Units and Measurement 1.
PHYSICAL QUANTITY, UNITS AND DIMENSIONS
Physical quantity: A quantity that can be measured by instrument, clearly defined and has proper units is called physical quantity. Physical quantities are classified as fundamental and derived quantities. Fundamental units: The physical quantity which does not depend on any other physical quantity is called a fundamental physical quantity such as length; mass and time are called fundamental units. Derived units: The units that can be obtained from fundamental units are called derived units. System of units: There are three systems of units. Name of system Fundamental unit of Length Mass Time F.P.S. Foot Pound Second C.G.S. Centimetre Gram Second M.K.S. (S.I.) Meter Kilogram Second In physics SI system is based on seven fundamental and two supplementary units. (i) Fundamental units: S.No. Basic Physics Fundamental Unit Symbol 1. Mass kilogram kg 2. Length meter m 3. Time second s 4. Electric current ampere A 5. Temperature Kelvin or Celsius K or C 6. Luminous intensity candela Cd 7. Quantity of matter mole Mol (ii) Supplementary units: S.No. Supplementary Quantities 1. Plane angle 2. Solid Angle
Physical Supplementary unit Radian Steradian
Symbol rad sr
Unit & Dimensions & Vectors DEFINITIONS OF BASE UNITS: (i) Meter: The currently accepted definition of meter is the length of path travelled by light in vacuum in 1/299,792,458th second. (ii) Kilogram:
Kilogram is the fundamental unit of mass. It is defined as the mass of a specific cylinder of platinum - iridium kept at the International Bureau of Weights and Measures in Paris. . (iii) Second: Second is the fundamental unit of time. It is defined as 86,400th part of a mean solar day. Second is accurately measured by an atomic clock. (iv) Coulomb: Coulomb is the fundamental unit of charge. It is defined as the charge required to obtain 9109 Newton of force between two equal charges separated at a distance of one meter in vacuum. (v) Candle: Candle is the fundamental unit of luminous intensity. It is defined as luminous intensity observed from a source of monochromatic light of frequency 5401012 Hz, that has an intensity of 1/683 watt per steradian. (vi) Kelvin: Kelvin is the fundamental unit of temperature. It has value of zero where the molecular activity of gases cease. (vii) Mole: Mole is the fundamental unit of quantity of matter. It is defined as amount of substance of a system that contains as many elementary particle as there are in 0.012 kg of carbon-12 (C12).
1.1
BASIC PHYSICAL QUANTITIES
PHYSICAL QUANTITY
SYMBOL DIMENSION MEASUREMENT UNIT
UNIT
Length
s
L
Meter
m
Mass
M
M
Kilogram
Kg
Time
t
T
Second
Sec
Electric charge
q
Q
Coulomb
C
luminous intensity Temperature Angle
I T
C K none
Candela Kelvin Radian
Cd o K None
Mechanical Physical Quantities (derived) PHYSICAL QUANTITY SYMBOL DIMENSION EQUATION Area
A
L2
MEASURMENT SI) square meter
(in UNIT m2
Volume
V
L3
cubic meter
m3
velocity
v
L/T
meter per second
m/sec
angular velocity
1/T
radians per second
1/sec
acceleration
a
L/T2
meter
angular acceleration
1/T2
Force
F
ML/T2
per square m/sec2 second radians per square 1/sec2 second Newton Kg m/sec2
Energy
E
ML2/T2
Joule
Kg m2/sec2
Work
W
ML2/T2
Joule
Kg m2/sec2
Heat
Q
ML2/T2
Joule
Kg m2/sec2
Torque
ML2/T2
Newton meter
Kg m2/sec2
Power
P
ML2/T3
watt or joule/sec
Kg m2/sec3
Density
D or
M/L3
pressure
P
impulse
p
ML/T
kilogram per Kg/m3 cubic meter Newton per square Kg m-1/sec2 meter Newton second Kg m/sec
Inertia
I
ML2
Kilogram square meter Kg m2
luminous flux illumination
C
E
C/L2
entropy
S
lumen (4Pi candle for cd sr point source) lumen per cd sr/m2 square meter joule per degree Kg m2/sec2K
Volume rate of flow kinematic viscosity dynamic viscosity
Q
L3/T
L2/T
ML-1/T2
ML2/T2K
M/LT
cubic meter per second square meter per second Newton second per square meter
m3/sec m2/sec Kg/m sec
specific M/L2 T2 weight Electrical Physical Quantities (derived) Electric I Q/T current emf, voltage, E ML2 /T2 Q potential resistance or R ML2 /TQ2 impedance Electric TQ2 /M2L2 conductivity capacitance C T2 Q2 /ML2 inductance L ML2 /Q2 Current density J Q/TL2
Kg m-2/sec2
Newton per cubic meter Ampere
C/sec
Volt
Kg m2/sec2C
ohm
Kgm2 /secC2
mho
secC2/Kg m3
Farad Henry ampere per square meter coulomb per meter weber per square meter
sec2C2/Kgm2 Kg m2 /C2 C/sec m2
Charge density
Q/L3
magnetic flux, Magnetic induction
B
M/TQ
magnetic intensity magnetic vector potential Electric field intensity Electric displacement
H
Q/LT
ampere per meter
C/m sec
A
ML/TQ
weber/meter
Kg m/sec C
E
ML/T2 Q
D
Q/L2
permeability
ML/Q2
volt/meter or Kg m/sec2 C newton/coulomb coulomb per square C/m2 meter henry per meter Kg m/C2
permittivity, dielectric constant frequency
K f or
T2Q2/ML3 M0L0T0 1/T
farad per meter None Hertz
sec2C2/Kgm3 None sec-1
angular frequency Wave length
1/T L
radians per second Meters
sec-1 M
1.
cubic C/m3 Kg/sec C
Meter: Since 1983, the standard metre is defined as the length of the path travelled
1 th part of a second. 299 , 792458
by light in vacuum in 2.
Kilogram: Now a days the standard kilogram is the mass of a cylinder made of platinum-iridium alloy and stored in a special vault in the International Bureau of Standards in France.
3.
Second: At present second is defined as the time interval of 9,192,631,770 vibrations of electromagnetic radiation in cesium-133 atom (corresponding to the transition between two specific hyperfine levels).
Limitations of Dimensional Analysis 1. Dimension does not depend on the magnitude. Therefore, a dimensionally correct equation need not be actually correct. Eg – dimension of 2.
1 2 and are same. T T
Dimensional method cannot be used to derive relations other than product of
physical parameters. e.g. - r ro ut
1 2 at or y = a cos(t kx) can not be derived 2
by their method. 3.
This method cannot be applied to derive formula if in mechanics a physical quantity depends on more than 3 physical quantities (mass, length, time). e.g. – T = 2 cannot be derived by theory of dimensions.
Dimensions of Physical Quantities Quantity Dimensions Acceleration LT2 Angular acceleration T2 Angular displacement -Angular frequency/ speed T1 Angular Momentum ML2T1 Angular velocity T1 Area L2 Displacement L Energy (Total /Kinetic /potential/ Internal) ML2T2 Force MLT2 Frequency T1 Gravitational Field strength LT2
Quantity Dimensions Velocity LT1 Heat/ Temperature -Capacitance M1L2T2Q2 Charge Q Conductivity M1L3TQ2 Current T1Q Current Density L2T1Q Electric dipole moment LQ Electric field Strength MLT2Q1 Electric Flux ML3T2Q1 Electric Potential ML2T2Q1 Electromotive force ML2T2Q1 Inductance ML2Q2 Magnetic dipole moment L2T1Q
I mgh
Gravitational potential L2T2 Length L Mass M Mass density ML3 Momentum MLT1 Period T Power ML2T3 Pressure ML2T2 Rotational Inertia ML2 Time T Torque ML2T2 Velocity LT1
2.
Magnetic field Strength MT1Q Magnetic Flux ML2T1Q1 Magnetic Induction MT1Q1 Permeability MLQ2 Permittivity M1L3T2Q2 Resistance ML2T1Q2 Resistivity ML3T1Q2 Voltage ML2T2Q1 Volume L3 Wavelength L Entropy ML2T2
APPLICATIONS OF DIMENSIONAL ANALYSIS
(i)
To find the unit of a physical quantity Example-1 G = [M-1L3T-2]. Its SI unit is m3kg-1s-2 or Nm2kg-2. (ii) To convert a physical quantity from one system of units to another system of units n1u1 = n2u2 … (1) ( Where ni and ui are numerical constant unit and dimension in a particular system)
Example-2 Let us convert value of g (i.e. 9.8 m/s2) from SI system to CGS system From eq. no. 1 [ n1u1]in SI = [n2u2]in CGS [u1]in SI [n2]CGS = [n1 ]in SI [u2 ]in CGS L = n1 1 L 2
T 1-2 T 2-2 -2 1m 1sec -2 1cm 1 sec
= 9.8 m/sec2
-2 100cm 1sec = 9.8 -2 1cm 1 sec = 980
(iii)
To check the correctness of a given physical relation Based on principle of homogeneity, the dimensions on two sides must be same for a given relation.
Example-3 Check dimensionally mv 2 where,[F ] [MLT 2 ];[v ] [LT 1] Therefore, LHS [MLT 2 ] r [M ][LT 1 ]2 RHS [MLT 2 ] L If dimensions are same on both sides then the relation is dimensionally correct otherwise incorrect. F
(iv)
To derive a relation
Example-4 Derive Planck’s length in terms of G, c and h, where G is gravitation constant, c velocity of light and h is plank constant. x y z L= f(G, c, h), L = KG c h [L] = [M-1L3T2]x [LT-2]y [ML2T-1]z -x + y = 0, 3x + y + 2z = 1 and –2x – y – z = 0 1 3 1 x ,y and z 2 2 2 Gh Thus, L = K C3 If K = 1 then L - 10-35 m. The importance of Plank’s length is yet to be established. Limitations of dimensional Analysis: (i)
The dimensional analysis cannot be applied to derive relations other than product of power functions, for example, s = ut + 1at2 or y = y0 cos t and so on, cannot be 2 derived directly.
(ii)
The dimensional analysis cannot be applied to derive those relations that involve more than 3 unknowns, however, we can use them to check the correctness of a relation even if variables are more than 3.
(iii)
Even if a physical quantity depends upon 3 quantities, out of which two have same dimension then dimensional analysis cannot be applied to derive such a formula but can be used to check the relation.
(iv)
Numerical constants, trigonometric ratios and ratios which are dimensionless cannot be derived. Physical quantities having same dimensions may not be the same. For example [ML2T-2] is a dimensional relation for torque as well as work or energy.
Polygon Rule of vector addition This an extension of triangle law of addition of vectors. According to this rule ‘n’ number of vector can be added by drawing a diagram in which first vector is taken at the origin and other vectors are added one by one in which tail of next vector coincides with the heat of previous vector. The vector joining the tail of a n ) is first vector ( a1 ) to heat of last vector ( the resultant vector R R (Resultant vector) = a1 + a2 + . . . . . . + a5 = OE Example: There are four forces acting at a common point as shown in figure. Find the resultant F using polygon law of vector addition. (given F1 = 10 N, F2 = 10 2N, F3 = 15 2N and F4 = 20 N)
Solution: Form Polygon law Resultant force
F = OA AB BC CD OD
F
= F1 F2 F3 F4 = [20(cos30 ˆi +sin30 ˆj ) + 102{cos45( ˆi ) + sin45 ˆj } + 152 {cos45( ˆi ) + sin45( ˆj )} + 20 {cos60 ˆi + sin60( ˆj )} ] 1 3 ˆ 1 ˆ 1 ˆ 1 ˆ 1 ˆ 1 ˆ 13 ˆ i j 10 2 i j 15 2 i j 20 ˆi j 2 2 2 2 2 2 2 2 = [(53 ˆi +5 ˆj ) + (10 ˆi + 10 ˆj ) + (15 ˆi 15 ˆj ) + {10 ˆi 103 ˆj )] = [(15 53)( ˆi ) + 103() ˆj
F = 10
F = [(15 53) ˆi + 103 ˆj ] F = 15 5 3 2 10 Let = angle between F and x-axis as shown in the figure. tan =
10 3 195 5 3
= 2.73,
= 63.89
F = 18.45 N at an angle (69.89) with negative x-axis.
3
2
=18.45 N
3.
SOLVED PROBLEMS
SP1. The time period T of oscillation of a gas bubble from an explosion under water is proportional to pa db Ec, where p is the static pressure, d is the density of water and E is total energy of explosion. Find the value of a, b, and c? Key concept: Value a, b and c can be calculated by using dimension of time, pressure, density and energy. Solution: Given T = k pa db Ec, where k is a proportionally constant. Substituting the dimensions of T, p, d and E, we have (T) = (ML-1T-2) a (ML-3) b (ML2T -2) c Equating power of M, L, and T on both sides, we have a+b+c=0 –a –3b + 2c = 0 –2a – 2c = 1 5 1 1 Solving these equations, we get a , b and c 6 2 3 SP2. The frequency(n) of a tuning fork depends upon the length (l) of its prong, the density (d) and Young’s modulus (Y) of its material. Using dimensional consideration, find a relation of n in terms of l, d and Y? Solution: Let n = k la db Yc, where k is a dimensionless constant. Putting the dimensions of n, l, d and Y we have (T-1) = (L)a (ML-3)b (ML-1 T-2)c Equating power of M, L and T on both sides, we get b+c=0 a –3b – c = 0 –2c = –1 These give c = ½, b = –1/2 and a = –1
n = kl –1d-1/2 Y 1/2 or n = k Y l d This is the required relation for the frequency of a tuning fork. SP3. The period of revolution (T) of a planet moving round the sun in a circular orbit depends up on the radius (r) of the orbit, mass (M) of the sum and gravitational constant (G). Using dimensional considerations, obtain Kepler’s third law (the law of periods) of planetary motion? Solution: Given T = k raMbGc Where k is a dimensionless constant. Substituting the dimensions of all the quantities on both sides, we have (T) = (L)a (M)b (M-1L3T-2)c Equating the powers of T, M and L, we get 1 = –2c 0=b–c 0 = a + 3c which give a =
3 2
,b=
1 1 and c = 2 2
Hence T = kr3/2M -1 / 2 G -1 / 2 or T 2
k 2r 3M G
Since k, M and G are constants, T2 r3. This is Kepler’s third law of planetary motion. SP6. Find the dimensions of (a) self inductance (b) resistance (c) potential? Key concept: Use that formula in which you know the dimensions of each physical quantity except finding one. Solution: 1 2 E [ML2L2 ] Li E or L [ML2T 2 A 2 ] (a) 2 2 2 i [A ]
(b) R =
V [ML2T 3 A 1 ] [ML2T 3 A2 ] I [ A]
E [ML2T 2 ] [ML2T 3 A 1 ] (c) QV = E or V Q [ AT ] Keep remember that dimension of current can be taken as Q/T also. SP7. Find the dimensions of compressibility, thermal conductivity, resistivity and coefficient of viscosity. Solution: Compressibility =
1 A L2 = M-1LT+2 Bulk modulus F MLT -2
Thermal conductivity K dQ KAQ dQ 1 ML2T 3L or K = MLT-3K-1 2 dt 1 dt AQ LK Coefficient of viscosity () F A SP8.
dv dx
Fdx [MLT 2L ] 2 [ML1T 1 ] 1 Adv [L ][LT ] Find the dimensions of a and b in van der waal’s equation.
a P V 2 (V b ) nRT Key concept: Quantity having same dimensions can only be subtracted or added so P and a/V 2 has same dimensions. a = PV2 = ML-1T-2 (L6) = ML5T-2 a Since 2 has a dimension of Pressure, V
Solution:
Similarly b has a dimension of volume i.e. is L3.
4.
CONVERSION FACTORS
(i)
1 A.U = 1.4961011m 1X-ray unit = 10-13m 1foot = 30.48 cm 1Chandra Shekhar limit (CSL) = 1.4 times the mass of sun 1 metric Ton = 1000kg 1pound = 0.4537kg 1 atomic mass unit (a.m.u) = 1.67 10-27kg 1shake = 10-8kg 1 year = 365.25d = 3.156107s 1 carat = 200mg 1 bar = 0.1 M Pa = 105Pa 1curie = 3.71010s-1 1 roentgen = 2.58 10-4 C/kg 1quintal = 100kg 1barn = 10-28m2 1standard atmospheric pressure = 1.013105 Pa or N/m2 1mm of Hg = 133N/m2 1horse power = 746w Gas constant, R = 8.36j/mol k = 8.3610-7erg/mol k = 2cal/mol 1 Weber = 108 maxwell 1 tesla = 1wb/m2 = 104 gauss 1amp turn/meter = 410-3 oersted 1electron volt (eV) = 1.6 10-19J 1calorie = 4.19J 1watt-hour = 3.6 103J
(ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) (xii) (xiii) (xiv) (xv) (xvi) (xvii) (xviii) (xix) (xx) (xxi) (xxii) (xxiii) (xxiv) (xxv)
Example-5 The density of water is equal to Solution: Ideally speaking, the examiner should specify the temperature in this question. This is because the density of water varies with temperature. It is maximum (103 kg m-3) at 4C. Example: 5eV in joule?
What will be equivalent energy of
Solution:
5eV = 5×1.6 10-19J = 8.0 ×10-19J joule is the equivalent of ----------
Exercise 2: calorie.
One
5.
SIGNIFICANT FIGURES
All measurements have some degree (quantity) of uncertainty; how great the uncertainty depends on both the accuracy of the measuring device and the skill of its operator. For example, on a compass box scale, the diameter of circle can be measured to the nearest (least measurement) 0.1 cm; diameter differences less than this cannot be detected on this scale. We might therefore indicate the diameter of a cylinder measured on this scale as 15.2 ± 0.1 cm; the ±0.1 (read plus or minus 0.1) is a measure of the accuracy of the measurement of scale. It is important to have some indication of how accurately any measurement is made; the ± notation is one way to accomplish this. It is common to drop the ± notation with the understanding that there is uncertainty of at least one unit in the last digit of the measured quantity; that is, measured quantities are reported in such a way that only the last digit is uncertain. All of the digits, including the uncertain one, are called significant digits or, more commonly, significant figures. The number 2.2 has 2 significant figures, while the number 2.2405 has 5 significant figures.
6.
PERCENTAGE ERRORS
Measured physical quantities are always approximate and cannot be expressed in exact numbers; the difference between the true and the measured values of a quantity is called error. Error may be positive or negative. If there is an error L in measurement of any physical quantity L, then error and
L 100 is called percentage error. L
Combination of Errors (a)
Sum and difference of quantities:
x=ab x = (a + b) (b)
Products and quotients of quantities:
x = a b,
x = a/b
For both x a b x b a (c)
Powers of quantities: n
a bm Inx = nIna – mInb x=
L is called fraction L
differentiating db dx da m = n a b x For errors, Maximum fractional error in x,
x a b n m x a b
SP 13.Let us consider following relation between the function Y and variable a,b and c such that by ; where x, y Y x z a c and z R If there are small error a' , b' and z' to calculate in a, b, and c respectively, then what will be resulting error in Y? Calculate percentage error also. Differentiate the relation Y
Solution:
by , axcz
then we get logY y log b x log a z log c Y b a c y. x. z Y b a c To get maximum permissible error in Y, taking all errors to affect the result in one direction only, we get Y b a c y. x. z Y b a c Percentage error in Y= SP13.
Y b a c 100 y . x. z 100 Y b a c
The sides of a rectangle are (10.5 0.2) cm and (5.2 0.1) cm. Calculate its perimeter with error limit.
Solution: Here, l = (10.5 0.2) cm b = (5.2 0.1) cm
P = 2(l + b) = 2 (10.5 5.2) = 31.4cm P = 2(l + b) = 0.6 Hence perimeter = (31.4 0.6) cm SP14.
The following measurements were taken for an unknown resistance x, with a P.O.box. Rate (P) 10 10 10 10 10 10
Arms(Q) 10 10 100 100 1000 1000
Rheostat Arm(R) 15 16 152 153 1524 1525
Galvanometer Deflection Left Right Left Right Left Right
Find the value of X and the error in X. Solution: The resistance x satisfies (i) 15 < x <16 (ii) 15.2 < x, 15.3 (iii) 15.24 < x < 15.25 The error
15.25 15.24 = 0.005 2
SP15. P
A physical quantity P is related to four measurable quantities a, b, c, and d as 3 2 ab cd The percentage errors in the measurements of a, b, c, and d are 1%, 3%, 4% and 2% respectively. What is the percentage error in the quantity P? If the value of P obtained from the above relation turns out to be 3.763, to what value should you round off the result?
Key concept: Power of physical quantity is multiplier factor of error for that physical quantity. Solution:
Given, P = Proportional error in P is given by
a3 b 2 cd
= a3b2c-1/2d-1
P a b 1 c d 3 2 P a b 2 c d Maximum proportional error in P is a b 1 c d P 3 2 P a b 2 c d max Percentage error in P is b 1 c d P a 100 3 100 2 100 100 100 P a b 2 c d max = 3 (percentage error in a) + 2 (percentage error in b) +
1 2
(percentage error in c) + (percentage error in d) 1 = 3 1% + 2 3% + 4% + 2% 2 = 3% + 6% + 2% + 2% = 13 % Now, value of P = 3.763. The percentage error in P is 13%. Error in P is 3.763 13 = 0.48919 = 0.5 100 Since the error in P is in the first decimal place, the value of P = 3.763 must be rounded off to the first decimal place. The value of P up to appropriate significant figures is 3.8. P =
Example-6SP16.
The potential difference across a wire is measured with a voltmeter having least count 0.2 volt and the current in the wire is measured with an ammeter having least count 0.1 ampere. The following readings were obtained. Voltmeter reading (V) = 6.4volt Ammeter reading (I) = 2.0 ampere Find the value of the resistance of the wire with maximum permissible error. Also find the maximum percentage error.
Solution: is given by
From Ohm’s law, the resistance of the wire
V 6.4 = 3.2 ohms I 2.0 The maximum proportionate error in R is R V I 0.2 0.1 = 0.03125 + 0.05 = 0.08125 R V I 6.4 2.0 R=
R = 0.08125 R = 0.08125 3.2 = 0.26 Since the error is in the first decimal place, we must round off the value of R to the first decimal place, giving R = 0.3. Thus the value of R with maximum permissible error is R + R = (3.2 0.3) ohm The maximum percentage error is R 100 = 0.08125 100 = 8.125% = 8% R SP17. One atmospheric pressure is equal to Solution: 1 atmospheric pressure = 76 cm of Hg = 76 13.6 981 dyne cm-2 = 1.01 106 dyne cm-2 = 1.01 105 N m-2Two resistance R1 = 10.0 0.1 and R = 5.0 0.1 are connected (a) in series (b) in parallel. Calculate % error. Solution: (a) In series Rs = R1 + R2 Rs R1 R2 0.11 0.12 100 100 100 = 1.33% Rs R1 R2 10 5 (b) In Parallel R1R2 RP = R1 R2 Percentage error =
= 4.3%
RP R1 R2 R1 R2 100 RP R2 R1 R2 R1 0.1 0.1 0.1 0.1 100 = 5 10 5 10
Example-7 If C is the capacity and R is the resistance, then the dimensional formula of 1 is CR Solution:(B) CR is time constant of CR circuit.SP18. A student measured the length of the pendulum 1.211m using a meter scale and time for 25 vibrations as 2 minutes 20 second using his wristwatch. Find % error and absolute error in g. Solution: 0.84 + 1.43 = 2.27%
% error
g L 2T 100 100 g T L
=
L 2T 0.01 2 1 g 9.8 = T L 1.21 140
g (0.0227 9.8) = 0.22 ms-2
7.
EXPERIMENTS BASED ON VERNIER CALIPERS & SCREW GAUGE
A meter scale can measure accurately up to one–tenth part of one cm. Its least measurement 0.1 cm, is called least count of scale. There is limitation of meter scale that the meter scale cannot measure the value less than 0.1 cm. For greater accuracy measurement we have devices such as, (i) Vernier Callipers (ii) Screw Gauge (i) Vernier Callipers: A vernier callipers provides with an auxiliary (or vernier) scale in addition to the main scale. The vernier scale can slide along the main scale. The vernier scale is so graduated (or marked) that the length of total number of divisions on it is smaller by length of one division on main scale. The least count of vernier scale is calculated by using the following formula value of 1 main scale division Least count of vernier scale (or vernier constant)= Total number of division on vernier scale n or Least count (vernier constant) = 1 M.S.D.(Main scale division) – 1 V.S.D.(vernier scale division). Exercise 4: depend on ---------------
Least
count
of
screw
gauge
Example-8SP19.
If N division of vernier coincides with (N – 1) division of main scale. Given one main scale division is equal to ‘a unit’, find the least count of the vernier.
Solution:
N 1 a Vernier constant = 1MSD – 1VSD = 1 MSD = , N N
Generally, the value of 1 main scale division on vernier callipers is 0.1 cm and there are 10 divisions on the vernier scale, i.e., x = 0.1 cm and n = 10. Least count of vernier callipers =
0.1 cm = 0.01 cm. 10
Zero error of vernier callipers: If the zero marking of main scale and vernier callipers does not coincide, necessary correction has to be made for this error which is known as zero error of the instrument. If the zero error of the vernier scale is to the right of the zero of the main scale the zero error is said to be positive & the correction will be negative otherwise vice versa. (ii) Screw gauge: Screw gauge works on the principle of screw. It has a linear scale called the main scale, and another scale called the circular scale. The circular scale can be rotated by a head screw. On turning the screw, the circular scale advances linearly on the main scale. The distance moved by the tip of screw when it is given one complete rotation, is called the pitch of the screw. Dividing the pitch of screw by the total number of division on the circular scale, we get the distance which the screw advances on rotating the screw by 1 division on its circular scale. This distance is called the least count of the instrument. Thus Pitch Least count = Total number of division n on the circular scale Generally, in a screw gauge the screw advances by 1 or ½ division on main scale when the screw is given one rotation. If there are total 100 division on its circular scale and the 0.1 cm value of 1 division on main scale is 0.1 cm, then least count = = 0.001 cm 100 Zero error of screw gauge: In a perfect instrument the zeros of the main scale and circular scale coincides with each other, In this condition screw gauge has zero-error, otherwise the instrument is said to have zero-error which is equal to the cap reading with the gap closed. This error is positive when zero line or reference line of the cap lies above the line of graduation and corresponding corrections will be just opposite otherwise viceversa.
Example-9SP20.
What
will
be
the
measurement
of
following screw gauge position?
0
5
5 0 45 40
Solution: Reading = Main scale reading + Number of circular scale division(or screw gauge reading)least count 3mm + 45 0.01mm = 3.45 mm
Example-10SP21.
What will be the measurement of
following screw gauge position?
0
Solution:
5
25 20 15 10
Reading = Main scale reading + Number of circular scale division(or screw gauge reading)least count 5.5mm + 16 0.01mm = 5.66 mm
SP22.
What will be measurement of following vernier callipers gauge position? 0
0.5
1 Main scale
Vernier scale
Coinciding division Solution: Reading = main scale reading + vernier scale positionleast count = 0.1cm + 4 (due to fourth line of vernier matches with main scale) .01cm = 0.1+ 0.04= 0.14cm SP23. Find the absolute error in density of a sphere of radius 10.01 cm and mass 4.692kg. Least count of vernier callipers and balance is 0.01cm and 0.001kg respectively. M 4.692 3 3 6 4 Solution: = = 1.12 10-3 kg m-3 r 3 4 3.14 (10.01) 10 3 M 3r M r M 3r 0.001 3 01 1.12 103 = 3.97 kgm-3 r 4.692 10.01 M SP24. In an experiment for determining the acceleration due to gravity (g), the length of the string is measured with the help of a meter scale having a least count of 0.1 cm, the diameter of the bob is measured with a venire calipers having
venire constant 0.01 cm and the time period is measured using a stop watch having a least count of 0.1s. the following observations were obtained. Length of the string (l) = 98.0 cm Diameter of the bob (d) = 2.56 cm Time for 10 oscillations (t) = 20.0s Find the maximum permissible error in g and express the result in the form (g + g). use the formula T = 2
L g
Where L = l + r, where r = radius of the bob =
d 2.56 = 1.28 cm 2 2
t 20.0 = 2.0 s 10 10 4 (3.142)2 (98.0 1.28) g= = 980.1 cm–2 = 9.801 ms–2 (2.0)2 In terms of measured quantities, g is giving by (l r ) t l r 2 t 2 g= 2 t /100 t (l r t 0.1 0.01 2 0.1 = = 0.0011 + 0.01 = 0.0111 (98.0 1.28) 20.0 Error g = 0.0111 g = 0.0111 9.801 = 0.108 ms–2 Thus we find that the error in g expressed in ms –2 is in the first decimal place. Hence we must round off g to the first decimal place, giving g = 0.1 ms–2. Since the error in g is in the first decimal place, the value of g = 9.801 ms –2. Hence the result of the experiment is expressed as g = (9.8 0.1) ms–2. The maximum percentage error is
Solution:
T=
g 100 = 0.0111 100 = 1.11% g
OBJECTIVE 1.
The mass of a box measured by a grocer’s balance is 2.3kg. Two gold pieces of mass 20.15g and 20.17 g are added to the box. The total mass of the box to correct significant figure is (A) 2.3kg (B) 2.34032 kg (C) 2.34 kg (D) 2.3403 kg Which of the following sets cannot enter into the list of fundamental quantities in any system of units?
(A) length, time and mass (B) mass, time and velocity (C) length, time and velocity (D) gravitational constant Solution:(C) Since velocity is derivable from length and time therefore it cannot be grouped with length and time as fundamental quantity. 2.
Sleman is S.I unit for (A) Specific-ConductanceResistance (C) Capacitance Solution 2: (A)
(B) (D)
Inductance Pressure
3.
A science student takes 100 observations in an experiment. Second time he takes 500 observations in the same experiment. By doing so the possible error becomes (A) 5 times (B) 1/5 times (C) Unchanged (D) None of these Solution 3: (B) (1/5 times)
4.
The unit of surface energy per unit area may be expressed (A) Nm–2 (B) Nm–1 (C) Nm (D) Nm2 Solution 4: (B) Surface energy per unit Energy Surface energy per unit area = Area Force displacement Area N m Nm1 m2 5.
Density of a liquid is 13.6 gcm–3. Its value in SI units is (A) 136.0kgm–3 (B) 13600kgm–3 (C) 13.60kgm–3 (D) 1.360kgm–3 Solution 5: (B) Density = 13.6 g cm–3 13.6 10 3 kg = (102 m)3 = 13600 kg m–3 [Q 1 g = 10–3 kg, 1 cm = 10–2 m]
6.
The diameter of a cylinder is measured with vernier callipers having least count 0.01cm. The diameter is 1.95 cm. The radius to correct significant figures will be (A) 0.975cm (B) 0.98cm (C) 1.0cm (D) 1cm If the size of a unit be represented by k and is numerical value as n, then
1 k 1 (C) n k2 (D) n k Solution:(B) Value = nk. Since value is fixed therefore nk = constant. (A)
nk
(B)
n
7.
The SI unit of the universal gas constant R is (A) Erg K–1 MOL–1 (B) -1 -1 (C) Newton MOL (D) Joule Solution 7: (B) (mol k)
Watt K-1 MOL-1 Jule–1 MOL–1
8.
The maximum error in the measurement of mass and density of the cube are 3% and 9% respectively. The maximum error in the measurement of length will be (A) 9% (B) 3% (C) 64% (D) 2% Mass(M) M Solution 8: (C) Density = = Volume(V) V M V= M l3 = Max. fractional error 3l M l M Percentage error 3 % = 3% + 9% l l 12 % % l 3 l %= 4 % l 9.
What is the value of 600m + 600mm with due regard to the significant digits (A) 601m (B) 600m (C) 600.600m (D) 600.6m The SI unit of electrochemical equivalent is (A) kg C (B) C kg-1 -1 (C) kg C (D) kg2C-1
Solution:(C) According to Faraday's first law of electrolysis, m = ZQ or Z
m . So, SI unit of Q
Z is kg C-1 10.
Which of the following has a dimensional constant (A) Refractive index (B) Passion’s ratio (C) Relative velocity (D) Gravitation of constant Solution 10: (D) All of physical quantity has no dimension except gravitational force so correct. The dimensions of surface tension length are (A) ML0T-2 (B) -1 –2 (C) ML T (D) 0 –2 -2 Solution 11: (B) ML T L= MLT 11.
Pick the odd man out (A) Weight (C) Electromotive force Solution 12: (C)
MLT-2 ML–2T–2
12
(D)
Dimension formula for luminous flux is (A) ML2T-2 (C) ML2T–1 Solution 13: (D)
(B) Thrust Force
13
14.
(B) (D)
ML2T-3 None of these
If w, x, y and z are mass, length, time and current respectively, then dimensional formula same as (A) electric potential (C) electric field
(B) (D)
x2w has y3 z
capacitance permittivity
Sol. : (A) x 2 w ML2 3 T3 A y z x 3 w ML2T 2 [Work] 3 AT [Charge] y z
y 3 w [Work] [Potential] 14 The dimension of potential difference are 3 [Charge] y z
(A) ML2T-3I-1 (C) ML2T-2I
(B) MLT-2I-1 (D) MLT-2I
MLT-1 T-1 are the dimension of (A) Power (C) Force Solution 15: (C) 15.
(B) (D)
Momentum Couple
16.
The unit of impulse is the same as that of (A) Moment of force (B) Linear momentum (C) Rate of change of linear momentum (D) Force Solution 16: (B) Impulse = Force time = MLT–2 T = MLT–1 i.e. Dimension of linear momentum 17.
The dimensions of capacitance are (A) M-1L-2TI2 (C) ML-2T-2I-1 Q Solution 17: (D) C = V Q Q2 T 2 T2 2 I C= W / Q W T2 W 2 T 2I2 = ML2T 2 = M–1L–2T4I2
(B) (D)
M-1L-2T2I-2 M-1L-2T4I2
The dimension of angular momentum length are (A) MLT-1 ML2TML3T-1 (C) ML-1T 0 -2 ML T Solution 18: (B) ML3T–1 18.
19.
(B) (D)
The SI unit of the universal gas constant R is (A) 1
(B)
watt
mol-1
(C) joule K-1mol-1 Solution 19: (D) 20.
erg K-1mol-1 newton K-1mol-1
The dimension of plank’s are the same as those of (A) energy (C) angular frequency angular momentum
(D)
(B)
power (D)
K-
Solution 20: (D) E = h E ML2T 2 Dimension of (h) = 1/ T 2 –1 h = ML T Dimension of angular momentum = ML 2T–1 Plank’s constant h =
21.
The volume V of water passing any point of a uniform tube during t seconds is related to the cross-sectional area A of the tube and velocity u of water by the relation V Aut Which one of the following will be true? (A)
==
(B)
(C)
=
(D)
= Solution 21: (B) V = k. u t L3 = k (L2). (LT–1). (T) L3 = k . L(2 + ) T– + 2 + = 3 – + = 0 = , 2 + = 3 so are can conclude that = 22.
Which one of the following relations is dimensionally consistent where h is height to which a liquid of density rises in a capillary tube of radius, r, T is the surface ension of the liquid, the angle of contact and g the acceleration due to gravity
2T cos r g 2 g cos h (C) 2Tr Solution 22: (A) (A)
23
h
The dimension of calories are (A) ML2T-2 (C) ML2T–1
(B) (D)
(B) (D)
2Tr g cos 2Tr g h cos h
MLT-2 ML2T-1
Solution 23: (A) Calories is unit of energy so dimension of calories is = ML 2T–2 The dimension of potential difference length are (A) ML2TML3T-3I-1 (B) MLT-2I-1 2 -2 (C) ML T I (D) MLT-2I work ML2 T 2 length L Solution 24: (A) V = charge Q T T 3 –3 –1 = ML T I 24.
The mass of an electron is 9.1110–31kg. The speed of light is 2.9979108ms–1. Calculate the energy corresponding to the mass of the electron, from the Einstein relation E = mc2, correct to appropriate significant figures. (A) 81.9 10–15J (B) 801.9 10–12J (C) 81.9 10–12J (D) 81 10–15J What is the power of a 100 W bulb in cgs units? (A) 106 erg/s (B) 107 erg/s (C) 109 erg/s (D) 1011 erg/s Solution:(C) 100 W = 100 J s-1= 100 107 ergs-1. 25.
v , where 0 is the permittivity of free space, l is the t length, v is a potential difference and t is a interval. The dimensional formula for X is the same as that of (A) Resistance (B) Charge (C) Voltage (D) Current 2 v N M C = C That isunit of current C Solution: (D) x = 0l = 2 M t sec M N sec 26.
A quantity X is given by 0l
27.
Let (0) denote the dimensional formula for the permittivity of the vacuum, and (0) that of the permeability of the vacuum. If M = mass, L = length, T = time and A = electric current (A) [0] =[M–1L–3T2A] (B) [0] = [M–L–3T4A2]
(C) Solution: 28.
[0] = [MLT–2A–2] (B)
The dimensions of (A) (C)
(A2L3T4M–4) (A0M0L0T0)
(D)
[0] = [M– 1L–3T–2A]
(B) (D)
(A–2T–4L3M) (AT2L–3M–1)
e2 are 2 0 (hc )
Solution:
(C)
C2 C2 (N m2 ) Joule sec meter sec 1
If a measurement is made as 24.5 0.1cm, what is the percentage error? (A) 0.2% (B) 0.4% (C) 0.41% (D) 0.1% Density of liquid is 15.7 g cm-3. Its value in the International System of Units is (A) 15.7 kg m-3 (B) 157 kg m-3 -3 (C) 1570 kg m (D) 15700 kg m-3 Solution:(D) 15.7 g cm-3 = 15.7 10-3kg(10-2m)-3= 15700 kg m-3 29.
630.
On the basis of dimensional equation, the maximum number of unknown that can be found is (A) One (B) Two (C) Three (D) Four Solution: (C)
731.
If v stands for velocity of sound, E is elasticity and d the density, then find x in the x d equation v = E (A) 1 (B) ½ (C) 2 (D) –½
Solution:
d (D) V = E
x
ML 3
LT–1 =
MLT
2
x
/ L2
x
LT–1 = L–3x + x T2x –1 = 2x 1 x= 2 The dimension of ½0E2 (0 is permittivity of free space and E is electric field) are (A) MLT-1 (B) ML2T-2 (C) ML-1T-2 (D) ML2T-1 2 –2 Solution: (B) Dimension of energy = ML T 832.
933.
The measurement of a radius of a circle has error of 1%. The error in measurement of its area is (A) 1% (B) 2% (C) 3% (D) None of these
A weber is equivalent to (A) A m-2 (C) A m2 Solution:(D) 1 T = 1 Wb m-2
(B) A m-1 (D) T m-2
1034. With the usual notation, the equation tan is (A) (B) (C) (D) Solution:
rg said to give the angle of banking v2
Numerical correct only Dimensionally correct only Both numerical & dimensionally correct Neither numerical nor dimensionally correct (C)
1135. When light travels through glass, the refractive index is found to vary with the wavelength as = A + B/2 , what is dimension of B ? (A) L (B) L2 -1 (C) L (D) L-2 Solution: (B) Dimension of wavelength = L2 Dimension of refractive index = M0L0T0 Dimension of B is L2 1236. The dimension of ½0E2 (0 is permittivity of free space and E is electric field) are (A) MLT-1 (B) ML2T-2 (C) ML-1T-2 (D) ML2T-1 Solution: (B) 137.
A travelling wave in a stretched string is described by the equation y = A sin (kx-t) The dimension of k is (A) M0L-1T0 (B) M0L0T0 0 2 0 (C) MLT (D) MLT-1 2 Solution: (A) k = Dimension of k = L–1 1438. Dimension formula of Stefan’s constant (A) ML2T-2-4 (C) ML0T-3-4 Solution: (C)
(B) (D)
ML2T-3-4 M0LT-1
1539. Of the following quantities, which one has dimensions different from the remaining three (A) Energy per unit volume
(B) (C) (D) Solution:
Force per unit area Product of voltage and charge per unit volume Angular momentum (C)
1640. The dimension equation for magnetic flux is (A) ML2T-2I-1 (C) ML-2T-2I-1 Solution: (A) = B. A MLT 2 L2 F = .A= IL I L = ML2T–2I–1 417.1 The dimension of the Rydberg constant are (A) M0L-1T (C) M0L–1T0ML-1T0 Solution: (C) M0L–1T0
(b) (d)
ML2T-2I-2 ML-2T-2I-2
(B) (D)
MLT1 ML0T2
1842. The pairs of physical quantities which have same dimension are (A) Reynolds number and coefficient of friction (B) Latent heat and gravitational potential (C) Curie and frequency of light wave (D) Plank’s constant and torque Solution: (B) 1943. In the relation x = 3yz2, x and z represent the dimensions of capacitance and magnetic induction respectively. What will be the dimension of y (a) M-3L-2T4Q4 (B) M2L-2T4Q4 (c) M-2L-2T4Q4 (D) M-3L-2T4Q 2 Solution: (A) x = 3yz x dimension of y = 2 z Capici tance = (Magnetic induction)2 M1L2T 2Q2 (MT 1Q1 )2 = M-3 L-2 T+4 Q4 =
2044. A sphere has a mass of 12.2 kg .1kg and radius 10cm 0.1 cm. The maximum % error in density (A) 3.83% (B) 3.84% (C) 3.74% (D) 3.94%
A sextant is used to measure (A) area of hill (B) height of an object (C) breadth of a tower (D) volume of the building. Solution:(B) The height of a tree, building, tower, hill etc. can be determined with the help of a sextant. 2145. A cube has side 1.210-2m. Its volume will be recorded as (A) 1.72810-6m3 (B) 1.7210-6m3 (C) 1.710-6m3 (D) 0.7210-6m3 What is the dimensional formula of coefficient of linear expansion? (A) [ML2T-2K-1] (B) [MLT-2K-1] (C) [M0L0TK-1] (D) [M0L0T0K-1] lt l0 Solution:(D) lt = l0(1 + t) or l0 t 2246.
In the measurement of a physical quantity X
A2B C1/ 3D3
The percentage errors
introduced in the measurement of the quantities A, B,C and D are 2%, 2%, 4% and 5% respectively. Then the minimum amount of percentage of error in the measurement of X is contributed by (A) A (B) B (C) C (D) D A pressure of 106 dyne cm-2 is equivalent to (A) 105 N m-2 (B) 104 N m-2 (C) 106 N m-2 (D) 107 N m-2 Solution:(A) Remember the conversion factor of 10. 2347. The Vander Waal’s equation for a gas is (P+a/v2)(V-b) = nRT. The ratio b/a will have the following dimensional formula (A) M-1L-2T2 (B) M-1L-1T-1 2 2 (C) ML T (D) MLT-2 Solution: (A) Dimension of (b) = L3 Dimension of (a) = ML5 T-2 b L3 Dimension of = M1L2 T 2 a ML5 T 2 2448. If the time period of a drop of liquid of density d, radius r, vibrating under surface tension s is given by the formula t (d a r bs c )1/ 2 and if a = 1, c = -1, then b is (A) 1 (B) 2 (C) 3 (D) 4 Solution: (C) T = (M L-3)a/2 Lb/2 (ML0T-2)c/2 M0L0 T = Ma/2 + c/2 L-3a/2+b/2 T-c
3a b 0 2 2 - 3 a + b =0 b=3a b = 3 1 b=3 2549. If P represents radiation pressure, C represents speed of light and Q represents radiation energy striking a unit area per second, then the non-zero integers x, y and z, such that PxQyCz is dimensionless are (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 0 0 0 x y z Solution: (B) M L T = P Q C = (M L-1 T-2)x (ML2T-2)y (LT-1)z x + y = 0 x = -y -x + 2y +z = 0 -2x – 2y –z = 0 x = -y
kEZ e 2650. In the relation ,P is pressure Z is distance k is Boltzman constant and is the temperature. The dimension formula of will be P
M0L2T0 (B) M1L2T-1 ML0T-1 (D) M0L2T1 z Solution 26: (A) is dimension less quantity k k dimension of = z 2 2 1 ML T K = K L = MLT–2 Dimension of is equal to dimension of pressure P (A) (C)
P=
ML–1T–2 =
MLT 2
MLT 2 ML1T 2 = M0L2T0 =
2751. Velocity v, acceleration a and force f are taken as fundamental quantities, then angular momentum will have the dimension (A) fv2a-2 (B) f2v2a-2 (C) fv3a-2 (D) None of thesef2 Solution: (D) Angular momentum (L) vx ay fz MLT–2 = LT 1
52.
x
LT MLT 2
y
2
MLT–2 = Mz Lx +y + z T–x –2y –2z 1=z 1=x+y+z –2 = – x – 2y –2z z = 1 x+y=0 x = –y –2 = –x + 2x –2 0 = x, y = 0 Angular momentum(L) = f Fund the unit of acceleration time? (A) ms–1 (C) ms+1
Solution:
z
(B) (D)
ms–3 ms+2
m Velocity V m (A) Acceleration = C time t s 2 s ms 1 time S s s
What is the unit of current Resistance. (A) amps (B) (C) coulomb (D) Solution: (B) 53.
volt farad
54.
What will be equivalent energy of 5eV in joule? (A) 8.0 × 10–22J (B) 8.0 × 10–19J (C) 8.0 × 10–25J (D) 8.0 × 10–26J Solution: (B) 55.
One joule is the equivalent of ? 1 calorie (A) 5.19 1 calorie (C) 6.19 1 calorie Solution: (B) 4.19 56.
(B) (D)
Least count of screw gauge depend on ?
1 calorie 4.19 1 calorie 8.19
(A) (B) (C) (D) Solution:
Main scale division circular scale no. of circular scale division Main scale division & no. of circular scale division (D)
57.
Least count of vernier calipers depend on? (A) Main scale division (B) vernier scale (C) no. of vernier scale division (D) Main scale division & no. of vernier scale division Solution: (D) 58.
Least count of spherometer depend on ? (A) Main scale division (B) circular scale (C) no. of circular scale division (D) Main scale division & no. of circular scale division Solution: (D) What is the dimension of angular frequency time? (A) Dimension less (B) sec–2 –3 (C) sec (D) sec+1 Solution: (A) 59.
What is the dimension of wave length Frequency? (A) M (B) LT–1 (C) T (D) MT–1 Solution: (B) 60.
