eamcet_pb_physics_jr inter physics_01_01units and dimensions.pdf

7/27/2019 EAMCET_PB_Physics_Jr Inter Physics_01_01UNITS AND DIMENSIONS.pdf

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MEASUREMENT AND UNITS & DIMENSIONS

PREVIOUS EAMCET BITS

1. When a wave traverses a medium the displacement of a particle located at x at a time t is

given by ( )y = asin bt - cx , where a, b, c are constants of the wave, which of the following is a

quantity with dimensions ? [EAMCET 2009 E]

1) y/a 2) bt 3) cx 4) b/c

Ans: 4

Sol : 1.y Displacement L

= = = Noa amplitude L

dimensions.

2. bt = No dimension. Since angle has no dimensions.

3. cx = No dimensions. Since angle has no dimensions.

4. Angle has no dimensions, according to the principle of homogeneity of dimensions.

-1b x Lbt = cx = = = LT = velocityc t T

2. The energy [ ]E , angular momentum ( )L and universal gravitational constant ( )G are chosen as

fundamental quantities. The dimensions of universal gravitational constant in the dimensional

formula of planks constant ( )h is. [EAMCET 2008 E]

1) 0 2) -1 3)5

34) 1

Ans: 1

Sol : From the given problema b c

h = E L G Where a, b, c are constants

From the principle of homogeneity of dimensions

Dimensions of 2 -1

h = ML T

Dimensions of 2 -2

E = ML T

Dimensions of 2 -1

L = ML T

Dimensions of -1 3 -2

G = M L T

a b c2 -1 2 -2 2 -1 -1 3 -2ML T = ML T ML T M L T

.................1 a+b-cM = M a + b - c = 0 1

...................2 2a+2b=3cL = L 2a + 2b + 3c = 0 2

...............-1 -2a-b-2cT = T - 2a - b - 2c = 1 3

Solving 1, 2 & 3 we get c = 0

3. Some physics constants are given in List I and their dimensional formula are given in List II.

Match the following: [EAMCET 2007 E]

List I List II

1) Plancks constant i) 1 2ML T


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Measurement and Units & $ dimensions

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2) Gravitational constant ii) 1 1ML T

3) Bulk modulus iii)2 1

ML T

4) Coefficient of viscosity iv)1 3 2

M L T

The correct answer is

1 2 3 4 1 2 3 4

1) iv iii ii i 2) ii i iii iv

3) iii ii i iv 4) iii iv i ii

Ans: 4

Sol : 1. Plancks constant2 -2

2 -1

-1

Energy ML T= = = ML T

Frequency T

2.Gravitational constant ( ) 2 -1 3 -21 2

FrG = = M L T

m m

3.Bulk modulus -1 -2Bulk Stress F/A= = = ML TdvVolume strain

v

4.Co-efficient of viscosity F dv= from F = adv dx

A.dx

-2-1 -1

-12

MLT= = ML T

LTL

L

2006 :

4. If C, R, L and I denote capacity, resistance, inductance and electric current respectively, the

quantitates having the same dimensions of time are [EAMCET 2006 E]

a) CR b) L/R c) LC d) 2LI

1) a and b only 2) a and c only 3) a and d only 4) a, b and c only

Ans: 4

Sol : Dimensional formula of capacity -1 -2 4 2

= M L T A

Dimensional formula of Resistance 2 -3 -2= M L T A

Dimensional formula of Inductance 2 -2 -2

= M L T A

Dimensional formula of Current 0 0 0

= M L T A

a. ( )( )-1 -2 4 2 2 -3 -2CR = M L T A ML T A 0 0 0= M L TA = Time

b.2 -2 -2

2 -3 -2

L ML T A= = T = Time

R ML T A

c. ( )( )2 -2 -2 -1 -2 4 2LC = ML T A M L T A = T = Time

d. 2 2 -2 -2 2 2 -2LI = ML T A x A = ML T = energy


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5. Names of units some physical quantities are given in List I and their dimensional formulae are

given in List II. Match the correct pairs in the lists [EAMCET 2005 E]

List I List II

a) Pa s (e)

2 2 1

L T K

b) Nm K

1(f) 3 1MLT K

c) 1 1J kg K (g) 1 1ML T

d) 1 1Wm K (h) 2 2 1ML T K

a b c d a b c d

1) h g e f 2) g f h e

3) g e h f 4) g h e f

Ans: 4

Sol :

a. Dimensional formula of Pa s = [ ] -1 -2ML T T =

-1 -1ML T

b. Dimensional formula of [ ] -1 -2 -1Nm - k = MLT L K =

2 -2 -1ML T K

c. Dimensional formula of -1 -1J kg k = 2 -2 -1 -1ML T M K =

2 -2 -1L T K

d.

Dimensional formula of

-1 -1 2 -3 -1 -1

Wm k = ML T L K =

-3 -1MLT K

6. The position of a particle at time t is given by the equation x(t) = ( )At0V

1 eA

V0 = constant and

A > 0. Dimension of V0 and A respectively are [EAMCET 2004 E]

1) 1M LT and T 2) 1 2M LT and LT 3) 1M LT and T 4) 1 1M LT and T

Ans: 4

Sol : Since exponential functions has no dimensions

0 0 0 -1

AT = M L T A = T

According to the principle of homogeneity of dimensions

-10

0

Vx = V = xA = LT

A


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7. In planetary motion the areal velocity of position vector of a planet depends on angular velocity

() and the distance of the planet from sun (r). If so the correct relation for areal velocity is[EAMCET 2003 E]

1)dA

rdt 2)2

dArdt 3)

2

dArdt 4)

dArdt

Ans : 3

Sol : In the problem it is given that Areal velocitydA

dtdepends on angular frequency ( ) and distance

of the planet from the sun (r)

a bdA

rdt

Where a and b are dimensions of and r .

a bd A

= K r

d t

K is proportionality constant writing dimensional formulas on both sides [ ] a b2 -1 -1

L T = T L


2 bL = L b = 2

a-1 -aT = T = 1

8. The Vander Waals equation for a gas is ( )2a

P V b nRTV

+ =

where P, V, R, T and n

represent the pressure volume, universal gas constant, absolute temperature and number of moles

of a gas respectively, a and b are constant. The ratio b/a will have the following dimensional

formula : [EAMCET 2002 E]

1) 1 2 2M L T 2) 1 1 1M L T 3) 2 2ML T 4) 2MLT

Ans: 1

Sol : According to the principle of homogeneity of dimensions physical quantities having same

dimensional formula can be added or subtracted.

2

2 -1 -2 3

2

aP = a = Pv = ML T L

v 5

-2= ML T

33 -1 -2 +2

5 -2

b Lv = b = L = = M L T

a ML T

9. In C.G.S system the magnitude of the force is 100 dynes. In another system where thefundamental physical quantities are kilogram, meter and minute, the magnitude of the force is

[EAMCET 2001 E]

1) 0.0.36 2) 0.36 3) 3.6 4) 36

Ans: 4

Sol : From 1 1 2 2n u n u= 2 2

1 1 1 1 1 2 2 2n M L T n M L T =

2

1 1 12 1

2 2 2

M L Tn n

M L T

=


5/9


5

2

2

g cm sn 100

kg m min

=

2g cm s

100 1000g 100cm 60s

=

2n 3.6=

10. The fundamental , physical quantities that have same dimensions in the formulae of torque and

angular momentum are [EAMCET 2000 E]

1) mass, time 2) time, length 3) mass, length 4) time, mole

Ans: 3

Sol : Dimensional formula of Torque 2 -2

= ML T

Dimensional formula of angular momentum = mvr 2 -1

= ML T

Mass and length have same dimensions11. If pressure P, velocity V and time T are taken as fundamental physical quantities, the dimensionalformula of the force is [EAMCET 2000 E]

1) 2 2PV T 2) 1 2 2P V T 3) 2PVT 4) 1 2P VT

Ans: 1

Sol : From the given problem a b cFP v T

Where a, b, c are dimensionsa b cF = K P v T

K is dimensionless proportionality constant

Dimensional formula of Force

-2= MLT

Dimensional formula of Pressure -1 -2

= ML T

Dimensional formula of Velocity -1= LT

Dimensional formula of Time [ ]= T

[ ] a b c-2 -1 -2 -1

MLT = ML T LT T

[ ] [ ] -a+b -2a-b+c-2 a

MLT = M L T


1 aM = M a = 1

1 -a+bL = L - a + b = 1

-2 -2a-b+cT = T - 2a - b + c = -2

Solving a = 1, b = 2, c = 2.


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MEDICAL12. If energy E , velocity V and time T are taken as fundamental quantitates, the dimensional formula

for surface tension is [EAMCET 2009 M]

1)1 2 2

E V T

2)2 1 2

E V T

3)2 2 1

E V T

4)2 2 1

E V T

Ans 1

Sol. From the problem surface tension (s) a b cE V T a b cS KE V T=

where a, b, c are dimensions, K is a proportionality constant

( ) ( ) ( )a b c0 2 2 2 1ML T ML T LT T =

equating the powers

a = 1, b = - 2 and c = 2 1 2 2s E V T

13. If power (p)surface tension (T) and plancks constant (h) are arranged so that the dimensions oftime in their dimension formulae are in ascending order, then which of the following is correct

[EAMCET 2008 M]

1) P, T, h 2) P, h, T 3) T, P, h 4) T, h, P

Ans : 1

Sol : Dimensional formula of power 2 -3Work

= = ML Ttime

Dimensional formula of surface tension -2Force= = MT

Length

Dimensional formula of plancks constant

2 -1Energy= = ML T

frequency

Ascending order of time is -3, -2, -1

14. If force (F), work (W) and velocity (V) are taken as fundamental quantities then the dimensional

formula of time (T) is [EAMCET 2007 M]

1) 1 1 1W F v 2) 1 1 -1W F v 3) -1 -1 -1W F v 4) 1 -1 -1W F v

Ans : 4

Sol : From the given problem a b cTF w v

Where a, b, c are dimensions

a b cT = K F w v

K is a dimensionless constant

a b c-2 2 -2 -1

T = MLT ML T LT

[ ] [ ] [ ]a+b a+2b+c -2a-2b-c

T = M L T

Comparing the powers we get

a + b = 0 .....1

a + 2b+c = 0 .. 2

-2a -2b - c = 1. 3

Solving equations 1, 2 & 3 we get

a = -1, b = 1, c = -1 -1 1 -1T = F w v


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15. Some physics constants are given in List I and their dimensional formula are given in List II.

Match the following: [EAMCET 2006 M]

List I List II

1) Plancks constant i)

1 2

ML T

2) Gravitational constant ii) 1 1ML T

3) Bulk modulus iii) 2 1ML T

4) Coefficient of viscosity iv) 1 3 2M L T

The correct answer is

1 2 3 4 1 2 3 4

1) iv iii ii i 2) ii i iii iv

3) iii ii i iv 4) iii iv i ii

Ans: 4

Sol : 1. Plancks constant2 -2

2 -1

-1

Energy ML T= = = ML T

Frequency T

2. Gravitational constant ( )2

-1 3 -2

1 2

FrG = = M L T

m m

3. Bulk modulus -1 -2Bulk Stress F/A

= = = ML TdvVolume strain

v

4. Co-efficient of viscosityF dv

= from F = adv dx

A.dx

-2-1 -1

-12

MLT= = ML T

LTL

L

16. According to Bermoullis theorem2P V

ghd 2

+ + = constant. The dimensional formula of the

constant is (P = Pressure, d= density, h= height, V = velocity, g = acceleration due to gravity)

[EAMCET 2005 M]

1)0 0 0M L T

2)0 0M LT

30 2 2M L T

4)0 2 4M L T

Ans: 3

Sol : According to Bernoullis theorem2

p v+ + gh = constant

d 2

According to the principle of homogeneity of dimensions only same quantities can be added or

subtracted.

Dimensional formula ofp

=d

Dimensional formula of2

v

2

= Dimensional formula of gh

= Dimensional formula of constant


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= 0 2 -2M L T

17. The correct order in which the dimension of Length increases in the following physical

quantities is [EAMCET 2004 M]

a) permittivity b) resistance c) magnetic permeability d) stress1) a, b, c, d 2) d, c,b, a 3) a, d, c, b 4) c, b, d, a

Ans: 3

Sol : a) From coulombs law .

1 2

2

0

Q Q1F =

4 r

( ) ( )2

1 20 -2 2

IT x ITQ Q= =

F4 r MLT xL

-1 -3 4 2= M L T I

b) From ohms law =v w w

v = iR R = =i Q i it.i

2 -2

2ML T=

I T

2 -3 -2= ML T I

c) From coulombs inverse square law

2

0 1 202

1 2

m m Fd 4F = =

4 d m m

Where1 2

m ,m = pole strength = [ ]I L

-2 2-2 -2

0 2 2

MLT xL = = MLT II L

d) Stress

-2-1 -2

2

Force MLT= = = ML T

area L

As < 3 < -1 < 1 < 2 A, D, C, B18. The dimensional equation for magnetic flux is [EAMCET 2003 M]

1) 2 2 1ML T I 2) 2 2 2ML T I 3) 2 2 1ML T I 4) 2 2 2ML T I Ans: 1

Sol : Magnetic flux = BA but F = mB orF

B =m

=-2 2FA Force x Area MLT x L

= =m PoleStrength I L

2 -2 -1

= ML T I

19. The dimensional formula of coefficient of kinematic viscosity is [EAMCET 2002 M]

1) 0 1 1M L T 2) 0 2 1M L T 3) 2 1ML T 4) 1 1ML T

Ans: 2

Sol : Coefficient of kinematic viscosity = 0 2 1M L T


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20. The dimensional formula of the product of two physical quantities P and Q is 2 2ML T . The

dimensional formula of P/Q is MT2

. P and Q respectively are [EAMCET 2001 M]

1) Force, Velocity 2) Momentum , Displacement

3) Force , Displacement 4)Work, VelocityAns: 3

Sol : Check all the options which satisfies 2 -2PQ = ML T and -2P / Q = M L

1. Let force = P Velocity = Q

-2 -1 2 -3PQ = MLT xLT =ML T 2. PQ = Momentum x displacement

-1 2 -1= MLT xL = ML T

3. PQ = force x displacement = -2 2 -2

MLT xL = ML T

-2-2P MLT

= = M TQ L

As both the conditions are satisfiedP = Force Q = displacement

21. The physical quantity which has dimensional formula as that ofEnergy

mass lengthis [EAMCET 2000M]

1) Force 2) Power 3) Pressure 4) Acceleration

Ans: 4

Sol :2 -2

-2Energy ML T= = L T =

mass x length M x L

acceleration

22. The dimensional formula of magnetic induction is [EAMCET 2000 M]

1) 1 1MT A 2) 2 1MT A 3) 1MLA 4) 2MT A

Ans: 2

Sol : From -2

-2 -1F Force MLTF = mB B = = = = MT A

m pole strength AL

777

eamcet_pb_physics_jr inter physics_01_01units and dimensions.pdf

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