Formula in Physics1

8/12/2019 Formula in Physics1

1/66

Mohammed

Asif

Name :

Roll No. :

Topic : Electric Field &Potential

Ph : 939132!"#

$%!"

2. MEA'REMENT (F P)*+A, -'ANT+T+E

1. If a physical quantity P. is given by

P = x + y

m nn m

AbsoluteerrorinP P X y

x y x yPRelativeerror !errorinP 1""

P x y x y

#. IfP x y P x y$

x yP P!errorinp 1""

P x y P

%. IfP x y orx & y .

m x n yPP x y

= +

+ + = = + +

= = + +

= =

=

= +

i.e. ! error in P = m. ! error in x + n. ! error in y.

'. a b cP ( ) * . =

If an arethepercentageerrorsinthemeasurementxof() an*

thenpercentage errorinP a b c .

= + +

,ignificant figurei- all non/ero igits are ,ignificant 0x. #%.2 3o. of s. f 4 'ii- 5ero in bet6een t6o non/ero igits are ,ignificant 0x. #%"'" 3o. of s. f 4 2iii- All /eros to the left of the first non/ero are not ,ignificant 0x. ""#%2 ".""#%2 3o. of s. f 4 'iv- All /eros to the right of the last non/ero igit in a measure value are ,ignificant

7lat 3o.%"1 Ru88us 9 :ellus Arcae 7ever ;ospital Roa


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0x. #%2"" m 3o. of s. f 4 2

(ean eviation 1 # % n+ + + ...+

@eviation -n

+ + += =

,tanar eviation# # #

1 # n ... @6hennislarge.-n

+ + + =

Probable error = r = ".2>' = *ruevaluex X r Probableerrorr n

1



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3. E/T(R

M 0 +: Manitde and direction of a ector:

$ $= + +

= + +

$

# # #

1. IfA ai b B c8

*hen@a-magnitue A a b c .

@b- Cnit vector in the irection of =uur AA isgivenbyA .

A

= = =

+ + =

+ + =

ur

ur

r

# # #

# # #

@c-If an aretheanglesmaebyA6ithxy an/axesthen

a b cDos Dos Dos .

A A A

AlsoDos Dos Dos 1.

@-IfAis aunit vector then a b c 1.@e-EectorparalleltoAanhavingmagnitueequal

toanothervector


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Q

P

R

r

M 0 +++: Paralleloram la4 of ectors

2 2

2 2

If R P Q(sumof the vectors)

Magnitudeof ther esul t an t R P Q 2PQCos .

Q sin P sinDirection tan tan

P Q cos Q P cos

If R P Q (difference) R P Q 2PQCos

Q sin P sintan tan

P Q cos Q P cos

If P Q then

= +

= + +

= =

+ +

= = +

= =

+ +

> = =

Q

R

P

M 0 +: Resoltion of ectors:

If a vector P ma8es an angle 6ith hori/ontal then its hori/ontal componentPx = Pcos $ Eertical component = Py = P sin

2 2

x yP p p= +

= = +$ $x x yy

Ptan P P i P BP

Py

Px

P

M 0 : Pol5on la4 of ectors:

n equal force each of magnitue 7 are acting simultaneously on a particle.

0ach force ma8es an angle 6ith one then

n

%2"a- If then theresul tan tforceis/ero.=

n

%2"b- If 1thentheresultantforceis7. =



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Area of parallelogram A < A an < are the t6o aBacent sies of aparallelogram

1Area of triangle A < A an < are the t6o aBacent sies of atriangle.

#

=

= r rur ur

$ $x y / x y /

f- Cnit vector prepen+icular to vectors A an+ < is

A T(N? ,A> (F M(T+(N@

1. 7orce 7 @m - m ma 6hen mass is constant

t t m

7 @m - if m is changes in the case of a variable masst t

= = =

= =

,ystem li8e a roc8et.#. a- Fhen a lift moves up6ars 6ith acceleration a @or- o6n6ars 6ith a

eceleration a then apparent 6eight of a boy in the lift 6 = m @g + a-.

b- Fhen a lift moves up6ars 6ith eceleration a @or- o6n6ars 6ith anacceleration a then apparent 6eight of a boy in the lift 6 = m @g a-. c- Fhen the lift is stationary or moving 6ith uniform velocity the 61= mg.

- Percentage change in 6eighta

1"".g

=

e- If h is the epth of 6ater in a buc8et place in the lift then pressure at thebottom of the buc8et is P = h @g a-.

%. Fhen a machine gun fires NnG bullets in a time t then the force require to holthe gun is

mn

7 t

= .

'. a- Fhen a metallic plate of mass m is hel in mi air by firing NnG shots&secon 6ith avelocity u if the bullets stop ea after stri8ing the plate then mnu = mg.

b- If the shot comes bac8 6ith a velocity E then nm @v+u- = mg.. If 6ater from a narro6 pipe hits a 6all hori/ontally 6ith a velocity v an stops ea

force acting on the 6all#

# H7 A H 6hereH avA

= = = =

@volume flo6 rate-.A = area of cross section of the pipe

= ensity of 6ater.If 6ater bounces bac8 6ith same spee 7 = # Av#

2. If a ball hits a 6all normally 6ith a velocity v an rebouns 6ith same velocity changein momentum = # mv.



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7orce acting on the 6all#m

t

= 6here NtG is the time of collision.

>. a- A boy of mass m moving 6ith a velocity v hits 6all an rebouns 6ith samevelocity then change in momentum perpenicular to the 6all p #mvsin = force

on the 6all#m sin

7 @ is the angle ma+e 6ith the 6all-t

=

b- Dhange in momentum parallel to 6all = ".J. 0xtra force require to 8eep the conveyor belt moving 6ith a velocity E 6hen

,an @any mass- is falling on it the rate ofm m

is given by 7 E.t t

=

?. If hot gases are coming out of the roc8et at the rate of

mt

6ith a velocity E

then

a- *hrust on the roc8et =

m7 .

t

b- If the roc8et moves up 6ith uniform velocity then 7 = (g(mass of the roc8et.

c- If the roc8et moves up 6ith uniform acceleration a then 7 = (@g + a-.If gravity is neglecte then 7 = (a

- Acceleration of the roc8et after t sec a

+m

+t@neglecting gravity-.

+m( t

+t

=

,A> (F /(NERAT+(N (F ,+NEAR M(MENT'M AN6 /(,,++(N

M 0 +: ,a4 of conseration of momentm:

A boy of mass m1 moving 6ith a velocity u1collies 6ith another boy of mass m#moving 6ith a velocity u# if v1an v#are their final velocities then Accoring to la6 ofconservation of momentumm1u1 + m#u# = m1v1 + m#v#If the t6o boies stic8 together after collision then common velocity

=

+1 1 # #

1 #

m u m uE

m m

1. If boies are moving in mutually perpenicular irections before collision then

common Eelocity( ) ( )+

=+

# #

1 1 # #

1 #

mu m uE .

m m

#. In case of obliquic collision i.e. after collision if the irection of motion of boies



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ma8es an angle 6ith initial irectiona- 1 1 # #mu m u+ = @ 1 1 # # xi xf@m v m v -cos @P P -+ =

b- 1 1 # # yi yfm v m v . @P P -= =

%. If a bullet of m come out 6ith a velocity from a gun of mass ( the Recoil velocity

of gunm

E .(

=

bullet

gun

0 ( .0 m

= Eelocity of bullet relative to gun = v + u.

'. If a stationary shell brea8s into t6o fragments they 6ill move in oppositeirections 6ith velocities in the inverse ratio of their masses.

1 #

# 1

E m.

E m=

. In the above case the inetic energy of the t6o fragments is inversely proportionalto their masses.

1 #

# 1

0 m0 m=

2. *he total energy release in the explosion # #1 1 # #1 1

m m# #

= +

>. If a shell at rest exploes into three fragments having masses in the ratio m 1 m#(%if first t6o fragments travel in mutually perpenicular irection 6ith velocitiesv1an v#then the velocity of the thir fragment is given by

( ) ( )

# #

1 1 # #%

%

mu m uE .

m

+=

J. If a shell of mass m moving 6ith a velocity E exploes into t6o fragments havingmasses m1 an m# an if 1 #E anE are the velocities of the fragments then

1 #1 #m m m = +

M 0 ++: Elastic collision:

1. 7or one imensional elastic collision v#4 v1= u14 u#.

1 # # #1 1

1 # 1 #

1# 1

# #1 # 1 #

m m #m uu

m m m m

#mum m

um m m m

= + + +

= + + + ,pecial cases

a- Fhen m1= m#$v# = u1v1 = u#.b- If u# = " @i.e. secon boy is at rest-then



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If m1= m# v# = u1v1 = "If m# OO m1 v1 = u1v# = "If m1 OO m# v1 = u1v# = # u1.

#. In case of elastic collision if the secon boy is at rest *he fraction of .0 transferre or lost by the first boy

( )

1 #

##1 #

'm m 'n

.@m m - n 1= =+ + Fhere n = m1 & m#

%. 7raction of .0 retaine by the first boy is

# #

# 1 f

1 # i

0m m 1 n

m m 0 1 n

= =

+ + Fhere n = m1 & m#

If m1= m#fraction of 0 transferre is maximum an that is 1"" !.M 0 +++: +n elastic collision:

1. a- 1 1 # #mu m u+ = 1 1 # #m v m v+

b-# 1

1 #

e

u u

=

7inal velocity can be foun by using above t6o equations

#. If u# = " i.e. if the secon boy is at rest ratio of final velocities1

#

1 e.

1 e

= +

( ) ( )1 11 #u u

1 e 1 e .# #

= = +

# # # #1 1 # # 1 1 # #

1 1 1 1%. )oss of 0 mu m u m m

# # # # = + +

( )

( ) ( )# #1 #

1 #1 #

1 m mu u 1 e

# m m=

+

'. A bloc8 of mass ( is suspene freely. A bullet of mass m is fire in to the bloc86ith a velocity . If the bullet gets embee in to the bloc8 then

a- Eelocity of bloc8 after collisionm

M m

=

+

b- ;eight rise by the bloc8

=#

#;

g

c- (aximum angle mae by the bloc8 6ith vertical cos#

#gl

=

- )oss of .0( )

#

#(m

( m =

+

M 0 +: /oefficient of Restittion:



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1. Doefficient of restitution# 1

1 #

.eu u

=

a- Fhen a boy falls from a height h an rises to a height h1after first collision

6ith the groun then 1h

eh

=

b- ;eight rise after n bounces hn = e#n. h

c- Eelocity after n bounces vn = en. E = en

#gh

#. If a boy falls from a height h the total istance travele before coming to rest.#

#.

ex h

e

1 += 1

%. If a boy falls from a height h the total time ta8en to come to rest.

# 1

.1

h et

g e

+ =

'. If a boy hits the groun 6ith a velocity v1an rebouns 6ith a velocity v#then

## #1 # 1

p I0e

p I0

= = =

. Percentage loss of velocity after n bounces = @1 en- X 1"".2. Percentage loss of momentum after n bounces = @1 en- X 1"".

>. Percentage loss of 0 after n bounces = @1 4 e#n- X 1"".J. A boy hits groun 6ith a velocity u ma8ing an angle 1 6ith vertical an rebounces

6ith a velocity v at an angle # then

#

1

coscos

eu

=

If it is a smooth surface then1

#

tan .tan

e

=



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. >(R 0 P(>ER ENERB*

M 0 +: >orC:

1. a- If a force N7G acts on a boy ma8ing an angle 6ith hori/ontal an isplaces the

boy through a istance N,G 6or8 one "F 7., 7,cos .6hen ?" F ".= = = =

b-F Dos @If 7 an , are same-.#c-F Dos

@If 7 an t are same-.#. For8 one against gravity F = mgh.

%. For8 one in imparting velocity to a boy #1

F m#

=

'. For8 one against friction on a level surface. F 7x, mgs. = = coefficient of friction.

. For8 one in compressing or elongating a spring#

#1 7 1F 8x 7x.# #8 #

= = =

4 spring constant x 4 elongation2. For8 one in moving a boy up on to a smooth incline plane is F = @mg sin - ,.>. For8 one to compress or expan a gas F Px E= at constant pressure.J. a- For8 one in pulling the bob of a simple penulum asie through an angle from

the vertical is

( ) #F mgl 1 cos #mglsin .# = =

m 4 mass of the bob l length of penulumb- For8 one in isplacing the penulum from an angle 1 to an angle # is

( )1 1 #F mg cos cos=

?. For8 one in pulling asie a bar through an angle from the vertical is

( ) #1

F mgl 1 cos mglsin# #

= =

m 4 mass of the ro l length of the ro



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1". *he 6or8 one in lifting a boy of mass NmG an ensity NsG in a liqui of ensity N1Gthrough a height NhG uner gravity is

1

s

F mgh 1 .

=

11. For8 one by external resultant force on a boy is equal to change in 0 of theboy

( )# #1

F m u 0.#

= =

1#. For8 one in lifting 6ater from a 6ell from a 6ell of epth h is "m

F m gh#

= +

m(ass of 6ater m"(ass of rope1%. A ro of mass m an length l lies hori/ontally on a NfloorG. *he 6or8 one in rotating

the ro through an angle about one ege in a vertical plane is givenF = K mgl sin .

1'. In the above point the 6or8 one in bringing the ro to vertical position is

F = K mgl.1. A uniform chain of length l an mass m is place on a friction less table such that

1&nth of itslength hanging over the ege. *he 6or8 one in pulling the hanging part

on to the table is #.#

mglF

n=

M 0 ++: Po4er:

1. Average po6er P = F&t.#. Instantaneous po6er = 7 x E = . .7 E%. *he po6er of machine gun firing NnG bullets each of mass NmG in one secon 6ith

velocity NvG is #1. .#

P n m=

'. A motor sens a liqui 6ith a velocity NEG in a tube of cross section NAG an NG is the

ensity of the liqui then the po6er of the motor is %1

.#

P A+E =

. A motor lifts m 8g of 6ater to a height h in t sec. *hen the po6er of the motor ismgh

Pt

= .

If is the efficiency of the motor then .mgh

Pt

=

If the 6ater comes at of the pipe 6ith a velocity then#1

# .mgh m

Pt

+=

2. 0xtra po6er require to 8eep a conveyor belt moving 6ith constant velocity if



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gravel is falling the belt at a rate of m&t is#.

+mP

+t

=

>. A vehicle moves 6ith a constant velocity E on a rough hori/ontal roa. *he frictionalforce acting on the vehicle is f then the po6er o f the engine P= f x v.

M 0 +++: Ener5:

1. 0 = K mv#.#. P0 = mgh.%. P0 of a liqui in a capillary tube P0 = @- @ah- g h = agh# & #.

4 ensity of liquia 4 area of cross section of the capillary tube.h 4 height of the liqui in the capillary tube.

'. P0 of a bloc8 place on the groun =h

lbh g. .#

. P.0 in a spring = K 8x# .

M 0 +: ,a4 of conseration of ener5:1. A boy is roppe freely from a height h uring collision 6ith groun it looses x !

of energy an then rises to a height NhG then( ) 11"" x mgh mgh $

1""

=

@1""x- remaining energy$1 1"" xh h.

1""

#. In the above question if the boy is thro6n vertically o6n6ars 6ith a velocity Ethen

# 11"" x 1mgh m mgh

1"" #

+ =

%. A boy of mass m is roppe from a height h on to a spring of spring constant .If x is the compression prouce in the spring then

( ) #1

mg h x 8x#

+ =

If x Q Q Q h mgh = K 8x#

'. A boy of mass m moving 6ith a velocity E collies 6ith a spring of spring constant an comes to rest. If x is the compression prouce in the spring then

K mv# = K 8x#

M 0 : >orC 0 Ener5 theorem:1. For8 one = Dhange in 0$

7 x , = K m @E# u# -.#. A boy of mass m is roppe from a height h it reaches the groun 6ith a velocity

( )E E #gh< 6or8 one by air resistance F = mgh 4 K mv# .



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%. A 8nife ege of mass m is roppe from a height h on to a table. It penetrates aistance x into the table. If the resistance offere by the table is 7 then7 x x = mg @h+x-. if x Q Q Q h then 7x= mgh

M 0 +: Relation 8et4een E and Momentm:

( )

( )

#

#

P 11. 0 0 if P is same .

#m m

0 P if m is constant .

=

( )

( )

0#. P #m8 P m If0 is same

P 0 If m is same .

=

%. *6o boies of ifferent mass are moving 6ith same momentum if same brea8

force is applie on them. *hen

1 #

# 1

10 Accoringly6.e.theoremF. 0.

m1 , m

a-7x, 0 " 0 , 0m , m

=

= =

)ighter boy 6ill travel more istance bet6een coming target.b- In pulse = change in momentum

7 x t = P 4 " ,in 7 an P are same t 6ill7 x t = P t1 t#= 1 1.

,o both ta8e same time to come to rest.'. *6o boies of ifferent mass are moving 6ith same 0. If same brea8 force isapplie on them then

1 #

P m

a-7 x, 0 ,ince7 an 0$are same

,6illbesame, , 1 1.

=

=

,o both travel same istance before coming to rest.

b- 7 x t = Pi 4 Pf it P m

i # #t t m m=

;eavier boy ta8es more time to come to rest.



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". /ENTRE (F MA

M 0 +: /oordinates of center of mass:

1. Particles istribute in space. If @x1y#/1- @x#y#/#- 4 are the position coorinatesof particles of masses m1 m#4 the position coorinates of their center of mass

are

1 1 # # n ncm

1 # n

1 1 # # n ncm

1 # n

1 1 # # n ncm

1 # n

m x m x ..... m xx

m m ..... m

m y m y ..... m yy

m m ..... m

m / m / ..... m //

m m ..... m

+ + +=

+ + +

+ + +=

+ + +

+ + +=

+ + +

#. In vector notation. If r1 r# r%.. Are the position vectors of particles of

masses m1 m# m% *hen the position vector of their center of mass is1 1 # # n n

1 # n

m r m r ..... m rr

m m ..... m

+ + +=

+ + +

M 0 ++: elocit5 /enter of Mass:

%. a- Eelocity of center of mass

If 1 # % nEE E .....E are the velocities of particle of masses.

m1 m# m% mnthe velocity of their center of mass is

1 1 # # % % n n 1 # % ncm

1 # % n 1 # % n

m m m ..... m P P P ...PE

m m m ..... m m m m .....m

+ + + + + += =

+ + + + +

cm 1 # nb- (E P P ... P= + + + i.e. total momentum of the system is the prouct of massof the 6hole system an the velocity of the center of mass.

c- If E1an E#are the magnitues of velocities of t6o particles of masses m1an m#then the magnitue of velocity of c.m.



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F r i c t i o n

N

m g

F s i n

F c o s

( )

6sinmg7 . angleoffriction

cos sin cos

= =

+

7 6ill be minimum if = .

min #

mg7 6sin .

1

= =

+

J. In case of pushing3 = mg + 7 sin .

7 = Lmg + 7 sin M87cos fa .

m

=

(inimum pushing force require to isplace the boy

( )

6sinmg7 .

cos sin cos

=

+ "if ?"+ 7 becomes infinity.

;ence ?" ?"+ < < N

m g

F s i n

F c o s

F

?. Acceleration 87 f

am

= @If force is applie hori/ontally-.

1". Acceleration of a freely falling boyf

a g .m=

11. eceleration of a vertically proBecte boyf

a g .m

= +

@ffriction force ue to air-.



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1#. 7or a vertically proBecte boy au mu

t .g f&m mg f

= =+ +

@If air resistance is ta8en into account-#u

;f

# gm

= +

Eelocity before reaching grounmg f

umg f

=

+

1%. (inimum force require to isplace a boy is applie an the same force is

continue then the acceleration of the boy ( )s 8a g.=

1'. A chain of uniform length N)G is place on a rough hori/ontal table. *hecoefficient of friction bet6een the chain an table is then the maximumfractional length of chain that can be hung freely from the ege of the table is.x

) 1

=

+

(inimum fraction of length of chain that can be on the table is1

.1 +

1. bloc8 slips or slies on the floor. 7rictional force on the bloc8

8mg.= *he acceleration @a- of the bloc8 relative to lorry is1

8a a g=

12. Dar moving on an umban8e circular roaa- Fhen a car goes aroun an unban8e circular roa the static frictional force

bet6een the 6heels an the roa provies the necessary centripetal force.

b- (aximum spee 6ith 6hich curve can be negotiate 6ithout s8iing ismax sE rg=

s = coefficient of static frictional bet6een 6heels an the roa.c- *he maximum angular velocity is



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smax

gF

r

=

1>. a- If a bloc8 having initial velocity u slies on a rough hori/ontal surface ancomes to rest the acceleration of the bloc8 is

8a g=

b- istance travele by the bloc8 before coming to rest is#

#

8

u, , C# g

=

c- *ime ta8en by the bloc8 to come to rest is

8

ut t C

g=

7od5 on ertical srface.

1J. Fhen a boy of mass m is presse against a vertical surface 6ith a force P thena- 3ormal reaction 3 =P.

b- )imiting frictional force )f P.=

c- 7rictional force bet6een the boy an surface P or mg 6hich ever is less.- (inimum force require to slie the boy up6ars.

7 = mg + P

e- (inimum force require to prevent the boy from sliing o6n 7 = mg P

P N

m g

f

1?. A boo8 of mass m is presse bet6een t6o hans by applying a hori/ontal force P

6ith each han the boo8 oes not slie ifmg

.#P

=

#". A vehicle is moving on a hori/ontal surface. A bloc8 of mass NmG is stuc8 on thefront part of the vehicle. *he coefficient of friction bet6een the truc8 an thebloc8 is . *he minimum acceleration 6ith 6hich the truc8 shoul travel sothat the boy oes not slie o6n is

ga .=

Lmg = maM

mooth inclined plane:



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- If the boy is sliing o6n 6ith uniform velocity then

8 8a " f mgsin tan .= = =

e- 3et contact force acting on the boy = # #f 3 .+

mg

sin

m g mg

cos

N

f

#%. If the bloc8 slie o6n from the top of the incline plane. Eelocity at thebottom of the plane is

( ) ( )8 8E #gl sin cos #gh 1 cot= =

#'. In the above case time of escent is

( )8#)t g sin cos=

#. *he time ta8en by a boy to slie o6n on a rough incline plane is NnG times thetime ta8en by it to slie o6n on a smooth incline plane of same inclination anlength then coefficient of friction is

#

1tan 1 .

n =

#2. If a bloc8 is proBecte up a rough incline plane the acceleration of the bloc8 is

8a g sin cos = +

#>. a- *he istance travele by the bloc8 up the plane before the velocity becomes/ero is

( )

#

8

u,

#g sin cos=

+

b- *he time of ascent is ( )8

ut .

g sin cos=

+

c- If the boy slies o6n 6ith uniform velocity on the same plane then#

8

C Csin cos then, t

'gsin #gsin = = =

#J. In the above case if time of ecent is n times the time of ascent then#

#

n 1tan .

n 1

= +

#?. a- 7orce neee to be applie parallel to the plane to move the bloc8 up 6ith



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constant velocity is

( )8 87 mgsin mgcos mg sin cos .= + = +

b- 7orce neee to be applie parallel to the plane to move the bloc8 up 6ith anacceleration a is

87 mgsin mgcos ma.= + =

%". If bloc8 has a tenency to slie the force to be applie on the bloc8 parallel anup the plane to prevent the bloc8 from sliing is

( )s s7 mgsin mgcos mg sin cos .= =

%1. If a force P is applie on the boy up the plane then the acceleration of theboy is sin .

( )8P mg sin cosa .m

+ =

%#. An air craft of mass m travels through a istance , on a run6ay an ta8e off6ith a velocity E in t sec. *hen.

a- 7orce applie by the engine#m

7 mg#,

= +

b- For8 one by the engine #1

F mgs m#

= +

c- Po6er of the engine#1mgs m

#P .t

+ =



32/66


33/66

M ++: Motion of a 8od5 in a ertical circle:

Dase I If a boy of m is tie to a string of length 1 an rotate in a vertical circle

6ith uniform spee ( )v. v g1 .>

1". *ension in the string 6hen it ma8es an angle 6ith vertical is given by#m

* mgcos .

l

=

a- *ension at the highest point @ = 1J"" -

( )#

#minh

m* mg ml mg * .

l

= = =

b- *ension 6hen the string is in hori/ontal position ( )"?"=#

#m* mll

= =

c- *ension at the lo6est point ( )"" . =

( )#

#

) max

m* mg ml mg * .

l

= + = + =

- ifferent bet6een maximum an minimum tensions.

max min* * #mg. =

e-#

max

min

n 1*If nthen .

* n 1 1g

+ = =

f- *ime perio ( )# l l

* # Ifv g1 .g

= = =

g- *otal energy at the highest point#

; ;h

10 0 P0 mv #mgr.#= + = +

h- *otal energy at the lo6est point

i- # #) ) )1 1

0 0 P0 mv " mv .# #

= + = + =

B- ; ) ; ) ; )0 0 0 "$ 0 0 0 #mgr$ P0 P0 P0 #mgr. = = = = = =

/ase 0++:

11. If the boy rotate 6ith non uniform spee. If u is the velocity at the highestpoint an v is the velocity at the lo6est point then.

a- #u 'gl. = +

b-#

; min

mu* mg * u gl

l = = >

#

max

m*) mg *

l

= + =



34/66

c-#

max#

min

u gl*.

* u gl

+=

-#

max#

min

*u n 1If nthen .

* n

= =

/ase 0+++:

1#. If the boy is revolving 6ith critical speea- (inimum or critical spee of the boy 6hen the string ma8es an angle 6ith

vertical is given by

( )gr % #cos . = +

At the lo6est position ( )"1 gr " = =Q

At the hori/ontal position ( )"1 %gr ?" = =Q

At the highest point ( )"% gr 1J" = =Q

1 # %

1 # %

v v v % 1

0 0 0 % 1.

=

=b- *ension in the string 6hen it ma8es an angle 6ith vertical * = %mg @1 + cos-.

At the lo6est position ( )"1 max* 2mg * " .= = =Q

Fhen the string is in hori/ontal position ( )"# max* %mg * ?" .= = =Q

At the highest position ( )"% min* " * 1J" .= = =Q

max min* * 2mg. =

c- *otal energy at the highest position.

; ; ;

mgr 0 0 P0 #mgr mgr.

# #

= + = + =

*otal energy at the lo6est position

) ) )

0 0 P0 mgr " mgr.

# #= + = + =

) ; ; ) ; )0 0 0 #mgr$ 0 0 0 "$ P0 P0 P0 #mgr. = = = = = =

/ase 0+:

1%. ,imple penulum is given a hori/ontal velocity u at the lo6est position @meanPosition- then

a- Ifu #gr< the boy oscillates about A.

b- If #gr u gr< < the boy leaves the 6ithout completing the circle.c- Ifu gr= the boy completes the circle.

- ;eight at 6hich velocity u = ". is h = u#g.



35/66

e- ;eight at 6hich tension#u rg

* "ish .%g

+=

f- Angle 6ith vertical at 6hich velocity#u

v ".isDos 1 .#gr

= =

g- Angle 6ith vertical at 6hich the tension #* ".isDos #&% u &%gr.= =

h- *ension in the string at an angular isplacement

6ith vertical is#* mu &r mg@# %cos -.=

1'. ,afe spee of a car going on a convex brige to travel in contact 6ith the brige

is v gr.=

1. A ball of mass NmG is allo6e to slie o6n from rest from the top of a incline ofheight NhG. 7or the ball to loop in a loop of raius NrG

a- (inimum height of incline h = r.b- NhG is inepenent of mass of the ball.

12. A ball of mass N(G is suspene vertically by a string of length NlG. A bullet of mass

NmG is fire hori/ontally 6ith a velocity NuG on to the ball stic8s to it. 7or thesystem to complete the vertical circle the minimum value of NuG is given by

( mu gl.

m+

=

1>. If the bob of the simple penulum is given an initial isplacement then

a- Eelocity at the lo6est position ( )#gl 1 cos . =

b- *ension at the lo6est position * mg % #cos .= 1J. A boy is place on the top of a hemispherical bo6l an it is given a hori/ontal

velocity v then

a- *hrust on the bo6l 6hen the raius vector turns through an angle

( )#m

3 mg # %cosr

= +

*he boy looser contact at# #

Dos .%rg %

= +

b- If v = " then 3 = mg L# 4 % cosM.

*he boy looser contact at ( )1Dos #&% or h #r&%. = =

1?. 0quation of motion#

# 1 1

# # 1 ## 1

1

a-6 6 t c- 6 t t#6 6

b-6 6 # - t#

= + = + + =

Also 6 = # n @if n is r.p.s-F= # n&2" @if n is in rpm-.



36/66

#". *orque )# 1l l

I .t t

= = =

#1. For8 one 6 = x .

##. Po6er P = x6.

#%. 0 = K I6#.

#'. For8 one ( )

# #

# 1

16 I 6 6 .

#=

#. If a boy of mass m is revolving in a circular path of raius r 6ith a velocity vthen angular momentum of the boy about origin is

) = mvr = mr# 6 = I6 6here I = mr#

#2. A boy of mass m is moving 6ith a velocity v along a straight line p representeby ax + by + c = " then the angular momentum of the boy about origin

# #

c) m .

a b=

+#>. Angular momentum of a proBectile about point of proBection 6hen it is at its

highest point.# #u sin

) mucos ; mucos .#g

= =

#J.#)

0#I

=

#?. (oment of inertial # #I mr (8 .= =

Parallel axes theorem #I Ig (x .= +

Perpenicular axes theorem x y /I I I .+ =

%". 7or a uniform circular ring.# %

# %

I (R aR +

a area of cross section

+ +ensity.

i.e.I R if m is same$ I + if R is same. I R if + is same

%1. 7or a uniform circular isc.

( )#

# '

#

mI (R R t t thic8ness

t

i.e. I R if m is same.

I R if is same.

%#. 7or a sphere& %

# #& %

mI (R R t

.

#i.e. I R if m is same.



37/66

I R if is same.

%%. Rolling of a boy 6ithout slipping on a hori/ontal surface.a- *he boy possesses both transnational an rotational 8inetic energy.b- *otal energy of the boy

( ) ( )# # # # # # #*1 1 1

0 mv I .0 mv 1 &r .0 1 &r .# # #

= + = + = +

* *ransalational 0 of the boy.c- 7raction of energy associate 6ith translation motion ( )# #1& 1 &r .= +

- 7raction of energy associate 6ith rotational motion ( )# #R&0 1& 1 r &8 .= +

%'. Rolling of a boy o6n an incline 6ithout slipping.a- Acceleration of the boy

( )# #gsin

a inclinationoftheplane.1 &r

=

+b- Eelocity acquire by boy on reaching the bottom

# ##ghE .1 8 &r= +h height of the incline.c- *ime ta8en to reach the bottom

#

#

8#h 1

rt 1&sin .

g

+

=

%. Fhen a boy rolls o6n 6ithout sliing on an incline plane 6ith an inclinationan height NhG.

,.3o. g sin 1"gh&>

#. isc K r# #&% g sin 'gh&%

%. ,oli cyliner K r# #&% g sin 'gh&%

'. ;ollo6 sphere #&% r# %& g sin 2gh&

. Ring r# K g sin gh

2. ;ollo6 cyliner r# K g sin gh



38/66

M(MENT (F +NERT+A (F (ME REB',AR 7(6+E

,.3o. & (r#

2. ;allo6sphere

a- Any iameterb- Any tangent

#&% (r#

&% (r#

>. *hinuniform ro

a- Passing through its centre an normal to lengthb- Passing through one en an normal to its length

(1#&1#(1#&%



39/66

1%. BRA+TAT+(N

M 0 +: epler?s ,a4s:

1. If * is the time perio of plant an R is average istance of the planet from ,un*hen

( )#

# # # %'* R or * RT(

=

#. If v1 is the velocity of the planet 6hen its istance from ,un is an v#is thevelocity 6hen the istance is #then accoring to la6 of conservation of angularmomentum.

# #1 1 # # 1 # #+ v + v + 6 + 6 .= =

M 0 ++: Ne4ton?s ,a4 of raitation:

%. Travitational force bet6een t6o boies of manes m1 an m# separate byistance

is given by. 1 ##Tm m

7 .

=

'. Fhen t6o similar spheres each of raius r are in contact then the gravitation

force bet6een then 7 'r . . Travitational force acting on a boy of mass m place at a istance x istance x

from the centre @x Q R- is given by %T(m

7 xR

=

M 0 +++: Relation 8et4een and B:

2. #T( '

g . g TR R %

= = ensity of earth

M 0 +: ariation of 4ith heiht:

>. Acceleration ue to gravity at a height h above the surface of the earth is given


40/66


41/66

- *ime perio of the satellite%

"

# R R %* # # #

6 g T( ' T

= = = =

M 0 +++: Escape elocit5:

11. 0scape velocity#

#c #Tm J#gR TRR % = = =

c "v # v=

M 0 +: Ener5 of a satellite:

1#.T(m T(m

P0 0r #r

= = @r = raius of the orbit$ r = R + h-.

*otal energyT(m

0 P0 0 .#r

= + =

1%. For8 one in shifting a satellite from orbit of r1to another orbit of raius r#isgiven by

# 11 #

T(m 1 16 0 0

# r r

= =

1'. For8 one by lifting a boy to a height h @h is comparable to R" = change in P0-

h T(m h6 mgR

R h R R h = = + +

1. If a boy is proBecte 6ith a velocity v from the surface of the earth then theheight raise by the boy can be foun by

#e e

1 hm mgR . v comparable to v . If v v $ h

# r h

= = = = +

If v O vethen velocity at infinity can be foun using la6 of conservation ofenergy i.e.

# #1 #

1 T(m 1m m .

# R #

+ =

v1 = Eelocity of proBection.v#= Eelocity at infinity.



42/66

11. .).M

M 0 +: 6isplacement# elocit5 and Acceleration:1. isplacement after t sec y = r sin ( )6t . If particle starts from mean

position the "$y rsin6t. = =

If particle starts from extreme position ?"$y rcos6t. = =

#. Eelocity after t secons E r 6 cos 6t.=

Eelocity 6hen the isplacement is # #y 6 r y . =

At mean position v = r6 @maximum-At extreme position v = " @minimum-

%. Acceleration a = r6#sin 6t

i.e. a = 6#y.at 0.P a = r6#@maximum-.At (.P a = " @minimum-.

'. 7orce 7 = ma = mr6# sin 6t.= m6#y

at (.P = 7= " at 0.P 7 = mr6#

. *ime perioy#

* # .6 a

= =

2. 7requency

1#1 6 a

n .* # y

= = =

( )

( )

>. In y r sin 6t

6t is calle phase

is calle phase constant.

=

M 0 ++: Ener5 of the particles:

( )

( )

( )

# # #

# #

# #max min

J. 0 1m6 r y

1At (.P 0 m6 r maximum

#

At0.P 0 " minimum0 0 1

Average 0 m6 r# '

=

=

=+

= =

?. P0= K m6#y#

At (P P0 = " @minimum-



43/66

Average P0 = U m6# r#At 0P P0 = K m6# r#@minimum-.

1". *otal energy 0 = K m6# r#

# ## # # # #

max max #

1 #m r0 P0 0 m6 r #m n r .

# *

= = = = =

M 0 +++: Time period of a simple pendlm:

11. *ime perio of a simple penuluml* # .g

=

1* l *

g

1# # #

1 1 1 #

g* l *

* l * g = =

1#. 7or a simple penulum in a lift moving up 6ith an acceleration or coming o6n

Fith a ecelerationl

* #

g a

=

+1%. In a lift moving up 6ith eceleration or coming o6n Fith a acceleration time

perio of simple penuluml

* #g a

=

In a freely falling lift effg = ". ,o the penulum oes not oscillate. *he timeperio is infinity.

1'. 7or a penulum of infinite length @or- for a boy roppe in tunnel along the

iameter the time perioR

* # J'.2min.g

= =

1. *ime perio of a secons penulum is #s.

*ime perio remains same if1 #

1 #

l l.

g g=

Its length is #g

l .=

A secons penulum is ta8en from one place to another place then the Dhange in

the length of the seconsG penulum # 1#g g

l .

=

,o that time perio remains constant

12. If the simple penulum is in a cart moving hori/ontally 6ith an acceleration a

then time perio # #1

* #g a

= +

If the cart is moving along a circular path of raius r 6ith a spee v then a = v#&r



44/66

If the cart is sliing o6n an incline plane incline at angle 6ith the

hori/ontall

* # .gcos

=

1>. Fhen t6o simple penulum of lengths l1 an l#are get into vibration in thesame irection at the same instant 6ith same phase.Again they 6ill be in same phase after he shorter penulum has complete noscillations. *o fin the value of n

n *s= @n 1- *1an * 1

( ) 1

s s

n * nor .

n 1 * n 1 l = =

l

l

, = shorter$ l= longer1J. *6o penulum of lengths l1an l#@l1O l#- start vibrating from the mean position in

the same phase. *hey 6ill be again in the same phase at the mean position after

larger penulum completes n oscillations an the shorter one completes @n + 1-oscillations. *hen

( )

1 1 # #

## 1

#

n l n l .

n ln l n 1 l .

n 1 l

=

= + =+

1?. *6o simple penulum of time perio *1 an *# start vibrating from the meanposition in same phase. *he phase ifference bet6een them after t sec. is givenby

t

# 1# 1 # 1

# # 1 16 t 6 t # t .* * * *

= = =

*hey 6ill be in same phase after a time1 #

1 #

**t .

* *=

M 0 +: Time period of a loaded sprin:

#". *ime perio of a loae spring(

#8

= 6here 8 = 7&x.

*a8ing the mass of the spring m into consierationm

(%* # .

8

+ =

*ime perio of the springx

* #g

= 6here x is elongation uner the loa.

#1. If a spring is cut in n equal parts the spring constant of each part is 81= n8.*hen time perio of each part uner the same loa is given by



45/66

1 1** 3 n3.n

= =

##. If a spring of spring constant NG an length N l G is cuts in to t6o springs oflengths N l1G an N l#G then the spring constants of the t6o parts is

( ) ( )1 # 1 #1 #

1 #

an .

+ += =

l l l l

l l

#%. Fhen t6o spring of force constants 1 an # are connecte in series. *hen theeffective force constant is

1 #

1 #

.

=

+

#'. Fhen t6o spring of force constants 81 an 8# are connecte in parallel. *heeffective spring constant is = 81 + 8# .

#. * 1an *#are the time perios of t6o springs uner same loa.If the springs are connecte in series an the same loa is attache.

*ime perio # #1 #* * * .= +

If the springs are in parallel *ime perio1 #

# #1 #

*** .

* *=

+



46/66

12. E,AT+/+T*

Re storing 7orce 7

1. ,tress .Area A= =

#. a- )ongituinal strainl

l

= percentage change in length = longituinal

strain x 1"".b- ,hear strain

)ateral +isplacement bet6een t6o layers l

Perpen+icular +istance bet6een t6o layers l

= = =

c- Eolume strain .

=

,hear strain = # x longituinal strain volume strain = % x longituinal strain.M 0 +: *on?s modls:

%. :oungGs moulus #)ongituinal stress mgl7l

: .)ongituinal strain Ae r e

= = =

7 Applie force$ A Area of cross section) Sriginal length$ e 0longationR Raius of the 6ire.

'. 0longation

#1 1 # #

## # 1 1

7l e1 7 l r :e .

Ay e# 7 l r :

= =

If 7 an : are same

#

# '

l l me .

r m r If youngGs moulus = stress then strain = 1 an final length

= # x initial lengthm mass of the 6ire.

( )

( )

( )

# '1 1 #

# '# # 1

1 1

# #

1 #

# 1

e l rif m is same

e l r

e mif r is same

e me m

if l is samee m

= =

=

=

Fhen t6o 6ires are connecte in series an same force is applie thena- ,tress is same in both 6ires



47/66

b- Ratio of the strains = :#&:1c- *otal elongation e = e1+ e#

Fhen t6o 6ires are connecte in parallel an same force is applie thena- ,train is same in both the 6iresb- Ratios of the stresses = :1&:#c- *otal force 7 = 71+ 7#

. Fhen a rubber cor is suspene vertically from a support it elongates uner itso6n 6eight.

#l gelangation e .

#:=

2.


48/66

@ = ensity of material-1". Fhen a pressure p is applie on a substance its ensity changes. 3e6 ensity

G = 8& @8 p -11. Dompressibility *he reciprocal of bul8 moulus is calle Dompressibility

1D

8=

7or incompressible substances D = " 8 = M 0 +: Poisson?s ration & relation 8et4een *# n and C:

( )lateral contraction strain

1#. Poissions Ratiolongituinal elongation strain

rtransverse strain ! ecrease in raiusr

llongituinal strain ! increase in lengthl

=

=

= =

i- PoissonGs ratio has no unit an no imensionsii- *heoretical limits of 1 to".

iii- Practical limit of " to".

iv- If ". = the substance is perfectly incompressible.1%. Relation among elastic constants : n

( )

( )( )

? 1 %i- ii- y #n 1

y 8 n

%8 #niii- y %8 1 # iv-

# n %8

= + = +

= =

+

1'. Percentage change in volume of a 6ire ue to elongation@!E- = @!1- +#@!r-! change in volume = ! change in length + # @! change in raius-

M 0 : train Ener5

1. For8 one ins stretching 6ire = strain energy store in the 6ire= K x 7orce x elongation = K 7 x e

#

#

1 :Ae :Ae6 ,.0 7

# l l

1 7l 7le# AE lA

= = =

= =

Q

Q

12. If t6o material of same material are stretche by same amount i.e. e is same



49/66

( )

# '

#

' #1 1 1 #

' ## # # 1

1 # 1 1

# 1 # #

A r r m6

l l m l

6 r 6 lIf m is same

6 r 6 l

6 m 6 mif r is same if 1 is same.

6 m 6 m

= =

= =

1>. If t6o 6ires of same material are stretche by same force then#

# '

l l l m6

A r m r

1J. 0lastic strain energy = K stress x strain x volume,train energy per unit volume = K stress x strain

( )( )

##stress :

strain .#: #

= =

1?. If l1an l#are the lengths of a 6ire uner tensions *1 an *# the actual length

of the 6ire1 # # 1

# 1

l* l *

.* *

= *hermal stress evelope in a boy ue to change in temperature : t=



50/66

13. 'RFA/E TEN+(N

M 0 +: Force de to .+:

1. ,urface tension 7orce 7* .)ength l= =

# 1

6or8 one 6increase in Area A A

= =

( )# 16 * A A =

#. a- 7orce in aition to 6eight require to lift a thin 6ire of length l from liquisurface is 7 = #l x * = #*l.b- 7orce require to lift a thin8 circular plates of raius R from liqui surface is7 # Rx* # R*.= =

c- 7orce in aition to 6eight require to pull a circular ring from 6ater surfaceis ( )1 #7 # * r r= +

r1an r#are internal an external raius.If it is a thin ring then7 ' r* = .- 7orce require to pull a thin8 rectangular plate of length 1 an breath b is

from liqui surface is ( )7 #1x* #bx* #* 1 b .= + = +

e- 7orce require to pull a thin8 rectangular frame from liqui surface is

( ) ( )7 # x #* 1 b '* 1 b .= + = +

f- 7orce require to pull a capillary tube from 6ater surface is 7 ' R*.=

%. If a small rop of 6ater is squee/e bet6een t6o plates the force require toseparate the plates

#*7 A.

=

A= Area of 6ater la6yer. = *hic8ness of 6ater layer.'. A metallic 6ire of ensity floats hori/ontal in 6ater. *he maximum raius of

the 6ire so that the 6ire may not sin8 6ill be @surface tension of 6ater = *-.

#*a-

g

. A metallic 6ire of ensity is laying hori/ontal on the surface of 6ater. *hemaximum length of 6ire so that it may not sin8 6ill be@surface tension of 6ater = *-



51/66

# *a-

g

M 0 ++: rface ener5:

2. ,urface tension6or8 one 6

* 6 * Aincrease in area A

= = = =

*his 6or8 6ill be store as surface energy0 *. A= .>. a- For8 one in forming a liqui rop of raius #R ' R *.= b- For8 one in forming an air bubble of raius r is #6 J r *.= c- For8 one in increasing the raius of a liqui rop from R1to R#

( ) ( )# ## 1 # 16 * A A ' * R R= =

- For8 one in forming of a soap film of si/e l x b is = #* lbe- For8 one in increasing the raius of a soap bubble from R1to R#

( ) ( )# ## 1 # 16 * A A J * R R= =

J. Fhen a big rop of raius R is split in to n ientical rops each of raius r.

a- R = n1&% r.b- Increase in surface area ( ) ( )# # # 1 %A n ' r ' r ' R n 1 .= = =

c- For8 one in splitting the rop ( )# 1&%' R * n 1 .=

- If this process ta8es place uner aiabatic conitions the temperature of V

the rop ecreases by%* 1 1

t .s r R

= ?. Fhen n ientical small rops each of raius r combine to from a big rop of

raius R then

a- R = n1&%

r.b- ecrease in surface area ( ) ( )# # # 1&%n' r ' R ' R n 1 .= =

c- 0nergy release in this process ( )# 1&%' R * n 1 .=

- If this process ta8es place uner aiabatic conitions then the temperature

of the rop increases by%* 1 1

t .s r R

= ( )# 1&%L' R * n 1 ms *M =

e- If the energy converts into 0 then ( )# 1&% #' R * n 1 1mv . =

2* 1 1E .

r R

=

1". *6o rops of raii r1an r#combine in vacuum uner isothermal conitions. *hen

if r is the raius of the rop forme % % %1 #r r r .= +



52/66


53/66

If P"is outsie pressure total pressure insie =P"+ excess pressure

1. If t6o bubbles of raii r1an r# are in contact the raius of curvature of the

interface is1 #

# 1

rrr .r r

=

12. An air bubble of raius r is at a epth h is 6ater.#*

Pr

=

Insie pressure i a#*

P P hg .r

= + +

1>.

1&% 1

1 # # #

1 1 1 1

P r E A.

P r E A

= = =

1J. A bubble of raius r1 is insie another bubble of raius r#. *he raius of a singlebubble 6hose excess pressure is equal to ifference in pressure bet6een insieof inner bubble an out sie the outer bubbles is

1 # 1 #

1 1 1 '* '* '* R r r R r r

= + = +

1?. A long capillary tube of raius r is fille 6ith 6ater an place vertically. *he

height of 6ater column remaining the capillary tube'*

.rg

=



54/66

M 0 +: iscos force:

1. Eiscous force 7 A Eelocity graient.x x

=

7orce require to rag a plate place on an oil layer of thic8ness t 6ith a

velocity v is given by 7 A

t

=

Area of the plate.

M 0 ++: Poiseille?s e;ation:

2. Eolume of liqui flo6ing through a capillary tube in unit time is given by

( ) '' '1 1 1 # #'

# # # 1 1

hg rPr E h r lE .

J l J l E h r l

= = =

P Pressure ifference across the tube. P = hg.r Raius of the capillary$ 1 )ength of the tube.

'rE

l

@if P an are same-.

Also 'P J l

E 6here RR r

= =

it is 8no6n as flui resistance.

%. Fhen t6o capillary tubes are connecte in series then

1 #

1 #

E E E$

P P P

= =

= +

1P Pressure ifference across the first tube.#P Pressure ifference across the secon tube.

'

1 1 #

# # 1

P l r.

P l r

=

If the t6o tubes are replace by a single tube of raius r then



55/66

1 #' '

1 #'

1 # # ' '1 #'

l l lr r r

1 1 1if 1 1 1 then

r r r

= +

= = = +

'. Fhen t6o capillary tubes are connecte in parallel

1 #.

1 #

P P

E E E

=

= +

'1 1 #

'# # 1

E r l.

v r l

=

If the t6o tubes are replace by a single tube of raius r

'rl

=' '

1 #

1 #

r r.

l l

@If r1= r#= r-

*hen1 #

1 1 1

l l l

= +

M 0 +++: toCes formlas:

. Fhen a spherical boy of raius r is moving through a flui 6ith a velocity Ethen the viscous force acting on boy is given by 7 2 rv.=

1&% 1&%7 r E m

After some time the boy attains constant velocity calle terminalvelocity in this case.

( )*

# #&%

*

mg 2 rv if buoyancy is neglecte

m

E r E .r

=

If n ientical rops moving 6ith terminal velocity v. Dombine to form a signal mpthen terminal velocity of the big rop E = n#&% v

M 0 +: /ontinit5 e;ation:

2. #1 1 # # 1 1 # #1

E a v a v r v r v .A

= =

Eolume of liqui flo6ing through a tube in unit time @or- volume flux q = av.

M 0 : 7o5ant force:>. If a boy of volume E ensity sis immerse in a liqui of ensity 1then

a-


56/66

1 1 s 1 1 sE g E E E . = =

c- Apparent 6eight of the boy 61= mg 4 7b.

( )

( )

1s 1

s 1 1 s 1

1 s

i- If+ + $ 6 ".

ii-If+ + $6 vg + +

mg 1 + & + .

=

> =

=

3ote Fith increase in temperature tbecreases hence 61

increases.M 0 +: 7ernolli?s theorem:

J.# #

1 1 # #1 #

P E P Egh gh

# #+ + = + + @all are in ,I units-

#P 3&m also P hg. =

If the liqui is flo6ing hori/ontally then# #

1 1 # #

1

P E P E # #

+ = +

Dhange in 0 per ay unit mass# #

# 1 1 #E E P P .#

= =

?. If P1an P#are pressures at the top an bottom of the 6ings of the aero plane

then the ynamic lift on the aero plane ( )# 17 P P A$A Area of the 6ings.= +# #

# 1# 1

E EP P

#

=

Fhere E1 an E#are the velocities at the top an bottom of

the 6ings an is ensity.

M 0 : Torricelli?s theorem:

1". A vessel fille 6ith a liqui up to a height ; has a small hole near the bottom. Ifh is epth of the hole then.

a- Eelocity of efflux from the orifice is v #gh=

b- Eolume efflux H = aE

c- *ime ta8en by the 6ater to touch the base level t( )# ; h

tg

=

- ;ori/ontal range ( )R # h ; h=

e- R is maximum if h = ;Rmax= ;

f- *ime ta8en for emptying the tan8A #;

t ha g

=

g- *ime after 6hich the level falls from h1to h#is ( )1 #A #t h ha g=



57/66

1!. EPAN+(N (F (,+6

1. )inear expansion 1 1 t. =

( ) ( )# 1

# 1# 1

l l

1 1 1 tl1 t t

= = +

If same amount of heat given thenH l H

t then lms ms

= =

( )

( ) ( )

( ) ( )

#

# 1# 1

1 # 1

# 1# 1

1 # 1

l 1l r raius of the ro .

m rAral expansion A A t.

A AA A 1 t

A t t

Eolume expansion E E t.

E E 1 tt t

=

= = +

=

= = +

! change in length tx1""=

! change in area # tx1""=

! change in Eolume % tx1""=

#. )oss or gain of a penulum cloc8 ue to a change in temperature per ay is

( )1

*# *1 J2'""sec.#

%. If t6o ros of ifferent material have equal lengths of 1 an at temperature *1as the temperature is increase to *# ifferent in their final length is

( )( )# 1 # 11 1 * * . =

'. 7or an anisotropic substance if x y / an are the coefficients of linear

expansion in three perpenicular irections then x y / . = + +



58/66

. Fhen a bimetallic strip is heate it bens. *hen the raius of curvature of thebimetallic strip

( ) ( )# # 11

R .

* *=

Fhere is the thic8ness of each strip.

( )# 1* * is change in temperature

1 # are coefficients of linear expansion of the t6o metals.

2. " " " "

in per 7 in per in per 7 in per D.? '

= =

>. *hermal stress evelope in a ro heate 6hen it is prevente from expansion

is calle thermal stress ( )# 1: * * .=

7orce evelope ( )# 1:A * * .=

J. A 6ire is bent in the form of a ring 6ith a small gap of length "1 1x at t D. on

heating to "#t D if the gap increase to x#in length then the coefficient of linear

expansion of the 6ire material

( )# 1

1 # 1

x x

x t t

=

?. If t6o ros of ifferent materials have the ifference bet6een their lengths at

all temperatures then 1 1 # # 1 # # 11 1 or 1 &1 & = =

# 11 #

1 # 1 #

x xl an l

= =

: :

1". *he change in moment of inertia of a boy ue to rise in temperatureI #I t =

11. (etal tapesa- (etal tape sho6s correct reaing only at a temperature at 6hich it isconstructe.

b- A metal tape is grauate at "1t D an is use at"

#t D.

i- # 1If t t istance bet6een the ivisions ecreases then

a- Sbserve reaing ) is less than actual value

b- Dorrection to be applie ( )# 1D ) t t =

c- Dorrect reaing ) D.= + ii- It t# Q t1 istance bet6een the ivisions ecreases then

a- Sbserve reaing ) is greater than actual valueb- D ) t =

c- Dorrect reaing ) D.= 1#. Eariation of ensity 6ith temperature

( )( ) ( )" t t " 1 or 1 t= + =



59/66

1%. A mercury barometer has a metal scale 6hich is calibrate at ""D the pressureas rea by it at t"D is ;t. *hen the pressure at "

"D 6ill be given by

" R; ;t 1 @ -t= R is coefficient of real expansion of mercury

is coefficient of linear expansion of metal

1. EPAN+(N (F ,+-'+6

1. R a a R a% relation bet6een an . = + = +

#. 7rom specific gravity bottle experiment

( )

( )( ) ( )

( )

a

# % 1 #a a

% 1 # 1 # # 1

mass of liqui expelle

mass of the liqui remaine change in temperature

6 6 m mor

6 6 * * m t t

=

= =

m1is nearly equal to m#an hence it can be 6ritten ( )1 #

R# # 1

m m.

m t t

=

%. If 1an #are the ensities at temperature t1an t#

( )

( )

( )

1 # R # 1

" t R

"t

1 #R

1 # # 1

1 * *

1

1 t

t t

= +

= +

=

=

'. *he unoccupie volume in a container remains same if the volume expansion ofthe container an that of liqui are same. *he conition for this is 1 1 s sE E . =

. *he temperature at 6hich the ensity of liqui is x! less than at

( )"

R

x" D is .1"" x

2. If a liqui column of height h1at temperature t1balances another liqui columnof height h# at temperature t#.



60/66

( )# 1

a1 # 1

h h.

h t t

=

1". EPAN+(N (F BAE

M 0 +: olme and pressre coefficient of a as.

1. Eolume coefficient ( )"

""

E E E E 1 tE t = = +

# 1

1 # # 1

E EEt E t

=

@at constant pressure-.

#. Pressure coefficient ( )"

" t"

P PP P 1

P t

= = +

# 1

1 # # 1

P P

Pt P t

==

@ at constant valume-.

%. ( )"1>% ".""%22& c. for any gas = = =

M 0 ++: 7o5le?s la4:

'. A* constant temperature ( ) 1 1 # #1

P . or PE D PE PEE

= =

. *6o vessels of volume E1 an E# fille 6ith a gas at pressure P1 an P# are

connecte the common pressure( )1 1 # #

1 #

PE PEP .

E E+

=+

2. If the pressure of the gas is increase by x! then the ! ecrease in volume at

constant temperature is

1""x

.1"" x+>. If 11an 1#are the length of the air column 6ith open en up6ars an

o6n6ars respectively then ( ) ( ) # 11 ## 1

l l; h 1 ; h 1 . ; h

l l

++ = =



61/66

; atm. Pressure.h length of mercury threa.

J. Fhen an air bubble at the epth h in a la8e rises to the top its volumeincreases. Assuming the temperature to be constant

If the volume becomes n timesh = ; @n 4 1- @; + h- E1= ;E#

An if raius becomes n times( ) ( )% % %1 #h ; n 1 ; h r ;r= + =

Fhere ; = 1 atm 1" m epth of 6ater = >2 cm of ;g.

If temperature is constant then( ) 1 #

1 #

; h E ;E* *

+=

M 0 +++: /harle?s la4:

?. At constant volume P * @for a given man of a gas-

( ) 1 #

1 #

P P Por constant .

* * *

= =

1". At constant pressure E * @for a given man of a gas-

( ) 1 #

1 #

E E Eor constant .

* * *= =

M 0 +: +deal as e;ation:

11. If one gram of gas is consiere.PE = r* @r = gas constant-.If one mole of gas is consiere

1 1 # #

1 #

PE PE RPE R* . r

* * (

= = = for n moles PE = nR*.

1#. ifference forms of ieal gas equation.a- PE = nR*b- PE = @m&(-R*c- P = @R*-&(m 4 mass of the gas 4 ensity of the gas ( 4 molecular 6eight.7or a given gas.

1 #

1 #

P Pm m

= @if E an * are constant-.

1 #

1 1 # #

P Pm* m * = @if E is constant-.

1 #

1 1 # #

P P

+* + * =

If * is constant1 #

1 #

P P.

=

If P is constant 1 1 # #* * .=



62/66

1%. ! change in * = ! change P + ! change in E.1'. Fhen n1moles of a non reacting gas in the state @P1E1*1- is mixe 6ith n#moles

of gas in state @P#E#*#- if the resultant mixture is in the state @P E *-.

1 1 # #

1 #

PE PE PE.

* * *+ =

1. If t6o ifferent gases of same mass occupy same volume at same pressure then

(1&*1= (#&*#@(1 (# are molecular 6eights-

1D. T)ERM(6*NAM+/

M 0 +: Hole?s la4:

1. *he 6or8 one on a system @6- is irectly proportional to heat prouce in thesystem.

6 H or 6 WH$ W '.# B&cal W 1 in ,I system = = =

#. A boy of mass falls from height h an if it is potential energy converts intoheat then

( )gh

mgh ms t t ,. shoul+ be in ,.I unitss

= =

If x ! of energy is converte into heat then

x mgh ms t.1""

=

%. An ice bloc8 of mass ( falls from a height h an if its energy converts into heatthen

(gh(gh m) m

)) ,I Cnits.

m mass of the ice melte+

=

*he ice completely melts if h = %%.2 8m.

'. Fhen a boy of mass m moving 6ith velocity v is stoppe an all its 0 convertsinto heat energy then

( )#

#1 m ms t t , ,.I units .# #s

= =

. If a boy of mass m is ragge on a rough surface through a istance x then



63/66

gxmg ms t t

,

= =

If it is an ice bloc8 then1mgx m ). =

(ass of the ice melte 1mgx

m .)

=

M 0 ++: First la4 of thermod5namics:2. H = u + 6.

H + ve if heat is supplie to the system. ve if heat is absorbe from the system.

u + ve if temperature increases. ve if temperature ecreases.( )# 16 P v v= = + ve if 6or8 is one by the gas.

ve 6or8 is one on the gas.

M 0 +++: vpD D and relation 8et4een them:

>. A gas at constant

,pecific heat at constant pressure p1 +

cm +t

= @at constant pressure-.

(olar ,pecific heat at constant pressure p1

Dn t

=

,pecific heat at constant pressure1 +

cm +t

=

(olar ,pecific heat at constant pressure1

D .n t

=

p p v vD (c $ D (c .= =J. p v p vD D R c c R&(. = =

?. *he follo6ing table sho6 the values of p vD 9D .

T5pe of the as vD pD

(onoatomic %R#

R

#

1.2>%

=

iatomic R#

>R

#>

1.'

=

Polyatomic@*rihigher-

2R#

JR#

' 1.%%%

=

M 0 +: /onstant olme process: =+sochoric@



64/66

b-1 #

1 #

P P* *

=

c- 6 = ".- H = C = n Dv*.

M 0 : /onstant pressre process: =+so8aric@

1". a-1 #

1 #

E E* *

=

b- u = nDv *.

c- 6 = ( )# 1P E E

- H = nDp *.11. *he fraction of heart energy supplie 6hich is store as internal energy

u 1.

H=

1#. *he fraction of heat energy supplie 6hich is use in oing external 6or8

6 11 .H

=

M 0 +: +sothermal process:

1%. a- 1 1 # #PE PE .=

b- u = * = ".

# 1

1 #

#

1

#

1

v Pv P

e e

vv

evv

1"

c- 6 nR* log nR* log

PE log

#.%"% nR* log .

= = =

=

=

- ( ) ( )Dp Dv R$ Dp&Dv r$ Dv R& 1 $ Dp R 1 = = = =

e- ,lope of isothermal curve = P&E.f- Isothermal bul8 moules = .P.

M 0 ++: Adia8atic process:



65/66

( )

1 1 # #

1 1 11 1 # #

1 1 11 1 # #

p v

1 #1 1 # #

1'. a- PE cons tan t PE P E

b-*E Dons tant *E * E

c-* P Donstant * P * P

+-+H " D D "

nR * * PE PEe-+6 +C .1 1

f-A+iabatic bul8 mo+ulus P.

,lope of a+iabatic curve P&E.

= =

= =

= =

= = =

= = =

=

=

1. 7or a mixture of t6o gases1 1 # # 1 1 # #

1 # 1 #

n Dp n Dp n D n DDp D

n n n n

+ + = =

+ +Dp&Dv =

12. ;eat capacity or thermal capacity or 6ater equivalent @ms- = H& t

19. T)ERMA, RA6+AT+(N

1. AbsorptivitytHHa Hra Reflectivity r *ransimitivity t

H H H= = =

a + r + t = 1

#. 0nergy emitte by a bloc8 boy from unit surface area in unit time is given by'0 * .= 0 is also calle emissive po6er.

7or the boies '0 e * 6here 0 emissivity.=

%. *otal energy emitte by a boy in a time t is given by'0 e A* t.=

7or a sphere ( )# '0 e ' R * t.=

i.e. # '0 R * .

'. *he net heat lost or gaine by a boy at temperature * in unit time

( )'s'0 e A * * .=

. 3e6tonGs la6 of cooling ( )< s+H

* * .+t


66/66

s* *emperature of the surrounings.2. If the temperature of a boy ecrease from *1to *#in a time t then accoring

to 3e6tonGs la6 of cooling.

1 # 1 #s

* * * * * .

t #

+ = ". If m is the 6avelength corresponing to maximum monochromatic emissive

po6er of a boy at temperature * the ( )m m1 or * b*

= @b is a constant-.

b = #.?% x 1"%m8.J. 7or a spherical boy at temperature *

c- Rate of heat loss ' #H H

e A* R .t t

= @If * is same-.

- Rate of cooling is given by'

'ms* * e A* * 1e A*t t ms t R

= =

@If * is same-

Formula in Physics1

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