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टेलीग्राम चैनल का लोगो jeemainguru — JEEMAIN.GURU| JEE materials download J
टेलीग्राम चैनल का लोगो jeemainguru — JEEMAIN.GURU| JEE materials download
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नवीनतम संदेश 4

2021-09-21 07:24:25
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2.1K viewsedited  04:24
ओपन / कमेंट
2021-09-20 07:35:07 Notes on s-Block Elements:

Physical Properties of Alkali Metals:

These are soft in nature and can be cut with the help of knife except the lithium.
The atoms of alkali metals have the largest size in their respective periods.
The first ionization energy of the alkali metals are the lowest as compared to the elements in the other group.
The alkali metals show +1 oxidation state.
The alkali metals have low values of reduction potential (as shown in table-I) and therefore have a strong tendency to lose electrons and act as good reducing agents.
The melting and boiling points of alkali metals are very low because the intermetallic bonds in them are quite weak.

Hydroxides of Alkali Metals:

a)All the alkali metals, their oxides, peroxides and superoxides readily dissolve in water to produce corresponding hydroxides which are strong alkalies.

2Na + 2H2O → 2NaOH + H2

Na2O + 2H2O 2NaOH

Na2O2 + 2H2O → 2NaOH + H2O2

2KO2 + 2H2O → 2KOH + H2O2 + O2

b) The basic strength of these hydroxides increases as we move down the group Li to Cs.
c) All these hydroxides are highly soluble in water and thermally stable except lithium hydroxide.
d) Alkali metals hydroxides being strongly basic react with all acids forming salts.

NaOH + HCI → NacI + H2O
2NaOH + H2 SO4 → Na2SO4 + 2H2O

Halides of Alkali metals:
M2O + 2HX → 2MX + H2O
MOH + HX → MX + H2O
M2CO3 + 2HX → 2MX + CO2 + H2O (M = Li, Na, K, Rb or Cs)

(X = F, Cl, Br or I)
ll the alkali metals form ionic (electrovalent) compounds.
The alkali metals are good conductors of heat and electricity.
Alkali metals (except Li) exhibit photoelectric effect
The alkali metals and their salts impart a characteristic colour to flame.
858 views04:35
ओपन / कमेंट
2021-09-20 07:34:05
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777 views04:34
ओपन / कमेंट
2021-09-19 08:07:21
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1.8K views05:07
ओपन / कमेंट
2021-09-18 08:23:48 Revision Notes on Vectors


Some Basic Results of Vector Calculus:
1) Vectors in the same direction can be added by simply adding their magnitudes. But if the vectors to be added are in opposite directions, then their magnitudes are subtracted and not added.

2) Column vectors can be added by simply adding the values in each row.

3) You can find the magnitude of a vector in three dimensions by using the formula a2 = b2 + c2 + d2, where a is the magnitude of the vector, and b, c, and d are the components in each direction.

4) If l1a + m1b = l2a + m2b then l1 = l2 and m1 = m2

5) Collinear Vectors are also parallel vectors except that they lie on the same line.

6) When two vectors are parallel, the dot product of the vectors is 1 and their cross product is zero.

7)Two collinear vectors are always linearly dependent.

8) Two non-collinear non-zero vectors are always linearly independent

9) Three coplanar vectors are always linearly dependent.

10) Three non-coplanar non-zero vectors are always linearly independent

11) More than 3 vectors are always linearly dependent.

12) Three vectors are linearly dependent if they are coplanar that means any one of them can be represented as a linear combination of other two.
1.3K views05:23
ओपन / कमेंट
2021-09-17 15:47:14 Some Important Increasing Order :

1. Bond Angle :

(i) CH4 , C2H4 , C2H2
(ii) H2O, NH3 , CH4 , CO2
(iii) H2O, NH3 , CH4 , BH3
(iv) NO2—, NO2 , NO2+
(v) H2Se, H2S, H2O
(vi) AsH3 , PH3 , NH3
(vii) PF3 , PCl3 , PBr3 , PI3
(viii) NF3 , NCl3
(ix) NF3 , NH3 , NCl3
(x) OF2 , OH2 , Cl2O

2. Melting Point :

(i) Cs, Rb, K, Na, Li
(ii) Mg, Ba, Sr, Ca, Be
(iii) CaI2 , CaBr2 , CaCl2 , CaF2
(iv) BeCl2 , MgCl2 , CaCl2 , SrCl2 , BaCl2
(v) NaI, NaBr, NaCl, NaF
(vi) CsCl, RbCl, KCl, NaCl
(vii) AlCl3 , MgCl2 , NaCl

3.Boiling Point :

(i) PH3 , AsH3 , NH3 , SbH3
(ii) H2S, H2Se, H2O
(iii) HCl, HBr, HI, HF
(iv) NH3 , HF, H2O
(v) He, Ne, Ar, Kr
(vi) H2O, D2O
(vii) H2 , Cl2 , Br2

4.Dipole moments :

(i) CCl4 , CHCl3 , CH2Cl2 ,CH3Cl
(ii) NF3 , NH3 , H2O, HF
(iii) Cis-chloropropene, Trans-chloropropene
(iv) p, m, o-dichlorobenzene
(v) CH3I, CH3Br, CH3F, CH3Cl
(vi) NH3 , SO2 , H2O, HF
(vii) H2S, H2O
(viii) HI, HBr, HCl, HF
(ix) PH3 , ASH3 , SbH3 , NH3
(x) H2O, H2O2


°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
2.2K viewsedited  12:47
ओपन / कमेंट
2021-09-17 07:13:37
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1.1K views04:13
ओपन / कमेंट
2021-09-16 07:40:00 Notes on Quadratic Equations


In order to solve a quadratic equation of the form ax2 + bx + c, we first need to calculate the discriminant with the help of the formula D = b2 – 4ac.

The solution of the quadratic equation ax2 + bx + c= 0 is given by x = [-b ± √ b2 – 4ac] / 2a

If α and β are the roots of the quadratic equation ax2 + bx + c = 0, then we have the following results for the sum and product of roots:

α + β = -b/a

α.β = c/a

α – β = √D/a

It is not possible for a quadratic equation to have three different roots and if in any case it happens, then the equation becomes an identity.

Nature of Roots:

Consider an equation ax2 + bx + c = 0, where a, b and c ∈ R and a ≠ 0, then we have the following cases:

D > 0 iff the roots are real and distinct i.e. the roots are unequal

D = 0 iff the roots are real and coincident i.e. equal

D < 0 iffthe roots are imaginary

The imaginary roots always occur in pairs i.e. if a+ib is one root of a quadratic equation, then the other root must be the conjugate i.e. a-ib, where a, b ∈ R and i = √-1.

Consider an equation ax2 + bx + c = 0, where a, b and c ∈Q and a ≠ 0, then

If D > 0 and is also a perfect square then the roots are rational and unequal.

If α = p + √q is a root of the equation, where ‘p’ is rational and √q is a surd, then the other root must be the conjugate of it i.e. β = p - √q and vice versa.

If the roots of the quadratic equation are known, then the quadratic equation may be constructed with the help of the formula
x2 – (Sum of roots)x + (Product of roots) = 0.

So if α and β are the roots of equation then the quadratic equation is

x2 – (α + β)x + α β = 0

For the quadratic expressiony = ax2 + bx + c, where a, b, c ∈ R and a ≠ 0, then the graph between x and y is always a parabola.

If a > 0, then the shape of the parabola is concave upwards

If a < 0, then the shape of the parabola is concave upwards

Inequalities of the form P(x)/ Q(x) > 0 can be easily solved by the method of intervals of number line rule.

The maximum and minimum values of the expression y = ax2 + bx + c occur at the point x = -b/2a depending on whether a > 0 or a< 0.

y ∈[(4ac-b2) / 4a, ∞] if a > 0

If a < 0, then y ∈ [-∞, (4ac-b2) / 4a]

The quadratic function of the form f(x, y) = ax2+by2 + 2hxy + 2gx + 2fy + c = 0 can be resolved into two linear factors provided it satisfies the following condition: abc + 2fgh –af2 – bg2 – ch2 = 0

In general, if α1,α2, α3, …… ,αn are the roots of the equation

f(x) = a0xn +a1xn-1 + a2xn-2 + ……. + an-1x + an, then

1.Σα1 = - a1/a0

2.Σ α1α2 = a2/a0

3.Σ α1α2α3 = - a3/a0

……… ……….

Σ α1α2α3 ……αn= (-1)n an/a0

Every equation of nth degree has exactly n roots (n ≥1) and if it has more than n roots then the equation becomes an identity.

If there are two real numbers ‘a’ and ‘b’ such that f(a) and f(b) are of opposite signs, then f(x) = 0 must have at least one real root between ‘a’ and ‘b’.

Every equation f(x) = 0 of odd degree has at least one real root of a sign opposite to that of its last term.
2.4K views04:40
ओपन / कमेंट
2021-09-16 07:29:18
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1.9K views04:29
ओपन / कमेंट
2021-09-15 05:10:38 Notes on Gravitation and Projectile


Gravitation:-
Kepler’s first law (law of elliptical orbit):- A planet moves round the sun in an elliptical orbit with sun situated at one of its foci.

Kepler’s second law (law of areal velocities):- A planet moves round the sun in such a way that its areal velocity is constant.

Kepler’s third law (law of time period):- A planet moves round the sun in such a way that the square of its period is proportional to the cube of semi major axis of its elliptical orbit.Keplers Law of Planetary Motion

T2 ∝ R3

Here R is the radius of orbit.

T2 = (4π2/GM)R 3

Newton’s law of gravitation:-

Every particle of matter in this universe attracts every other particle with a forcer which varies directly as the product of masses of two particles and inversely as the square of the distance between them.

F= GMm/r2

Here, G is universal gravitational constant. G = 6.67 ´10 -11 Nm2 / kg2

Dimensional formula of G: G = Fr2/Mm =[MLT-2][L2]/[M2] = [M-1L3T-2]

Acceleration due to gravity (g):- g = GM/R2

Variation of g with altitude:- g' = g(1- 2h/R), if h<
Variation of g with depth:- g' = g(1- d/R). Here g' be the value of acceleration due to gravity at the depth d.

Variation with latitude:-

At poles:- θ = 90°, g' = g

At equator:- θ = 0°, g' = g (1-ω2R/g)

Here ω is the angular velocity.

As g = GMe/Re2 , therefore gpole > gequator

Gravitational Mass:- m = FR2/GM

Gravitational field intensity:-

E = F/m

= GM/r2

Weight:- W= mg

Gravitational intensity on the surface of earth (Es):-

Es = 4/3 (πRρG)

Here R is the radius of earth, ρ is the density of earth and G is the gravitational constant.

Gravitational potential energy (U):- U = -GMm/r

(a) Two particles: U = -Gm1m2/r

(b) hree particles: U = -Gm1m2/r12 – Gm1m3/r13 – Gm2m3/r23

Gravitational potential (V):- V(r) = -GM/r

At surface of earth,

Vs= -GM/R

Here R is the radius of earth.

Escape velocity (ve):-

It is defined as the least velocity with which a body must be projected vertically upward in order that it may just escape the gravitational pull of earth.

ve = √2GM/R

or, ve = √2gR = √gD

Here R is the radius of earth and D is the diameter of the earth.

Escape velocity (ve) in terms of earth’s density:- ve = R√8πGρ/3

Orbital velocity (v0):-

v0 = √GM/r

If a satellite of mass m revolves in a circular orbit around the earth of radius R and h be the height of the satellite above the surface of the earth, then,

r = R+h

So, v0 = √MG/R+h = R√g/R+h

In the case of satellite, orbiting very close to the surface of earth, then orbital velocity will be,

v0 = √gR

Relation between escape velocity ve and orbital velocity v0 :- v0= ve/√2 (if h<
Time period of Satellite:- Time period of a satellite is the time taken by the satellite to complete one revolution around the earth.

T = 2π√(R+h)3/GM = (2π/R)√(R+h)3/g

If h<
Height of satellite:- h = [gR2T2/4π2]1/3 – R

Energy of satellite:-

Kinetic energy, K = ½ mv02 = ½ (GMm/r)

Potential energy, U = - GMm/r

Total energy, E = K+U

= ½ (GMm/r) + (- GMm/r)

= -½ (GMm/r)

Gravitational force in terms of potential energy:- F = – (dU/dR)

Acceleration on moon:-

gm = GMm/Rm2 = 1/6 gearth

Here Mm is the mass of moon and Rm is the radius of moon.

Projectile:-
Projectile fired at angle α with the horizontal:- If a particle having initial speed u is projected at an angle α (angle of projection) with x-axis, then,

Time of Ascent, t = (u sinα)/g

Total time of Flight, T = (2u sinα)/g

Horizontal Range, R = u2sin2α/g

Maximum Height, H = u2sin2α/2g

Equation of trajectory, y = xtanα-(gx2/2u2cos2α)

Instantaneous velocity, V=√(u2+g2t2-2ugt sinα)

and

β = tan-1(usinα-gt/ucosα)

Projectile fired horizontally from a certain height:-

Equation of trajectory: x2 = (2u2/g)y

Time of descent (timer taken by the projectile to come down to the surface of earth), T = √2h/g

Horizontal Range, H = u√2h/g. Here u is the initial velocity of the body in horizontal direction.
1.9K views02:10
ओपन / कमेंट