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A tank of uniform crosssection is completely filled with ice. The height of ice is H and its mass is m. When the entire ice melts, the work done by gravity is
(ρ_{ice} = 0.9 gm/cc, ρ_{water} = 1 g/cc and g represents acceleration due to gravity)
H height of frozen water becomes 0.9 H height of liquid on melting.
Thus U_{i} = m g(H/2) = 0.5 mgH, U_{f} = mg(0.9H/2) = 0.45 mgH
W_{G} = U_{i} – U_{f} = 0.5mgH − 0.45mgH = 0.05 mgH
A hydrogen like species having atomic number Z = 2, in ground state, is excited by means of electromagnetic radiation of frequency 1.315 × 10^{16} Hz. How many spectral lines will be observed in the emission spectrum?
(Planck’s constant h = 4.14 × 10^{15} eVs)
E = hν = 4.14 × 10^{15} × 1.315 × 10^{16} = 54.4 eV
Energy of the given species in ground state is
Ionization energy of the given species is = E_{1} = 54.4 eV
Thus the electron will get ionized and therefore become free.
So no spectral lines will be observed.
Velocity of a point on the equator of a rotating spherical planet is v. The angular velocity of the planet is such that the apparent value of acceleration due to gravity ‘g’ at the equator is half of that at the poles. The escape velocity of a particle from the surface of the planet is
Apparent value of g at pole is equal to
A positive charge q is projected from origin with a velocity along positive xaxis in a region having uniform magnetic field directed towards negative yaxis. If T is the time period of circular motion then the velocity vector of charge q at some instant t where
The charge executes circular motion in xz plane with its center lying on the negative zaxis. At time t where the charge has completed (3/4)th of the circle and so its velocity vector is pointing towards 1^{st} quadrant in xz plane.
⇒ v_{x} is positive, v_{z} is positive and vy is zero.
Four charges q_{1}, q_{2}, q_{3} and q_{4} are placed at the positions as shown in the figure, given q_{1} +q_{2}+ q_{3} + q_{4} = 0 . The electric field on zaxis
As q_{1} +q_{2} + q_{3} + q_{4} = 0 , z axis is an equipotential line. Thus electric field is either zero or perpendicular to the zaxis. When q_{2} = q_{4} and q_{3} = q_{1} electric field is zero at origin and in all other cases it is perpendicular to zaxis.
For the circuit as shown in the figure q_{1} and q_{2} be the charges on 3μF and 6μF capacitors respectively, then
At any time t the current distribution in the circuit will be as shown. If i_{2} is the current going out of upper loop through 3 μF capacitor then current i_{2} must also come back to this through 6μF capacitor to make the current in that loop i1.Thus the current through both the capacitors will be same at all time instants. Thus they will have equal charge.
A ball is released from position A and drops 10 m before striking a smooth incline. The coefficient of . If the time taken by the ball to strike the incline again is t then find the value of t^{2} [in (second)^{2}]. (g = 10 m/s^{2})
Just after collision component of velocity perpendicular to the incline is
Component of g perpendicular to incline is
∴ time after which the ball strikes the incline again is
Two identical beads P and Q of mass 1 kg each are connected by an inextensible massless string and they can slide along the two arms AB and BC of a rigid smooth wire frame in vertical plane. If the system is released from rest and vQ is the speed of bead Q when they have both moved by a distance of 0.1 m then find the value of (in m/s). (g = 10 m/s^{2})
Velocity of P and Q along the string should be same.
From conservation of mechanical energy, decrease in potential energy of Q = increase in kinetic energy of both
A solid uniform sphere rotating about its axis with kinetic energy E_{o} is gently placed on a rough horizontal plane. The coefficient of friction on the plane varies from point to point. After some time, the sphere begins pure rolling with total kinetic energy equal to E. Then find the value of
From conservation of angular momentum about point of contact
A variable voltage V = 2t is applied across an inductor of inductance L = 2H as shown in figure. Then find the rate at which magnetic potential energy stored in the inductor is increasing at t = 1 s (in J/s). Take the current through the inductor at t = 0 as zero.
Potential energy stored in the inductor is
A calorimeter of mass m contains an equal mass of water in it. The temperature of water and calorimeter is t_{2}. A block of ice of mass m and temperature t_{3} < 0^{o}C is gently dropped into the calorimeter. Let C_{1}, C_{2} and C_{3} be the specific heats of calorimeter, water and ice respectively and L be the latent heat of fusion of ice.
Q.
The whole mixture in the calorimeter becomes ice if
For whole mixture to become ice t_{f} ≤ 0^{o}C
Let t_{f} = −t_{o} then
From heat lost = heat gained we get
A calorimeter of mass m contains an equal mass of water in it. The temperature of water and calorimeter is t_{2}. A block of ice of mass m and temperature t_{3} < 0^{o}C is gently dropped into the calorimeter. Let C_{1}, C_{2} and C_{3} be the specific heats of calorimeter, water and ice respectively and L be the latent heat of fusion of ice.
Q.
The whole mixture in the calorimeter becomes water if
For whole mixture to convert to water t_{f} ≥ 0^{o}C
Let t_{f} = t_{o} then
From heat lost = heat gained we get
A calorimeter of mass m contains an equal mass of water in it. The temperature of water and calorimeter is t_{2}. A block of ice of mass m and temperature t_{3} < 0^{o}C is gently dropped into the calorimeter. Let C_{1}, C_{2} and C_{3} be the specific heats of calorimeter, water and ice respectively and L be the latent heat of fusion of ice.
Q.
Water equivalent of calorimeter is
Let water equivalent of calorimeter be m_{o} then m_{o}C_{2} = mC_{1} ⇒
A parallel plate capacitor has its plate horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown. The distance between the plates is d_{1}. Now the capacitor is connected with an electric source having voltage V. The separation between the plates changes to d_{2} at equilibrium. The mass of lower plate is ‘m’ and crosssectional area of each plate is A.
Q.
The spring constant k is
When capacitor was uncharged, for equilibrium of lower plate of mass m kx_{o} = mg where x_{o} is the compression in the spring. When voltage source is connected and lower plate again reaches equilibrium then compression in the spring is x_{o} – (d_{1} – d_{2}).
Electric force between capacitor plates is
A parallel plate capacitor has its plate horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown. The distance between the plates is d_{1}. Now the capacitor is connected with an electric source having voltage V. The separation between the plates changes to d_{2} at equilibrium. The mass of lower plate is ‘m’ and crosssectional area of each plate is A.
Q.
The maximum voltage V_{m} for a given k for which an equilibrium exists is
A parallel plate capacitor has its plate horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown. The distance between the plates is d_{1}. Now the capacitor is connected with an electric source having voltage V. The separation between the plates changes to d_{2} at equilibrium. The mass of lower plate is ‘m’ and crosssectional area of each plate is A.
Q.
When lower plate is slightly displaced about equilibrium position, time period T of small oscillations is
Let the lower plate be displaced upward by small distance x, then FBD of the plate is as shown.
Net restoring force acting on the lower plate is
Comparing with standard equation F = −Kx for SHM, we get
Two thin symmetrical lenses of different nature have equal radii of curvature R = 20 cm. The lenses are placed in contact and then immersed in water. The focal length of the system is found to be 24 cm. If the refractive indices of the two lenses are μ_{1} and μ_{2} respectively, then find the magnitude of 9(μ_{1}  μ_{2}). Refractive index of water is 4/3.
where are the focal lengths of the two lenses in water
A certain weak acid has a dissociation constant 1.0 x 10^{4}. The equilibrium constant for its reaction with strong base is
This is nucleophilic addition followed by dehydration.
The oxidation state of molybdenum in [(η^{7}tropylium)Mo(CO)_{3}]^{+} is
Tropylim cation has one positive charges and CO is neutral ligand.
In metalolefin interaction, the extent of increase in metal ⎯→ olefin πbackdonation would
The highest oxidation state of an element in the following compound that behaves as an acid in H_{2}SO_{4} is
AcOH, HNO_{2}, HNO_{3}, H_{2}O, HClO, HClO_{4}
HClO_{4} is stronger acid than H_{2}SO_{4}
How many unpaired electrons are present in O_{2} molecule?
One mole of Pb_{3}O_{4} is separately reacted with excess of HCl and HNO_{3}. The difference in moles of HCl and HNO_{3} is
Pb_{3}O_{4} is a mixture of PbO_{2} and PbO. Only PbO reacted with HNO_{3} since Pb^{4+} is a very good oxidizing agent. HCl is a good reducing agent that can reduces Pb^{4+} into Pb^{2+}.
Pb_{3}O_{4} + 8HCl ⎯⎯→ 3PbCl_{2} + Cl_{2} + 4H_{2}O
Pb_{3}O_{4} + 4HNO_{3} ⎯⎯→ 2Pb(NO_{3})_{2} + PbO_{2} + 2H_{2}O
How many moles of phenyl hydrazine are used in the formation of osazone from glucose?
If a concentrated solution of copper sulphate is placed at the bottom of a beaker of water or that of a dilute solution of copper sulphate is carefully poured over it there will be a two distinct layer visible. However, after some time the boundaries will disappear. This property is called diffusion. If we now consider a solution which is separated from the pure solvent by a semipermeable membrane then the solvent particles move from the pure solvent region through the SPM to the solution region. This phenomenon is called osmosis.
Q.
for an indefinitely dilute boundary then dμ_{1} is
If a concentrated solution of copper sulphate is placed at the bottom of a beaker of water or that of a dilute solution of copper sulphate is carefully poured over it there will be a two distinct layer visible. However, after some time the boundaries will disappear. This property is called diffusion. If we now consider a solution which is separated from the pure solvent by a semipermeable membrane then the solvent particles move from the pure solvent region through the SPM to the solution region. This phenomenon is called osmosis.
Q.
Which of the following solution has highest osmotic pressure?
If a concentrated solution of copper sulphate is placed at the bottom of a beaker of water or that of a dilute solution of copper sulphate is carefully poured over it there will be a two distinct layer visible. However, after some time the boundaries will disappear. This property is called diffusion. If we now consider a solution which is separated from the pure solvent by a semipermeable membrane then the solvent particles move from the pure solvent region through the SPM to the solution region. This phenomenon is called osmosis.
Q.
Which of the following is correct?
Thionyl chloride can be synthesized by chlorinating SO_{2} using PCl_{5}. Thionyl chloride is used to prepare anhydrous ferric chloride starting from its hexahydrated salt. Alternatively, the anhydrous ferric chloride can also be prepared from hexahydrated salt by treating with 2,2dimethoxypropane.
Q.
Consider the following reaction
The compound X is
SO_{2} + PCl_{5} ⎯→ SOCl_{2} + POCl_{3}
Thionyl chloride can be synthesized by chlorinating SO_{2} using PCl_{5}. Thionyl chloride is used to prepare anhydrous ferric chloride starting from its hexahydrated salt. Alternatively, the anhydrous ferric chloride can also be prepared from hexahydrated salt by treating with 2,2dimethoxypropane.
Q.
In the preparation of anhydrous FeCl_{3} from hexahydrated salt, SOCl_{2} acts as
SOCl_{2} acts as dehydrating agent.
Thionyl chloride can be synthesized by chlorinating SO_{2} using PCl_{5}. Thionyl chloride is used to prepare anhydrous ferric chloride starting from its hexahydrated salt. Alternatively, the anhydrous ferric chloride can also be prepared from hexahydrated salt by treating with 2,2dimethoxypropane.
Q.
In the following reaction
FeCl_{3}.6H_{2}O + 6MeC(OH)(OH)Me ⎯⎯→FeCl_{3} + Y + Z
FeCl_{3}.6H_{2}O + 6MeC(OH)(OH)Me ⎯⎯→ FeCl_{3} + 6 Acetone + 12 MeOH
Each question has four statements (A, B, C and D) given in Columns I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II.
Nylon6,6 is a condensation product of hexamethylenediamine and adipic acid, it is a poly amide fiber polymer. SBR is the addition polymer of styrene and butadiene; it is a synthetic rubber polymer which is also known as BunaS. Nylon6 is a polyamide polymer of Caprolactam, it if also an example of fiber polymer. PTFE (Polytetrafluro ethylene) is the addition polymer of Tetrafluoroethylene.
Each question has four statements (A, B, C and D) given in Columns I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II.
How many lone pair of electrons at Xe atom are present in XeOF_{4}?
Matrices of order 3 × 3 are formed by using the elements of the set A = {−3, −2, −1, 0, 1, 2, 3}, then probability that matrix is either symmetric or skew symmetric is
Probability that matrix is symmetric
Again that matrix is skew symmetric
One matrix containing all elements = 0; is common in both type of matrices.
∴ Required probability
Through any point (x, y) of a curve which passes through the origin, lines are drawn parallel to the coordinate axes. The curve, given that it divides the rectangle formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of
Let P(x, y) be the point on the curve passing through the origin O(0, 0), and let PN and PM be the lines parallel to the x and yaxes, respectively. If the equation of the curve is y = y(x), the area POM equals and the
This solution represents a parabola. We will get a similar result if we had started instead with 2(PON) = POM.
is neither an even function nor an odd function. {0}, where [.] is greatest integer function.
S_{3}: f(x) = sgn (x) and g(x) = sgn(sgn(x)) is not a pair of identical function.
S_{4}: The sum of two non periodic function is always a non periodic function.
If the number of subsets X of {1, 2, 3, …. 10} such that X contains at least two elements and no two elements of X differ by 1 is K, then sum of digits of K is equal to
Every allowable kelements subset corresponds to a way of choosing k out of a row of 10 objects so that no two are adjacents, remove (k − 1) unselected objects, one from each gap.
This establishes, for each k ≥ 2 , a one to one correspondence between allowable subsets of {1, 2, …… 10} containing k elements, and the number of ways of choosing k out (10 − k+ 1) objects. It follows that there are
Allowable subsets.
A rectangle, HOMF is constructed with sides HO = 11 and OM = 5. The triangle ABC has orthocenter H, circumcentre O, M is the mid point of BC and F is foot of altitude from A. If length of BC = l , then l/7 is equal to
The centroid G of triangle is collinear with H and O, and G lies two thirds of way from A to M. Therefore H is two thirds of the way from A to F, so AF = 3 × OM = 15.
Since the triangle BFH and AFC are similar, hence,
⇒ BF. FC = FH.AF = 75.
Now, BC^{2} = (BF + FC)^{2} = (BF – FC)^{2} + 4BF.FC
But, FC – BF = (FM + MC) – (BM – FM) = 2FM = 2 HO = 22
If the smallest integer with exactly 24 divisors is N, then N/40 is equal to
If n is the required number and
Then divisor of n : T(n) = (α_{1} + 1)(α_{2} + 2)…….(α_{k} + 1)
But 24 can be written as the product of 2 or 3 or 4 factors.
Corresponding to each factorization; we can get a smallest composite number.
∴ The smallest number having 24 divisors is 360.
In a certain town, the probability that, all the outgoing telephone lines are jammed, in the telephone exchange is 1/4. The probability that the customer will attempt to telephone is 3/20. If a customer telephones and it fails to get connected, the probability that he would replace his telephone by a cell phone is 3/4. If P is the probability that the customer replaces the telephone by a cell phone, when the outgoing lines of the exchange are jammed then is equal to
(where [ ] denotes the greatest integer function)
Let A : be the event that the outgoing telephone lines are jammed.
B : be the event that the customer attempt telephone.
C : be the event that the customer replaces the telephone by a cell phone due to the failure caused by the jamming of telephone lines.
Here A and B are independent events, but A∩B and C are mutually dependent. Required probability
The terms of are all integers (where a, x > 0). If K is the least composite odd integral value of ‘a’, then K/5 is equal to
∴ The terms of the AP are :which are integers
∴ a = 5n^{2}, n∈N
for n = 1, a = 5 which is not composite
for n = 2, a = 20 which is composite but not odd.
for n = 3, a = 45 which is the least composite odd.
Let z_{1} and z_{2} be complex numbers such that and the roots α and β of x^{2} + z_{1}x + z^{2} + m = 0 for some complex number m satisfies
Q.
The locus of the complex number m is a curve
Let z_{1} and z_{2} be complex numbers such that and the roots α and β of x^{2} + z_{1}x + z^{2} + m = 0 for some complex number m satisfies
Q.
The maximum value of m is
Let z_{1} and z_{2} be complex numbers such that and the roots α and β of x^{2} + z_{1}x + z^{2} + m = 0 for some complex number m satisfies
Q.
The minimum value of m is
If functions f(x) and g(x) are continuous on the interval [a, b] and g(x) retain the same sign on [a, b] then there is c ∈ (a , b) such that . This is known as MeanValue Theorem. This result can be used to estimate some definite integrals. Other results which can be used for estimation are
(i) If f increases and has a concave graph in the interval [a, b] then
(ii) If f increases and has a convex graph in the interval [a, b] then
Q.
Using MeanValue Theorem, the best upper bound of
If functions f(x) and g(x) are continuous on the interval [a, b] and g(x) retain the same sign on [a, b] then there is c ∈ (a , b) such that . This is known as MeanValue Theorem. This result can be used to estimate some definite integrals. Other results which can be used for estimation are
(i) If f increases and has a concave graph in the interval [a, b] then
(ii) If f increases and has a convex graph in the interval [a, b] then
Q.
Using (i) or (ii) (above), the best upper bound of
If functions f(x) and g(x) are continuous on the interval [a, b] and g(x) retain the same sign on [a, b] then there is c ∈ (a , b) such that . This is known as MeanValue Theorem. This result can be used to estimate some definite integrals. Other results which can be used for estimation are
(i) If f increases and has a concave graph in the interval [a, b] then
(ii) If f increases and has a convex graph in the interval [a, b] then
Q.
Using (iii) (above), the best upper bound of
(A) Period of sin^{2} π x is 1 and period of x – [x] is equal to 1 Thus period of f(x) is 1.
⇒ f(x) takes 6 integer values
⇒ k = 6.
(C) f ′(x) = 3x 2+ 2mx + m
for f(x) to be invertible it must be one–one
Hence any point on L_{1} and L_{2} can be A (λ, λ −1, λ) and B(2μ  1, μ, μ)
In a Δ A BC, A ≡ (α, β ) , B ≡ (1, 2), C ≡ (2, 3) and point A lies on the line y = 2x + 1 where α,β∈ integer. Area of triangle ABC is Δ such that [Δ] = 2, where [.] denotes greatest integer function. Find number of all possible coordinates of A.
Since ‘α’ is a n integer
⇒ α = −5, −4, 4, 5
∴ A ≡ (−5, − 9 ) , ( −4, −7 ) , ( 4, 9) , (5,11)
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