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Three charges e, e and  2e are placed at the corners of the triangle as shown. The net dipole moment of the system is
At A, the xcomponents of the two dipole moments cancel and ycomponents add.
∴ P_{y 2}P_{1 }cos α
A mass is attached to one end of a spring of spring constant k. The spring is stretched and then released such that its amplitude of oscillation is A. For a displacement y from the mean position, if the kinetic energy is 44% of its potential energy, then y in terms of A is
At a given instant there are 25% of undecayed radioactive nuclei in a sample. After 20s, the number of undecayed nuclei reduces to 12.5%. The extra time required in which the number of undecayed nuclei will further reduce to 3.125% of the sample is
Halflife of the radioactive sample, t_{1/2} = time in which the number of undecayed nuclei becomes half = 20 sec (given).
The material further reduces to 3.125% in n halflives is given by
where N_{f }is % of undecayed nuclie after n half lifes , and A is % of undecayed nuclie initially.
here N_{f }= 3.125 %
and N_{i} = 12.5 %
so,
so time taken t = nt_{1/2 }= 2 x 20 = 40 second
The escape velocity for an atmospheric particle 2000 km above the earth's surface is (Radius of the earth = 6.4 x 10^{6} m and g = 10 m s^{ 2} )
Escape velocity v_{e }=
At the above the Earth,s surface = and
A circular disc of radius r and thickness r/6 has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and recasted to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is
Volume oc circular disc = surface of area x thickness
The mass of the circular disc and sphere are the same
where R is the radius of the sphere,
Moment of interia of a circluar disc
Moment of interia of solid sphere
When an UV light of 10^{15} Hz and intensity 2 W/m^{2} is directed at a metal surface, photoelectrons emitted were found to have a maximum kinetic energy of 1.6 eV. If the work functions for different materials are as follows: Potassium 2.2 eV, Sodium 2.3 eV, Lithium 2.5 eV and Calcium 3.2 eV, identify the metal in the given problem.
E = hv = 6.63 x 10^{34 }x^{ }10^{15}
=^{ }6.63 x 10^{19 }J
= 4.1 eV
( dividing by e = 1.6 x 10^{19})
E_{max }=_{ }hv  hv_{0}
⇒ hv_{0 }= 4.1  1.6
= 2.5 eV
⇒ Lithium
Two sources of sound S_{1} and S_{2} each emitting waves of wavelength A are kept symmetrically on either side of the centre 0 of a circle ABCD such that S_{1}O = S_{2}O = K . When the detector is moved along the circumference of the circle, the number of maxima recorded by the detector in one revolution is
For the point A, the path difference is zero and for the point B, it is 2 λ. In between A and B there will be a point E where the path difference is λ.
Hence at the points A, E and B there will be maximum sound due to S_{1} and S_{2}.
In a circular path, it will be easier to locate points like F, C, G, D and H so that the total number of maxima amounts to 8.
A boat capable of a speed v in still water wants to cross a river of width d. The speed of the water current increases linearly from zero at either bank to a maximum of u at the middle of the river. When the boat is rowed at right angles to the bank, its downstream drift is
Let us choose a coordinate system with its origin as the starting point of the boat, the + xaxis points downstream and + yaxis points at right angles to the bank of the river.
xmotion of the boat is due to the water current velocity vcurrent while the ymotion is caused solely by the velocity v of the boat.
The above two motions are independent of each other and can be treated separately.
Assuming that the boat starts at time t = 0, the y coordinate after a time t is y = v t ... (i) The speed of water current is a function of y and is given by
Substituting for y from equation (i)
By symmetry, its value at the middle of the river is D/2.
The time required to reach the middle of the river is
Separating the variable in equation (iii) and intergrating
Figure shows the plot of a potential energy function of a conservative system U versus x. Which of the following statements is correct ?
Force F versus x is drawn in figure. At points E_{1 }points E_{1},E2,E_{3 }and E_{4}
F=0
F has the greatest magnitude nearer to point E_{1}.
If U is minimum i.e., = positive, equilibrium is stable.
If U is maximum i.e.,
negative, equilibrium is stable.
If U is constant i.e., equilibrium is neutral.
If U is constant i.e.,equilibrium is neutral.
A sphere with some cavity has outer radius R. It rolls down an inclined plane without slipping and attains a speed v at the bottom. When this sphere slides down without rolling on the frictionless inclined plane of same height its speed at the bottom is 5v/4. The radius of gyration of the sphere is
Let k be the radius of gyration and M the mass of the sphere. The kinetic energy should be the same in both the cases
The antenna current of an AM transmitter is 8 A when only carrier wave is sent but the current increases to 8.88 A when the carrier wave is sinusoidally modulated. The percentage of modulation is
I_{T} = 8.88 A, I_{C }= 8 A; m_{a }= ?
A square metal wire loop of side 20 cm and resistance 1 Ω is moved with a constant velocity Vo in a uniform magnetic field of induction B = 4 Wb/m^{2}. The magnetic field lines are perpendicular to the plane of the loop and directed inwards. The loop is connected to a network of resistors each of value 2 Ω . The resistance of load wires AB and CD are negligible. To get a current of 2 mA in the loop, the speed of motion of the loop is
The network of resistors in a balanced Wheatstone's network. Hence the resistance EF is ineffective. The equivalent resistance R' of the network is
The resistance of the square loop is 1 Ω
∴ effective resistance of the circuit R = 2 + 1 = 3 Ω
The emf induced in the loop
∴ speed of the loop v_{0
}
i = 2 mA = 2 x 10^{3} A ; I = 0.2 m B = 4 Wb/m^{2}
Speed of the loop
Four identical hollow cylindrical columns of steel, support a big structure of mass 60,000 kg. The inner and outer radii of each column are 40 cm and 50 cm respectively. When the load distribution is uniform, the compressional strain on each column is (Young’s modulus of steel is 2 x 10^{11} Pa)
Total mass supported by the columns = 60,000 kg
Total weight supporte d = 60,000 x 10 N
Compressional force one achcolumn ,
Crosssectional area of each colum n,
a = 0.283 m^{2}
Young modulus,
∴ compressional strain
= 0.265 x 10^{5}
= 2.65 x 10^{6}
A steel ball of mass 60 g falls from a height 0.8 m on the horizontal surface of a massive slab. The coefficient of restitution between the ball and the slab is e = 0.4; the total momentum imparted to the slab by the ball after numerous bounces is
By Newton’s law of collision, the velocity v′ after the first impact is v′ = ev (upward direction).
In nuclear reaction, energy released per fission is 200 MeV. When uranium 235 is used as nuclear fuel in a reactor having a power level of 1 MW, the amount of fuel needed in 30 days will be
Energy produced by the reactor in 1 day = 10^{6} x 86400 J
Energy released per fission = 200 x 10^{6} x 1.6 x 10^{19} J
No. of fissions required (i.e.,) no. of ^{235}U atoms fissioned in a month
Mass of ^{235}U having the requisite no. of atoms
The temperature at which the speed of sound in oxygen will be same as the speed of sound in nitrogen at 25 °C
Since the speeds of sound equal
∴ t = 340.6  273 = 67.6 °C
An alternating voltage having frequency of 50 cycles/sec and maximum voltage 220 V is supplied to a circuit containing a pure inductance of 0.02 H and a pure resistance of 10 Ω in series. The value of maximum current in circuit is
Impendance of L.R series circut
Value of current
A beam of light traveling in water strikes a glass plate which is also immersed in water. When the angle of incidence is 50 ° the reflected beam is found to be plane polarised. The refractive index of water is 4/3. The refractive index of the glass plate is (Given tan 50 ° = 1.198)
by brewster's law,
This is the refractive index of glass w.r.to water.
∴ refractive index of glass
A reversible heat engine converts one fourth of heat input into work. When the temperature of the sink is reduced by 200 K, its efficiency is doubled. The temperature of the source is
⇒ T_{1 }= 800 K
A man pulls a loaded sledge of mass 60 kg along a horizontal surface at constant velocity as shown in figure. The coefficient of kinetic friction between the sledge and the surface is 0.15. The tension in the rope during pulling when it makes an angle of cp = 40 ° with the horizontal is (cos 40 0 = 0.766, sin 40 ° = 0.643 and g = 10 ms'2)
Free body diagram
T cos φ = f_{K }μ_{K}R...(1)
T sin φ + R = mg...(2)
Gravitational acceleration on the surface of planet is where is the gravitational acceleration on the surface of earth. The average mass density of the planet is 2/3 times that of the earth. If escape speed on the surface of the earth is taken to be 11kms^{–1}, escape speed on the surface of the planet in kms^{–1}, will be:
let the gravitational acceleration on the surface be g' and desity be ρ'
A point object ‘O’ is placed in a medium of refractive index μ_{1} = 1.4. S_{1} and S_{2} are two concentric spherical surfaces of radii 1m and 2m. To the right of ‘O’ contains a medium of refractive index μ_{2} = 1.5 between the interfaces S_{1} and S_{2}. Find the object distance of O form S_{1} (in meter) so that an image of ‘O’ as seen by observer from air, coincides with O.
R = d tan C
A radioactive substance A decays in to B. It is known that only A is present at t = 0. Find the number of half lives at which the probability of getting B in the mixture is 15 times that of finding A, if we pick out randomly from the sample.
A single electron orbits round a stationary nucleus of charge Ze, where Z is a constant and  e is the electron charge. It is observed that it requires 47.2 eV to excite the electron from second Bohr orbit to third orbit. Then to ionise the atom from ground state, required wavelength of electromagnetic(EM) radiations is found to be 4nÅ. Find the value of n. [Take energy of electron in ground state of hydrogen atom is  13.6 eV.]
An inductor L = 50 mH carrying an initial current l_{0} = 2.5 amp is connected across a non linear resistor. The voltage across resistor is related to current as V = 10 l^{2}. After how much time (in ms) current through inductor becomes 1.25 amp.
The reaction is an example of cine subsitution or elimination addition
This test is meant for detecting the coloured cations. When borax is heated on a platinum loop it swells up into a white porous mass first which on subsequent heating melts into a shining transparent glass bead (B_{2}O_{3}).
When this glass bead is heated with coloured compounds, a characteristic colour of metaborate is formed.
However, in a reducing flame, a different colour is obtained.
Acidic character depends on stability of conjugate base. More stable conjugate base, more will be the acidic character
Electronegativity of Cl > Br > I So stability order of conjugate bases CIO > BrO > IO Hence acidic character HCIO > HBrO > HIO
In the formal charge on each oxygen atom and P  0 bond order respectively are
on exist ;n 4 resonating structures. Bond order of
3 negative charges of the ion is beings h are d by 4 oxygen atoms,
∴ formal charge on an O atom = 3/4
Aromaticity of a compound can be decided by Huckel's rule.
Huckel's rules are as follows:
1. Compound must be cyclic
2. The cyclic compound must have resonance and complete delocalization of electrons i.e., there must be conjugation.
3. The total number of pi electrons must fit the formula = 4n + 2, where n is a non  negative integer.
The important summary of this concept is as follows: (Remember this summary)
The actinoids exhibit more number of oxidation states in general than the lanthanoids because
shielding effect of 5f < shielding effect of 4f
so Z_{eff} (z effect) in actinoidis < Zeff (z effect) in lanthanoid
so extraction of electrons in actinoid is easy than lanthanoid which has 4f and better shielding.
The compound when kept in pentane converts to a constitutional isomer. Identify that
isomer.
The vapour pressure of water at 20 °C is 17.5 torr. What will be the no. of moles of water present in one litre of air at 20 °C and 40% relative humidity.(R =0.082 L atm/molk)
Relative humidity (RH) =
∴ Partial pressure of H_{2}O = RH × Vapour pressure of H_{2}O
0.400 g of impure CaSO_{4} (Molar mass = 136) solution when treated with excess of barium chloride solution, gave 0.617 g of anhydrous BaSO^{4} (Molar mass = 233). The percentage of CaSO^{4} present in the sample is
At temperature 300 K for reaction C_{2}H_{4}(g) + 3O_{2}(g)→ 2CO_{2}(g) + 2H_{2}O(l); ΔH = − 336.2 kcal. The approximate value of ΔU at 300 K for the same reaction (R = 2 cal degree^{− 1} mol^{− 1}) will be
C_{2}H_{4}(g) + 3 O_{2}(g)→2 CO_{2}(g) + 2 H_{2}O(l)
At 300 K ΔH = − 336.2 kcal
Δn_{g} = 2 − (1 + 3) = − 2
T = 300 K, ΔH = − 336.2 kcal,
R = 2 × 10^{−3} kcal degree^{−1} mol^{−1}
ΔH = ΔU+ Δn_{g}RT
− 336.2 = ΔU + (− 2) × 300 × 2 × 10^{−3}
ΔU = − 336.2 + 1.2 = − 335.0 kcal
20 g helium is compressed isothermally and reversibly at 27 °C from a pressure of 2 atmosphere to 20 atmosphere. The change in the value of heat energy change during the process is (R = 2 cal kelvin^{1} mol^{1})
Which of the following type of compound is used to increase the hardness of the silicone polymer.
Hydrolysis followed by condensation polymerization of RSiCl_{3} produces 3D crosslinked silicones which are hard.
Consider the reaction 2A + B →products, when the concentration of A alone was doubled, the halflife of the reaction did not change. When the concentration of B alone was doubled, the rate was not altered. The unit of rate constant for this reaction is
When concentration of B doubled rate does not changes
So,
Now when Concentration of A doubled half life does not change
it means reaction is first order so
order of reaction =1
so unit of rate constant = s ^{1}
0.066 g of a metal was deposited when a current of 2 ampere passed for 100 seconds through an aqueous solution of its metal ion. The equivalent mass of the metal is
Total charge passed in Coulombs = It = 2 X100 = 200 Coulombs
200 Coulombs of charge deposits 0.066 g of metal.
Equivalent mass of the metal is that mass of metal deposited by 1 Farad or 96500 Coulombs.
Equivalent mass of metal
A mixture of gaseous nitrogen and gaseous hydrogen attains equilibrium with gaseous Ammonia according to the reaction
It is found that at equilibrium when temperature is 673 K , atmospheric pressure is 100, moles of gaseous nitrogen and gaseous hydrogen are in ratio 1:3 and mole percent of gaseous ammonia is 24 . The equilibrium constant Kp for the reaction at the above condition is
Partial pressure of
=24.0 atm
Pressure of N_{2} + H_{2} = 100 – 24 = 76 atm
Partial pressure of
Partial pressure of N_{2}
New carboncarbon bond is formed is Friedal Craft reaction and ReimerTeimann reaction
Friedal Craft’s reaction
Riemer Teimann Reaction
I, IV, VII, VIII
The coordination number of central ion of the complex obtained in the sodium nitroprusside test of sulphide ion is
Na_{4}[Fe(CN)_{4}NOS]
A buffer solution is formed by mixing 100 mL 0.01 M CH_{3}COOH with 200 mL 0.02 M CH_{3}COONa. If this buffer solution is made to 1.0 L by adding 700 mL of water. pH will change by a factor of
No change in p^{H}
CH_{3} (CHOH)_{2} — COOH in this compound number of optical isomer is
In this compound there are 4 optical isomers; this is calculated by multiplying the number of chiral centres by two.
The coefficient of x^{5} in (1+2x+3x^{2}+……….up to infinite term)^{3/2} is
(1+2x+3x^{2}+………)^{–3/2}
=((1–x)^{–2})^{–3/2}
= (1 − x)^{3} = 1 − 3x + 3x^{2} − x^{3}
∴ coefficient of x^{5} = 0.
We have
Limit is of the form 1^{∞}, so we have
= (ln a + ln b + ln c) 2/3
= 2/3(ln abc)
∴ Desired limit =
If the progressions 3, 10, 17, ....... and 63, 65, 67, ....... are such that their n^{th} terms are equal, then n is equal to
n^{th} term of 1^{st} series =n^{th} term of 2^{nd} series
⇒ 3 + (n − 1) 7 = 63 + (n − 1) 2
⇒ (n − 1)5 = 60
⇒ n 1=12
⇒ n = 13
If the latus rectum of a hyperbola through one focus, subtends 60 ° angle at the other focus, then its eccentricity is
Taking only positive value of e as eccentricity cannot be negative.
Find the solution of the differential equation
given that y=4 at x=3
Integrating both sides
⇒Putting x=3 and y=4, we get
⇒Thus we get the solution as
If the mean of a binomial distribution is 25, then its standard deviation lies in the interval
Mean, np = 25 and q < 1
⇒0 ≤ σ< 5
The equation of the normal to the ellipse at the positive end of the latus rectum is
The equation of the normal at (x_{1}, y_{1}) to the given ellipse is
Here x_{1} = ae and y_{1} = b^{2}/a
So the equation of the normal at positive end of the latus rectum is
The tangent to the circle x^{2} + y^{2} = 5 at (1, − 2) also touches the circle x^{2} + y^{2} − 8x + 6y + 20 = 0. Then the point of contact is
Tangent at (1, − 2) to x^{2} + y^{2} = 5 is
x − 2y − 5 = 0 ... (i).
Centre and radius of
x^{2} + y^{2} − 8x + 6y + 20 = 0 are
C (4, − 3) and radius r = √5
Perpendicular distance from
C (4, − 3) to (i) is radius.
∴ (i) is also a tangent to the second circle.
Let P (h, k) be the foot of the drawn circle from C (4, − 3) on (i)
∴ (h, k) = (3, −1)
The equation x^{3} – 3x + [a] = 0, will have three real and distinct roots if –
(where [ ] denotes the greatest integer function)
f(x) = x^{3} – 3x + [a]
Let [a] = t (where t will be an integer)
f(x) = x^{3} – 3x + t ……….(i)
⇒ f ’(x) = 3x^{2} – 3
⇒ f ‘(x) = 0 has two real and distinct solution which are x = 1 and x = 1
so f(x) = 0 will have three distinct and real solution when f (1). f(1) < 0
……………. (ii)
Now,
f(1) = (1)^{3} 3(1) + t = t – 2
f(–1) = (–1)^{3} – 3 (–1) + t = t + 2
From equation (ii)
(t –2) (t + 2) < 0
⇒ t ∈ (2, 2)
Now t = [a]
Hence [a] ∈ (2, 2)
⇒ a ∈ [1, 2)
Taking magnitude and squaring on both sides, we get
If g(x) satisfies the conditions of Rolle’s theorem in [1, 2] and g′(x) = f(x), then is equal to
As g(x) satisfies the condition of Rolle’s theorem in [1, 2], g(x) is continuous in the interval and
g(1) = g(2)
= g(2) − g(1) = 0
The equation of the plane through the line of intersection of planes x + y + z + 3 = 0 and 2x − y + 3z + 1 =0 and parallel to the line is
Any plane through the given line is 2x − y + 3z + 1 + λ (x + y + z + 3) = 0 If this plane is parallel to the linethen the normal to the plane is also perpendicular to the above line.
∴(l_{1}l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0
⇒ (2 + λ) 1 + (λ −1) 2 + (3 + λ) 3 = 0
⇒ λ= 3/2
and the required plane is
2x − y + 3z + 1 + (3/2) (x + y + z + 3) = 0
⇒ x − 5y + 3z − 7 = 0
or
x − 5y + 3z = 7
Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated is
The total number of words that can be formed is 10^{5} and number of these words in which no letters are repeated is ^{10}P_{5}.
Hence the required number
= 10^{5} − ^{10}P_{5}
= 100000 − 10 × 9 × 8 × 7 × 6
= 69760
∼(p ⇒ q) ≡ p ∧ ∼q
∴ ∼(∼p ⇒ q) ≡ ∼p ∧ ∼q
Doing R_{3}→ R_{3}−xR_{2} and
R_{2}→R_{2}− xR_{1}
we get
So,
f(2x) − f(x)
= a[(a + 2x)^{2} − (a + x)^{2}]
= a (a + 2x − a − x) (a + 2x + a + x)
= ax (2a + 3x)
So, f(2x)  f(x)=ax(2a+3x)
Thus, f(2x) f(x) is divisible by a, x and (2a+3x).
B^{1027} = (B^{4})^{256} . B^{3}
= (I)^{256} . B^{3}
= I . B^{3}
= B^{3}
= B . B^{2
}
Two vertical poles AL and BM of heights 20 m and 80 m respectively stand apart on a horizontal plane. If A and B be the feet of the poles and AM and BL intersect at P, then the height of P is equal to
⇒Z = 16 m
ABC is a right angle triangle (right angled at B) inscribed in the parabola y^{2} = 4x. The minimum length of the intercept cut off by the tangents at A and C to the parabola from yaxis is
Let A(t_{1}), B(t_{2}) and C(t_{3}) be the points on the parabola
also length of intercept cut off by tangents from yaxis is t_{1}  t_{3}≥ 4
From a point on the line y= x + c, c (parameter), tangents are drawn to the hyperbola such that chords of contact pass through a fixed point (x_{1}, y_{1}). Then x_{1}/y_{1} s equal to
If the sum of the coefficients in the expansion of (ℓ^{2}x^{2}−2ℓx+1)^{51} vanishes then ℓ is equal to
A sevendigit number made up of all distinct 8, 7, 6, 4, 2, x and y is divisible by 3. Then possible number of order pair (x, y) is
8, 7, 6, 4, 2, x and y
Any number is divisible by 3 if sum of digits is divisible by 3 i.e. x + y + 27 is divisible by 3
x and y can take values from 0, 1, 3, 5, 9
possible pairs (5, 1) (3, 0) (9, 0) (9, 3) & (1, 5), (0, 3) (0, 9) (3, 9).
If z_{1}, z_{2}, z_{3} are complex numbers such that z_{1}=z_{2}=z_{3}=1, then z_{1}−z_{2}^{2}+z_{2}−z_{3}^{2}+z_{3}−z_{1}^{2} cannot exceed
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