A to Z of Physics

Problems-Bodies in vertical motion

2009-01-29T02:34:00.000-08:00

6.A body starting from rest slides down an inclined plane.Find the velocity after it has descended vertically a distance of 5metres. (g=9.8 ).

Soln: The velocity of the body when it touches the ground sliding down an inclined plane ,will be same as when the body vertically falls freely from height ‘h”.

From the problem height S=h=5m; acceleration due to gravity g=9.8; Initial velocity u=0 ; V=?

substitute the values in the equation = 2gs ,

we get = 2(9.80) (5) ; = 98

V =9.899 m/sec.

7.A ball projected vertically upwards returns to ground after 15sec.Calculate i)maximum height to which it rises ii)velocity with which it is projected and iii)its position after 6 seconds.

Soln:From the data given in the problem,

Velocity of projection u=?

acceleration due to gravity g=9.8,

time of flight T =15 sec

substitute the values of T,g in the equation T =

=15 ; 2u = 15 (9.80) = 147

ii )Velocity with which it is projected u= 73.5 m/sec.

i)Maximum height =,

substitute the values of u,g in the equation we get =

= =275.625 m

1iii) Let Its position after t=6sec be S

substitute the values of u,t and g in the equation S=ut -

S = (73.5)(6) - =441-176.4 = 264.6 m high from the ground.

8.A stone was dropped from a rising baloon at a height of 150m above the ground and it reaches the ground in 15 seconds.Find the velocity of the baloon at the instant the stone was dropped.

Soln: From the data given in the problem,

Acceleration due to gravity g=9.8 ,

Height of the balloon when the stone is dropped from it h=150m,

Time of flight T=15 sec,

Let the velocity of the baloon when the stone is dropped from it is u (say).

Substitute the values of g,t and h in the equation h = -ut

we get 150 = -u(15) = (4.9)(225) - 15u,

150 = 15[(4.9)(15)- u],

10=73.5 - u ; u = 73.5- 10 =63.5 m/sec.

9. From the top of a tower of 200 metres high, a stone is projected vertically upwards with a velocity 49m/sec.Calculate the i)maximum height traveled by it from ground level, ii)the velocity with which it strikes the ground and the iii) time it takes to reach the ground.

Soln: From the data given in the problem,Height of the tower is h=200m,

velocity of projection u = 49 m/sec,

Max height of the stone from top of the tower be S= X (say),

velocity at maximum height v = 0,

i) substitute the values of u,v and g in equation = -2gs,

we get = -2(9.8)X,

X = = 122.5 m from the top of the tower

Maximum height from ground level H = X +h =122.5 + 200 = 322.5 m.

ii) Let the velocity with which it reaches the ground be say,

substitute the values in the equation =

= ,

= 79.50 m/sec.

iii) Time of flight T =?

Substitute the values u,g and h in the equation h = -ut,

we get 200 = -49T,

200 = - 49T

- 10T = ,

by solving this quadratic equation for T we get T= 13.11 sec

Problems- Bodies in vertical motion.

2009-01-19T22:56:00.000-08:00

1. A stone is projected vertically upwards with a velocity 29.4m/sec. i)Calculate the maximum height to which it rises ii) calculate the time taken to reach maximum height iii) Find the ratio of velocities after 1sec, 2 sec, 3sec of its journey.

Soln: From the data given in the problem,

Initial velocity of stone u=29.4 m/sec,

Acceleration due to gravity g=9.8 ,

i) = =

= = 14.7 m.

ii) Time of ascent = = = 3sec.

iii) Ratio of velocities of a vertically projected up body after 1 sec, 2 sec, 3sec , . . . . . . . . . . . . . . of its journey will be :: . . . . . . . . . . . . : = (u-g) : (u-2g) : (u-3g) : . . . . . . . . : (u-ng).

:: = (29.4-9.8 ) : (29.4- 19.6) : (29.4 - 29.4)

::= 19.6 : 9.8 : 0 .

2. A stone is dropped from the top of a tower.The stone touches the ground after 5sec.Calculate the i)height of the tower and the ii) velocity with which it strikes the ground.

Soln: From the data given in the problem,

Initial velocity of the stone u = 0 m/sec,

Acceleration due to gravity g = 9.8

Time of descent = 5 sec.

i) Height of the tower be S=H (say)

substitute the above values in the equation S=ut+

H = 0(5)+ = 0 + = (4.9)(25) = 122.5 m.

ii) Let the velocity when it touches = v (say),

substitute the above values in the equation v=u +g

v = 0+(9.8)(5) = 49 m/sec.

3. A stone is projected vertically upwards with an initial velocity 98 m/sec . Calculate i) maximum height the body reaches ii ) find its velocity when it is exactly at the mid point of it’s journey iii) time of flight.

Soln : From the data given in the problem,

Initial velocity of the stone u = 98 m/sec,

acceleration due to gravity a = 9.8 ,

i) Maximum height = = = 490 m.

ii) When it is in the middle S= =245 m

Let the velocity = v (say),

substitute the values in the equation = 2gS,

we get = 2(-9.8)(245),

= 9604 - 4802 =4802 ,

v =69.30 m/sec

iii) time of flight T = = = 20 sec.

4.Estimate the following.
(a) how long it took King Kong to fall straight down from the top of a 360 m high building?_________ seconds
(b) his velocity just before “landing”___________ m/s. ( Question by ELISE in Yahoo answers).

Soln: From the data given in the problem,

Height of building S=H = 360m,

Acceleration Due to gravity g=10 ,

i) time of descent be = ?

substitute the values in the formula S=ut+

we get 360 = (0)() + ,

360 = 5 ; = 360/5=72

Therefore = 8.49 sec.

ii )Velocity of king kong just before touching the ground v =g,

v = (10)(8.49) =84.9 m/sec.

5. A baseball is hit straight up into the air with a speed of 23 m/s.
(a) How high does it go?____ m
(b) How long is it in the air?_____ s. ( Question by ELISE in Yahoo answers).

Soln: From the data given in the problem,

Acceleration Due to gravity g=10 ,

a) Maximum height = =

= 529/20 = 26.45 m.

b ) Time of flight of base ball T = ,

T= , = 46/10 = 4.6 sec.

Andhra Pradesh Inter Exams Time table March2009.

2009-01-13T08:35:00.000-08:00

timetablegenm09.pdf (application/pdf Object)

Problems Linear motion.

2009-01-09T22:58:00.000-08:00

In this post and in few of my posts to come, I would like to solve problems on linear motion,freely falling bodies,vertically projected up bodies and projectiles .

1.An object accelerates from rest to a velocity 20m/sec in 4seconds.If the object has uniform acceleration, find its acceleration and displacement in this time.

Soln: From the data given in the problem we have,

Initial velocity = u =0,

final velocity v=20 m/sec,

Time of journey t=4sec,

Acceleration a = = = 5

Displacement S = ut + = 0 (4)+

S = 40m.

2.An object starting from rest moves with uniform acceleration of 3 for 6sec.Find its velocity and displacement after 6seconds.

Soln: From the data given in the problem we have,

Initial velocity of the object u = 0,

Acceleration of the object a= 3,

Time of journey t =6sec.

Final velocity of the object v=u+at =0 + 3(6) = 18 m/sec.

Displacement S= ut + = 0 (6)+

S= 54m.

3.An object starting from rest moves with uniform acceleration of 4.Find its displacement i) 5seconds ii) in 5th second iii) 8th second.

Soln : From data given in the problem

Initial velocity of the object u=0,

Acceleration of the object a=4,

i) time t=5 seconds,

Displacement of the object S=ut + = 0 (5)+

Displacement of the object in 5seconds S= 50m.

ii) Displacement of the body in 5th second =?

Let us substitute n=5 in the formula =u+a(n-1/2)

=0+4(5-1/2) = 4(4.5) =18m.

iii)Displacement of the body in 8th second =?

Let us substitute n=8 in the formula =u+a(n-1/2)

=0+4(8-1/2) = 4(7.5) =30m.

4.An object started moving with an initial velocity of 10m/sec, after traveling a distance of 5m gets a velocity 20m/sec.Find its i) acceleration ii) time taken for 5m displacement.

soln: From the data given in the problem,

Initial velocity of the object u=10m/sec,

Final velocity of the object v= 20m/sec,

Displacement S=5m,

i) acceleration a=?

Substitute the values of u,v and S in the equation =2as,

we get =2a(5)

300 = 10a or a = 300/10=30.

ii)Time t=?

Substitute the values of u,v and a in the equation v=u+at,

we get 20=10+(30)t ; 10=30t

t =10/30 = 1/3 = 0.333 sec.

5.When an observer started observing a car it’s velocity was x m/sec , if it travels for 10sec with uniform acceleration 2.5 and its velocity increases to 75m/sec.Find i) Initial velocity of the car ii) Displacement of the car in 10sec iii) Displacement of the car in first 5sec and last 5sec, what is your inference.

Soln: From the data given in the problem,

Initial velocity of the car u = x (say),

Final velocity = 75m/sec,

Acceleration a = 2.5,

Time of journey t=10sec.

i) Substitute the values of u,v,a and t in the equation v=u+at

we get 75 = x+2.5(10) ; x=50 m/sec.

ii) Let the displacement of the car in 10sec be S

Substitute the values of u,a and t in the equation S=ut +

We get S = 50(10)+ ;

S = 500+125 ; s=675 m

ii) Let the displacement in first 5sec be

Substitute t=5 sec and thve values of u and a in the equation =ut +

we get =50(5) +

= 250 + 31.25 = 281.25 m.

Let the displacement in next 5sec be

= S -

= 675 - 281.25 = 393.75 m

We can observe that, even though the time of journey is same , >

Displacement of the body in second half is greater than in the first half time of its journey।

6 .A cheetah can accelerate from rest to 24.0 m/s in 6.70 s.

Assuming constant accelaration, how far has the cheetah run in this time?(Question by suzi in yahoo answers).

Soln: From the data given in the problem,

Initial velocity of the cheetah u=0,

Final velocity of cheetah v=24 m/sec,

Time t = 6.7sec

Distance traveled by cheetah be S=?

The equations of motion are = 2aS- - - - - (1)

and a = - - - - - (2)

From equations (1) and (2) we get =2S

simplifying it we get S = =

Substitute the values of u,v and t we get S = =12(6.7)=80.4 m

7) A car covers first halt of the distance between two places at a speed of 30 km/hr and the second half at a 90km/hr.What will be the average speed of the car?

Soln: Method I: Let the total distance between the places be S.

Time taken to cover First half = = hours.

Time taken to cover Second half = = hours.

Total time t = + =+= = .

Average speed = Total distance /total time.

= S/t =

= 45km/hr.

Method II (Short cut): If the body covers 1st half of distance with a speed x and the second half with a speed y,then the average speed = .

Average speed = latex \frac{2700}{60}$ = 45km/hr.

8) A body starts from rest and acqires a velocity of 400m/sec in 10seconds.Calculate the acceleration and distance traveled.

Soln: From the data given in the problem

Initial velocity of the body u=0,

Time of journey t=10sec,

Final velocity v=400m/sec,

Acceleration a=? and distance traveled in 10sec s=?

Substitute the values of u,v and t in the equation v=u+at,

we get 400= (0)(10) + a (10) ; 10a=400

a= 40

Substitute the values of u,a and t in the equation S= ut +

S=(0)(10)+ = 0+2000 = 2000m.

9) A body moving with uniform acceleration covers 6m in second and 16m in second.Calculate the initial velocity,acceleration and distance moved in in $latex 6^{th} second.

Soln: From the data given in the problem

Distance moved in second = 6m,

Distance moved in second = 16m,

- = a(4-2) =2a

therefore 2a = - =16-6 =10

a = 5 m/

Substitute the values of ,a and n=2 in the equation =u+a(n -1/2)

we get 6=u+5(2-1/2) ; 6=u+7.5

u=-7.5 +6 = -1.5 m/sec.

Distance moved in the $latex 6^{th} second = u+a(6-1/2),

Substitute the values of u,a in the above equation we get $latex 6^{th} second=-1.5+5(5.5)

$latex 6^{th} second =-1.5+27.5 =26m.

10) An object started traveling with a velocity 2m/sec moves with an uniform acceleration of 3 m/.

i) Find the ratio of displacements in a) ,, seconds b) ,, and seconds .

ii)Find the ratio of velocities a) ,, seconds b) ,, seconds .

Soln: From the data given in the problem

Initial velocity u=2 m/sec,

Acceleration a =3 m/,

i- a) The ratio of displacements in ,, seconds

From the formula :: = (2u+a) : (2u+5a) : (2u+9a)

:: = (4+3) : (4+15) : (4+27)

:: = 7:19:31 .

i-b) The ratio of displacements in ,, seconds

From the formula :: = (2u+3a) : (2u+7a) : (2u+11a)

:: = (4+9) : (4+21) : (4+33)

:: = 13 : 25 : 37 .

ii -a) From the formula ::: . . . . . . . . := (u+a) : (u+2a) : (u+3a) : . . . . . . . . . : (u+na).

The ratio of velocities a) ,, seconds

:: = (u+a) : (u+3a) : (u+5a) = 5 :11 : 17 .

ii- b) The ratio of velocities a) ,, seconds

:: = (u+2a) : (u+4a) : (u+6a) =8 :1 4 :20 .

11) A body travels 200cm in the first two seconds and 220cm in the next four seconds.What will be the velocity at the end of the seventh second from the start?

Soln: The displacement of the body in first 2 sec =200cm,

Let the initial velocity = u(say) ,

Acceleration = a(say), time =2sec

= +,

Substitute the value of in above equation, we get

= 2u+,

= 2u +2a ; 2(u+a) =200

Therefore u+a = 100 - - - - - - - - - - - - - - - - - - - (1)

Given that the body travels 220cm in next 4sec.That is from the start it displaces 200+220 = 420 cm in 6sec.

Displacement =420 cm,

time =6 sec,

substitute theses values in the equation = u+,

= 6u+=6u+18a,

6(u+3a) = 420 ; u+3a = 70 - - - - - - - - - - - - - - - - - - - (2)

Solving equations (1) and (2) or Eq (2) - Eq(1)

we get 2a= -30 ; a=-15 cm/,

Substitute value of a in Eq(1) we get u-15 = 100,

u = 115 cm/sec.

The velocity at the end of second v=u+a

we get v= 115+(-15)(6) =115-90 =25cm/sec.

Therefore final velocity v=25cm/sec.

12.A subway train starts from rest at a station and accelerates at a rate of 16.5 for 13.1 sec.It runs at constant speed for 69.7s and slows down at a rate of 3.45 until it stops at the next station.What is the total distance covered?

Soln: From the data given in the problem,

Initial velocity of the train u = 0 m/sec,

Acceleration a = 16.5 ,

time t = 13.1 sec,

Velocity after 13.1 sec v=?

and the distance traveled =?

Substitute the values in the equation v =u+at

we get v= 0+(16.5)(13.1) = 216.15 m/sec.

Substitute in equation = ut +

we get = (0)(13.1)+

= 0+1415.78 =1415.78 m

After that the train travels with velocity v=216.15 m/sec for 69.7sec. calculate distance traveled during this time.

v=216.15 m/sec, t = 69.7 sec =?

substitute the values in equation = vt = (216.15)(69.7)=15065.66 m

Finally the trains decelerates at the rate of 3.45 and comes to rest.find distance traveled before coming to rest.

Acceleration a = -3.45 ,

Initial velocity u = 216.15m/sec,

Final velocity v=0,

distance traveled =?

substitute the values in = 2as,

we get = 2(-3.45)

-6.90 = -46720.82

= 6771.13.

Total distance traveled by train S = ++ = 1415.78+15065.66+6771.13=23252.57 m or 23.252 km.

MultipleChoiceQuestions,Units-Dimensions

2009-01-08T05:22:00.000-08:00

41)Which of the following is dimensionally correct formula.

a)V=ut + at b) v+u = at c) V/u =at d) vt = u -a

Soln: The dimensions of L and T are same in all the terms of V+u=at.Hence, according to priciple of homogeneity it a dimensionally correct equation.

42) The dimensional formula of coefficient of kinematic viscosity is

a) b) c) d)

43) Which of the following is not the unit of energy?

a) joule b) Nm c) W d)

44) Dimensional formula of latent heat is

a) b) c) d)

45) Pair of quantities having same dimensional formula are

a) velocity , Impulse b)Force,Weight C)Impulse ,Inertia d) Angular momtntum, Linear momentum

46) If R is resistance and L is the inductance,then the dimensions of will be same as the dimensions of . . . .

a)time b) speed c) frequency d) acceleration

Soln: Dimensional formula of resistance R = and dimensional formula of Inductance is L= .

= = which is the dimensional formula of frequency.

47) If the unit of power is 100 erg/minute,the unit of force is 100 dyne and the unit of time is 100 seconds, the unit of length is

a) 5/3 cm b) 2/3 cm c) 1/3 cm d) none

Soln: P = = 100 erg/min =(100/60) erg/sec

F= = 100dyne T= = 100 sec

P= =

L = = (100/60)(1/100)(100) =5/3

48 ) If is the dimensional formula of Electric power, find the value of 5a+2b-6c.

a) 25 b) 27 c) 30 d)-9

Soln: Dimensional formula of Electric power = =

Comparing the powers of M,L and T we get a=1,b=2 and c=-3 .

Therefore 5a+2b-6c = 5(1)+2(2)-6(-3) =5+4+18 =27.

49) If is the dimensional formula of Magnetic moment, fine the value of a-2b+3c -d

a) 5 b) 7 c)3 d)-5

Soln: Dimensional formula of Magnetic moment is =

Comparing the powers of M,L,T and I we get a=0,b=2, c=0 and d=1 ;

Therefore a-2b+3c-d = 0-2(2)+3(0)-1 = -4-1 = -5.

50) If is the dimensional formula of magnetic pole strength, fine the value 5a+b+3c+2d

a) 3 b) 2 c) -2 d)0

Dimensional formula of Magnetic pole strength is =

Comparing the powers of M,L,T and I we get 5a=0,b=1,3c=0 and 2d=1 ;

Therefore 5a+b+3c+2d = 0+1+0+1 = 2.

Kinematics - Important points.

2009-01-05T09:09:00.000-08:00

1.The terms Rest & Motion are relative terms.

There is no object in the universe which is at absolute rest.Ex: A text book in a book rack may be relatively at rest, with respect to the immediate surroundings. But, the earth is revolving around sun, therefore every particle on earth will be in motion including the text book.Hence, the text book also will be in motion with respect to Sun.
There is no object in the universe which is in absolute motion. Ex:A person traveling in a train will be in motion as the train is moving.But, with respect to the fellow passengers and all the non moving objects in the train the person will be relatively at rest.

2. The shortest distance between initial and final positions of a moving body is called its displacement.

Ex: If a object starts its journey from a point A, travels in a circular path and again reaches the point A. Then its displacement is Zero,because its initial and final points are same. But, the distance traveled is d = 2r = length of the circumference of circular path, i.e the distance (d) traveled by a object may not be equal to its displacement (S). When the object moves in a curved path or zig zag path d>S.
If the object travels in a specified direction along straight line path from a point A to a point B. In this case the distance d=AB and displacement S=AD will have same magnitudes i.e d=S.
Displacement is a vector and distance is a scalar quantity.
An object started from rest,travels with uniform acceleration.Find the ratio of its displacements in ,,. . . . . seconds of its journey. ::: . . . . . . . . := 1:3:5: . . . . . . . . . : (2n-1).
If body starts with an initial velocity u, and travels with uniform acceleration a .Find the ratio of its displacements in ,,. . . . . seconds of its journey. ::: . . . . . . . . := (2u+a) : (2u+3a) : (2u+5a) : . . . . . . . . . : {2u+(2n-1)a}.

Speed (v) : The distance traveled by a body in unit time is called Speed of the body.Speed is a scalar quantity.

Speed (v) =

Velocity (v) : Displacement of a body in unit time is called Velocity (or) Rate of change of displacement of a body is called its velocity. Velocity is a vector .

Velocity (v) =

Units: Both speed and velocity have same units. C.G.S unit is cm / sec;, F.P.S unit is ft / sec; & M.K.S (or) S.I unit is m / sec.
Relation between magnitude of velocity and speed will be speed velocity.
For a object moving in a circular path with a constant speed, the magnitude of velocity remains constant.But,the direction of motion the object continuously changes.
Velocity of an object is said to be constant if a)its magnitude of velocity is constant and b)its direction of motion remains unchanged.
If the velocity of an object is varying (not constant) such object will possess acceleration, i.e the condition for an object to possess acceleration is v constant.
An object started from rest and traveling with uniform acceleration a.Find the ratio of velocities at the ,,. . . . . seconds of its journey.::: . . . . . . . . := 1:2:3: . . . . . . . . . : n.
If body starts with an initial velocity u, and travels with uniform acceleration a .Find the ratio of its velocities at ,,. . . . . seconds of its journey. ::: . . . . . . . . := (u+a) : (u+2a) : (u+3a) : . . . . . . . . . : (u+na).

Acceleration ( a ) : Rate of change of velocity of a body is known as its acceleration. (or) Change in velocity of a body in unit time is called its acceleration. Acceleration is a vector.

Acceleration in a body may be due to a) change in magnitude of velocity of a body (or) b) change in direction of velocity of the body or c) change in both magnitude and direction of velocity of the body.
Ex: For a vertically projected body or for a freely falling body the acceleration is due to change in the magnitude of velocity.
Ex:For a body moving along a circular path with constant speed, the acceleration is due to change in the direction of velocity .
Units : C.G.S unit is cm/ ; F.P.S unit is ft/ and M.K.S or S.I unit is m/ .
Direction of acceleration : a) For freely falling body the direction of velocity and acceleration are same, in this case acceleration is considered to be positive( + ) b) For a body moving up vertically , direction of acceleration will be opposite to direction velocity of the body,in this case the acceleration is considered to be negative ( - ) c)For the body in circular motion the direction of acceleration will be perpendicular to the direction of velocity at every point.
Acceleration of a body may not be zero, even if the velocity of the body is zero. Ex: For a vertically projected body , when it is at Maximum height its velocity becomes zero but its acceleration is equal to ” g “.

Freely falling body : Any object falling under the influence of gravitational force, with acceleration due to gravity is called freely falling body.

Ratio of displacements of a freely falling body after 1sec,2sec, . . . . . . . . . . . . . . . . n sec of its journey will be 1 : 4 : 9 : 16 . . . . . . . . . .
Ratio of displacements of a freely falling body 1st,2nd,3rd …… nth sec of its journey will be 1 : 3 : 5 : 7 : 9 . . . . . . . . (2n-1).
Ratio of velocity of a freely falling body after 1 sec, 2 sec, 3sec , . . . . . . . . . . . . . . of its journey will be 1 : 2 : 3 : 4 :
Ratio of velocity of a freely falling body after 1st , 2nd , 3rd , 4th . . . . . . . . . . . . . . . . . of its journey will be 1 : 1 : 1 : . . . . . . . . . . i.e its velocity in every interval is equal to “g”.
For a freely falling body velocity will continuously change (Increases) and the acceleration ( g ) remains constant.

Vertically projected up body: For a body projected vertically up with an initial velocity u,

Velocity at maximum height will be zero.
At any point during its upward journey, direction of velocity will be opposite to direction of acceleration.
Velocity of the object at any point in its path is same in magnitude,when it is going up and coming down.
It reaches ground with the velocity(u) with which it is projected.
Displacement in nth second of its up ward journey = u-
For a vertically projected up body velocity will continuously change (decrease) and the acceleration ( -g ) remains constant.
Ratio of velocity of a vertically projected up body after 1 sec, 2 sec, 3sec , . . . . . . . . . . . . . . of its journey will be :: . . . . . . . . . . . . : = (u-g) : (u-2g) : (u-3g) : . . . . . . . . : (u-ng).

Projectile : A body projected with an angle other than (≠ ) is called a projectile .

Path of a projectile is parabola.
For a projectile as no force acts on it in the horizontal direction, it does not possess acceleration in horizontal direction.
The horizontal component of velocity remains constant through-out its journey.
gravitational force acts on it vertically downwards,hence it will possess acceleration equal to acceleration due to gravity and vertical component of velocity will be different at different points.
Path of a body projected horizontally from top of a tower is a parabola.
Path of an object dropped from airplane flying at certain height will be a) parabola with respect to a stationary observer, but b ) path of that object is a straight line with respect to an observer in the airplane.
If a body is projected horizontally from the top of a tower and another body dropped freely from the same height at the same time, both will reach the ground at the same time.
At maximum height of a projectile : a)It will possess only horizontal velocity b)Horizontal component of velocity = c)vertical component of velocity is = 0 d) K.E = e)P.E = f) velocity is minimum, hence K.E also will be minimum G) P.E is maximum as it is at maximum height.

Question by gloom striken in Yahoo Answers.

2009-01-03T23:11:00.000-08:00

Q: A particle is thrown vertically upwards with speed 20 meters per second. Two seconds later, another particle is thrown vertically upwards from exactly the same place. Its initial velocity is also 20 meters per second. How long after the first particle was projected, do the two particles collide? ( Question asked by gloom striken in Yahoo answers).

Ans : Initial velocity of the 1st particle =20 m /sec assume acceleration due to gravity g = 10m/.

Maximum height traveled by this particle = = .

= ; = 20m.

Time of ascent = u/g =20/10 =2 sec.

i.e when the second object is projected from ground, first object is at maximum height .

When a body is dropped from a Maximum height or tower of height h and another body is thrown up vertically with a velocity u then they will meet after t=h/u seconds.

t =20/20=1 sec

They will collide 1sec after the II particle is projected or 3sec ater the I particle is projected.

Multiple choice questions, Units - Dimensions

2009-01-03T07:22:00.000-08:00

31 ) If the unit of length is doubled, unit of time is halved and unit of momentum is quadrupled, the unit of work would change by . . . . . times.

a) 1/8 b) 1/16 c)16 d) 8

soln: Let the original unit of work be W = ;W= [][][]

But = Therefore W= - - - - - - - - - (1)

Let the new units of work, momentum, length and time be W’,p’,L’ and T’ respectively.

Given that p’=4 ; L’ = 2 and T’= (1/2)

The new unit of work W’ = ;

Substituting the values of p’,L’ and T’ in above equation we get W’ =

W’ = 16 = 16W .

32 ) If the unit of force were 20N,that of power were 1MW and that of time were 1 millisecond then the unit of length would be

a) 20m b)50m c) 100m d) 1000m

Soln: Given that unit of power P=[] = W; F= []=20N;T=sec

P= =

Substitute the values of P,F and T in above equation = 20(L)

L= = 50m

33 ) If the unit of force is 10N,that of length is 2m and that of velocity is 100m/sec, then the unit of mass is

a)0.002kg b) 2kg c ) 20kg d) 0.2kg

Soln : Given that unit of F=10N = []; unit of length L= 2m ; unit of velocity V=100m/sec=[

The unit of F=[] = [][] [L]

F= ; M =

M = = = 0.002 kg

34) A force 100N acts on a body.If the units of mass and length are doubled and unit of time is halved,then the force in the new system changes to

a)160N b) 1.6 N c) 16N d) 1600N

Soln: Let the original unit of force F=100N=[]

Let the new unit of force,length,mass and time be F’, L’, M’ and T’ respectively.

Given that units of mass and length are doubled i.e L’=2L and M’=2M and unit of time is halved T’=T/2

The new unit of force F’=[] = [ = 16 []=16F =1600N

35) The unit of energy is 10J,if the unit of mass is tripled,unit of acceleration is doubled and the unit of length is halved. What will be the new unit of energy.

a)15J b) 30J c) 300J d) 3J

Soln: Original unit of power is E==10J

E= = =

Let the new units of mass, acceleration and length be M’,a’ and L’ respectively

Given that L’= L/2 , a’=2a and M’=3M

The new unit of energy E = = =

E = = 3 = 3E = 30J

36) The power of a moter is 150W.If the unit of force is doubled,unit of velocity is tripled what will be the new unit of power.

a) 600W b)750W c) 900W d) 300w

Soln: The original unit power of is P=150W=[]

The Unit of force F=[] and unit of velocity V=[]

P=[][] = F V

Let the new unit of power,force and velocity be P’,F’ and V’ respectively.

Given that F’ = 2F and V’ = 3V

P’ = []= []=F’V’

P’ = (2F)(3V) = 6FV = 6P =6(150) = 900W

37) The electric resistance of a conductor is 54 ohm.If the unit of mass and length are tripled, units of time and electric current are doubled.Then the value of new electric resistance.

a)540 ohm b) 1080 ohm c) 1620 ohm d)1944 ohm

Soln: The original unit of electric resistance R =[[]=54 ohm

Let the changed units of resistance,length,mass,time and current are R’,L’,M’,T’ and I’ respectively.

Given that L’=3L , M’ =3M , T’ = 2T and I’=2I

R’ = [[] ; R’ =[[]

R’ = 36[[] = 36R =36(54) = 1944 ohm.

38)The unit of angular momentum is 25 . If the momentum is doubled and length is quadrupled, what will be the new unit of angular momentum.

a)200units b) 150 units c) 400 units d)600units

Soln: let the angular momentum X = [] = 25

momentum p =[]

X =[[] ; X = p (L)

let the new Angular momentum, momentum and length be X’,p’ and L’ respectively.

Given that p’ = 2p and L’ = 4L

X’ = [] = []

X’ = p’ (L’) = (2p)(4L) = 8pL =8 X = 8(25) = 200 units

39) The ratio of C.G.S unit of gravitational constant to S.I unit is

a) b) c) d)

Soln: Let the S.I unit of gravitational constant G =[] and C.G.S unit be G’ =[] ;

M = 1000 M’ ; L= 100L’ and T = T’

Substituti0n these values in above equation we get =

= =

40 ) If the units of mass,length and time are doubled how does the unit of coefficient of linear expansion will changes

a) becomes 8times b) becomes 16 times c) becomes 1/2 d) remains same

Soln: Coefficient of linear expansion = [] .From the dimensional formula it is clear that it doen not depend on mass,length and time.even if the these physical quantities changes the Coefficient of linear expansion will remain same.

Definitions - Kinematics.

2009-01-03T01:11:00.000-08:00

Kinematics : The study of motion of bodies, does not taking in to consideration the cause for motion is called Kinematics.

Rest: If the position of an object remains to be same for any length of time with respect to its surroundings, then they object will be at rest with respect to those surroundings.

Motion: If the position of a body is continuously changing with respect its surroundings with time , then the body is said to be in motion.

Uniform motion : If a body moving along a straight line path travels equal distances in equal interval of time, then the body is said to be in uniform motion.

Non Uniform motion : If a body moving along straight line travels unequal distances in equal intervals of time, the body is said to be in Non uniform motion. (or) If a body has unequal displacements in in equal intervals of time, it will be in Non uniform motion.

Displacement ( S ) : The change in the position of a body in a specific direction, is called Displacement. Displacement is a vector quantity.

Speed (v) : The distance traveled by a body in unit time is called Speed of the body.Speed is a scalar quantity.

Speed (v) = .

Uniform speed : If the distance traveled by a body in equal intervals of time are same, for any length of time then the body is said to be in uniform speed.

Non-uniform speed : If the distance traveled by a body in equal intervals of time are different,the body is said to be in non uniform speed.

Velocity (v) : Displacement of a body in unit time is called Velocity (or) Rate of change of displacement of a body is called its velocity. Velocity is a vector .

Velocity (v) = .

Uniform velocity (v) : If the displacements of a body in equal intervals of time are same, for any length of time then the body is said to be in uniform motion.

Non-uniform velocity : If the displacements of a body in equal intervals of time are different,the body is said to be in non uniform motion.

Instantaneous Velocity: The velocity of a body at a particular instant of time is known as instantaneous velocity.

Average Velocity : If the displacement of a body in a small interval of time dt is ds then, the ratio of the displacement ds to the time interval dt is called average velocity. Average velocity v = (ds)/dt.

Initial velocity ( u ) : Velocity of the body in the beginning of the observation is called initial velocity.

Final velocity ( v ) : Velocity of the body at the end of the observation is called its final velocity.

Acceleration ( a ) : Rate of change of velocity of a body is known as its acceleration. (or) Change in velocity of a body in unit time is called its acceleration. Acceleration is a vector.

Deceleration : If the velocity of a body is continuously decreasing with time, the body will possess negative acceleration. It is also called deceleration.

Uniform acceleration: If the change (increase or decrease ) in velocity of a body is constant in equal intervals of time, the body will possess uniform acceleration.

Non uniform acceleration : If the change (increase or decrease ) in velocity of a body is different in equal intervals of time, the body will possess non-uniform acceleration.

Acceleration due to gravity ( g ): The acceleration in a freely falling body due to gravitational force acting on it is called acceleration due to gravity. This is a vector. The value of g in C.G.S system is 980 and in M.k.s & S.I systems is 9.8 .

Freely falling body : Any object falling under the influence of gravitational force, with acceleration due to gravity is called freely falling body.

Maximum Height ( H) : The distance traveled by a body vertically projected up, just before coming to rest is called, maximum height.

Time of ascent ( ) : The time taken by a vertically projected body to reach the maximum height is called time of ascent.

Time of descent ( ) : The time taken by a body to reach the ground from its maximum height is called time of descent.

Time of flight ( T ) :The total time during which the vertically projected body will remain in air is called time of flight.

( or ) Total time taken by a projectile to reach the same horizontal plane from which it is projected is called called time of flight.

Time of flight T =( ) + ( ) .

Projectile : A body projected with an angle other than (≠ ) is called a projectile .

Horizontal Range ( R ) :Maximum horizontal distance traveled by the body before it touches the ground (or) Maximum horizontal distance traveled by the body before it touches the point on the same level of projection is called horizontal range ( R ).

Important Formulas in Physics

2009-01-03T00:45:00.000-08:00

Kinematics :

- Average Speed =
- If the body covers 1st half of distance with a speed x and the second half with a speed y,then the average speed =
- If the body covers 1st 1/3rd of a distance with a speed x , and 2nd 1/3 with a speed y , and the 3rd 1/3rd distance with a speed z, then average speed =
- Average velocity =
- If a body travels a displacement in seconds and a displacement in seconds, in the same direction then Average velocity = .
- If a body travels a displacement with velocity , and displacement with velocity in the same direction then Average velocity =
- If a body travels first half of the displacement with a velocity and next half of the displacement with a velocity in the same direction , then Average velocity = .
- If a body travels a time with velocity and for a time with a velocity in the same direction then Average velocity = .
- If the body travels 1st half of the time with a velocity next half of the time with a velocity in same direction , then Average velocity =
- For a body moving with uniform acceleration if the velocity changes from u to v in t seconds, then Average velocity = (u+v)/2 .
Equations of motion of a body moving with uniform acceleration along straight line.
a) V=u+at b) S=ut+ c) - =2as
- Distance traveled in the nth second =u+a(n-1/2)
- Equations of motion for a freely falling body ( Note: we can obtain these equations by substitution of u=0 and a=g in above equations . a) v=gt b) S=1/2 c) = 2gs and the equation for the distance traveled in nth second changes to =g(n-1/2)
- Equations of motion of a body projected up vertically :(we will obtain these equations by substitution a=-g in equations of motion)
- a) v=u-gt b) S=ut-1/2 c) =-2gs and =u-g(n-1/2)
- Equation for maximum height reached =
- Time of ascent =; u
- Time of descent =; u
- Time of flight T=2u/g
- When a body is thrown up from top of a tower or released from a rising baloon,with velocity u.Displacement traveled before reaching ground S=-ut+1/2. (t= time during which the object is in the air and S=h=height of the tower).
- When a body is dropped from a tower of height h and another body is thrown up vertically with a velocity u then they will meet after t=h/u seconds.
- When a body is dropped from a tower of height h . Its velocity when it reaches ground v=
- If the displacements of a body in seconds of its journey.Then the uniform acceleration of the body a=
- From the above equation we can observe that by substituting n=1,2,3,4,…. we get a===……….. = =a.
- A body projected up with velocity u from the top of a tower reaches ground in seconds.If it is thrown down with the same velocity u it reaches ground in seconds.Then, when it is dropped freely the time taken to reach the ground will be t= and h=1/2 g and =2 .
- Projectile motion :Let us suppose that a projectile is projected with an initial velocity u making an angle with x axis. a)Horizontal component of velocity , and ,which will be constant through out the flight of the projectile as horizontal component of acceleration = 0. b)Vertical component of velocity of the projectile . Vertical component of velocity at any time of its journey =-gt or = -gt. c)Magnitude of the resultant velocity V = and the angle x made by v with the horizontal is given by =
- Time of ascent =
- Time of descent =
- Time of flight =
- Maximum height reached = ; = when u is same
- Horizontal Range R= = a)R is maximum when = b) = c) If T is the time of flight, R= d)For given velocity of projection R is same for the angles of projections and Ex: 25 and 65 i.e the Range for those two angles will be same whose sum is )

QuestionBy Creev in-Yahoo-answers

2009-01-01T07:50:00.000-08:00

Q:A car accelerates uniformly from rest to 20.0 m/s in 5.6 s along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is 9.0 x 10^3 N, and (b) the weight of the car is 1.4 x 10^4 N.

Ans : Read the problem note the given physical quantities with symbols from it. Given that Initial velocity u=o m/sec , Final velocity v=20m/sec , Time t = 5.6 sec , let g=10m/ .

Acceleration of the car a =(v-u)/t =(20-0)/5.6 =(20/5.6)m/

a) Mass of the car m=

m=(9.0 x )/10 ; m = 900 kg

Force on car F = = = N = 3214.28N

Average power P = ; P= = 64285.6watt or 64.2856 kwatt.

b)Mass of the car m = (1.4 x )/10 =1400kg

Force on car F = = = N = 5000N

Average power P = ; P= = watt or 100kwatt.

Q :Please define and/or explain impulse. Please say more than just product of F and t, and also the change of momentum. (Question by Tim C in Yahoo answers)

Ans : i) The effect of a large force acting on a body for a very short duration of time is called Impulse. ii) Change in momentum of a body is called Impulse iii ) The product of force applied on a body and the time for which it is applied is called Impulse. i.e Impulse = F x t (X is not vector product it is normal multiplication )

But if you try to understand the term Impulse from the equation Impulse = F x t , you will be confused. Because from this equation we come to a conclusion that, Impulse is directly proportional to time ( t ) i.e longer the period of time more will be the Impulse.Actually the Impulse will be less if the time period increases.

Smaller the time larger will be the impulse.

Ex: i)When a cricketer hits the ball hard, the time of contact (time of force applied ) of the ball is very, very short .Hence the effect will be large.More harder he hits the ball , the time of contact decreases further ,therefore the Impulse will further increase.

ii)when a person jumps on a hard surface from certain height,the moment his feet touches the floor they will be brought to rest, i.e the time of application of force is very small.Hence, the effect(Impulse) is high.But, if he jumps from the same height in to sand the foot will come to rest after some time, in this case even though the force is same but the time of application of force is increased.Hence, the effect of the force(Impulse) will be less.

Multiple choice questions, Units - Dimensions

2009-01-01T03:56:00.000-08:00

22.Unit used to measure nuclear diameter is

Ans : 1. picometer 2. Fermi 3. Micron 4. millimeter

23. S.I unit of Electric Intensity is

Ans : 1. coulomb / m 2. Henry 3. V / m 4. watt

24. Which is the dimensional formula of the physical quantity whose dimensional S.I unit is Siemen .

Ans : 1. 2. 3. 4.

25. Which is the physical quantity whose dimensional formula is is

Ans : 1. Bulk Modulus 2. Electric conductance 3. Surface tension 4. Frequency

26 . If is the dimensional formula of resistance, then the value of 4a+5b+c-2d

Ans : 1. 15 2. 10 3. 8 4. 12

27. If is the dimensional formula of force, find the value of 2a-b-c

Ans : 1. 8 2. -4 3. 3 4. 6

28. If is the dimensional formula of momentum, and dimensional formula of energy find the value of ax+by-cz .

Ans : 1. -3 2. -1 3. 2 4. 7

29.Which of the following is a derived unit ?

Ans : 1. ampere 2. mole 3. candela 4. newton

30. Which of the following is not a physical quantity ?

Ans : 1. kelvin 2. candela 3. henry 4. all the above

Multiple choice questions, Units - Dimensions

2008-12-30T11:13:00.000-08:00

1 . 1KWH is unit of

Ans : 1.Time 2. Power 3. Energy 4. Stress

2. Unit of Intensity of magnetic induction field is

Ans : 1.N/Am 2. Tesla 3.Wb 4. All above

3. Which of the following has no units?

Ans : 1. Thermal capacity 2. Magnetic susceptibility 3. Angular acceleration 4. Moment of a magnet

4.Which one of the following units is a fundamental unit?

Ans : 1. watt 2. joule/sec 3. ampere 4. newton

5. Fermi is equal to

Ans : 1. 1 meter 2. 100 micron 3. 1angstrom unit 4. 1 mm

6. kg m/sec is the unit of

Ans : 1. Impulse 2. Angular acceleration 3 . Capacity of condenser 4. Acceleration.

7. candela is the unit of

Ans : 1. Magnetic flux 2. Intensity of electric field 3. Luminous intensity 4. Charge

8. If 10 newton = X dynes, the value of x is

Ans : 1. 2. 3. 4.

9. 1 KWh is equal to

Ans : 1. 360 J 2. 1800 J 3.J 4. J

10. Which of the following is a common unit of a physical quantity in M.K.S & S.I systems.

Ans : 1. ampere 2.kelvin 3. mole 4. joule/sec

11. The fundamental unit which is common in F.P.S and M.K.S systems is

Ans : 1. foot 2. sec 3. kilo gram 4. pound

12. Which of the following is Unit of time?

Ans 1. Lunar Month 2. Kelvin 3. candela 4. Light year

13. Boltzman’s constant and planck’s constant differ in the dimensions of

Ans : 1. Time and temperature 2. Mass and temperature 3. Length and mass 4. Length and time .

14. Magnetic induction and magnetic flux differ in the dimensions of

Ans : 1 Time 2. Mass 3.Electric current 4.Length

15.Which of the following is a fundamental quantity in M.K.S and C.G.S systems.

16 . rad / sec is the unit of

Ans : 1.Angular displacement 2. Angular velocity 3. Angular acceleration 4. Angular momentum .

17 . The ratio of S.I unit of K.E to C.G.S unit of K.E is

Ans : 1. 2. 3. 4.

18. If and are the units of a physical quantity and ‘ are the numerical values, then

Ans : 1. 2. 3. 4. None of the above

19. which one of the following is the unit of energy.

Ans : 1. newton 2. N/sec 3. N - sec 4.None of the above

20. Which of the following is not a unit of power .

Ans : 1. Watt 2. joule/hr 3. Nm/sec 4. N/sec

21 . The physical quantity having units of mass is

Ans: 1. Density 2. Momentum 3. Inertia 4. Moment of force

units-Dimensions ( QA)

2008-12-30T09:51:00.000-08:00

23. What are the dimensions of electric conductivity in mass , length and current.

Ans : Electric conductivity has -1,-3 and 2 dimensions in mass,length and current respectively.

24. What is the unit of electric conductivity in C.G.S and S.I systems?

Ans : It has no unit in C.G.S system ; its unit in S.I system is Siemen/meter or S/m.

25.What are the uses of Dimensional methods?

Ans : To convert units from one system to another. ii )To check the correctness of equations connecting physical quantities iii )To derive the expressions connecting physical quantities.

26. Which is the physical quantity whose S.I unit is Am ?

Ans: Magnetic pole strength.

27. V/m or N/Coulomb are the units of ……. Physical quantity.

Ans : These are the units of Electric field strength.

28.Name five physical quantities which neither have dimensions nor units.

Ans : Refractive Index , specific gravity,susceptibility,dielectric constant, coefficient of friction.

29. If V = Xt+Y ; V is the velocity , t is time.What are the dimensional formulas of X and Y ?

Ans : According to principle of homogeneity of dimensions, the dimensions of M,L and T in every term should be same.

Therefore = X → X = ; X → and Y→

30.Which physical quantities does not possess dimensions in mass ?

Ans :Area,volume, velocity, acceleration,angular displacement, angular velocity, angular acceleration.

units-Dimensions ( QA)

2008-12-30T05:11:00.000-08:00

1. What is a physical quantity?

Ans : Any quantity which is measurable is called physical quantity.

2. Explain the term Fundamental Physical quantity.

Ans: The physical quantity which is independent or which can not be derived from any other physical quantity is called fundamental physical quantity. EX: Mass, Length and Time.

3.Explain the term Derived physical quantity.Give examples.

Ans :The physical quantity which is dependent on other physical quantity or which is derived from other physical quantity is called derived physical quantity. Ex : Area, Electric charge, Magnetic field strength, power etc.

4.How many fundamental quantities are there in C.G.S; F.P.S and M.K.S systems? What are they?

Ans : There are 3 fundamentals quantities in C.G.S; F.P.S and M.K.S systems, they are mass, length and time.

5.How many fundamental quantities are there in S.I systems? What are they?

Ans : In S.I system 7 fundamental quantities are there,they are i) Mass ii)Length iii)Time iv)Electric current v)Intensity of light vi) Thermodynamic temperature vii) Quantity of matter.

6.How many supplementary quantities are there in S.I system? What are they?

Ans : In S.I system there are 2 supplementary quantities, they are i) Plane Angle ii) Solid Angle.

7. What are the units of length in C.G.S ; F.P.S and M.K.S systems.

Ans : The units of length are cm,foot and meter respectively in C.G.S ; F.P.S and M.K.S systems .

8. what are the units of fundamental quantities in S.I system?

Ans : Mass → kg ; Length → m ; Time → sec ; Electric current → Amp Thermodynamic temperature → kelvin ;

Intensity of light → candela ; Quantity of matter → mole .

9.what are the units of supplementary quantities in S.I system?

Ans : Plane angle → radian ; Solid angle → steradian .

10. Name the physical quantities whose dimensional formula is ?

Ans : The physical quantities are i)Distance or length ii) displacement iii)wave length

11. Name the physical quantities whose dimensional formula is ?

Ans : The physical quantities are i) speed ii) velocity

12.Name the physical quantities whose dimensional formula is ?

Ans : The physical quantities are i)Thermal capacity ii) Entropy

13. Name the physical quantities whose dimensional formula is ?

Ans : The physical quantities are i)Momentum ii) impulse .

14.Name the physical quantities whose dimensional formula is ?

Ans : The physical quantities are i)force ii ) Tension iii) weight .

15.Name the physical quantities whose dimensional formula is ?

Ans : The physical quantities are i) Work ii) Energy iii) Heat iv)Moment of force Iv) Torque .

16.Name the physical quantities whose dimensional formula is ?

Ans : The physical quantities are i) pressure ii ) stress iii) Young’s modulus iv) Rigidity modulus v) Bulk modulus .

17.Name the physical quantities whose dimensional formula is ?

Ans : The physical quantities are i) frequency ii) Decay constant iii)Angular velocity .

18 . Name the physical quantities whose dimensional formula is ?

Ans : The physical quantities are i )angular momentum ii )Plank constant .

19. Name the physical quantities whose dimensional formula is ?

Ans : The physical quantities are i )Force constant ii )surface tension .

20. Which physical quantity has negative dimensions in mass ?

Ans : Gravitational constant (G) .

21. State few constants which have dimensions ?

Ans : i) Plnak’s constant (h) ii)Velocity of light in vacuum (c) iii)Permeability of free space () iv) Permittivity of free space () v)Universal gravitational constant (G) vi) Universal gas constant (R)

vii)Boltzmann constant (k) .

22 .which physical quantities have the unit henry ?

Ans : self Inductance and Mutual Inductance have the unit henry .

Comments (0)

Table of Units,dimensional Formulas of physical quantities.

2008-12-29T23:06:00.000-08:00

Fundamental Physical Quantities:

S.No	Fundamental Physical Quantity	Formula	Dimensional Formula	S.I Unit of physical quantity
1.	Mass	Amount of matter in the object	M	kg
2.	Length		L	meter
3.	Time		T	sec
4.	Electric current		I or A	ampere
5.	Amount of substance		N	mole(mol)
6.	Luminous intensity		J	candela(cd)
7.	Temperature		K or	Kelvin

Derived Physical Quantities:

S.No	Derived Physical Quantity	Formula	Dimensional Formula	S.I Unit of physical quantity
1.	Area		[]
2.	Volume		[]
3.	Density		[]
4.	Specific Gravity		[]	No units
5.	Frequency		[]	hertz
6.	Angle			No units
7.	Velocity			m/sec
8.	Speed			m/sec
9.	Areal velocity
10.	Acceleration
11.	Linear momentum			kg m/sec
12.	Force			kg-m/ or Newton
13.	Weight	w=mg		kg-m/ or Newton
14.	Moment of force/Torque/Couple			kg
15.	Impulse			kg m/sec or Ns
16.	Pressure			N/ or Pa
17.	Work			Nm or Joule
18.	Kinetic Energy			joule
19.	Potential Energy	mgh		joule
20.	Gravitational constant
21.	Gravitational field strength
22.	Gravitational Potential
23.	Force constant (k)
24.	Power			W or J/sec
25.	Moment of Inertia ( I )			kg
26.	Stress			N/ or Pa
27.	Strain			No units
28.	Modulus of Elasticity			N/ or Pa
29.	Poission’s Ratio	σ =-1		No units
30.	Velocity gradient
31.	Coefficient of dynamic viscosity			kg(or) N-sec/$latex \m^2$ (or)pascal-sec (or)poiseuille
32.	Surface Tension			,N/m
33.	Angular displacement ()			no Units
34.	Angular velocity(ω)			rad/sec
35.	Angular acceleration(α)			rad/
36.	Angular momentum	Iω
37.	Angular Impulse	Iω
38.	Temperature		or K	kelvin or degree Celsius
39.	Coefficient of linear expansion(α)			/kelvin
40.	Specific heat
41.	Latent heat
42.	Entropy
43.	Thermal capacity
44.	Gas constant
45.	coefficient of thermal conductivity
46.	Pole strength			Am
47.	Magnetic Moment
48.	Magnetic flux			weber ; ;J/Amp
49.	Magnetic field,magnetic flux density (B)			Tesla;
50.	Permeability of free space
51.	Magnetic susceptibilty also called volumetric or bulk susceptibility χ_m	χ_m = μ_r − 1		no units
52.	Electric Charge			Amp sec , coul
53.	Electric potential			Volt
54.	E.M.F			Volt
55.	Electric Capacity			Farad
56.	Electric Resistance			Ohm (Ω) or volt/amp
57.	Resistivity			Ohm mt (Ω-m)
58.	Conductivity	1/		Siemens/m
59.	Permittivity			farad/m
60.	Electric conductance			Siemens (or) mhos
61.	Electric power			Watt
62.	Electrical Impedance(Z)			Ohm (Ω) or volt/amp
63.	Electrical admittance	1/Z(Reciprocal of electric impedance)		Siemens (or) mhos
64.	Self Inductance(L)			weber/amp or Henry
65.	Boltzmann’s constant			J/kelvin
66.	Stefan’s constant
67.	Co-efficient of friction	=,N=Normal reaction	dimension less scalar	no units
68.	Dielectric constant	It is also called relative permittivity	dimension less	no units
69.	Planck’s constant			J.sec (or) eV.sec
70.	Refractive index	μ		no units
71.	Focal length(f)	Distance between center of the lens(mirror) to its focus	L	meter
72.	Power of a lens (P)	The reciprocal of the focal length of a lens in meters is called power of a lens; p=1/f		diaptors
73.	Wave number	No.of waves/distance
74.	Wave length	Length of a wave	L	meter

Dimensions-Dimensional formulae.

2008-12-29T22:42:00.000-08:00

Dimensions: Dimensions of a physical quantity are,the powers to which the fundamental units are raised to get one unit of the physical quantity.

The fundamental quantities are expressed with following symbols while writing dimensional formulas of derived physical quantities.

Mass →[M] ; Length→[L]; Time→[T]; Electric current →[I] ; Thermodynamic temperature →[K] ;Intensity of light →[cd] ; Quantity of matter →[mol] .

Dimensional Formula :Dimensional formula of a derived physical quantity is the “expression showing powers to which different fundamental units are raised”.

Ex : Dimensional formula of Force F →[]

Dimensional equation:When the dimensional formula of a physical quantity is expressed in the form of an equation by writing the physical quantity on the left hand side and the dimensional formula on the right hand side,then the resultant equation is called Dimensional equation.

Ex: Dimensional equation of Energy is E = [] .

Question : How can you derive Dimensional formula of a derived physical quantity.

Ans : We can derive dimensional formula of any derived physical quantity in two ways

i)Using the formula of the physical quantity : Ex: let us derive dimensional formula of Force .

Force F→ma ; substitute the dimensional formula of mass m →[M] ; acceleration →[]

we get F → [M][]; F →[] .

ii) Using the units of the derived physical quantity. Ex: let us derive the dimensional formula of momentum.

Unit of Momentum ( p ) → [] ;

kg is unit of mass → [M] ; is unit of length → [L] ; sec is the unit of time →[T]

Substitute these dimensional formulas in above equation we get p →[].

• Quantities having no units, can not possess dimensions: Trigonometric ratios, logarithmic functions, exponential functions, coefficient of friction, strain, poisson’s ratio, specific gravity, refractive index, Relative permittivity, Relative permeability. All these quantities nighter possess units nor dimensional formulas.

• Quantities having units, but no dimensions : Plane angle,angular displacement, solid angle.These physical quantities possess units but they does not possess dimensional formulas.

• Quantities having both units & dimensions : The following quantities are examples of such quantities.

Area, Volume,Density, Speed, Velocity, Acceleration, Force, Energy etc.

Physical Constants : These are two types

i) Dimension less constants (value of these constants will be same in all systems of units): Numbers, pi, exponential functions are dimension less constants.

ii)Dimensional constants(value of these constants will be different in different systems of units): Universal gravitational constant (G),plank’s constant (h), Boltzmann’s constant (k), Universal gas constant (R), Permittivity of free space() , Permeability of free space (),Velocity of light (c).

Principle of Homogeneity of dimensions: The term on both sides of a dimensional equation should have same dimensions.This is called principle of Homogeneity of dimensions. (or) Every term on both sides of a dimensional equation should have same dimensions.This is called principle of homogeneity of dimensions.

Uses of Dimensional equations : dimensional equations are used i) to convert units from one system to another,

ii)to check the correctness of the dimensional equations iii)to derive the expressions connecting different physical quantities..

Limitations of dimensional method: The limitations of dimensional methods are

i)The value of dimensionless constants can not be calculated using dimensional methods,

ii)We can not analyze the equations containing trigonometrical, exponential and logarithmic functions using method of dimensions.

iii)If a physical quantity is sum or difference of two or more than two physical quantities, such physical quantities can not be derived with dimensional methods,

iv)If any equation having dimensional constants like, G, R etc can not be derived using dimensional methods,

v)If any equation is involving more than three fundamental quantities in it, such expressions can not be derived using dimensional methods.

Special Units Of Physical Quantities.

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1. Length: a) Micron( )= m = cm

b) Angstrom (A) = m = cm

c) Fermi = m = cm

d)Astronomical Unit (A.U) = 1.5 m = 1.5 cm

e)X ray unit (X.U) = m (wave length of X-Rays)

f) Light year = Distance traveled by light in one year= m =km

g ) parsec = 3.26 light years = m

2.Time :

a ) Solar day def: The time taken by earth to complete one rotation about its own axis with respect to sun is called solar day. ( Average value for all the days of one year is Mean solar day).

b)Siderial day : It is 4.1min shorter than Mean solar day .

c )siderial year :365.26 Mean solar day d ) Solar year = 365.24 Mean solar day

e )Lear year = The year in which February month has 29 days is called leap year.It is divisible by 4.

f)Lunar month :Time taken by moon to complete one rotation around earth is lunar month = 27.3 days.

3. Mass : a ) Atomic mass Unit ( a.m.u) : = of mass of atom = = gr = gr = kg.

4.Pressure : a ) Atmosphere =760 mmHg = dyne/ = dyne/ = or pa.

b ) Bar = 750 mmHg = dyne/= dyne/ = or pa.

c) Torr =1 mm Hg = dyne/=1333 dyne/=133N/ or pa.

5.Area : Barn: this is unit of area,it is used to measure cross section of nuclei.

Barn =

6.Horse Power : It is the British Unit of power =746 w.

Multiples,sub multiples of units.

2008-12-28T21:10:00.000-08:00

Multiples and sub multiples of Units in S.I system :

Depending upon the magnitudes of physical quantities we measure, we have to use different multiplication factors suitable for that particular case.Here let us see some widely used multiplication factors.

Multiplication Factor	Prefix	symbol
	deci	d
	centi	c
	milli	m
	micro
	nano	n
	pico	p
	femto	f
	deca	da
	hecta	h
	kilo	K
	mega	M
	giga	G
	tera	T
	peta	P

Units in different systems.

2008-12-28T21:04:00.000-08:00

Generally we can use any convenient unit to measure a physical quantity depending on how much magnitude we are measuring or in which system of units we want to measure it.

What kind of unit we should use?

The unit i) must be accepted internationally.

ii) Should be reproducible.

iii) Should be invariable.

iv) Should be easily available.

v) Should be consistent.

vi) Should be large, if the physical quantity to be measured is a big quantity.

Ex: To measure larger lengths we use units like Km, mt etc, to measure large magnitude of time we use units like hour , day ,week, month , year etc.

vii) Should be small if the physical quantity to be measured is small.

Ex: To measure small time we use units like millisecond, microsecond etc

To measure small lengths we use units like millimeter, centimeter etc.

Types of physical Quantities.:

We can broadly divide the physical quantities in to two types i)Fundamental Physical quantities ii)Derived physical quantities.

Fundamental physical quantities: A physical quantity which can exist independently is called Fundamental physical quantity.

Ex: Length, mass and time etc.

Derived physical quantities: A physical quantity which can not exist independently is called derived physical quantity. (Or) A physical quantity which is dependent or derived from any other physical quantity is called derived physical quantity.

Ex : Area, volume, density, speed, acceleration, force, energy etc.

Like the physical quantities we can divide the units in to two types. I)Fundamental units ii)derived units.

Fundamental units : The units of fundamental physical quantities are called fundamental units, (or) The units which are independent or can not derived from any other unit is called fundamental unit.

Ex:Every unit of length is fundamental unit (irrespective of the system to which it belongs);millimeter, centimeter, meter, kilometer etc.

Every unit of time is a fundamental physical quantity ; microsecond, millisecond, second, minute, hour, day etc.

Derived units: The units of derived physical quantities are called derived units. Units of area, volume, speed, density, energy etc are derived units.

Ex: Every unit of speed is a derived unit ; m/sec, cm/sec, km/hr etc.

Every unit of density is a derived unit; kg/m³, gr/cm³ etc.

Every unit of acceleration is a derived unit; m/sec², cm/sec², km/hr² etc.

Systems of units: To measure the fundamental physical quantities Length, Mass and time we have three systems of units, they are i) C.G.S System (Metric system)ii)F.P.S System (British system) and iii)M.K.S System. In all these three systems only three physical quantities length, mass and time are considered to be fundamental quantities.

But, in systems International (S.I) system there are seven fundamental physical quantities. Which are i)Length ii)Mass iii)Time iv)Electric current v)Thermodynamic temperature vi)Luminous intensity vii)Quantity of substance.

In addition to these two more quantities were added as supplementary physical quantities. They are i)Plane angle ii)Solid angle.

Systems,Fundamental physical quantities and their units:In
C.G.S system: Length (centimeter); Mass (gram); Time (second).

F.P.S system :Length (foot);Mass(pound);Time (second).

C.G.S system: Length (meter); Mass (kilogram); Time (second).

S.I System:Length (meter); Mass (kilogram); Time (second); Electric current (ampere); Thermodynamic temperature (kelvin); Intensity of light (candela); Quantity of matter (mole). The units of suplimentary quantities are Plane angle( radian); Solid angle(Steradian).

What are Units.

2008-12-20T10:20:00.000-08:00

In my previous post, I explained with examples how to solve the problems in physics.

To solve problems and to under stand the basics of the Physics it is very important to know what is a physical quantity, types of physical quantities.What is a unit, what are the units of different physical quantities, types of units, symbols of units.

There is"one and only one" branch of science which measures a physical quantity, that branch of science is “Physics”. Measurements have an important role not only in physics, but also in every branch of science and everywhere in our day-to-day life.

To measure physical quantities we need units. Let’s try to understand necessity of measurements and "units of measurements" in Physics.

The information about a physical quantity, by description of its external properties like color, taste etc is incomplete with out knowing its temperature, size (dimensions), which depends on measurement, i.e. with out measurements it is impossible to know completely about the external properties of any object. So, it becomes necessary to measure it.

As we know, to measure a physical quantity we require a unit. Different physical quantities will have different units.

What is unit? A standard reference of the same physical quantity is essential to measure any physical quantity. That standard which we use to measure a physical quantity is called unit.

Let me put it this way, if we want to measure length of a table, we have to select a standard length (length of our hand), and by comparing the table’s length with the standard length we can measure the length of the table. If the table is 3.5 times that of standard length, i.e. length of our hand then we can write the result as “length of table = 3.5 times the length of our hand or 3.5 units. In this example length of hand is taken as standard length or unit to measure the table’s length.

Like that we can define any convenient standard or unit to measure a physical quantity.

But, if we choose a standard as in the above example which is not consistent, can not be reproduced. Because of such undefined units, errors and confusion in measurements will creep in. To avoid such confusion, instead of taking any undefined reference as a standard, well-defined and universal standards are used. Such a well-defined reference taken a standard is generally called a well defined unit or unit.

Measurement of every physical quantity will have two parts, a number (n) followed by a unit (u).
There fore n u = constant.

Ex: If the length of a table is 1.2 meters.In this measurement number n= 1.2 and unit is meter.

→ length (L)=n₁u₁ = 1.2 meters
→ length (L)=n₂u₂ = 120 centimeters
→ length (L)=n₃u₃= 1200 millimeters
From the above data we can understand that

i) we can measure a physical quantity in different units.what ever may be the unit it’s value is same.
→ L = n₁u₁= n₂u₂ = n₃u₃
ii) If the unit chosen smaller ,the multiple number will be greater.
→ u₁>u₁>u₁ ==> n₁<n₂<n₃
nu = constant ==> n₁u₁=n₂u₂ or u proportional to 1/n
or n proportional to 1/u

==>n1 /n2 = u2/ u1

Generally we can use any convenient unit to measure a physical quantity depending on how much magnitude we are measuring or in which system of units we want to measure it.

What kind of unit we should use?
The unit i) must be accepted internationally.ii) Should be reproducible.iii) Should be invariable.iv) Should be easily available.v) Should be consistent.vi) Should be large, if the physical quantity to be measured is a big quantity.

Ex: To measure larger lengths we use units like Km, mt etc, to measure large magnitude of time
we use units like hour , day ,week, month , year etc.

vii) Should be small if the physical quantity to be measured is small.
Ex: To measure small magnitude time, we use units like millisecond, microsecond etc
To measure small lengths we use units like millimeter, centimeter etc.
Types of physical Quantities.:
We can broadly divide the physical quantities in to two types i)Fundamental Physical quantities ii)Derived physical quantities.

Fundamental physical quantities: A physical quantity which can exist independently is called Fundamental physical quantity.
Ex: Length, mass and time etc.
Derived physical quantities: A physical quantity which can not exist independently is called derived physical quantity. (Or) A physical quantity which is dependent or derived from any other physical quantity is called derived physical quantity.
Ex : Area, volume, density, speed, acceleration, force, energy etc.

Like the physical quantities, we can Divide the units in to two types.
i)Fundamental units ii)derived units.

Fundamental units : The units of fundamental physical quantities are called fundamental units, (or) The units which are independent or can not derived from any other unit is called fundamental unit.

Ex:Every unit of length is fundamental unit (irrespective of the system to which it belongs); millimeter, centimeter, meter, kilometer etc.
Every unit of time is a fundamental physical quantity ; microsecond, millisecond, second, minute, hour, day etc.

Derived units: The units of derived physical quantities are called derived units. Ex: Units of area, volume, speed, density, energy etc are derived units.
Ex: Every unit of speed is a derived unit ; m/sec, cm/sec, km/hr etc.
Every unit of density is a derived unit; kg/m³, gr/cm³ etc.
Every unit of acceleration is a derived unit; m/sec², cm/sec², km/hr² etc.

Systems of units: To measure the fundamental physical quantities Length, Mass and time we have three systems of units, they are i) C.G.S System (Metric system) ii)F.P.S System (British system) and iii)M.K.S System. In all these three systems only three physical quantities length, mass and time are considered to be fundamental quantities.

But, in systems Internationale (S.I) system there are seven fundamental physical quantities. Which are i)Length ii)Mass iii)Time iv)Electric current v) Thermodynamic temperature vi)Luminous intensity vii)Quantity of substance.

In addition to these two more quantities were added as supplementary physical quantities. They are i)Plane angle ii)Solid angle.

Systems,Fundamental physical quantities and their units:In
C.G.S system: Length (centimeter); Mass (gram); Time (second).

F.P.S system :Length (foot);Mass(pound);Time (second).

C.G.S system: Length (meter); Mass (kilogram); Time (second).

S.I System:Length (meter); Mass (kilogram); Time (second); Electric current (ampere); Thermodynamic temperature (kelvin); Intensity of light (candela); Quantity of matter (mole). The units of suplimentary quantities are Plane angle( radian); Solid angle(Steradian).

How to solve the problems in Physics.

2008-12-19T03:14:00.000-08:00

Many of the students feel Physics as a difficult subject. Because, it involves many calculations,mathematical equations .

To solve the problems in Physics, it is essential to have knowledge of Algebra,Trigonometry,Calculus and integration.

Let me explain few steps to solve problems of physics.When a problem is given to you,
# Read the question as many times you want, until you understand it.

# Try to recollect to which branch of the Physics it belongs i.e does it belong to Kinematics or heat or electricity.

# Note down every physical quantity involved in the question, including that physical quantity which has to be calculated in the question, with their symbols Ex:Time as(t); acceleration as (a);mass as (m) .

# Check whether all the physical quantities involved in the question, are in the same system of units C.G.S or F.P.S or S.I system of unit or not.

# If all the physical quantities are not in the same system of units, then convert them in to any one system i.e into either C.G.S or S.I systems of units.

# write an equation relating all the known physical quantities and the physical quantity which has to be determined in the equation.

# Substitute the values of all the physical quantities and constants in the equation.

# Solve the equation for the unknown physical quantity.

# After getting the result write the proper units.

Let me explain with few examples:
Ex 1) A person of mass 50kg in a lift.Calculate the apparent weight of the person when he moves ( i ) up with acceleration of 3 m/sec² ( ii ) moves down with an acceleration of 4 m/sec² ,( g = 10m/sec² ).

Soln: After going through the problem, we can understand that the physical quantities involved in the problem are a)Mass of the person (m) = 50Kg ; b)acceleration of the lift (a) = 3 m/sec² c) acceleration due to gravity (g) = 10m/sec² , d) apparent mass of the person ( R ) =?

^{Part i) Now assume a formula which is connecting m,a,g and R when the lift is moving up, which is R = m(g + a), as all the given physical quantities are in the same system of units i.e S.I system we can directly substitute values in the above equation.

⇒ R = 50 ( 10 + 3) ;R=50(13) ⇒R = 650 N

ii) When the lift is moving down the formula for apparent weight is R =m(g-a)

⇒R = 50(10 - 4) ; R = 50(6) ⇒R =300N.}^{Ex 2)A 25Kg block is in motion on a rough horizontal surface.A horizontal force of 75N is required to keep the body moving with constant speed.Find the co-efficient of kinetic friction.(g=10} m/sec² ).
^{Soln: After going through the question,we can understand that the physical quantities in the problem are a) fk =75N m=25Kg g=}10 m/sec² µk =?
^{This is a problem of friction, the formula for co-efficient of kinetic friction}µk⁼^{fk/mg =75 /25 × 10

⇒}µk^{= 75/250 = 0.3}