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Å (^) All the quantities which are used to describe the laws of physics are called physical quantities, e.g. length, mass, volume, etc.
Å (^) Physical quantities are of two types, first is fundamental quantities which are independent of other physical quantities and second is derived quantities which can be derived from the fundamental quantities.
Units
Å (^) Measurement of any physical quantity involves comparison with a certain basic, arbitrarily chosen, internationally accepted reference standard called unit.
Å (^) The units which are used to represent fundamental quantities are called fundamental or base units. However, the unit of derived quantities which are represented in terms of fundamental units are called derived units.
Å (^) Some of the commonly used systems of units for measurement with the base units for length, mass and time are given as (i) CGS system (Centimetre, Gram and Second, respectively) (ii) FPS system (Foot, Pound and Second, respectively) (iii) MKS system (Metre, Kilogram and Second, respectively)
International System of Units (SI)
Å (^) The system of units which is at present internationally accepted for measurement is the international system of units. Å (^) This system contains 7 fundamental units and 2 supplementary units which are tabulated as follows. Fundamental units Fundamental quantity Fundamental unit Symbol Length metre m Mass kilogram kg Time second s Electric current ampere A Temperature kelvin K Luminous intensity candela cd Amount of substance mole mol
Supplementary units Supplementary quantity Supplementary unit Symbol Plane angle radian rad Solid angle steradian sr
Units and
Measurements
CHAPTER > 02
KEY NOTES
Measurement of Length
Å (^) There are two methods for the measurement of length as follows. (i) Direct Method In this method, measurement of length involves the use of (a) a metre scale (10 −^3 to 10^2 m) (b) Vernier callipers (upto 10 −^4 m) (c) screw gauge and spherometer (upto 10 −^5 m) (ii) Indirect Method This method is used to measure large distances such as the distance of planet or a star from the earth. e.g Parallax method , etc.
Å (^) The apparent shift in the position of an object with respect to another when we shift our eye sidewise is called parallax. The distance between two points of observation is called the basis.
Å (^) While measuring the distance D of a far away planet S by the parallax method observing from two different positions ( A and B ) as shown below
We get, D
b
θ
, where θ is called the parallax angle or
parallactic angle.
Å (^) Some special units of length for measurement of short lengths are as follows 1 fermi ( )f = 10 − 15 m 1 angstrom ( Å )= 10 − 10 m
Å (^) Some special units of length for measurement of large lengths are as follows 1 astronomical unit (1 AU) = 1 496. × 1011 m 1 light year (ly) = 9 46. × 1015 m 1 parsec = 3 08. × 1016 m
Measurement of Mass
Å (^) For measuring mass of atoms and molecules, we use unified atomic mass unit (u), where 1 atomic mass unit (u) = 1 66. × 10 − 27 kg
Other units that are used for measuring mass are (i) Pound = 0 4536. kg (ii) Slug = 1459. kg Å (^) Large masses in universe like planets, stars, etc. based on Newton’s law of gravitation can be measured by using gravitational method. Å (^) Small masses of atomic/subatomic particles, etc. can be measured by the use of mass spectrograph.
Measurement of Time
Å (^) To measure any time interval, we use atomic standard of time, which is based on the periodic vibrations produced in a cesium atom. This is the basis of the cesium clock, called atomic clock. Å (^) This clock has very high accuracy of part in 10^3.
Accuracy, Precision of Instruments
and Errors of Measurement Å (^) The accuracy of a measurement is a measure of how close the measured value is to the true value, while precision tells us to what resolution the quantity is measured. Å (^) Difference in the true value and the measured value of a quantity is called error of measurement. Å (^) The errors in measurement can be broadly classified as systematic and random errors (i) The systematic errors are those errors that tend to be in one direction, either positive or negative. Some sources of systematic errors are given as (a) Instrumental Errors It arises due to imperfect design or calibration of the measuring instrument. (b) Imperfection in Experimental Technique or Procedure It occurs due to external conditions such as change in temperature, humidity, wind velocity, pressure, etc. during experiment. (c) Personal Errors It arises due to an individual’s bias, lack of proper setting of the apparatus or individual’s carelessness in taking observations without observing proper precautions. (ii) The random errors are those errors which occur irregularly and hence are random with respect to sign. Å (^) The least count error is the error associated with the resolution of the instrument. It occurs with both random and systematic errors.
KEY NOTES
S
D D
A b
B
θ
CHAPTER 02 >Units and Measurements^13
Å (^) Rules for arithmetic operations with significant figures are as follows (i) In multiplication or division , the final result should retain as many significant figures as are there in the original number with the least significant figures. (ii) In addition or subtraction , the final result should retain as many decimal places as are there in the number with the least decimal places.
Rounding Off
Å (^) The process of omitting the non-significant digits and retaining only the desired number of significant digits, in corporating the required modifications to the last significant digit is called rounding off the number.
Å (^) Rules for rounding off a measurement
(i) If the number lying to the right of cut-off digit is less than 5, then the cut-off digit is retained as such. However, if it is more than 5, then the cut-off digits is increased by 1. e.g. x = 5 34. is rounded off to 5 3. to two significant digits and x = 5 328. is rounded off to 5 33. to three significant digits. (ii) If the insignificant digit to be dropped is 5, then the rule is (a) If the preceding digit is even, the insignificant digit is simply dropped, e.g. x = 6 265. is rounded off to to x = 6 26. to three significant digits. (b) If the preceding digit is odd, the preceding digit is raised by 1. e.g. x = 6 275. is rounded off to x = 6 28. to three significant digits.
Å (^) Rules for determining the uncertainty in the results of arithmetic calculations (i) If a set of experimental data is specified to n -significant figures, a result obtained by combining the data will also be valid to n -significant figures.
(ii) The relative error of a value of number specified to significant figures depend not only on number of significant figures but also on the number itself. (iii) Intermediate results in a multi-step computation should be calculated to one and more significant figures in every measurement than the number of digits in the least precise measurement.
Dimensions Analysis
Å (^) Dimensions of any physical quantity are those powers which are raised on fundamental units to express the unit of that physical quantity. Å (^) Dimensional symbols of seven fundamental quantities are given as Length → [L], Mass → [M], Time → [T], Electric current (^) → [A], Temperature (^) → [K], Luminous intensity → [cd], Amount of substance → [mol]. Å (^) Dimensional formula of a physical quantity is an expression which shows how and which of the fundamental quantities represent the dimensions. Å (^) An equation obtained by equating a physical quantity with its dimensional formula is called the dimensional equation of the physical quantity. Å (^) Principle of Homogeneity of Dimensions Only those physical quantities can be added or subtracted which have same dimensions. Å (^) Applications of dimensions (i) To check the correctness of any physical quantity. (ii) To convert any physical quantity from one system of units to another system of units. n u 1 1 (^) = n u 2 2 where, n 1 & n 2 are the magnitudes and u 1 & u 2 are the units of any physical quantity in two systems of units. (iii) To find a relation between interdependent physical quantities.
1 Amongst the following physical quantities which one
has the same unit in all three system of units?
(a) Length (b) Mass (c) Time (d) None of the above
2 The solid angle subtended by the periphery of an area
1 cm 2 at a point situated symmetrically at a distance
of 5 cm from the area is
(a) 2 × 10 −^2 sr (b) 4 × 10 − 2 sr (c) 6 × 10 −^2 sr (d) 8 × 10 −^2 sr
3 If the value of work done is 10^10 gcm 2 s −^2 , then its
value in SI units will be
(a) 10 kg-m^2 s−^2 (b) 10^2 kg-m^2 s−^2 (c) 10^4 kg-m^2 s−^2 (d) 10^3 kg-m^2 s−^2
4 The unit of thermal conductivity is NEET 2019
(a) J m−^1 K−^1 (b) W mK−^1 (c) W m−^1 K−^1 (d) J mK−^1
5 The damping force on an oscillator is directly
proportional to the velocity. The unit of the constant
of proportionality is CBSE AIPMT 2012
(a) kg-ms−^1 (b) kg-ms−^2 (c) kgs−^1 (d) kg-s
6 The moon is observed from two diametrically
opposite points A and B on earth. The angle θ
subtended at the moon by the two directions of
observation is 1°54′; given that the diameter of the
earth to be about 1.276×10^7 m. Compute the distance
of the moon from the earth.
(a) 4.5 × 10 9 m (b) 3.8 3 × 10 8 m (c) 2.5 × 10 4 m (d) 4 × 10 7 m
7 The ratio of the volume of the atom to the volume of
the nucleus is of the order of
(a) 10^15 (b) 10^25 (c) 10^20 (d) 10^10
8 Which of the technique is not used for measuring
small time intervals?
(a) Electrical oscillator (b) Atomic clock (c) Spring oscillator (d) Decay of elementary particles
9 Age of the universe is about 10^10 yr, whereas the
mankind has existed for 10^6 yr. For how many
seconds would the man have existed, if age of
universe were 1 day?
(a) 9.2 s (b) 10.2 s (c) 8.6 s (d) 10.5 s
10 To reduce the least count error, instruments need
higher
(a) precision (b) accuracy (c) mean value (d) true value
11 A device which is used for measurement of length to
an accuracy of about 10 −^4 m, is
(a) screw gauge (b) spherometer (c) vernier callipers (d) Either (a) or (b)
TOPIC 2 ~ Measurement of Length, Mass and Time
MULTIPLE CHOICE QUESTIONS
Mastering NCERT
TOPIC 1 ~ International System of Units
TOPIC 3 ~ Accuracy, Precision of Instruments
and Errors in Measurement
24 In 4700 m, significant figures are
(a) 2 (b) 3 (c) 4 (d) 5
25 The number of significant figures in the numbers
48000. × 104 and 48000.50 are respectively,
(a) 5 and 6 (b) 5 and 7 (c) 2 and 7 (d) 2 and 6
26 Each side of a cube is measured to be 7.203 m. What
are the total surface area and the volume of the cube
to appropriate significant figures?
(a) 373.7 m , 311.3 m^2 3 (b) 311.3 m , 373.7 m^2 (c) 273.4 m , 342.4 m^2 3 (d) 423.4 m , 437.4 m^2
27 5.74 g of a substance occupies 1.2 cm 3. Express
its density by keeping the significant figures in
view.
(a) 4.9 g cm−^3 (b) 5.2 g cm−^3 (c) 4.8 g cm−^3 (d) 4.4 g cm−^3
28 Find the value of 12.9 g − 7.0 6 g. Keeping significant
figures in consideration.
(a) 5.84 g (b) 5.8 g (c) 5.86 g (d) 5.9 g
29 If 3.8 × 10 −^6 is added to 4.2 × 10 −^5 , then the result in
significant figures will be
(a) 4.58 × 10 −^5 (b) 4.6 × 10 −^5 (c) 45 × 10 −^5 (d) None of these
30 The numbers 5355. and 5345. on rounding off to
3 significant figures will give
(a) 5 35. and 5 34. (b) 5 36. and 5 35. (c) 5 35. and 5 35. (d) 5 36. and 5 34.
31 The relative error in the value of a number specified
to significant figures depends on
(a) number of significant figure (b) number itself (c) Both (a) and (b) (d) None of the above
32 The quantity having same dimension as that of
Planck’s constant is
(a) work (b) linear momentum (c) angular momentum (d) impulse
33 Which set of physical quantities has same
dimensions?
(a) Force and power (b) Torque and energy (c) Torque and power (d) Force and torque
34 Dimensions of force are JIPMER 2018
(a) [ M LT^2 −^1 ] (b) [ MLT −^2 ] (c) [ M L^2 − 1 T^ −^2 ] (d) [ MLT −^1 ]
35 If mass M , distance L and time T are fundamental
quantities, then dimensions of torque are JIPMER 2019
(a) ML T^2 −^2 [ MLT −^2 ] (c) [ MLT] (d) [ MLT T^2 ]
36 A quantity f is given by f
hc
G
5
, where c is speed of
light, G universal gravitational constant and h is the
Planck’s constant. Dimension of f is that of
JEE Main 2020 (a) area (b) volume (c) momentum (d) energy
37 Obtain the dimensional formula for universal gas
constant.
(a) M L^2 T −^2 mol −^1 K −^1 [ML^3 T −^1 mol −^2 K −^2 ] (c) M 2 LT −^1 mol −^1 K −^1 [M 3 LT −^2 mol −^1 K −^2 ]
38 The dimension of stopping potential V 0 in
photoelectric effect in units of Planck’s constant h ,
speed of light c and gravitational constant G and
ampere A is JEE Main 2020
(a) h −^ 2 3/^ c −^ 1 3/^ G^ 4 3/^ A −^1 (b) h 1 3/^ G^ 2 3/^ c 1 3/^ A −^1 (c) h G^2 3 2/^ c^ 1 3/^ A −^1 (d) h 2 3/^ c^ 5 3/^ G^ 1 3/^ A −^1
39 If speed ( V ), acceleration ( A ) and force ( F )are
considered as fundamental units, the dimension of
Young’s modulus will be JEE Main 2019
(a) [ V− 4 A −^2 F^ ] (b) [ V− 2 A F^2 2 ] (c) [ V− 2 A F^2 −^2 ] (d) [ V− 4 A F^2 ]
40 If surface tension ( S ), moment of inertia ( ) I and
Planck’s constant ( ) h , were to be taken as the
fundamental units, the dimensional formula for linear
momentum would be JEE Main 2019
(a) S 1 2/^ I 1 2/^ h^ −^1 (b) S 3 2/^ I^ 1 2/ h^0 (c) S 1 2/^ I 1 2/^ h^0 (d) S 1 2/^ I^ 3 2/^ h^ −^1
TOPIC 5 ~ Dimensional Analysis
TOPIC 4 ~ Significant Figures
41 The force of interaction between two atoms is given
by F
x
kT
exp
2
; where x is the distance, k is
the Boltzmann constant and T is temperature and α
and β are two constants. The dimension of β is
JEE Main 2019 (a) [MLT−^2 ] (b) [M L T^0 2 −^4 ] (c) [M LT^2 −^4 ] (d) [M L T^2 2 −^2 ]
42 The potential energy of a particle varies with distance
x from a fixed origin as U
A x
x B
, where A and B are
constants. The dimensions of AB are
(a) [ML5/ 2 T −^2 ] (b) [ML^2 T −^2 ] (c) [M 3/ 2L^3 T −^2 ] (d) [ML7/ 2 T −^2 ]
43 A physical quantity of the dimensions of length that can
be formed out of c, G and
e^2
is [ c is velocity of
light, G is universal constant of gravitation and e is
charge] NEET 2017
(a)
2
0
1 2
c
G
e πε
/ (b) c G 2 e^2 0
1 2
4 πε
/
(c)
2
0
1 2
c
e G πε
/ (d)
2
c 0
G
e πε
44 In the formula, X = 3 YZ 2 , X and Z have dimensions
of capacitance and magnetic induction. The
dimensions of Y in MKSQ system are AIIMS 2018
(a) [ M −^3 L− 2 T^4 Q^4 ] (b) [ ML T Q^2 8 4 ]
(c) [ M −^2 L− 3 T Q^2 4 ] (d) [ M −^2 L^ − 2 TQ^2 ]
45 If energy E , velocity v and time T are chosen as the
fundamental quantities, the dimensional formula of
surface tension will be CBSE AIPMT 2015
(a) [Ev −^2 T−^1 ] (b) [Ev −^1 T^ −^2 ] (c) [Ev −^2 T−^2 ] (d) [E v^2 −^1 T^ −^3 ]
46 The expression for viscous force F acting on a tiny
steel ball of radius r moving in a viscous liquid of
viscosity η with a constant speed v with the help of
the method of dimensional analysis is
(a) kr η v (b) kr^2 η v (c) kr η v 3 2/^ (d) kr η^2 v
47 If dimensions of critical velocity vc of a liquid
flowing through a tube are expressed as [ η ρ x^ y^ rz ],
where η ρ, and r are the coefficient of viscosity of
liquid, density of liquid and radius of the tube
respectively, then the values of x , y and z are given by
CBSE AIPMT 2015 (a) 1, − 1 , − 1 (b) − 1 , −1 1, (c) − 1 , − 1 , − 1 (d) 1 1 1, ,
48 Given that the amplitude of the scattered light is
(i) directly proportional to amplitude of incident light
(ii) directly proportional to the volume of the scattering
dust particle
(iii) inversely proportional to its distance from the
scattering particle and
(iv) dependent upon the wavelength λ of the light.
Then, the relation of intensity of scattered light with
the wavelength is
(a)
λ 2 (b)^
λ^4
(c)
λ 6 (d)^
λ^7
49 The density of a material in CGS system is
10 g cm −^3. If unit of length becomes 10 cm and unit
of mass becomes 100 g, the new value of density will
be
(a) 10 units (b) 100 units (c) 1000 units (d) 1 unit
50 Find the value of power of 60 J per min on a system
that has 100 g, 100 cm and 1 min as the base units.
(a) 2.16 × 104 units (b) 2.16 × 106 units (c) 3 × 104 units (d) 4 × 107 units
51 When 1 m, 1 kg and 1 min are taken as the
fundamental units, the magnitude of the force is
36 units. What will be the value of this force in CGS
system?
(a) 10^5 dyne (b) 10^3 dyne (c) 10^8 dyne (d) 10^4 dyne
52 In a new system of units, unit of mass is α kg, unit of
length is β metre and unit of time is γ second. In this
system, 10 J will be represented as
(a) 10 α −1β^2 γ^2 (b) 10 α −2^ β −^1 γ−^2 (c) 10 α −1β^ −2^ γ^2 (d) 10 αβ γ 2 −^2
(a) Error in a measurement is equal to the sum of true value and measured value of the quantity. (b) Systematic errors occur in both directions, positive and negative. (c) Random errors occur irregularly and at random, in magnitude and direction. (d) In constant errors, errors affect each observation by the different amount.
66 Choose the incorrect statement.
(a) Change in unit does not change the number of significant figure. (b) In 4 700. m = 4700 mm a change of significant figure occur from 4 to 2 due to change in unit. (c) In 4 700. m = 4 700. × 103 cm, there is no change in the number of significant figures. (d) None of the above
III. Matching Type
67 Match the Column I (physical quantity) with
Column II (unit) and select the correct answer from
the codes given below.
Column I Column II A. Capacitance 1. (^) Am 2 B. Inductance 2. Wb C. Magnetic flux 3. Coulomb (volt)−^1 D. Magnetic moment 4. Ohm-second
A B C D (a) 3 4 2 1 (b) 1 2 4 3 (c) 3 2 1 4 (d) 4 1 3 2
68 Match the Column I (unit) with Column II (value)
and select the correct answer from the codes given
below.
Column I Column II A. 1 barn 1. 200 mg B. 1 are 2. (^1013). × 105 Pa C. 1 bar 3. (^102) m^2 D. 1 carat 4. 10 −^28 m^2
A B C D (a) 3 2 1 4 (b) 3 4 2 1 (c) 4 3 2 1 (d) 4 3 1 2
69 Match the Column I (type of errors) with Column II
(cause) and select the correct answer from the codes
given below.
Column I Column II A. Instrumental errors 1. External conditions such as change in temperature, humidity, etc. B. Imperfection in experimental technique
- Person’s carelessness
C. Personal errors 3. Either positive or negative in one direction. D. Systematic errors 4. Imperfect design or calibration of the measuring instruments.
A B C D (a) 4 1 2 3 (b) 4 2 3 1 (c) 2 1 4 3 (d) 4 1 3 2
70 Match the Column I (number) with Column II
(number of significant figures) and select the correct
answer from the codes given below.
Column I Column II A. 0.004608 1. 3 B. 8.9000 2. 5 C. 186 3. 6 D. 2.00891 4. 4
A B C D (a) 1 4 2 3 (b) 4 2 1 3 (c) 3 4 1 2 (d) 4 2 1 3
71 Match the Column I (physical quantity) with
Column II (dimension) and select the correct answer
from the codes given be low.
Column I Column II A. Boltzmann constant ( ) k 1. [ML T^2 −^1 ]
B. Coefficient of viscosity (η) 2. (^) [ML− 1 T −^1 ]
C. Planck constant ( ) h 3. (^) [MLT −^3 K −^1 ]
D. Thermal conductivity ( K ) 4. (^) [ML T^2 −^2 K −^1 ]
A B C D (a) 3 2 1 4 (b) 4 2 1 3 (c) 4 1 3 2 (d) 1 4 2 4
NCERT
72 If G = 667. × 10 − 11 N m- 2 kg −^2 , then the value of G in
CGS system of units is
(a) 6 67. × 10 −^8 (b) 6 67. × 10 −^9 (c) 6 67. × 10 −^10 (d) 6 67. × 10 −^11
73 A new unit of length is chosen in which length is
equals to the speed of light in vacuum is taken as
unity. If light takes 8 min 20 s to reach from the
surface of the sun to the earth, find the distance in
terms of new unit.
(a) 200 units (b) 300 units (c) 400 units (d) 500 units
74 Which of the following is the most precise device for
measuring length?
(a) A vernier callipers with 20 divisions on the sliding scale (b) A screw gauge of pitch 1 mm and 100 divisions on the circular scale (c) An optical instrument that can measure length to within a wavelength of visible light (d) Both (b) and (c)
75 The number of significant figures in 0.007 m 2 ,
0.2370 g cm −^3 and 6.032 Nm−^2 respectively are
(a) 3, 4, 4 (b) 1, 4, 4 (c) 4, 5, 4 (d) 1, 3, 4
76 The length, breadth and thickness of a rectangular
sheet of metal are 4.234 m, 1.005 m and 2.01 cm,
respectively. Give the area and volume of the sheet to
correct significant figures.
(a) 8.7 m , 0.092 m^2 3 (b) 8.4 m , 0.095 m^2 (c) 9.2 m , 0.001^2 3 (d) 8.72 m , 0.0855 m^2
77 The mass of a box measured by a grocer’s balance is
2.3 kg. Two gold pieces of masses 20.15 g and
20.17 g are added to the box. What is (i) the total
mass of the box (ii) and the difference in the mass of
the pieces to correct significant figures?
(a) 2.8 kg, 0.08 g (b) 2.9 kg, 0.02 g (c) 2.3 kg, 0.02 g (d) 3.0 kg, 0.02 g
78 A physical quantity P is related to four observations
a b c , , and d as follows : P = a b^3^2 / c ⋅ d
The percentage errors of measurement in 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
calculated using the above relation turns out to be
3 763. , to what value should you round off the result?
(a) 10%, 3.76 (b) 13%, 3. (c) 10%, 3.8 (d) 13%, 3.
79 A book with many printing errors contains four
different formulae for the displacement y of a particle
under going a certain periodic motion.
I. y a
t
T
= sin
II. y = a sin vt
III. y
a
T
= t a
^
^
sin ( / )
IV. y
a t
T
t
T
sin cos
where, a = maximum displacement of the particle,
v = speed of the particle, T = time period of motion.
Which are the correct formulae on dimensional
grounds?
(a) I, II and III (b) Both III and II (c) Both I and IV (d) Both II and I
80 A parsec is a convenient unit of length on the
astronomical scale. It is the distance of an object that
will show a parallax of 1′′(second of arc) from
opposite ends of a base line equal to the distance from
the earth to the sun. How much is parsec in terms of
metres?
(a) 1495. × 1011 m (b) 3 08. × 1016 m (c) 4 85. × 10 −^6 m (d) 2 05. × 1016 m
81 The sun is a hot plasma (ionised matter) with its inner
core at a temperature according 10^7 K and its outer
surface at a temperature of about 6000 K. At these
high temperatures, no substance remains in a solid or
liquid phase. In what range do you expect the mass
density of the sun to be? (Take, mass of the sun
= 20. × 1030 kg and radius of the sun = 70. × 108 m)
(a) 14. × 103 kg m− 3 (b) 16. × 103 kg m−^3 (c) 2 × 103 kg m− 3 (d) 4 × 103 kg m−^3
82 When the planet jupiter is at a distance of
824 7. million kilometres from the earth, its angular
diameter is measured to be 35 72. s of arc. The
diameter of jupiter is
(a) 3 5. × 106 km (b) 2 7. × 106 km (c) 14. × 105 km (d) 17. × 107 km
83 It is claimed that two cesium clocks, if allowed to run
for 100 years without any disturbance may differ by
only about 0.02 s. What is the accuracy of the clock
in measuring a time interval of 1s?
(a) 10 −^10 (b) 10 −^11 (c) 10 −^5 (d) 10 −^8
MULTIPLE CHOICE QUESTIONS
NCERT &NCERT Exemplar
2 ( b ) Given, dA = 1cm^2 and r = 5 cm
∴Solid angle, d
dA r
= 0 04. sr = 4 × 10 − 2 sr
3 ( d ) Given, work done, W = 1010 gcm^2 s−^2 which is in CGS system of units.
In SI unit, W = (^10102)
g cm^2 s
− − (^10 )
10 ( 3 kg )( 4 m^2 ) s = 103 kg-m s^2 −^2 4 ( c ) The coefficient of thermal conductivity is given by
K
L
A T
dQ dt
where, L = length of conductor, A = area of conductor, ∆ T = change in temperature and
dQ dt
= rate of flow of heat.
∴Unit of K = ×
× = −^ −
metre metre (kelvin)
watt Wm 1 K^1 ( )^2
5 ( c ) Given, damping force ∝ velocity F ∝ v ⇒ F = kv where, k = constant of proportionality.
⇒ k
F
v
∴ Unit of k
F
v
− −
Unit of − Unit of
kg -ms ms
kg s
2 1
1
6 ( b ) We have, θ = °1 54 ′ = ( 60 + 54 )′ = 114 ′ = ( 114 × 60 )′′ Since, (^1) ′′ = 4 85_._ × 10 − 6 rad = ( 114 × 60 ) ′′ × (4.85 × 10 − 6 )rad
= 3.33 × 10 − 2 rad Also, diameter of earth, b = 1 276_._ × 107 m Hence, the earth-moon distance is given as
D = b /θ = × × −
7 2 =^ 3.83^ ×^10 m
8
7 ( a ) We know that, radius of atom, ra = 10 − 10 m Radius of nucleus, rn = 10 − 15 m
∴ Ratio,
r r
a n
− −
10 15
5
Ratio of volume = =
3
3
π^3
π
r
r
r r
a
n
a n
= ( 10 5 )^3 = 1015
8 ( c ) Since, frequency of spring oscillator is smallest among all the given options, hence small time interval cannot be measured with spring oscillator. 9 ( c ) Given, age of mankind = 106 yr and age of universe = 1010 yr
Magnification in time (^) = Age of mankind Age of universe
6 10
4
∴Apparent age of mankind = 10 − 4 × 1 day = 10 − 4 × 86400 s = 8.64 s ≈ 8.6 s 12 ( b ) The range of variation of time over the seven days of observations is 162 s for clock 1 and 31 s for clock 2. The average reading of clock 1 is much closer to the standard time than the average reading of clock 2. The important point is that a clock’s zero error is not as significant for precision work as its variation. It is because the zero error can be easily corrected. Hence, clock 2 is to be preferred over clock 1. 13 ( a ) The mean period of oscillation of the pendulum,
T mean
= 2.624 = 2.62 s
The absolute errors in the measurements are ∆ T 1 = 2.63 s −2.62 s = 0.01 s ∆ T 2 = 2.56 s − 2.62 s = −0.06 s ∆ T 3 = 2.42 s − 2.62 s = −0.20 s ∆ T 4 = 2.71 s −2.62 s = 0.09 s ∆ T 5 = 2.80 s −2.62 s = 0.18 s The arithmetic mean of all the absolute errors is
∆ T ∆ Ti i
mean =
5
= [(0.01 + 0.06 + 0.20 + 0.09 + 0.18)] / 5
= 0.54 / 5 = 0.108 −~0.11 s 14 ( a ) Mean of five observations,
O mean = =
μ
Absolute errors in the observations are ∆ O 1 = 80 0. −μ ∆ O 2 = 80 5. −μ ∆ O 3 = 810. −μ ∆ O 4 = 815. −μ ∆ O 5 = 82 − μ ∴Arithmetic mean of all the absolute errors,
∆
O
O
n
i i
mean =
1
5
μ μ μ μ μ
Substituting the value of μ, we get
∆ O mean =
∴Mean percentage error = ×
∆ O
O
mean mean
= ×
Hints & Explanations
Hints & Explanations
15 ( b ) As we know, time period of oscillation is T
= 2 π
L
g
So, g = 4 π 2 L T^ /^2 Therefore, relative error in g is ( ∆ g g / ) = ( ∆ L L / ) + 2 ( ∆ T T / ) Given, ∆ L = 1mm = 01. cm, L = 20 cm, ∆ T = 1s and T = 90 s
⇒
∆ g g
Thus, the percentage error in g is
= ×
∆ g g
100% = 0 027. × 100 % = 2 7. % ~ −^3 %
16 ( b ) The error in the measurement of mass 1.02 g is ± 0.01 g, whereas that of another measurement 9.89 g is also ± 0.01g. ∴The relative error in 1.02 g = [ ± 0.01/ 1.02) ×100% = ± 0 98. % ≅ ±1% Similarly, the relative error in 9.89 g = ( ± 0.01/ 9.89) ×100% = ± 0.1% The relative errors in measurement of two masses are ± 1% and ± 01. %.
17 ( c )Q Density, ρ = =
Mass Volume
M
L^3
or ρ =
M
L^3
⇒ Error in density^
∆ρ ρ
M ∆
M
L
L
So, maximum % error in measurement of ρ is ∆ρ ρ
× =
× +
100 100 ×
M
M
L
L
or % error in density = 15. + 3 × 1 % error = 4 5. %
18 ( a ) By ascent formula, we have surface tension
T
rhg = × 2
N
m
= ×
dhg 4
N
m
Q r
d =
∆ T ∆ ∆
T
d d
h h
= + [given, g is constant]
So, percentage = × = +
×
∆ T ∆ ∆
T
d d
h h
×
×
×
×
×
− −
− −
2 2
2 2
∆ T
T
× 100 = 15. %
19 ( c ) Given, length of pendulum, l = 25 0. cm So, there is an uncertainty of 0.1 cm in measurement of length. Resolution of stopwatch is 1s. So, uncertainty in measurement of time is 1s.
Now using, T
l g
= 2 π or g
l T
4 π^2 2
We have,
g ∆ g
l l
T
T
×
g g
^
×
l l
T
T
Accuracy in measurement of g is ∆ × = +
×
×
g g
= ( .0 004 + 0 04. )×100 %
20 ( b ) Given, (^) R 1 = 100 & (^) ∆ R 1 = 3 Ω, and R 2 = 200 & ∆ R 2 = 4 Ω The equivalent resistance of parallel combination, 1 1 1 R ′ R 1 (^) R 2
= + …(i)
R
R R
R R
×
1 2 1 2
Then from Eq. (i), we get ∆ R ∆ ∆ R
R
R
R
R
2 =^1 +
1
2
2 2
2
R R
R
R
R
R
R
1
2
(^2 ) 2
2
×^ +
×^ =^ ≅
2 2
.. ..
∴Equivalent resistance, R ′ = ( 66 7. ± 18. ) Ω 21 ( b ) Given, R = 65 Ω ∆, R = 1 Ω , l = 5 mm = 5 × 10 − 3 m, ∆ l = 01. mm = 0.1 × 10 − 3 m , d = 10 mm = 10 × 10 − 3 m, ∆ d = 0 5. mm = 0 5. × 10 − 3 m
∴ Resistivity, ρ =
RA
l
⇒ ρ
π π = =
R d l
Rd l
2 2
∴ ∆^ ρ ∆^ ∆^ ∆ ρ
=^ R + +
R
d d
l l
Substituting the given values, we get ∆ρ ρ
×
×
+^
×
×
− −
− −
3 3
3 3
∆ρ ρ
∆ρ ρ
So, % error in calculation of resistivity = 01354. × 100 % = 13.5% ≈ 13%.
22 ( a ) Given, X
A B
C D
2 1 2 1 3 3
/ / The percentage error in X is given by ∆ X ∆ ∆ X
A
A
B
B
× =
×^ +^
100 2 100 ×
×^ +^
×
∆ C ∆
C
D
D
% % …(i) Hints & Explanations
35 ( a ) Dimensions of torque, (^) τ = Dimension of force, F
× Dimensions of displacement, r = [ MLT− 2 ] [ L]= [ ML T^2 −^2 ]
36 ( d ) Dimensions of quantity f are
[ ]
[ ] [ ]
[ ]
f
h c
G
1 2
5 2 1 2
…(i)
As, h
E
ν
[ h ] = [ML T^2 − 2 ] [T]= [ML T^2 − 1 ] c = [LT− 1 ]
and G
F r m
2
⇒ [ G ]=
[MLT^ − ] [L ]
[M ]
2 2 2 =^
[M− 1 L T^3 − 2 ]
So, dimensions of f using Eq. (i),
[ ]
[ ] [ ]
f =
− −
− −
ML T LT
[M L T ]
2
1 3 2
1 2
1
1 2 1
5 2
1 2
1 2
5 2
3 2
1 1 2
5 2
2 (^2) = [ML T^2 − 2 ]
Thus, it is the dimensions of energy.
37 ( a ) According to ideal gas equation, i.e. pV = nRT , where
n is the number of moles of gases.
∴Universal gas constant, R
p V n T
Dimensional formula of R =
[ML^ − T− ] [L ]
[mol] [K]
1 2 3
= [ML T^2 − 2 mol− 1 K −^1 ]
38 ( * ) Let V (^) 0 = ( h ) a^ ⋅ ( c ) b^ ⋅ ( G ) c^ ⋅( A ) d …(i)
Then, [ V 0 ]= [potential] =
potential energy charge
= =
− [ML T − − [AT]
ML T A
2 2 ]
[ 2 3 1 ]
[ h ] =
Energy Frequency
− −
[ ML T ]
[T ]
2 2 1 =^
[ML T^2 −^1 ]
[ ] c = [Speed] = [ LT− 1 ]
[ ]
G =
×
Force Distance) Mass
2 2 =
[MLT^ − ][L
[M
2 2 ]
2 ]
= [ M− 1 L T^3 −^2 ]
Substituting the dimensions of V h 0 , C G , and A in Eq. (i) and equating dimension on both sides, we get [ ML T^2 −^3 A −^1 ] = [ ML T^2 −^1 ] a^ ×[ LT−^1 ] b × [ M− 1 L T^3 − 2 ] c^ ×[ A] d
⇒ a − c = 1 …(ii) 2 a + b + 3 c = 2 …(iii) − a − b − 2 c = − 3 …(iv) d = − 1 …(v) On solving above equations, we get a = 0 , b = 5 , c = − 1 , d = − 1 Substituting these values in Eq. (i), we get V (^) 0 = h^0 ⋅ c^5 ⋅ G −^1 ⋅ A −^1 None of the given options matches with the result. 39 ( d ) Dimensions of speed are [ V] = [ LT −^1 ] Dimensions of acceleration are [ A] = [ LT− 2 ] Dimensions of force are [ F] = [ MLT− 2 ] Dimension of Young modulus is [Y] = [ ML− 1 T^ −^2 ] Let dimensions of Young’s modulus is expressed in terms of speed, acceleration and force as; [ Y] = [ V] α^ [ A ] [β^ F ] γ …(i) Then substituting dimensions in terms of M, L and T we get, [ML− 1 T^ −^2 ] = [ LT −^1 ] [α^ LT −^2 ] [β^ MLT−^2 ]γ = [ M Lγ^ α +β^ +^ γ^ T−^ α^ −^2 β^ −^2 γ] Now comparing powers of basic quantities on both sides we get, γ = 1 α + β + γ= − 1 and −α − 2 β − 2 γ = − 2 Solving these we get; α = −4, β = 2, γ = 1 Substituting α β, , & γ in (i) we get; [Y] = [ V− 4 A F^2 1 ]
40 ( c ) Suppose, linear momentum ( p )depends upon the Planck’s constant ( h )raised to the power ( a ), surface tension( S ) raised to the power ( b ) and moment of inertia ( I )raised to the power ( c ). Then, p ∝ ( h ) ( a^ S ) ( b^ I ) c or p = kh Sa^ bI^ c where, k is a dimensionless proportionality constant. Thus, [ p] = [ h ] [ a^ S] [ ] b^ I c^ …(i) Then, the respective dimensions of the given physical quantities, i.e. [ p] = [mass × velocity] = [ MLT− 1 ] [ ]I = [mass × distance^2 ] = [ ML T^2 0 ] [ S] = [force × length] = [ ML T^0 −^2 ] [ h] = [ML T^2 −^1 ] Then, substituting these dimensions in Eq. (i), we get [MLT −^1 ] = [ML T^2 −^1 ] [MT a^ −^2 ] [ML^ b^ +^2 ] c For dimensional balance, the dimensions on both sides should be same. Thus, equating dimensions, we have a + b + c = 1 2 ( a + c )= 1 or a + c =
− a − 2 b = − 1 or a + 2 b = 1
Hints & Explanations
Solving these three equations, we get
a = 0, b =
, c =
∴ p = h S^0 I
1 2
1 (^2) or p = S I h
1 2
1 2 0
41 ( c ) Force of interaction between two atoms is given as F = αβ exp ( − x^2 / α kT ) As we know, exponential terms are always dimensionless, so
dimensions of
x kT
2 0 0 0 α
[ M L T]
⇒ Dimensions of α = Dimension of ( x^2 / kT ) Now, substituting the dimensions of individual term in the given equation, we get
= (^) −
[M L T
[M L T
0 2 0 1 2 2
]
]
{Q Dimensions of kT equivalent to the dimensions of energy = [ M L T^1 2 −^2 ]} = [ M− 1 L T^0 2 ] …(i) Now from given equation, we have dimensions of F = dimensions of α × dimensions of β
⇒ Dimensions of β = Dimensions of
F
α
− −
[M L T
[M L T
1 1 2 1 0 2
]
]
[Qusing Eq. (i)]
= [ M L T2 1^ −^4 ]
42 ( d ) Given, U
A x x B
Dimensions of U = Dimensions of potential energy = [ML T^2 − 2 ] According to the principle of homogeneity, Dimensions of B = Dimensions of x = [M LT ]^0 ∴ Dimensions of A =
Dimensions of × Dimensions of ( + )
Dimensio
U x B
ns of x
=
[ML T^ − ] [M LT ]
[M L T ]
2 2 0 0 0 1/ 2 0 =^
[ML5/ 2 T− 2 ]
Hence, dimensions of AB = [ML5/ 2 T− 2 ] [M LT ]^0 = [ML7/ 2 T^ − 2 ]
43 ( a ) As force between two charges,
F
e r
2
0 4 πε 2 ⇒^
ε ρ Φ
2
0
2 4πε
Putting dimensions of r and F, we get
e^2 0
2 2 4 πε
=^
[ L ][ MLT− ] ...(i)
Also, force between two masses, F
Gm r
2 2
⇒ G
Fr m
2 2 ⇒^ [^ G^ ]=
[MLT^ − ] [L ]
[M ]
2 2 2
⇒ [ G ]= [M− 1 L T^3 − 2 ] ...(ii)
and
c^2
[L T^2 2 ]
= [L− 2 T ]^2 ...(iii)
Now, checking optionwise for option (a)
2
0
1 2
c
Ge πε
/
= [ L− 2 T^2 ]{[ M− 1 L T^3 − 2 ][ L^2 ][ MLT−^2 ]}1 2 /
= [L− 2 T ] [L T^2 6 − 4 1/ 2]^ = [ ]L
For option (b),
c G 2 e^2 0
1 2 5 4 4 πε
=^ =
− − − −
/ [L 2 T ]^2 1 [L T^6 4 1/ 2] [L T ]
For option (c), 1 (^24)
2 1 2^2
c
e G πε 0
=^
−
− − −
/ [L T ]
[L ][MLT ]
[M L T ]
2 2
2 1 3 2
1/ 2
= [L− 2 T ][M ]^2 2 1/ 2^ = [ML− 2 T ]^2
For option (d), 1 4
2
c 0
G
e πε
= [L− 2 T ][M^2 − 1 L T^3 − 2 ][L ][MLT^2 −^2 ]
= [L T^4 − 2 ]
∴Physical quantity
2
0
1 2
c
G
e πε
/ has the dimensions
of length. Hence, option (a) is correct. 44 ( a ) According to question, [ X ]= Dimensions of capacitance = [M −^1 L^ − 2 T Q ]^2 and [ Z ] = Dimensions of magnetic induction. = [MT −^1 Q^ − 1 ] Given, X = 3 YZ 2 , ∴ [ ]
[ ]
[ ]
Y
X
Z
⇒ [ ]
[M L T Q ]
[M T Q ]
1 2 2 2 Y = 2 2 2
− − − − =^
[M −^3 L− 2 T Q ]^4
45 ( c ) We know that, Surface tension, S
F
L
Force [ ] Length [ ]
So, [ ]
[ ]
[ ]
S = =[ ]
− MLT − L
ML T
2 0 2
Energy, E = Force ×Displacement ⇒ [ E] = [ MLT− 2 ][ L ] =[ ML T^2 −^2 ]
Velocity, v =
Displacement Time
⇒ [ ] v = [ LT− 1 ]
As, S ∝ E a^ v b^ Tc , where, a b , and c are constants. From the principle of homogeneity, [LHS] = [RHS]
Hints & Explanations
Hints & Explanations
Energy, [ E ] = [ ML T^2 −^2 ] Q a = 1, b = 2, c = − 2
∴ n n
a b c 2 1
1 2
1 2
1 2
M
M
L
L
T
T
n 2
2 = 10 ×
− M M
L
L
T
T
1 2
1 2
2 1 2
= ×
^
− 10
2 2
α β γ
∴ n 2 = 10 α −^1 β −^2 γ^2 53 ( c ) Barn is used in nuclear physics for measuring the cross-sectional area of nuclei. One barn is equal to 10− 28 m.^2 Therefore, Assertion is correct but Reason is incorrect. 54 ( a ) Parallax method is used for measuring distances of nearby stars only. If D is a distance of a far away star from Earth, then D
b
θ where, θ is called parallactic angle and b is the distance between the two different positions on Earth from where the star is being observed. ∴With increase in the distance of star, parallactic angle becomes too small to be measured accurately. Therefore, Assertion and Reason are correct and Reason is the correct explanation of Assertion. 55 ( b ) When you hold a pencil infront of you against some specific point on the background (a wall) and look at the pencil first through your left eye (closing the right eye) and then look at the pencil through your right eye (closing the left eye), you would notice that the position of the pencil seems to change with respect to the point on the wall. This is called parallax. The distance between the two points of observation is called the basis, e.g. the basis is the distance between the eyes. Therefore, Assertion and Reason are correct but Reason is not the correct explanation of Assertion. 56 ( b ) Random errors are those errors, which occur irregularly and hence are random with respect to sign and size. These can arise due to random and unpredictable fluctuations in experimental conditions, personal (unbiased) errors by the observer taking readings, etc. Therefore, Assertion and Reason are correct but Reason is not the correct explanation of Assertion. 57 ( d ) In all mathematical operations, the errors are of additive nature. When a quantity appears with a power n greater than one in an expression, its error contribution to the final result increases n times. So, quantities with higher power in the expression should be measured with maximum accuracy. Therefore, Assertion is incorrect but Reason is correct. 58 ( a ) The method of dimensions can only test the dimensional validity but not the exact relationship between physical quantities in any equation. This is because it does not distinguish between the physical quantities having same dimensions.
Therefore, Assertion and Reason are correct and Reason is the correct explanation of Assertion. 59 ( b ) The arguments of special functions, such as the trigonometric, logarithmic and exponential functions must be dimensionless. A pure number, ratio of similar physical quantities, such as angle as the ratio (length/length), refractive index as the ratio of (speed of light in vacuum/speed of light in medium), etc. has no dimensions. Therefore, Assertion and Reason are correct but Reason is not the correct explanation of Assertion. 60 ( b ) Statements I and III are correct, but II is incorrect and it can be corrected as The unit chosen for measuring any physical quantity, should be easily reproducible, i.e. replicas of the unit should be available easily. 61 ( a ) Statements I and II are correct but III is incorrect and it can be corrected as The terminal or trailing zero(s) in a number without a decimal point are not significant. 62 ( a ) Statements I and II are correct but III is incorrect and it can be corrected as Heat capacity has unit cal/kg, while gravitational potential has unit Jkg−^1 , i.e. both the quantities will have same dimensions [L T^2 −^2 ]. 63 ( d ) Statement given in option (d) is incorrect and it can be corrected as While dealing with atoms, kilogram is an inconvenient unit. In this case, there is an important standard unit of mass called unified atomic mass unit (u), which has been established for expressing the mass of atom. Rest statements are correct. 64 ( a ) Precision refers to the limit to which the quantity is measured. It is determined by the least count of the measuring instrument. ∴The smaller the least count, greater is the precision. Thus, the statement given in option (a) is correct, rest are incorrect. 65 ( c ) Statement given in option (c) is correct but rest are incorrect and these can be corrected as Error in a measurement is equal to the difference of the true value and measured value of a quantity. Systematic errors occur only in one direction, either positive or negative. In constant errors, errors affect each observation by the same amount. 66 ( b ) As per the rule for determining the number of significant figures, there is no change in number of significant figures on changing the units. ∴In 4.700 m = 4700 mm, there is no change in significant figures i.e. it remains same as 4. Thus, the statement given in option (b) is incorrect, rest are correct. 68 ( c ) Barn is the unit of area. It is used to measure small cross-sectional area. 1 barn = 10 − 28 m^2 Are is also unit of area, 1 are = 102 m^2
Atmospheric pressure is measured in SI unit of bar. 1 bar = 1013. × 105 N/m^2 = 1013. × 105 Pa Carat is the unit of mass. i.e. 1 carat = 200 mg Hence, Α → 4, B → 3 , C → 2 and D → 1.
70 ( b ) A. If the number is less than 1, the zeros on the right of decimal point out to the left of the first non-zero digit are not significant. So, in 0 .00 4608, underlined zero (s) are not significant. So, it has 4 significant figures. B. The trailing zero (s) in a number with decimal points are significant. Thus, 8 9000. has 5 significant figures. C. All non-zero digits are significant, so 186 has 3 significant figures. D. All the zero (s) between two non-zero digits are significant, no matter where decimal points. So, 2.00891 has 6 significant figures. Hence, A → 4, B → 2, C → 1and D → 3.
71 ( b ) A. As, internal energy,^ U^ =^ kT
⇒ [ ML T^2 −^2 ] = [ k ] [ K] ⇒ [ k ] = [ ML T^2 −^2 K^ −^1 ]
B. Viscous force, F A
dv dx
= η
⇒ [ ]
[ ]
[ ]][ ][ ]
η =
− − −
MLT
L LT L
2 2 1 1 =^
[ ML− 1 T^ −^1 ]
C. Energy, E = h ν ⇒ [ ML T^2 2 ] = [ h ] [ T−^1 ] ⇒ [ h ] = [ ML T^2 −^1 ]
D. dQ dt
k A l
∆θ ⇒ [ ]
[ ]
[ ]
k =
ML T^ − L
L K
2 3 2
= [ MLT− 3 K−^1 ] Hence, A → 4, B → 2, C → 1 and D → 3. 72 ( a ) Given, G = 6 67. × 10 − 11 N-m 2 kg−^2 = 6 67. × 10 − 11 kg-ms −^2 m^2 kg−^2 = 6 67. × 10 − 11 m s^3 −^2 kg−^1 = 6 67. × 10 − 11 ( 10 2 cm )^3 s −^2 ( 1000 g)−^1 = 6 67. × 10 − 8 cm^3 s− 2 g−^1 So, value of G in CGS system of units = 6 67. × 10 − 8 73 ( d ) Given, time t = 8 min 20 s = ( 8 × 60 +60 s) Distance of sun to earth = time × speed = ( 8 × 60 + 20 )× c = ( 480 + 20 ) × c = 500 c = 500 units (Q c = 1 unit)
74 ( c ) The device that has minimum least count will be more precise for measuring length. So, let us calculate the least count of the given conditions in the options. Least count of vernier callipers = 1 MSD – 1VSD = 1 MSD –^19 20
MSD =^1
MSD
= × =
mm cm = 0 005. cm
Least count of screw gauge
Pitch Number of divisions on circular scale = =
. mm cm = 0 001. cm
Least count of optical instrument = Wavelength of visible (red) light = 6000 Å = 6000 × 10 − 8 cm = 0 00006. cm Hence, the most precise device for measuring length is the given optical instrument. 75 ( b ) In 0.007 m^2 , significant figure is only number 7. In 0.2370 g cm−^3 , significant figures are 2, 3, 7, 0. In 6.032 Nm−^2 , significant figures are 6, 0, 3, 2. So, significant figures in the given numbers are 1, 4 and 4, respectively. 76 ( d ) Given, length, l = 4.234 m, Breadth, b = 1.005 m and thickness, t = 2.01 cm =0.0201 m ∴ Area of sheet, A = 2 ( l × b + b × t + t × l ) = 2 [(4.234 × 1.005) + (1.005 ×0.0201)
- (0.0201 ×4.234)] = 2 ×4.3604739 = 8.7209478 m^2 As thickness has least number of significant figures, i.e. 3, therefore rounding off area upto three significant figures, we get Area of sheet, A = 8.72 m^2 Volume of sheet, V = l × b × t = 4.234 × 1.005 ×0. = 0.0855289 m^3 Similarly, rounding off upto three significant figures, we get volume of the sheet = 0.0855 m^3 77 ( c ) Given, mass of the box, m = 2.3 kg Mass of first gold piece, m 1 = 20.15 g = 0.02015 kg Mass of second gold piece, m 2 = 20.17 g = 0.02017 kg Total mass of the box, M = m + m 1 (^) + m 2 = 2.3 + 0.02015 +0. = 2.34032 kg As the mass of the box has least decimal places 1, therefore total mass of the box can have only 1 decimal place. So, rounding off the total mass of the box upto 1 decimal place, we get Total mass of the box, M =2.3 kg Similarly, difference in masses of gold pieces, ∆ m = m 2 (^) − m 1 = 20.17 −20.15 = 0.02 g Since, the masses of two gold pieces has two decimal places, therefore the final result is corrected upto two decimal places.
78 ( d ) Given, P
a b cd
3 2
The percentage error in the quantity P is given by ∆ P ∆ ∆ P
a a
b b
× 100 % = 3 × 100 % + 2 × 100 %
+ × + ×
∆ c ∆ c
d d
Hints & Explanations
