eamcet formula in physics1

Mohammed Asif

Name :

Roll No. :

Topic : Ph : 9391326657, 64606657

2. MEASUREMENT OF PHYSIAL QUANTITIES

1. If a physical quantity P. is given by P = x + y

i.e., % error in P = m. % error in x + n. % error in y.4.

Significant figurei) all non-zero digits are Significant Ex. 23.56 No. of s. f – 4ii) Zero in between two nonzero digits are Significant Ex. 230405 No. of s. f – 6iii) All zeros to the left of the first nonzero are not Significant Ex. 002356, 0.002356 No. of s. f – 4iv) All zeros to the right of the last nonzero digit in a measured value are SignificantEx. 235600 m No. of s. f – 6

Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-00065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 1

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Mean deviation

Standard deviation

Probable error = r = 0.6745

3. VECTORS

M – I: Magnitude and direction of a vector:



(b) Unit vector in the direction of

M – II: Triangle law of vectors: 2. A simple pendulum of weight w and length 1 is pulled aside by a horizontal

force F through a distance r so that the string makes an angle Ө with vertical. Then

b) Lami’s theorem: If three coplanes forces acting on a particle as shown in the fig keep it equilibrium then



M – III: Parallelogram law of vectors:

M – IV: Resolution of vectors:If a vector makes an angle with horizontal then its horizontal

component Px = Pcos ; Vertical component = Py = P sin

M – V: Polygon law of vectors:

n equal force each of magnitude F are acting simultaneously on a particle. Each force makes an angle Ө with one then



M – VI: Dot product:

M – VII: Cross product:



M – VIII: Application of triangle law:Boat and river:If the man wants to reach the exactly opposite point on the other bank (or) to cross the river along the shortest path, he has to swim along AC.

2. If the man wants to cross the river in shortest time.



a) Direction of motion of boat → Along the normal (or) 900 with direction of flow.

Rain and umbrella:Rain is falling vertically down wards with a velocity Vr and a man is moving horizontally with a velocity Vm (or wind is flowing with a velocity Vm). To protect himself from rain the man has to hold his umbrella at an angle Ө with vertical

Change in velocity:are the initial and final velocities of a body and Ө is the angle between

Relative velocityare the velocities if two bodies A and B respectively then



4. KINEMATICS



M – I: Freely falling body:1. Equations of motion:

M – II: Vertically projected body:1. Equations of motion:

(a) v = u – gt (b) h = ut - ½ gt2 (c) V2 = u2 – 2gh d) Sn =

2. Time of ascent.

5. A body thrown vertically upward with a velocity U crosses a point p in its path after

t1 sec while going up and after t2 sec while coming down, then

M – III: Body projected vertically upwards from the top of a tower:1. Equations of motion:(a) v = -u + gt (b) H = -ut + ½ gt2 (c) V2 = u2 + 2gH (d) Sn =

2. A balloon is moving upwards with uniform velocity U. A body is dropped from it

when it is it at a height H above the ground then,a) Velocity of the body after t sec V = -U + gtb) Time taken by the body to reach ground cane be found from H = -ut + ½

gt2

c) Separation between body and balloon after t sec = ½ gt2

d) Height reached by the body above ground =

3. A body thrown vertically upwards with a velocity u from the top of a tower reaches the ground in t1 sec. Another body thrown vertically downwards with

same



velocity from same tower reaches the ground in t2 sec. Then(a) Height of the tower H = ½ gt1 t2

(b) Time taken by a body dropped freely from same tower to reach ground is

M – IV: Body thrown vertically downwards with initial velocity u: 1. Equations of motion:

(a) V = u + gt (b) H = ut + ½ gt2 (c) V2 = u2 + 2gH (d) Sn =

M – V: Oblique projectile:

Velocity of the projectile after t sec.: Vx = u cos Ө [Horizontal component] Vy = u sin Ө - gt. [Vertical component]

Magnitude of Resultant velocity

5. Displacement after t seconds.Horizontal displacement X = u cos Ө. T

Vertical displacement Y = u sin Ө t - ½ gt2

Resultant displacement =

6. Range is max. if Ө = 450 and Rmax =

7. Range is same for angle of projection Ө and (90 - Ө).

8.



9. Two bodies are projected with same velocity u so that their horizontal ranges

are same (i.e., Angles of projection are complementary).If H1 and H2 are the max. Heights reached by them and T1 and T2 are

their time of flights and R is range. Then

10. Path of the projectile is represented by Y = Ax – Bx2 where A = TanӨ,

11. Initial energy of the projectile E = ½ mu2

At maximum height K.E = E cos2 Ө, PE = E Sin2Ө. 12. Minimum velocity of a projectile = u cos Ө (at max height)

M – VI: Horizontal Projectile: If a body is thrown horizontally with a velocity U from a tower of height H then

1. Velocity after t sec Vx = u,

2. Displacement after t sec. Horizontal displacement x = ut, Vertical displacement y = ½ gt2.

3. Time of descent

5. Velocity just before reaching ground

6. The angle at which it hits the ground tan

Resultant displacement

7. The time after which the velocity is perpendicular to initial direction t = u/g sinӨ

8. Two bodies are projected horizontally in opposite directions with a velocities V1

and V2 from a tower



i) The time after which the displacement vectors are perpendicular

ii) The time after which the velocity vectors are perpendicular

5. DYNAMIC (NEWTON’S LAWS OF MOTION)



System like a rocket. 2. a) When a lift moves upwards with acceleration a (or) downwards with a

deceleration a then apparent weight of a body in the lift w = m (g + a). b) When a lift moves upwards with deceleration a (or) downwards with an

acceleration a, then apparent weight of a body in the lift w = m (g - a). c) When the lift is stationary or moving with uniform velocity the w1 = mg.

d) Percentage change in weight

e) If h is the depth of water in a bucket placed in the lift then pressure at the bottom of the bucket is P = hd (g a).

3. When a machine gun fires ‘n’ bullets in a time t then the force required to hold the gun is

.

4. a) When a metallic plate of mass m is held in mid air by firing ‘n’ shots/second with a

velocity u, if the bullets stop dead after striking the plate, then mnu = mg. b) If the shot comes back with a velocity V then nm (v+u) = mg.5. If water from a narrow pipe hits a wall horizontally with a velocity v and stops dead,

force acting on the wall

(volume flow rate).A = area of cross section of the piped = density of water.If water bounces back with same speed F = 2 Adv2

6. If a ball hits a wall normally with a velocity v and rebounds with same velocity change

in momentum = 2 mv.

Force acting on the wall where ‘t’ is the time of collision.

7. a) A body of mass m, moving with a velocity v hits wall and rebounds with same

velocity, then change in momentum perpendicular to the wall force

on the wall

b) Change in momentum parallel to wall = 0.8. Extra force required to keep the conveyor belt moving with a velocity V when



Sand (any mass) is falling on it the rate of

9. If hot gases are coming out of the rocket at the rate of with a velocity V

then

a) Thrust on the rocket

b) If the rocket moves up with uniform velocity then F = MgM → mass of the rocket.

c) If the rocket moves up with uniform acceleration a then F = M(g + a).If gravity is neglected then F = Ma

d) Acceleration of the rocket after t sec a

LAW OF CONSERVATION OF LINEAR MOMENTUM AND COLLISIONS

M – I: Law of conservation of momentum:A body of mass m1 moving with a velocity u1 collides with another body of mass m2 moving with a velocity u2, if v1 and v2 are their final velocities then According to law of conservation of momentumm1 u1 + m2 u2 = m1 v1 + m2 v2

If the two bodies stick together after collision then common velocity

1. If bodies are moving in mutually perpendicular directions before collision then

common Velocity

2. In case of obliquic collision i.e., after collision, if the direction of motion of bodies

makes an angle with initial directiona) = (

b) 3. If a bullet of m come out with a velocity from a gun of mass M the Recoil velocity

of gun

Velocity of bullet relative to gun = v + u.



4. If a stationary shell breaks into two fragments, they will move in opposite directions, with velocities in the inverse ratio of their masses.

5. In the above case, the Kinetic energy of the two fragments is inversely proportional

to their masses.

6. The total energy released in the explosion

7. If a shell at rest explodes into three fragments having masses in the ratio m1 : m2 : M3 if first two fragments travel in mutually perpendicular direction with velocities v1 and v2 then the velocity of the third fragment is given by

8. If a shell of mass m moving with a velocity explodes into two fragments having

masses m1 and m2 and if are the velocities of the fragments then

M – II: Elastic collision:1. For one dimensional elastic collision v2 – v1 = u1 – u2.

Special casesa) When m1 = m2; v2 = u1, v1 = u2.b) If u2 = 0, (i.e. second body is at rest)then

If m1 = m2 v2 = u1, v1 = 0If m2 >> m1 v1 = -u1, v2 = 0If m1 >> m2 v1 = u1, v2 = 2 u1.

2. In case of elastic collision, if the second body is at rest The fraction of K.E transferred or lost by the first body

Where n = m1 / m2

3. Fraction of K.E retained by the first body is



Where n = m1 / m2

If m1 = m2 fraction of KE transferred is maximum and that is 100 %.M – III: In elastic collision:1. a) =

b)

Final velocity can be found by using above two equations

2. If u2 = 0 i.e., if the second body is at rest ratio of final velocities

4. A block of mass M is suspended freely. A bullet of mass m is fired in to the block with a velocity . If the bullet gets embedded in to the block then

a) Velocity of block after collision

b) Height rised by the block

c) Maximum angle made by the block with vertical cos

d) Loss of K.E

M – IV: Coefficient of Restitution:

1. Coefficient of restitution

a) When a body falls from a height h, and rises to a height h1 after first collision

with the ground then

b) Height rised after n bounces hn = e2n . h

c) Velocity after n bounces vn = en. V = en



2. If a body falls from a height h the total distance traveled before coming to rest.

3. If a body falls from a height h, the total time taken to come to rest.

4. If a body hits the ground with a velocity v1 and rebounds with a velocity v2 then

5. Percentage loss of velocity after n bounces = (1 - en) X 100.6. Percentage loss of momentum after n bounces = (1 - en) X 100.

7. Percentage loss of KE after n bounces = (1 – e2n) X 100.8. A body hits ground with a velocity u making an angle with vertical and rebounces

with a velocity v at an angle then

If it is a smooth surface then

6. WORK – POWER - ENERGY



M – I: Work:1. a) If a force ‘F’ acts on a body making an angle with horizontal and

displaces the body through a distance ‘S’ work done,

(If F and S are same).

(If F and t are same).2. Work done against gravity W = mgh.

3. Work done in imparting velocity to a body

4. Work done against friction on a level surface.

5. Work done in compressing or elongating a spring

K – spring constant x – elongation6. Work done in moving a body up on to a smooth inclined plane is W = (mg sin

) S.7. Work done to compress or expand a gas at constant pressure.8. a) Work done in pulling the bob of a simple pendulum aside through an angle

from the vertical is

m – mass of the bob l- length of pendulum b) Work done in displacing the pendulum from an angle to an angle is

9. Work done in pulling aside a bar through an angle from the vertical is

m – mass of the rod l- length of the rod10. The work done in lifting a body of mass ‘m’ and density ‘ds’ in a liquid of density ‘d1’ through a height ‘h’ under gravity is

11. Work done by external resultant force on a body is equal to change in KE of the body



12. Work done in lifting water from a well from a well of depth h is

m→ Mass of water m0→ Mass of rope13. A rod of mass m and length l lies horizontally on a ‘floor’. The work done in rotating

the rod through an angle about one edge in a vertical plane is given W = ½ mgl sin .

14. In the above point, the work done in bringing the rod to vertical position is W = ½ mgl.15. A uniform chain of length l and mass m is placed on a friction less table such that

1/nth of its length hanging over the edge. The work done in pulling the hanging part

on to the table is

M – II: Power:1. Average power P = W/t.2. Instantaneous power = F x V = 3. The power of machine gun firing ‘n’ bullets, each of mass ‘m’ in one second with

velocity ‘v’ is

4. A motor sends a liquid with a velocity ‘V’ in a tube of cross section ‘A’ and ‘d’ is the

density of the liquid, then the power of the motor is

5. A motor lifts m kg of water to a height h in t sec. Then the power of the motor is

.

If is the efficiency of the motor then

If the water comes at of the pipe with a velocity then

6. Extra power required to keep a conveyor belt moving with constant velocity if

gravel is falling the belt at a rate of dm/dt is

7. A vehicle moves with a constant velocity V on a rough horizontal road. The frictional



force acting on the vehicle is f then the power o f the engine P= f x v.

M – III: Energy:1. KE = ½ mv2.2. PE = mgh.3. PE of a liquid in a capillary tube PE = (d) (ah) g h/2 = dagh2 / 2.

d – density of liquida – area of cross section of the capillary tube.h – height of the liquid in the capillary tube.

4. PE of a block placed on the ground =

5. P.E in a spring = ½ kx2 .M – IV: Law of conservation of energy:1. A body is dropped freely from a height h, during collision with ground it looses x %

of energy and then rises to a height ‘h’ then

(100-x) remaining energy;

2. In the above question if the body is thrown vertically downwards with a velocity V

then

3. A body of mass m is dropped from a height h on to a spring of spring constant K.

If x is the compression produced in the spring then

If x < < < h mgh = ½ kx2

4. A body of mass m moving with a velocity V collides with a spring of spring constant

K and comes to rest. If x is the compression produced in the spring then½ mv2 = ½ kx2

M – V: Work – Energy theorem:1. Work done = Change in KE;

F x S = ½ m (V2 - u2 ).2. A body of mass m is dropped from a height h it reaches the ground with a velocity

work done by air resistance W = mgh – ½ mv2 .

3. A knife edge of mass m is dropped from a height h on to a table. It penetrates a



distance x into the table. If the resistance offered by the table is F thenF x x = mg (h+x). if x < < < h then Fx = mgh

M – VI: Relation between KE and Momentum:

3. Two bodies of different mass are moving with same momentum if same break

force is applied on them. Then

Lighter body will travel more distance between coming target.b) In pulse = change in momentum

F x t = P – 0 Sin F and P are same t willF x t = P t1 : t2 = 1 : 1.So both take same time to come to rest.

4. Two bodies of different mass are moving with same KE. If same break force is

applied on them then

So both travel same distance before coming to rest.b) F x t = Pi – Pf

Heavier body takes more time to come to rest.



7. CENTRE OF MASS

M – I: Coordinates of center of mass:1. Particles distributed in space. If (x1y2z1) - (x2y2z2) – are the position coordinates

of particles of masses m1, m2 – the position coordinates of their center of mass

are

2. In vector notation. If r1, r2, r3….. Are the position vectors of particles of masses m1, m2, m3…… Then the position vector of their center of mass is

M – II: Velocity Center of Mass:3. a) Velocity of center of mass

If are the velocities of particle of masses.m1, m2, m3…… mn the velocity of their center of mass is

i.e., total momentum of the system is the product of mass

of the whole system and the velocity of the center of mass.c) If V1 and V2 are the magnitudes of velocities of two particles of masses

m1 and m2 then the magnitude of velocity of c.m.

+ ve for same direction

- ve for opposite direction.



d) If two particles of masses m1 and m2 are moving with velocities V1 and V2 at

right angles to each other, then the velocity of their center of mass is given

by

M – III: Acceleration of center of mass:4. a) if a1, a2, a3 ….an are the acceleration of particles of masses m1, m2, m3…… mn

Then the acceleration of their centre of mass is

If a1 and a2 are the magnitude of acceleration of two particles of masses m1 and m2 then the magnitude of acceleration of their centre of mass.

(If they are moving at right angles to each other)

M – IV: Two particle system:5. Position of centre of mass:

Two particles of masses m1 and m2 are separated by a distance ‘d’. If x1

and x2 are the distance of their center of mass from m1 and m2 thenm1 x1 = m2 x2

If m1 is moved through a distance towards the c.m and m2 is also moved

through a distance towards the c.m the position of c. m remains same if

m1 = m2

6. Out of a uniform circular disc of radius R, if a circular sheet of radius r is removed, the centre of mass of the remaining part shits by a distance

if r =R/2 then x = R/6

7. Out of a uniform solid sphere of radius R, if a sphere of radius r is removed, the



centre of mass of the remaining part, shifts by

if r = R/2 then x = R/14

8. When a person walks on a boat in still water centre of mass of person, boat system is not displaced.

a) If the man walks a distance L on the boat, the boat is displaced in the opposite direction relative to shore or water by a distance.

M – mass of the man M – mass of the boatb) Distance walked by the man relative to shore or water is (L – x).

9. If a square sheet of side a is cut in a square sheet of side A then the distance

of centre of mass of remaining portion from the centre of origin sheet is

10. A uniform wire of length l is bent at the mid point into L shape then the

coordinates of c. m from the bent point is (x, y)

Distance of c. m from bent point is



M – I: Motion on the horizontal plane:1. Limiting friction = fL = fs (max) =

Normal reaction N = mgKinetic friction fk = Rolling friction fR = . FL > fK > fR

2. If F = 0, f = fs = 0 → Body does not slide.F < fL f = fs = F→ Body does not slide.F = fL f = fL = F – Body is just ready to slideF>fL f = fk – Body slides.

Here F = applied force (or) force that tries to displace the body, f = frictional force.3. Minimum force required to displace a body = fL= .4. minimum force required in displacing a body placed in a lift moving with

acceleration. if the lift is falling freely f = 0

5. If a body is moving with uniform velocity then F = fk

6. Work done against frictional force = fk x S = x S7. When applied force makes an angle with horizontal (pulling).Normal reaction N = mg – F sinFrictional force f = [mg - Fsin ]

Acceleration of body

Minimum pulling force required to displace the body


8. FRICTION..


F will be minimum if = .

8. In case of pushing:N = mg + F sin .F = [mg + F sin ]

Minimum pushing force required to displace the body

F becomes infinity.

Hence

9. Acceleration (If force is applied horizontally).

10. Acceleration of a freely falling body

11. Deceleration of a vertically projected body

(f → friction force due to air).

12. For a vertically projected body

(If air resistance is taken into account)

Velocity before reaching ground

13. Minimum force required to displace a body is applied and the same force is continued then the acceleration of the body



14. A chain of uniform length ‘L’ is placed on a rough horizontal table. The coefficient of friction between the chain and table is then the maximum fractional length of chain that can be hung freely from the edge of the table is.

Minimum fraction of length of chain that can be on the table is

15. Block on a lorry:a) The maximum acceleration of the lorry for which block beings to slide

on the floor of the lorry is

b) If a block of mass m is placed on a lorry moving with uniform acceleration a

The force acting on the block F = ma.c) If a g block does not slide and friction force on the block is f = ma.

If block slips or slides on the floor. Frictional force on the block

The acceleration (a) of the block relative to lorry is

16. Car moving on an umbanked circular road:a) When a car goes around an unbanked circular road, the static frictional

force between the wheels and the road provides the necessary centripetal

force. b) Maximum speed with which curve can be negotiated without skidding is

coefficient of static frictional between wheels and the road.c) The maximum angular velocity is

17. a) If a block having initial velocity u slides on a rough horizontal surface and

comes to rest, the acceleration of the block is

b) Distance traveled by the block before coming to rest is



c) Time taken by the block to come to rest is

Body on vertical surface.18. When a body of mass m is pressed against a vertical surface with a force P then

a) Normal reaction N =P.b) Limiting frictional force c) Frictional force between the body and surface or mg which ever is

less.d) Minimum force required to slide the body upwards.

F = mg + e) Minimum force required to prevent the body from sliding down F = mg-

19. A book of mass m is pressed between two hands by applying a horizontal force P

with each hand the book does not slide if

20. A vehicle is moving on a horizontal surface. A block of mass ‘m’ is stuck on the front part of the vehicle. The coefficient of friction between the truck and the block is The minimum acceleration with which the truck should travel, so that the body does not slide down is

[mg = ma]

Smooth inclined plane:

21. Block on a smooth inclined plane:a) N = mg cosb) Acceleration of sliding block a = g sinc) If 1 is the length of the inclined plane, h is the height and x is base. The time

taken to slide down starting from rest from the top is



d) t will be minimum if

e) Velocity of the block at the bottom of the inclined plane is same as the speed attained if block falls freely from the top of the inclined plane.

f) If a block is projected up the plane with a velocity u, the acceleration of the block is

g) Distance traveled up the plane before its velocity becomes zero is

h) Time of ascent is

i) Horizontal force required to prevent the body from sliding down F = mgtan .j) The minimum acceleration of the inclined plane so that a body placed on it does not

slide down is given by a = gtan .Rough inclined plane:

22. A body of mass m is placed on a rough inclined plane of inclination . If is angle

of repose. N = mgcos .

d) If the body is sliding down with uniform velocity then

e) Net contact force acting on the body =



23. If the block slide down from the top of the inclined plane. Velocity at the bottom of the plane is

24. In the above case time of descent is

25. The time taken by a body to slide down on a rough inclined plane is ‘n’ times the time taken by it to slide down on a smooth inclined plane of same inclination and length, then coefficient of friction is

26. If a block is projected up a rough inclined plane, the acceleration of the block is

27. a) The distance traveled by the block up the plane before the velocity becomes

zero is

b) The time of ascent is

c) If the body slides down with uniform velocity on the same plane, then

28. In the above case if time of decent is n times the time of ascent, then

29. a) Force needed to be applied parallel to the plane to move the block up with

constant velocity is



b) Force needed to be applied parallel to the plane to move the block up with an

acceleration a is

30. If block has a tendency to slide, the force to be applied on the block parallel and

up the plane to prevent the block from sliding is

31. If a force P is applied on the body up the plane then the acceleration of the body is sin

32. An air craft of mass m travels through a distance S on a runway and take off

with a velocity V in t sec. Then.

a) Force applied by the engine

b) Work done by the engine

c) Power of the engine



9. ROTATORY MOTION

M -I: Horizontal circular motion:

1. Linear velocity

2. Linear acceleration angular acceleration

3. Time period

4. In a circular path centripetal acceleration or radial acceleration

5. Tangential acceleration in non-uniform circular motion

6. The resultant acceleration in non-uniform circular motion

7. Centripetal force

Tangential force

8. When a stone attached to a string is whirled in a horizontal circle with constant

speed v, tension in the string

Since If T is the maximum tension that the string can withstand then,

9. In an unbanked road the centripetal force is balanced by frictional force,

When a body on a rotating platform is just about to fly off,

M -II: Motion of a body in a vertical circle:



Case I: If a body of m is tied to a string of length 1 and rotated in a vertical circle

with uniform speed

10. Tension in the string when it makes an angle with vertical is given by

a) Tension at the highest point ( = 1800 )

b) Tension when the string is in horizontal position

c) Tension at the lowest point

d) Different between maximum and minimum tensions.

e)

f) Time period

g) Total energy at the highest point

h) Total energy at the lowest point

i)

j) Case –II:11. If the body rotated with non uniform speed. If u is the velocity at the

highest point and v is the velocity at the lowest point then.

a)

b)



c)

d)

Case –III:12. If the body is revolving with critical speed

a) Minimum or critical speed of the body when the string makes an angle with

vertical is given by

At the lowest position

At the horizontal position

At the highest point

b) Tension in the string when it makes an angle with vertical T = 3mg (1 + cos ).

At the lowest position

When the string is in horizontal position

At the highest position

c) Total energy at the highest position.

Total energy at the lowest position

Case –IV:13. Simple pendulum is given a horizontal velocity u at the lowest position

(mean Position) then

a) the body oscillates about A.

b) the body leaves the without completing the circle.

c) the body completes the circle.

d) Height at which velocity u = 0. is h = u2/2g.



e) Height at which tension

f) Angle with vertical at which velocity

g) Angle with vertical at which the tension h) Tension in the string at an angular displacement with vertical is

14. Safe speed of a car going on a convex bridge to travel in contact with the bridge

is

15. A ball of mass ‘m’ is allowed to slide down from rest, from the top of a incline of height ‘h’. For the ball to loop in a loop of radius ‘r’

a) Minimum height of incline h = 5r/2.b) ‘h’ is independent of mass of the ball.

16. A ball of mass ‘M’ is suspended vertically by a string of length ‘l’. A bullet of mass ‘m’ is fired horizontally with a velocity ‘u’ on to the ball sticks to it. For the system to complete the vertical circle, the minimum value of ‘u’ is given by

17. If the bob of the simple pendulum is given an initial displacement then

a) Velocity at the lowest position

b) Tension at the lowest position 18. A body is placed on the top of a hemispherical bowl and it is given a

horizontal velocity v then a) Thrust on the bowl when the radius vector turns through an angle

The body looser contact at

b) If v = 0 then N = mg [2 – 3 cos ]. The body looser contact at

19. Equation of motion:



Also w = 2 n (if n is r.p.s)W= 2 n/60 (if n is in rpm).

20. Torque

21. Work done w =22. Power P = 23. KE = ½ Iw2.

24. Work done

25. If a body of mass m is revolving in a circular path of radius r, with a velocity v then angular momentum of the body about origin is

L = mvr = mr2 w = Iw where I = mr2

26. A body of mass m is moving with a velocity v along a straight line p represented by ax + by + c = 0 then the angular momentum of the body about origin

27. Angular momentum of a projectile about point of projection when it is at its highest point.

28.

29. Moment of inertial

Parallel axes theorem

Perpendicular axes theorem30. For a uniform circular ring.

31. For a uniform circular disc.



32. For a sphere

33. Rolling of a body without slipping on a horizontal surface.a) The body possesses both transnational and rotational kinetic energy.b) Total energy of the body

Transalational KE of the body.

c) Fraction of energy associated with translation motion

d) Fraction of energy associated with rotational motion

34. Rolling of a body down an incline without slipping.a) Acceleration of the body

b) Velocity acquired by body on reaching the bottom

h - height of the incline.c) Time taken to reach the bottom

35. When a body rolls down without sliding on an inclined plane with an inclination and height ‘h’.

S.No.

Body K2 Acceleration Velocity at the bottom

1. Solid sphere 2/5 r2 5/7 g sin

2. Disc ½ r2 2/3 g sin

3. Solid cylinder ½ r2 2/3 g sin



4. Hollow sphere 2/3 r2 3/5 g sin

5. Ring r2 ½ g sin

6. Hollow cylinder r2 ½ g sin

MOMENT OF INERTIA OF SOME REGULAR BODIESS.No

.Body Axis Moment of

Inertia1. Circular

ringa) Passing through centre and normal to its plane b) Any diameterc) Any tangent in its plane

Mr2

Mr2/23Mr2/2

2. Circular Disc

a) Passing through centre and normal to its plane b) Any diameterc) Any tangent in its plane

Mr2/2Mr2/45Mr2/4

3. Solid Cylinder

a) About its axisb) About an axis through its CG and normal to length

Mr2/2M[12/12+ r2/4]

4. Hallow Cylinder

About its axis Mr2

5. Solid sphere

a) Any diameterb) Any tangent

2/5 Mr2

7/5 Mr2

6. Hallow sphere

a) Any diameterb) Any tangent

2/3 Mr2

5/3 Mr2

7. Thin uniform rod

a) Passing through its centre and normal to lengthb) Passing through one end and normal to its length

M12/12M12/3



10. GRAVITATION

M – I: Kepler’s Laws: 1. If T is the time period of plant and R is average distance of the planet from Sun Then

2. If v1 is the velocity of the planet when its distance from Sun is d, and v2 is the velocity when the distance is d2 then according to law of conservation of angular momentum.

M – II: Newton’s Law of gravitation:3. Gravitational force between two bodies of manes m1 and m2 separated by

distance

d is given by.

4. When two similar spheres each of radius r are in contact then the gravitation force between then F

5. Gravitational force acting on a body of mass m placed at a distance x distance x

from the centre (x < R) is given by

M – III: Relation between g and G:

6. Density of earth

M – IV: Variation of g with height:7. Acceleration due to gravity at a height h above the surface of the earth is

given By

a)

b) Percentage change in weight of a body or (% change in g value).



Percentage change in weight of a body

M – V: Variation of g with depth:

8. Acceleration due to gravity at a depth d below the surface of the earth is given by

Percentage change in weight of a body

M – VI: Variation of g with latitude:

9. a) Acceleration due to gravity at a latitude is given by W angular velocity of earth.At equator So due to rotation of the earth the value of g at equator decreases

by Rw2 = 0.034 ms-2. If earth stops rotating the value of g at equator increase by Rw2.

At poles So at poles rotation rotation of earth does not have any effect on g

value.b) If the weight of a body at equator becomes zero then

M – VII: Orbital velocity:

10. a) orbital velocity

h height of the orbit.



d) Time period of the satellite

M – VIII: Escape velocity:

11. Escape velocity

M – IX: Energy of a satellite:

12. (r = radius of the orbit; r = R + h).

Total energy

13. Work done in shifting a satellite from orbit of r1 to another orbit of radius r2

is given by

14. Work done by lifting a body to a height h (h is comparable to R0 = change in PE)

15. If a body is projected with a velocity v from the surface of the earth, then the height raised by the body can be found by

If v > ve then velocity at infinity can be found using law of conservation of energy i.e.,

v1 = Velocity of projection.v2 = Velocity at infinity.



11. S.H.M

M – I: Displacement, Velocity and Acceleration:1. Displacement after t sec, y = r sin If particle starts from mean

position the If particle starts from extreme position

2. Velocity after t seconds

Velocity when the displacement is

At mean position v = rw (maximum)At extreme position v = 0 (minimum)

3. Acceleration a = -rw2 sin wt i.e., a = - w2y.at E.P a = rw2(maximum).At M.P a = 0 (minimum).

4. Force F = ma = mrw2 sin wt.= mw2y

at M.P = F= 0 at E.P F = mrw2

5. Time period

6. Frequency

M – II: Energy of the particles:

9. PE= ½ mw2 y2

At MP PE = 0 (minimum)



Average PE = ¼ mw2 r2 At EP PE = ½ mw2 r2 (minimum).

10. Total energy E = ½ mw2 r2

M – III: Time period of a simple pendulum:

11. Time period of a simple pendulum

12. For a simple pendulum, in a lift moving up with an acceleration or coming down

With a deceleration,

13. In a lift moving up with deceleration or coming down With a acceleration, time

period of simple pendulum

In a freely falling lift = 0. So the pendulum does not oscillate. The time period is infinity.

14. For a pendulum of infinite length, (or) for a body dropped in tunnel along the

diameter the time period

15. Time period of a seconds pendulum is 2s.

Time period remains same if

Its length is

A seconds pendulum is taken from one place to another place then the Change in

the length of the seconds’ pendulum

So that time period remains constant



16. If the simple pendulum is in a cart moving horizontally with an acceleration

a then time period

If the cart is moving along a circular path of radius r with a speed v then a = v2/r

If the cart is sliding down an inclined plane, inclined at angle with the

horizontal,

17. When two simple pendulum of lengths 1 and 2 are get into vibration in the same direction at the same instant with same phase.Again they will be in same phase after he shorter pendulum has completed

n oscillations. To find the value of n,

n Ts = (n - 1) T1 and T

S = shorter; = longer18. Two pendulum of lengths l1 and l2 (l1 > l2) start vibrating from the mean

position in the same phase. They will be again in the same phase at the mean position after larger pendulum completes n oscillations and the shorter one completes (n + 1) oscillations. Then

19. Two simple pendulum of time period, T1 and T2 start vibrating from the mean position in same phase. The phase difference between them after t sec. is given by

They will be in same phase after a time

M – IV: Time period of a loaded spring:

20. Time period of a loaded spring where k = F/x.

Taking the mass of the spring m into consideration



Time period of the spring, , where x is elongation under the load.

21. If a spring is cut in n equal parts the spring constant of each part is k1 = nk.

Then time period of each part under the same load is given by

22. If a spring of spring constant ‘K’ and length ‘ ’ is cuts in to two springs of lengths ‘ 1’ and ‘ 2’ then the spring constants of the two parts is

23. When two spring of force constants K1 and K2 are connected in series. Then the effective force constant is

24. When two spring of force constants k1 and k2 are connected in parallel. The effective spring constant is K = k1 + k2 .

25. T1 and T2 are the time periods of two springs, under same load.If the springs are connected in series and the same load is

attached.

Time period

If the springs are in parallel, Time period



12. ELASTICITY

2. a) Longitudinal strain percentage change in length = longitudinal

strain x 100.b) Shear strain

c) Volume strain

Shear strain = 2 x longitudinal strain, volume strain = 3 x longitudinal strain.

M – I: Young’s modulus:

3. Young’s modulus

F Applied force; A Area of cross sectionL Original length; e ElongationR Radius of the wire.

4. Elongation

If F and Y are same

If young’s modulus = stress then strain = 1 and final length

= 2 x initial lengthm mass of the wire.

When two wires are connected in series and same force is applied then a) Stress is same in both wiresb) Ratio of the strains = Y2/Y1



c) Total elongation e = e1 + e2

When two wires are connected in parallel and same force is applied thena) Strain is same in both the wiresb) Ratios of the stresses = Y1/Y2

c) Total force F = F1 + F2

5. When a rubber cord is suspended vertically from a support it elongates under its own weight.

6. Breaking stress

Breaking force

Maximum possible elongation

Length of a wire required to break under its own weight

Rigidity modulus

7. Rigidity modulus

Shearing strain8. A tangential force F is applied on a cube of side a. if a is the lateral

displacement of the top face w.r.t the bottom face

Then Shear stress

Shear strain

M – II: Bulk Modulus:

9. Bulk Modulus

i) If a block of coefficient of cubical expansion V is heated through a rise in temperature of the pressure to be applied on it to prevent its expansion

where K is its bulk modulus.ii) When a rubber ball of volume V bulk modulus K is taken to a depth h in

water

decrease in volume

(d = density of material)10. When a pressure is applied on a substance its density changes. New

density d’ = dk/ (k - )



11. Compressibility: The reciprocal of bulk modulus is called Compressibility

For incompressible substances C = 0, k = : M – IV: Poisson’s ration & relation between Y, n and k:

i) Poisson’s ratio has no unit and no dimensionsii) Theoretical limits of iii) Practical limit of

iv) the substance is perfectly incompressible.13. Relation among elastic constants Y, n, K,

14. Percentage change in volume of a wire due to elongation(%V) = (%1) +2(%r)% change in volume = % change in length + 2 (% change in radius)

M – V: Strain Energy;15. Work done ins stretching wire = strain energy stored in the wire

= ½ x Force x elongation = ½ F x e

16. If two material of same material are stretched by same amount i.e. e is same



17. If two wires of same material are stretched by same force then

18. Elastic strain energy = ½ stress x strain x volumeStrain energy per unit volume = ½ stress x strain

19. If l1 and l2 are the lengths of a wire under tensions T1 and T2, the actual

length of the wire

Thermal stress developed in a body due to change in temperature

13. SURFACE TENSION



M – I: Force due to S.I:

1. Surface tension

2. a) Force in addition to weight required to lift a thin wire of length l from liquid surface is F = 2l x T = 2Tl.b) Force required to lift a think circular plates of radius R from liquid surface is

c) Force in addition to weight required to pull a circular ring from water surface is r1 and r2 are internal and external radius.If it is a thin ring then .d) Force required to pull a think rectangular plate of length 1 and breadth b is from liquid surface is e) Force required to pull a think rectangular frame from liquid surface is

f) Force required to pull a capillary tube from water surface is 3. If a small drop of water is squeezed between two plates the force required

to separate the plates

A= Area of water lawyer. d= Thickness of water layer.4. A metallic wire of density d floats horizontal in water. The maximum radius

of the wire so that the wire may not sink, will be (surface tension of water = T).

5. A metallic wire of density d is laying horizontal on the surface of water. The maximum length of wire so that it may not sink, will be (surface tension of water = T)

M – II: Surface energy:

6. Surface tension

This work will be stored as surface energy .



7. a) Work done in forming a liquid drop of radius b) Work done in forming an air bubble of radius r is c) Work done in increasing the radius of a liquid drop from R1 to R2

d) Work done in forming of a soap film of size l x b is = 2T lbe) Work done in increasing the radius of a soap bubble from R1 to R2

8. When a big drop of radius R is split in to n identical drops each of radius r.

a) R = n1/3 r.

b) Increase in surface area

c) Work one in splitting the drop

d) If this process takes place under adiabatic conditions, the temperature

of ` the drop decreases by

9. When n identical small drops each of radius r combine to from a big drop of radius R thena) R = n1/3 r.

b) Decrease in surface area

c) Energy released in this process

d) If this process takes place under adiabatic conditions then the temperature

of the drop increases by

e) If the energy converts into KE then

10. Two drops of radii r1 and r2 combine in vacuum under isothermal conditions. Then if r is the radius of the drop formed,

11. When two bubbles of radii r1 and r2 in vacuum combine, under isothermal

conditions, the radii of the bubbles formed

A film of water is formed between two straight parallel wires each of length 1. The work done in increasing the separation between then by xm is w = 2Tlx.

M – III: Capillary rise:9. When a capillary tube of radius r is dipped in a liquid of surface tension T

and density d then height of the liquid in the capillary tube is given by



Note: in gravity free space the liquid rises to the full lengths of the capillary to be but it will not over flow.

10. When a capillary tube is dipped vertically the capillary rise is h. when it is

inclined by an angle to the vertical, capillary rise =

If is taken with horizontal capillary rise

Weight of liquid in capillary tube = force due to surface tension.11. Weight of the liquid in the capillary tube 12. A vessel has a small hole of radius r at the bottom. The maximum height

to which water can be poured inside the vessel without leakage in

13. A U tube has two vertical limbs of radii r1 and r2 when liquid is poured in to

it the difference in levels in

M – IV: Excess of pressure:

14. Excess of pressure inside a liquid drop of radius

Excess of pressure inside a soap bubble

Where

Excess of pressure inside a bubble in a liquid

If P0 is outside pressure, total pressure inside =P0 + excess pressure

15. If two bubbles of radii r1 and r2 are in contact, the radius of curvature of the

interface is

16. An air bubble of radius r is at a depth h is water.



Inside pressure

17.

18. A bubble of radius r1 is inside another bubble of radius r2. The radius of a single bubble whose excess pressure is equal to difference in pressure between inside of inner bubble and out side the outer bubbles is

19. A long capillary tube of radius r is filled with water and placed vertically.

The height of water column remaining the capillary tube

M – I: Viscous force:

1. Viscous force


14. VISCOSITY


Force required to drag a plate placed on an oil layer of thickness t with a

velocity v is given by

M – II: Poiseuille’s equation:2. Volume of liquid flowing through a capillary tube in unit time is given by

Pressure difference across the tube. P = hdg. Radius of the capillary; 1 Length of the tube.

(if P and are same).

Also it is known as fluid resistance.

3. When two capillary tubes are connected in series then

Pressure difference across the first tube. Pressure difference across the second tube.

If the two tubes are replaced by a single tube of radius r then

4. When two capillary tubes are connected in parallel

If the two tubes are replaced by a single tube of radius r

(If r1 = r2 = r)

Then

M – III: Stokes formulas:



5. When a spherical body of radius r is moving through a fluid with a velocity V then the viscous force acting on body is given by

After some time the body attains constant velocity called terminal velocity in this case.

If n identical drops moving with terminal velocity v. Combine to form a signal dmp

then terminal velocity of the big drop V = n2/3 v

M – IV: Continuity equation:

6.

Volume of liquid flowing through a tube in unit time (or) volume flux q = av.

M – V: Buoyant force:7. If a body of volume V, density ds is immersed in a liquid of density d1 then,

a) Buoyant force acting on the body V1d1 g where V1 is the volume of the body inside the liquid.

b) If the body is floating then V1d1 g = mg. c) Apparent weight of the body w1 = mg – Fb.

Note: With increase in temperature tb decreases hence w1 increases.M – IV: Bernoulli’s theorem:

8. (all are in SI units)

If the liquid is flowing horizontally then

Change in KE per day unit mass



9. If P1 and P2 are pressures at the top and bottom of the wings of the aero plane then the dynamic lift on the aero plane

Where V1 and V2 are the velocities at the top and bottom

of the wings and d is density.

M – V: Torricelli’s theorem:10. A vessel filled with a liquid up to a height H has a small hole near the

bottom. If h is depth of the hole then.

a) Velocity of efflux from the orifice is

b) Volume efflux Q = aV

c) Time taken by the water to touch the base level t

d) Horizontal range

e) R is maximum if h = H/2Rmax = H

f) Time taken for emptying the tank

g) Time after which the level falls from h1 to h2 is

15. EXPANSION OF SOLIDS

1. Linear expansion

If same amount of heat given then



% change in length % change in area % change in Volume

2. Loss or gain of a pendulum clock due to a change in temperature per day is

3. If two rods of different material have equal lengths of 1 and at temperature T1 as the temperature is increased to T2 different in their final length is

4. For an anisotropic substance if are the coefficients of linear

expansion in three perpendicular directions then 5. When a bimetallic strip is heated it bends. Then the radius of curvature of

the bimetallic strip

Where d is the thickness of each strip. is change in temperature,

are coefficients of linear expansion of the two metals.

6.

7. Thermal stress developed in a rod heated, when it is prevented from expansion is called thermal stress

Force developed

8. A wire is bent in the form of a ring with a small gap of length on

heating to if the gap increase to x2 in length, then the coefficient of linear expansion of the wire material



9. If two rods of different materials have the difference between their lengths at all temperatures, then

10. The change in moment of inertia of a body due to rise in temperature

11. Metal tapes:a) Metal tape shows correct reading only at a temperature at which it is

constructed.b) A metal tape is graduated at and is used at

i) distance between the divisions decreases, thena) Observed reading L is less than actual valueb) Correction to be applied, c) Correct reading ii) It t2 < t1, distance between the divisions decreases, thena) Observed reading L is greater than actual valueb) c) Correct reading

12. Variation of density with temperature

13. A mercury barometer has a metal scale which is calibrated at 00C, the pressure as read by it at t0C is Ht. Then the pressure at 00C will be given by

is coefficient of real expansion of mercury is coefficient of linear expansion of metal

16. EXPANSION OF LIQUIDS

1.2. From specific gravity bottle experiment,



m1 is nearly equal to m2 and hence it can be written

3. If d1 and d2 are the densities at temperature t1 and t2,

4. The unoccupied volume in a container remains same if the volume expansion of the container and that of liquid are same. The condition for this is

5. The temperature at which the density of liquid is x% less than at

6. If a liquid column of height h1 at temperature t1 balances another liquid column of height h2 at temperature t2.

17. EXPANSION OF GASES

M – I: Volume and pressure coefficient of a gas.

1. Volume coefficient

(at constant pressure).

2. Pressure coefficient



( at constant valume).

3.

M – II: Boyle’s law:

4. AT constant temperature

5. Two vessels of volume V1 and V2 filled with a gas at pressure P1 and P2

are connected the common pressure

6. If the pressure of the gas is increased by x% then the % decrease in

volume at constant temperature is

7. If 11 and 12 are the length of the air column with open end upwards and

downwards respectively, then

H atm. Pressure.h length of mercury thread.

8. When an air bubble at the depth h in a lake rises to the top, its volume increases. Assuming the temperature to be constant

If the volume becomes n timesh = H (n – 1) (H + h) V1 = HV2

And if radius becomes n times,

Where H = 1 atm 10 m depth of water = 76 cm of Hg.

If temperature is constant then

M – III: Charle’s law:9. At constant volume (for a given man of a gas)

10. At constant pressure (for a given man of a gas)

M – IV: Ideal gas equation:11. If one gram of gas is considered.

PV = rT (r = gas constant).If one mole of gas is considered



for n moles PV = nRT.

12. Difference forms of ideal gas equation.a) PV = nRTb) PV = (m/M)RTc) P = (dRT)/Mm – mass of the gas, d – density of the gas, M – molecular weight. For a given gas.

(if V and T are constant).

(if V is constant).

If T is constant

If P is constant 13. % change in T = % change P + % change in V.14. When n1 moles of a non reacting gas in the state (P1 V1 T1) is mixed with n2

moles of gas in state (P2 V2 T2), if the resultant mixture is in the state (P, V, T).

15. If two different gases of same mass occupy same volume at same pressure then M1/T1 = M2/T2 (M1, M2 are molecular weights)

18. THERMODYNAMICS

M – I: Joule’s law:1. The work done on a system (w) is directly proportional to heat produced in

the system.

2. A body of mass falls from height h and if it is potential energy converts into heat then

If x % of energy is converted into heat then



3. An ice block of mass M falls from a height h and if its energy converts into heat then

The ice completely melts if h = 33.6 km.4. When a body of mass m moving with velocity v is stopped and all its KE

converts into heat energy then

5. If a body of mass m is dragged on a rough surface through a distance x then

If it is an ice block then

Mass of the ice melted

M – II: First law of thermodynamics:6. dQ = du + dw.

dQ + ve if heat is supplied to the system.- ve if heat is absorbed from the system.

du + ve if temperature increases. -ve if temperature decreases.

ve if work is done by the gas. - ve work is done on the gas.

M – III: and relation between them:7. A gas at constant

Specific heat at constant pressure (at constant pressure).

Molar Specific heat at constant pressure

Specific heat at constant pressure

Molar Specific heat at constant pressure



8.

9. The following table show the values of

Type of the gas

Monoatomic

Diatomic

Polyatomic(Tri-higher)

M – IV: Constant volume process: (Isochoric)

b)

c) dw = 0.d) dQ = dU = n CvdT.

M – V: Constant pressure process: (Isobaric)

10. a)

b) du = nCv dT.c) dw = d) dQ = nCp dT.

11. The fraction of heart energy supplied which is stored as internal energy

12. The fraction of heat energy supplied which is used in doing external work

M – VI: Isothermal process:13. a)

b) du = dT = 0.



d) e) Slope of isothermal curve = -P/V.f) Isothermal bulk modules = .P.

M – VII: Adiabatic process:

15. For a mixture of two gases

16. Heat capacity or thermal capacity or water equivalent (ms) =

19. THERMAL RADIATION

1. Absorptivity

a + r + t = 12. Energy emitted by a block body from unit surface area in unit time is given

by

E is also called emissive power.For the bodies

3. Total energy emitted by a body in a time t is given by

For a sphere

i.e., 4. The net heat lost or gained by a body at temperature T in unit time



5. Newton’s law of cooling

Temperature of the body.

Temperature of the surroundings.6. If the temperature of a body decrease from T1 to T2 in a time t then

according to Newton’s law of cooling.

7. If is the wavelength corresponding to maximum monochromatic

emissive power of a body at temperature T the (b

is a constant).b = 2.93 x 10-3 mk.

8. For a spherical body at temperature T

c) Rate of heat loss (If T is same).

d) Rate of cooling is given by

(If T is same)



eamcet formula in physics1

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