6
Prof. Muhammad Amin 1 Chapter-1 measurement 1.1 Name several repetitive phenomenon occurring in nature which could serve are responsible time standards. Ans. A natural phenomenon which repeats itself after equal intervals of time could be used as time standard e.g The rotation of earth about its own axis and also around the sun. The rotation of moon around the earth Atomic vibrations in solid. The change of shadow of an object in the sun. 1.2 Give the drawback to use the period of pendulum as time standard. Ans. We know that the time period of pendulum is 2 T= π g The time period ‘T’ of a pendulum depends upon its length’ ’ and value of ‘g’. These are following drawbacks to use the period of a pendulum as time standard. Air resistance affect the time period. The period of a pendulum varies with altitude because ‘g’ varies with altitude. The effect of temperature on the length of pendulum. 1.3 Why do we find it useful to have two units for the amount of substance, the kilogram and the mole? Ans. Both kilogram and mole are S.I units of mass. Kilogram is used for large mass when we require amount of substance without considering number of atoms/molecules. Mole is used for small mass when we require amount of substance by considering number of atoms/molecules. i.e 1 mole = 6.022 × 10 23 atoms/molecules 1.4 The period of simple pendulum is measure by a stop watch. What type of errors are possible in the time period? Ans. The following errors are possible. 1. Random Error Random error occurs when repeated measurements of a quantity give different values under the same conditions. It occurs due to some unknown reason 2. Systematic Error Systematic error occurs due to the fault in the instrument. it always ocures under some definite rule e.g zero error, poor calibration 3. Personnel Error Personal error occurs due to negligence or inexperience of a person. Measurements Chapter 1 Important short questions

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Prof. Muhammad Amin 1

Chapter-1 measurement

Type equation here.

1.1 Name several repetitive phenomenon occurring in nature which could serve are

responsible time standards.

Ans. A natural phenomenon which repeats itself after equal intervals of time could be used as

time standard

e.g

The rotation of earth about its own axis and also around the sun.

The rotation of moon around the earth

Atomic vibrations in solid.

The change of shadow of an object in the sun.

1.2 Give the drawback to use the period of pendulum as time standard.

Ans. We know that the time period of pendulum is

2T = πg

The time period ‘T’ of a pendulum depends upon its length’ ’ and value of ‘g’. These

are following drawbacks to use the period of a pendulum as time standard.

Air resistance affect the time period.

The period of a pendulum varies with altitude because ‘g’ varies with altitude.

The effect of temperature on the length of pendulum.

1.3 Why do we find it useful to have two units for the amount of substance, the kilogram

and the mole?

Ans. Both kilogram and mole are S.I units of mass.

Kilogram is used for large mass when we require amount of substance without

considering number of atoms/molecules.

Mole is used for small mass when we require amount of substance by considering

number of atoms/molecules.

i.e 1 mole = 6.022 × 1023 atoms/molecules

1.4 The period of simple pendulum is measure by a stop watch. What type of errors are

possible in the time period?

Ans. The following errors are possible.

1. Random Error

Random error occurs when repeated measurements of a quantity give different values

under the same conditions.

It occurs due to some unknown reason

2. Systematic Error

Systematic error occurs due to the fault in the instrument. it always ocures under

some definite rule

e.g zero error, poor calibration

3. Personnel Error

Personal error occurs due to negligence or inexperience of a person.

Measurements Chapter

1

Important short questions

Prof. Muhammad Amin 2

Chapter-1 measurement

1.5 Does a dimensional analysis give any information of constant of proportionality that

may appear in an algebraic expression?

Ans. The dimensional analysis does not give any information about the constant of

proportionality, however, it can be found by experiments.

Example

F = 6 r V

Here 6 = constant of proportionality the numerical value of the constant cannot be

determined by dimensional analysis. However, it can be found by experiments.

1.6 Write the dimensions of (i) Pressure (ii) Density

Ans. (i) By definition,

Pressure = Force

Area

Dimension of pressure

F= P =

A

2

2

MLTP =

L

or 1 2P ML T

(ii) As density =Mass

Volume

Dimension of density

3

M

L

or 3ML

1.7 Write any three uses of dimensional analysis?

Ans: The use of dimensions is called dimensional analysis. The following are the main uses of

dimensional analysis.

(i) It is used to find the relationship between different physical quantities.

(ii) It is used to convert one system of unit into another.

(iii) It is used to confirm the correctness of any physical equation

1.8 Write any two drawbacks of dimensional analysis?

Ans: The following are the draw backs of dimensional analysis.

(i) The dimensional analysis is unable to find the values of various constants.

(ii) It cannot be applied to physical quantities involving trigonometric and logarithmic

functions.

(iii) It cannot differentiate between terms having same dimensions. For example work

and Torque, stress and pressure.

1.9 What is meant by S.I system?

Ans: in 1960, the internationally adopted system of units used by all the scientists and almost

all the countries of the world is international system (SI) of units. It consists of seven

base units, two supplementary units and a number of derived units.

Prof. Muhammad Amin 3

Chapter-1 measurement

The S.I base units

QUANTITIES

SYMBOL

OF

QUANTITY

UNIT

NAME

SYMBOL

OF UNIT

Length L Meter m.

Mass M Kilogram Kg.

Time T Second S.

Electric

Current I Ampere A.

Temperature T Kelvin K.

Light

Intensity L Candela Cd.

Amount of

Substance N Mole mol.

1.10 Differentiate b/w radian and steradian?

Ans: Radian

It is unit of plane angle made at the centre of a circle by an arc

equal in length to the radius of the circle.

If Arc AB = r, then

< AOB = 1 Radian

Steradian

The solid angle made at the centre of a sphere by its surface of

urea equal to square of its radius, is called Steradian.

1.11 State the principle of homogeneity of dimensions.

Ans: In order to check the correctness of an equation, we have to show

that the dimensions of the quantities on both sides of the equation are the same,

irrespectively of the form of formula. This is called the principle of homogeneity of

dimensions.

e.g E = mc2 is dimensionally consistent since dimensions of both sides are equal

i.e [L.H.S] = [R.H.S]

[ML2T-2] = [ML2T-2]

1.12 Define the types of error and their remedy (correction).

Ans: The errors in any measurement are usually classified into two types.

(i) Random Errors

Random error occurs when repeated measurements of a quantity give different

values under the same conditions.

Cause: It occurs due to some unknown reason.

Remedy:

Random error can be reduced by taken average of several readings.

(ii) Systematic Errors

Systematic error occurs due to the fault in the instrument. It always occurs under

some definite rule

Cause: zero error, poor calibration

Prof. Muhammad Amin 4

Chapter-1 measurement

Remedy:

Systematic error can be removed by comparing the faulty apparatus with a standard.

1.13 Differentiate between precision and accuracy.

Ans: A precise measurement is the one which has less absolute uncertainty and an accurate

measurement is the one which has less fractional or percentage uncertainty or error.

1.14 Calculate the dimensions of pressure and work?

Ans: Dimensions of Pressure

F

P Dimensionof work W F dA

2

2 2

maW MLT L

A

W ML T

2

2

MLT

L

1 2P ML T

1.15 Distinguish between base units and derived units?

Ans:

Base Units Derived Unit

Those units which cannot be defined in

terms of any other units known as base

units.

e.g These are units of base quantities.

kilogram, meter, second etc.

The units of the physical quantities

defined in terms of base and

supplementary units are known as

derived units.

e.g These are units of derived

quantities.

Newton, m/s , Pascal etc.

Prof. Muhammad Amin 5

Chapter-1 measurement

1.1 A light year is the distance light travels in one year. How many meters are there in one light

year: (speed of light = 3.0x 08 ms-1).

Solution:

Data:

Time = t = 1 year = 365 days

t = 365 x 24 x 60 x 60 sec

t = 31536000 s

Speed of light = c = 3.0 x 108 ms-1

To Find:

Distance = S =?

Calculation:

We know that

S = v t

v = c

S = ct

Putting the values, we get

S = 3 x 108 x 31536000

or 15S = 9.5 × 10 m Ans.

Conclusion:- There are 9.5 x 1015 meters in one light year

1.2 a) How many seconds are there in 1 year?

b) How many nanoseconds in 1 year?

c) How many years in 1 second?

To find

(a) Seconds in one year =?

(b) Nanoseconds in one year =?

(c) Years in one second =?

Solution:

(a) We already know that

1 year = 365 days

1 day = 24 hours

1 hour = 60 minutes

1 minute = 60 seconds

Therefore,

1 year = 365 x 24 x 60 x 60 sec

= 31536000 s

= 3.1536 x 107 s Ans

(b) As we know that

One year = 3.1536 x 107s

One second = 109 nanosecond (:1 nanosecond = 10-9s)

One year = 3.1536 x 1007 x 109 nanoseconds

One year = 3.1536 x 1016 ns Ans.

(c) We know that

1 year = 3.1536 x 107 seconds

1 second = 7

1years

3.1536 × 10

1 second = 3.17 x 10-8 years Ans.

Numerical Problems

Prof. Muhammad Amin 6

Chapter-1 measurement

2.1 The length and width of a rectangular plate are measured to be 15.3 cm and 12.80 cm,

respectively. Find the area of the plate.

Solution:

Length of the rectangular plate = L = 15.3 cm.

Width of the rectangular plate = w = 12.80 cm

Area of rectangular plate = A =?

We know that

A = L x W

= 15.3 x 12.80

= 195.84 cm2

A = 196 cm2 nearly Ans.

2.2 Add the following masses given in kg upto appropriate precision. 2.189, 0.089, 11.8 and 5.32

Solution:

Four masses = 2.189 kg, 0.089 kg, 5.32 kg and 11.8 kg

Sum of masses upto appropriate precision =?

Total mass = 2.189 + 0.089 + 11.8 + 5.32 = 19.398 kg

Since mass of least precision is 11.8 kg, because it has one decimal place. Therefore, the

total mass should be one decimal place, which is the appropriate precision.

Total mass = 19.4 kg Ans.

2.3 What are the dimensions and units of gravitational constant G in the formula 1 2

2

m mF = G

r

Solution:

Data: Gravitational force = 1 2

2

m mF = G

r

To Find:

Dimensions of G =?

Units of G =?

Calculation:

Dimensions of Gravitational constant(Dimensions of force)(Dimension of Length)

=(Dimensions of mass)(Dimension of mass)

2[F][L ]

[G] =[M][M]

[F] = [MLT-2]

-2 2

2

[MLT ][L ][G] =

[M ]

-1 3 -2[G] = [M L T ] Ans.

UNIT OF G: As we know that the S.I units of force, length and mass are Newton, meter and kilogram

respectively. So using the formula 2

1 2

FrG =

m m

Unit of G = Nm2/kg2 Ans

2.4 Show that the famous “Einstein equation” E = mc2 is dimensionally consistent.

Solution:

E = mc2

Dimensions of L.H.S. = E = [M L2 T-2]

∵(E = W = Fd = ma x d = [M][LT-2][L])

Dimensions of R.H.S. = mc2

=[M][LT-1]2 = [M L2 T-2]

Since the dimensions of both sides are same, therefore the equation is dimensionally consistent.