Physics notes for class 11 KPK Board Chapter 1 | FSc Pre Engineering

FSc Pre Engineering chapter 1 kpk all obards Physics notes for class 11, Conceptual Questions, Comprehensive Questions, and Numerical Questions.

Conceptual Questions Physics notes for class 11 KPK

Q.1) For an answer to be complete, the units need to be specified. Why?

Answer:
In order to represent a physical quantity completely its unit must be specified along with the magnitude. Otherwise, it would be meaningless. For example, if we say the length of a meter rod is 10, it won’t serve the whole purpose as we don’t know units in which we are measuring the length. It can either be in meters or centimeters depending on the standard of the unit set for measurement. Thus for an accurate representation of physical quantities the specification of units is very important.

Physics notes for class 11 KPK Board Chapter 1 | FSc Pre Engineering
Physics notes for class 11 KPK Board Chapter 1 | FSc Pre Engineering

Q.2) What are the advantages of using International System of Units (SI)?

Answer:
Advantages of International System of Units:
1. SI units are used internationally i.e. this system is recognized and understood worldwide.
2. It is very helpful for the scientific community to work efficiently no matter where they are in the world e.g. if two scientists in two different countries are working on the same project under the same circumstances using the International system of units, they will end up with same results. But if they are using different systems they won’t get the same result even if they are measuring the same physical quantity. In that case, they have to spend more time finding the cause of the problem.
3. It significantly reduces the chances of errors in the transformation of information.
4. It is coherent i.e. the SI unit of power, Watt is equal to one joule per second, unlike the British Engineering system in which the unit of power, horsepower is equal to 550-foot pounds per sec or 746 watts.

Read more: Physics All chapters MCQs class 11 Pdf Download

Q.3) How many radians account for circumference of a circle? How many steradians account for circumference of a sphere?

Answer:
The circumference of circle is 2πr. Radians that account for circumference of circle can be found as;

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Number of radians in one revolution = Circumference of Circle / Radius of same circle
= 2πr / r
= 2π radians

Hence the radians account for the circumference of circle is 2π radians 

The area of sphere is 4πr2.Steradians in the sphere can be found as;

Number of steradians in sphere = Area of sphere /  squared radius of same sphere
= 4πr2. / r2
= 4π steradians

Hence the number of steradians in sphere is 4π steradians.

Q.4) What is least count error? How can least count error be reduced?

Answer:
Least count error:
The error that is associated with the resolution of the instrument is known as least count error.

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For Example, the least count of Vernier caliper is 0.01 cm i.e. the smallest value that can be measured by it is 0.01 cm. If a more precise reading is needed, it can’t be found using Vernier caliper which results in the error.

Reduction of Least Count Error:
Least count error can occur with both systematic and random errors. It can be reduced by using instruments with higher resolution or by taking the readings several times and finding its mean value.

Q.5) Why including more digits in answers, does not make it more accurate?

Answer:
Accuracy of a measurement depends on the number of significant figures i.e. the accurately known digits and the first doubtful digit. For example, two persons measure the length of a pencil using a ruler that has divisions/intervals of 1 cm. One person records the value to be 5.5 cm which has two significant figures. The other person records the value to be 5.53 cm. It also has two significant figure because the last digit ‘3’ cannot be justified according to the scale of this ruler. Thus by increasing the number of digits does not make it more accurate.

Q.6) What determines the precision of a measurement?

Answer:
The precision i.e. degree of exactness with which a measurement is made and stated can be determined by the position of the last significant figure of the measured value.

For example, if a measurement result comes out to be 2642 m, its precision is 1 m because the last significant figure is in the units place. Similarly, if a measurement result comes out to be 34000 km, its precision is 1000 km because in this value the significant figures are 3 and 4 only and the last significant figure ‘4’ is in the thousands place.

Q.7) If two quantities have different dimensions, is it possible to multiply and/or divide. Can we add and/or subtract them?

Answer:
Yes, two quantities with different dimensions can be multiplied or divided.
For example, velocity of an object is determined by dividing displacement over time.
                                                                 v = d/t
The dimension of displacement ‘d’ is [L] while the dimension of time ‘t’ is [T].

In order to add or subtract two physical quantities, their dimensions should be same. Otherwise these cannot be added or subtracted.
For example, length [L] can be added/subtracted to length only. It cannot be added to mass or time having dimension [M] and [T].

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Q.8) The human pulse and the swing of a pendulum are possible time units. Why are they not often used?

Answer:
Human pulse cannot be used as time units because it is not fixed and may vary from one moment to the other. Swing of pendulum is also not a good standard because it varies with different conditions. The length of pendulum can expand or contract depending on environmental conditions. Hence their frequency and period of swing can vary. The time period of pendulum is given by;

                                                              T =  2π√(l/g)

Longer pendulums take more time to swing than shorter pendulums. Similarly, value of ‘g’ also varies with height and depth. Hence it is not a good standard of time.

Read more: schedule of annual examinations of 10th and 12th class released | Peshawar

Q.9) If an equation is dimensionally correct, is that equation a right equation?

Answer:
If a physical equation is dimensionally correct, it does not always have to be physically correct. It may or may not be correct. The dimensionally correct equation does not prove the exactness of the equation. However, a dimensionally wrong equation is always wrong e.g. if we are finding the length of a rod, the answer should be in meters (for SI unit). It cannot be in kilograms or liters. 

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Comprehensive Questions for class 11 Physics Notes KPK

Q.1) Define Physics? Explain the scope and importance of physics in science, technology and society?

Answer:
Definition:
“Physics is a branch of science, which deals with the study of matter, energy and their mutual relationship.”

Scope and Importance:
Scope and importance of physics is very clear in everyday life. It gives us detailed information regarding several occurring events in universe. It is not only a study of smallest subatomic particles like electron, proton, neutron etc. However, it also covers the huge heavenly bodies like galaxies etc. It explains nature in terms of some fundamental laws e.g. Newton’s law, Archimedes law principle, Coulomb’s law etc. It has brought a great revolution in information technology and communication fields.

The information media and fast means of communication has brought the whole world into close contact. Global positioning system and tracking system of vehicles are also the latest achievements. Video mobile communication is now in common use. It helped us to get many answers to some very common questions like, what is moon? What is sun? How the universe formed?

Physics has also developed many useful instruments for precise and accurate measurements.

Q.2) What is system of units? In SI what is meant by base, derived and supplementary units?

Answer:
“A complete set of units for all physical quantities is called system of units”.
It is a system used to define and describe different physical phenomena by relating them to some selected standards.

In SI three systems of units are used named as;

  • Base units.
  • Derived units.
  • Supplementary units.

Base Units:
Base units are SI units, which are defined arbitrarily and cannot be defined in terms of other units. There are seven base units namely length, mass, time temperature, electric current, luminous intensity and amount of substance.

Derived Units:
These are units, used to measure all other physical quantities and they are derived from seven base units. Some derived units are newton, joule, watt etc.

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Supplementary Units:
There are some units which general conference on weights and measures has not yet either classified under base units or derived units. These units are supplementary units. This class contains only two units named as radian and steradian.

Q.3) What conventions are used in SI to indicate units?

Answer:
There are certain conventions that are used in SI to write unit symbols and names;
Conventions for Unit Symbols:

  • Unit symbols are printed in roman (upright) type regardless of the type used in the surrounding text. For example, m for metre, s for second and Pa for pascal.
  • A multiple or sub-multiple prefix, if used, is part of the unit and precedes the unit symbol without a separator. A prefix is never used in isolation, and compound prefixes are never used. For example nm, not mum or pm not ppm.
  • In forming products and quotients of unit symbols the normal rules of algebraic (multiplication or division) apply. Multiplication must be indicated by a space or a half-high (centred) dot (•). Division is indicated by a horizontal line, by a solidus (oblique stroke, /) or by negative exponents. For example N m or N • m, for a newton metre and m/s or m s-1 for metre per second.
  •  It is not permissible to use abbreviations for unit symbols or unit names, such as sec (for either s or second), sq. mm (for either mm2 or square millimetre).
  • When multiple of unit is raised to the power the power applies to the whole multiple not just the unit.

Read more: KPK Class 10 Chemistry Notes Chapter #16 | (Chemical Industries)

Conventions for Writing Unit names:

  • The names of units start with a lower-case letter (even when the symbol for the unit begins with a capital letter), except at the beginning of a sentence or in capitalized material such as a title. For example joule J, hertz Hz, second s and ampere A.
  • In case of writing unit name instead of the unit symbol, the unit name should be spelled out in full. For example, 2.1 metres per second. 
  • When the name of a unit is combined with the name of a multiple or sub-multiple prefix, no space or hyphen is used between the prefix name and the unit name.  For example, milligram, but not milli-gram or milli gram and kilopascal, but not kilo-pascal or kilo pascal.

Q.4) What are errors? Differentiate between systematic and random errors?

Answer:
Errors:
“It is a difference or disagreement between measured and accepted standard value.” 

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or

“The doubts that exist about the result of any measurement are termed as errors.”

Systematic Errors
1. The errors that arise due to faulty apparatus and poor calibration of instrument itself are known as systematic error.

2. These errors can also arise due to zero error of the instrument. 
3. These errors tend to be in one direction, either positive or negative.
4. These errors can be removed either by using a more accurate instrument or by adding or subtracting the zero error from the calculated reading.

Random Errors
1. When repeated measurements of the same quantity under the same conditions result in different values the error is known as random error.
2. These errors can arise due to changes in temperature or humidity or some other unknown causes in an experimental environment.
3. These errors occur irregularly and so are random concerning sign and size.
4. Maintaining strict control conditions in the laboratory, repeating the measurements several times, and taking the mean of measured values can minimize this error.

Q.5) What is uncertainty in measurement? explain the propagation of uncertainty in addition, subtraction, multiplication and division?

Answer:
Uncertainty in Measurement:

“The quantification or magnitude of error or doubt in measurement is known as uncertainty.”

It estimates how large or small the error is. In order to represent a measurement successfully the propagation of uncertainty to this measured value must be calculated. Hence a measurement needs to be written in the form;

measurement = best estimate ± uncertainty

For example, if a measured value of length of rod is (50.0 ± 0.2) m. It means that the length lies between 49.8 m i.e. ( 50.0 – 0.2) m and 50.2 m i.e. (50.0 + 0.2) m.

 Uncertainties are of two types;
1- Absolute Uncertainty
2- Percentage Uncertainty

Propagation of uncertainty in Addition or Subtraction:
Suppose two physical quantities A and B have measured values A ± ΔA and B ± ΔB respectively where ΔA and ΔB are their absolute uncertainties. The following steps are followed for the result Z = Z ± ΔZ in their addition and subtraction.

Addition: Let Z = A + B and the measured values of A and B are A ± ΔA and B ± ΔB. By addition,
Z ± ΔZ = (A ± ΔA) + (B ± ΔB).
Z ± ΔZ = (A +B) ± (ΔA + ΔB). ∴ absolute uncertainties are added
The maximum possible uncertainty in Z ± ΔZ = ΔA + ΔB.

Subtraction: Let Z = A – B and the measured values of A and B are A ± ΔA and B ± ΔB. By subtraction,
Z ± ΔZ = (A ± ΔA) – (B ± ΔB)
Z ± AZ = (A – B) ± (ΔA + ΔB). ∴ absolute uncertainties are added
The maximum value of the uncertainty ΔZ is again ΔA + ΔB i.e. absolute uncertainties are added in, both, addition and subtraction.

Propagation of uncertainty in Multiplication and Division:
In case of multiplication and division absolute uncertainties are converted into percentage uncertainties which are added. Suppose two physical quantities A and B have measured values A ± ΔA, B ± ΔB respectively where ΔA and ΔB are their absolute uncertainties and ΔA% and ΔB% are their percentage uncertainties. The following steps are followed for the result Z = Z ± ΔZ in their multiplication and division.

Read more: KPK Class 10 Chemistry Notes Chapter #15 | (Environmental Chemistry II: Water)

Multiplication:
Suppose Z = A.B and the measured values of A and B are A ± ΔA and B ± ΔB. Then
Z ± ΔZ = (A ± ΔA) (B ± ΔB)
Convert fractional uncertainty to percentage uncertainty;
Z ± ΔZ = (A ± ΔA%).(B ± ΔB%)
Multiply the product and add percentage uncertainties such as;
Z ± ΔZ = (A.B) ± (ΔA% + ΔB%)
Z ± AZ = (A.B) ± (ΔA + ΔB)%
Convert back to absolute uncertainty
Z ± ΔZ = (A.B) ± (ΔZ)

Division:
Suppose Z = A/B and the measured values of A and B are A + ΔA and B + ΔB. Then
Z ± ΔZ = (A ± ΔA) / (B ± ΔB)
Convert fractional uncertainty to percentage uncertainty
Z ± ΔZ = (A ± ΔA%) / (B ± ΔB%)
Divide the ratios and add percentage uncertainties
Z ± ΔZ = (A / B) ± (ΔA% + ΔB%)
Z ± ΔZ = (A / B) ± (ΔA + ΔB)%
Convert back to fractional uncertainty
Z ± ΔZ = (A / B) ± (ΔZ)

Q.6) What are significant figures? What are the rules for determining significant figures in the final result after addition, subtraction, multiplication and division?

Answer:
Significant Figures:

“A significant figure is one that is reliably known. In any measurement the accurately known digits and the first doubtful digit are collectively called significant figures.”

Rules for Determining Significant Figures:

  • Addition and Subtraction:
    When two or more quantities are added or subtracted, the result is as precise as the least precise of the quantities. After adding or subtracting, round the result by keeping only as many decimal places as are in the figure containing least decimal places of the quantities that were added or subtracted.
    For Example:

    44.56005 + 0.0698 + 1103.2 = 1147.82985.

    We do not want to write all of those digits in the answer.
    Rounding to the nearest tenth of the figure, the sum is written = 1147.8. Thus the significant figures in the final result after addition are 5.
  • Multiplication and Division:
    When quantities are multiplied or divided, the result has the same number of significant figured as the quantity with the smallest number of significant figures. For example:

    45.26 × 2.41 = 109.0766.

    Since the answer should have only three significant figures, we round the answer to 45.26 × 2.41 = 109. Thus the significant figures in the final result after multiplication are 3.
    In scientific notation, we write 1.09 x 102 .

Q.7) Differentiate between precision and accuracy in the measurement?

Answer:
Difference between precision and accuracy:

                            Precision
1. Precision of a measurement is associated with the least count of the measuring instrument.
2. Smaller the least count of measuring instrument, better will be its precision.
3. It can be referred to “how close the two or more measurements are with each other”.
4. Precision ensures the reliability and consistency of the measured value.

                                Accuracy
1. Accuracy is associated with the magnitude of fractional error or relative error in a measurement.
2. Smaller the magnitude of fractional or relative error, better will be its accuracy.
3. It can be referred to “how close the measured value is with the standard value”
4. Accuracy ensures the correctness of the measured value.

Q.8) What is meant by dimensions of physical quantities? What are limitations and applications of dimensional analysis?

Answer:
“The powers positive or negative to which the fundamental units of first system must be raised to give the unit of a given physical quantity are called dimensions of that quantity.” 

Dimensions of physical quantity:
In physics, a dimension represents the physical nature of a quantity. The dimension of each basic measurable physical quantity is represented by specific symbol and written within square bracket.  For example, different quantities such as length, breadth, wavelength etc. are all measured in meter and have the dimension [L], similarly dimension of time is [T] and that of mass is [M].

Dimensions of derived quantities are product of dimensions of base quantities. For example, dimension of velocity is given as
v = [LT-1]
In addition as acceleration, “a” is velocity per unit time square so its dimensions will be,
a = [LT-2]

Limitation of Dimensional Analysis:

  1. Even If a physical equation is dimensionally correct, it does not always have to be physically correct. Dimensionally correct equation does not prove the exactness of equation. However, a dimensionally wrong equation is always wrong e.g. if we are finding length of a rod, the answer should be in meters (for SI unit). It cannot be in kilograms or liters. 
  2.  Dimensional constants can not be obtained by using dimensional analysis method of deducing the relation among interdependent physical quantities.
  3. It does not distinguish between the physical quantities having same dimensions.

Applications of dimensional analysis:

  1.  We can check the homogeneity of an equation with dimensional analysis.
    For example, we can check the homogeneity of following equation,
11
Physics notes for class 11 KPK Board Chapter 1 | FSc Pre Engineering 8

Putting dimensions, we get

Putting dimensions we get
Physics notes for class 11 KPK Board Chapter 1 | FSc Pre Engineering 9

As 3/2 is a dimensionless constant so above equation gives us, [L] = [L], which tells that equation is homogeneous.

      2.   It is useful to drive a formula for a physical problem.

Numerical Questions for class 11th kpk boards

Q.1)  A circular pizza is cut into 3 equal parts, one piece of pizza is taken out. Estimate the degree measure of the single piece of pizza and convert the measure into radians. What is the radian measure of the angle of the remaining part of pizza?

Answer:
Total angle in a circular pizza = 360°
It is cut into 3 equal parts so,
No of degrees in single piece of pizza= 360°/ 3 = 120°
Converting it into radians;

1°= 0.01745 rad
⇒ 120° = 120 × 0.01745 rad
= 2.09 rad

Hence the no of radians in single piece of pizza are 2.09 rad

Pieces of cake left after taking one piece out = 2
No of degrees in the remaining part of pizza = 2 ×120° = 240°
In radians;

1°= 0.01745 rad240° = 240° × 0.01745 = 4.19 rad

Hence the no of radians in the remaining two pieces of pizza are 4.19 rad

Q.2) The length of a pendulum is (1.5 ± 0.01) m and the acceleration due to gravity is taken into account as (9.8 ± 0.1) m s-2. Calculate the time period of the pendulum with uncertainty in it.

Answer:
Given Data:
Length of pendulum = ℓ ± Δℓ = (1.5 ± 0.01) m
                   Acceleration due to gravity = g ± Δg = (9.8±0.1) m s-2

To Find:
                   Time period of pendulum = T ± ΔT =?
Solution:
                    
As time period of pendulum is given by,

T = 2π √ (ℓ /g)
= 2π (ℓ /g)1/2

Hence                 T ± ΔT = 2π [(ℓ ± Δℓ) / (g ± Δg)]1/2
T ± ΔT = 2π [(1.5 ± 0.01) m / (9.8±0.1) m s-21/2           ………. (i)

First finding percentage uncertainties in ‘ℓ’ and ‘g’
Percentage uncertainty in length = (0.01 / 1.5) × 100% = 0.7%
Percentage uncertainty in gravitational acceleration = ( 0.1 / 9.8) × 100% = 1.0%

Writing eq (i) in terms of percentage uncertainties;
T ± ΔT = 2π [(1.5 m ± 0.7%) / (9.8 m s-2 ± 1.0%)] 1/2

In division the quantities i.e. length ‘ℓ’ and gravitational acceleration ‘g’ are multiplied and percentage uncertainties in them are added up so;
T ± ΔT = 2π [(1.5 m / 9.8 m s-2) ± (0.7% +  1.0%)] 1/2
T ± ΔT = 2π (0.15 ± 1.7%)1/2
In case of power the percentage uncertainties are multiplied by power

y
Physics notes for class 11 KPK Board Chapter 1 | FSc Pre Engineering 10

      (T ± ΔT) = 2.4 ± 0.9%

Q.3) Determine the area of a rectangular sheet with length (l ± Δl) = (1.50 ± 0.02) m and width (w ± Δw) = (0.20 ± 0.01) m. Calculate the area (A ± ΔA).

Answer:
Given Data:

Length of rectangular sheet = (l ± Δl) = (1.50 ± 0.02) m 
Width of rectangular sheet = (w ± Δw) = (0.20 ± 0.01) m

To Find:

Area of rectangular sheet = (A ± ΔA) =?

Solution:

As area of rectangle is given by,

                                 Area = length × width
A = l × w
or (A  ± ΔA) = (l ± Δl) × (w ± Δw)
⇒ (A  ± ΔA) = (1.50 ± 0.02) × (0.20 ± 0.01) m2 ……………. (i)

First finding the percentage uncertainties in length and width;
Percentage uncertainty in length of rectangular sheet = (0.02 / 1.50 ) × 100% = 1.33%
Percentage uncertainty in width of rectangular sheet = (0.01 / 0.20) × 100% = 5%

Now writing percentage uncertainties in eq (i) so,
(A  ± ΔA) = (1.50 m ± 1.33%) × (0.20 m ± 5%)

In multiplication the quantities i.e. length ‘l’ and width ‘w’ are multiplied and percentage uncertainties in them are added up;
(A  ± ΔA) = (1.50 m× 0.20 m) ± (1.33% + 5%)
= 0.30 m2 ± 6.33%

Now converting the percentage uncertainty back to fractional uncertainty;

Thus area of rectangular sheet
Physics notes for class 11 KPK Board Chapter 1 | FSc Pre Engineering 11
Thus area of rectangular sheet 1
Physics notes for class 11 KPK Board Chapter 1 | FSc Pre Engineering 12

Thus area of rectangular sheet = (A  ± ΔA) = (0.30 ± 0.02) m2.

Q.4) Calculate the answer up to appropriate numbers of significant digits


(a) 246.24 + 238.278 + 98.3 
(b) 1.4 x 2.639 + 117.25 
(c ) (2.66 x 104) – (1 .03 x 103)
(d) (112 x 0.456) / (3.2 x 120) 
(e) 168.99 x 9
(f) 1023 + 8.5489

Answer:
(a) 246.24 + 238.278 + 98.3 = 582.818
Since the quantity 98.3 contains least decimal places so the sum should be = 582.8

(b) 1.4  ×  2.639 + 117.25 =120.9446
Since the quantity 1.4 contains least decimal places so the sum should be = 120.9

(c ) (2.66 x 104) – (1 .03 x 103) = 2.5579 × 104
Since the no of decimal places are 2 in both quantities So, the answer should be =2.56 × 104

(d) (112 × 0.456) / (3.2 × 120) = 0.133
Since the quantity 3.2 has smallest no of significant figures i.e. 2 so the result of division should have 2 significant figures i.e. = 0.13

(e) 168.99 ×  9 = 1520.91 = 1.52091 × 103
The quantity 168.99 has 5 significant figures while the quantity 9 has 1 significant figure so the result should have one significant figure so the answer = 2 × 102000

(f) 1023 + 8.5489 = 1031.55
For appropriate significant figure the answer should have decimal places equal to the decimal places of the quantity containing the least decimal places which, in this case, is 1023 so, the answer = 1032

Q.5) Calculate the answer up to appropriate numbers of significant digits (a) The ratio of mass of proton ‘mp‘ to the mass of electron ‘me‘ 

m/ m= 1.67 ×  10-27 kg / 9.1096 ×  10-31 kg 
(b) The ratio of charge on electron ‘qe‘ to mass of electron ‘me‘ 
qe / me = 1.6 × 1019 C / 9.1096 × 10-31 kg 

Answer:
(a) m/ m= 1.67 ×  10-27 kg / 9.1096 ×  10-31 kg
= 1.833 × 103
According to the rule of significant figures formultiplication and division,the answer should have the same no of significant figures as the quantity with the smallest no of significant figures.  In this case mp has smallest no of significant figures i.e. 3 so, the no of significant figures in the answer should be 3. Hence we round off the answer as;
m/ m= 1.83 × 103

(b) qe / me = 1.6 × 1019 C / 9.1096 × 10-31 kg
                   = 1.76 × 1049 C/kg
Since the given quantity 1.6 × 1019 C has the smallest significant figures i.e. 2. So, the answer should have 2 significant figures. Thus we round off the answer as;
qe / me = 1.8 × 1049 C/kg

Q.6) Find the dimensions of 

(a) planck’s constant ‘h’ from formula E = hf 
Where E is the energy and f is frequency. 
(b) gravitational constant ‘G’ from the formula F = G m1m2 / r2 
Where ‘F’ is force, ‘m1‘ and `m2‘ are masses of objects and ‘r’ is the distance between centers of objects.

Answer:
(a) Dimensions of planck’s constant ‘h’:
      Since                                             E = hf
⇒                                                      h = E / f   ………………..  (i)

As energy is equivalent to work so,
                                                          E = W = F.d

Substituting value of E in eq (i);
                                                          h = F.d / f

Writing dimensions of force, displacement and frequency;
Since force is given as F = ma and frequency is f = 1/t so their dimensions can be written as;

h = [MLT-2][L] /  [T-1]
h = [ML2T-1]

Hence the dimension of h is [ML2T-1]

(b) Dimensions of Gravitational Constant ‘G’:
As, gravitational force is given by,

F = (Gm1m2)/ r2

Where ‘G’ is gravitational constant, ‘m1’ and  ‘m2’ are the masses of objects an ‘r’ is the distance between their centers. So, we get, 

G = (Fr2) / (m1m2)

Substituting dimensions of force, distance and masses, we get. 

gh
Physics notes for class 11 KPK Board Chapter 1 | FSc Pre Engineering 13

So dimensions of gravitational constant will be,
                                                            G = [M-1L3T-2

Q.7) Show that 

(a) KE = (1/2)mvand  
(b)  PEg = mgh are dimensionally correct.

Answer:
(a) KE = (1/2)mv2  ………………… (i)
Since KE is the energy of an object due to its motion. So whenever work is done on an object, the KE stored in it is equivalent to work and can be written as;;

KE = W = F.d

Substituting KE = F.d in eq (i);

F.d = (1/2)mv2    

Since 1/2 is a constant. It is dimensionless. Now writing dimensions of force, displacement, mass and velocity;

[MLT-2][L] = [M][LT-1]2
[ML2T-2] = [M][L2T-2]
[ML2T-2] = [ML2T-2]

Since both sides of above equation are same, the equation is dimensionally correct.

(b)  PEg = mgh  ………………… (ii)
Since PE is the energy of an object due to its position. So whenever work is done on an object, the PE stored in it is equivalent to work and can be written as;

PE = W = F.d

Substituting PE = F.d in eq (ii);

F.d = mgh

Now writing dimensions of force, displacement, mass, gravitational acceleration and height;

 [MLT-2][L] = [M][LT-2][L]
[ML2T-2] = [ML2T-2]

Since both sides of the above equation are equal, the equation is dimensionally correct.

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