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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 Q_{1} = mL = 4 × 540 = 2160 cal.
Heat required to raise temperature of water from 46°C t 100°C is Q_{2} = mS Δθ = 20 × 1 × 54 = 1080 J
Hence all steam is not converted to water only half steam shall be converted to water
So 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 ( α_{1}, α_{2} = linear expansion coefficient A_{1}, A_{2} = Area of rods Y_{1}, Y_{2} = Young modulus )
For a gas sample with N_{0} number of molecules, function N(V) is given by : N(V) = = V V_{0} and N(V) = 0 for V > V_{0}. Where dN is number of molecules in speed range V to V+ dV. The rms speed of the molecules is :
The coefficient 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 λ_{0}. If at another temperature T' the power radiated is
P' and wavelength at maximum intensity is then
Thermal coefficient of volume expansion at constant pressure for an ideal gas sample of n moles having pressure P_{0}, volume V_{0} and temperature T_{0} is
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 ]
During an experiment, an ideal gas is found to obzey a condition The
gas is initially at temperature T, pressure P and density ρ. The gas expands such that density changes to .
When the temperature of a copper coin is raised by 80 ^{o}C, its diameter increases by 0.2%,
Graph shows a hypothetical speed distribution for a sample of N gas particles (for V > V_{0 ;} =0)
Since PV = nRT therefore P and V both can change simultaneously keeping temperature constant.
STATEMENT–1 : A gas is kept in an insulated cylinder with a movable piston, in compressed state. As the piston is released, temperature of the gas decreases.
STATEMENT–2 : According to the kinetic theory of gas, a molecule colliding with the piston must rebound with less speed than it had before the collision. Hence average speed of the molecules is reduced.
Statement1 is True, Statement2 is True; Statement2 is a correct explanation for Statement1
Statement1 : As the temperature of the blackbody increases, the wavelength at which the spectral intensity (E_{λ}) is maximum decreases.
Statement2 : The wavelength at which the spectral intensity will be maximum for a black body is proportional to the fourth power of its absolute temperature.
From Wein's law λ_{m}T = constant i.e., peak emission wavelength
. Hence as T increase λ_{m} decreases.
The atmospheric lapse rate
For small volumes of gas, according to kinetic theory of gases, all parts of the gas are at the same temperature. But for huge volumes of gas like atmosphere, assumption of a uniform temperature throughout the gas is not valid. Different parts of the atmosphere are at different temperatures. Apart from the surface of the earth, variations also occur in temperature at different heights in the atmosphere.
The decrease in temperature with height called the atmospheric lapse rate is similar at various locations across the surface of the Earth. By analyzing the data collected at various locations, it is found that average global lapse rate is – 6.7 °C/Km.
The linear decrease with temperature only occurs in the lower part of the atmosphere called the troposphere. This is the part of the atmosphere in which weather occurs and our planes fly. Above the troposphere is the stratosphere, with an imaginary boundary separating the two layers. In the stratosphere, temperature tends to be relatively constant.
Absorption of sunlight at the Earth’s surface warms the troposphere from below, so vertical convection currents are continually mixing in the air. As a parcel of air rises, its pressure drops and it expands. The parcel does work on its surrounding, so that its internal energy and therefore, its temperature drops. Assume that the vertical mixing is so rapid as to be adiabatic and the quantity TP^{(1 – λ)/λ} has a uniform value through the layers of troposphere.
(M is molecular mass of the air, R is universal gas constant, g is gravitational acc., P and T are pressure and temperature respectively at the point under consideration and y is height.)
Q. If behaviour of the mixing of parcels of air is approximately assumed to be adiabatic then lapse rate can be expressed as :
The atmospheric lapse rate
For small volumes of gas, according to kinetic theory of gases, all parts of the gas are at the same temperature. But for huge volumes of gas like atmosphere, assumption of a uniform temperature throughout the gas is not valid. Different parts of the atmosphere are at different temperatures. Apart from the surface of the earth, variations also occur in temperature at different heights in the atmosphere.
The decrease in temperature with height called the atmospheric lapse rate is similar at various locations across the surface of the Earth. By analyzing the data collected at various locations, it is found that average global lapse rate is – 6.7 °C/Km.
The linear decrease with temperature only occurs in the lower part of the atmosphere called the troposphere. This is the part of the atmosphere in which weather occurs and our planes fly. Above the troposphere is the stratosphere, with an imaginary boundary separating the two layers. In the stratosphere, temperature tends to be relatively constant.
Absorption of sunlight at the Earth’s surface warms the troposphere from below, so vertical convection currents are continually mixing in the air. As a parcel of air rises, its pressure drops and it expands. The parcel does work on its surrounding, so that its internal energy and therefore, its temperature drops. Assume that the vertical mixing is so rapid as to be adiabatic and the quantity TP^{(1 – λ)/λ} has a uniform value through the layers of troposphere.
(M is molecular mass of the air, R is universal gas constant, g is gravitational acc., P and T are pressure and temperature respectively at the point under consideration and y is height.)
Q. Mechanical equilibrium of the atmosphere requires that the pressure decreases with altitude according to . Assuming free fall acceleration to be uniform, then lapse rate is given by
The atmospheric lapse rate
For small volumes of gas, according to kinetic theory of gases, all parts of the gas are at the same temperature. But for huge volumes of gas like atmosphere, assumption of a uniform temperature throughout the gas is not valid. Different parts of the atmosphere are at different temperatures. Apart from the surface of the earth, variations also occur in temperature at different heights in the atmosphere.
The decrease in temperature with height called the atmospheric lapse rate is similar at various locations across the surface of the Earth. By analyzing the data collected at various locations, it is found that average global lapse rate is – 6.7 °C/Km.
The linear decrease with temperature only occurs in the lower part of the atmosphere called the troposphere. This is the part of the atmosphere in which weather occurs and our planes fly. Above the troposphere is the stratosphere, with an imaginary boundary separating the two layers. In the stratosphere, temperature tends to be relatively constant.
Absorption of sunlight at the Earth’s surface warms the troposphere from below, so vertical convection currents are continually mixing in the air. As a parcel of air rises, its pressure drops and it expands. The parcel does work on its surrounding, so that its internal energy and therefore, its temperature drops. Assume that the vertical mixing is so rapid as to be adiabatic and the quantity TP^{(1 – λ)/λ} has a uniform value through the layers of troposphere.
(M is molecular mass of the air, R is universal gas constant, g is gravitational acc., P and T are pressure and temperature respectively at the point under consideration and y is height.)
Q. The value of theoretical lapse rate on the earth is (use g = 9.8 m/s^{2} ; R = 8.3 J/molk and M = 29 g/mol)
The figures given below show different processes (relating pressure P and volume V) for a given amount
of an ideal gas. ΔW is work done by the gas and ΔQ is heat absorbed by the gas.
ColumnI ColumnII
(A) In Figure (i) (p) ΔQ > 0.
(B) In Figure (ii) (q) ΔW < 0.
(C) In Figure (iii) (r) ΔQ < 0.
(D) In Figure (iv) (for complete cycle) (s) ΔW > 0.
(t) ΔU < 0.
The thermal conductance (reciprocal of the thermal resistance) for axial flow of a truncated cone of length
the radius of the two ends are r_{1} and r_{2} assuming that thermal conductivity of the material is K is
. Find the value of n2.
Consider a vertical tube open at both ends. The tube consists of two parts, each of different crosssections
and each part having a piston which can move smoothly in respective tubes. The two pistons are joined together by an inextensible wire. The combined mass of the two piston is 5 kg and area of crosssection of the upper piston is 10 cm^{2} greater than that of the lower piston. Amount of gas enclosed by the pistons is one mole. When the gas is heated slowly, pistons move by 50 cm. Find rise in the temperature of the gas, in the form (X/R) K where R is universal gas constant. Use g = 10 m/s^{2 }and outside pressure = 10^{5}N/m^{2}). Fill value of X in the answer sheet.
One mole of monoatomic ideal gas undergoes a process ABC as shown in figure. The maximum temperature of the gas during the process ABC is in the form . Find X.
One mole of an ideal gas is kept enclosed under a light piston (area=10^{2}m^{2}) connected by a compressed
spring (spring constant 200 N/m). The volume of gas is 0.83 m^{3} and its temperature is 100K. The gas
is heated so that it compresses the spring further by 0.1 m. Find the work done by the gas in the
process in joules ? (Take R = 8.3 J/Kmole and suppose there is no atmosphere).
(C) The hydrogen atoms are in a vertical plane with the axial fluorine atoms.
Hydrogen atoms lie in the CSF_{2} axial plane. We know that the π bond involving a porbital on the carbon atom must lie in the equatorial plane of the molecule.
The hybridisation and geometry of the anion of ICl_{3} in liquid phase is :
In tetramer, four square pyramidal XeF_{5}^{+} ions are joined to two similar ions by means of two bridging F^{} ions. The
XeF distances are 1.84 Å on the square pyramidal units and 2.23 Å and 2.60 Å 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, O_{2}SC(NH_{2})_{2} ? (consider the Lewis 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 ?
Orthosilicate ions,(SiO_{4}^{4}) undergo polycondensation to forms pyrosilicate, [O_{3}Si – O – SiO_{3}]^{6}, in presence
of :
Select the correct statement(s) with respect to the pπ–dπ dative bond.
(A) Steric repulsions of bulkier groups and pπ–dπ dative bonding favour for a linear Si–O–Si group.
(B) Due to stabillization of the conjugate base anion by O(pπ) → Si(dπ) bonding motion.
(C) It is pyramidal because pπ–dπ bonding is not effective due to the bigger size of phosphorus atom.
(D) Most effective pπ–dπ overlapping due to small size of chlorine.
(A) Nitrogen is more electronegative than phosphorus.
So, dipole moment of trimethylamine is greater than trimethy phosphine.
In trisilyl ether the lone pair of electron on oxygen atom is less easily available for donation because of pπdπ
delocalisation due to presence of the vacant dorbital with Si. This however is not possible with carbon in CH_{3}–O–CH_{3} due to the absence of dorbital making it more basic.
(C) Bond order of C_{2}
and O_{2} are same i.e., 2. In C_{2} molecules both bonds are πbonds whereas, there is one σ and one πbond in O_{2} molecule
C_{2} = 131 pm ; O_{2} = 121 pm.
Highly pure dilute solution of sodium in liquid ammonia :
Blue colour of the solution is due to ammoniated electrons and
good conductor of electricity because of both ammoniated cations and ammoniated electrons.
(change of oxidation states from 0 to – 1).
Statement1 : The melting points of the various fluorides of xenon decrease in the order XeF_{2} > XeF4 > XeF_{6}.
Statement2 : Xe – F bond length in XeF_{2}, XeF_{4} and XeF_{6} decreases in the order XeF_{2} > XeF4 > XeF_{6}.
S1 : The melting point depends on the packing of the molecules. Symmetrical molecules are more closely packed as compared to unsymmetrical molecules. Hence symmetrical molecules will have higher melting points as compared to unsymmetrical molecules.
S2 : This may be owing to the increase in charge on xenon atom with increasing number of F atoms.
XeF_{2} XeF_{4} XeF_{6 }
2.00 1.95 1.89
Statement1 : Ethers behave as bases in the presence of mineral acids.
Statement2 : It is due to the presence of lone pair of electrons on the oxygen atom.
The elements of group 1 describe, more clearly than any other group of elements, the effects of increasing the size of atoms or ions on the physical and chemical properties. The chemical and physical properties of the elements are closely related to their electronic structures and sizes. These metals are highly electropositive and thus form very strong bases, and have quite stable oxosalts. In the manufacturing of sodium hydroxide, chlorine and sodium carbonate, the sodium chloride is used as starting material.
Basic strength of the oxides increase in the order Li_{2}O < Na_{2}O < K_{2}O < Rb_{2}O < Cs_{2}O. The increase in basic strength is due to the decrease in I.E. and increase in electropositive character.The melting points of the halides decrease in the order NaF > NaCl > NaBr > NaI, as the size of the halide ion increases. The decrease in melting point is due to increase the covalent character with increase in the size of anion according to Fajan's rule.
The elements of group 1 describe, more clearly than any other group of elements, the effects of increasing the size of atoms or ions on the physical and chemical properties. The chemical and physical properties of the elements are closely related to their electronic structures and sizes. These metals are highly electropositive and thus form very strong bases, and have quite stable oxosalts. In the manufacturing of sodium hydroxide, chlorine and sodium carbonate, the sodium chloride is used as starting material.
Q. Which of the following acts as an oxidising as well as reducing agent ?
NaNO2 (sodium nitrite) acts both as oxidising agent and reducing agent because the nitrogen atom in it is in +3 oxidation state (+3 is intermediate oxidation state for nitrogen)
Oxidising property : 2NaNO_{2} + 2KI + 2H_{2}SO_{4 }→ NaSO_{4} + K_{2}SO_{4} + 2NO + 2H_{2}O + I_{2}
Reducing property : NaNO_{2} + H_{2}O_{2} → NaNO_{3} + H_{2}O.
The elements of group 1 describe, more clearly than any other group of elements, the effects of increasing the size of atoms or ions on the physical and chemical properties. The chemical and physical properties of the elements are closely related to their electronic structures and sizes. These metals are highly electropositive and thus form very strong bases, and have quite stable oxosalts. In the manufacturing of sodium hydroxide, chlorine and sodium carbonate, the sodium chloride is used as starting material.
Q. Sodium hydroxide is manufacture by the electrolysis of brine solution. The reaction byproducts are :
Match the entries in Column I with the correctly related characteritic(s) in Column II.
Column – I Column – II
(A) Cyclic silicates (p) Thortveitite
(B) Single chain silicates (q) Beryl
(C) Pyro silicates (r) General formula is (SiO_{3})_{n}^{2n –}
(D) Sheet silicates (two dimensional) (s) Two oxygen atoms per tetrahedron are shared
(t) Mica (muscovite)
(A) Two oxygen atoms per tetrahedron are shared forming rings. They have general formula (SiO_{3})_{n}^{2n}.
Be_{3}Al_{2}[Si_{6}O_{18}], beryl.
(B) Two oxygen atoms per tetrahedron are shared forming a chain of tetrahedron. They have general
formula (SiO_{3})_{n}^{2n}.
(C) One oxygen atom per tetrahedron is shared. They have general formula Si_{2}O_{7}^{2} . Thortveitite.
(D) Three oxygen atoms per tetrahedron are shared. They have general formula (Si_{2}O_{5})_{n}^{2} .
Mica (muscovite).
Give the atomic number of the inert gas atom in which the total number of delectrons is equal to the difference in the number of total p and selectrons.
How many maximum atoms lie in the same plane in the sructure of methylenesulphurtetrafluoride (CH_{2}SF_{4})?
There will be three different flourineflourine distances in molecule CF_{2}(C)_{2}CF_{2}. Assuming ideal bond angles for a particular hybridisation (assume no distortion due to double bonds) find out the smaller flourineflourine distance and fill the result (in pm) in the increasing order in your answer sheet. Round off your answer to the nearest whole number.
(Given that C–F bond length = 134 pm, C=C bond length = 134 pm √3 = 1.73)
Answer the following questions with respect to the compound NO[BF_{4}].
(P) Bond order of the part underlined.
(Q) Total number of σ bonds in the compound.
(R) Total number of π bonds in the compound.
(S) Number of hybrid orbitals involved in the hybridisation of boron.
NO^{+}[BF4]^{}
(P) Molecular orbital electronic configuration ;
NO^{+} is derivative of O_{2}, so MOT configuration is :
σ1s^{2} σ*1s^{2} σ2s^{2} σ*2s^{2} σ2p_{x}^{2π}2p_{y}^{2} = π2p_{z}^{2}
(x is taken as molecular axis)
Bond order = = 3
One σ bond and two ^{π} bonds.
(Q) Number of σ bonds in NO^{+} is one and in BF_{4}^{} are four. So total number of σ bonds are five.
(R) Number of σ bonds in NO^{+} are two.
(S)
Steric number of central atom boron is 4 + 0 = 4 ; so its hybridisation is sp^{3} and thus the number of hybrid orbitals
involved in sp^{3} hybridisation is four.
From a point P(1, 2) pair of tangent’s are drawn to a hyperbola ‘H’ in which one tangent to each arm of hyperbola. Equation of asymptotes of hyperbola H are √3x – y + 5 = 0 & √3x + y – 1 = 0 then eccentricity of ‘H’ is
Length of latus rectum of parabola which touches the line 3x + 4y + 7 = 0 having focus at (2, 3) and tangent at vertex passes through a point (1, 1), is
An ellipse whose focii are (2, 4) & (14, 9) and touches xaxis then its eccentricity is
If tangent and normal at P to ellipse intersect major axis at T and N in such a way that ratio of area of PTN and PSS` is 91/60, then area of PSS` is (S and S` are focii)
An ellipse E has its centre at (1, 2), focus at (6, 2) and passes through the point P(4, 6). A hyperbola H is concentric with the ellipse and its transverse axis coincides with the major axis of the ellipse. If E and H intersect each other at a right angle at P, then the equation of the hyperbola is
Let ‘P’ be a point which does not lie outside the triangle ABC, A = (3, 2), B = (0, 0), C = (0, 4) which satisfy d (P, A) < maximum {d (P, B), d (P, C)} then maximum distance of P from side BC, where d (P, A) gives the distance between P & A, is
Let P, Q, R and S be the feet of the perpendiculars drawn from a point (1, 1) upon the lines x + 4y = 12, 4y – x = 4 and their angle bisectors respectively, then equation of the circle which passes through Q, R, S is
Equation of the straight line meeting the circle with centre at origin and radius equal to 5 in two points at equal distances of 3 units from the point A(3, 4), is
Let R(x_{1}, y_{1}) and S(x_{2}, y_{2}) be the end points of latus rectum of parabola y^{2} = 4x. The equation of ellipse with latus rectum RS and eccentricity 1/2 are (a > b)
If equation of tangents at P, Q and vertex A of a parabola are 3x + 4y – 7 = 0, 2x + 3y – 10 = 0 and x – y = 0 respectively, then
If H = (3, 4) and C = (1, 2) are orthocentre and circumcentre of ΔPQR and equation of side PQ is x – y + 7 = 0, then
If a variable tangent of circle x^{2} + y^{2} = 1 intersects the ellipse x^{2} + 2y^{2} = 4 at points P and Q then the locus of the point of intersection of tangents at P and Q is :
Statement1 : Let L_{1} = 0 & L_{2} = 0 are two non perpendicular intersecting straight lines in two dimensional plane and if for λ = 1, L_{1} + λL_{2} = 0 gives line passes through acute angular region between the lines then for λ = – 4, L_{1} + λL_{2} = 0 gives line passes through obtuse angular region.
Statement2 : Points (x_{1}, y_{1}) and (x_{2}, y2) are on the same / opposite sides of a line ax + by + c = 0 if and only if ax_{1} + by1 + c and ax_{2} + by_{2} + c are of same/opposite signs respectively.
Statement1 : Normal drawn at a fixed point P(t_{1}), t1 0, on the parabola y^{2} = 4ax again intersects the
parabola at point t_{2} for all nonzero real values of t_{2}.
Statement2 : Normal drawn at a point P(t_{1}), t1 0, on the parabola y^{2} = 4ax again intersects the parabola
at the point t_{2}, where t_{2} = – t_{1} – .
An ellipse has semimajor axes of lengths 2 and semiminor axis of length 1. It is slipping between the coordinate axes in the first quadrant while maintaining contact with both xaxis and yaxis.
Q. The locus of centre of the ellipse is
The coordinate axes are tangent to the ellipse at any position. So,
origin lies on the director circle of the ellipse.
Let C be (α, β). The equation of director circle depends only on the location of centre of ellipse and its semimajor and minor axes, not on their orientation.
Hence, director circle is (x – α)^{2} + (y – β)^{2} = a^{2} + b^{2} = 2^{2} + 1^{2} = 5
Since it passes through (0, 0)
α^{2} + β^{2} = 5
so locus is x2 + y2 = 5
An ellipse has semimajor axes of lengths 2 and semiminor axis of length 1. It is slipping between the coordinate axes in the first quadrant while maintaining contact with both xaxis and yaxis.
Q. The locus of foci of the ellipse is
An ellipse has semimajor axes of lengths 2 and semiminor axis of length 1. It is slipping between the coordinate axes in the first quadrant while maintaining contact with both xaxis and yaxis.
Q. When the ordinate of the centre of ellipse is 1, a light ray passing through origin and centre of ellipse strikes the tangent at the vertex of the parabola y^{2} + 4x – 10y – 15 = 0. The reflected ray cuts yaxis at the point
You may use the fact that the product of perpendicular distances of any tangent on ellipse from its foci is
always equal to square of semiminor axis.
Since the path of centre is x^{2} + y^{2} = 5 when y = 1, x = 2.
The incident ray passes through origin and (2, 1) ⇒ its equation is y = 1/2x
Given parabola is (y – 5)^{2} = – 4 (x – 10)
whose vertex (10, 5) lies on y = 1/2x.
The tangent at the vertex is the line x = 10.
Figure shows that OP = 2 (OR) = 10
so reflected ray cuts yaxis at (0, 10)
Given two circle let the radius of the third circle, which
is tangent to the two given circles and to their common diameter, be , then find the value of p.
Let O_{1} and O_{2} be the centres of the circles of radius R = 5 and r = 3 respectively and O_{3} be the centre of the third circle. Let x be the radius of the third circle. and P be the point of tangency of the circle and the diameter O_{1} O_{2}.
If (α, β) is the centroid of a triangle PQR where P, Q and R three points lying on the parabola
(y – 4)^{2} = 15 (x + 3) and normals at P and Q meet at R. Find the largest value of m satisfying
α + β > m for all possible values of α and β.
Consider a family of circles cutting the circles x^{2} + y^{2} + 2x – 4y – 4 = 0 and x^{2} + y^{2} – 4x + 4y + 4 = 0
orthogonally. Show that the chords in which the circle 2x^{2} + 2y^{2} – 5x – 6y – 2 = 0 cuts the member of
the family are concurrent at a point. If the coordinates of this point are (a, b), then find the value of
50(a + b).
An isosceles ΔABC is inscribed in the circle x^{2} + y^{2} = 16. From points A, B and C three ordinates are drawn
which cut the ellipse at the points P, Q and R respectively such that A and P are on the
opposite sides, Q and B are on the same side and C and R are on the same side of the major axis. If ΔPQR
is right angled at P and θ is the smallest angle of the ΔABC, then find the value of 16tan^{2}θ + 8 tan θ + 9.
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