1. IITJEE Syllabus
Units and dimensions, least count, significant figure; Methods of measurement and error analysis for physical quantities pertaining to the following experiments: Experiments based on using vernier calipers and screw gauge (micrometer), determination of g using simple pendulum, Young’s modulus by Searle’s method, Specific heat of a liquid using calorimeter, focal length of a concave mirror and a convex lens using u-v method, speed of sound using resonance column, verification of Ohm’s law using voltmeter and ammeter, and specific resistance of material of a wire using meter bridge and post office box.
2. Physical Quantity
A physical quantity is a quantity that can be measured i.e. a physical quantity is properly defined, has proper units, and its value can be measured by an instrument.
Physical quantities are classified as fundamental and derived quantities. Fundamental Quantities
Fundamental quantities are those that are defined directly by the process of
measurement only. They are not defined in terms of other quantities; their units are not defined in terms of other units.
In mechanics we treat length, mass and time as basic or fundamental quantities. Derived Units
The units of all other physical quantities, which can be obtained from fundamental units, are called derived unit.
System of Units
Some common system of units used in mechanics are given below: Name of System Fundamental unit of
Length Mass Time
F.P.S. Foot Pound Second
C.G.S Centimeter Gram Second
M.K.S. (SI System) Meter Kilogram Second
Illustration 1: Find the unit of speed.
Solution : Speed = m/s ms 1
s m time
distance
2.1 Definitions of Base Units:
1. Meter:
Since 1983, the standard metre is defined as the length of the path travelled by light in vacuum in 458 , 792 , 299 1 th part of a second. 2. Kilogram:
Nowadays the standard kilogram is the mass of a cylinder made of platinum-iridium alloy and stored in a special vaule in the International Bureau of Weights and Measures at Sevres in France.
3. Second:
At present second is defined on the basis of an atomic clock, which uses the energy difference between the two lowest energy states of the cesium atom. When bombarded by microwaves of precisely the proper frequency, cesium atoms undergo a transition from one of these states to other. One second is defined as the time required for 9,192,631,770 cycles of this radiation.
In physics SI system is based on seven fundamental and two derived units.
Sl.No. Basic Physical Quantities Fundamental Unit Symbol
1. Mass kilogram kg
2. Length meter m
3. Time second s
4. Temperature kelvin or Celsius K or C
5. Electric current ampere A
6. Luminous intensity candela cd
7. Quantity of matter mole mol
Sl. No. Supplementary Physical Quantities Supplementary unit Symbol
1. Plane angle radian rad
2. Solid Angle steradian Sr
Prefixes to the Power of 10
The physical quantities whose magnitude is either too large or too small can be expressed more compactly by the use of certain SI prefixes.
Factor of 10 Prefix Symbol
10-1 deci d 10-2 centi c 10-3 milli m 10-6 micro 10-9 nano n 10-12 pico p 103 kilo k 106 mega M 109 giga G 1012 tera T
Illustration 2: Fill in the blank by suitable conversion of units 1 kg m2s-2 = ________ g cm2s-2
Solution : 1kg m2s-2 = 1103 g(102cm)2s-2 = 107g cm2s-2 Exercise 1: (i) What is the value of one micron in centimeter?
2.2 Dimensional Analysis
Dimension of a physical quantity is defined as the power to which the fundamental units have to be raised to represent the derived unit of that quantity.
Uses of Dimensional equations
(i) Conversion of one system of units into another. (ii) Checking the accuracy of various formulae. (iii) Derivation of formula
Illustration 3: Check the accuracy of the relation =
m T 2
1
, where is the frequency, is length, T is tension and m
is mass per unit length of the string.
Solution : The given relation is = m T 2 1
Writing the dimensions on either side, we get LHS = = [T-1] = [MLT-1] RHS = m T 2 1 = 1 1 2 T ML MLT L 1 As LHS = RHS
Dimensionally the formula is correct. Exercise 2: What is the dimension of the practical unit Calorie?
Illustration 4: Convert 1 N into dyne.
Solution: Dimensional formula of force is F = [M1L1T-2]
Now we have to convert M.K.S system into C.G.S system
M = 1kg L = 1m T = 1s (M.K.S)
M = 1g L = 1cm T = 1s (C.G.S)
F = M1L1T-2
F (SI) = [1 kg] [1m] [s2] = [103g] [102] [s2] = 105 dynes.
Limitations of Dimensional Analysis
1. Dimension does not depend on the magnitude of the quantity involved. Therefore, a dimensionally correct equation need not be actually correct. e.g. :– dimension of
T 1 and
T
2. Dimensional method cannot be used to derive relations other than those involving products of physical parameters. e.g. : 2
o at 2 1 t u r r or
y = a cos(t kx) can not be derived using this method.
3. This method cannot be applied to derive formula if in mechanics a physical quantity depends on more than three physical quantities (mass, length, time). e.g. :– T = 2 mgL cannot be derived by using dimensions.
Exercise 3: “The dimension of torque is equal to the dimension of work”. Yet, the two quantities are different. Explain.
Dimensional Formula of Some Physical Quantities
Quantity Dimensions Quantity Dimensions
Acceleration LT2
Angular acceleration T2
Angular frequency/ speed T1
Angular Momentum ML2T1
Angular velocity T1
Area L2
Displacement L
Energy ML2T2
(Total /Kinetic /potential/ Internal)
Force MLT2
Frequency T1
Gravitational Field strength LT2
Gravitational potential L2T2 Length L Mass M Density ML3 Momentum MLT1 Power ML2T3 Pressure ML1T2 Rotational Inertia ML2 Time T Torque ML2T2 Heat ML2T2 Capacitance M1L2T4I2 Charge IT Conductivity M1L3T3I2 Current IT Current Density L2TI1
Electric dipole moment LIT Electric field Strength MLT3I1
Electric Flux ML3T3I1
Electric Potential ML2T3I1
Electromotive force ML2T3I1
Inductance ML2I2I2
Magnetic dipole moment L2T0I
Magnetic field Strength MT0I1
Magnetic Flux ML2T2I1 Magnetic Induction MT2I1 Permeability MLI2I2 Permittivity M1L3T4I2 Resistance ML2T3I2 Resistivity ML3T3I2 Voltage ML2T3I1 Volume L3 Wavelength L Work /Energy ML2T2
Velocity LT1
3.
Errors in Measurement
Significant figure
The number of significant figures in the measured value of a physical quantity gives the accuracy of its value.
The number of digits in a measurement about which we are reasonably sure, plus the one additional digit which is uncertain are significant.
Common rules of counting significant figure a) All non-zero digits are significant.
b) All zeros occurring between two non-zero digits are significant, no matter where the decimal point is, if at all.
c) In a number less than one, all zeros to the right of decimal point and to the left of the first non-zero digit are not significant. [In 0.002308, the underline zeroes are not significant] d) The terminal or trailing zeroes in a number without a decimal point are not significant.
[Thus 123 m = 12300 cm = 123000 mm has three significant figures, the trailing zeroes being not significant].
e) All zeros on the right of the last non-zero digit in the decimal part are significant. [The numbers 3.500 or 0.06900 has four significant figures each]
Significant figures in calculations
Significant figures in Addition and subtraction
The accuracy of a sum or a difference is limited to the accuracy of the least accurate observation.
Rule: Do not retain a greater number of decimal places in a result computed from addition and subtraction than in the observation, which has the fewest decimal places.
Illustration 5: Add and subtract 428.5 and 17.23 with due regards to significant figures
Solution: we have 428.50 428.50
17.23 17.23
Sum 445.73 Difference 411.27
Rounding off the results of the above sum and difference to the first decimal, we have
Correct sum 445.7 and correct difference 411.3 Significant figures in Multiplication and division
When the values of different observations are multiplied or divided, the number of digits to be retained in the answer depends upon the number of significant figures in the weakest link.
Rule: Do not retain a greater number of significant figures in a result computed from multiplication and or division than the least number of significant figures in the data from which the result is computed.
Illustration 6: Multiply 312.65 and 26.4 with due regards to significant figures.
Solution: 312.65 26.4 = 8253.960
But as the weakest link i.e. the data 26.4 has only three significant figures, the correct result of multiplication will be 8250. This is because in 8250, there are three significant figures. Hence, 312.65 26.4 = 8250
Errors
The difference between the true and the measured values of a quantity is the error. Propagation of Errors
(a) Sum and difference of quantities: x = a
bx = (a + b)
(b) Products and quotients of quantities: x = a b x = a/b For both b b a a x x (c) Powers of quantities: x = m n b a lnx = nlna – mlnb differentiating b db m a da n x dx For errors,
Maximum fractional error in x,
b b m a a n x x
(d) When taking the mean () of several uncorrelated measurements of the same quantity, the error is: =
n x ... x1 n = n x , for n measurements.
Illustration 7: 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, = (10.5 0.2) cm b = (5.2 0.1)cm
P = 2 ( + b) = 0.6
Hence perimeter = (31.4 0.6) cm.
Exercise 4: A hypothetical old standard for 1m refers to it as a simple fraction of the distance between the equator and the north-pole. What is this fraction? What would be the practical difficulties in using this as a standard?
4.
Experiments in Physics
4.1 Measurement of Length
The simplest method measuring the length of a straight line is by means of a meter scale. But there exists some limitation in the accuracy of the result:
(i) the dividing lines have a finite thickness.
(ii) naked eye cannot correctly estimate less than 0.5 mm For greater accuracy devices like
(a) Vernier callipers (b) micrometer scales (screw gauge) are used . (a) Vernier Callipers :
It consists of a main scale graduated in cm/mm over which an auxiliary scale (or Vernier scale) can slide along the length. The division of the Vernier scale being either slightly longer and shorter than the divisions of the main scale.
Least count of Vernier Callipers:
The least count or Vernier constant (v.c.) is the minimum value of correct estimation of length without eye estimation. If N division of vernier coincides with (N1) division of main scale, then
Vernier constant = 1 ms 1vs = 1 N 1 ms N = 1 ms
N , which is equal to the value of the
smallest division on the main scale divided by total number of divisions on the vernier scale. Zero error :
If the zero marking of main scale and vernier callipers do not coincide, necessary correction has to be made for this error which is known as zero error of the instrument. If the zero of the vernier scale is to the right of the zero of the main scale the zero error is said to be positive and the correction will be negative and vice versa.
Illustration 8: Consider the following data:
10 main scale division = 1cm, 10 vernier division = 9 main scale divisions, zero of vernier scale is to the right of the zero marking of the main scale with 6 vernier divisions coinciding with main scale divisions and the actual reading for length measurement is 4.3 cm with 2 vernier divisions coinciding with main scale graduations. Estimate the length.
Solution: In this case, vernier constant = (1mm/10) = 0.1 mm Zero error = 6 0.1 = + 0.6 mm
Actual length = (4.3 + 2 0.01) + correction = 4.32 0.06 = 4.26 cm
(b)Screw Gauge (or Micrometer Screw)
In general vernier callipers can measure accurately upto 0.01 cm and for greater accuracy micrometer screw devices e.g. screw gauge, spherometer are used. These consist of accurately cut screw which can be moved in a closely fitting fixed nut by turning it axially. The instrument is provided with two scales:
(i) The main scale or pitch scale M graduated along the axis of the screw.
(ii) The cap-scale or head scale H round the edge of the screw head.
Constants of the Screw Gauge
(a) Pitch : The translational motion of the screw is directly proportional to the total rotation of the head. The pitch of the instrument is the distance between two consecutive threads of the screw which is equal to the distance moved by the screw due to one complete rotation of the cap. Thus for 10 rotation of cap 5 mm, pitch = 0.5 mm
(b) Least count: In this case also, the minimum (or least) measurement (or count) of length is equal to one division on the head scale which is equal to pitch divided by the total cap divisions. Thus in the aforesaid Illustration, if the total cap division is 100, then least count = 0.5mm/100 = 0.005 mm
Zero Error: In a perfect instrument the zero of the main scale coincides with the line of graduation along the screw axis with no 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 vice-versa. The corresponding corrections will be just opposite.
4.2 Measurement of g using a simple pendulum
A small spherical bob is attached to a cotton thread and the combination is suspended from a point A. The length of the thread (L) is read off on a meter scale. A correction is added to L to include the finite size of the bob and the hook. The corrected value of L is used for further calculation. The bob is displaced slightly to one side and is allowed to oscillate, and the total time taken for 50 complete oscillations is noted on a stop-watch. The time period (T) of a single oscillation is now calculated by division.
Observations are now taken by using different lengths for the cotton thread (L) and pairs of values of L and T are taken. A plot of L vs T2, on a graph, is linear. g is given by
g = 42
2
T L
(a) Systematic: Error due to finite amplitude of the pendulum (as the motion is not exactly SHM). This may be corrected for by using the correct numerical estimate for the time period. However the practice is to ensure that the amplitude is small.
(b) Statistical: Errors arising from measurement of length and time.
T T 2 L L g g
The contributions to L, T are both statistical and systematic. These are reduced by the process of averaging.
The systematic error in L can be reduced by plotting several values of L vs T2 and fitting
to a straight line. The slope of this fit gives the correct value of L/T2
4.3 Determination of Young’s Modulus by Searle’s Method The experimental set up consists of two identical wires P and Q of uniform cross section suspended from a fixed rigid support. The free ends of these parallel wires are connected to a frame F as shown in the figure. The length of the wire Q remains fixed while the load L attached to the wire P through the frame F is varied in equal steps so as to produce extension along the length. The extension thus produced is measured with the help of spirit level SL and micrometer screw M attached to the frame on the side of the experimental wire.
On placing the slotted weights on the hanger H upto a permissible value (half of the breaking force) the wire gets extended by small amount and the spirit level gets disturbed from horizontal setting. This increase in length is measured by turning the micrometer screw M upwards so as to restore the balance of the spirit level. If n be the number of turns of the micrometer screw and f be the difference in the cap reading, the increase in length is obtained by
= n pitch + f least count
The load on the hanger is reduced in the same steps and spirit level is restored to horizontal position. The mean of these two observations gives the true increase in length of the wire corresponding to the given value of load.
From the data obtained, a graph showing extension () against the load (W) is plotted which is obtained as a straight line passing through the origin. The slope of the line gives tan = W l Mgl Now, stress = 2 r Mg and strain = L Y = Stress/ strain = L r Mg 2 = r tan L 2
With known values of initial length L, radius r of the experimental wire and tan, Young’s modulus Y can be calculated.
4.4
Specific Heat of a liquid using a calorimeter
The principle is to take a known quantity of liquid in an insulated calorimeter and heat it by passing a known current (i) through a heating coil immersed within the liquid for a known length of time (t). The mass of the calorimeter (m1) and the combined mass of the
calorimeter and the liquid (m2) are measured. The potential drop across the heating coil
is V and the maximum temperature of the liquid is measured to 2.
The specific heat of the liquid (S) is found by using the relation
(m2 m1)S(2o) + m1Sc(2o) = i.V.t
or, (m2 m1)S + m1Sc = i.V.t/(2o) . . . (1)
Here, o is the room temperature, while Sc is the specific heat of the material of the
calorimeter and the stirrer. If Sc is known, then S can be determined.
On the other hand, if Sc is unknown : one can either repeat the experiment with water or
a different mass of the liquid and use the two equations to eliminate m1Sc.
The sources of error in this experiment are errors due to improper connection of the heating coil, radiation, apart from statistical errors in measurement.
The direction of the current is reversed midway during the experiment to remove the effect of any differential contacts, radiation correction is introduced to take care of the second major source of systematic error.
Radiation correction: The temperature of the system is recorded for half the length of time t, (i.e. t/2, where t is the time during which the current was switched on) after the current is switched off. The fall in temperature , during this interval is now added to the final temperature 2 to give the corrected final temperature:
2 = 2 +
This temperature is used in the calculation of the specific heat, S.
Error analysis:
After correcting for systematic errors, equation (i) is used to estimate the remaining errors.
4.5 Focal length of a concave mirror and a convex lens using the u-v method. In this method one uses an optical bench and the convex lens (or the concave mirror) is placed on the holder.
The position of the lens is noted by reading the scale at the bottom of the holder.
A bright object (a filament lamp or some similar object) is placed at a fixed distance (u) in front of the lens (mirror).
The position of the image (v) is determined by moving a white screen behind the lens until a sharp image is obtained (for real images).
For the concave mirror, the position of the image is determined by placing a sharp object (a pin) on the optical bench such that the parallax between the object pin and the image is nil. A plot of |u| versus |v| gives a rectangular hyperbola. A plot of
| v |
1
vs |u1| gives a straight line.
Error: The systematic error in this experiment is mostly due to improper position of the object on the holder. This error may be eliminated by reversing the holder (rotating the holder by 180 about the vertical) and then taking the readings again. Averages are then taken.
The equation for errors gives:
v u v u v v u u f f
The errors u, v correspond to the error in the measurement of u and v. 4.6 Speed of sound using resonance column
A tuning fork of known frequency (f) is held at the mouth of a long tube, which is dipped into water as shown in the figure.
The length (1) of the air column in the tube is adjusted until it resonates
with the tuning fork. The air temperature and humidity are noted.
The length of the tube is adjusted again until a second resonance length (2) is found (provided the tube is long).
Then, 2 1 = /2, provided 1, 2 are resonance lengths for adjacent
resonances.
= 2(2 1), is the wavelength of sound.
Since the frequency f, is known; the velocity of sound in air at the temperature () and humidity (h) is given by c = f = 2(2 1)f
It is also possible to use a single measurement of the resonant length directly, but, then it has to be corrected for the “end effect”:
(fundamental) = 4(1 + 0.3d), where d = diameter
Errors: The major systematic errors introduced are due to end effects in (end correction) and also due to excessive humidity.
Random errors are given by
1 2 1 2 1 2 1 2c
c
4.7 Verification of Ohm’s law using voltmeter and ammeter A voltmeter (V) and an ammeter (A) are connected
in a circuit along with a resistance R as shown in the figure, along with a battery B and a rheostat, Rh Simultaneous readings of the current i and the potential drop V are taken by changing the resistance in the rheostat (Rh). A graph of V vs i is plotted and it is found to be linear (within errors). The magnitude of R is determined by either (a) taking the ratio
i V
and then
(b) fitting to a straight line : V = iR, and determining the slope R. Errors:
Systematic errors in this experiment arise from the current flowing through V (finite resistance of the voltmeter), the Joule heating effect in the circuit and the resistance of the connecting wires/ connections of the resistance. The effect of Joule heating may be minimized by switching on the circuit for a short while only, while the effect of finite resistance of the voltmeter can be overcome by using a high resistance instrument or a potentiometer. The lengths of connecting wires should be minimized as much as possible.
Error analysis:
The error in computing the ratio R = i V is given by i i V V R R
where V and i are of the order of the least counts of the instruments used. 4.8 Specific resistance of the material of a wire using a meter bridge A known length () of a wire is connected in one of
the gaps (P) of a metre bridge, while a Resistance Box is inserted into the other gap (Q). The circuit is completed by using a battery (B), a Rheostat (Rh), a Key (K) and a galvanometer (G).
The balance length () is found by closing key k and momentarily connecting the galvanometer until it gives zero deflection (null point). Then,
Q P = 100 . . . (1)
using the expression for the meter bridge at balance. Here, P represents the resistance of the wire while Q represents the resistance in the resistance box. The key K is open when the circuit is not in use.
The resistance of the wire, P = 2 r L = P L r2 . . . (2)
where r is the radius of wire and L is the length of the wire, r is measured using a screw gauge while L is measured with a scale.
Errors: The major systematic errors in this experiment are due to the heating effect, end corrections introduced due to shift of the zero of the scale at A and B, and stray resistances in P and Q, and errors due to non-uniformity of the meter bridge wire.
Error analysis: End corrections can be estimated by including known resistances P1
and Q1 in the two ends and finding the null point:
1 1 Q P = 1 1 100
. . . (2), where and are the end corrections. When the resistance Q1 is placed in the left gap and P1 in the right gap,
2 2 1 1 100 P Q . . . (3)
which give two linear equation for finding and .
In order that and be measured accurately, P1 and Q1 should be as different from
each other as possible. For the actual balance point,
Q P = 100 = 2 1
,Errors due to non-uniformity of the meter bridge wire can be minimised by interchanging the resistances in the gaps P and Q.
2 2 1 1 P P
where, 1 and 2 are of the order of the least count of the scale.
The error is, therefore, minimum if 1 = 2 i.e. when the balance point is in the middle of
the bridge. The error in is
P P L L r r 2
4.9 Measurement of unknown resistance using a P.O. Box A P.O. Box can also be used to measure an
unknown resistance. It is a Wheatstone Bridge with three arms P, Q and R; while the fourth arm(s) is the unknown resistance. P and Q are known as the ratio arms while R is known at the rheostat arm.
At balance, the unknown resistance
S = R Q P . . . (1)
The ratio arms are first adjusted so that they carry 100 each. The resistance in the rheostat arm is now adjusted so that the galvanometer deflection is in one direction, if R = Ro (Ohm) and in the opposite direction,
this implies that the unknown resistance, S lies between Ro and Ro + 1 (ohm).
Now, the resistance in P and Q are made 100 and 1000 respectively, and the process is repeated.
Equation (1) is used to compute S.
The ratio P/Q is progressively made 1 : 10, and then 1 : 100. The resistance S can be accurately measured.
Errors: The major sources of error are the connecting wires, rusted resistance plugs, change in resistance due to Joule heating, and the insensitivity of the Wheatstone bridge.
These may be removed by using thick connecting wires, clean plugs, keeping the circuit on for very brief periods (to avoid Joule heating) and calculating the sensitivity.
In order that the sensitivity is maximum, the resistance in the arm P is close to the value of the resistance S.
* * *
7
.Answers to Exercise
1. (i)10-4cm (ii) 105N/m2
2. ML2T-2
3. Work is scalar while torque is vector. 4. (1/107), Earth is not a perfect sphere.
8.
Solved Problems
8.1 Subjective
Problem 1: If nth division of main scale coincides with (n+1)th divisions of vernier scale. Given one main scale division is equal to ‘a’ units, find the least count of the vernier.
Solution: (n + 1) division of vernier scale = n division of main scale
one Vernier division = 1 n
n
main scale division
Least count = 1 M.S.D. – 1 V.D. = 1 n 1 M.S. D. = n 1 a
Problem 2: The following measurements were taken for an unknown resistance X, with
a P.O. box: in Ohms
Rate (P) Arms(Q) Rheostat
Arm (R) Galvanometer Deflection 10 10 15 left 10 10 16 right 10 100 152 left 10 100 153 right 10 1000 1524 left 10 1000 1525 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 ~ 2 24 . 15 1525 = 0.005
Problem 3: Two consecutive lengths of a resonance column taken with a tuning fork of frequency 480 Hz at 250C are 54 cm and 88.4 cm respectively. Find the velocity of sound in air.
Solution: If wavelength = ,
/2 = (88.4 54) cm = 34.4 cm = 68.8 cm
c (velocity of sound in air) = 0.688 480 m/s 330 m/s
Problem 4: If all measurements in an experiment are taken upto same number of significant figures then mention two possible reasons for maximum error. Solution: The maximum error will be due to (i) measurement, which is least
(ii) measurement of the quantity which has maximum power in formulas. Problem 5: The initial and final temperature of water as recorded by an observer are
(40.6 0.2)C and (78.3 0.3 )C. Calculate the rise in temperature with proper error limit.
Solution: Let1 = 40.6C, 1 = 0.2 C
2 = 78.3C, 2 = 0.3 C
= 2 – 1 = 78.3 – 40.6 = 37.7C
& = (1 + 2)= (0.2 + 0.3) = 0.5C
Hence rise in temperature = (37.7 0.5)C
Problem 6: A physical quantity x is calculated from the relation x =
d c
b a2 3
. If percentage error in a, b, c and d are 2%, 1%, 3% and 4% respectively. What is the percentage error in x? Solution: As x = d c b a2 3 d d Δ 2 1 c c Δ b b Δ 3 a a Δ 2 x x Δ = [2 2% + 31% + 3% + 2 1 4%= 12%
Problem 7: In the equation y = A sin(t – kx) obtain the dimensional formula of and k. Given x is distance and t is time.
Solution: The given equation is y = A sin(t – kx)
The argument of a trigonometrical function i.e. angle is dimensionless i.e. t = = T 1 T 1 t = [M0L0T-1] Also kx = k = L 1 x = L-1 = [M0L-1T0]
Problem 8: In the Vander Walls equation
2 v a
P (v – b) = RT. Find the dimensions of
a and b.
Solution: The Vander walls equation
2 v a P (v – b) = RT As pressure can be added only to pressure therefore 2
v a
represents pressure P.
i.e. 2 v
a
= P or a = Pv2
[a] = [ML-1T-2] [L3]2 = M1L5T-2
Again, from volume V we can subtract only a volume, therefore b must be representing volume only i.e.
b = v = [L3] = M0L3T0
Problem 9: Find the dimensions of a and b in the relation P =
at 2 x
b where P is power, x is distance and t is time.
Solution: The given relation is P =
at x b 2
As x2 is subtracted from b therefore the dimensions of b are of x2
i.e. b = L2
We can rewrite relation as P =
at L2 = at L2 a =
T T ML L 3 2 2 = M-1L0T2Problem 10: Given that F = at + bt2, where F denotes force and t, time. Find the
dimensions of a and b are respectively. Solution: F = at
2 MLT a T a = MLT3 b = 4 2 F MLT t 8.2 Objective
Problem 1: Dimensional formula of Planck’s constant is
(A) MLT1 (B) ML2T1 (C) ML2T2 (D) none of these. Solution: E = h h =
2 2 2 1 1 ML T E ML T v T Hence (B) is correct.Problem 2: Dimensional formula of Stefan’s constant (A) ML T2 24 (B) ML T2 34 (C) ML T 34 (D) none of these Solution: = 3 4 4 E ML T At Hence (C) is correct.
Problem 3: A physical quantity x is calculated from the relation x =
2 3
a b c d
If percentage error in a, b, c and d are 2%, 1%, 3% and 4% respectively. What is the percentage error in x?
(A)11% (B)13% (C)12% (D)14% Solution: x 2 a 3 b c d x a b c 2d = (22 + 31 + 3 + 1 4 2 ) = 12% Hence (C) is correct.
Problem 4: A cube has side 1.2 102 m. Its volume will be recorded as
(A) 1.728 10 m 6 3 (B) 1.72 10 m 6 3
(C)1.7 10 m 6 3 (D) 0.72 10 m 6 3
Solution: v l3 1.728 10 6
Length has two significant figure v = 1.7 10 m 6 3
Hence (C) is correct.
Problem 5: The dimension of pressure gradient are
(A) ML-2T-2 (B) ML-2T-1
Solution: P = gh 3 2 T L L M g dh dP = 2 2 T L M Hence (A) is correct.
Problem 6: Which of the following equations is dimensionally correct? (A) Pressure = energy per unit volume
(B) pressure = energy per unit area (C) Pressure = force per unit volume (D) pressure=momentum per unit volume
Solution: Volume Energy = volume mv 2 1 2 3 2 2 L T ML = ML-1 T-2
Hence (A) is correct.
Problem 7: An athlete’s coach told his team that muscle times speed equals power. What dimensions does he view for “muscle”?
(A) MLT2 (B) ML2 T-2
(C) MLT–2 (D) L
Solution: Power = force velocity = muscle times speed muscle represents force muscle = [MLT-2]
Hence (C) is correct.
Problem 8: Which of the following is a dimensional constant? (A) refractive index (B) dielectric constant (C) relative density (D) gravitational constant Solution: A gravitational constant G has a constant value and dimensions
Hence (D) is correct.
Problem 9: If force, length and time would have been the fundamental units what would have been the dimensional formula for mass?
(A) FL1 T2 (B) FL1T2 (C) FLT2 (D) F Solution: Let M = K FaLbTc = [MLT-2]a [Lb] Tc = [MaL(a+b)T(-2a+c)] a = 1, a + b = 0 & – 2a + c = 0 a = 1, b = – 1, c = 2 Hence (B) is correct.
Problem 10: The dimensions of the Rydberg constant are (A) M L-1 T (B) MLT-1 (C) ML-1 T (D) MLT2 Solution: From 2 2 2 1 n 1 n 1 R 1 dimension? R = L 1 L 1 = ML-1 T Hence (C) is correct.
9.
Assignment (Subjective Problems)
LEVEL-I
1. 10 rotations of the cap of a screw gauge is equivalent to 5 mm. The cap has 100 dimensions. Find the least count. A reading taken for the diameter of wire with the screw gauge shows 4 complete rotations and 35 on the circular scale. Find the diameter of the wire.
2. A certain pendulum clock with a 12hr dial happens to gain 1 min/day. After setting the clock to the correct time how long it will take to indicate correct time again? 3. The mass of a block is 87.2g and its volume is 25cm3. What are its density upto
correct significant figures?
4. The radius of a sphere is (5.3 0.1) cm. Find the percentage error in its volume. 5. The Van-der-Waal’s interaction between two molecules separated by a distance r
is given by the energy E = 6 r A + 12 r B
LEVEL-II
1. In an experiment for finding the specific heat of alcohol, as copper calorimeter of mass 190 gm is filled with alcohol, and the total mass is found to be 390 gm. When it is heated by using a 50 W heater for 9 minute the following reading were recorded with a thermometer and stop watch.
Time (minute) Temperature (C)
Find the specific heat capacity of alcohol. It is not required to take the radiation correction into account.
2. Suppose, the torque acting on a body, is given by = KL + MI
Where L = angular momentum, I = moment of inertia & = angular speed What is the dimensional formula for KM?
3. When a current of (2.5 0.1)A flows through a wire it develops a potential difference of (20 1)V. What is the resistance of wire?
4. What is the fractional error in g calculated from T = 2 g . Given fractional errors in T and are x and y respectively?
5. A planet of mass m rotates around a star of mass M. The time period of revolution is T, while the average distance of the planet from the star is a. It is known that there exists a relationship between them: find it. Assume the dimensional expression for G.
10. Assignment (Objective Problems)
LEVEL- I1. ML T1 2
is the dimensional formula of
(A) force (B) coefficient of friction
(C) modulus of elasticity (D) energy 2. The dimensional formula of coefficient of viscosity is
(A) MLT1
(B) M L T1 2 2
(C) ML T1 1
(D) none of these
3. On the basis of dimensional equation, the maximum number of unknown that can be found, is
(A) one (B) two
(C) three (D) four
4. If v stands for velocity of sound, E is elasticity and d the density, then find x in the equation x d v E . (A) 1 (B) ½ (C) 2 (D)1/2
5. The multiplication of 10.610 with 0.210 upto correct number of significant figure is
(A) 2.2281 (B) 2.228
(C) 2.22 (D) 2.2
6. The measurement of 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
7. Dimensional formula of latent heat
(A) M0L2T-2 (B) MLT-2
(C) ML2T-2 (D) ML2T-2
8. In case of measurement of ‘g’, if error in measurement of length of pendulum is 2%, the percentage error in time period is1 %. The maximum error in
measurement of g is
(A) 1 % (B) 2 %
(C) 4 % (D) no error.
9. If length of pendulum is increased by 2%. The time period will (A) increases by 1% (B) decreases by 1% (C) increases by 2% (D) decreases by 2%
10. If radian correction is not considered in specific heat measurement. The measured value of specific heat will be
(A) more than its actual value. (B) less than its actual value. (C) remains same as actual value. (D) none of these.
11. The S.I. unit of universal gas constant is
(A) Watt K-1mol-1 (B) N K-1mol-1
(C) JK-1mol-1 (D) erg K-1mol-1
12. The dimensional formula of couple
(A) ML2T-2 (B)MLT-1
(C) ML-1T-1 (D) M1L1T-2
13. An experiment measures quantities a, b, c and x is calculated from x = ab2/c3. If
the maximum percentage error in a, b and c are 1%, 3% and 2% respectively, the maximum percentage error in x will be
(A) 13% (B) 17%
(C) 14% (D) 11%
14. Dimensional formula of thermal conductivity is
(A) ML2T-3-1 (B) ML2T-2-4
(C) ML2T-2-1 (D) MLT-3-1
15. Three measurements 7.1J, 7.2J and 6.7J are made as experiment the result with correct number of significant figures is
(A) 7.1 J (B) 7.06 J
(C) 7.0 J (D) 7J
16. If P represents radiation pressure, c represents speed of light and Q represents radiation energy striking a unit area per second, then non-zero integers x, y and z, such that PxQycz is dimensionless, may be
(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
17. A spherical ball of mass m and radius r is allowed to fall in a medium of viscosity . The time in which the velocity of the body increases from zero to 0.63 times the terminal velocity is called time constant (). Dimensionally can be represented by (A) 6 mr2 (B) g2 mr 6 (C) r 6 m (D) none of these.
18. Which of the following is a possible dimensionless quantity? (A) Velocity gradient (B) Pressure gradient (C) Displacement gradient (D) Force gradient
19. In specific resistance measurement of a wire using a meter bridge, the key k in the main circuit is kept open when we are not taking readings. The reason is (A) the emf of cell will decrease.
(B) the value of resistance will change due to joule heating effect. (C) the galvanometer will stop working.
(D) none of these.
20. In the experiment of verification of Ohm’s law the error in the current measurement is 1%, while that in the voltage measurement is 2%. The error in the resistance has a maximum value of
(A)1% (B) 2%
(C) 3% (D) none of these.
LEVEL - II
1. Which of the following are not the dimensions of calorie?
(A) [ML2T-2] (B) [MLT-2] (C) [ML-2T-1] (D) [ML2T-1]
2. Which of the following are not the dimensional formula for kinetic energy? (A) [M2L2T] (B) [ML2T-2] (C) [M0L-1] (D) [ML2T]
3. Which of the following are dimensionally wrong? (A) Pressure = Energy per unit area
(B) Pressure = Energy per unit volume (C) Pressure = Force per unit volume
(D) Pressure = Momentum per unit volume per unit time 4. Given that: F Adv
dx ,
where F is force, A is area and dv
dx velocity gradient, then which of the following are not the dimensional formula of ?
(A) [ML-1T-1] (B) [ML-1T] (C) [ML-2T-2] (D) [ML2T-1]
5. A particle moving along a straight line with uniform acceleration has velocities 7 m/s at P and 17 m/s at Q,R is the mid point of PQ. Then:
(A) the average velocity between R and Q is 15 m/s
(B) the ratio of time to go from P to R and that from R to Q is 3:2 (C) the velocity at R is 10 m/s
LEVEL - III
MATCH THE FOLLOWING
1. Column I Column II
(A) Amount of substance (P) Second
(B) Time (Q) Kelvin
(C) Temperature (R) Mole
(D) Electric current (S) Ampere
(T) Kilogram 2. Column I Column II (A) Speed (P) M L T1 1 2 (B) Force (Q) M L T1 1 2 (C) Pressure (R) M L T1 2 2 (D) Work (S) M L T2 2 2 (T) M L T0 1 1
Integer Answer Type Questions
3. The unit of force and length are doubled, the unit of energy will be ____ times.
4. The heat dissipated in a resistance can be obtained by the measurement of resistance, the current and time. If the maximum error in the measurement of these quantities is 1%, 2% and 1% respectively, the maximum error in the determination of the dissipated heat is ___%.
5. The relative density of a material is found by weighing the body first in air and then in water. If the weight in air is (10.0 0.1) gm and weight in water is (5.0 0.1) gm then the maximum permissible percentage error in relative density is ____%