Fiitjee Study Material

2021-10-01 07:56:51 USES OF INERT GASES :

(1) He is non-inflammable and light gas, so it is used in filling balloons for meteorological observations.
(2) He is used in gas cooled nuclear reactors.
(3) Liquid He is used as cryogenic agent.
(4) He is used to produce powerful superconducting magnets.
(5) Ne is used in discharge tubes.
(6) Ar is used as inert atmosphere in metallurgical process.
(7) Xenon and Krypton are used in light bulbs designed for special purposes.

IMPORTANT ORDER :

(1) He < Ne < Ar < Kr < Xe Atomic radius
(2) He > Ne > Ar > Kr > Xe Ionisation energy
(3) He < Ne < Ar < Kr < Xe < Rn Density
(4) He < Ne < Ar < Kr < Xe < Rn mpt. bpt Open / Comment

2021-10-01 07:56:51 Notes on Wave Optics

Huygens Principle:- Wave-front of a wave, at any instant , is defined as the locus of all the particles in the medium which are being disturbed at the same instant of time and are in the same phase of vibration.

(a) Each point on a wave front acts as a source of new disturbance and emits its own set of spherical waves called secondary wavelets. The secondary wavelets travel in all directions with the velocity of light so long as they move in the same medium.

(b) The envelope or the locus of these wavelets in the forward direction gives the position of new wave front at any subsequent time.

Determination of Phase Difference:-
The phase difference between two waves at a point will depend upon

(a) The difference in path lengths of the two waves from their respective sources.
(b) The refractive index of the medium
(c) Initial phase difference between the source if any.
(d) Reflections, if any, in the path followed by waves.

Reflection of plane wave at plane surface (Laws of reflection):-

a) The incident ray, the reflected ray and normal to the reflecting surface at the point of incidence, all lie in one plane and that plane is perpendicular to the reflecting surface.
b) The angle of incidence is equal to the angle of reflection.

So, ∠i = ∠r

This signifies angle of incidence is equal to the angle of reflection.

Refraction of light:-
Refraction is the phenomena by virtue of which a wave going from one medium to another undergoes a change in velocity.

(a) The sine of the angle between the incident ray and the normal bears a constant ratio to the sine of the angle between refracted ray and the normal.

sin i/sin r = v1/v2 = 1µ2 = constant

Here, v1 and v2 are the velocities of sound in first and second medium respectively.1µ2 is the refractive index of the second medium with respect to first.

(b) The incident ray, the refracted ray and the normal to the refracting surface lie in the same plane.

Interference:- The modification in the distribution of light energy obtained by the superposition of two or more waves is called interference.

Principle of superposition:- It states that a number of waves travelling, simultaneously, in a medium behave independent of each other and the net displacement of the particle, at any instant, is equal to the sum of the individual displacements due to all the waves.

Displacement equation:- y = R sin 2π/λ (vt+x/2)

Amplitude:- R = 2a cos πx/λ

Intensity:- I = K4a2 cos2 (πx/λ) [I = KR2]

Maxima:- A point having maximum intensity is called maxima.

x = 2n (λ/2)

A point will be a maxima if the two waves reaching there have a path difference of even multiple of λ/2.

Imax = 4Ka2 = 4i (Here, i = Ka2)

Minima:- A point having minimum intensity is called a minima.
x = (2n+1) (λ/2)

A point will be a minima if the two waves reaching there have a path difference of odd multiple of λ/2.

Imin = K. 4a2×0 = 0

Condition for constructive interference:-
Path difference = (2n)λ/2

Phase difference = (2n)π

Condition for destructive interference:-
Path difference = (2n+1)λ/2

Phase difference = (2n+1)π

Coherent Sources:- Coherent sources are the sources which either have no phase difference or have a constant difference of phase between them.
Conditions for interference:-
(a) The two sources should emit, continuously, waves of same wavelength or frequency.

(b) The amplitudes of the two waves should be either or nearly equal

(c) The two sources should be narrow.

(d) The sources should be close to each other.

(e) The two sources should be coherent one.

Young’s double slit experiment:-
Path difference, x = yd/D

Maxima, y = nλD/d

Here, n = 0,1,2,3….

Minima, y = (2n+1) λD/d

Here, n = 0,1,2,3….

Fringe Width:- It is the distance between two consecutive bright and dark fringes.
β = λD/d

Displacement of fringes due to the introduction of a thin transparent medium:-
(a) Shift for a particular order of fringes:-

?y = (β/λ) (µ-1)t

(b) Shift across a particular point of observation:-

µ = (mλ/t) +1

Lloyd’s single mirror:-
?λ = β .2a/D

Power of lens:- P = 100/f Open / Comment

2021-10-01 07:56:51 Revision Notes on Vectors


The length or the magnitude of the vector = (a, b, c) is defined by w = √a2+b2+c2

A vector may be divided by its own length to convert it into a unit vector, i.e. ? = u / |u|. (The vectors have been denoted by bold letters.)

If the coordinates of point A are xA, yA, zA and those of point B are xB, yB, zB then the vector connecting point A to point B is given by the vector r, where r = (xB - xA)i + (yB – yA) j + (zB – zA)k , here i, j and k denote the unit vectors along x, y and z axis respectively.

Some key points of vectors:
1) The magnitude of a vector is a scalar quantity

2) Vectors can be multiplied by a scalar. The result is another vector.

3) Suppose c is a scalar and v = (a, b) is a vector, then the scalar multiplication is defined by cv = c (a, b) = (ca, cb). Hence each component of vector is multiplied by the scalar.

4) If two vectors are of the same dimension then they can be added or subtracted from each other. The result is gain a vector.

If u, v and w are three vectors and c, d are scalars then the following results of vector addition hold true:
1) u + v = v + u (the commutative law of addition)

2) u + 0 = u

3) u + (-u) = 0 (existence of additive inverses)

4) c (du) = (cd)u

5) (c + d)u = cu + d u

6) c(u + v) = cu + cv

7) 1u = u

8) u + (v + w) = (u + v) + w (the associative law of addition)

Some Basic Rules of Algebra of Vectors:
1) a.a = |a|2 = a2

2) a.b = b.a

3) a.0 = 0

4) a.b = (a cos q)b = (projection of a on b)b = (projection of b on a) a

5) a.(b + c) = a.b + a.c (This is also termed as the distributive law)

6) (la).(mb) = lm (a.b)

7) (a ± b)2 = (a ± b) . (a ± b) = a2 + b2 ± 2a.b

8) If a and b are non-zero, then the angle between them is given by cos θ = a.b/|a||b|

9) a x a = 0

10) a x b = - (b x a)

11) a x (b + c) = a x b + a x c

Any vector perpendicular to the plane of a and b is l(a x b) where l is a real number.

Unit vector perpendicular to a and b is ± (a x b)/ |a x b|

The position of dot and cross can be interchanged without altering the product. Hence it is also represented by [a b c]

1) [a b c] = [b c a] = [c a b]

2) [a b c] = - [b a c]

3) [ka b c] = k[a b c]

4) [a+b c d] = [a c d] + [b c d]

5) a x (b x c) = (a x b) x c, if some or all of a, b and c are zero vectors or a and c are collinear.

Methods to prove collinearity of vectors:
1) Two vectors a and b are said to be collinear if there exists k ? R such that a = kb.

2) If p x q = 0, then p and q are collinear.

3) Three points A(a), B(b) and C(c) are collinear if there exists k ? R such that AB = kBC i.e. b-a = k (c-b).

4) If (b-a) x (c-b) = 0, then A, B and C are collinear.

5) A(a), B(b) and C(c) are collinear if there exists scalars l, m and n (not all zero) such that la + mb+ nc = 0, where l + m + n = o

Three vectors p, q and r are coplanar if there exists l, m ? R such that r = lp + mq i.e., one can be expressed as a linear combination of the other two.

If [p q r] = 0, then p, q and r are coplanar.

Four points A(a), B(b), C(c) and D(d) lie in the same plane if there exist l, m ? R such that b-a = l(c-b) + m(d-c).

If [b-a c-b d-c] = 0 then A, B, C, D are coplanar.

Two lines in space can be parallel, intersecting or neither (called skew lines). Let r = a1 + μb1 and r = a2 + μb2 be two lines.

They intersect if (b1 x b2)(a2 - a1) = 0

The two lines are parallel if b1 and b2 are collinear.

The angle between two planes is the angle between their normal unit vectors i.e. cos q = n1 . n2

If a, b and c are three coplanar vectors, then the system of vectors a', b' and c' is said to be the reciprocal system of vectors if aa' = bb' = cc' = 1 where a' = (b xc) /[a b c] , b' = (c xa)/ [a b c] and c' = (a x b)/[a b c] Also, [a' b' c'] = 1/ [a b c]

Dot Product of two vectors a and b defined by a = [a1, a2, ..., an] and b = [b1, b2, ..., bn] is given by a1b1 + a2b2 + ..., + anbn . Open / Comment

2021-10-01 07:56:51 Revision Notes on Vectors


The length or the magnitude of the vector = (a, b, c) is defined by w = √a2+b2+c2

A vector may be divided by its own length to convert it into a unit vector, i.e. ? = u / |u|. (The vectors have been denoted by bold letters.)

If the coordinates of point A are xA, yA, zA and those of point B are xB, yB, zB then the vector connecting point A to point B is given by the vector r, where r = (xB - xA)i + (yB – yA) j + (zB – zA)k , here i, j and k denote the unit vectors along x, y and z axis respectively.

Some key points of vectors:
1) The magnitude of a vector is a scalar quantity

2) Vectors can be multiplied by a scalar. The result is another vector.

3) Suppose c is a scalar and v = (a, b) is a vector, then the scalar multiplication is defined by cv = c (a, b) = (ca, cb). Hence each component of vector is multiplied by the scalar.

4) If two vectors are of the same dimension then they can be added or subtracted from each other. The result is gain a vector.

If u, v and w are three vectors and c, d are scalars then the following results of vector addition hold true:
1) u + v = v + u (the commutative law of addition)

2) u + 0 = u

3) u + (-u) = 0 (existence of additive inverses)

4) c (du) = (cd)u

5) (c + d)u = cu + d u

6) c(u + v) = cu + cv

7) 1u = u

8) u + (v + w) = (u + v) + w (the associative law of addition)

Some Basic Rules of Algebra of Vectors:
1) a.a = |a|2 = a2

2) a.b = b.a

3) a.0 = 0

4) a.b = (a cos q)b = (projection of a on b)b = (projection of b on a) a

5) a.(b + c) = a.b + a.c (This is also termed as the distributive law)

6) (la).(mb) = lm (a.b)

7) (a ± b)2 = (a ± b) . (a ± b) = a2 + b2 ± 2a.b

8) If a and b are non-zero, then the angle between them is given by cos θ = a.b/|a||b|

9) a x a = 0

10) a x b = - (b x a)

11) a x (b + c) = a x b + a x c

Any vector perpendicular to the plane of a and b is l(a x b) where l is a real number.

Unit vector perpendicular to a and b is ± (a x b)/ |a x b|

The position of dot and cross can be interchanged without altering the product. Hence it is also represented by [a b c]

1) [a b c] = [b c a] = [c a b]

2) [a b c] = - [b a c]

3) [ka b c] = k[a b c]

4) [a+b c d] = [a c d] + [b c d]

5) a x (b x c) = (a x b) x c, if some or all of a, b and c are zero vectors or a and c are collinear.

Methods to prove collinearity of vectors:
1) Two vectors a and b are said to be collinear if there exists k ? R such that a = kb.

2) If p x q = 0, then p and q are collinear.

3) Three points A(a), B(b) and C(c) are collinear if there exists k ? R such that AB = kBC i.e. b-a = k (c-b).

4) If (b-a) x (c-b) = 0, then A, B and C are collinear.

5) A(a), B(b) and C(c) are collinear if there exists scalars l, m and n (not all zero) such that la + mb+ nc = 0, where l + m + n = o

Three vectors p, q and r are coplanar if there exists l, m ? R such that r = lp + mq i.e., one can be expressed as a linear combination of the other two.

If [p q r] = 0, then p, q and r are coplanar.

Four points A(a), B(b), C(c) and D(d) lie in the same plane if there exist l, m ? R such that b-a = l(c-b) + m(d-c).

If [b-a c-b d-c] = 0 then A, B, C, D are coplanar.

Two lines in space can be parallel, intersecting or neither (called skew lines). Let r = a1 + μb1 and r = a2 + μb2 be two lines.

They intersect if (b1 x b2)(a2 - a1) = 0

The two lines are parallel if b1 and b2 are collinear.

The angle between two planes is the angle between their normal unit vectors i.e. cos q = n1 . n2

If a, b and c are three coplanar vectors, then the system of vectors a', b' and c' is said to be the reciprocal system of vectors if aa' = bb' = cc' = 1 where a' = (b xc) /[a b c] , b' = (c xa)/ [a b c] and c' = (a x b)/[a b c] Also, [a' b' c'] = 1/ [a b c]

Dot Product of two vectors a and b defined by a = [a1, a2, ..., an] and b = [b1, b2, ..., bn] is given by a1b1 + a2b2 + ..., + anbn .

@arvind_arora_vedantu_chemistry Open / Comment

2021-10-01 07:56:51 Notes on Chemical Coordination and Regulation


Properties of hormones
(a) These are secreted by endocrine gland (biogenic in origin).

(b) Their secretions is released directly into blood (except local hormones e.g. gastrin).

(c) These are carried to distantly locate specific organs, called target organ.

(d) These have specific physiological action (excitatory or inhibatory). These co-ordinate different physical, mental and metabolic activities and maintain homeostasis.

(e) The hormones have low molecular weight e.g. ADH has a molecular weight of 600–2000 daltons.

(f) These act in very low concentration e.g. around10–10 molar.

(g) Hormones are non antigenic.

(h) These are mostly short-lived. So have a no camulative effect.

(i) Some hormones are quick acting e.g. adrenalin, while some acting slowly e.g. ostrogen of ovary.

(j) Some hormones secreted in inactive form called Prohormone e.g. Pro-insulin.

(k) Hormones are specific. They are carriers of specific information to their specific target organ. Only those target cell respond to a particular hormone for which they have receptors.

@arvind_arora_vedantu_chemistry Open / Comment

2021-10-01 07:56:50 Notes on Coordination Compounds


Ligands: an ion or molecule capable of donating a pair of electrons to the central atom via a donor atom.

Unidentate ligands: Ligands with only one donor atom, e.g. NH3, Cl-, F- etc.
Bidentate ligands: Ligands with two donor atoms, e.g. ethylenediamine, C2O42-(oxalate ion) etc.
Tridentate ligands: Ligands which have three donor atoms per ligand, e.g. (dien) diethyl triamine.
Hexadentate ligands: Ligands which have six donor atoms per ligand, e.g. EDTA.

Chelating Ligands:

Multidentate ligand simultaneously coordinating to a metal ion through more than one site is called chelating ligand. Example: Ethylenediamine (NH2CH2CH2NH2)
These ligands produce a ring like structure called chelate.
Chelation increases the stability of complex.

Werner’s Theory:

Metals possess two types of valencies i.e. primary (ionizable) valency and secondary (nonionizable) valency.
Secondary valency of a metal is equal to the number of ligands attached to it i.e. coordination number.
Primary valencies are satisfied by negative ions, while secondary valencies may be satisfied by neutral, negative or positive ions.

Secondary valencies have a fixed orientation around the metal in space.

[Co(NH3)6]Cl3

Primary Valencies = 3 Cl-

Secondary Valencies = 6 NH3

Coordination Sphere = [Co(NH3)6]3-


Nomenclature of Complexes:

Positive ion is named first followed by negative ion.
Negative ligands are named by adding suffix - o.
Positive ligands are named by adding prefix – ium.
Neutral ligands are named as such without adding any suffix or prefix.
Ligands are named in alphabetical order.
Name of the ligands is written first followed by name of metal with its oxidation number mentioned in roman numbers in simple parenthesis.
Number of the polysyllabic ligands i.e. ligands which have numbers in their name, is indicated by prefixes bis, tris etc,
Number and name of solvent of crystallization if any, present in the complex is written in the end of the name of complex.
When both cation and anion are complex ions, the metal in negative complex is named by adding suffix-ate.

In case of bridging ligands:
[Name of the groups to the left of bridging ligand (Oxidation state)] –μ – [Name of the groups to the right of bridging ligand (Oxidation state)] – [Name of negative ion] Open / Comment

2021-10-01 07:56:50 #Revision

Blood Group

Introduction
Based on the presence and absence of antibodies, the blood is classified into different groups.
Further, while classification, the presence and absence of the inherited antigenic substances also considered.
The types of blood groups are inherited and represent contributions from both the father and the mother.

ABO Blood Group System
In human blood, usually, there are two antigens and antibodies.
The two antigens are antigen A and antigen B.
The two antibodies are antibody A and antibody B.
The antigens are remaining in the red blood cells, whereas the antibodies are found in the serum.
Based on the antigen property, the blood group of all human beings can be classified as −
➭ Blood Group A − antigen A and antibody B
➭ Blood Group B − antigen B and antibody A
➭ Blood Group AB − antigen A and antigen B and no antibody
➭ Blood Group O − no antigen, but antigen A as well as antibody B
Consideration of the ABO system is the most imperative while transfusion of human blood.
The ABO blood group systems were first discovered by Karl Landsteiner in 1901.

Rh Blood Group System
The Rh system (the meaning of Rh is Rhesus) is another significant blood-group system. It is very important to match Rh system while blood transfusion.
Rh antigen first studied in Rhesus monkeys; therefore, its name is given Rh factor/system.
The person who does not have Rh antigen is known as Rh negative (Rh-ve) and the person who has the Rh antigen is known as Rh positive (Rh+ve). Open / Comment

2021-10-01 07:56:50 Chemical Properties of Alkali Earth Metals:



1. Reaction with water :

Mg + H2O → MgO + H2

or, Mg + 2H2O → Mg (OH)2 + H2

Ca + 2H2O → Ca(OH)2 + H2

2. Formation of oxides and nitrides

Be + O2 (air) +Δ→ 2BeO

3Be + N2 (air) +Δ → Be3N2

Mg + air + Δ → MgO + Ng3N2

3. Formation of Nitrides

3M + N2 + Δ → M3N2

Be3N2 + Δ → 3Be + N2

Ba3N2 + 6H2O + Δ → 3Ba (OH)2 + 2NH3

Ca3N2 + 6H2O + Δ → 3Ca (OH)2 + 2NH3

4. Reaction with hydrogen:

M + H2 + Δ → MH2

Both BeH2 and MgH2 are covalent compounds having polymeric structures in which H – atoms between beryllium atoms are held together by three
centre – two electron (3C - 2e) bonds as shown below:

5. Reaction with carbon – (Formation of carbides)

When BeO is heated with carbon at 2175 – 2275 K a brick red coloured carbide of the formula Be2C is formed

2BeO +2C \xrightarrow[]{2175 - 2275 K}Be_2C +2CO

It is a covalent compound and react water forming methane.

Be2C + 4H2O → 2Be (OH)2 + CH4

6. Reaction with Ammonia:

Like alkali metal, the alkaline earth metals dissolve in liquid ammonia to give deep blue black solution from which ammoniates [ M (NH3)6 ]2+ can be recovered.

Anamolous Behaviour of Beryllium:
Be is harder than other members of its group.

Be is lighter than Mg.

Its melting and boiling points are higher than those of Mg & other members.

Be does not react with water while Mg reacts with boiling water.

BeO is amphoteric while MgO is weakly basic.

Be forms covalent compounds whereas other members form ionic compounds.

Beryllium carbide reacts with water to give methane whereas carbides of other alkaline earth metals gives acetylene gas.

Be2C + 4H2O → 2Be (OH)2 + CH4

Mg2C2 + 2H2O → Mg (OH)2 + C2H2

CaC2 + 2H2O → Ca (OH)2 + C2H2

Beryllium does not exhibit coordination number more than four as it has four orbitals in the valence shell. The other members of this group has coordination number

Diagonal relationship of Be with Al:
Unlike groups – 2 elements but like aluminium, beryllium forms covalent compounds.

The hydroxides of Be, [Be(OH)2] and aluminium [Al(OH)3] are amphoteric in nature, whereas those of other elements of group – 2 are basic in nature.

The oxides of both Be and Al i.e. BeO and Al2O3 are high melting insoluble solids.

BeCl2 and AlCl3 have bridged chloride polymeric structure.

The salts of beryllium as well as aluminium are extensively hydrolysed.

Carbides of both the metal reacts with water liberating methane gas.

Be2C + 4H2O → 2Be (OH)2 + CH4

AI4C3 + 12H2O → 4Al (OH)3 + 3CH4

The oxides and hydroxides of both Be and Al are amphoteric and dissolve in sodium hydroxide as well as in hydrochloric acid.
BeO + 2HCI → BeCI2 + H2O

BeO + 2NaOH → Na2BeO2 + H2O

Al2O3 + 6HCI → 2AICI3 + H2O

AI2O3 + 2NaOH → 2NaAIO2 + H2O

Like Al, Be is not readily attacked by acids because of the presence of an oxide film.
Calcium Carbonate (CaCO3):
It occurs in nature as marble, limestone, chalk, coral, calcite, etc. It is prepared as a white powder, known as precipitated chalk, by dissolving marble or limestone in hydrochloric acid and removing iron and aluminium present by precipitating with NH3, and then adding ammonium carbonate to the solution; the precipitate is filtered, washed and dried.

CaCl2 + (NH4)2CO3 →CaCO3 + 2NH4Cl
It dissolves in water containing CO2, forming Ca(HCO3)2 but is precipitated from solution by boiling.

CaCO3 + H2O + CO2 Ca(HCO3)2

@arvind_arora_vedantu_chemistry Open / Comment

2021-10-01 07:56:50 1. Formulas related to force:
F = ma
F = kx
F = m(vf² - vi²/2S)
F = mv/t
F = md/t²
F = m(vf - vi) /t
F = Area × density × velocity²
F = 1/2 mv²/d
F = 1/2 Pv/d
F = Power/velocity
Fc = mv²/r
Fc = mrw²
Fc/2 = mv²/2r
Fc = 2K.E/r
F = Area × Stress
F = pir² × stress
F = YA × Strain
F = YAl/L
F = pressure × area
F = change in momentum × time interval
F = - 2mVx × Vx/2l
F2 = F1/A1 × A2
F = qE
F = kQ/r²
F = ILB sintheta
F = q (v × B)
F = qE + q(v × B)

2. Formulas related to energy and work
Fd = k.e
mgh = 1/2 mv²
E = 1/2 kx²
E = Ve
E = nhf
E = nhc/lambda
E = Pc
K.e = hf - work function = hf - hf° = hf - hc/w° (here w° is cutt off wavelength)
E = 1/2 Pv
mv²/2r= Fc/2
K.E/r = Fc/2
K.E = Fc×/r/2
K.e = 1.5 KT
E = VQ
E = Power × time
E = Fvt
% loss in K.e = v1² - v2²/v1² × 100
% loss in P.e = h1² - h²/h1² × 100
Energy lost due to air friction(Fh) = 1/2mv² - mgh (when body is thrown upward)
Energy lost due to air friction(FS) = mgh - 1/2mv² (when body is thrown downward)
E = 1/2 CV² (capacitor)
E = R × hc (R is Rydberg' constant)
J = m-¹ × Js ms-¹
hf kalpha x rays = EL - Ek
hf kbeta x rays = EM - Ek
Binding energy = mass defect × c²
W = Fd Costheta
W = nmgh (when person is climbing stairs)
W = n(m+m) gh (when person is climbing stairs with some load)
W = 0mgh + 1mgh + 2mgh + 3mgh ....... (in case of stacking bricks. For ist brick h=0. For 2nd brick h=1. For 3rd brick h=2 and so on)
W = Fd = PA × change in V
W = Q - change in U
Q = mc × change in T
T/273.16 = Q/Q3 (Thermodynamic scale)
W = I²Rt
W = emf×charge
W = VQ
W = 1/2 lF
W = YAl²/2L
W = StressAl²/2Strain
W = PressureAl²/2Strain
W = Fl²/2Strain

3. Formulas related to Power
P = Fv
P = E/t
P = n(mgh/t)
P = Fd/t
P = mv²/2t

4. Formulas related to distance, displacement, velocity and accelration
d = vt
d = at²
d = (vf + vi/2) ×t
d = 5t² (for distance in 'n' seconds)
d = 5(2tn - 1) (for distance in 'nth' second)
d = 1/2 mv²/F
d = vit + 5t²
d = v × underroot 2H/g
d = vt = x°wt = x°2pi/T × t = x°2pift
x = x° Sin wt
x = x° Sin (underroot k/m) t
vf = vi + at
2as = vf² - vi²
2as = (vi + at)² - vi²
2as = vf² - (vf - at) ²
v = underroot Vfx² + Vfy²
v = Power/Force
v = 2×K.E/momentum (k.e = 1/2 Pv)
v² = 2×Power×time/mass (P = mv²/2t)
v = underroot 2as
v = underroot gr (speed at highest point in a verticle circle)
v = underroot 5gr (speed at lowest point in a verticle circle)
v² = 2FS/m
v² = 2E/m
v² = 2Ve/m
v = eBr/m (velocity of particle under action of magnetic force along circular path)
v² = Force/Area.Density
v = w underroot x°² - x²
v = underroot k/m × underroot x°² - x²
v = x°w (at mean position where x=0)
v = x° underoot k/m
v = v° underroot 1 - x²/x°² (for determining ratio b/w inst. Velocity and maxi. Velocity)
v= x°2pif = x°2pi/T
a = x°w² = x°w.w = vw = v.2pif
Common velocity = m1v1/m1+m2
vi² = Rg/Sin2theta
v = underoot Tension×length/mass
V = 2pi ke²/nh (speed of e- in nth orbit)
Vn = V/n
v = nh/2pimr (lambda = 2pir and lambda=h/p)
ma = kx
a = kx/m (SHM)
a = - gx/l (Simple pendulum)
ac = v²/r

5. Formulas related to wavelength 'w'
w = v/f
w = 1/wave number
w1 = 2l (when pipe is opened at both ends)
w1 = 4l (when pipe is opened at one end)
Delta w = Us/f (doppler shift)
Wavelength for obs. = w - delta w = v/f - Us/f
w = hc/Ve
w = hc/E
w = h/mv
w = h/P as P = underroot 2mE so
w = h/underroot 2mE (de Broglie wavelength)
w = underroot 150/V A° (short method for de Broglie wavelength. This formula is applicable only for e-)
1/w = RH (1/p²-1/n²)
Wmaxi/Wmini = n²/n²-p² (for determining ratio b/w maxi. Wavelength to mini. Wavelength for series of atomic spectrum)
w = 2pir/n (n is no. of loops in a circle)
h/mv = 2pir

@arvind_arora_vedantu_chemistry Open / Comment

2021-10-01 07:56:50 Top tips for the night before, and the morning of NEET exam

1. Use your moments wisely

The few minutes before you switch off the light is a fantastic time for memorisation. Learning before sleep significantly improves memory retension - so when you're tucked up in bed, have a quick read through of the most important facts, equations or vocabulary. Then switch off.

2. Eat a great meal

Make a special effort to cook yourself a decent meal. I'm not suggesting “brain food” such as nuts or oily fish, but a meal you’d have if you were celebrating and wanted to spoil yourself, such as steak and chocolate tart. It’ll make you feel happy and give you a boost ready for the next day.

3. Laugh

Do one fun thing like watching a comedy show, having a kickabout or fiddling around with photos on instagram. Laughing will relax you, lower stress and help you get a good night’s sleep. 

4. Make sure you wake up

Set an alarm. Set two. Get a family member or friend to check you're up if you're still worried. Just make sure you get to the exam in plenty of time.

5. Know when to stop  

In the morning, eat a hearty breakfast and go over your most important facts before you leave the house. Then again on the way to the exam. When you arrive at the exam hall, stop using your brain. If you don’t know your stuff by now, you never will! Too much last minute cramming minutes before, can send your brain into a spin. 

6. Use the space

When in the exam room, sit comfortably and spread out. It’s proven that when we physically make ourselves larger, by spreading our arms or leaning back on the chair, this releases a type of hormone which makes you more confident. It works - we promise.

7. Don't drink too much

Don’t drink too much water, a sip every hour is all you need. Dehydration is not going to be a problem during the exam, but drinking many pints of water and needing a toilet break every five minutes is. The sip is just for a quick diversion to give your brain and hand a breather.

8. Do what works for you

Finally, everyone is different. Some people like to be on their own around exams, others blare motivational music from their earphones. Maybe you're someone who likes to wear their shiniest pair of high heels for luck. Do whatever puts you in the best frame of mind and you'll be set up perfectly to ace the exam.  Open / Comment