.1 mm c. 5 ìm d. 1 mm

Who perfected a magnetic lens in 1927 a. Gabor b. Broglie c. Busch d. None of these

Pseudomonas aeruginosa was first named a. Schroeter and Gessard b. Robert Koch c. Louis Pasteur d. Edward Jenner

Mycobacterium lepree was discovered by a. Robert Koch b. Hansen c. Edward Jenner d. Louis Pasteur

Source of light in fluorescence microscopy is from a. Mercury lamp b. Sunlight c. Both a and b d. None of these

Rh factor of the blood was discovered by scientist a. Louis Pasteur b. Landsteiner and Weiner c. Janskey d. Moss e. None of these