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Interview study book NCERT Solutions Physics Class 11th of Nipendra Bhatnagar - ISBN: 9789351416340 (Physics formula)

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Formulae S he e t f o r P h ys i c s w w w .c o n c e p t s - o f -p h y s i c s . c om | pg. 1

Physics formulas from Mechanics, Waves, Optics, Heat and Motion in a straight line with constant a:
Thermodynamics, Electricity and Magnetism and Modern
Physics. Also includes the value of Physical Constants. Helps v = u + at, s = ut + 21 at2 , v 2 − u2 = 2as
in quick revision for CBSE, NEET, JEE Mains, and Advanced.
Relative Velocity: ~vA/B = ~vA − ~vB
0.1: Physical Constants

Speed of light c 3 × 108 m/s
u
Planck constant h 6.63 × 10−34 J s y




u sin θ
hc 1242 eV-nm H
x
Projectile Motion:
Gravitation constant G 6.67×10−11 m3 kg−1 s−2 θ
Boltzmann constant k 1.38 × 10−23 J/K O u cos θ
Molar gas constant R 8.314 J/(mol K) R

Avogadro’s number NA 6.023 × 1023 mol−1
Charge of electron e 1.602 × 10−19 C x = ut cos θ, y = ut sin θ − 21 gt2
Permeability of vac- µ0 4π × 10−7 N/A2 g
y = x tan θ − 2 x2
uum 2u cos2 θ
Permitivity of vacuum 0 8.85 × 10−12 F/m 2u sin θ u2 sin 2θ u2 sin2 θ
T = , R= , H=
Coulomb constant 1
4π0 9 × 109 N m2 /C2 g g 2g
Faraday constant F 96485 C/mol
Mass of electron me 9.1 × 10−31 kg 1.3: Newton’s Laws and Friction
Mass of proton mp 1.6726 × 10−27 kg
Mass of neutron mn 1.6749 × 10−27 kg Linear momentum: p~ = m~v
Atomic mass unit u 1.66 × 10−27 kg
Atomic mass unit u 931.49 MeV/c2 Newton’s first law: inertial frame.
Stefan-Boltzmann σ 5.67×10−8 W/(m2 K4 )
Newton’s second law: F~ = d~
p
dt , F~ = m~a
constant
Rydberg constant R∞ 1.097 × 107 m−1 Newton’s third law: F~AB = −F~BA
Bohr magneton µB 9.27 × 10−24 J/T
Bohr radius a0 0.529 × 10−10 m Frictional force: fstatic, max = µs N, fkinetic = µk N
Standard atmosphere atm 1.01325 × 105 Pa
v2 v2
Wien displacement b 2.9 × 10−3 m K Banking angle: rg = tan θ, rg = µ+tan θ
1−µ tan θ
constant
mv 2 v2
Centripetal force: Fc = r , ac = r
2

1 MECHANICS Pseudo force: F~pseudo = −m~a0 , Fcentrifugal = − mv
r

Minimum speed to complete vertical circle:
1.1: Vectors
p p
vmin, bottom = 5gl, vmin, top = gl
Notation: ~a = ax ı̂ + ay ̂ + az k̂
q
Magnitude: a = |~a| = a2x + a2y + a2z θ
l
q
l cos θ
Conical pendulum: T = 2π θ T
Dot product: ~a · ~b = ax bx + ay by + az bz = ab cos θ g


ı̂ mg
a × ~b
~ ~b
Cross product:
θ k̂ ̂
~
a
1.4: Work, Power and Energy
~a ×~b = (ay bz − az by )ı̂ + (az bx − ax bz )̂ + (ax by − ay bx )k̂ Work: W = F~ · S
~ = F S cos θ, W =
R
F~ · dS
~

|~a × ~b| = ab sin θ Kinetic energy: K = 12 mv 2 = p2
2m

Potential energy: F = −∂U/∂x for conservative forces.
1.2: Kinematics
Ugravitational = mgh, Uspring = 21 kx2
Average and Instantaneous Vel. and Accel.:

~vav = ∆~r/∆t, ~vinst = d~r/dt Work done by conservative forces is path indepen-
dent and depends only on initial and final points:
~aav = ∆~v /∆t ~ainst = d~v /dt H
F~conservative · d~r = 0.

Work-energy theorem: W = ∆K



Get Formulas www.concepts-of-physics.com Get Our Book
c 2020 by Jitender Singh Ver. 2020 1

, Formulae S he e t f o r P h ys i c s w w w .c o n c e p t s - o f -p h y s i c s . c om | pg. 2

Mechanical energy: E = U + K. Conserved if forces are Rotation about an axis with constant α:
conservative in nature.
ω = ω0 + αt, θ = ωt + 21 αt2 , ω 2 − ω0 2 = 2αθ
Power Pav = ∆W
∆t , Pinst = F~ · ~v

mi ri 2 , r2 dm
P R
Moment of Inertia: I = i I=
1.5: Centre of Mass and Collision
P R
R xdm
2
Pxi mi ,
1
2 mr m(a +b )
2 2
Centre of mass: xcm = xcm = mr 2 1 2 2 2 2 2 1 2 mr 2
mi dm 2 mr 3 mr 5 mr 12 ml 12


b
a
CM of few useful configurations: ring disk shell sphere rod hollow solid rectangle

m1 r m2
1. m1 , m2 separated by r: C
m2 r m1 r Ik Ic
m1 +m2 m1 +m2 2
Theorem of Parallel Axes: Ik = Icm + md d
cm
h
2. Triangle (CM ≡ Centroid) yc = 3 h
C
h
3
z y
Theorem of Perp. Axes: Iz = Ix + Iy
2r
x
3. Semicircular ring: yc = π
C
2r
r π
p
Radius of Gyration: k = I/m
4r
4. Semicircular disc: yc = 3π C 4r
r ~ = ~r × p~, ~ = I~
3π Angular Momentum: L L ω
r y
5. Hemispherical shell: yc = 2 C r ~ P θ ~
r 2
Torque: ~τ = ~r × F~ , ~τ = dL
dt , τ = Iα F
~
r x
O
3r
6. Solid Hemisphere: yc = 8 C 3r
r 8 ~ ~τext = 0 =⇒ L ~ = const.
Conservation of L:
7. Cone: the height of CM from the base is h/4 for Equilibrium condition:
P~
F = ~0,
P
~τ = ~0
the solid cone and h/3 for the hollow cone.
Kinetic Energy: Krot = 12 Iω 2

Motion of the CM: M =
P
mi Dynamics:
P
mi~vi F~ext ~τcm = Icm α
~, F~ext = m~acm , p~cm = m~vcm
~vcm = , p~cm = M~vcm , ~acm = 1 2 1 2 ~
M M ~ + ~rcm × m~vcm
K = 2 mvcm + 2 Icm ω , L = Icm ω

Impulse: J~ = F~ dt = ∆~
R
p
1.7: Gravitation
Before collision After collision
Collision: m1 m2 m1 m2 m1 F F m2
Gravitational force: F = G mr1 m
2
2

v1 v2 v10 v20 r
Momentum conservation: m1 v1 +m2 v2 = m1 v10 +m2 v20
2
Elastic Collision: 12 m1 v1 2+ 12 m2 v2 2 = 12 m1 v10 + 12 m2 v20
2 Potential energy: U = − GMr m
Coefficient of restitution: GM
Gravitational acceleration: g = R2
−(v10 − v20 )

1, completely elastic
e= = h

v1 − v2 0, completely in-elastic Variation of g with depth: ginside ≈ g 1 − R

2h

Variation of g with height: goutside ≈ g 1 −
If v2 = 0 and m1
m2 then v10 = −v1 . R

If v2 = 0 and m1
m2 then v20 = 2v1 . Effect of non-spherical earth shape on g:
Elastic collision with m1 = m2 : v10 = v2 and v20 = v1 . gat pole > gat equator (∵ Re − Rp ≈ 21 km)

Effect of earth rotation on apparent weight:
1.6: Rigid Body Dynamics
∆θ dθ
Angular velocity: ωav = ∆t , ω= dt , ~ × ~r
~v = ω
∆ω dω
Angular Accel.: αav = ∆t , α= dt , ~ × ~r
~a = α



Get Formulas www.concepts-of-physics.com Get Our Book
c 2020 by Jitender Singh Ver. 2020 1

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