A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is:
In an ideas gas internal energy
Maxwell’s velocity distribution curve is given for two different temperatures. For the given curves.
Higher is the temperature greater is the most probable velocity.
The ratio of translational and rotational kinetic energies at 100 K temperature is 3 : 2. Then the internal energy of one mole gas at that temperature is [R = 8.3 J/mol-K]
According to law of equipartition of energy, energy equally distributes among its degree of freedom, Let translational and rotational degree of freedom be f1 and f2.
Hence the ratio of translational to rotational degrees of freedom is 3:2. Since translational degrees of freedom is 3, the rotational degrees of freedom must be 2.
∴ Internal energy (U) =
= U 2075 J
12 gm He and 4 gm H2 is filled in a container of volume 20 litre maintained at temperature 300 K. The pressure of the mixture is nearly :
PV = nRT
= 6.25 x 105 Pa
Which of the following will have maximum total kinetic energy at temperature 300 K.
Total KE = U = f/2nRT
In case of H2 degree of freedom is greatest and number of moles n is highest. So this is the case of maximum kinetic energy.
A ring shaped tube contains two ideal gases with equal masses and atomic mass numbers M1 = 32 and M2 = 28. The gases are separated by one fixed partition P and another movable conducting partition S which can move freely without friction inside the ring. The angle a as shown in the figure in equilibrium is:
P1 = P2 T1 = T2
In an experiment the speeds of any five molecules of an ideal gas are recorded. The experiment is repeated N times where N is very large. The average of recorded values, is :
When speed of 5 molecules which are selected randomly are recorded, then the average is most likely to be equal to the most probable speed.
∴ The average of these values is most likely equal to
Temperature at which Fahrenheit and Kelvin pair of scales give the same reading will be:
= ⇒ 5θ - 32 x 5 = 9θ - 273 x 9
= θ = 574.25
20 gm ice at –10 ºC is mixed with m gm steam at 100 ºC. The minimum value of m so that finally all ice and steam converts into water is : (Use sice = 0.5 cal/gmºC,swater = 1 cal/gmºC,L (melting) = 80 cal/gm and L (vaporization) = 540 cal/gm)
For minimum value of m, the final temperature of the mixture must be 0°C.
∴ = m 540 + m.1. 100
An ice block at 0°C is dropped from height ‘h’ above the ground. What should be the value of ‘h’ so that it melts completely by the time it reaches the bottom assuming the loss of whole gravitational potential energy is used as heat by the ice ? [Given : Lf = 80 cal/gm]
Applying energy conservation : mgh = mLf
= 33.6 km
n moles of a gas filled in a container at temperature T is in thermodynamic equilibrium initially. If the gas is compressed slowly and isothermally to half its initial volume the work done by the atmosphere on the piston is:
Work done by atmosphere = Patm ΔV =
Initially gas in container is in thermodynamic equilibrium with its surroundings.
∴ Pressure inside cylinder = Patm
& PV = nRT ⇒ Patm V = nRT or
In a process the pressure of an ideal gas is proportional to square of the volume of the gas. If the temperature of the gas increases in this process, then work done by this gas:
As p ∝ V2
T ∝ V3
i.e. if temperature increases, volume also increases hence work done will be positive.
A vessel contains an ideal monoatomic gas which expands at constant pressure, when heat Q is given to it. Then the work done in expansion is:
For process at constant pressure
5 moles of Nitrogen gas is enclosed in an adiabatic cylindrical vessel. The piston itself is a rigid light cylindrical container containing 3 moles of Helium gas. There is a heater which gives out a power 100 cal/sec to the nitrogen gas. A power of 30 cal /sec is transferred to Helium through the bottom surface of the piston. The rate of increment of temperature of the nitrogen gas assuming that the piston moves slowly:
Net power given to N2 gas = 100 - 30 = 70 calls
The nitrogen gas expands isobarically.
The gas law PV/2 =constant for a given amount of a gas is true for :
As PV = nRT
For n = constant : = constant for all changes Hence (C)
All the rods have same conductance ‘K’ and same area of cross section S. If ends A and C are maintained at temperature 2T0 and T0 respectively then which of the following is/are correct:
IAB = IBC & IAD = IDC
∴ No current in BO and OD
∴ TB = TO = TD
A hollow copper sphere and a hollow copper cube, of same surface area and negligible thickness, are filled with warm water of same temperature and placed in an enclosure of constant temperature, a few degrees below that of the bodies. Then in the beginning :
[Newton's law of cooling]
∴ Rate of loss of heat is same
The colour of a star indicates its :
The colour of an object indicates the rate at which energy is emitted and hence indicates the temperature.
Two identical solid spheres have the same temperature. One of the sphere is cut into two identical pieces. The intact sphere radiates an energy Q during a given small time interval. During the same interval, the two hemispheres radiate a total energy Q'. The ratio Q'/Q is equal to :
Heat radiated (at temp same temp) ∝ A
⇒ Q ∝ 4πR2 and Q' ∝ (4πR2 + 2 x πR2)
Here πR2 is extra surface area of plane surface of the hemisphere
A calorimeter contains 50 g of water at 50°C. The temperature falls to 45°C in 10 minutes. When the calorimeter contains 100 g of water at 50°C, it takes 18 minutes for the temperature to become 45°C. then the water equivalent of the calorimeter is :
assuming rate of heat transfer to be constant.
s (m1 + w) ΔT = Q1 where s = specific heat of water = 1 g/cc s
s (m2 + w)ΔT = Q2 w = water equivalent of calorimeter
Heat is flowing through two cylindrical rods made of same materials whose ends are maintained at similar temperatures. If diameters of the rods are in ratio 1 : 2 and lengths in ratio 2 : 1, then the ratio of thermal current through them in steady state is :
for same temperature difference, thermal current
A balloon containing an ideal gas has a volume of 10 litre and temperature of 17ºC. If it is heated slowly to 75ºC, the work done by the gas inside the balloon is (neglect elasticity of the balloon and take atmospheric pressure as 105 Pa)
Since elasticity of balloon is negligible, pressure inside ballon outside baloon = Patm.
= Vin = 10 litre.
In the P-V diagram shown. The gas does 5 J of work in isothermal process a b and 4 J in adiabatic process b c. What will be the change in internal energy of the gas in straight path c to a?
ΔU in a - b process = 0 (isothermal process)
in b - c process Q = 0 (adiabatic process)
∴ ΔU = - W = - 4 J
ΔU in cyclic process = 0
∴ ΔU in c - a process = 4J
A monoatomic ideal gas is filled in a nonconducting container. The gas can be compressed by a movable nonconducting piston. The gas is compressed slowly to 12.5% of its initial volume. The percentage increase in the temperature of the gas is
Let initial temperature and volume be T0 and V0. Since the process is adiabatic, the final temperature and volume is (γ = 5/3 for monoatomic gas)
∴ percentage increase in temperature of gas is 4 x 100 - 100 = 300%
A diatomic ideal gas is heated at constant volume until the pressure is doubled and again heated at constant pressure until volume is doubled. The average molar heat capacity for whole process is:
Let initial pressure, volume, temperature be P0, V0, To indicated by state A in P-V diagram. The gas is then isochorically taken to state B (2P0, V0, 2T0) and then taken from state B to state C (2P0, 2V0, 4T0) isobarically.
Total heat absorbed by 1 mole of gas
Total change in temperature from state A to C is ΔT = 3T0
∴ Molar heat capacity
The ratio of final root mean square velocity to initial root mean square velocity of nitrogen molecules if nitrogen gas is compressed adiabatically from a pressure of one atmosphere to a pressure of two atmosphere is :
= = 21/7
Four particles have velocities 1, 0, 2, 3 m/s. The root mean square velocity of the particles is: (in m/s)
V–T diagram for a process of a given mass of ideal gas is as shown in the figure. During the process pressure of gas.
PV = nRT P(aT - b) = nRT
now as T increases then increases hence P decreases
A slab X of thickness ‘t’, thermal conductivity ‘K’ and area of cross-section ‘A’ is placed in thermal contact with another slab Y which is 2n2 times thicker, 4n times conductive and having n times larger cross section area. If the outside face of X is maintained at 100°C, the outside face of Y at 0°C, then the temperature of the junction θ is represented by the graph (n > 0) :
At equilibrium :
⇒ θ = 100/3.
Two conducting movable smooth pistons are kept inside a non conducting, adiabatic container with initial positions as shown. Gas is present in the three parts A, B & C having initial pressures as shown. Now the pistons are released. Then the final equilibrium position length of part A will be
Moles in 'A' initially & finally will be same
Moles in 'B' remain same
Divided (1) by (2)
⇒ x = L/4
4 gms of steam at 100°C is added to 20 gms of water at 46°C in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation = 540 cal/gm. Specific heat of water = 1 cal/gm-°C.
Heat released by steam inconversion to water at 100°C is Q1 = mL = 4 x 540 = 2160 cal.
Heat required to raise temperature of water from 46°C t 100°C is Q2 = mS Δθ = 20 x 1 x 54 = 1080 J
Hence all steam is not converted to water only half steam shall be converted to water
∴ Final mass of water = 20 + 2 = 22 gm
A diatomic ideal gas undergoes a thermodynamic change according to the P–V diagram shown in the figure. The total heat given to the gas is nearly (use ln2 = 0.7) :
Two rods are joined between fixed supports as shown in the figure. Condition for no change in the lengths of individual rods with the increase of temperature will be (α1, α2 = linear expansion co-efficient A1, A2 = Area of rods Y1, Y2 = Young modulus)
In both the rods tension will be same so T1 = T2
Hence Y1 A1 α1 ΔT = Y2 A2 α2 ΔT
For a gas sample with N0 number of molecules, function N(V) is given by : < V < V0 and N(V) = 0 for V > V0. Where dN is number of molecules in speed range V to V+ dV. The rms speed of the molecules is :
The co-efficient of thermal expansion of a rod is temperature dependent and is given by the formula α = a T, where a is a positive constant and T in ºC. If the length of the rod is l at temperature 0 ºC, then the temperature at which the length will be 2 l is:
A black body emits radiation at the rate P when its temperature is T. At this temperature the wavelength at which the radiation has maximum intensity is λ0. If at another temperature T' the power radiated is P' and wavelength at maximum intensity is then
For a block body, wavelength for maximum intensity :
& PαT4 ⇒
⇒ P' = 16P. ∴ P'T' 32PT ⇒ P' = 16P.
Thermal coefficient of volume expansion at constant pressure for an ideal gas sample of n moles having pressure P0, volume V0 and temperature T0 is
For given temperature T0,
There are two thin spheres A and B of the same material and same thickness. They emit like black bodies. Radius of A is double that of B. A and B have same temperature T. When A and B are kept in a room of temperature T0 (< T), the ratio of their rates of cooling (rate of fall of temperature) is: [assume negligible heat exchange between A and B]
The rate of heat loss by a thin hollow sphere of thickness ‘Δx’, mean radius and made of density ‘ρ’ is given by
⇒ is independent of radius
Hence rate of cooling is same for both spheres.
A piston can freely move inside a horizontal cylinder closed from both ends. Initially, the piston separates the inside space of the cylinder into two equal parts each of volume V0 , in which an ideal gas is contained under the same pressure p0 and at the same temperature. What work has to be performed in order to increase isothermally the volume of one part of gas 2 times compared to that of the other by slowly moving the piston ?
= work done on both gases
= - (work done by the gases)
A uniform pressure P is exerted by an external agent on all sides of a solid cube at temperature tºC. By what amount should the temperature of the cube be raised in order to bring its volume back to its original volume before the pressure was applied if the bulk modulus is B and co-efficient of volumetric expansion is γ ?
Two rods of same dimensions, but made of different materials are joined end to end with their free ends being maintained at 100ºC and 0ºC respectively. The temperature of the junction is 70ºC. Then the temperature of the junction if the rods are interchanged will be equal to :
Assume a sample of an ideal gas in a vessel. Where velocity of molecules are between 2 m/sec to 5 m/sec and velocity of molecules (v) and number of molecules (n) are related as n = 7v – v2 – 10. The most probable velocity in sample is. Where v is measured in m/sec.
n = 7v - v2 - 10
for most probable velocity
∴ n is maximum at this velocity
If specific heat capacity of a substance in solid and liquid state is proportional to temperature of the substance, then if heat is supplied to the solid initially at – 20°C (having melting point 0°C) at constant rate. Then the temperature dependence of substance with time will be best represented by :
or S = K1T. (K1 being proportionality constant)
⇒ ⇒ Hence, the graph will be parabolic
In the figure shown the pressure of the gas in state B is:
A hot black body emits the energy at the rate of 16 J m–2 s–1 and its most intense radiation corresponds to 20,000 Å. When the temperature of this body is further increased and its most intense radiation corresponds to 10,000 Å, then the energy radiated in Jm–2 s–1 will be :
Wein's displacement law is : λm . T = b
Here, λm becomes half.
∴ Temperature doubles.
Also E = σT4
The statement that is not correct for the periodic classification of elements is :
The d-sub shells are not filled with electrons monotonically with increase in atomic number. There are some exceptions like Cr, Cu etc.
The increasing order of acidic nature of Li2O, BeO, B2O3, CuO is
On moving left to right in periodic table, acidic nature of oxide increases. So, metallic oxides are basic, non metallic oxides are acidic and BeO is amphoteric.
Which of the following statements is wrong for the transition elements ?
Transition elements starts from 4th period as they have valence shell configuration (n – 1)d1 - 10 ns0–2. Thus for n = 3, there is no d-orbital available.
Which of the following is not isoelectronic series ?
The number of electrons present are not same in (D) options N3– = 10 electrons, S2– = 18 electrons, Cl– = 18 electrons. So this group does not represent the isoelectronic species.
The first ionisation energy of Al is smaller than that of Mg because :
Penetration of p-subshell electron is less than s-subshell electrons. In case of Mg, the first electron is to be removed from completely filled 3s2 valence shell configuration as compared to partially filled 3p1 of Al. These two factors collectively accounts for the higher ionisation energy of Mg than that of Al. Therefor, (C) option is correct.
Which of the following properties is affected by stable configuration of an atom ?
(b) Ionisation potential
(c) Electron affinity
Correct answer is :
The addition of extra electron is difficult to the atom having stable configuration and so electron affinity will be less or zero. Similarly the removal of electron is quite difficult for stable configuration and so ionisation energy is higher. Hence electron affinity and ionisation energy both are affected by stable configuration.
Fluorine has the highest electronegativity among the ns2 np5 group on the Pauling scale, but the electron affinity of fluorine is less than that of chlorine because :
There is more interelectronic repulsion in 2p-subshell of fluorine than 3p-subshell of chlorine due to its small atomic size. So extra electron will be added easily in 3p-subshell of chlorine as compared to 2p-subshell of fluorine.
Which of the following orders is incorrect ?
Option A is incorrect.
All are isoelectric species so with increasing effective nuclear charge :
(i) size decreases (ionic size ∝ 1/nuclear charge).
(ii) ionisation energy increases as size decreases and nuclear charge increases.
(iii) electron affinity increases as size decreases and nuclear charge increases.
The correct order of the increasing ionic character is :
According to Fajan's rule as size of cations increase, their polarising power decrease and thus ionic character increase. So option (A) is correct.
In which of the following sets the central atom of each member involves sp3 hybridisation ?
Hybridisation is calculated by steric number rule.
Steric no. = no. of atoms attached to central atom + no. of lone pairs present on central atom.
Steric no. for IO4- = 4 + 0 = 4 sp3 ; Steric no. for ICI4- = 4 + 2 = 6 sp3d2
Steric no. for IF4+ = 4 + 1 = 5 sp3d ; Steric no. for SO3 = 3 + 0 = 3 sp2
Steric no. for SO42- = 4 + 0 = 4 sp3 ; Steric no. for SO32- = 3 + 1 = 4 sp3
Steric no. for PCI4+ = 4 + 0 = 4 sp3 ; Steric no. for BF4- = 4 + 0 = 4 sp3
Steric no. for XeO3 = 3 + 1 = 4 sp3 ; Steric no. for XeO4 = 4 + 0 = 4 sp3
Steric no. for XeF4 = 4 + 2 = 6 sp3d2 ; Steric no. for CIO4- = 4 + 0 = 4 sp3
The average charge on each O atom and average bond order of S–O bond in SO42– is :
Which is the right structure of XeF4 ?
Lone pair - lone pair repulsion is maximum.So to minimize the repulsion the lone pairs occupy the positions which are trans to each other in the given octahedral geometry. Therefore, (C) option is correct.
According to Molecular orbital theory which of the following is correct ?
M.O configure for
As π2py and π2pz electrons participate in bonding so both the bonds between two carbon atoms are π bonds.
A diatomic molecule has a dipole moment of 1.2 D. If its bond distance is equal to 2.0Å then the % of ionic character in the diatomic molecular is :
Assuming complete charge transfer then dipole moment = (4.8 x 10-10 esu) (10-8 cm) x 2 = 9.6 D
so % ionic character = 12.5%
Correct order of bond energy is:
Bond order is directly proportional to the bond energy.
Bond order of N2 = 3 , N2+, N2- = 2.5 N22- = 2
But N2- has more electrons in antibonding MO's and thus N2+ is more stable than N2-. So correct order of bond energy will be N2> N2+ > N2-> N22- Therefore, (A) option is correct
Molecule XF4 has non-zero value of dipole moment. Then X is :
Which of the following hydrogen bonds is the strongest ?
Strength of H–bond depends on the size and electronegativity of the atom.
The correct order of boiling point is :
(i) H2O has highest boiling point because of H-bonding.
(ii) Boiling point also depends on the magnitude of van der Waal's force of attraction, which in turns depends on molecular weight of the compounds. Thus the correct order is H2O > H2Te > H2Se > H2S.
Which of the following statements is true for all the alkali metals ?
(A) 4 LiNO3 → 2Li2O + 4NO2 + O2
2NaNO3 → 2NaNO2 + O2 (similar decomposition with the nitrates of K, Rb and Cs)
(B) Only LiCI is deliquescent and crystallises as a hydrate LiCI.2H2O
(C) 2M + 2H2O → 2M+ + 2OH- + H2 (M = an alkali metal)
(D) Halides of Li are covalent in nature.
Property of all the alkaline earth metals that increases with their atomic number is :
(A), (C) and (D) decreases down the group but (B) increases down the group with increasing metallic character.
Sodium and potassium react with water much more vigorously than lithium because :
When sodium and potassium react with water, the heat evolved causes them to melt, giving a larger area of contact with water, lithium on the other hand, does not melt under these condition and thus reacts more slowly.
Li Na K
Melting point (ºC) 180 98 64.
On dissolving moderate amount of sodium metal in liquid NH3 at low temperature, which one of the folloiwng does not occur ?
liquid NH3 solution is paramagnetic due to the presence of solvated unpaired electrons.
KO2 (potassium super oxide) is used in space and submarines because it :
Which of the following statements is incorrrect ?
(A) Bigger anion is stabilised by bigger cation through lattice energy effect.
(B) Because of their high reactivity towards air and water.
(C) True Statement
(D) In concentrated solution, unpaired electrons with opposite spins paired up-forming the solution diamagnetic.
On commercial scale, sodium hydroxide is prepared by which one of the following methods ?
(A) Used for extraction of Mg from sea water.
(B) Used for the manufacture of Na2CO3 using NaCl as raw material.
(C) Used for manufacture of NaOH by electrolysis of brine(concentrated NaCl solution).
(D) Used for the extraction of aluminium from bauxite.
A metal (M) burns with dazzling brilliance in air to give a white powder. The white powder reacts with water to form a white precipitate and a colourless gas with a characteristic smell. The metal (M) decomposes hot water but not cold water, liberating the inflammable hydrogen gas. The metal(M) is :
Hydrogen is liberated when Mg reacts with hot water. Rb, K and Ca gives hydrogen gas even with cold water.
On the addition of mineral acid to an aqueous solution of borax, the compound formed is:
Na2B4O7 + H2SO4 + 5H2O → Na2SO4 + 4H3BO3 (orthoboric acid)
B3+ cannot exist in aqueous solution becuase of its :
Being covalent in nature because of small size and large charge it gets hydrolysed in aqueous solution according to the following reaction
BCl3 + 3H2O H3BO3 + 3HCl.
The decrease stability of higher oxidation state in p-block with increasing atomic number is due to :
The inert pair effect is the tendency of the electrons in the outermost atomic s orbital to remain unionized or unshared in compounds of post-transition metals. Down the group in the p-block, the inert pair effect becomes more pronounced on account of enhanced increase in effective nuclear charge. Therefore, due to reluctance of electrons of s-subshell to participate in chemical bonding, there occurs a decrease in stability of higher oxidation states in p-block with increasing atomic number.
The structural unit present in pyrosillicates is :
In pyrosilicate, one oxygen atom is shared between two SiO44- tetrahedra.
Select the incorrect statement about the XeF6.
Capped octahedron (or distorted octahedron)
In tetramer, four square pyramidal XeF5+ ions are joined to two similar ions by means of two bridging F- ions. The Xe-F distances are 1.84 A on the square pyramidal units and 2.23 A and 2.60 A in the bridging groups. The solid has various crystalline forms, of which three are tetrameric and a fourth has both hexamers and tetramers.
Based on the Valence Shell Electron Pair Repulsion (VSEPR) model, what is the geometry around the sulphur, carbon and nitrogen in the thiourea – S , S – dioxide, O2SC(NH2)2 ? (consider the Lewis structure with zero formal charges on all atoms).
Structure with zero formal charges on all atoms.
The following flow diagram represents the manufacturing of sodium carbonate
Which of the following options describes the reagents, products and reaction conditions (given in parentheses)?
When boron is fused with potassium hydroxide which pair of species are formed ?
B2H6 + NH3 → Addition compound In the above sequence Y and Z are respectively :
Orthosilicate ions, (SiO44-) undergo polycondensation to froms pyrosilicate [O3Si - O - SiO3]6-, in presence of:
2SiO44- + H2O → [O3Si - O - SiO3]6- + 2OH- ; 2SiO44- + 2H+ → [O3Si - O - SiO3]6- + H2O.
Which of the following statements is correct ?
(C) The hydrogen atoms are in a vertical plane with the axial fluorine atoms.
Hydrogen atoms lie in the CSF2 axial plane. We know that the it bond involving a p-orbital on the carbon atom must lie in the equatorial plane of the molecule.
The hybridisation and geometry of the anion of ICl3 in liquid phase is :
Which of the following orders is correct for the size ?
(1) Mg2+ < Na+ < F– < Al
(2) Al3+ < Mg2+ < Li+ < K+
(3) Fe4+ < Fe3+ < Fe2+ < Fe
(4) Mg > Al > Si > P
(1) Mg2+, Na+ and F- are isoelectronic and thus follows the order 12Mg2+ < 11Na+ < gF-. Al belongs to third period and has no charge so it is largest. Na+ = 102 pm; Mg2+ = 72 pm ; Al = 143 pm, F- = 133 pm.
(2) K+ has more number of shells than Mg2+ and Al3+. Al3+ and Mg2+ are isoelectronic but Al3+ has higher nuclear charge so Al3+ < Mg2+. Mg2+ and Li+ has diagonal relationship. But due to +2 charge in Mg2+, the Mg2+ is smaller than Li+ Hence Al3+ is the smallest one. K+ =1.38 Å, Li2+ = 0.76 Å Mg2+ = 0.72 Å and Al3+ = 0.535 Å.
(3) As the number of electrons are lost, the attraction between valence shell electrons and nucleus increases. As a consequence of this the electrons are pulled closer to the nucleus leading to the contraction in size of ions.
(4) Across the period the nuclear charge increases and thus the size of atoms decreases. Mg = 160 pm; Al = 143 pm; Si = 118 ; P = 110 pm.
Which reactions involves a change in the electron–pair geometry for the under lined atoms ?
Anhydrous AlCl3 is covalent. From the data given below
Lattice Energy = 5137 KJ/mol.
ΔH hydration for Al3+ = – 4665 KJ/mol
ΔH hydration for Cl– = – 381 KJ/mol
Identify the correct statement.
Hence AlC13 will dissolve and solution consists of hydrated Al3+ and CI- ions. In other words higher hydration energy compensates the higher ionisation energy. As a result, it will exist as hydrated Al3+ and hydrated CI- in aqueous solution.
In which of the following molecules / ions are all the bond angles not equal ?
Remaining molecules / ions have symmetrical structures ; so have all bonds equal.
When sodium nitrate is heated at 800ºC :
In which of the following reaction carbon monoxide is obtained as one of the products ?
(A) Na2Cr2O7 + 2C Na2CO3 + Cr2O3+ CO
(B) Ba(COO)2 BaCO3 + CO (C) [Fe(CN)6]4- + 6 H2SO4 + 6H2O Fe2+ + 6 CO + 6 NH4+ + 6 SO42-
The repeating structural units in silicone is :
Consider the following statement A to E –
(A) PS II and PSI both present in mesophyll cell in C-3 plant
(B) Mesophyll cell of C-4 plant lack rubisco
(C) Bundle sheath cell in C-4 plant lack PEPcase
(D)In C-4 plant calvin cycle is absent
(E)In CAM plant calvin cycle is absent
How many are correct-
Explanation- calvin cycle occur in all photosynthetic plant cell
At high light intensity in summer which of the following can not be limiting factor-
at high light intensity other factor is limiting
Which of the following is an organic acid not form in C-4 plant during photosynthesis-
Except alpha ketoglutaric acid all other form in C-4 plant
Mark the incorrect statetment-
Lamellar thylakoid lack PSII in mesophyll cell
Colour for chl a is
Chlorophyll is a green pigment that gives most plants their color.The reason that it is green is because it absorbs other colors of light such as red and blue, so in a way the green light is reflected out since the pigment does not absorb it.
Which can be absent in C-4 plants-
phosphoglycolate not form in C-4 plant as photorespiration absent
Consider the following-
(a) F0 is transmembrane channel which is responsible for facilitated diffusion of proton from lumen to matrix
(b) Chemiosmosis not require ATPase
How many are correct
chemiosmosis require a membrane, a proton pump, a proton gradient and ATPase
Which is incorrect regarding photorespiration-
PGA also form during photosynthesis
Mark the incorrect statement –
Current level of CO2 is limiting for C-3 plant not for C-4 plant
Mineral involve in chlorophyll biosynthesis -
Magnesium is part of chlorophyll and iron require for biosynthesis
Mark the incorrect statement-
micro-elements require in less than 10m mole per kg of dry matter
In which type of Cell can not have fix sugar in all -
mesophyll cell of C-4 plant lack calvin cycle and hence sugar
At CO2 concentraion of below 360ppm ,in C-4 plant rate of photosynthesis surely depend upon-
CO2 conc. below 360 ppm rate of photostynthesis depend upon CO2 conc. for C-4 plant as saturation conc. for C-4 plant is 360ppm
Which of the following is not role of boron-
Which of the following is not chemoautotroph-
all bacteria involve in nitrification are chemoautotroph and thiobaccilus are involve in denitrification
Mineral is component of nitrogenase and nitrate reductase both-
Deficiency of potassium and sulphur involve in both-
Deficiency of N, K, S and Mo causes stunted growth because of inhibition of cell division.
In which of the following iron not required -
Iron is required for catalase and peroxidase both
Mineral not absorb in form of anion-
Amination of which amino acid give amide –
Aspargine and glutamine are already amide and OAA amination give amino acid
Mark the correct statement –
Leghaemoglobin synthesis require both plant and bacteria , bacteria establish connection with both xylem and phloem and association of rhizobium and leguminous plants are non-obligatroy
Mark the incorrect statement-
nitrogenase is present for N2 fixation , which is perform by both bacteria and cyanobacteria both is prokaryotes
Consider the following statement A to E –
(A) Water potential values can be positive, negative and zero
(B) In an open beaker water potential is equal to solute potential
(C) Addition of solute decrease solute potential
(D) Maximum value of solute potential is zero
(E) pressure potential can not be zero
How many are correct-
Explanation- In open beaker pressure potential is zero and value of water potential is equal to solute potential and if no solute is present in beaker than value of solute potential is maximum which is equal to zero
Which of the following statement is not true -
channels are controlled through hormones and potential
Which of the following is passive process-
Explanation – all process not require energy
Few match the following is given, mark the wrongly matched –
simple diffusion is not selective it facilitated diffusion which is selective
Diffusion of any molecule across membrane depend upon on all except-
Diffusion is passive not require energy
Consider the following –
(B) osmotic pressure
(C) root pressure
(D) transpiration pull
(E) Osmotic potential
(F) Suction pressure
How many are positive pressure-
osmotic potential, suction pressure and transpiration pull are negative pressure
Which of the following can contribute as a force for transport of water in plant-
apoplast and symplast are pathway not force for transport of water
Which is correct regarding the transport in phloem-
since sieve tube is living so movement is symplast and pressure is positive due to entry of water
Correct Reason for root pressure –
Root pressure develop in roots due to accumulation of solute in Root xylem
Mark the incorrect -
Explanation- cytoplasmic streaming help symplast movement
During opening of stomata all occur except –
Cellulose microfibril orient radially
Which is involve in growth of root and root hair formation -
Consider the following statement-
(1) ABA act as antagonist against GA
(2) adventitious root growth promoted by auxin
How many correct-
Hormone involve in adventitious shoot formation-
Mutrient mobilization help in delay in senescence promoted by -
Mark the wrongly matched-
Kinetin is furfuryl aminopurine not zeatin
Mark the incorrect statement-
Differentiation in plants are also open
Which of the following is not product of redifferentiation-
pimary xylem is product of differentiation
Maximum number of plasmodesmatal connection present in -
Consider the following –
(1) two leaf having different relative growth can have same absolute growth
(2) two leaf having different absolute growth can have same relative growth
How many correct-
If two leaf have different intial surface area can have same absolute growth with different relative growth and vice versa
Mark the incorrect stataement-
In geometric growth increase is double than relatice growth rate is constant
If critical day length of LDP is 14hr than , mark the correct-
(I) flower when 11hr countinuos night is given
(II) flower when 8hr night is given than countinuos day
(III) flower when 10 hr day followed by 2hr darkness and again 10hr day
Mark the correct statement with respect to flowering-
If 14hr critical day length , it will flower when countinuos night of less than 10hr is given
Glycerol enter into respiratory pathway at –
Glycerol can enter at PGAL level
Tripalmitin produce … Number of CO2 in complete oxidation
Oxalo acetic acid is precursor of which amino acid-
OAA on amination give aspartate
Consider the following statement and mark correct –
In glycolysis catabolism of glucose occur
Number of oxidative decarboxylation in complete oxidation of fructose in cell is-
all carbon lost in aerobic respiration by oxidative decarboxylation
Excess of amino acid can give energy by converging at –
Mobile electron carrier which get oxidise by complex II is –
UQ is reduced by complex II
Which of the following reaction not occur in mitochondria –
Complex IV is reduced by cyt c not oxidised
Which of the following can not be respiratory substrate –
cell lack machinery to break coenzyme A for energy
Consider the following statement-
(1) oxidation of NADH occur fast in anaerobic than aerobic respiration
(2) Complex I of ETS is reduced by NADH
How many correct-
Number of ATP produced by triose phosphate in anaerobic respiration
triose enter into PGAL and produce 2ATP and 1 NADH
Which is correct order in decreasing percentage of elements present in human body-
Which of the following group of compounds obtained in acid insoluble pool when chemicals extracted in tricholoroacetic acid.
Explanation- linolenic acid and palmitic acid are present in acid soluble pool
Consider the following statements-
(A) t-RNA are heteropolymer
(B) all polysaccharides are heteropolymer
Mark the correct statement-
nucleic acid and protein are heteropolymer
Consider the following statements-
(A) Ribozyme are made up of nucleic acid
(B) Both DNA and RNA have polynucleotide chain
Mark the correct statement-
Co-factors affect enzyme by–
Mark the incorrectly matched cofactor –
Mark the correct statement –
Which of the following is covalently linked-
Which of the following is heterocyclic-
Which of the following is secondary metabolite-
Non- competitive inhibition is responsible for –
actual number of enzyme decrease and hence Vmax decrease
Mark the incorrect –
Amount of DNA and ploidy of diploid cell in G2 phase-
In G2 amount is double but ploidy is same
Histones are synthesise in –
During S phase of the cell cycle, the histone proteins are synthesized in a eukaryotic cell. The major event in S phase is DNA replication. The goal of this process is to create exactly two identical semi-conserved chromosomes. The cell prevents more than one replication from occurring by loading pre-replication complexes onto the DNA at replication origins during G1 phase which is dismantled in S phase as replication begins.
In a diploid somatic cell have 10 pair of chromosome present than total number of chromatids in G1 and G2 are respectively-
number of chromatid is double of chromosome and in G2 it is 4 time to pair of chromosome
Kinetochore is chemically is –