Units And Measurements

Rules for Counting Significant Figures

(i) All the non-zero digits are significant. In 1.248, the

number of significant figures is 4.

(ii) All the zeroes between two non-zero digits are

significant, no matter where the decimal point is, if at

all. As examples, 406 and 9.001 have 3 and

4 significant figures respectively.

(iii) If the measurement of number is less than 1, the

zero(es) on the right of decimal point and to the left

of the first non-zero digit are non-significant.

In 0.00606, first three underlined zeroes are

non-significant and the number of significant figures

is only 3.

(iv) The terminal or trailing zero(es) in a number

without a decimal point are not significant. Thus,

12.3 =1230cm=12300 mm has only 3 significant

figures.

(v) The trailing zero(es) in number with a decimal point

are significant. Thus, 3.800 kg has 4 significant

figures.

(vi) A choice of change of units does not change the

number of significant digits or figures in a measurement.

So, here using rule (iii) and (v) we get, in 0.06900

-

0.06900 the underlined part is not significant and the bold part is significant.

Therefore, the number of significant figures is 4.

Rules for Adding Significant Figures

(i) Count the number of significant figures in the decimal portion of each given number.

(ii)The digits to the left of the decimal place are not used to determine the number of decimal places in the final answer.

(iii) Add or subtract the given numbers in the normal way.

(iv) Round the answer to the LEAST number of places in the decimal portion of any number in the problem.

Here,

436. 32

+ 227. 2

+ 0. 301

663. 821

Here, 436.32 has 2 digits after decimal, 227.2 has 1 digit after decimal and 0.301 has 3 digits after decimal. According to rules the final result will have the least no. of digit after decimal i.e. 1 here.

Therefore, the final sum is 663.8

RULES OF DIVISION IN SIGNIFICANT FIGURES:

(i)In division the answer is rounded off to the same number of significant figures as possessed by the least precise term used in the calculation.

(ii)The final result must contain as many significant figures as are in the original number with the least significant figures.

Here,

No. of significant figures in 4.237 is 4 and in 2.5 is 2. Here 2.5 is the least precise term used. Therefore, the answer must contain 2 significant figures.

Now, applying the rules and rounding off the answer we get

RULES OF ROUNDING OFF MEASUREMENT:

1. If the digit after the digit to be rounded off is less than 5, then the preceding digit remains same.

2. If the digit after the digit to be rounded off is more than 5, then the preceding digit increases by one.

3. If the digit after the digit to be rounded off is 5 followed by digits other than zero, then the preceding digit increases by one.

4. If the digit after the digit to be rounded off is 5 and the no. to be rounded off is even, then the number to be rounded off remains same.

Example- 1.285 becomes 1.28 after rounding off

5. If the digit after the digit to be rounded off is 5 and the no. to be rounded off is odd, then the number to be rounded off increases by one.
Example- 1.215 becomes 1.22 after rounding off

Now, here in the given question,

According to rule (4),

2.745 after rounding off becomes 2.74 as 4 is even number.

And

According to rule (5),

2.735 after rounding off becomes 2.74 as 3 is odd number.

(i)In multiplication the answer is rounded off to the same number of significant figures as possessed by the least precise term used in the calculation.

(ii)The final result must contain as many significant figures as are in the original number with the least significant figures.

No. of significant figures in 16.2 is 3 and in 10.1 is 3. Here both are the least precise terms used. Therefore, the answer must contain 3 significant figures.

So, rounding off the answer to 3 significant figures we get

Now error,

If is the error in the area, be the error in length and be the error in breadth, then relative error is calculated as .

(For the formula part refer to

NCERT class 11 part 1 unit

and measurement chapter)

In dimensional analysis, in multiplication the answer is rounded off to the same number of significant figures as possessed by the least precise term used in the calculation. Here 0.1 has least precision and has 1 significant figure. Therefore,

(By rounding off to one significant figure)

Area,

Most physical quantities can be expressed in terms of combination of five basic dimensions. These are,

1.mass(M)

2.length(L)

3.time(S)

4.electrical current(I) and,

5.temperature ()

For,

(a) Work and torque

From the table above we can see that only pair having different dimensional formula is tension and surface tension.

Given,

Let, X=AB

X=AB=

Precision can be calculated by,

If is the error in the X, be the error in A and be the error in B, then relative error is calculated as .

Therefore now,

(For the formula part refer to

NCERT class 11 part 1 unit

and measurement chapter)

Rounding off 0.075 to two significant figures we get,

Therefore, the value of AB is .

Let

Let be the error in X.

From error analysis we know that,

(For the formula part refer to

NCERT class 11 part 1 unit

and measurement chapter)

Putting the known values we get,

In dimensional analysis, in multiplication the answer is rounded off to the same number of significant figures as possessed by the least precise term used in the calculation. Here 0.1 has least precision and has 1 significant figure. Therefore,

(By rounding off to one significant figure)

Precision means closeness of two or more measurements.

If measured with an instrument of minimum least count the measurement is most precise.

Here mm scale has the minimum least count and hence is most precise.

A measurement is said to be accurate if its value is closest to the real value of the physical quantity.

Here,

4.9 has the least difference. Hence 4.9 is most accurate.

Given,

Substituting the values of and we get,

According to question, momentum area and time are taken as fundamental quantities.

Now, we know that dimension of

Momentum(p)=

Area(A)=

Time (T)=

Energy=

Let us suppose,

E=

Now using principal of homogeneity, which states that the dimensions on both sides of a dimensional equation is same, we get

=

Equating the like terms we get

a=1, a+2b=2, -a+c=-2

1+2b=2, -1+c=-2

2b=1, c=-1

a=1, b=1/2, c=-1

Therefore,

E=

Dimensional formula of displacement =

From study of dimensional analysis we know that,

Numbers have no dimension.

Trigonoetric functions are dimensionless.

Constants are dimensionless.

According to above rules

Dimension of

Dimension of

Dimension of

Dimension of

So, only option is C that doesn’t have dimension that of displacement.

Given,

P, Q, R have different dimensions.

From dimensional analysis we know that physical quantities having different dimensions cannot be added or subtracted.

From the above arguments we can conclude that, the combinations in option (a) and (e) are not dimensionally possible.

On the other hand,

In option (b) the product of P and Q may result into dimension of R.

Option (c) is also dimensionally possible as physical quantities with different dimensions can multiplied or divided.

In option (d) the product of P and R may result into dimension of Q².

According to plank, the energy of photon is given by the expression,

Where,

f= frequency of light

h= Planck’s constant

Dimension of E=

Dimension of f=

So, dimension of h,

Now,

Linear impulse=forcetime

Angular impulse= torque time

=

=

Linear momentum= mass velocity

=

=

Angular momentum= moment of inertia angular velocity

=

=

According to plank, the energy of photon is given by the expression,

Where,

f= frequency of light

h= Planck’s constant

Dimension of E=

Dimension of f=

So, dimension of h,

As we can see in the question, we have to relate M, L and T in terms of h and c as fundamental quantities.

There is no charge or ampere dimension so option (c) is eliminated.

We have to check for option a, b and d.

Dimensions of m_e, m_p and G are , and respectively.

For m_e and m_p,

For mass, let

m

By substituting the dimension for each quantity

Now using principal of homogeneity, which states that the dimensions on both sides of a dimensional equation is same, we get

=

Equating like terms and solving above we get,

a+b=1, 2b+c=0, -b-c=0

a=1, b=0, c=0

Substituting the values in equation,

m

m=m_e

Similarly for length,

l

=

Solving we get,

and for time,

This holds true for both m_e and m_p.

And similarly for G,

Also we know,

Dimension of G=

Let,

m ...........(1)

By substituting the dimension for each quantity

Now using principal of homogeneity, which states that the dimensions on both sides of a dimensional equation is same, we get

=

Equating like terms and solving above we get,

a-c=1, 2a+b+3c=0, -a-b-2c=0

a=1/2, b=1/2, c=-1/2

Substituting the values in equation,

m

Similarly for length and time,

and

Dimensional formula of pressure=

Thus, A and B have dimensions of pressure.

Second – It is a unit of time.

Parsec – It is a unit of distance which is equal to 3.26 light year.

Year – It is a unit of time which is equal to 31536000 seconds.

Light year- It is a unit of distance. It is defined as distance covered by light in 1 year. i.e.

We have different units for same physical quantity for our convenience.

Let us suppose we have to measure a distance between two atoms in a molecule we will probably work in Angstrom(Å) and in another scenario we have to calculate distance between stars so we will probably work in light year.

Therefore in both cases the physical quantity used is length but the units are different. So, the point is that the units can vary for different measurements but dimensions remains same for the physical quantity.

According to question,

Radius of nucleus is of the order 1 Fermi i.e.

We know,

From classical physics, the shape of atom and nucleus is considered to be spherical.

And we know that the volume of sphere =

So, we get that atom is times bigger than nucleus.

Mass spectrometer is used for measuring the mass of atoms and molecules.

The method used for measuring the mass is known as mass spectrometry.

Fig 1. Mass Spectrometer

One unified atomic mass unit is defined as mass of one atom of C-12.

We know that molar mass of C-12 atom=12 gm/mole

mass of 1 atom=

And 1unified mass unit=

=

In dimensional analysis, according to rule a physical quantity can be added or subtracted if and only if the physical quantities have same dimension.

For example, we can add 5 kg apple and 5 kg watermelon but cannot add 5 kg apple and 100 metre race because they have different dimensional unit.

Similarly, in the given question 1 has no unit and dimension. Therefore, θ should also not have any unit and dimension in order to be added with 1. Hence θ is a dimensionless quantity.

In mechanics we have to deal with all the physical changes happening around us like force, pressure, momentum which are not fundamental quantities.

They are obtained from some physical quantities which are fundamental quantities.

Length mass and time are chosen as base quantities due to following reasons:

1. These quantities cannot be derived from other quantities.

2. They are the simplest units.

3. All other physical quantities can be written in terms of these units.

(a) Given:

Distance between earth and moon = 60 x earth’s radius

The distance between the earth and moon can be treated as a radius of arc that is subtended by the diameter of the earth from the moon. Let θ be the angle subtended by the arc, R be the earth-moon distance and R_E be the earth’s radius. Then,

2R_E = length of the arc.

And distance between earth and moon = 60R_E

Therefore the angle subtended by the arc

Therefore the diameter of earth as seen from earth will be 2°.

(b) Given:

Diameter of moon as seen from earth =

As calculated before the diameter of earth as seen from moon is approximately 2° and if moon is as seen from earth, then,

Therefore, earth’s diameter is 4 times that of moon’s diameter.

(c) Given:

Distance between sun and earth = 400 x earth-moon distance.

let r_{m �} be the moon’s distance from earth and r_S be the sun’s distance from earth, then we have

Let D_s, D_{m �} and D_e be the diameters of sun, moon and earth respectively. Since we know that the sun and the moon, appear to be the same size from the earth, we have

Also

Therefore, the sun’s diameter is 100 times larger than earth.

The correct answer is (d). An atomic clock is based on the characteristic frequencies of radiation emitted by atoms of certain elements which have a precision of about 1s in 3 x 10⁵ years. However, a wall clock can measure correctly up to one second, a stop watch can measure correctly about a fraction of a second and digital clock can also measure correctly up to about a fraction of second.

Given:

Distance of galaxy from earth = 10²⁵ m

We know that the speed of light in vacuum or space is 3 x 10⁸ m/s.

Therefore, the time taken by light to reach earth from the galaxy can be calculated as follows

Therefore, the order of magnitude of time taken by light to reach is 10¹⁶s.

Given:

Number of Vernier scale divisions (VSD) = 50

Number of Main scale divisions (MSD) = 49

Least count of main scale = 0.5 mm

As per the question 50 VSD coincide with 49 MSD, therefore

The minimum accuracy in measurement is given as

But, 1 MSD = 0.5 mm

Therefore, the minimum accuracy in measurement is 0.01 mm

During an eclipse, the moon entirely covers the sphere of the sun, therefore it can be said that both moon and the sun appear the same size form earth or they subtend the same solid angle on earth. Let ω be the solid angle subtended on earth, in which case we have,

where A_sun and A_moon, are the projected areas of the sun and moon as seen from earth and R_se, and R_{me �} are the distances of sun and moon from earth. Let R_sun and R_moon be the radii of the sun and the moon.

Then we have

Therefore, the distances of sun and moon are relative to their sizes.

Given:

A unit of force = 100 N

A unit of length = 10 m

A unit of time = 100 s

The dimensional formula for Force is, F = [MLT^-2].

Therefore mass, M = [FL^-1T²].

Substituting the values force, length and time in the above relation we have

Therefore, in this system of units the unit of mass is 10⁵ kg

(a) The physical quantity plane angle θ = has a unit radian but no dimensions as it is defined as relation of Length over radius which both have the same dimensions.

(b) Specific density is defined as the ratio of density of a substance to density of water. Thus, both quantities having the same dimensions and units, specific gravity has neither unit nor dimensions.

(c) The gravitational constant G = 6.67408 x 10^-11 m³ kg^-1 s^-2, thus has a unit.

(d) The constant Reynold’s number (R) used in fluid mechanics has no unit.

Where ρ is the density of the fluid, v is the velocity, d is the diameter of the pipe in which the fluid is flowing and μ is the viscosity of the fluid.

Given:

Radius of circle R = 31.0 cm

Angle subtended by arc L =

We know that the angle θ subtended by an arc of length L at the centre of circle of radius R is given by

Therefore, the length of the arc is 16.23 cm

Given:

Area of periphery =1 cm²

Radial distance of point = 5 cm

We know that the solid angle ω subtended by an object is given by

Therefore, the solid angle subtended is 0.04 steradian.

Given:

Equation for wave y = A sin(ωt – kx)

We know that the argument for the sine function is always in degrees or radians and therefore dimensionless. And we know that only dimensionally like quantities can be added or subtracted, therefore the quantities ωt and kx must be dimension less, therefore

but t =

Similarly,

But x =

Therefore, the dimensional formulas are (i) ω = [M⁰L⁰T^-1] and (ii) k =[M⁰L^-1T⁰]

Given:

Time measurement for 20 oscillations of a pendulum

t₁=39.6s

t₂=39.9s

t₃=39.5s

Since, all the measurements have a maximum of on decimal place, the lease count of the instrument used to measure the time is 0.1s.

Since precision in measurement is equal to the least count of the instrument, therefore

Now the mean value of time period of the pendulum for 20 oscillations:

Now mean time period for one oscillation:

Now accounting for the precision in measuring time for one oscillation we have the measured mean time period

Now the accuracy in measurement is given by the maximum observed error from the measured value, which is

Given

New unit of mass = α kg

New unit of length = β m

New unit of time = γ s

Magnitude of energy in original system n = 5J

The dimensional formula for energy is given as

Now let the magnitude of U in the original system of units be n, and in the new system of unit be n’

Then

Where [M₁L₁² T₁^-2] and [M₂L₂² T₂^-2] are the dimensions in the original system of units and new system of units respectively.

Now in the original system of units M₁ = 1 kg , L₁ = 1 m and T₁ = 1 s. In the new system of units we have M₂ = α kg , L₂ = β m and T₂ = γ s.

Thus, 5 J of energy will measure in the new system of units.

Given:

The equation for volume flow per second V=

The for the equation to be dimensionally correct the the dimensional formula of the equation must be equal to the dimensional formula of the volume flow per second.

Thus dimensional formula of

Dimensional formula of pressure

Dimensional formula of coefficient of viscosity

Dimensional formula of length

Dimensional formula of radius

Thus dimensional formula of RHS of the equation

And dimensional formula of LHS

Therefore, LHS = RHS

Thus, the given equation is dimensionally correct.

Given:

The equation of X = a² B³ c^5/2 d^-2

Percentage error in measurement of a = 1%

Percentage error in measurement of b = 2%

Percentage error in measurement of c = 3%

Percentage error in measurement of d = 4%

Calculated value of X = 2.763

The percentage error in X is given as , where ∆X is the error in X.

Similarly,

The percentage error in a =

The percentage error in b =

The percentage error in c =

The percentage error in d =

Therefore, from the given equation of X , the maximum percentage error in X can be calculated as follows

Therefore, the percentage error in measurement of X is 23.5%.

The absolute error in X is then = ±0.235 = ±0.24 after rounding off to two significant digits. Then, the value of X should be rounded off to two significant digits i.e. to 2.8.

Given:

Equation of P = E l² m^-5 G^-2

The dimensional equation of energy E is [M²L¹T^-2], for mass m is [M¹L⁰T⁰], for length l is [M⁰L¹T⁰] and for gravitational constant G is [M^-1L³T^-2].

Now to prove that P is a dimensionless quantity we have to find, the dimensional formula for P. For this, we substitute the dimensional formulae of the above quantities in the expression for P

Hence, P is a dimensionless quantity.

The dimensional formula for velocity of light c is [M⁰L¹T^-1], for plank’s constant h is [ML²T^-1] and for gravitational constant G is [M^-1L³T^-2].

1. Let the dimensional formula for mass m = c^ah^bG^c, then substituting the dimensional formula for c, h and G we have

Equating the exponents of like quantities on both sides we have

Adding the three equations and solving for b we have

Using this value of b and substituting in the above equations, we have

Now, substituting these equations in the expression for mass m we have

2. Let the dimensional formula for length l = c^ah^bG^c, then substituting the dimensional formula for c, h and G we have

Equating the exponents of like quantities on both sides, we have

Adding the three equations and solving for b we have

Using this value of b and substituting in the above equations, we have

Now, substituting these equations in the expression for length l we have

3. Let the dimensional formula for time t = c^ah^bG^c, then substituting the dimensional formula for c, h and G we have

Equating the exponents of like quantities on both sides, we have

Adding the three equations and solving for b we have

Using this value of b and substituting in the above equations, we have

Now, substituting these equations in the expression for length l we have

According to Kepler’s Third law, the square of the time period of a satellite revolving around a planet, is proportional to the cube of the radius of the orbit of the satellite. Let T be the time period of revolution and r is the radius of the orbit, then

Since, the revolution of the satellite follows the laws of gravitation the time period of revolution depends on acceleration due to gravity g and the radius of the planet R,

Where k, is a dimensionless constant of proportionality.

Now the dimensional formula for time period [T] = [M⁰L⁰T¹], for radius of planet [R] = [M⁰L¹T⁰], for radius of orbit of satellite [r] = [M⁰L¹T⁰] and for acceleration due to gravity [g] = [M⁰L¹T^-2]. Therefore,

Equating the exponents of like terms on both sides, we have

Substituting these values in the expression for T we obtain,

(a) Oleic acid is hydrophobic (water-hating) in nature and thus it cannot be dissolved in water. Therefore, to reduce its concentration it is dissolved in alcohol in which it is highly soluble. This alcohol is then dissolved in water to form a thin layer of oleic acid, and since it is hydrophobic the oleic acid molecules get organized next to each other in a single layer which is one molecule thick.

(b) Lycopodium powder is used to distinguish the region of oleic acid and water by forming a boundary between the two. It also weakens the surface tension of the oleic acid and allows the layer to expand to its maximum size.

(c) Given:

Volume of oleic acid in given solution = 1 ml

Volume of alcohol in given solution = 19 ml

Volume of additional alcohol added in the solution = 20 ml

We know that, 1 ml of oleic acid is present in 20 ml of the solution. Therefore, in 1ml of solution,

Therefore, in every 1 ml of solution there would be 0.05 ml of oleic acid. Now since this 1 ml solution is diluted by adding further 20 ml of alcohol, the final solution,

(d) For determining the volume of n drops of this solution an instrument called burette will be used. The burette will be filled to a desired volume as marked on the burette scale. The burette allows the solution to be released drop by drops and thus n drops will be counted, the final value of the volume will be measure through burette scale and then subtracted from the initial value which will be the volume of the no of drops.

Volume of n drops = Initial value – Final value

(e) Let there be n drops in 1 ml of this solution which will be measured by using the burette. We know that in 1 ml of this solution there is 0.0025 ml of oleic acid, then,

(a) Parsec is defined as the distance at which 1 A.U. long arc subtends an angle of 1s. The angle subtended by an arc of length L and radius r is given as:

Here L = 1 A.U., θ = 1s and r = 1 parsec. Therefore

(b) Given:

Sun’s diameter as seen from earth =

Star’s distance from earth = 2 parsecs

Magnification of the telescope = 100

Minimum resolution eye due to atmosphere = 1’

The size of the star as seen from the earth can be considered by placing the sun at the same distance from earth as the star since it is sun like. Thus if sun appears to be in diameter from earth at 1 A.U. then for 2 parsecs or as calculated in (a) from 2 x 2.06 x 10⁵ A.U. its angular diameter will be,

When seen from the telescope, the star will be 100 times magnified and therefore,

Converting to arc minutes the angular diameter will be 7.25 x 10^-5 arc minutes. Since due to atmospheric fluctuations, the eye cannot resolve below 1 arc minute, the sun like star appears to be 1 arc minute in angular diameter.

(c) Given:

Ratio between mar’s diameter and earth’s diameter =

Let D_e, D_ma and D_s be the diameters of earth, mars and the sun. Then we have,

Also, we know that

Therefore, we have,

To find the size of mars as seen from earth, we first place mars at distance equal to the earth-sun distance form earth i.e. 1 A.U. At 1 A.U. the sun appears to be in diameter, therefore for mars,

So, at mars’ original distance of A.U. mars will have a diameter of

With the telescope used in the previous problem,

Therefore, after magnification mars appears to be 30 arc minutes which is more than 1 arc minute and hence is not influenced by atmospheric fluctuations and is magnified. Thus, the distance of stars are so large in comparison to planets that stars cannot be magnified on earth while planets can be. To study stars, the stars are made brighter by allowing more light to enter the telescopes by using different designs.

(a) By using Einstein’s mass energy relationship we have,

For one atomic mass unit m = 1 u = 1.67 x 10^-27 kg.

Therefore, the energy equivalent,

Or

(b) Given:

Incorrect relation 1 u = 931.5 MeV

Frome Einstein’s equation we have,

For, m = 1 u , E = 931.5 MeV

Therefore, the dimensionally correct relation would be