Chapter 1 Units and Measurements – Physics free study material by TEACHING CARE online tuition and coaching classes
Physical Quantity.
A quantity which can be measured and by which various physical happenings can be explained and expressed in form of laws is called a physical quantity. For example length, mass, time, force etc.
On the other hand various happenings in life e.g., happiness, sorrow etc. are not physical quantities because these can not be measured.
Measurement is necessary to determine magnitude of a physical quantity, to compare two similar physical quantities and to prove physical laws or equations.
A physical quantity is represented completely by its magnitude and unit. For example, 10 metre means a length which is ten times the unit of length 1 kg. Here 10 represents the numerical value of the given quantity and metre represents the unit of quantity under consideration. Thus in expressing a physical quantity we choose a unit and then find that how many times that unit is contained in the given physical quantity, i.e.
Physical quantity (Q) = Magnitude × Unit = n × u
Where, n represents the numerical value and u represents the unit. Thus while expressing definite amount of physical quantity, it is clear that as the unit(u) changes, the magnitude(n) will also change but product ‘nu’ will remain same.
i.e. n u = constant, or
n u = n u = constant ; \ n µ 1
1 1 2 2 u
i.e. magnitude of a physical quantity and units are inversely proportional to each other .Larger the unit, smaller will be the magnitude.
Types of Physical Quantity.
 Ratio (numerical value only) : When a physical quantity is a ratio of two similar quantities, it has no
e.g. Relative density = Density of object/Density of water at 4^{o}C Refractive index = Velocity of light in air/Velocity of light in medium
Strain = Change in dimension/Original dimension
Note : @Angle is exceptional physical quantity, which though is a ratio of two similar physical quantities (angle = arc / radius) but still requires a unit (degrees or radians) to specify it along with its numerical value.
 Scalar (Magnitude only) : These quantities do not have any direction g. Length, time, work, energy etc.
Magnitude of a physical quantity can be negative. In that case negative sign indicates that the numerical value of the quantity under consideration is negative. It does not specify the direction.
Scalar quantities can be added or subtracted with the help of following ordinary laws of addition or subtraction.
 Vector (magnitude and direction) : g. displacement, velocity, acceleration, force etc.
Vector physical quantities can be added or subtracted according to vector laws of addition. These laws are different from laws of ordinary addition.
Note : @There are certain physical quantities which behave neither as scalar nor as vector. For example, moment of inertia is not a vector as by changing the sense of rotation its value is not changed. It is also not a scalar as it has different values in different directions (i.e. about different axes). Such physical quantities are called Tensors.
Fundamental and Derived Quantities.
 Fundamental quantities : Out of large number of physical quantities which exist in nature, there are only few quantities which are independent of all other quantities and do not require the help of any other physical quantity for their definition, therefore these are called absolute These quantities are also called fundamental or base quantities, as all other quantities are based upon and can be expressed in terms of these quantities.
 Derived quantities : All other physical quantities can be derived by suitable multiplication or division of different powers of fundamental These are therefore called derived quantities.
If length is defined as a fundamental quantity then area and volume are derived from length and are expressed in term of length with power 2 and 3 over the term of length.
Note : @ In mechanics Length, Mass and time are arbitrarily chosen as fundamental quantities. However this set of fundamental quantities is not a unique choice. In fact any three quantities in mechanics can be termed as fundamental as all other quantities in mechanics can be expressed in terms of these. e.g. if speed and time are taken as fundamental quantities, length will become a derived quantity because then length will be expressed as Speed ´ Time. and if force and acceleration are taken as fundamental quantities, then mass will be defined as Force / acceleration and will be termed as a derived quantity.
Fundamental and Derived Units.
Normally each physical quantity requires a unit or standard for its specification so it appears that there must be as many units as there are physical quantities. However, it is not so. It has been found that if in mechanics we choose arbitrarily units of any three physical quantities we can express the units of all other physical quantities in mechanics in terms of these. Arbitrarily the physical quantities mass, length and time are choosen for this purpose. So any unit of mass, length and time in mechanics is called a fundamental, absolute or base unit. Other units which can be expressed in terms of fundamental units, are called derived units. For example light year or km is a fundamental units as it is a unit of length while s^{–1}, m^{2} or kg/m are derived units as these are derived from units of time, mass and length respectively.
System of units : A complete set of units, both fundamental and derived for all kinds of physical quantities is called system of units. The common systems are given below –
 CGS system : The system is also called Gaussian system of In it length, mass and time have been chosen as the fundamental quantities and corresponding fundamental units are centimetre (cm), gram (g) and second (s) respectively.
 MKS system : The system is also called Giorgi system. In this system also length, mass and time have been taken as fundamental quantities, and the corresponding fundamental units are metre, kilogram and
 FPS system : In this system foot, pound and second are used respectively for measurements of length, mass and In this system force is a derived quantity with unit poundal.
 I. system : It is known as International system of units, and is infact extended system of units applied to whole physics. There are seven fundamental quantities in this system. These quantities and their units are given in the following table
Quantity  Name of Unit  Symbol 
Length  metre  m 
Mass  kilogram  kg 
Time  second  s 
Electric Current  ampere  A 
Temperature  Kelvin  K 
Amount of Substance  mole  mol 
Luminous Intensity  candela  cd 
Besides the above seven fundamental units two supplementary units are also defined – Radian (rad) for plane angle and Steradian (sr) for solid angle.
Note : @Apart from fundamental and derived units we also use very frequently practical units. These may be fundamental or derived units
e.g., light year is a practical unit (fundamental) of distance while horse power is a practical unit (derived) of power.
@ Practical units may or may not belong to a system but can be expressed in any system of units
e.g., 1 mile = 1.6 km = 1.6 × 10^{3} m.
S.I.
Prefixes.
In physics we have to deal from very small (micro) to very large (macro) magnitudes as one side we talk about the atom while on the other side of universe, e.g., the mass of an electron is 9.1 ´ 10^{–31} kg while that of the sun is 2 ´ 10^{30} kg. To express such large or small magnitudes simultaneously we use the following prefixes :
Power of 10  Prefix  Symbol 
1018  exa  E 
1015  peta  P 
1012  tera  T 
10^{9}  giga  G 
10^{6}  mega  M 
10^{3}  kilo  k 
10^{2}  hecto  h 
10^{1}  deca  da 
10–1  deci  d 
10–1  centi  c 
10–3  milli  m 
10–6  micro  m 
10–9  nano  n 
10–12  pico  p 
10–15  femto  f 
10–18  atto  a 
Standards of Length, Mass and Time.
 Length : Standard metre is defined in terms of wavelength of light and is called atomic standard of
The metre is the distance containing 1650763.73 wavelength in vacuum of the radiation corresponding to orange red light emitted by an atom of krypton86.
Now a days metre is defined as length of the path travelled by light in vacuum in 1/299,7792, 458 part of a second.
 Mass : The mass of a cylinder made of platinumiridium alloy kept at International Bureau of Weights and Measures is defined as 1 kg.
On atomic scale, 1 kilogram is equivalent to the mass of 5.0188 ´ 10^{25} atoms of _{6}C^{12} (an isotope of carbon).
 Time : 1 second is defined as the time interval of 9192631770 vibrations of radiation in Cs133 This radiation corresponds to the transition between two hyperfine level of the ground state of Cs133.
Practical Units.
(1) Length :
 1 fermi = 1 fm = 10^{–15} m
 1 Xray unit = 1XU = 10^{–13} m
(iii) 1 angstrom = 1Å = 10^{–10} m = 10^{–8} cm = 10^{–7} mm = 0.1 mmm
(iv) 1 micron = mm = 10^{–6} m
(v) 1 astronomical unit = 1 A.U. = 1. 49 ´ 10^{11} m » 1.5 ´ 10^{11} m » 10^{8} km
 1 Light year = 1 ly = 46 ´ 10^{15} m
 1 Parsec = 1pc = 26 light year
(2) Mass :
 Chandra Shekhar unit : 1 CSU = 4 times the mass of sun = 2.8 ´ 10^{30} kg
 Metric tonne : 1 Metric tonne = 1000 kg
 Quintal : 1 Quintal = 100 kg
 Atomic mass unit (amu) : amu = 67 ´ 10^{–27} kg mass of proton or neutron is of the order of 1 amu
(3) Time :
 Year : It is the time taken by earth to complete 1 revolution around the sun in its
 Lunar month : It is the time taken by moon to complete 1 revolution around the earth in its orbit. 1 M. = 27.3 days
 Solar day : It is the time taken by earth to complete one rotation about its axis with respect to Since this time varies from day to day, average solar day is calculated by taking average of the duration of all the days in a year and this is called Average Solar day.
1 Solar year = 365.25 average solar day
or average solar day = 1 365.25
the part of solar year
 Sedrial day : It is the time taken by earth to complete one rotation about its axis with respect to a distant 1 Solar year = 366.25 Sedrial day = 365.25 average solar day
Thus 1 Sedrial day is less than 1 solar day.
 Shake : It is an obsolete and practical unit of 1 Shake = 10^{–} ^{8} sec
Dimensions of a Physical Quantity.
When a derived quantity is expressed in terms of fundamental quantities, it is written as a product of different powers of the fundamental quantities. The powers to which fundamental quantities must be raised in order to express the given physical quantity are called its dimensions.
To make it more clear, consider the physical quantity force
Force = mass × acceleration = mass ´ velocity
time
= mass ´ length/time time
= mass × length × (time)^{–2}…….. (i)
Thus, the dimensions of force are 1 in mass, 1 in length and – 2 in time.
Here the physical quantity that is expressed in terms of the base quantities is enclosed in square brackets to indicate that the equation is among the dimensions and not among the magnitudes.
Thus equation (i) can be written as [force] = [MLT^{–2}].
Such an expression for a physical quantity in terms of the fundamental quantities is called the dimensional equation. If we consider only the R.H.S. of the equation, the expression is termed as dimensional formula.
Thus, dimensional formula for force is, [MLT ^{–} ^{2}].
Important Dimensions of Complete Physics.
Mechanics
S. N. Quantity  Unit  Dimension 
(1) Velocity or speed (v)  m/s  [M^{0}L^{1}T ^{–1}] 
(2) Acceleration (a)  m/s^{2}  [M^{0}LT ^{–2}] 
(3) Momentum (P)  kgm/s  [M^{1}L^{1}T ^{–1}] 
(4) Impulse (I)  Newtonsec or kgm/s  [M^{1}L^{1}T ^{–1}] 
(5) Force (F)  Newton  [M^{1}L^{1}T ^{–2}] 
(6) Pressure (P)  Pascal  [M1L–1T –2] 
(7) Kinetic energy (E_{K})  Joule  [M^{1}L^{2}T ^{–2}] 
(8) Power (P)  Watt or Joule/s  [M^{1}L^{2}T ^{–}^{3}] 
(9) Density (d)  kg/m^{3}  [M^{1}L^{–} ^{3}T ^{0}] 
(10) Angular displacement (q)  Radian (rad.)  [M^{0}L^{0}T ^{0}] 
(11) Angular velocity (w)  Radian/sec  [M^{0}L^{0}T ^{–} ^{1}] 
(12) Angular acceleration (a)  Radian/sec^{2}  [M^{0}L^{0}T ^{–} ^{2}] 
(13) Moment of inertia (I)  kgm^{2}  [M^{1}L^{2}T^{0}] 
(14) Torque (t)  Newtonmeter  [M^{1}L^{2}T ^{–}^{2}] 
(15) Angular momentum (L)  Joulesec  [M^{1}L^{2}T ^{–}^{1}] 
(16) Force constant or spring constant (k)  Newton/m  [M^{1}L^{0}T ^{–2}] 
(17) Gravitational constant (G)  Nm^{2}/kg^{2}  [M–1L3T – 2] 
(18) Intensity of gravitational field (E_{g})  N/kg  [M^{0}L^{1}T ^{–} ^{2}] 
(19) Gravitational potential (V_{g})  Joule/kg  [M^{0}L^{2}T ^{–} ^{2}] 
(20) Surface tension (T)  N/m or Joule/m^{2}  [M^{1}L^{0}T ^{–} ^{2}] 
(21) Velocity gradient (V_{g})  Second^{–1}  [M^{0}L^{0}T ^{–} ^{1}] 
(22) Coefficient of viscosity (h)  kg/ms  [M1L– 1T – 1] 
(23) Stress  N/m^{2}  [M1L– 1T – 2] 
(24) Strain  No unit  [M^{0}L^{0}T ^{0}] 
(25) Modulus of elasticity (E)  N/m^{2}  [M1L– 1T – 2] 
(26) Poisson Ratio (s)  No unit  [M^{0}L^{0}T ^{0}] 
(27) Time period (T)  Second  [M^{0}L^{0}T^{1}] 
(28) Frequency (n)  Hz  [M^{0}L^{0}T ^{–1}] 
Heat
S. N. Quantity  Unit  Dimension 
(1) Temperature (T)  Kelvin  [M^{0}L^{0}T^{0}q ^{1}] 
(2) Heat (Q)  Joule  [ML^{2}T^{–} ^{2}] 
(3) Specific Heat (c)  Joule/kgK  [M0L2T– 2q –1] 
(4) Thermal capacity  Joule/K  [M1L2T – 2q –1] 
(5) Latent heat (L)  Joule/kg  [M^{0}L^{2}T ^{–} ^{2}] 
(6) Gas constant (R)  Joule/mol–K  [M1L2T– 2q – 1] 
(7) Boltzmann constant (k)  Joule/K  [M1L2T– 2q – 1] 
(8) Coefficient of thermal conductivity (K)  Joule/ms–K  [M1L1T– 3q – 1] 
(9) Stefan’s constant (s)  Watt/m^{2}–K^{4}  [M1L0T– 3q – 4] 
(10) Wien’s constant (b)  MeterK  [M^{0}L^{1}T^{o}q^{1}] 
(11) Planck’s constant (h)  Joules  [M^{1}L^{2}T^{–1}] 
(12) Coefficient of Linear Expansion (a)  Kelvin^{–1}  [M0L0T0q –1] 
(13) Mechanical eq. of Heat (J)  Joule/Calorie  [M^{0}L^{0}T^{0}] 
(14) Vander wall’s constant (a)  Newtonm^{4}  [ML^{5}T^{–} ^{2}] 
(15) Vander wall’s constant (b)  m3  [M^{0}L^{3}T^{0}] 
Electricity
 N. Quantity Unit Dimension
 Electric charge (q) Coulomb [M^{0}L^{0}T^{1}A^{1}]
 Electric current (I) Ampere [M^{0}L^{0}T^{0}A^{1}]
 Capacitance (C) Coulomb/volt or Farad [M^{–1}L^{–} ^{2}T^{4}A^{2}]
 Electric potential (V) Joule/coulomb M^{1}L^{2}T^{–}^{3}A^{–}^{1}
 Permittivity of free space (e_{0})
 Dielectric constant (K)
Coulomb^{2}
Newton – meter ^{2}
Unitless
[M–1L–3T4A2]
[M^{0}L^{0}T^{0}]
 Resistance (R)
 Resistivity or Specific resistance (r)
Volt/Ampere or ohm [M^{1}L^{2}T^{–} ^{3}A^{–} ^{2}]
Ohmmeter [M^{1}L^{3}T^{–} ^{3}A^{–} ^{2}]
 Coefficient of Selfinduction (L)
volt – second ampere
or henery or ohmsecond
[M1L2T– 2A– 2]
 Magnetic flux (f) Voltsecond or weber [M^{1}L^{2}T^{–2}A^{–1}]
 Magnetic induction (B)
newton ampere – meter
Joule ampere – meter ^{2}
[M1L0T– 2A– 1]
volt – second
meter ^{2}
or Tesla
 Magnetic Intensity (H)
 Magnetic Dipole Moment (M)
Ampere/meter [M^{0}L^{–} ^{1}T^{0}A^{1}]
Amperemeter^{2} [M^{0}L^{2}T^{0}A^{1}]
Quantities Having Same Dimensions.
 N. Dimension Quantity
 [M^{0}L^{0}T^{–1}] Frequency, angular frequency, angular velocity, velocity gradient and decay constant
 [M^{1}L^{2}T^{–2}] Work, internal energy, potential energy, kinetic energy, torque, moment of force
 [M^{1}L^{–1}T^{–2}] Pressure, stress, Young’s modulus, bulk modulus, modulus of rigidity, energy density
 [M^{1}L^{1}T^{–1}] Momentum, impulse
 [M^{0}L^{1}T^{–2}] Acceleration due to gravity, gravitational field intensity
 [M^{1}L^{1}T^{–2}] Thrust, force, weight, energy gradient
 [M^{1}L^{2}T^{–1}] Angular momentum and Planck’s constant
 [M^{1}L^{0}T^{–2}] Surface tension, Surface energy (energy per unit area)
 [M^{0}L^{0}T^{0}] Strain, refractive index, relative density, angle, solid angle, distance gradient, relative permittivity (dielectric constant), relative permeability
 [M^{0}L^{2}T^{–2}] Latent heat and gravitational potential
 [M^{0}L^{2}T^{–2}q^{–1}] Thermal capacity, gas constant, Boltzmann constant and entropy
(12) [M^{0}L^{0}T^{1}]
, where l = length

g = acceleration due to gravity, m = mass, k = spring constant
(13) [M^{0}L^{0}T^{1}] L/R,
(14) [ML^{2}T^{–2}]
, RC where L = inductance, R = resistance, C = capacitance


where I = current, t = time, q = charge,
L = inductance, C = capacitance, R = resistance
Application of Dimensional Analysis.
 To find the unit of a physical quantity in a given system of units : Writing the definition or formula for the physical quantity we find its Now in the dimensional formula replacing M, L and T by the fundamental units of the required system we get the unit of physical quantity. However, sometimes to this unit we further assign a specific name, e.g., Work = Force ´ Displacement
So [W] = [MLT^{–2}] ´ [L] = [ML^{2}T^{–2}]
So its units in C.G.S. system will be g cm^{2}/s^{2} which is called erg while in M.K.S. system will be kg m^{2}/s^{2} which is called joule.
Problem 1. The equation æ P + a ö (V – b) = constant. The units of a is [MNR 1995; AFMC 1995]
(a)
ç

è
Dyne ´ cm^{5}
V 2 ÷
(b)
Dyne ´ cm ^{4}
(c)
Dyne / cm ^{3}
(d)
Dyne / cm^{2}

Solution : (b) According to the principle of dimensional homogenity [P] = é a ù
ë V ^{2} û
Þ [a] = [P] [V ^{2}] = [ML^{–}^{1}T ^{–}^{2}][L^{6}] = [ML^{5} T ^{–}^{2} ]
or unit of a = gm × cm^{5} ´ sec^{–2}= Dyne ´ cm^{4}
Problem 2. If x = at + bt ^{2} , where x is the distance travelled by the body in kilometre while t the time in seconds, then
the units of b are [CBSE 1993]
 km/s (b) kms (c) km/s^{2} (d) kms^{2}
Solution : (c) From the principle of dimensional homogenity [x] = [bt ^{2}] Þ
 = éx ù

ë t ^{2} û
\ Unit of b = km/s^{2}.
Problem 3. The unit of absolute permittivity is [EAMCET (Med.) 1995; Pb. PMT 2001]
 Farad – meter (b) Farad / meter (c) Farad/meter ^{2} (d) Farad
Solution : (b) From the formula C = 4pe_{0}R
\ e = C

4pR
By substituting the unit of capacitance and radius : unit of e _{0} = Farad/ meter.
Problem 4. Unit of Stefan’s constant is [MP PMT 1989]
(a)
Js 1
Jm2 s 1 K 4
Jm2
 Js
Solution : (b) Stefan’s formula Q
At
= sT ^{4}
\ s =
Q AtT ^{4}
\Unit of s =
Joule =
m^{2} ´ sec´ K^{4}
Jm2 s 1 K 4
Problem 5. The unit of surface tension in SI system is
[MP PMT 1984; AFMC 1986; CPMT 1985, 87; CBSE 1993; Karnataka CET (Engg/Med.) 1999; DCE 2000, 01]
(a)
Dyne / cm^{2}
 Newton/m (c) Dyne/cm (d) Newton/m^{2}
Solution : (b) From the formula of surface tension T = F
l
By substituting the S.I. units of force and length, we will get the unit of surface tension = Newton/m
Problem 6. A suitable unit for gravitational constant is [MNR 1988]
(a)
kg metre sec ^{–}^{1}
(b)
Newton metre ^{–}^{1} sec (c)
Newton metre ^{2}kg ^{–}^{2}
(d)
kg metre sec ^{–}^{1}
Solution : (c) As F = Gm1m2
r 2
\ G =
Fr ^{2}
m1m2
Substituting the unit of above quantities unit of G =
Problem 7. The SI unit of universal gas constant (R) is
Newton metre ^{2}kg ^{–}^{2} .
[MP Board 1988; JIPMER 1993; AFMC 1996; MP PMT 1987, 94; CPMT 1984, 87; UPSEAT 1999]
 Watt K ^{–}^{1}mol ^{–}^{1}
 Newton K ^{–}^{1}mol ^{–}^{1}
 Joule K ^{–}^{1}mol ^{–}^{1}
Erg K ^{–}^{1}mol ^{–}^{1}
Solution : (c) Ideal gas equation PV = nRT \
= [P][V] = [ML^{–}^{1}T ^{–}^{2} ][L^{3} ] =
[ML^{2}T ^{–}^{2} ]
[R]
So the unit will be Joule K^{–}^{1}mol^{–}^{1} .
[nT]
[mole] [K]
[mole] ´ [K]
 To find dimensions of physical constant or coefficients : As dimensions of a physical quantity are unique, we write any formula or equation incorporating the given constant and then by substituting the dimensional formulae of all other quantities, we can find the dimensions of the required constant or
 Gravitational constant : According to Newton’s law of gravitation
F = G m1m2
r 2
or G =
Fr ^{2}
m1m2
Substituting the dimensions of all physical quantities [G] = [MLT 2 ][L2 ] = [M ^{1} L^{3}T ^{2} ]
[M][M]
 Plank constant : According to Planck
E = hn
or h = E
n
Substituting the dimensions of all physical quantities [h] = [ML2 T 2 ] = [ML^{2}T ^{1} ]
[T ^{1} ]
 Coefficient of viscosity : According to Poiseuille’s formula
dV = ppr ^{4}
dt 8hl
or h =
ppr ^{4}
8l(dV / dt)
Substituting the dimensions of all physical quantities [h] = [ML1T 2 ][L4 ] = [ML^{1}T ^{1} ]
[L][L^{3} / T]
Problem 8.
X = 3YZ ^{2} find dimension of Y in (MKSA) system, if X and Z are the dimension of capacity and magnetic field respectively [MP PMT 2003]
(a)
M 3 L2 T 4 A 1
(b)
ML2
(c)
M 3 L2 T 4 A 4
(d)
M 3 L2 T 8 A 4
Solution : (d)
X = 3YZ ^{2}
\ [Y] = [X]
[Z ^{2}]
= [M 1 L2 T 4 A 2 ] = [M ^{–}^{3} L^{–}^{2}T ^{8} A ^{4} ] .
[MT 2 A 1 ]2
Problem 9. Dimensions of
1
m _{0}e _{0}
, where symbols have their usual meaning, are [AIEEE 2003]
(a)
[LT ^{–}^{1} ]
(b)
[L^{–}^{1}T]
(c)
[L^{–}^{2}T ^{2} ]
(d)
[L^{2}T ^{–}^{2} ]
Solution : (d) We know that velocity of light C = 1
\ 1
m_{0}e_{0}
= C^{2}
\ So
é 1 ù = [LT ^{–}^{1}]^{2}

m_{0}e_{0}
= [L^{2}T ^{–}^{2}] .
ë û
Problem 10. If L, C and R denote the inductance, capacitance and resistance respectively, the dimensional formula for
C ^{2} LR
is [UPSEAT 2002]
(a)
[ML^{–}^{2}T ^{–}^{1} I ^{0} ]
(b)
[M ^{0} L^{0} T ^{3} I ^{0} ]
(c)
[M 1 L2 T 6 I 2 ]
(d)
[M ^{0} L^{0} T ^{2} I ^{0} ]
2 é 2
_{2} R ù é
_{2}æ R öù
Solution : (b)
[C LR] = êC L L ú = ê(LC) ç L ÷ú
ë û ë è øû
and we know that frequency of LC circuits is given by f = 1 1 i.e., the dimension of LC is equal to [T ^{2}]
and é L ù gives the time constant of L – R
êë R úû
circuit so the dimension of
 is equal to [T].
R
By substituting the above dimensions in the given formula
é LC)^{2}æ R öù = [T ^{2}]^{2}[T ^{–}^{1}] = [T ^{3}] .
ê( ç L ÷ú
ë è øû
Problem 11. A force F is given by F = at + bt ^{2} , where t is time. What are the dimensions of a and b
[BHU 1998; AFMC 2001]
(a)
MLT ^{–}^{3} and ML^{2}T ^{–}^{4}
(b)
MLT ^{–}^{3} and MLT ^{–}^{4} (c)
MLT ^{–}^{1} and MLT ^{0}
(d)
MLT ^{–}^{4}
and MLT ^{1}


Solution : (b) From the principle of dimensional homogenity [F] = [at] \ [a] = é F ù = é MLT 2 ù = [MLT ^{–}^{3} ]
Similarly [F] = [bt ^{2}] \ [b] = é F ù = é MLT 2 ù = [MLT ^{–}^{4} ] .
êë t úû ê T ú


êë t ^{2} úû
êê T 2 ú
Problem 12. The position of a particle at time t is given by the relation
x(t) = æ v0 ö(1 – c ^{–a} ^{t} ), where v is a constant and
ç a ÷ _{0}
è ø
a > 0 . The dimensions of v_{0}
and a are respectively [CBSE 1995]
(a)
M ^{0} L^{1}T ^{–}^{1} and T ^{–}^{1}
(b)
M 0 L1T 0
and T ^{–}^{1}
(c)
M ^{0} L^{1}T ^{–}^{1} and LT ^{–}^{2}
(d)
M ^{0} L^{1}T ^{–}^{1} and T
Solution : (a) From the principle of dimensional homogeneity [a t] = dimensionless \ [a ] = é1ù = [T ^{–}^{1}]
êë t úû
Similarly [x] = [v0 ]
[a]
\ [v0 ] = [x][a] = [L][T ^{–}^{1}] = [LT ^{–}^{1} ] .
Problem 13. The dimensions of physical quantity X in the equation Force = X
Density
is given by [DCE 1993]
(a)
M 1 L4 T 2
(b)
M 2 L2 T 1
(c)
M 2L2T 2
(d)
M 1 L2 T 1
Solution : (c) [X] = [Force] × [Density] = [MLT ^{–}^{2}]´[ML^{–}^{3}] = [M ^{2}L^{–}^{2}T ^{–}^{2}] .
Problem 14. Number of particles is given by
n = –D n2 – n1
x 2 – x1
crossing a unit area perpendicular to X– axis in unit time,
where
n1 and
n_{2} are number of particles per unit volume for the value of x meant to
x 2 and
x1 .
Find
dimensions of D called as diffusion constant [CPMT 1979]
(a)
M ^{0} LT ^{2}
(b)
M 0 L2 T 4
(c)
M 0 LT 3
(d)
M 0 L2 T 1
Solution : (d) (n) = Number of particle passing from unit area in unit time =
No. of particle = [M ^{0} L^{0} T ^{0} ] = [L2T 1]
[n1] = [n2] = No. of particle in unit volume = [L^{–}^{3}]
A ´ t
[L^{2} ] [T]
Now from the given formula [D] = [n][x2 – x1] = [L2 T 1 ][L] = [L^{2}T ^{–}^{1} ] .
[n2 – n1]
[L^{–}^{3} ]
Problem 15. E, m, l and G denote energy, mass, angular momentum and gravitational constant respectively, then the
dimension of
El ^{2}
m5 G2
are
[AIIMS 1985]
 Angle (b) Length (c) Mass (d) Time
Solution : (a)
[E] = energy = [ML^{2}T ^{–}^{2}] , [m] = mass = [M], [l] = Angular momentum = [ML^{2}T ^{–}^{1}]
[G] = Gravitational constant = [M ^{–}^{1}L^{3}T ^{–}^{2}]
Now substituting dimensions of above quantities in
i.e., the quantity should be angle.
El ^{2}
m5G2
= [ML^{2}T ^{–}^{2}]´[ML^{2}T ^{–}^{1}]^{2} = [M 0 L0T0 ]
[M ^{5}]´[M ^{–}^{1}L^{3}T ^{–}^{2}]^{2}
Problem 16. The equation of a wave is given by Y = A sin w æ x – k ö
where w is the angular velocity and v is the linear

ç ÷
è ø
velocity. The dimension of k is [MP PMT 1993]
(a) LT (b) T (c)
T 1
(d) T ^{2}
Solution : (b) According to principle of dimensional homogeneity [k] = é x ù = é L ù = [T] .
êë v úû
êë LT ^{–}^{1} úû
Problem 17. The potential energy of a particle varies with distance x from a fixed origin as U =
A x ,
where A and B
are dimensional constants then dimensional formula for AB is
x^{2} + B
 ML^{7/2}T ^{–}^{2}
(b)
ML11 / 2T 2
(c) M 2L9 / 2T 2
(d) ML13 / 2T 3
Solution : (b) From the dimensional homogeneity [x ^{2} ] = [B] \ [B] = [L^{2}]
As well as [U] = [A][x1 / 2 ]
[x ^{2} ] +[B]
Þ [ML^{2}T
2 ]
= [A][L^{1/} ^{2} ] [L^{2} ]
\ [A]
= [ML^{7} ^{/} ^{2}T
2 ]
Now [AB] =[ML^{7} ^{/} ^{2}T ^{–}^{2} ] ´ [L^{2} ] = [ML11 / 2T 2]
Problem 18. The dimensions of
1 e_{0}E^{2}
2
( e _{0} = permittivity of free space ; E = electric field ) is [IITJEE 1999]
(a)
MLT ^{–}^{1}
 ML ^{2} T ^{–}^{2}
 ML ^{–}^{1} T ^{–}^{2}
ML2T 1
Solution : (c) Energy density = 1 e
E 2 = Energy
= é ML2 T 2 ù = [ML^{–}^{1}T ^{–}^{2} ]
2 ^{0} Volume
êê L3 ú


Problem 19. You may not know integration. But using dimensional analysis you can check on some results. In the integral


dx = a^{n} sin^{–}^{1}æ x – 1ö the value of n is
ò (2ax – x^{2})^{1} ^{/} ^{2} ç ÷
(a) 1 (b) – 1 (c) 0 (d) 1
2
Solution : (c) Let x = length \ [X]=[L] and [dx] =[L]
By principle of dimensional homogeneity é x ù = dimensionless \ [a] = [x] =[L]
êë a úû
By substituting dimension of each quantity in both sides: [L] = [L^{n} ] \ n = 0
[L2 – L2 ]1 / 2
Problem 20. A physical quantity P = B2l 2
m
where B= magnetic induction, l= length and m = mass. The dimension of P is
(a)
MLT ^{–}^{3}
(b)
ML^{2}T ^{–}^{4} I^{–2} (c)
M 2 L2 T 4 I
(d)
MLT 2 I 2
Solution : (b) F = BIL \ Dimension of [B] = [F] = [MLT 2 ] =[MT ^{–}^{2} I ^{–}^{1} ]
[I][L] [I][L]
Now dimension of [P] = B 2l 2
m
= [MT ^{–}^{2} I ^{–}^{1} ]^{2} ´ [L^{2} ] = [ML2T 4 I 2 ]
[M]
Problem 21. The equation of the stationary wave is y= 2a sin æ 2pct ö cosæ 2px ö , which of the following statements is wrong

ç ÷ ç ÷
è ø è l ø
 The unit of ct is same as that of l (b) The unit of x is same as that of l
(c) The unit of 2pc /l is same as that of 2px /lt (d) The unit of c/l is same as that of x / l
Solution : (d) Here,
2pct
as well as
2px
are dimensionless (angle) i.e.
é 2pct ù = é 2px ù = M 0 L0 T 0
l l êë l úû
ëê l úû
So (i) unit of c t is same as that of l (ii) unit of x is same as that of l (iii)
é 2pc ù = é 2px ù
ëê l úû ê l t ú
ë û
and (iv) x is unit less. It is not the case with c .
l l
 To convert a physical quantity from one system to the other : The measure of a physical quantity is nu = constant
If a physical quantity X has dimensional formula [M^{a}L^{b}T^{c}] and if (derived) units of that physical quantity in
two systems are [M^{a} L^{b}T^{c} ] and [M ^{a} L^{b} T^{c} ] respectively and n_{1} and n_{2} be the numerical values in the two systems
1 1 1 2 2 2
respectively, then n1 [u1 ] = n2 [u2 ]




Þ n1 [M^{a} L^{b}T^{c} ] = n
[M^{a} L^{b} T^{c} ]



é M ù a é L ù b é T ù c
Þ n2 = n1 ê 1 ú
ë M 2 û
ê 1 ú
ë L2 û
ê 1 ú
ë T2 û
where M_{1}, L_{1} and T_{1} are fundamental units of mass, length and time in the first (known) system and M_{2}, L_{2} and T_{2} are fundamental units of mass, length and time in the second (unknown) system. Thus knowing the values of fundamental units in two systems and numerical value in one system, the numerical value in other system may be evaluated.
Example : (1) conversion of Newton into Dyne.
The Newton is the S.I. unit of force and has dimensional formula [MLT^{–2}]. So 1 N = 1 kgm/ sec^{2}
é M ù a é L
ù b é T ù c
é kg ù1 é m ù^{1} é sec ù ^{2}
é10^{3} gm ù1 é10^{2} cm ù1 é sec ù ^{2}
By using n2 = n1 ê 1 ú
ê 1 ú
ê 1 ú
= 1 ê
ú ê ú ê
ú = 1 ê
ú ê ú ê
ú = 10^{5}
ë M 2 û
\ 1 N = 10^{5} Dyne
ë L2 û
ë T2 û
ë gm û
ë cm û
ë sec û
ë gm û ë
cm û
ë sec û
(2) Conversion of gravitational constant (G) from C.G.S. to M.K.S. system
The value of G in C.G.S. system is 6.67 ´ 10^{–8} C.G.S. units while its dimensional formula is [M^{–1}L^{3}T^{–2}] So G = 6.67 ´ 10^{–8} cm^{3}/g s^{2}
é M ù a é L
ù b é T ù c
é gm ù 1 é cm ù ^{3} é sec ù ^{2}
By using n2 = n1 ê 1 ú
ê 1 ú
ê 1 ú
= 6.67 ´ 10 ^{8} ê
ú ê ú ê ú
ë M 2 û
ë L2 û
ë T2 û
ë kg û
ë m û
ë sec û
é gm
ù 1 é
cm ù ^{3} é sec ù ^{2}
= 6.67 ´ 10 ^{8} ê
ú ê ú ê
ú = 6.67 ´ 10 ^{11}
\ G = 6.67 ´ 10^{–11} M.K.S. units
ë10^{3} gm û
ë10^{2} cm û
ë sec û
Problem 22. A physical quantity is measured and its value is found to be nu where n =
numerical value and u =
unit.
Then which of the following relations is true [RPET 2003]
(a)
n µ u ^{2}
(b)
n µ u
(c)
n µ (d)
n µ 1
u
Solution : (d) We know P = nu = constant \ n1u1 = n2u2 or n µ 1 .
u
Problem 23. In C.G.S. system the magnitude of the force is 100 dynes. In another system where the fundamental physical quantities are kilogram, metre and minute, the magnitude of the force is [EAMCET 2001]
(a) 0.036 (b) 0.36 (c) 3.6 (d) 36
Solution : (c)
n_{1} = 100 ,
M_{1} = g , L_{1} = cm , T_{1} = sec
and
M2 = kg ,
L_{2} = meter , T_{2} = minute , x = 1 , y = 1 , z = 2
é M ù x é L ùy é T ù2
By substituting these values in the following conversion formula n2 = n1 ê 1 ú
ë M2 û
ê 1 ú
ë L2 û
ê 1 ú
ë T2 û
é gm ù1 é
cm ù^{1} é
sec
ù2
n2 = 100ê kg ú
êë meter úû êë minute úû
ë û
é gm
ù1 é cm ù^{1} é
sec
ù2
n2 = 100ê103 gm ú
êë10^{2} cm úû
êë 60 sec úû
= 3.6
ë û
Problem 24. The temperature of a body on Kelvin scale is found to be X K. When it is measured by a Fahrenheit thermometer, it is found to be X F. Then X is [UPSEAT 200]
(a) 301.25 (b) 574.25 (c) 313 (d) 40
Solution : (c) Relation between centigrade and Fahrenheit
K – 273 = F – 32
According to problem
X – 273 = X – 32 \
5 9
X = 313 .
5 9
Problem 25.  Which relation is wrong  [RPMT 1997]  
(a) 1 Calorie = 4.18 Joules  (b)  1Å =10^{–10} m  
Solution : (d) 
(c) 1 MeV = 1.6 × 10^{–13} Joules
Because 1 Newton = 10^{5} Dyne. 
(d)  1 Newton =10^{–5} Dynes 
Problem 26. To determine the Young’s modulus of a wire, the formula is Y = F . L ; where L= length, A= area of cross
A Dl
section of the wire, DL = Change in length of the wire when stretched with a force F. The conversion factor to
change it from CGS to MKS system is
(a) 1 (b) 10 
(c) 0.1 
[MP PET 1983]
(d) 0.01 

Solution : (c)  We know that the dimension of young’s modulus is  [ML^{–}^{1}T ^{–}^{2} ]  
C.G.S. unit : gm cm^{–}^{1} sec ^{–}^{2} and M.K.S. unit : kg.  m^{–1} sec^{–2} . 
é M ù1 é L
ù 1 é T
ù 2
é gm ù1 é
cm ù ^{–}^{1} ésec ù ^{–}^{2}
By using the conversion formula: n2 = n1 ê 1 ú
ê 1 ú
ê 1 ú = ê ú ê
ú ê ú
ë M 2 û
ë L2 û
ë T2 û
ë kg û
ë meter û
ësec û
n é gm
ù1 é cm ù 1 é sec ù 2 1
\Conversion factor ^{ } ^{2} = ê
ú ê ú ê ú =
= 0.1
n1 êë10 3 gm úû
ë10^{2} cm û
ë sec û 10
Problem 27. Conversion of 1 MW power on a new system having basic units of mass, length and time as 10kg, 1dm and 1 minute respectively is
(a)
2.16 ´ 10^{12} unit
(b)
1.26 ´ 10^{12} unit
(c)
2.16 ´ 10^{10}unit
(d)
2 ´ 10^{14} unit
Solution : (a)
[P] = [ML^{2}T ^{–}^{3} ]
é M ù x é L
ù y é T ù z
é 1kg ù1 é 1m ù 2 é 1s
ù 3
Using the relation n2 = n1 ê 1 ú
ê 1 ú
ê 1 ú
= 1 ´ 10^{6} ê
ú ê ú ê
ú [As
1MW = 10^{6} W ]
ë M 2 û
ë L2 û
ë T2 û
ë10 kg û
ë1 dm û
ë1 min û
é 1kg ù é10 dm ù 2 é 1sec ù ^{–}^{3}
= 106 ê ú ê ú ê ú
= 2.16 ´ 10^{12} unit
ë10kg û ë 1 dm û ë 60sec û
Problem 28. In two systems of relations among velocity, acceleration and force are respectively v = a 2 v , a = aba
and
2 b 1 2 1

F = F1 . If a and b are constants then relations among mass, length and time in two systems are
ab
(a)
 = aM , L
= a 2 L , T
= a ^{3}T1
(b)
M = 1 M , L
= a 3 L , T
= T a
2 b 1 2
b 2 1 2 b
2 a 2b 2 1 2
b 3 1 2
1 b 2
(c)
M = a 3 M , L
= a 2 L , T
= a T
(d)
M = a 2 M , L
= a
L , T
= a 3 T
2 b 3 1 2
b 2 1 2 b 1
2 b 2 1 2
b 2 1 2
b 3 1
Solution : (b)
v = v a 2 Þ [L T ^{–}^{1}] =[L T ^{–}^{1}] a 2
……(i)
2 1 b 2 2 1 1 b
a2 = a1ab
Þ [L T ^{–}^{2} ] = [L T ^{–}^{2} ]ab
……(ii)
2 2 1 1
and F = F1
Þ [M
L T ^{–}^{2} ] =[M
L T ^{–}^{2} ]´ 1
……(iii)
^{2} ab
2 2 2
1 1 1 ab
Dividing equation (iii) by equation (ii) we get M 2 =
M1
(ab )ab
= M1
a 2 B2
Squaring equation (i) and dividing by equation (ii) we get
L = L a 3
Dividing equation (i) by equation (ii) we get T2 = T1 a
b 2
2 1 b 3
Problem 29. If the present units of length, time and mass (m, s, kg) are changed to 100m, 100s, and
1 kg then
10
 The new unit of velocity is increased 10 times (b) The new unit of force is decreased
1 times
1000
(c) The new unit of energy is increased 10 times (d) The new unit of pressure is increased 1000 times
Solution : (b) Unit of velocity = m/sec ; in new system =
100m =
100 sec
m
sec
(same)
Unit of force = kg ´ m ; in new system = 1 kg ´ 100m =
1 kg ´ m
sec ^{2}
10 100 sec ´100 sec
1000
sec ^{2}
Unit of energy = kg ´ m2
sec ^{2}
; in new system = 1 kg ´ 100m ´ 100m
10 100 sec ´100 sec
= 1 kg ´ m^{2}
10 sec ^{2}
Unit of pressure =
kg ; in new system =
m ´ sec ^{2}
1 kg ´
10
1 m ´
100
1
100 sec ´ 100 sec
= 10 ^{–}^{7}
kg
m ´ sec ^{2}
Problem 30. Suppose we employ a system in which the unit of mass equals 100 kg, the unit of length equals 1 km and the unit of time 100 s and call the unit of energy eluoj (joule written in reverse order), then
 1 eluoj = 10^{4} joule (b) 1 eluoj = 10^{3} joule (c) 1 eluoj = 10^{4} joule (d) 1 joule = 10^{3} eluoj
Solution : (a)
[E] = [ML^{2}T ^{–}^{2} ]
1 eluoj = [100kg] ´ [1km]^{2} ´ [100 sec]^{–}^{2} = 100kg ´ 10^{6} m^{2} ´ 10^{–}^{4} sec ^{–}^{2} = 10^{4} kg m^{2} ´ sec ^{–}^{2}
= 10^{4} Joule
Problem 31. If 1gm cms^{–1} = x Ns, then number x is equivalent to
(a) 1´ 10^{–}^{1}
3 ´ 10^{–}^{2}
6 ´ 10^{–}^{4}
1´ 10^{–}^{5}
Solution : (d)
gm – cm s ^{–}^{1}
= 10^{–}^{3} kg ´ 10^{–}^{2} m ´ s ^{–}^{1} = 10^{–}^{5} kg ´ m ´ s ^{–}^{1} = 10^{–5} Ns
 To check the dimensional correctness of a given physical relation : This is based on the ‘principle of homogeneity’. According to this principle the dimensions of each term on both sides of an equation must be the
If X = A ± (BC)^{2} ± ,
then according to principle of homogeneity [X] = [A] = [(BC)^{2}] = [ DEF ]
If the dimensions of each term on both sides are same, the equation is dimensionally correct, otherwise not. A dimensionally correct equation may or may not be physically correct.
Example : (1) F = mv ^{2} / r ^{2}
By substituting dimension of the physical quantities in the above relation – [MLT ^{2} ] = [M][LT ^{1} ]^{2} /[L]^{2}
i.e. [MLT ^{2} ] = [MT ^{2} ]
As in the above equation dimensions of both sides are not same; this formula is not correct dimensionally, so can never be physically.
(2) s = ut – (1 / 2)at ^{2}
By substituting dimension of the physical quantities in the above relation – [L] = [LT^{–1}][T] – [LT^{–2}][T^{2}]
i.e. [L] = [L] – [L]
As in the above equation dimensions of each term on both sides are same, so this equation is dimensionally
correct. However, from equations of motion we know that s = ut + (1 / 2)at ^{2}
Problem 32. From the dimensional consideration, which of the following equation is correct [CPMT 1983]
(a)
T = 2p
(b)
T = 2p
(c)
T = 2p
(d)
T = 2p
Solution : (a)
T = 2p
= 2p
= 2p
[As GM = gR^{2}]
Now by substituting the dimension of each quantity in both sides.
é L ù1 / 2
[T] = êë LT 2 úû
= [T]
L.H.S. = R.H.S. i.e., the above formula is Correct.
Problem 33. A highly rigid cubical block A of small mass M and side L is fixed rigidly onto another cubical block B of the
same dimensions and of low modulus of rigidity h such that the lower face of A completely covers the upper
face of B. The lower face of B is rigidly held on a horizontal surface. A small force F is applied perpendicular to one of the side faces of A. After the force is withdrawn block A executes small oscillations. The time period of which is given by [IITJEE 1992]
 2p
 2p
 2p
 2p
Solution : (d) Given m = mass = [M], h = coefficient of rigidity = [ML^{–}^{1}T ^{–}^{2} ] , L = length = [L]
By substituting the dimension of these quantity we can check the accuracy of the given formulae
æ [M]
ö1 / 2 é M
ù1 / 2
[T] = 2p ç ÷
[h][L]
= êë ML^{–}^{1}T ^{–}^{2} L úû
= [T].
è ø
L.H.S. = R.H.S. i.e., the above formula is Correct.
Problem 34. A small steel ball of radius r is allowed to fall under gravity through a column of a viscous liquid of coefficient
of viscosity. After some time the velocity of the ball attains a constant value known as terminal velocity
vT .
The terminal velocity depends on (i) the mass of the ball. (ii) h (iii) r and (iv) acceleration due to gravity g. which of the following relations is dimensionally correct [CPMT 1992; CBSE 1992; NCERT 1983; MP PMT 2001]
(a)
v µ mg

hr
(b)
vT µ hr
mg
(c)
vT µ hrmg
(d)
v µ mgr

h
Solution : (a) Given vT = terminal velocity = [LT ^{–}^{1} ] , m = Mass = [M], g = Acceleration due to gravity = [LT ^{–}^{2} ]
r = Radius = [L], h = Coefficient of viscosity = [h]

By substituting the dimension of each quantity we can check the accuracy of given formula v µ mg
hr
\ [LT ^{–}^{1} ] =
[M][LT ^{–}^{2} ] =
[ML^{–}^{1}T ^{–}^{1}][L]
[LT
1 ]
L.H.S. = R.H.S. i.e., the above formula is Correct.
Problem 35. A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity h
through a tube of radius r and length l and having a pressure difference p across its end, is
flowing per second
(a)
V = ppr ^{4}
8hl
(b)
V = phl
8 pr ^{4}
(c)
V = 8 phl
pr ^{4}
(d)
V = pph
8lr ^{4}
Solution : (a) Given V = Rate of flow = Volume = [L^{3}T ^{–}^{1}] , P = Pressure = [ML^{–}^{1}T ^{–}^{2}] , r = Radius = [L]
sec
h = Coefficient of viscosity = [ML^{–}^{1}T ^{–}^{1}] , l = Length = [L]
By substituting the dimension of each quantity we can check the accuracy of the formula V = pP r 4
8hl
\ [L^{3}T ^{–}^{1}] = [ML1T 2][L4 ] = [L^{3}T ^{–}^{1}]
[ML^{–}^{1}T ^{–}^{1}][L]
L.H.S. = R.H.S. i.e., the above formula is Correct.
Problem 36. With the usual notations, the following equation St
= u + 1 a(2t – 1) is
2
 Only numerically correct (b) Only dimensionally correct
 Both numerically and dimensionally correct (d) Neither numerically nor dimensionally correct
Solution : (c) Given St = distance travelled by the body in t^{th} sec.= [LT ^{–}^{1}] , a = Acceleration = [LT ^{–}^{2}] ,
v = velocity = [LT ^{–}^{1}] , t = time = [T]
By substituting the dimension of each quantity we can check the accuracy of the formula
St = u + 1 a (2t – 1) 2
\ [LT ^{–}^{1}] = [LT ^{–}^{1}] + [LT ^{–}^{2}] [T] Þ [LT ^{–}^{1}] = [LT ^{–}^{1}] + [LT ^{–}^{1}]
Since the dimension of each terms are equal therefore this equation is dimensionally correct. And after deriving this equation from Kinematics we can also proof that this equation is correct numerically also.
Problem 37. If velocity v, acceleration A and force F are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of v, A and F would be
(a)
FA ^{–}^{1} v (b)
Fv 3 A2
(c)
Fv 2 A1
(d)
F 2v 2 A1
Solution : (b) Given, v = velocity = [LT ^{–}^{1}] , A = Acceleration = [LT ^{–}^{2}] , F = force = [MLT ^{–}^{2}]
By substituting, the dimension of each quantity we can check the accuracy of the formula [Angular momentum] = Fv ^{3} A^{–}^{2}
[ML^{2}T ^{–}^{1} ] = [MLT ^{–}^{2} ] [LT ^{–}^{1} ]^{3} [LT ^{–}^{2} ]^{–}^{2}
= [ML^{2}T ^{–}^{1}]
L.H.S. = R.H.S. i.e., the above formula is Correct.
Problem 38. The largest mass (m) that can be moved by a flowing river depends on velocity (v), density ( r ) of river water and acceleration due to gravity (g). The correct relation is
(a)
m µ r 2v4
g 2
(b)
m µ rv^{6}
g 2
(c)
m µ rv^{4}
g 3
(d)
m µ rv^{6}
g 3
Solution : (d) Given, m = mass = [M], v = velocity = [LT ^{–}^{1}] , r = density = [ML^{–}^{3} ] , g = acceleration due to gravity = [LT^{–2}]
By substituting, the dimension of each quantity we can check the accuracy of the formula
m = K r v6
g 3
Þ[M] = [ML^{–}^{3} ][LT ^{–}^{1} ]^{6}
[LT ^{–}^{2} ]^{3}
= [M]
L.H.S. = R.H.S. i.e., the above formula is Correct.
 As a research tool to derive new relations : If one knows the dependency of a physical quantity on other quantities and if the dependency is of the product type, then using the method of dimensional analysis, relation between the quantities can be
Example : (i) Time period of a simple pendulum.
Let time period of a simple pendulum is a function of mass of the bob (m), effective length (l), acceleration due to gravity (g) then assuming the function to be product of power function of m, l and g
i.e., T = Km^{x}l ^{y} g ^{z} ; where K = dimensionless constant
If the above relation is dimensionally correct then by substituting the dimensions of quantities –
[T] = [M]^{x} [L]^{y} [LT^{–2}]^{z} or [M^{0}L^{0}T^{1}] = [M^{x}L^{y+z}T^{–2}^{z}]
Equating the exponents of similar quantities x = 0, y = 1/2 and z = – 1/2 So the required physical relation becomes T = K
The value of dimensionless constant is found (2p ) through experiments so T = 2p
(ii) Stoke’s law : When a small sphere moves at low speed through a fluid, the viscous force F, opposing the motion, is found experimentally to depend on the radius r, the velocity of the sphere v and the viscosity h of the fluid.
So F = f (h, r, v)
If the function is product of power functions of h, r and v,
F = Kh ^{x}r ^{y}v^{z} ; where K is dimensionless constant.
If the above relation is dimensionally correct [MLT ^{2} ] = [ML^{1}T ^{1} ]^{x} [L]^{y} [LT ^{1} ]^{z}
or [MLT 2 ] = [M x L– x +y+ z T – x –z ]
Equating the exponents of similar quantities x = 1; – x + y + z = 1 and – x – z = – 2 Solving these for x, y and z, we get x = y = z = 1
So eq^{n} (i) becomes F = Khrv
On experimental grounds, K = 6p; so F = 6phrv
This is the famous Stoke’s law.
Problem 39. If the velocity of light (c), gravitational constant (G) and Planck’s constant (h) are chosen as fundamental units, then the dimensions of mass in new system is [UPSEAT 2002]
(a)
c1 / 2 G1 / 2 h1 / 2
(b)
c1 / 2 G1 / 2 h1 / 2
(c)
c1 / 2 G1 / 2 h1 / 2
(d)
c 1 / 2 G1 / 2 h1 / 2
Solution : (c) Let m µ c ^{x} G^{y}h^{z}
or m = K c ^{x} G^{y} h^{z}
By substituting the dimension of each quantity in both sides
[M 1 L0 T 0 ] = K[LT 1 ]x [M 1 L3 T 2 ]y [ML2 T 1 ]z = [M –y+ z Lx +3y+ 2z T – x 2y–z ]
By equating the power of M, L and T in both sides :
–y + z = 1 ,
x + 3y + 2z = 0 , –x – 2y – z = 0
By solving above three equations x = 1 / 2 , y = 1 / 2 and z = 1 / 2 .
\ m µ c1 / 2 G1 / 2 h1 / 2
Problem 40. If the time period (T) of vibration of a liquid drop depends on surface tension (S), radius (r) of the drop and density (r) of the liquid, then the expression of T is [AMU (Med.) 2000]
(a)
T = K
(b)
T = K
(c)
T = K
 None of these
Solution : (a) Let T µ S^{x}r ^{y} r ^{z}
or T = K S^{x}r ^{y} r ^{z}
By substituting the dimension of each quantity in both sides
[M ^{0} L^{0} T ^{1} ] = K [MT ^{–}^{2} ]^{x} [L]^{y} [ML^{–}^{3} ]^{z} = [M ^{x}^{+}^{z} L^{y}^{–}^{3}^{z} T ^{–}^{2}^{x} ]
By equating the power of M, L and T in both sides x + z = 0 , y – 3z = 0 , 2x = 1
By solving above three equations \ x = 1 / 2 , y = 3 / 2 , z = 1 / 2
So the time period can be given as, T = K S ^{–}^{1/} ^{2}r ^{3} ^{/} ^{2} r ^{1/} ^{2} = K .
Problem 41. If P represents radiation pressure, C represents speed of light and Q represents radiation energy striking a unit
area per second, then nonzero integers x, y and z such that P^{x}Q^{y}C^{z}
is dimensionless, are
[AFMC 1991; CBSE 1992; CPMT 1981, 92; MP PMT 1992]
(a)
x = 1, y = 1, z = 1
(b)
x = 1, y = 1, z = 1
(c)
x = 1, y = 1, z = 1
(d)
x = 1, y = 1, z = 1
Solution : (b)
[P ^{x} Q^{y} C^{z} ] = M ^{0} L^{0} T ^{0}
By substituting the dimension of each quantity in the given expression
[ML^{–}^{1}T ^{–}^{2} ]^{x} [MT ^{–}^{3} ]^{y} [LT ^{–}^{1} ]^{z}
= [M x +y L– x + z T 2x 3y–z ] = M 0 L0 T 0
by equating the power of M, L and T in both sides: by solving we get x = 1, y = 1, z = 1 .
x + y = 0 , –x + z = 0 and 2x – 3y – z = 0
Problem 42. The volume V of water passing through a point of a uniform tube during t seconds is related to the crosssectional area A of the tube and velocity u of water by the relation V µ A^{a} u ^{b} t ^{g} , which one of the following will be true
(a)
a = b = g
(b)
a ¹ b = g
(c)
a = b ¹ g
(d)
a ¹ b ¹ g
Solution : (b) Writing dimensions of both sides [L^{3} ] = [L^{2} ]^{a} [LT ^{–}^{1} ]^{b} [ T]^{g} Þ [L^{3}T ^{0} ] = [L^{2}^{a} ^{+b} T ^{g} ^{–b} ]
By comparing powers of both sides 2a + b = 3 and g – b = 0
Which give b = g
and a = 1 (3 – b ) i.e. a ¹ b = g .
2
Problem 43. If velocity (V), force (F) and energy (E) are taken as fundamental units, then dimensional formula for mass will be
(a)
Solution : (d) Let
V 2F 0 E
M = V^{a} F^{b} E^{c}
(b)
V ^{0}FE^{2}
(c)
VF 2 E0
(d)
V 2F 0 E
Putting dimensions of each quantities in both side [M] = [LT ^{–}^{1} ]^{a} [MLT ^{–}^{2} ]^{b} [ML^{2}T ^{–}^{2} ]^{c}
Equating powers of dimensions. We have b + c = 1,
a + b + 2c = 0
and –a – 2b – 2c = 0
Solving these equations, a = 2, b = 0 and c = 1
So M = [V ^{–}^{2} F ^{0} E]
Problem 44. Given that the amplitude A of scattered light is :
 Directly proportional to the amplitude (A_{0}) of incident
 Directly proportional to the volume (V) of the scattering particle
 Inversely proportional to the distance (r) from the scattered particle
 Depend upon the wavelength ( l ) of the scattered then:
(a)
Solution : (b) Let
A µ 1
l
A = KA0Vl^{x}
r
(b)
A µ 1
l2
(c)
A µ 1
l3
(d)
A µ 1
l4
By substituting the dimension of each quantity in both sides
Þ[ L] = [L].[ L^{3} ][L^{x} ]
[L]
\ [L] = [L^{3}^{+}^{x} ] ; Þ
3 + x = 1 or
x = 2
\ A µ l^{–}^{2}
Limitations of Dimensional Analysis.
Although dimensional analysis is very useful it cannot lead us too far as,
 If dimensions are given, physical quantity may not be unique as many physical quantities have same For example if the dimensional formula of a physical quantity is [ML^{2}T ^{2} ]it may be work or energy or torque.
 Numerical constant having no dimensions [K] such as (1/2), 1 or 2p etc. cannot be deduced by the methods of
 The method of dimensions can not be used to derive relations other than product of power functions. For example,
s = u t + (1 / 2)a t ^{2} or y = a sinw t
cannot be derived by using this theory (try if you can). However, the dimensional correctness of these can be checked.
 The method of dimensions cannot be applied to derive formula if in mechanics a physical quantity depends on more than 3 physical quantities as then there will be less number (= 3) of equations than the unknowns
(>3). However still we can check correctness of the given equation dimensionally. For example not be derived by theory of dimensions but its dimensional correctness can be checked.
T = 2p
1 mgl can
 Even if a physical quantity depends on 3 physical quantities, out of which two have same dimensions, the
formula cannot be derived by theory of dimensions, e.g., formula for the frequency of a tuning fork cannot be derived by theory of dimensions but can be checked.
Significant Figures.
f = (d / L^{2} )v
Significant figures in the measured value of a physical quantity tell the number of digits in which we have confidence. Larger the number of significant figures obtained in a measurement, greater is the accuracy of the measurement. The reverse is also true.
The following rules are observed in counting the number of significant figures in a given measured quantity.
 All nonzero digits are
Example : 42.3 has three significant figures.
243.4 has four significant figures.
24.123 has five significant figures.
 A zero becomes significant figure if it appears between to nonzero
Example : 5.03 has three significant figures.
5.604 has four significant figures.
4.004 has four significant figures.
 Leading zeros or the zeros placed to the left of the number are never
Example : 0.543 has three significant figures.
0.045 has two significant figures.
0.006 has one significant figures.
 Trailing zeros or the zeros placed to the right of the number are
Example : 4.330 has four significant figures.
433.00 has five significant figures.
343.000 has six significant figures.
 In exponential notation, the numerical portion gives the number of significant
Example : 1.32 ´ 10^{–2} has three significant figures.
1.32 ´ 10^{4} has three significant figures.
Rounding Off.
While rounding off measurements, we use the following rules by convention:
 If the digit to be dropped is less than 5, then the preceding digit is left
Example :
x = 7.82
is rounded off to 7.8, again
x = 3.94
is rounded off to 3.9.
 If the digit to be dropped is more than 5, then the preceding digit is raised by
Example : x = 6.87 is rounded off to 6.9, again x = 12.78 is rounded off to 12.8.
 If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit is raised by
Example : x = 16.351 is rounded off to 16.4, again x = 6.758 is rounded off to 6.8.
 If digit to be dropped is 5 or 5 followed by zeros, then preceding digit is left unchanged, if it is
Example : x = 3.250 becomes 3.2 on rounding off, again x = 12.650 becomes 12.6 on rounding off.
 If digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one, if it is
Example : x = 3.750 is rounded off to 3.8, again x = 16.150 is rounded off to 16.2.
Significant Figures in Calculation.
In most of the experiments, the observations of various measurements are to be combined mathematically, i.e., added, subtracted, multiplied or divided as to achieve the final result. Since, all the observations in measurements do not have the same precision, it is natural that the final result cannot be more precise than the least precise measurement. The following two rules should be followed to obtain the proper number of significant figures in any calculation.
 The result of an addition or subtraction in the number having different precisions should be reported to the same number of decimal places as are present in the number having the least number of decimal places. The rule is illustrated by the following examples :
(i) 33.3 ¬ (has only one decimal place) 3.11
+ 0.313
36.723 ¬ (answer should be reported to one decimal place)
Answer = 36.7
(ii) 3.1421
0.241
+ 0.09 ¬ (has 2 decimal places)
3.4731 ¬ (answer should be reported to 2 decimal places)
Answer = 3.47
(iii) 62.831 ¬ (has 3 decimal places)
– 24.5492
38.2818 ¬ (answer should be reported to 3 decimal places after rounding off)
Answer = 38.282
 The answer to a multiplication or division is rounded off to the same number of significant figures as is possessed by the least precise term used in the The rule is illustrated by the following examples :
(i) 142.06
 23 ¬ (two significant figures)
32.6738 ¬ (answer should have two significant figures)
Answer = 33
(ii) 51.028
 31 ¬ (three significant figures) 66.84668
Answer = 66.8 (iii)
Answer = 0.21
0.90 = 0.2112676
4.26
Order of Magnitude.
In scientific notation the numbers are expressed as, Number
= M ´ 10 ^{x} . Where M is a number lies between 1
and 10 and x is integer. Order of magnitude of quantity is the power of 10 required to represent the quantity. For determining this power, the value of the quantity has to be rounded off. While rounding off, we ignore the last digit which is less than 5. If the last digit is 5 or more than five, the preceding digit is increased by one. For example,
 Speed of light in vacuum = 3 ´ 10^{8} ms ^{1} » 10^{8} m / s (ignoring 3 < 5)
(2) Mass of electron = 9.1 ´ 10 ^{31} kg » 10 ^{30} kg (as 9.1 > 5).
Problem 45. Each side a cube is measured to be 7.203 m. The volume of the cube up to appropriate significant figures is (a) 373.714 (b) 373.71 (c) 373.7 (d) 373
Solution : (c) Volume = a ^{3}
= (7.023)^{3}
= 373.715 m ^{3}
In significant figures volume of cube will be 373.7 m^{3}
because its side has four significant figures.
Problem 46. The number of significant figures in 0.007 m ^{2} is
(a) 1 (b) 2 (c) 3 (d) 4
Solution : (a)
Problem 47. The length, breadth and thickness of a block are measured as 125.5 cm, 5.0 cm and 0.32 cm respectively.
Which one of the following measurements is most accurate
 Length (b) Breadth (c) Thickness (d) Height
Solution : (a) Relative error in measurement of length is minimum, so this measurement is most accurate.
Problem 48. The mass of a box is 2.3 kg. Two marbles of masses 2.15 g and 12.39 g are added to it. The total mass of the box to the correct number of significant figures is
(a) 2.340 kg (b) 2.3145 kg. (c) 2.3 kg (d) 2.31 kg Solution : (c) Total mass = 2.3 + 0.00215 + 0.01239 = 2.31kg
Total mass in appropriate significant figures be 2.3 kg.
Problem 49. The length of a rectangular sheet is 1.5 cm and breadth is 1.203 cm. The area of the face of rectangular sheet to the correct no. of significant figures is :
(a) 1.8045 cm^{2}
(b) 1.804 cm^{2}
(c) 1.805 cm^{2}
(d) 1.8 cm^{2}
Solution : (d) Area = 1.5 ´ 1.203 = 1.8045 cm^{2} = 1.8 cm^{2} (Upto correct number of significant figure).
Problem 50. Each side of a cube is measured to be 5.402 cm. The total surface area and the volume of the cube in appropriate significant figures are :
(a) 175.1 cm^{2} , 157 cm^{2}
(c) 175 cm^{2} , 157 cm^{2}
(b) 175.1 cm^{2} , 157.6 cm^{3}
(d) 175.08 cm^{2} , 157.639 cm^{3}
Solution : (b) Total surface area = 6 ´(5.402)^{2} = 175.09 cm^{2} = 175.1cm^{2} (Upto correct number of significant figure)
Total volume = (5.402)^{3} = 175.64 cm^{3} = 175.6cm ^{3}
(Upto correct number of significant figure).
Problem 51. Taking into account the significant figures, what is the value of 9.99 m + 0.0099 m
(a) 10.00 m (b) 10 m (c) 9.9999 m (d) 10.0 m
Solution : (a)
9.99 m + 0.0099 m = 9.999 m
= 10.00 m (In proper significant figures).
Problem 52. The value of the multiplication 3.124 ´
4.576 correct to three significant figures is
Solution : (b)
(a) 14.295 (b) 14.3 (c) 14.295424 (d) 14.305
3.124 ´ 4.576 = 14.295 =14.3 (Correct to three significant figures).
Problem 53. The number of the significant figures in 11.118 ´ 10 ^{–}^{6} V is
(a) 3 (b) 4 (c) 5 (d) 6
Solution : (c) The number of significant figure is 5 as 10^{–}^{6}
does not affect this number.
Problem 54. If the value of resistance is 10.845 ohms and the value of current is 3.23 amperes, the potential difference is 35.02935 volts. Its value in significant number would be [CPMT 1979]
(a) 35 V (b) 35.0 V (c) 35.03 V (d) 35.025 V
Solution : (b) Value of current (3.23 A) has minimum significant figure (3) so the value of potential difference V(= IR) have only 3 significant figure. Hence its value be 35.0 V.
Errors of Measurement.
The measuring process is essentially a process of comparison. Inspite of our best efforts, the measured value of a quantity is always somewhat different from its actual value, or true value. This difference in the true value of a quantity is called error of measurement.
 Absolute error : Absolute error in the measurement of a physical quantity is the magnitude of the difference between the true value and the measured value of the
Let a physical quantity be measured n times. Let the measured value be a_{1}, a_{2}, a_{3}, ….. a_{n}. The arithmetic mean
of these value is am
= a1 + a2 +…. an
n
Usually, a_{m} is taken as the true value of the quantity, if the same is unknown otherwise. By definition, absolute errors in the measured values of the quantity are
Da1 = am – a1
Da2 = am – a2
………….
Dan = am – an
The absolute errors may be positive in certain cases and negative in certain other cases.
 Mean absolute error : It is the arithmetic mean of the magnitudes of absolute errors in all the measurements of the It is represented by Da. Thus
Da = Da1 +Da2 +…. Dan 
n
Hence the final result of measurement may be written as a = am ± Da
This implies that any measurement of the quantity is likely to lie between (am + Da) and (am – Da).
 Relative error or Fractional error : The relative error or fractional error of measurement is defined as the ratio of mean absolute error to the mean value of the quantity Thus
Relative error or Fractional error = mean absolute error = Da
mean value am
 Percentage error : When the relative/fractional error is expressed in percentage, we call it percentage Thus
Percentage error = Da ´ 100%
am
Propagation of Errors.
 Error in sum of the quantities : Suppose x = a + b
Let Da = absolute error in measurement of a
Db = absolute error in measurement of b
Dx = absolute error in calculation of x i.e. sum of a and b.
The maximum absolute error in x is Dx = ±(Da + Db)
Percentage error in the value of
x = (Da + Db) ´ 100%
a + b
 Error in difference of the quantities : Suppose x = a – b
Let Da = absolute error in measurement of a,
Db = absolute error in measurement of b
Dx = absolute error in calculation of x i.e. difference of a and b.
The maximum absolute error in x is Dx = ±(Da + Db)
Percentage error in the value of
x = (Da + Db) ´ 100%
a – b
 Error in product of quantities : Suppose x = a ´ b
Let Da = absolute error in measurement of a,
Db = absolute error in measurement of b
Dx = absolute error in calculation of x i.e. product of a and b.
The maximum fractional error in x is
Dx = ± æ Da + Db ö


x ç a b ÷
Percentage error in the value of x = (Percentage error in value of a) + (Percentage error in value of b)
 Error in division of quantities : Suppose Let Da = absolute error in measurement of a,
Db = absolute error in measurement of b
x = a
b
Dx = absolute error in calculation of x i.e. division of a and b.
The maximum fractional error in x is
Dx = ± æ Da + Db ö


x ç a b ÷
Percentage error in the value of x = (Percentage error in value of a) + (Percentage error in value of b)
 Error in quantity raised to some power : Suppose Let Da = absolute error in measurement of a,
Db = absolute error in measurement of b
Dx = absolute error in calculation of x
x = an
bm
The maximum fractional error in x is
Dx = ± æ n Da + m Db ö


x ç a b ÷
Percentage error in the value of x = n (Percentage error in value of a) + m (Percentage error in value of b)
Note : @The quantity which have maximum power must be measured carefully because it’s contribution to error is maximum.
Problem 55. A physical parameter a can be determined by measuring the parameters b, c, d and e using the relation
a = b^{a} c ^{b} / d^{g} e^{d} . If the maximum errors in the measurement of b, c, d and e are
b1 %,
c1 %,
d1 % and
Solution : (d)
e_{1} %, then the maximum error in the value of a determined by the experiment is [CPMT 1981]
(a) ( b1 + c1 + d1 + e1 )% (b) ( b1 + c1 – d1 – e1 )%
(c) ( ab1 + bc1 – gd1 – de1 )% (d) ( ab1 + bc1 + gd1 + de1 )%
a = b^{a} c ^{b} /d ^{g} e ^{d}
So maximum error in a is given by
æ Da ´ 100 ö = a . Db ´ 100 + b . Dc ´ 100 + g . Dd ´ 100 + d . De ´ 100

ç ÷
è ømax b
c d e
= (ab1 + bc1 + gd1 + de1 )%
Problem 56. The pressure on a square plate is measured by measuring the force on the plate and the length of the sides of the plate. If the maximum error in the measurement of force and length are respectively 4% and 2%, The maximum error in the measurement of pressure is [CPMT 1993]
(a) 1% (b) 2% (c) 6% (d) 8%
Solution : (d)
P = F = F
A l 2
, so maximum error in pressure (P )
æ DP ´ 100ö = DF ´ 100 + 2 Dl ´ 100 = 4% + 2 × 2% = 8%

ç ÷
è ømax F l
Problem 57. The relative density of material of a body is found by weighing it first in air and then in water. If the weight in air is (5.00 ±0.05 ) Newton and weight in water is (4.00 ± 0.05) Newton. Then the relative density along with the maximum permissible percentage error is
(a) 5.0 ± 11% (b) 5.0 ± 1% (c) 5.0 ± 6% (d) 1.25 ± 5%
Solution : (a) Weight in air = (5.00 ± 0.05) N
Weight in water = (4.00 ± 0.05) N
Loss of weight in water = (1.00 ± 0.1) N
Now relative density =
weight in air weight loss in water
i.e. R . D = 5.00 ± 0.05
1.00 ± 0.1
Now relative density with max permissible error = 5.00 ± æ 0.05 + 0.1 ö ´ 100 = 5.0 ± (1 + 10)%
= 5.0 ± 11%


1.00 ç 5.00 1.00 ÷
Problem 58. The resistance R = V
i
where V= 100 ± 5 volts and i = 10 ± 0.2 amperes. What is the total error in R
(a) 5% (b) 7% (c) 5.2% (d) 5 %
2
Solution : (b)
R = V
\ æ DR ´ 100ö
= DV ´ 100 + DI ´ 100 = 5 ´ 100 + 0.2 ´ 100
= (5 + 2)% = 7%

I ç R
÷
ømax V
I 100 10
Problem 59. The period of oscillation of a simple pendulum in the experiment is recorded as 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s respectively. The average absolute error is
(a) 0.1 s (b) 0.11 s (c) 0.01 s (d) 1.0 s
Solution : (b) Average value = 2.63 + 2.56 + 2.42 + 2.71 + 2.80
5
= 2.62 sec
Now
DT_{1} = 2.63 – 2.62 = 0.01
DT_{2} = 2.62 – 2.56 = 0.06
DT_{3} = 2.62 – 2.42 = 0.20
DT_{4} = 2.71 – 2.62 = 0.09
DT_{5} = 2.80 – 2.62 = 0.18
Mean absolute error DT = DT1 +DT2 +DT3 +DT4 +DT5  = 0.54 = 0.108 = 0.11sec
5 5
Problem 60. The length of a cylinder is measured with a meter rod having least count 0.1 cm. Its diameter is measured with venier calipers having least count 0.01 cm. Given that length is 5.0 cm. and radius is 2.0 cm. The percentage error in the calculated value of the volume will be
(a) 1% (b) 2% (c) 3% (d) 4%
Solution : (c) Volume of cylinder V = pr ^{2}l
Percentage error in volume
DV ´ 100 = 2Dr ´ 100 + Dl ´ 100
V r l
= æ 2 ´ 0.01 ´ 100 + 0.1 ´ 100ö
= (1 + 2)% = 3%


ç 2.0 5.0 ÷
Problem 61. In an experiment, the following observation’s were recorded : L = 2.820 m, M = 3.00 kg, l = 0.087 cm,
Diameter D = 0.041 cm Taking g = 9.81 m / s ^{2} using the formula , Y= 4 Mg , the maximum permissible error in Y is
pD^{2}l
(a) 7.96% (b) 4.56% (c) 6.50% (d) 8.42%
4 MgL
DY æ DM
Dg DL
2DD
Dl ö
Solution : (c)
Y = pD ^{2}l
so maximum permissible error in Y =
Y
´ 100 = ç
è M
+ + +
g L D
+ ÷ ´ 100
l ø
= æ 1 + 1 + 1 + 2 ´ 1 + 1 ö ´ 100
= 0.065 ´ 100 = 6.5%


ç 300 9.81 9820 41 87 ÷
Problem 62. According to Joule’s law of heating, heat produced H = I ^{2} Rt, where I is current, R is resistance and t is time.
Solution : (b)
If the errors in the measurement of I, R and t are 3%, 4% and 6% respectively then error in the measurement of H is
(a) ± 17% (b) ± 16% (c) ± 19% (d) ± 25%
H = I ^{2} Rt
\ DH ´ 100 = æ 2DI + DR + Dt ö ´ 100
= (2 ´ 3 + 4 + 6)%
= 16%

H ç I R t ÷
ø
Problem 63. If there is a positive error of 50% in the measurement of velocity of a body, then the error in the measurement of kinetic energy is
(a) 25% (b) 50% (c) 100% (d) 125%
Solution : (c) Kinetic energy
E = 1 mv^{2}
2
\ DE ´100 = æ Dm + 2Dv ö ´100



E ç v ÷
Here Dm = 0 and
Dv ´ 100 = 50%
v
\ DE ´ 100 = 2 ´ 50 = 100%
E
Problem 64. A physical quantity P is given by P=
P is
1
A ^{3} B 2
3
C 4 D 2
. The quantity which brings in the maximum percentage error in
 A (b) B (c) C (d) D Solution : (c) Quantity C has maximum So it brings maximum error in P.