A and C are concentric conducting spherical shells of radius a and c respectively. A is surrounded by a concentric dielectric medium of inner radius a, outer radius b and dielectric constant K as shown in the figure. If sphere A is given a charge Q, the potential at the outer surface of the dielectric is.
Charge on outer surface of C = - charge on inner surface of C
Hence potential at B due to charge on conductor C = 0 charge on outer surface of dielectric = - charge on inner surface of dielectric
∴ Potential at B due to charge on dielectric = 0
Potential at B due to charge on A
∴ net potential at B
In the given circuit the rms value of voltage across the capacitor C, inductor L and resister R1 are 12V, 10V and 5V respectively. Then the peak voltage across R2 is
In parallel combination each branch should have equal potential but in AC circuit in a branch
Two identical capacitors C1 and C2 are connected in series with a battery. They are fully charged. Now a dielectric slab is inserted between the plates of C2. The potential difference across C1 will :
Potential difference across ‘C1’
When dielectric is inserted, C2 will increase. So V1 will increase.
A solenoid having an iron core has its terminals connected across an ideal DC source. If the iron core is removed, the current flowing through the solenoid
Current will have to increase in order to oppose the cause (decrease in flux).
An electric field ‘E’ whose direction is radially outward varies as distance from origin ‘r’ as shown in the graph. E is taken as positive if its direction is away from the origin. Then the work done by electric field on a 2 C charge if it is taken from (1, 1, 0) to (3, 0, 0) is
work done on 2C charge
[∵ r for (1, 1, 0) = √2 & r for (3,0, 0) = 3]
= 2 x (area of E-r graph from r = √2 m to r = 3)
A negative test charge is moving near a long straight current carrying wire. A force will act on the test charge in a direction parallel to the direction of the current, if the motion of the charge is in a direction:
The magnetic force shall act on the negatively charged test particle in direction parallel to current carrying wire, if the velocity of the negative charge is as shown.
Potential difference between centre and the surface of sphere of radius R and having uniform volume charge density ρ within it will be :
The magnetic field at the origin due to the current flowing in the wire is
There exists a uniform electric field in the space as shown. Four points A, B, C and D are marked which are equidistant from the origin. If VA, VB, VC and VD are their potentials respectively, then
Four lines, perpendicular to lines of electric field and passing through A, B, C and D are drawn. These are equipotential lines. As potential decreases in the direction of electric field, therefore VA > VB > VD > VC
A current carrying wire AB of the length L is turned along a circle, as shown in figure. The magnetic field at the centre O.
Let r be the radius of new arc
∴ r(2π - θ) = L or
∴ Magnetic field at centre O is
Total electric force on an electric dipole placed in an electric field of a point charge is:
Electric field of a point charge is non-uniform hence net force can never be zero.
The magnitude of magnetic field at O (centre of the circular part) due to the current carrying coil as shown is :
Magnetic field due to circular segment
Due to one straight wire segment (sin 45° + sin 0°) =
Eight point charges (can be assumed as small spheres uniformly charged and their centres at the corner of the cube) having values q each are fixed at vertices of a cube. The electric flux through square surface ABCD of the cube is
Net charge enclosed inside the cube
net electric flux through cube
∵ charges are symmetrically placed so the flux through each face
A charged particle is moving in the region around a long current carrying wire. Due to wire, it may experience:
The charge particle will experience magnetic as will as gravitational force due to wire. Gravitation force is very small.
The figure shows a charge q placed inside a cavity in an uncharged conductor.
Now if an external electric field is switched on :
The distribution of charge on the outer surface, depends only on the charges outside, and it distributes itself such that the net, electric field inside the outer surface due to the charge on outer surface and all the outer charges is zero. Similarly the distribution of charge on the inner surface, depends only on the charges inside the inner surface, and it distributes itself such that the net, electric field outside the inner surface due to the charge on inner surface and all the inner charges is zero.
Also the force on charge inside the cavity is due to the charge on the inner surface. Hence only induced charge on outer surface will redistribute
Two infinite sheets carrying current in same direction (of equal current per unit length K) are separated by a distance ‘d’. A proton is released from a point between the plates with a velocity parallel to the sheets but perpendicular to the direction of current in the sheets. Then the path of the proton is
As the net force acting on the proton is zero, it will move on straight line
Three large parallel charged conducting plates are placed at a small distance d apart from each other. The surface charge density on the face B of the conductor is +5.0 εo coulomb/m2 and that on face E is + 6.0 εo columb/m2. Charges on other faces are not known. The electric field intensity at P (shown in the figure) between the plates is
From Gauss theorem surface charge density on C face is – σ .
A horizontal metallic rod of mass 'm' and length ' l ' is supported by two vertical identical springs of spring constant 'K' each and natural length l0. A current ' i ' is flowing in the rod in the direction shown. If the rod is in equilibrium then the length of each spring in this state is:
The force on the rod due to magnetic field and gravity is i ℓ B–mg (upwards)
Hence the extension in the springs is
(Note that effective spring constant is 2k)
Therefore the length of the spring is
A 150 m long metal wire connects points A and B. The electric potential at point B is 50 V less than that at point A. If the conductivity of the metal is 60 × 106 mho/m, then magnitude of the current density in the wire is equal to :
(or as J = σ E)
using values j = 20 x 106 A/m2
A toroid of mean radius ' a ' , cross section radius ' r ' and total number of turns N. It carries a current ' i '. The torque experienced by the toroid if a uniform magnetic field of strength B is applied :
The resultant magnetic dipole moment of toroid is zero. of small parts of toroid turn along a circle and hence there resultant is zero.
∴ Torque acting on it is zero.
The equivalent resistance between A and B will be (in ?)
As C & D are at same potential by symmetry of circuit °
It is balanced wheat-stone bridge Hence the circuit has equivalant resistance 7/3 Ω
As shown in figure, a permanent magnet and current carrying coil are placed. If the coil is moved towards magnet, then current in coil (Magnet is symmetrical) :
No change in the flux occurs due to the described motion of the magnet. Hence no current will be induced in the coil.
In the figure shown:
Current through resistance woll be from
A to B if
20 – ε > 2 ⇒ ε < 18
B to A if 20 – ε < 2
A super conducting loop having an inductance 'L' is kept in a magnetic field which is varying with respect to time. If is the total flux, e = total induced emf, then:
For any loop εtotal , = iR, For superconductor, R = 0
∴ εtotal = 0 ⇒ ⇒ φ = constant
A battery of internal resistance ' r ' and e.m.f. ε is connected to a variable external resistor AB. If the sliding contact is moved from A to B, then terminal potential difference of battery will :
V = E - i r. Now, if we go from A → B, then resistance of the circuit increases, ' i ' decreases
∴ V increases
Figure shows a conducting horizontal rod of resistance r is made to oscillate simple harmonically with a fixed amplitude in a uniform and constant magnetic field B, directed inwards. The ends of rod always touch two parallel fixed vertical conducting rails. The ends of rails are joined by an inductor and a capacitor having self inductance and capacitance 1/π Henry and 1/π farad respectively. The amplitude of current in the circuit depends on the frequency of oscillation of rod. The amplitude of the current will be maximum when the time period of rod is : (do not consider self inductance anywhere other than in the inductor)
Here oscillating rod is an AC source because emf induced in it is (vBℓ) ; which varies sinusoidally because v varies sinusoidally.
Maximum current will flow through the circuit under resonance condition. Therefore time period of oscillation of rod is
In the figure shown the equivalent capacitance between 'A' and 'B' is :
Equivalent circuit is
A very small circular loop of area 5 × 10–4 m2 and resistance 2 ohm is initially concentric and coplanar with a stationary loop of radius 0.1 m. If one ampere constant current is passed through the bigger loop and the smaller loop is rotated about its diameter with constant angular velocity ω. The current induced (in ampere) in the smaller loop will be :
The capacitor shown in figure 1 is charged completely by connecting switch S to contact a. If switch S is thrown to contact b at time t = 0, which of the curves in figure 2 above represents the magnitude of the current through the resistor R as function of time t ?
The potential difference across completely charged capacitor is V. As the switch is pushed to b, the initial current in the resistor R is V/R. Hence J is the correct curve.
At a given instant the current and self induced emf in an inductor are directed as shown in figure. If the induced emf is 17 volt and rate of change of current is 25 kA/s the correct statement is :
Current is decreasing and
In the figure shown a parallel plate capacitor has a dielectric of width d/2 and dielectric constant K = 2. The other dimensions of the dielectric are same as that of the plates. The plates P1 and P2 of the capacitor have area 'A' each. The energy of the capacitor is :
A rod of length l having uniformly distributed charge Q is rotated about one end with constant frequency ' f '. Its magnetic moment.
Charge on the differential element dx, equivalent current di = f dq
∴ magnetic moment of this element du = dμ = (di) NA (N = 1)
Two identical spheres of same mass and specific gravity (which is the ratio of density of a substance and density of water) 2.4 have different charges of Q and – 3Q. They are suspended from two strings of same length l fixed to points at the same horizontal level, but distant l from each other. When the entire set up is transferred inside a liquid of specific gravity 0.8, it is observed that the inclination of each string in equilibrium remains unchanged. Then the dielectric constant of the liquid is
Two infinitely long parallel wires are a distance d apart and carry equal parallel currents I in the same direction as shown in the figure. If the wires are located on y axis (normal to x-y plane) at y = d/2 and y = then the magnitude of x-coordinate of the point on x-axis where the magnitude of magnetic field is maximum is (Consider points on x-axis only)
The magnetic field at point P is are magnetic field at P due to wire 1 and 2.
where r2 = x2 + (d/2)2
∴ field is along +y direction at point P and its magnitude is
B is max when dB/dx = 0 i.e. at x = d/2.
Figure shows a uniformly charged hemispherical shell. The direction of electric field at point p, that is off-centre (but in the plane of the largest circle of the hemisphere), will be along
Let electric field at point . 'p' has both x and y component. So similar electric field will be, for other hemisphere (upper half). Now lets overlap both.
(Enet)p = 2 Ex and it should be zero (as E inside a full shell = 0).
So Ex = 0, So electric field at 'p' is purely in y direction.
A wooden stick of length 3l is rotated about an end with constant angular velocity w in a uniform magnetic field B perpendicular to the plane of motion. If the upper one third of its length in coated with copper, the potential difference across the copper coating of the stick is
When the rod rotates, there will be an induced current in the rod. The given situation can be treated as if a rod 'A' of length '3ℓ' rotating in the clockwise direction, while an other rod 'B. of length '2ℓ' rotating in the anticlockwise direction with same angular speed 'ω'.
Resultant induced emf will be :
The resistance of each straight section is r. Find the equivalent resistance between A and B.
From symmetry, the current distribution in branches LP, MP, NP and OP are as shown in figure 1. Therefore junction at P can be broken as shown in figure 2
Hence equivalent resistance is 3.5 r.
PQ is an infinite current carrying conductor. AB and CD are smooth conducting rods on which a conductor EF moves with constant velocity V as shown. The force needed to maintain constant speed of EF is.
⇒ Induced emf ⇒ Power dissipated
Also, power = F.V ⇒
When a galvanometer is shunted with a 4? resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with a 2? wire, the further reduction in the deflection will be (the main current remains the same).
When shunted with 4 Ω
Further reduction in current
Hence further reduction in deflection = 8/13 of the deflection when shunted with 4Ω only.
An inductor (xL = 2?) a capacitor (xC = 8?) and a resistance (8?) is connected in series with an A.C. source. The voltage output of A.C. source is given by v = 10 cos 100πt. The instantaneous potential difference between A and B, when it is half of the voltage output from source at that instant will be :
The current leads in phase by (∵Xc > XL)
φ = 37°
∴ = cos (100 π t + 37°)
The instantaneous potential difference across A B is
The instantaneous potential difference across A B is half of source voltage.
⇒ 6 cos (100 π t - 53°) = 5 cos 100 π t
solving we get
∴ instantaneous potential difference
Figure shows a system of three concentric metal shells A, B and C with radii a, 2a and 3a respectively. Shell B is earthed and shell C is given a charge Q. Now if shell C is connected to shell A, then the final charge on the shell B, is equal to :
From given conditions,
Using it in (1),
A ring of mass m, radius r having charge q uniformly distributed over it and free to rotate about its own axis is placed in a region having a magnetic field B parallel to its axis. If the magnetic field is suddenly switched off, the angular velocity acquired by the ring is :
Four infinite ladder network containing identical resistances of R ? each, are combined as shown in figure. The equivalent resistance between A and B is RAB and between A and C is RAC. Then the value of is :
Let the equivalent resistance of one infinite ladder be x. Then the complete network reduces to
Two small balls, each having equal positive charge Q are suspended by two insulating strings of equal length L from a hook fixed to a stand. If the whole set-up is transferred to a satellite in orbit around the earth, the tension in equilibrium in each string is equal to
A satellite is in a state of free fall & hence weightlessness. Thus only electric force is responsible for the tension
In a practical wheat stone bridge circuit as shown, when one more resistance of 100 ? is connected in parallel with unknown resistance ' x ', then ratio l1 / l2 becomes '2'. l1 is balance length. AB is a uniform wire. Then value of ' x ' must be :
∵ wheat stone bridge is in balanced condition
⇒ x = 100 Ω
If the anions (A) form hexagonal close packing and cations (B) occupy only 2/3rd octahedral voids in it, then the general formula of the compound is :
No of octahedral voids in hcp = 6 so formula A6B4 = A3B2.
In an ionic solid r(+) = 1.6 Å and r(+) = 1.864 Å. Use the radius ratio rule to determine the edge length of the cubic unit cell in Å.
So, it is CsCl type of unit cell
Three lines are drawn from a single corner of an FCC unit cell to meet the other corner such that they are found to pass through exactly only 1 octahedral void, no voids of any type and exactly 2 tetrahedral voids with 1 octahedral void. Identify the line in the same order.
Platinum crystallises in a face centered cube crystal with a unit cell length of 3.9231 Å. The density and atomic radius of platinum are respectively. [Atomic mass of Pt = 195]
for fcc lattice, 4r = a√2
A 0.2 molal aqueous solution of a weak acid (HX) is 20 percent ionised. The freezing point of this solution is (Given kf = 1.86º C kg mol–1 for water) :
ΔTf = 0.2 x 1.2 x 1.86 = 0.45 therefore freezing point = – 0.45°C
The vapour pressure of the solution of two liquids A(pº = 80 mm) and B(pº = 120 mm) is found to be 100 mm when xA = 0.4. The result shows that :
PTotal = 0.4 x 80 + 0.6 x 120 = 104 > 100 mm of Hg
Experimental Ptotal is lower then ideal Ptotal, So solution shows negative deviation.
A 0.004 M solution of Na2SO4 is isotonic with 0.010 M solution of glucose at same temperature. The apparent percentage dissociation of Na2SO4 is :
Consider equimolal aqueous solutions of NaHSO4 and NaCI with ΔTb and ΔT'b as their respective boiling point elevations. The value of
As m → 0 (infinite dilution) both electrolytes will be completely dissociated so
The reduction potential of hydrogen half cell will be negative if (T = 298 K) :
Aqueous solution of Na2SO4 containing a small amount of Phenolphthalein is electrolysed using Pt-electrodes. The colour of the solution after some time will
Remains colourless since solution is acidic and Phenolphthalein shows pink colour only in basic medium.
A very thin copper plate is electro-plated with gold using gold chloride in HCl. The current was passed for 20 minutes and the increase in the weight of the plate was found to be 2 gram [Au = 197]. The current passed was
w = Zit
A current is passed through two voltameters connected in series. The first voltameter contains XSO4 (aq.) and second has Y2SO4 (aq.) the relative atomic masses of X and Y are in the ratio of 2 : 1. The ratio of the mass of X liberated to the mass of Y liberated is :
Suooose atomic mass of X = 2m and Y = m
Ratio of masses of liberated X and Y = 2m : 2m = 1 : 1
(i) MnO4- + 8H+ + 5e- → Mn2+ + 4H2O E0 + x1 V
(ii) MnO2 + 4H+ + 2e- → Mn2+ + 2H2O E0 + x2 V
Find E0 for the following reaction :
MnO4- + 4H+ + 3e- → MnO2 + 2H2O
The specific conductivity of an aqueous solution of a weak monoprotic acid is 0.00033 ohm–1cm–1 at a concentration 0.02 M. If at this concentration the degree of dissociation is 0.043, then calculate the value of (in ohm–1 cm2 /eqt) :
and λm = 0.00033 x 50 x 103 = 33 x 50 x 10-2
Which of the following ions will be most effective in coagulating the As2S3 sol :
According Hardy-Schulze rule.
For the cell prepared from electrode A and B, electrode and electrode E0red. = 0.77 V, which of the following statement is not correct?
A acts as cathode and B acts as anode because reduction potential of A is more.
Anode is B ⇒ is negative electrode.
⇒ electrons flow from B to A.
⇒ e.m.f. = (1.33 – 0.77) V
6I- (aq) + BrO3-(aq) + 6H+ (aq) → 3I2 (aq) + Br- (aq) + 3H2O(ℓ)
These data were obtained when this reaction was studied.
What are the units of the rate constant for this reaction ?
Order of reaction is 2 therefore unit of K is mole-1 Ls-1.
In a gaseous phase reaction. The increase in pressure from 100 mm to 110 mm is observed in 5 minute. The rate of disappearance of A2 in mm min-1 is :
rate of disappearence of A2 = 20/5 = 4
In a certain reaction, 10% of the reactant decomposes in one hour, 20 % in two hours, 30% in three hours and so on. The dimensions of the rate constant is :
It is a zero order reaction therefore units of K is mol L-1 s-1.
Which of the following is wrong :
The correct statement is : Enthalpy (numerical value) of physisorption is lower than that of chemisorption.
For the decomposition of HI the following logarithmic plot is shown :
[R = 1.98 cal/mol-K]
The activation energy of the reaction is about :
thus slope of graph will be
⇒ Ea = 2.303 x 1.98 x 104 = 45600 cal
Rate constant for the reaction NO2+ CO → NO + CO2 are 1.3 M-1S-1 at 700 K and 23 M-1S-1 at 800 K. What will 3 be the rate constant at 750 K ?
Given that the normal energy of the reactant and product are 40J and 20J respectively and threshold energy of the uncatalysed reaction is 120 J. If the rate of uncatalysed reaction at 400 K becomes equal to the rate of catalysed reaction at 300 K, then what will be the activation energy of the catalysed forward and backward reactions respectively?
In absence of catalyst = In presence of catalyst
A hypothetical reaction, A2 + B2 → 2AB
Follows mechanism as given below :
The order of overall reaction is :
Rate is governed by slowest step
If uranium (mass number 238 and atomic number 92) emits an α-particle, then mass number and atomic number of product are respectively :
An aqueous solution of NaCI freezes at -0.186°C. Given that kg mol-1 and kg mol-1, the elevation in boiling point of this solution is :
For a saturated solution of AgCI at 25°C, k = 3.4 x 10-6 ohm-1 cm-1 and that of H2O(ℓ) used is 2.02 x 10-6 ohm-1 cm-1. λ°m for AgCI is 138 ohm-1 cm2 mol-1, then the solubility of AgCl in moles per liter will be :
The vapour pressure of pure liquid solvent A is 0.80 atm. When a non-volatile substance B is added to the solvent, its vapour pressure drops to 0.60 atm. mole fraction of the component B in the solution is :
Osmotic pressure [in atm] of a 0.1 M solution of K4[Fe(CN)6], which undergoes 50% dissociation, will be ________ at 270C :
What weight of glucose dissolved in 100 grams of water will produce the same lowering of vapour pressure as one gram of urea dissolved in 50 grams of water, at the same temperature?
(half-life = 15 hrs.) is known to contain some radioactive impurity (half-life = 3 hrs.) in a sample. This sample has an initial activity of 1000 counts per minute, and after 30 hrs it shows an activity of 200 counts per minute. What percent of the initial activity was due to the impurity ?
Let the activity due to impurity be 'a' cpm.
∴ due to Na it is (1000 – a) cpm.
After 30 hrs 'a' would be reduced to a cpm and (1000 – a) would be reduced to cpm
∴ total activity after 30 hrs would be
= 200 (given)
solving we get
∴ ⇒ a = 200
Hence 20% activity was due to impurity.
For the cell (at 298 K)
Ag(s) | AgCl(s) | Cl– (aq) || AgNO3 (aq) | Ag(s)
Which of following is correct :
It [Ag+]3 = [Ag+]c then both the electrodes have same potential. [AT+] will increase in anodic compartment. AgCI(s) precipitate in anodic compartment will increase.
At 298K the standard free energy of formation of H2O(l) is –257.20 kJ/mole while that of its ionisation into H+ ions and OH– ions is 80.35 kJ/mole, then the emf of the following cell at 298 K will be (Take F = 96500 C] :
H2(g,1 bar) | H+ (1M) || OH¯ (1M) | O2 (g, 1bar)
Also we have
Hence for cell reaction
Consider the reaction, NH2NO2 (aq) ———? N2O(g) + H2O(l)
The concentration of nitramide as a function of time is shown below for a particular run.
Which line represents the correct tangent to the graph at the origin (t = 0) ?
In a hypothetical reaction
A(aq) 2B(aq) + C(aq) (1st order decomposition)
'A' is optically active (dextro-rototory) while 'B' and 'C' are optically inactive but 'B' takes part in a titration reaction (fast reaction) with H2O2. Hence the progress of reaction can be monitored by measuring rotation of plane of plane polarised light or by measuring volume of H2O2 consumed in titration.
In an experiment the optical rotation was found to be θ = 40° at t = 20 min and θ = 10° at t = 50 min. from start of the reaction. If the progress would have been monitored by titration method, volume of H2O2 consumed at t = 15 min. (from start) is 40 ml then volume of H2O2 consumed at t = 60 min will be:
As only A is optically active. So concentration of A at t = 20 min ∝ 40° While concentration of A at t = 50 min ∝ 10°, so t1/2 = 15 min.
So volume consumed of H2O2 at t = 15 min = t1/2 , is according to 50% production of B at t = 60 min. production of B = 94.75% (four half lives)
So volume consumed = 75 ml Ans.
How many m.moles of sucrose should be dissolved in 500 grams of water so as to get a solution which has a difference of 103.57°C between boiling point and freezing point ?
(Kf = 1.86 K Kg mol–1, Kb = 0.52 K Kg mol–1)
Boiling point of solution = boiling point + ΔTb = 100 + ΔTb
Freezing point of solution = freezing point – ΔTf = 0 – ΔTf
Difference in temperature (given) = 100 + ΔTb – (– ΔTf) 103.57 = 100 + ΔTb + ΔTf = 100 + molality x Kb + molality x Kf = 100 + molality (0.52 + 1.86)
and molality Moles of solute
When a graph is plotted between log x/m and log p, it is straight line with an angle 45° and intercept 0.6020 on y-axis. If initial pressure is 0.3 atm, what will be the amount of gas adsorbed per gram of adsorbent :
Diamond has face-centred cubic lattice. There are two atoms per lattice point, with the atoms at (000) and coordinates. The ratio of the carbon-carbon bond distance to the edge of the unit cell is:
Carbon atoms are at corners and are at alternate corners. So from geometry.
So required ratio
Assuming the formation of an ideal solution, determine the boiling point of a mixture containing 1560 g benzene (molar mass = 78) and 1125 g chlorobenzene (molar mass = 112.5) using the following against an external pressure of 1000 Torr:
20 mole C6H6, 10mole C6H5Cl
At t = 100°C
Decomposition of A follows first order kinetics by the following equation.
4A(g) → B(g) + 2C(g)
If initially, total pressure was 800 mm of Hg and after 10 minutes it is found to be 650 mm of Hg. What is half-life of A ? (Assume only A is present initially)
Pressure of A = 200, so
∴ 2 x t1/2 = 10 minutes
t1/2 = 5 minutes
For the following parallel chain reaction
what will be that value of overall half-life of A in minutes?
You are given the following cell at 298 K, = – 0.76 V. Which of the following amounts of NaOH (equivalent weight = 40) will just make the pH of cathodic compartment to be equal to 7.0 :
Anode Zn → Zn2+ + 2e-
Cathode 2H+ + 2e- → H2
Cell : ; E0cell = 0 - (- 0.76) = 0.76 V
∴ [H+] = 10-2 M
∴ NaOH required is 0.01 mole = 0.4 grams
Electrolysis of a solution of HSO4- ions produces S2O8– –. Assuming 75% current efficiency, what current should be employed to achieve a production rate of 1 mole of S2O8- - per hour ?
so required rate = 1 mole/hr
= 2 mole of e-/hr
so required current
In an f.c.c. unit cell, atoms are numbered as shown below. The atoms not touching each other are :
(Atom numbered 3 is face centre of front face).
Atoms along one edge or at corners do not touch each other in fcc cell.
In an f.c.c. crystal, which of the following shaded planes contains the following type of arrangement of atoms?
Shown arrangement is hexagonally closed pack plane and these type of planes are arranged perpendicular to
body diagonal of fcc unit cell as shown.
Which of the following not occur in 5’ to 3’ direction?
DNA replication, transcription and translation occur in 5’ to 3’ direction
Which of the following process not have basis complimentary base pairing-
translation is sequencing amino acid according to sequence of Nitrogenous bases
Few properties of Nucleic acid is given , how many belong to RNA- Reactive, catalytically active, 2’-H present, able to mutate , 2’-OH , thermostable
RNA is not thermostable and have 2’-OH group
DNA of nucleoid is present in form of –
Mark the correctly matched-
DNA polymerase polymerise in 5’ to 3’ direction
Which of the following is correctly matched-
(A) mRNA – provides template
(B) tRNA – reads genetic code in protein synthesis
(C ) rRNA – strucutural role in protein synthesis
(D ) rRNA- catalytic role
(E) tRNA – brings amino acid
How many of the following is correctly matched-
Mark the incorrect –
The template with 3’ to 5’ polarity show countinuos DNA replication –
Considering DNA replication is semiconsevative if two cell with N15 DNA on both strand is grown on source with N14 DNA for three generation than percentage of N14 N15 DNA is-
total 16 cell and 4 cell with N14 N15 DNA
The beads on sting structure in ………………. Is packaged to form ……………… that further condensed in metaphase stage
Which of the following not present in RNA-
RNA evolve into DNA by chemical modification and further evolving repair
At stop codon which of the following binds-
Which of the following is gene-
segment of DNA code for any type of RNA
Which of the following is not DNA dependent polymerase -
reverse transcriptase is RNA dependent DNA polymerase
Which of the following is not nucleotide-
adenylic acid is a nucleic acid and not a nucleotide. Thus, option (a) will be the answer.
Consider the following-
(a) origin is not replicated in DNA replication during s-phase
(b) minisatellite DNA mainly use for DNA fingerprinting
How many are correct
origin also get replicated during s-phase
Which of the following can make polynucleotide chain without template –
polynucleotide phosphorylase is for template free RNA synthesis
Consider the following –
(a) DNA ligase can join 5’ to 3’ phosphodiester linkage
(b) Both strand of DNA of structural gene can not be transcribed
How many is /are correct-
If lactose is absent in medium than which is probable correct for lac operon-
Small level of lac operon expression always present
The sequence of Chromosome I was completed only in –
Which is correct about DNA polymorphism-
Mark the wrongly matched –
High levels of phenylalanine, as seen in untreated PKU, cause brain damage and associated mental retardation. Early implementation of a low phenylalanine diet prevents the mental retardation associated with this condition.
In a cross between AABb and AaBb , percentage of progeny with AaBb Genotype is-
50 percent gamete receive Aa gametes
From given example which of the following is not according to mendel law of dominance-
(a) endosperm of gymnosperm
(b) aleurone layer in monocot seed
(c) nucellus of gymnosperm
Mark the correct answer-
for mendel law of dominance factor should be in pair
Law of segregation is not valid in case of –
(a) gametogenesis in algae
(b) sporogenesis in algae
(c) embyrogenesis in bryophytes
Mark the correct answer-
for law of segregation meiosis should occur
Which of the following amino acid have three codon in genetic code-
Mendal;s work was rediscovered in
Three gene polygene heterozygous is selfed give ………….number of progeny and………of phenotype classes respectively-
three gene polygene give 7 phenotypic classes
AaBbCc genotype form 3 type of different gamete this means –
de to linkage of two genes number of gamets reduce
Female heterogamety present in –
Two linked gene give non-parental gene combination due to
crossing increase non-parental gene combination
Which of the following is wrong statement-
A colour blind Man marries to normal female have one normal daughter and one colourblind son which is probability of next daughter to be colour blind-
mother is carrier so every time 50 percent chance is tha daughter is colour blind
In a population of 100 humans 4 show fuse Ear lobe , if it is autosomal recessive character and population is in genetic equlilibrium calculate frequency of dominant allele -
hardy Weinberg problem
AaBbCc individual is selfed than number of progeny with AABbCC phenotype is-
number of progeny is 2 out of 64 or 1 out of 32
If individual with genotype TtRr(tall round) is crossed with ttrr (dwarf wrinkled) than it give two type of genotype only-
In complete linkage only parental type obtained
Which of the following is intra- alleleic interaction-
intra alleleic interaction is between allele of same trait
Mark the incorrect-
In co-dominance dominant recessive relationship is absent
When one allele mask the expression of another allele at same loci on homologus chromosome than it is-
epistatis is at different gene loci
Which is true about hybridization experiment conducted by mendel-
Mendels finding was general rule of inheritance rather than unsubstainted data because-
Phenotype of organism not affected when modified allele present in homozygous state and produce-
phenotype of organism not affected when it produce normal enzyme
In some organism if only half the sperm have X-chromosome and other half lack X chromosome this than type of sex-determination is-
A female can be haemophilic , for this which is most necessary-
father has to be affected
Which is incorrect regarding t-RNA –
tRNA is functional in secondary strucuture
In lac operon which is incorrect-
Product of regulator gene is important it should not present close to lac genes
Mark the wrong statement-
DNA from different cell show different degree of polymorphism
In pBR322 restriction site for which of the following enzyme is absent-
Mark the incorrect statements-
non-functional origin give zero copy number
What is true about selectable marker-
selectable marker help to slect transfomants out of non-transformants
Which is correct order in genetic engineering-
(1) gene transfer for expression
(2) creation of recombinant DNA
(3) use of gene amplification
(4) isolation of DNA
Correct sequence is -
gene is isolated, recombinanat is created and amplified before transfer
What is true about plasmid-
small size plasmid will transfer easily
Consider the following statements and mark incorrect-
Any palindrome cut in centre produce blunt end
Two enzyme……… restrict the growth of bacteriophage is isolated from E.Coli was discovered in…….. year --
Consider the following statement, how many statements are correct -
(A) More than 900 restriction enzymes have been isolated from 230 strains of bacteria
(B) plasmid having functional origin can not use host DNA polymerase to replicate
(C) DNA runs towards cathode in gel electrophoresis when current is applied
(D) Recombinant plasmid will not obtain if source of gene and plasmid will cut with different R.E
Mark the correct answer-
Which of the following is correctly matched
(A) Biolistic - liver cell
(B) Microinjection - parenchyma
(C) Alter genotype - heterologus host
(D) Bioreactor - synthetic production
Consider the following statements
1) EcoR1 cuts after A of palindrome sequence 5’—GAATTC—3’
2) EcoR1 restriction enzyme obtained from RY15 strain.
3) It is necessary that alien DNA and Restriction site of cloning should be cut by same palindrome
how many statements are correct-
If host DNA and foreign DNA cut with same restriction site it will generate same sticky end
Mark the correct statement
IARI develops several variety for micronutrients ,which is wrong-
Which of the following is wrongly matched-
(1) hairy leaf - jassids
(2) smooth leaf and nectarless - aphids
(3) Flat bean - jassids
(4) Pusa swarnim - white rust
Source of SCP is -
Source of SCP is -
Transcription occur in nucleus
Disarmed retrovirus is develop for-
In a few plant cell which can be most useful cloning vector-
A divalent ion provided during heat shock technique–
Large size DNA not show transformation
Mark the correctly matched-
(1) Gene gun - silver and tungsten
(2) Cystic fibrosis – alpha-1- antitypsin
(3) Pusa sawani - shoot and fruit borer
(4) Pusa komal - chilli mosaic virus
RNA from DNA of eukaryotes is removed during isolation of DNA by –
Ethanol precipitation without RNase treatment will-
Ethanol can precipitate both RNA and DNA
Mark correctly matched
(1) DNA precipitation - spooling
(2) gel electrophoresis - isolation of DNA
(3) Emphysema - alpha-1- antitypsin
(4) blue white selection - tetracycline resistance
alpha-1- antitypsin Cure Emphysema
Which step involve phosphodiester bond formation in PCR-
In A test tube ‘PCR’ is carried out with primers , if DNA amplication or PCR cycle require 3 min and we start with single DNA piece than DNA amplification stops after –
PCR stop when nucleotide or Primer gent finish
Taq polymerase obtain from –
thermos aquaticus is archaebacteria
Recombinant protein is all except –
recombinant potein form by host genome
Mark the incorrectly matched- –
(1) Bioreactor - production of Enzyme
(2) bioreactor - ethanol fermentation
(3) Cultivar - develop new variety
(4) High yielding wheat – introduced into 1963
cultivar is variety which is under cultivation
Sparged in sparged stirred tank bioreactor help in-
A stirred tank bioreactor is usually cylindrical or with a curved base to facilitate the mixing of the reactants. It facilitates the oxygen availability and also has an agitator system, foam control system, temperature control system, pH control system and an oxygen delivery system so that small volumes of the culture can be withdrawn periodically. In the sparged stirred-tank bioreactor, the sparge (a porous ring made of metal or glass) facilitates the mixing and oxygen availability throughout the bioreactor.
Bt toxin activate in –