1. If C and R de note ca pac i tance and re sis tance

respectively, the di men sional for mula of CR is

(a) [M L T0 0 ] [1988]

(b) [M L T0 0 0]

(c) [M L T0 0 1− ]

(d) Not expressible in terms of [MLT]

2. The di men sional for mula for an gu lar

mo men tum is [1988]

(a) [M L T0 2 2− ] (b) [M T2 1L − ]

(c) [ ]MLT−1 (d) [ML T2 2− ]

3. Of the fol low ing quan ti ties, which one has

di men sions dif fer ent from the re main ing

three ? [1989]

(a) Energy per unit volume

(b) Force per unit area

(c) Product of voltage and charge per unit

volume

(d) Angular momentum

4. Di men sional for mula of self-inductance is

(a) [MLT A2 2− − ] (b) [ML T A2 1 2− − ] [1989]

(c) [ML T A2 2 2− − ] (d) [ML T A2 2 1− − ]

5. If x at bt= + 2, where x is the dis tance trav elled

by the body in ki lo me ter while t is the time in

sec ond, then the unit of b is [1989]

(a) km/s (b) km-s

(c) km/s2 (d) km- s2

6. The di men sional for mula of torque is [1989]

(a) [ML T2 2− ] (b) [MLT 2− ]

(c) [ML T1 2− − ] (d) [ML T2 2− − ]

7. The di men sional for mula of pres sure is [1990]

(a) [MLT 2− ] (b) [ML T1 2− ]

(c) [ML T1 2− − ] (d) [MLT 2− ]

8. Ac cord ing to New ton, the vis cous force act ing

be tween liq uid lay ers of area A and ve loc ity

gra di ent ∆∆

v

z is given by F A

dv

dz= − η , where η is

con stant called co ef fi cient of vis cos ity. The

di men sional for mula of η is [1990]

(a) [ML T2 2− − ] (b) [M L T0 0 0]

(c) [ML T2 2− ] (d) [ML T1 1− − ]

9. The fre quency of vi bra tion f of a mass m

sus pended from a spring of spring con stant k is

given by a re la tion of the type f Cm kx y= , where

C is a di men sionless con stant. The values of x

and y are [1990]

(a) x y= =1

2

1

2, (b) x y= − = −1

2

1

2,

(c) x y= = −1

2

1

2, (d) x y= − =1

2

1

2,

10. A cer tain body weighs 22.42 g and has a

mea sured vol ume of 4.7 cc. The pos si ble er ror

in the measurement of mass and vol ume are

0.01 g and 0.1 cc. Then max i mum er ror in the

den sity will be [1991]

(a) 22% (b) 2%

(c) 0.2% (d) 0.02%

11. The di men sional for mula for per me abil ity of

free space, µ0 is [1991]

(a) [MLT A2 2− − ] (b) [ML T A1 2 2− − ]

(c) [ML T A1 2− −2 ] (d) [MLT A2− −1]

12. The di men sions of grav i ta tional con stant G are

(a) [MLT 2− ] (b) [ML T3 2− ] [1992]

(c) [M L T3 2− −1 ] (d) [M L T2 3− −1 ]

13. If p rep re sents ra di a tion pres sure, c rep re sents

speed of light and S rep re sents ra di a tion en ergy

strik ing unit area per sec. The non-zero in te gers

x y z, , such that p S cx y z is dimensionless are

(a) x y z= = =1 1 1, , [1992]

(b) x y z= − = =1 1 1, ,

(c) x y z= = − =1 1 1, ,

(d) x y z= = = −1 1 1, ,

14. The time de pend ence of phys i cal quan tity p is

given by p p= 0 exp ( )− αt 2 , where α is a

con stant and t is the time. The con stant α(a) is dimensionless [1992]

(b) has dimensions [ ]T−2

(c) has dimensions [ ]T2

(d) has dimensions of p

15. Tur pen tine oil is flow ing through a tube of

length l and ra dius r. The pres sure dif fer ence

be tween the two ends of the tube is p. The

vis cos ity of oil is given by

Physical World and Measurement

1

Physical World and Measurement

η = −p r x

vl

( )2 2

4

where v is the velocity of oil a distance x from

the axis of the tube. The dimensions of η are

(a) [M L T0 0 0] (b) [MLT 1− ] [1993]

(c) [ML T2 2− ] (d) [ML T1 1− − ]

16. In a par tic u lar sys tem, the unit of length, mass

and time are cho sen to be 10 cm, 10 g and 0.1 s

re spec tively. The unit of force in this sys tem will

be equiv a lent to [1994]

(a) 0.1 N (b) 1 N

(c) 10 N (d) 100 N

17. In a ver nier calli pers N di vi sions of ver nier scale

co in cide with N − 1 di vi sions of main scale (in

which length of one di vi sion is 1 mm). The least

count of the in stru ment should be [1994]

(a) N (b) N − 1

(c) 1

10N(d)

1

1( )N −

18. Which of the fol low ing is a di men sional

con stant ? [1995]

(a) Refractive index

(b) Poisson’s ratio

(c) Relative density

(d) Gravitational constant

19. The per cent age er rors in the measurement of

mass and speed are 2% and 3% re spec tively.

The er ror in ki netic en ergy ob tained by

mea sur ing mass and speed, will be [1995]

(a) 12% (b) 10%

(c) 8% (d) 2%

20. An equa tion is given as pa

Vb

V+

=

2

θ, where

p = pres sure, V = vol ume and θ = ab so lute

tem per a ture. If a and b are con stants, then

di men sions of a will be [1996]

(a) [ML T ]5 2− (b) [M L T ]1 5 2−

(c) [ML T5 1− − ] (d) [ML T]5

21. The den sity of a cube is mea sured by mea sur ing

its mass and length of its sides. If the max i mum

er ror in the measurement of mass and length

are 4% and 3% re spec tively, the max i mum er ror

in the mea sure ment of den sity will be [1996]

(a) 7% (b) 9%

(c) 12% (d) 13%

22. Which of the fol low ing will have the di men sions

of time ? [1996]

(a) LC (b) R

L

(c) L

R(d)

C

L

23. The force F on a sphere of radius r moving in a

medium with velocity v is given by F rv= 6π η .

The dimensions of η are [1997]

(a) [ML ]–3 (b) [MLT ]–2

(c) [MT ]–1 (d) [ML T ]–1 –1

24. The dimensional formula for magnetic flux is

(a) [ML T A2 –2 –1] (b) [ML T A ]3 –2 –2[1999]

(c) [M L T A0 –2 2 –2] (d) [ML T A ]2 –1 2

25. A pair of physical quantities having samedimensional formula is [2000]

(a) force and torque(b) work and energy(c) force and impulse(d) linear momentum and angular momentum

26. Planck’s constant has the dimensions of [2001]

(a) linear momentum(b) angular momentum(c) energy(d) power

27. The value of Planck’s con stant is [2002]

(a) 6 63 10 31. × − J-s

(b) 6 63 10 30. /× − kg-m s

(c) 6 63 10 32 2. × − kg-m

(d) 6 63 10 34. × − J-s

28. The unit of permittivity of free space, ε0 is(a) coulomb/newton-metre [2004]

(b) newton-metre /coulomb2 2

(c) coulomb / newton-metre2 2

(d) coulomb / (newton-metre)2 2

29. The dimensions of universal gravitationalconstant are [2004]

(a) [ ]M L T− −1 3 2 (b) [ ]ML T2 1−

(c) [ ]M L T− −2 3 2 (d) [M L T− −2 2 1]

30. The ra tio of the di men sions of Planck’s con stantand that of the mo ment of in er tia is thedi men sion of [2005]

(a) frequency(b) velocity(c) angular momentum (d) time

31. The ve loc ity v of a par ti cle at time t is given by

vb

t c= +

+at , where a b, and c are con stants.

2

Physical World and Measurement

The di men sions of a b, and c are respectively(a) [LT ],–2 [L] and [T] [2006]

(b) [L ],[T] and [LT2 2]

(c) [LT ], [LT] and [L]2

(d) [L], [LT] and [T2]

32. Di men sions of re sis tance in an elec tri cal cir cuit,in terms of di men sion of mass M, of length L, oftime T and of cur rent I, would be [2007]

(a) [ ]ML T I2 3 1− − (b) [ ]ML T2 2−

(c) [ ]ML T I2 1 1− − (d) [ ]ML T I2 3 2− −

33. If the error in the measurement of radius of asphere is 2%, then the error in thedetermination of volume of the sphere will be(a) 4% (b) 6% [2008]

(c) 8% (d) 2%

34. Which two of the following five physicalparameters have the same dimensions ? (i) Energy density (ii) Refractive index(iii) Dielectric constant(iv) Young’s modulus (v) Magnetic field [2008]

(a) (ii) and (iv) (b) (iii) and (v)(c) (i) and (iv) (d) (i) and (v)

35. If the dimensions of a physical quantity aregiven by M L Ta b c, then the physical quantity willbe [2009]

(a) pressure if a b c= = − = −1 1 2, ,(b) velocity if a b c= = = −1 0 1, ,(c) acceleration if a b c= = = −1 1 2, ,(d) force if a b c= = − = −0 1 2, ,

n Answers

n Hints & Solutions

1. Q Cq

V

q

W

q

q

W= = =

2

=⋅

= −( )

]

it

F x

2 [AT]

[ML T

2

2 2

= − −[M L T A ]1 2 4 2

and RV

i

W

qi

F x

i t= = = ⋅

2

=−[ML T

[AT][A]

2 2] = − −[ML T A ]2 3 2

∴ Dimensional formula of CR

= − − − −[M L T A ][ML T A1 2 4 2 2 3 2]

=[M L T]0 0

2. An gu lar mo men tum

L r p= × = ×r mv

∴ Dimensional formula for angular momentum

= −[L][M][LT 1]

= −[ML T2 1]

3. Di men sions of en ergy per unit volume

= dimensions of energy

dimensions of volume

= =−

− −[ ]]

ML T

[L ][ML T

31 2

2 2

Dimensions of force per unit area

= dimensions of force

dimensions of area

=−[MLT

[L

2

2

]

] = − −[ML T1 2]

Voltage × charge/volume

=

×W

qit

l

( )

3

= =−( )

( )

]W

l3[ML T

[L ]

2 2

3

= − −[ML T1 2]

Angular momentum

= ( )( )r p

= ( )( )r mv

= −[L][M][LT 1]

= −[ML T2 1]

So, dimensions of angular momentum is

different from other three.

3

1. (a) 2. (b) 3. (d) 4. (c) 5. (c) 6. (a) 7. (c) 8. (d) 9. (d) 10. (b)

11. (a) 12. (c) 13. (c) 14. (b) 15. (d) 16. (a) 17. (c) 18. (d) 19. (c) 20. (a)

21. (d) 22. (c) 23. (d) 24. (a) 25. (b) 26. (b) 27. (d) 28. (c) 29. (a) 30. (a)

31. (a) 32. (d) 33. (b) 34. (c) 35. (a)

4. From e Ldi

dt=

Ledt

di

W

q

dt

di= = ⋅

=−[ML T ][T]

[AT][A]

2 2

= − −[ML T A2 2 2]

5. As x at bt= + 2

∴ unit of x = unit of bt 2

∴ unit of bx

t= unit of

unit of 2

= km/s2

6. Torque τ = ×r F

Dimensions of τ = dimension of r × dimensions of F

= −[L][MLT 2]

= −[ML T2 2]

7. Pres sure = force

area

= =−F

A

[MLT

[L

2

2

]

]

= − −[ML T1 2]

8. As F Adv

dz= − η

∴ η = −

F

Adv

dz

As F = −[MLT 2], A =[ ]L2

dv = −[LT 1], dz =[L]

∴ η =−

−[MLT ][L]

[L ][LT

2

2 1]

= − −[ML T1 1]

9. As f Cm kx y=

∴ (Dimension of f ) = C (dimension of m x)

× (dimensions of k y)

[T ] [M MT1− −= C x y] [ ]2 …(i)

where force

lengthk =

Applying the principle of homogeneity ofdimensions, we get

x y+ = 0, − = −2 1y

or y = 1

2

∴ x = − 1

2

10. Den sity = mass

volume

ρ = m

V

∴ ∆ ∆ ∆ρρ

= +m

m

V

V

Here, ∆m = 0.01, m = 22.42

∆V = 0.1, V = 4.7

∴ ∆ρρ

= +

×0.01

22.42

0 1

4 7100

.

. = 2%

11. From Biot-Savart’s law

dBIdl

r= µ

πθ0

24

sin

µ πθ0

24= r dB

Idl

( )

sin

=− −[L ][MT A ]

[A][L]

2 2 1

= − −[MLT A2 2]

12. As F Gm m

r= 1 2

2

∴ GFr

m m=

2

1 2

=−[MLT ][L ]

[M ]

2 2

2

= − −[M L T1 3 2]

13. p = − −[ML T1 2]

c = −[LT 1]

S = =−

−[ML T

[L T][MT

2 2

23]]

Now, p S cx y z is dimensionless

∴ [M L T0 0 0] = p S cx y z

or [M L T ] [M L T ] [M T ] [L T ]0 0 0 1 1 2 1 3 1 1= − − − −x y z

or [M L T ] [M] [L] [T]0 0 0 = + − + − − −x y x z x y z2 3

From principle of homogeneity of dimensions

x y+ = 0 …(i)

− + =x z 0 …(ii)

− − − =2 3 0x y z …(iii)

Solving Eqs. (i), (ii) and (iii), we get

x y z= = − =1 1 1, ,

4

14. p p t= −02exp ( )α

As powers of exponential quantity is dimension- less, so αt 2 is dimensionless.

or αt 2 = dimensionless =[M L T0 0 0]

∴ α = = = −1 12 2

2

t [ ][ ]

TT

15. Pressure = = =−

− −force

area

[MLT

[L[ML T1 2

2

2

]

]]

v = −[LT 1]

From principle of homogeneity, the dimensionsof r2 and x2 is same.

The dimensions of viscosity

=− −

−[ML T ][L ]

[LT ][L]

1 2 2

1

= − −[ML T1 1]

16. Force F = −[MLT 2]

= −( )( )( )10 10 2g cm 0.1 s

Changing these units into MKS system

F = − − − −( )( )( )10 10 102 1 1 2kg m s

= −10 1 N

17. N NVSD MSD= −( )1

11

VSD MSD= −

N

N

LC = least count = −1 1MSD VSD

LC MSD= − −

1

1N

N

= =1

N NMSD

0.1cm = 1

10cm

N

18. A quan tity which has di men sions and also has acon stant value is called di men sional con stant.

Therefore, gravitational constant ( )G is adimensional constant.

19. K mv= 1

22

∴ ∆ ∆K

K

m

m× = ×100 100 + × ×2 100

∆v

v

Here,∆m

m× =100 2%

∆v

v× =100 3%

∴ ∆K

K× = + ×100 2 2 3% %

= 8%

20. From prin ci ple of ho mo ge ne ity of di men sions.

Dimensions of p = dimensions of a

V2

pa

V=

2

∴ a pV= 2

= − −[ML T ] L1 2 [ ]3 2

= −[ML T5 2]

21. As den sity ρ = =m

V

m

l3

∴ ∆ ∆ ∆ρρ

× = ± +

×100 3 100

m

m

l

l%

= ± + ×( )4 3 3 = ± 13%

22.L

R is time con stant of RL cir cuit so, di men sions

of L

R is of same as that of time.

Alternative

Dimensions of

Dimensions of

[ML T A

[ML T A

2 2 2

2 3

L

R=

− −

−]

− =2

T]

[ ]

23. Vis cous force on a sphere of ra dius r is

F rv= 6 πη

∴ ηπ

= F

rv6

[ ][ ]

[ ][ ]η = F

r v

= [MLT ]

[L][LT ]=[ML T ]

–2

–1–1 –1

NOTE The above expression is the Stokes’ law.

24. Mag netic flux ( )φ through a sur face of area (A) is

the to tal num ber of mag netic lines of in duc tionpass ing through that area nor mally.Math e mat i cally, mag netic flux

φ = BA …(i)

but magnetic force

F Bil=

or BF

il=

Putting the value of B in Eq. (i), we have

φ = F

ilA

Thus, dimensions of φ = [MLT ][L ]

[AL]

–2 2

=[ML T A ]2 –2 –1

5

25. (a) Force = mass × ac cel er a tion

or F ma= =[M][LT ]–2

=[MLT ]–2

Torque = moment of inertia

× angular acceleration

or τ α= ×I

=[ML ][T ]2 –2

=[ML T ]2 –2

(b) Work = force × displacement

or W F d= × =[MLT ][L]–2 =[ML T ]2 –2

Energy = × ×1

2mass (velocity)2

or K mv= 1

22

=[M][LT ]–1 2

=[ML T ]2 –2

(c) Force as discussed above

[ ]F =[MLT ]–2

Impulse = force × time-interval

∴ [ ]I =[MLT ][T]–2

=[MLT ]–1

(d) Linear momentum = mass × velocity

or p mv=

∴ [ ]p =[M][LT ]–1

=[MLT ]–1

Angular momentum = moment of inertia

× angular velocity

or [L] = ×[ ] [ ]I ω∴ [ ]L [ML ][T ]2 –1=

=[ML T ]2 –1

Hence, we observe that choice (b) is correct.

NOTE In this problem, the momentum of inertia and impulseare given same symbol l.

26. E h= ν

⇒ h = Planck’s constant = E

ν

∴ [ ]hE= =

−

−ν[ML T ]

[T ]

2 2

1

=[ML T2 –1]

(a) Linear momentum = mass× velocity

or p m v= ×

= −[M] LT[ ]1

= −[ ]MLT 1

(b) Angular momentum=moment of inertia

× angular velocity

or L I= × ω = mr2ω [ Q I mr= 2]

∴ [ ]L = =− −[M][L ][T ] [ML T ]2 21 1

(c) Energy [ ]E =[ML T–22 ]

(d) Power = force × velocity

or P F v= ×∴ [ ]P = −[MLT ][LT ]–2 1

=[ML T2 –3]

Hence, option (b) is correct.

NOTE According to homogeneity of dimensions, thedimensions of all the terms in a physical expression should be

same. For example, in the physical expression s ut at= + 1

2

2,

the dimensions of s ut, and 1

2

2at all are same.

27. The value of Planck’s con stant is 6 63 10 34. × − J-s.

28. By Coulomb’s law, the electrostatic force

Fq q

r= ×1

4 0

1 22πε

⇒ επ0

1 22

1

4= × q q

r F

Substituting the units for q r, and F, we obtain

unit of

ε0 = ×coulomb coulomb

newton – (metre)2

= (coulomb)

newton – (metre)

2

2

= C /N-m2 2

29. According to Newton’s law of gravitation, the

force of attraction between two masses m1 and

m2 separated by a distance r is,

FG m m

r= 1 2

2 ⇒ G

Fr

m m=

2

1 2

Substituting the dimensions for the quantities

on the right hand side, we obtain

dimensions of G =−[ ][ ]

[ ]

MLT L

M

2 2

2

= − −[ ]M L T1 3 2

6

30. E h= ν

⇒ h = Planck’s constant = E

ν

∴ [ ]h = [ML T ]

[T ]

2 –2

–1

=[ML T ]2 –1

and I = moment of inertia = MR 2

⇒ [ ]I =[ML ]2

Hence, [ ]

[ ]

h

I= [ML T ]

[ML ]=[T ]

2 –1

2–1

= =1

[T]dimension of frequency

Al ter na tive h

I

E

I= /ν

= × = ×E T

I

(kg-m /s ) s

(kg-m

2 2

2)

= =1 1

s time

= frequency

Thus, dimensions of h

I is same as that of

frequency.

31. The given expression is

v atb

t c= +

+

From principle of homogeneity

[ ][ ] [ ]a t v=

[ ][ ]

[ ]a

v

t= = =[LT ]

[T] [LT ]

–1–2

Similarly, [ ] [ ] [ ]c t= = T

Further, [ ]

[ ][ ]

b

t cv

+=

or [ ] [ ][ ]b v t c= +or [ ]b = = [LT ] [T] [L]–1

NOTE If a physical quantity depends on more than threefactors, then relation among them cannot be establishedbecause we can have only three equations by equalising thepowers of M, L and T.

32. Re sis tance

RV

i

W

qi= = =potential difference

current

(Q Potential difference is equal to work done per unit charge)

So, Dimensions of R

=×

dimensions of work

dimensions of charge dimensions of current

= =−

− −[ML T

IT I[ML T I

2 22 3 2]

[ ][ ]]

33. Volume of a sphere, V r= 4

33π

∴ ∆ ∆V

V

r

r× = × ×100

3100

Here ∆r

r× =100 2%

∴ ∆V

V× = × =100 3 2 6% %

34. Energy density = energy

volume

uE

V=

Dimensions of uE

V= dimensions of

dimensions of

= =−

− −[ ]

[ ][ ]

ML T

LML T

2 2

31 2

Refractive index is a dimensionless quantity.Dielectric constant is a dimensionless quantity.

Young’s modulus = longitudinal stress

longitudinal strain

= − −[ ]ML T1 2

Magnetic field =×

Force

charge velocity

= F

qv

=−

−[ ]

[ ][ ]

MLT

AT LT

2

1

= − −[ ]MT A2 1

35. (i) Dimensions of velocity = −[M L T ]0 1 1

Here, a b c= = = −0 1 1, ,

(ii) Dimensions of acceleration = −[M L T ]0 1 2

Here, a b c= = = −0 1 2, ,

(iii) Dimensions of force = −[ ]M L T1 1 2

Here, a b T= = = −1 1 2, ,

(iv) Dimensions of pressure = − −[ ]M L T1 1 2

∴ Here, a b c= = − = −1 1 2, ,

∴ The physical quantity is pressure.

7