1 Crore+ students have signed up on EduRev. Have you? 
A metal wire PQ slides on parallel metallic rails having separation 0.25 m, each having negligible resistance. There is a 2Ω resistor and 10V battery as shown in figure. There is a uniform magnetic field directed into the plane of the paper of magnitude 0.5 T. A force of 0.5N to the left is required to keep the wire PQ moving with constant speed to the right. With what speed is the wire PQ moving? (Neglect selfinductance of the loop)
Solving V = 16 m/s.
In the given figure a ring of mass m is kept on a horizontal surface while a body of equal mass 'm' attached through a string, which is wounded on the ring. When the system is released the ring rolls without slipping. Consider the following statements and choose the correct option.
(i) acceleration of the centre of mass of ring is g/3
(ii) acceleration of the hanging particle is 2g/3
(iii) frictional force (on the ring) acts along forward direction
(iv) frictional force (on the ring) acts along backward direction
Free body diagram of ring and mass is
Since the ring is in pure rolling so its above point acceleration will be 2a and similarly acceleration of mass will be 2a.
The equation of motion of the system is
mg – T = m(2a) ......(1)
T = ma .......(2)
Substitute the value of T in equation (1)
mg  ma = 2ma
3ma = mg
a = g/3
So the acceleration of the center of the ring is, g/3
And acceleration of the hanging mass is 2a = 2g/3
A concave mirror gives a real image magnified 6 times when the object is moved 6 cm the magnification of the real image is 4 times. What will be the focal length of the mirror?
As we know that the magnification
And in the second case,
A spring mass system with unequal masses is placed at rest on a smooth horizontal surface. The spring is initially kept compressed with a thread. When the spring is cut, the mass m moves so as to enter in a vertical circular loop of radius r. The minimum compression x in the spring so that the mass m may negotiate the vertical loop is
Let x be the compression in the spring, v and V be the velocities of m and M respectively, when the spring released.
According to the law of conservation of momentum and energy
MV = mv ... (1)
From (2) and (1)
To negotiate a loop, minimum velocity required
A plane electromagnetic wave of angular frequency ω propagates in a poorly conducting medium of conductivity σ and relative permittivity ε. Find the ratio of conduction current density and displacement current density in the medium.
J_{c} = σE_{0} sin (ωt – kx)
= ∈ ∈_{0} × A E_{0}ω cos(ωt–kx)
J_{d} = ∈ ∈_{0}E_{0}ω
A uniform electric field of magnitude 245 V/m is directed in the negative y direction as shown in figure. The coordinates of point P and Q are ( 0.5m,  0.8m) and (0.3m, 0.7m) respectively. Calculate the potential difference V_{Q}  V_{P} along the path shown in the figure,
Potential difference between points P and Q is
A cylinder of radius r = 1 m and height 3 m filled with a liquid upto 2 m high and is rotated about its vertical axis as shown in the Figure.
The speed of rotation of the cylinder when the point at the centre of the base is just exposed is
v_{0} = 0: H = 3 m; r = 1 m: g = 10 ms^{− 2}
A charge particle q of mass m_{0} is projected along the yaxis at t = 0 from origin with a velocity V_{0}. If a uniform electric field E_{0} also exists along the xaxis, then the time at which the deBroglie wavelength of the particle becomes half of the initial value is:
Force on charged particle is =qE_{0} in x direction.
At any time t velocity in y direction will be same as at t=0 as there is no force in that direction.
At time t velocity in x direction is given by
v_{x} =a_{x}t
v_{x} =(qE_{0}/m)t
so speed of particle at time t is
debroglie wavelength is inversly proportional to speed of particle and given by
(Where R is a constant)
So for half the wavelength speed should be doubled.
The beat frequency produced when the following two waves x_{1} = 12 sin (484πt − 7πx) and x_{2} = 12 sin (480πt − 7πx) are sounded together is
x_{1} = 12 sin (484 πt − 7 π x)
∴ frequency n_{1} = 242 Hz
x_{2} = 12 sin (480 πt − 7 πx)
∴ frequency n_{2} = 240 Hz
Beat frequency = n_{1} − n_{2} = 2
An 80 kg man standing on ice throws a 4 kg body horizontally at 6 ms^{−1}. The frictional coefficient between the ice and his feet is 0.02. The distance through which the man slips is (g = 10 ms^{−2})
While throwing the body, velocity gained by the man is
Liquid cools from 50ºC to 45 ºC in 5 minutes and from 45ºC to 41.5ºC in the next 5 minutes. The temperature of the surrounding is : [Assume newton's law of cooling is applicable]
From Newton's law of cooling
in second case
From (i) and (ii)
θ_{0} = 33.3^{o}C
A long solenoid of selfinductance L and area of cross section A not carrying any current is placed in a uniform magnetic field of strength B with its axis parallel to the field direction. The total magnetic flux linked with the solenoid is φ. The energy of the magnetic field stored in the solenoid is
Flux linked with each turn of the solenoid = BA
Total number of turns = nℓwhereℓ is the length of the solenoid
Total flux linked
Energy per unit volume
∴total energy of the magnetic field stored in the solenoid
All wires have same resistance and equivalent resistance between A and B is R. Now keys are closed, then the equivalent resistance will become:
As R_{0 = R/3}
R_{eq}= 7R/9
Steam at 100°C is passed into 1.4 kg of water kept in a calorimeter of water equivalent 0.03 kg at 20 °C till the temperature of the calorimeter and contents reach 80°C. Latent heat of steam is 2.26 × 10^{6} J kg ^{− 1} and sp.heat capacity of water is 4200 J /kg/°C. The mass of steam condensed in kilogram is nearly equal to
(M + M_{w}) C(8020) = mL + mC(100 − 80)
(1.4 + 0.03) 4200 (80 − 20)= m × 2.26 × 10^{6} + m × 4200 × 20
1.43 × 4200 × 60= m (2.26 × 10^{6} + 4200 × 20)
= 0.1537 kg ≈ 0.15 kg
A satellite orbiting around the earth of radius R is shifted to an orbit of radius 2R. The time taken for one revolution will increase nearly by
The time period of a satellite orbiting around the earth is
When the radius of the orbit is increased to 2R
T’= 2√2 T
T' = 2 × 1.414 T
T' = 2.828 T ≈ 2.8 T
Screw gauge shown in Figure has 50 divisions and in one complete rotation of circular scale the main scale moves 0.5 mm.
As shown in first figure the zero error of the screw gauge is:
ZE = 5×0.5 / 50 = 0.05mm
Least count of circular scale = 0.5 / 50 = 0.01mm
Circular scale reading = 25×0.01=0.25mm
Main scale reading = 2×0.5=1mm
Actual measurement is = MSR+CSR − ZE
1+0.25−0.05=1.20mm
The given graph shows the extension (Δl) of a wire of length 1.0 m suspended from the top of a roof at one end and loaded at the other end. If the cross sectional area of the wire is 10^{− 6}m^{2}, the Young’s modulus Y of the material of the wire is
In Young’s double slit experiment, a monochromatic light of wavelength 5500 Å is used. The slits are 2 mm apart. The fringes are formed on a screen placed 20 cm away from the slits. It is found that the interference pattern shifts by 16 mm when a transparent plate of thickness 0.4 mm is introduced in the path of one of the slits. The refractive index of the transparent plate is
Δy = 16 mm = 16 × 10^{− 3} m
d = 2 mm = 2 × 10^{− 3} m λ = 5500 Å
D = 20 cm = 0.2 m
t = 0.4 mm = 4 × 10^{− 4} m
∴ μ = 1 + 0.4 = 1.4
The average degree of freedom per molecule of a gas is 6. The gas performs 25 J work, while expanding at constant pressure. The heat absorbed by the gas is:
Change in the internal energy of gas is
For a constant pressure process. Work done (W) = nRΔT
So heat absorbed by the gas is
The radioactive sources A and B have half lives of 2hr and 4hr respectively, initially contain the same number of radioactive atoms. At the end of 2 hours, their rates of distintegration are in the ratio:
Given that the halflife of radioactive source A is, t_{1/2} = 2 hr
halflife of radioactive source B is, t_{1/2} = 4 hr
Decay constant of radioactive source A is,
Decay constant of radioactive source B is,
The rate of disintegration (A) of source A at the end 2 hours is,
The rate of disintegration (A) of source B at the end 2 hours is,
The ratio of rate of distintegration of source A to B is
A Young's double slit apparatus is immersed in a liquid of refractive index 1.33. It has slit separation of 1 mm and interference pattern is observed on the screen at a distance 1.33 m from plane of slits. The wavelength of light used in air is 6300A°. When one of the slit is covered by a glass sheet of thickness and refractive index 1.53, the position of maxima and minima get interchanged. Find the maximum possible value of n.
An oscillator of frequency 425 Hz drives two speakers. The speaker are fixed on a vertical pole at a distance 2.4m from each other. A person whose height (of ears) is same as that of lower speaker runs horizontally away from the two speakers. Find the maximum distance (in m) of person from the pole where he hears no sound, (velocity of sound in air 340 m/s)
Let x be maximum distance of person from speaker  when he hears no sound (i.e, minimal)
A particle of mass m = 1 kg and having charge Q = 2C is projected on a rough horizontal xy plane from (4, 0, 0) m with velocity and in the region there exist a uniform electric field and a uniform magnetic field The coefficient of friction between the particle and the horizontal surface is μ = 1/3 and the particle comes to rest by moving a distance 1000/n metres. Find n.
A wedge of inclination 45° is moved towards right with a constant acceleration of 2√6 m/s^{2} as shown in figure. Find magnitude of acceleration of the ball placed between an inclined wall and the wedge.
In the situation shown in figure, a particle having charge q = 2C and mass = 1 kg is projected from bottom of an inclined plane of inclination 45° at an angle 45° with horizontal, so that the particle strike the inclined plane normally due to a uniform horizontal electric field E. Find the value of E (in NC^{1})
Boyle Temperature T_{b}= a/R_{b}
Critical Temperature T_{c}= 8a/27R_{b}
Dividing both
T_{b}/T_{c} =27/8
Hence option (b) is correct.
A reducing sugar is one which is capable of acting as reducing agent.
All monosaccharaides are reducing sugars because all monosaccharaides have an aldehyde group (if they are aldoses) or can tautomerize in solution to form an aldehyde group (if they are ketoses) and can themselves get oxidized hence can act as reducing agent.
Galactose
All of them are monosaccharides.
Hence option (d) is correct.
Which of the following complex square having planner complex can exhibit geometrical isomerism?
Transisomer
Cis –isomer
Square planar complexes having symmetric bidentate ligands carrying one or more substituents can form geometrical isomers. Because, such ligand show kind of asymmetry. Here pn (propylenediamine) shows this kind of nature hence leading to geometrical isomerism.
All the rest options are of symmetrical in nature around metal ion so no geometrical isomerism is possible in these.
Hence option (a) is correct.
An element has a bcc structure with a cell edge of 288 pm. The density of the element is 7.2 g cm^{–3}. Number of atoms present in 208 g of the element ?
Volume of unit cell = (288 pm)^{3} = (288 × 10^{–12} m)^{3}
= (288 × 10^{–10} cm)^{3}
=2.39 × 10^{–23} cm^{3}
For bcc, z_{eff} = 2
∴ Aw = 7.2 × 3 × 10^{23} × 2.39 × 10^{–23} g
∴ 7.2 × 3 × 2.39 g of element contains = N_{A} atoms = 6 × 10^{23} atoms.
∴ 208 g of the element contains
= 24.17 × 10^{23} atoms
Alternatively
Volume of 208 g of the element
=28.88 cm^{3}
Number of unit cells in this volume
= 12.08 × 10^{23} unit cells
Since each bcc cubic unit contains 2 atoms, therefore the total number of atoms in 208 g
= 2 (atoms/unit cell) × 12.08 × 10^{23} unit cells
= 24.16 × 10^{23} atoms
Hence option (b) is correct.
The acidity of oxyacids of halogen can be compared by checking the oxidation state of halogen in the acid. Higher the oxidation state, higher the acidity. Here H_{5}IO_{6} has an oxidation state of +7 and hence is the strongest acid.
The Clark cell ZnZn^{2+}// Hg_{2}SO_{4}Hg is often employed as a standard cell since its emf is known exactly as a function of temperature. The cell emf is 1.423 V at 298 K and its temperature coefficient of voltage is 1.2 × 10^{4} V K^{1}. What is ΔS_{cell} at 298 K?
Before we commence, we note that the spontaneous cell reaction is
so the cell reaction is a twoelectron process.
G_{cell} = nF × emf .
Inserting values for the cell at 298 K gives
Hence option (a) is correct.
a) H_{2}O →OH + H^{+}
NH_{3}→NH_{2}^{} +H^{+}
Water being more polar than ammonia dissociates more and gives plenty amount of H^{+} and hence H_{2} gas is released as result. In ammonia sufficient amount of H^{+} is not obtained due to less polar nature (so less dissociation) hence H_{2} gas is not evolved and then ammoniation of electron happens
b) As we dilute further dissociation of ammonia increases (because degree of dissociation is inversely proportional to concentration) and we get more H^{+} and NH_{2}^{} ion. H^{+} combines with ammoniated electron and H_{2} gas is evolved and paramagnetism is gone.
c) NH_{3}→NH_{2}^{} +H^{+}
M →M^{+} +e^{}
M^{+} +xNH_{2}^{}→[M (NH_{2})x] type of complex
On adding d block metal ion we see that metal ion combines with NH_{2}^{} and forms metal complex. Complex formation leads to ammonia dissociation even more and as a result of ammonia dissociation more H^{+} is released and which combines with ammoniated electron and H_{2} gas is evolved. Now colour and magnetic nature may or may not change. But whatever it will be it will be due to this complex.
d) Metallic bonding increases at very high concentration M_{2} is formed.
Hence option (d) is correct.
In which of the following, colour can be explained due to ‘Ligand to Metal Charge Transfer’?
Condition for ‘ligand to metal charge transfer’ are
i) Metal should be in high oxidation state so that it has high ionization energy. And it should also be of smaller size with vacant orbitals having low energies.
ii) Ligand should have lone pair of electrons having high energy
In KMnO_{4,} Mn is in +7 oxidation state and have all the 3d orbitals vacant. Mn^{+7} ion is surrounded by four oxide ions. All oxide ions have filled 2p orbitals. There is transfer of an electron from filled 2p orbital of oxide ion to vacant d orbitals of Mn^{+7} ion. Due to this transfer of electron from oxides of KMnO_{4} to metal happens and as a result colouris intensely purple.
Trick: Almost all the metal ion of d block having d0 configuration tend to show colour due to ligand to metal charge transfer. Another example of this is K_{2}Cr_{2}O_{7}
Hence option (a) is correct.
Observe the following reaction
Which of the following statement is true regarding solvent of the above reaction
In water the product is almost all benzyl naphthol. However, in DMSO (dimethyl sulfoxide) the
major product is the ether. In water the oxyanion is heavily solvated through hydrogen bonds to
water molecules and the electrophile cannot push them aside to get close to O– (this is an entropy
effect). DMSO cannot form hydrogen bonds as it has no OH bonds and does not solvate the oxyanion, which is free to attack the electrophile.
DMSO Solvent:
Water Solvent:
Hence option (a) is correct.
Correct order is
benzene > thiophene > pyrrole > furan.
Take a look at the structures
Electronegativity of atoms in the ring
O>N>S
In furan oxygen is in ring. It will be less reluctant to share electrons of lone pair for resonance and hence aromatic character will be least in it.
In pyrrole nitrogen is in ring. It will be more willing than furan to share electrons for resonance hence more aromatic than furan.
Thiophene has sulphur which is less electronegative than nitrogen hence more aromatic character than pyrrole.
Benzen has carbon and has 6π electrons already so most aromatic.
Hence option (c) is correct.
Enthalpy is not a conclusive measurement of whether a reaction will be more spontaneous and favorable. Entropy is.
Taking a look at both reactions, in reaction A two molecules are combining and giving one molecule and in reaction B only one molecule is giving one molecule so decrease in entropy in reaction B is less. So, ∆S will be less negative in here hence more favorable.
ΔG = ΔHTΔS
Hence option (d) is correct.
a) (NH_{4})_{2}Cr_{2}0_{7}→ N_{2} + Cr_{2}O_{3} +4H_{2}O
b) 8NH_{3} + 3Br_{2}→N_{2} + 6NH_{4}Br
c) 2NH_{3} + 3CuO →N_{2} + 3Cu + 3H_{2}O
d) 3Ca + 6NH_{3}→3[Ca(NH_{2})_{6}]
Hence option (d) is correct.
The hydride ion H^{} has the same electron configuration as helium but is much less stable. Which of the following option explains the given statement
Hydrogen _{1}H : 1S^{1}
Hydride: _{1}H^{} :1S^{2}
Helium _{2}He : 1S^{2}
Helium has two proton in it and it can stabilize 2 electrons. But, in hydrogen as one more electron enters and it becomes hydride ion, now one proton has to account for two electrons, the stability is affected. In hydride one proton cannot stabilize two electrons so structural deformity arises which leads to its reactivity.
Hence option (b) is correct.
IUPAC name of the compound will be
2,7dimethyl3,5octadiyne2,7diol
Hence option (a) is correct.
XeF_{2} when dissolved in water produces three compound A, B, C. A is inert. Compound B forms strongest Hbond with its anion and exists as D. Compound C is used in combustion. Which of the following is correct?
Reaction mentioned is given as below
2XeF_{2} +2H_{2}O →2Xe + 4HF + O_{2}
HF forms Hbond with F^{} and exists as HF_{2}^{}
Hence option (d) is correct.
Which of the marked position is most susceptible to nucleophilic attack
Position a and c will be more susceptible to nucleophilic attack
Position a has bromine attached to it making it electron deficient or more electropositive and hence inviting for nucleophile
Position c has oxygen atom attached to it. High electronegativity of oxygen makes carbon highly electropositive and susceptible to nucleophilic attack.
Hence option (c) is correct.
Ester and amide are found in equilibrium which of the following is true regarding the equilibrium of following. K_{1} and K_{2} are equilibrium constants
1.As we increase the acidic conditions amine will be protonated and equilibrium will shift backward hence K_{1}>K_{2}
2. In basic medium phenol will be deprotonated and hence reaction will shift forward here K_{2}>K_{1}
Hence option (a) is correct.
E_{2} eliminations therefore take place from the antiperiplanar conformation.
Here, E_{2} elimination gives mainly one of two possible stereoisomers.
2Bromobutane has two conformations with H and Br antiperiplanar, but the one that is less hindered leads to more of the product, and the Ealkene predominates.
Another way to understand will be in any alkene formation major product is always most stable alkene. We can see alkene given in option a is trans alkene and in option b is cis alkene. Trans alkene being more stable on account of less steric repulsion will be major product.
Hence option (a) is correct.
Arrhenius relation is described as K= A e^{Ea/RT} which of the following statement is correct regarding activation energy
K=
Differentiating with respect to T
Here we can see that on increasing temperature those reaction having high value of E_{a} will be more susceptible to change in rate.
Hence, option (b) is correct.
Which of the following is correct about acidic nature of boric acid?
a) B(OH)_{3} +2 H_{2}O →B(OH)_{4}^{} +H_{3}O^{+}
This reaction is characterized as Lewis Acidity of boric acid
b) Due to hydrogen bond present in crystalline structure it becomes difficult for it to donate H+ and hence acidic character decreases.
c) B(OH)_{3} +2 H_{2}O →B(OH)_{4}^{} +H_{3}O^{+}
Cisdiols form complex with borate ion and shift the equation forward hence ionizing the boric acid to higher extent. And acidic character increases.
Hence option (c) is correct.
The difference in the oxidation state of chromium in 'B' and 'C' is
For the reaction NO_{2 }+ CO → CO_{2}_{ }+ NO. The experimental rate expression is Find the number of molecules of CO involved in the slowest step.
No CO molecules involved in rate equation
What is the total score for the correct statement(s) from the following?
I, II, III are correct
Give the answer of the following questions for the reaction product [X].
(P) In how many M–O bonds, the bond lengths are equal.
(Q) The oxidation state of the central metal ion.
(R) The magnetic moment of the central metal ion.
(S) The number of peroxide linkage in the product formed by the reaction of [X] with H_{2}O_{2} in acidic medium.
On moving across a period, the basic character of the oxides gradually changes first into amphoteric and finally into acidic character. The acidic character of oxides increases with increase in the oxidation states of the element that is combined with oxygen. Arrange the following oxides in order of increasing acidic character.
(1) SO_{3} (2) Cl_{2}O_{7} (3) N_{2}O_{5} (4) CO_{2}
Fill in the boxes provided below with suitable number. Least acidic oxide will come in first box and the strongest acidic oxide in the last box.
Acidic Character of oxides increase as nonmetallic character of the element that is combined with oxygen increase.
The ionisation energy is the measure of matallic and nonmetallic character
AS ionisation energy of element decreases the matallic character increases and nonmetallic charachter decreases and viceversa.
Hence the correct increasing order of acidic character is increasing nonmetallic character
The length of the common chord of the circles x^{2} + y^{2} + 2x + 3y + 1 = 0 and x^{2} + y^{2} + 4x + 2 = 0 is
Consider the circles S_{1}:
The centre of S_{1} is (1,3/2) and the radius is
and S_{2}:
The equation of the common chord is S_{2}  S_{1 }=0
⇒ 2x + 1 = 0
In a triangle ABC, the angle B is greater than angle A. If the values of the angle A and B satisfy the equation 3 sin x  4 sin^{3} x  k = 0, 0 < k < 1, then value of C is
Suppose a, b, c are in AP and a^{2}, b^{2}, c^{2} are in GP. If a < b < c & a + b + c = 3/2, then the value of a is
⇒ ac = 1/4 …(2)
or
ac =  1/4
From (1) & (2) a & c are roots of a quadratic
x^{2} (a+c)x + ac = 0
⇒ x^{2} x + (1/4) = 0
Thus, the equation has roots
Since, a<b<c, the above results are not satisfied.
If we consider ac=  1/4, then
x^{2} x  (1/4) = 0
Since, a<b<c , then
If log_{2}(4^{x+1 }+ 4)log_{2}(4^{x} + 1) = log_{2} 8, then x equals
⇒ x = 0
If ‘M’ and σ^{2} are mean and variance of random variable X, whose distribution is given by –
Then
The mean of a probability distribution is given as
M = Σ P_{i}x_{i} = (¼ × 0) + (0 × 1) + (¼ × 2 ) + (0 × 3) + (½ × 4) + (0 × 5)= 5/2
σ^{2} = Σ P_{i} (M – x_{i})^{2}
If a,b,c are three unit vectors such that and b is not parallel to c, then the angle between a and c is:
Comparing on both sides of the equation
⇒ θ = π/3
The equation represents the equation of a circle with ω, ω^{2} as extremities of a diameter, then λ is , (where ω, ω^{2} are cube roots of unity)
If P(z) lies on a circle whose diametrically opposite points are A(z_{1}) & B(z_{2}) then
If sin^{–1}x + sin^{–1}y + sin^{–1}z = π/2 then x^{2} + y^{2} + z^{2} + 2xyz is equal to 
Given sin^{–1}x + sin^{–1}y + sin^{–1}z = π/2
taking cos on both side,
cos (sin^{–1}x + sin^{–1}y) = cos
⇒ cos (sin^{–1}x) cos (sin^{–1}y) – sin (sin^{–1}x) sin (sin ^{1}y) = cos
⇒ (1x^{2}) (1y^{2}) = (xy + z)^{2}
⇒ x^{2} + y^{2} + z^{2} + 2xyz = 1
If b^{2} > 4ac for the equation 4x^{4} + bx^{2} + c = 0, then all the roots of the equation will be real if:
Substituting
Clearly, if all the roots of are real, then the equation must have both the roots positive. But
which is a quadratic with positive coefficient of t^{2}. Hence for both the roots to be positive:
⇒ a & care of same sign and a & b are of opposite sign
Coordinates of the orthocentre of the triangle whose sides are x = 3, y = 4 and 3x + 4y = 6 will be:
Sides of a triangle ABC are given by x = 3, y = 4, 3x + 4y = 6 It forms a right angle triangle ABC with B(3, 4)as right angle.
Hence B is the orthocentre as perpendiculars drawn from A and C meet at B.
An equilateral triangle is inscribed in the parabola y^{2 }= 4ax whose vertex is at the vertex of the parabola. The length of its side is
Thus, the point will satisfy the equation of the parabola
The projection of the line joining the points (3,4,5) and (4,6,3) on the line joining the points (1,2,,4) and (1,0,5) is
The direction ratios of the line joining are proportional to 2,2, 1
∴ Its direction cosines are
Thus, the projection of the line joining and on PQ is given by
Consider the following statements regarding the events E, A, B and C .
(i) Event E can take place due to the occurrence of any of the events A, B, C.
(ii) Events A, B and C are equiprobable, mutually exclusive and exhaustive.
(iii) Probability of occurrence of event E is 5/12.
Considering above data the value of P(E/C) is:
A, B, C are equiprobable ⇒
A, B, C are mutually exclusive and exhaustive
Using the law of total probability:
Truth table
From truth table
(P→q) ↔ (q V ~ P) is always true so it is a tautology
The equation of the hyperbola whose foci are (6, 5), ( 4, 5) and eccentricity 5/4 is
Center of the hyperbola is the midpoint of the line segment joining two foci. Therefore, coordinates of the center are (1, 5). Also since the foci have same y coordinate therefore the line joining foci is parallel to xaxis. Hence the transverse axis is parallel to xaxis. Clearly the equation of the hyperbola is of form
Again the distance between the foci = 10
Hence, the equation of the hyperbola is
One mapping is selected at random from all mappings of the set S = {1, 2, 3,..., n} into itself. If the probability that the mapping is one  one is 3/32 then the value of n is
Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five balls. Find the number of ways in which we can place the balls in the boxes so that no box remains empty.
Find the coefficient of x^{2009} in the expansion of (1 x)^{2008} (1 +x + x^{2})^{2007}
f(x)=(1−x)^{2008} ×(1+x+x^{2})^{2007}
f(x)=(1−x)^{1 }× ((1−x)(1+x+x^{2}))^{2007}
f(x)=(1−x)^{1 }× (1+x+x^{2}−x−x^{2}−x^{3})^{2007}
f(x)=(1−x)^{1 }× (1−x^{3})^{2007}
f(x)=(1−x^{3})^{2007 }− x(1−x^{3})^{2007}
All the terms in the expansion (1−x^{3})^{2007} is in the form x^{3r}
And in the −x(1−x^{3})^{2007} is of the form x^{3p+1}
So 3r = 2009 or 3p+1 = 2009
Where we do not get integer values of r,p.
So the coefficient of x^{2009} is 0
2 videos324 docs160 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
2 videos324 docs160 tests









