# Grade 11 Physics Solution

# Unit and measurement

### Question Answer

**Chapter – 1**

**Unit and measurement**

**A. Short questions**

**1.**

What is a unit?

A unit is defined as a convention to define an amount of physical property in a specific system of units. There are several systems of units with different conventions to express these units. There are units defined for every physical quantity and that physical quantity is expressed in terms of that specific unit. For example, in International System of units we have kilogram for mass, meter for length, seconds/minutes/hours for time, etc. Without a unit a physical property cannot be distinguished or described.

For writing, units are expressed in terms of alphabets, referred to as unit symbols. Like meter is written ‘m’, second as ‘s’, grams as ’g’, etc.

To expresses larger quantities of units in terms of power of tens, certain prefixes are used.

Examples: kilogram for 1000 grams, centimeters for 1/100 of a meter meters, milliseconds for 1/1000 of a second, etc.

2.

What is the international system of units?

Initially there were several systems of units in practice such as the MKS system which used meter for length, kilogram for mass and second for time, CGS system which used centimeter for length, gram for mass and second for time, British Engineering system which used foot for length, slugs for mass and second for time. The diversity in the use of system of units by different fields caused problems in coordination between the units and hence there was a need of a system of units to be used internationally. Therefore, the MKS system was promoted to be called as The International System of Units abbreviated as the SI system. The Si system has seven fundamental physical quantities given below:

Physical Quantity |
Unit |
Symbol |

mass |
kilogram |
kg |

length |
Meter |
m |

time |
second |
s |

temperature |
Kelvin |
K |

electric current |
ampere |
A |

amount of matter |
Mole |
mol |

luminous intensity |
candela |
Cd |

3.

What is meant by derived units? Give its examples.

Derived units are composed of the fundamental units and are originated from the products and ratios of the fundamental units. Derived units belong to physical quantities that are derived from basic physical properties. Examples are:

Velocity – a physical quantity derived from fundamental quantities of length and time

${\rm{velocity}} = \frac{{{\rm{length}}}}{{{\rm{time}}}} = \frac{{\rm{m}}}{{\rm{s}}}$

So the units of velocity is ms^{-1}

Acceleration – a physical quantity derived from velocity and time

${\rm{acceleration}} = \frac{{{\rm{velocity}}}}{{{\rm{time}}}} = \frac{{\rm{m}}}{{{{\rm{s}}^2}}}$

The units of acceleration is ms^{-2}

Force – a physical quantity derived from mass and acceleration

force=mass * acceleration = kg m s_{-1} or N

The units of force are kg.m.s^{-1} or N (Newton)

Work – a physical quantity derived from force and length

Work = force*length = N m or J

The units of work are Nm or J (Joules)

4.

Are dimensionless quantities always unitless?

Dimensionless quantities are those physical quantities which do not have any physical dimension associated with them. These usually result as a product or ratio of physical quantities such that their dimensions get cancelled out leaving a dimensionless entity.

Examples are:

Strain – ratio of change in length and original length

Strain = $\frac{{\Delta {\rm{l}}}}{{\rm{l}}} = \frac{{\rm{L}}}{{\rm{L}}}$

Refractive index – ratio of speed of light in vacuum and speed of light in a medium

refractive index = $\frac{{\rm{c}}}{{\rm{v}}} = \frac{{{\rm{L}}{{\rm{T}}^{ - 1}}}}{{{\rm{L}}{{\rm{T}}^{ - 1}}}}$

Although the dimension less quantities doesn’t have dimensions but they can have unit associated with them. These units are termed as dimension less units.

5.

What is principle of homogeneity of dimensions?

The principle of homogeneity of dimensions is used to verify the physical laws involving physical quantities. For a relation to be valid and correct the dimensional form of the physical law must be balanced, i.e. the dimensions at both side of the equation must be same. Every physical law that is valid obeys the principle of homogeneity of dimensions.

For example consider following equations:

Equations of motion:

${{\rm{v}}_{\rm{f}}} = {{\rm{v}}_{\rm{i}}} + {\rm{a\: t}}$

${\rm{L}}{{\rm{T}}^{ - 1}} = {\rm{L}}{{\rm{T}}^{ - 1}} + {\rm{L}}{{\rm{T}}^{ - 2}}{\rm{\: T}}$

${\rm{L}}{{\rm{T}}^{ - 1}} = {\rm{L}}{{\rm{T}}^{ - 1}} + {\rm{L}}{{\rm{T}}^{ - 1}}$

${\rm{s}} = {{\rm{v}}_{\rm{i}}}{\rm{\: t}} + \frac{1}{2}{\rm{a\: }}{{\rm{t}}^2}$

${\rm{L}} = {\rm{L}}{{\rm{T}}^{ - 1}}{\rm{\: T}} + \frac{1}{2}{\rm{L}}{{\rm{T}}^{ - 2}}{{\rm{T}}^2}$

${\rm{L}} = {\rm{L}} + \frac{1}{2}{\rm{L}}$

$2{\rm{\: a\: s}} = {\rm{v}}_{\rm{f}}^2 - {\rm{v}}_{\rm{i}}^2$

$2{\rm{\: L}}{{\rm{T}}^{ - 2}}{\rm{\: L}} = {\left( {{\rm{L}}{{\rm{T}}^{ - 1}}} \right)^2} - {\left( {{\rm{L}}{{\rm{T}}^{ - 1}}} \right)^2}$

$2{{\rm{L}}^2}{{\rm{T}}^{ - 2}} = {{\rm{L}}^2}{{\rm{T}}^{ - 2}} - {{\rm{L}}^2}{{\rm{T}}^{ - 2}}$

Mass – Energy equivalence:

E = m.c^{2}

J = kg.(ms^{-1})^{2}

ML^{2} T^{-2 }= ML^{2} T^{-2}

All the equations that are given above are dimensionally correct i.e. the dimensions are same at both sides of the equation. It is to be noted that all the constants are ignored in dimensional form and also dimensions are never to be canceled by subtraction, just like we have two similar terms with opposite signs in last equation of motion.

6.

Write the dimensional formula of power and force.

The dimensional formula of power can be found as,

${\rm{power}} = \frac{{{\rm{work}}}}{{{\rm{time}}}}$

${\rm{power}} = \frac{{{\rm{F}}.{\rm{d}}}}{{\rm{t}}}$

${\rm{power}} = \frac{{{\rm{ML}}{{\rm{T}}^{ - 2}}{\rm{\: L}}}}{{\rm{T}}}$

So the dimensional form of power becomes,

power=ML^{2} T^{-3}

The dimensional formula of force can be found as,

force= mass . acceleration

force = m.a

force = MLT^{-2}

So the dimensional form of force becomes,

force = MLT^{-2}

7.

State the use of dimensional analysis.

Dimensional analysis involves checking of a given physical law to check for its validity. Any physical law that is valid follows principle of dimensional homogeneity, and hence this analysis is used to check and verify physical laws by expressing them in the form of dimensional equations.It is a useful technique to check the validity of derived relations between physical quantities.

Dimensional analysis is also used to find out relationship between physical quantities. For example, let’s derive an expression for kinetic energy just with the knowledge that it depends upon the mass and velocity of a body,

K.E ∝ m^{x}v^{y}

K.E = k m^{x}v^{y}

Converting in dimensional form

J = k kg^{x} (ms^{-1})^{y}

ML^{2} T^{-2 }= k M^{x} L^{y} T^{-y}

Comparing powers of dimensions we get

Comparing M |
1 = x |
Equation – 1 |

Comparing L |
2 = y |
Equation – 2 |

Comparing T |
-2 = -y |
Equation – 3 |

Using values obtained through the above equations, the kinetic energy can be written as,

K.E = k m^{1} v^{2}

And kinetic energy is given as,

K.E = $\frac{1}{2}{\rm{\: m\: }}{{\rm{v}}^2}$

Hence the dimensional analysis is used to find relations between the physical quantities.

8.

What are supplementary units?

Supplementary units are dimension less physical quantities that are used along with fundamental units for derivation of derived physical quantities. There are two supplementary units.However, in 1995 these units were classified as derived units.

9.

What are the limitations of dimensional analysis?

Dimensional analysis has following limitations:

1. Through dimensional analysis we cannot find the values of dimensionless constants. Consider the example of derivation of kinetic energy through dimensional analysis,

K.E ∝ m^{x}v^{y}

K.E = k m^{x}v^{y}

Converting in dimensional form

J = k kg^{x} (ms^{-1})^{y}

ML^{2} T^{-2 }= k M^{x} L^{y} T^{-y}

Comparing powers of dimensions we get

Comparing M |
1 = x |
Equation – 1 |

Comparing L |
2 = y |
Equation – 2 |

Comparing T |
-2 = -y |
Equation – 3 |

Using values obtained through the above equations, the kinetic energy can be written as,

K.E = k m^{1} v^{2}

We have derived the correct relationship between the physical quantities but still the value of dimensionless ‘k’ is unknown.

2. If a physical quantity is dependent on more than 3 variables than we cannot find the relation between them. Consider a quantity ‘A’ with dimensions MLT depending on variables ‘m’, ‘n’, ‘o’ and ‘p’, with dimensions L, M, T and MLT. Using dimensional analysis we can get,

A∝m^{t}n^{x}o^{y}p^{z}

MLT = k L^{t}M^{x} T^{y} (MLT)^{z}

Now by comparing we will get,

Comparing M |
1 = x + z |
Equation – 1 |

Comparing L |
1 = t + z |
Equation – 2 |

Comparing T |
1 = y + z |
Equation – 3 |

Hence we end up with 3 equations involving 4 unknowns and therefore the values cannot be found.

3. If the equation involves more than a single terms in form of sum or difference, than we cannot apply dimensional analysis.

4. Dimensional analysis cannot be applied to physical quantities that are associated with exponential or trigonometric functions.

5. An equation which is dimensionally correct may not be always correct because several physical quantities have same physical dimensions.

10.

What is coherent system of unit?

A coherent system of units is the one which have its derived units made up of from base unit without any other numerical factor other than 1 or some power of ten. The Si system is the coherent system of units. For example consider the unit of power in SI system i.e. watt. A watt is expressed as,

$1{\rm{\: watt}} = \frac{{1{\rm{\: Joule}}}}{{1{\rm{\: second}}}}$

Hence the SI unit of power ‘watt’ is coherent.

The unit of volume ‘liter’ is expressed as,

1 liter (dm^{3}) = 1000 cm^{3}

Hence ‘liter’ is also a coherent unit.

Example of non-coherent unit is horsepower which is expressed as,

$1{\rm{\: horsepower}} = \frac{{550{\rm{\: foot}} - {\rm{pound}}}}{{1{\rm{\: second}}}}$ or 746 watts

As the numerical factor is not 1 neither some whole power of ten, it is considered as non-coherent unit.

**B. Long Questions**

1.

Distinguish between basic and derived units giving examples.

Basic or fundamental units are those seven units which are called as independent units. These seven fundamental units are length (l), mass (m), time (t), amount of substance (n), Electric Current (I), Thermodynamic temperature (T) and Luminous intensity (I_{v}). These seven units provide the base for deriving other units hence also called as base units. On the other side, derived units can be defined as the dependent units resulting in the ratio or product of the fundamental units. Derive units refer to those physical quantities derived from the seven basic physical quantities. Some of the derived physical quantities are velocity, acceleration, momentum, force, pressure, etc. These fundamental and derived physical quantities are the integral part of physics as nothing can be defined in physics without referring to the relevant unit of that quantity. Here are some examples of Basic or fundamental and derived units.

Velocity:

Velocity is a derived quantity resulting in the ratio of two base quantities that is Distance and Time, and can be represented as:

${\rm{velocity}} = \frac{{{\rm{distance}}}}{{{\rm{time}}}} = \frac{{\rm{m}}}{{\rm{s}}}$

Acceleration:

Acceleration is defined as the rate of change of velocity and can be represented as:

${\rm{acceleration\: }}\left( {\rm{a}} \right) = \frac{{{\rm{velocity}}}}{{{\rm{time}}}} = \frac{{\rm{m}}}{{{{\rm{s}}^2}}}$

Momentum:

Momentum is defined as the product of mass and velocity and its unit can be expressed as:

Momentum (p) = mass * velocity = kg.$\frac{{\rm{m}}}{{\rm{s}}}$

Force:

According to Newton’s second law of motion, force can be defined as the product of mass and acceleration. So, its unit can be expressed as

force (F) = mass * acceleration = kg * $\frac{{\rm{m}}}{{{{\rm{s}}^2}}} = {\rm{N}}\left( {{\rm{Newton}}} \right)$

Pressure:

Pressure can be defined as the normal force acting on a unit area. So, its unit can be expressed as:

pressure (P) = $\frac{{{\rm{force}}}}{{{\rm{area}}}} = \frac{{{\rm{newton}}}}{{{\rm{metr}}{{\rm{e}}^2}}} = \frac{{\rm{N}}}{{{{\rm{m}}^2}}}$

2.

What do you mean by the dimensions of a physical quantity? What is meant by dimensional formula and equations?

Dimensions of a physical quantity refer to the nature of quantity which defines it in some measurable terms. The physical dimensions are expressed in terms of length, mass and time represented by L, M and T respectively. It is to be noted that these dimensions are not to be confused with three-dimensions of space. Those are spatial dimensions and are represented in terms of Length i.e. ‘L’.

Dimensions of length (one dimensional quantity) = L

Dimensions of area (two dimensional quantity) = L * L = L^{2}

Dimensions of space (three dimensional quantity) = L * L * L = L^{3}

Therefore length is expressed in meters/feet, area in square meters/feet and space or volume in cubic meters/feet. It is to be noted that L can represent length expressed in system of units and is not confined to any specific one. Same is true for mass ‘M’ and time ’T’.

Dimensional formula:

A dimensional formula expresses a physical property in terms of fundamental physical quantities which are length, mass and time. To express a physical property in its dimensional form, it units must be known.

For example,

velocity (v) = ms^{-1 }= length * time^{-1} = LT^{-1}

acceleration (a) = ms^{-2 }= length * time^{-2 }= LT^{-2}

force (F) = m.a = mass * length * time^{-2} = MLT^{-2}

pressure (P)=FA^{-1 }= mass * length * time^{-2 }* (length^{2})^{-1}

=mass*length^(-1)*time^(-2)

=ML^(-1) T^(-2)

work (W) = F.d=mass * length*time^{-2 }* length

=mass * length^{2} * time^{-2}

=ML^{2} T^{-2}

Dimensional equations:

Dimensional equations are equations of physical laws expressed in terms of fundamental quantities. When correctly written, these equations are also dimensionally correct, i.e. the physical quantities are balanced at both sides of the equation. This method is known as dimensional analysis and is used to verify analytical relationship obtained from experimental data.

For example:

v_{f} = v_{i} + at

expressing each quantity in dimensional form

LT^{-1 }=LT^{-1 }+ LT^{-2} * T

LT^{-1} = LT^{-1 }+ LT^{-1}

Hence the quantities on both sides of the equation have same dimensions of LT^{-1}, therefore the equation is dimensionally correct. One might think that right side may be written as

LT^{-1 }= 2LT^{-1}

In this case the numbers will not prove the equations as invalid. In dimensional analysis the numbers are not significant and only dimensions are considered.

For another example, time period is given as,

${\rm{T}} = 2{\rm{\pi }}\sqrt {\frac{{\rm{l}}}{{\rm{g}}}} $

Expressing in dimensional form

${\rm{T}} = 2{\rm{\pi }}\sqrt {\frac{{\rm{L}}}{{{\rm{L}}{{\rm{T}}^{ - 2}}}}} $

Where g is acceleration due to gravity and has same dimensions as that of acceleration

T=2π T

Which is again dimensionally correct and constant term ‘2π’ is ignored.

3.

State and explain giving illustrations the uses of dimensional equations.

Dimensional equation is the equation used to describe the relation between the fundamental quantity and the derived quantity. It as a type of expression which shows the number of power that is required for obtaining the derived quantity. Following are the main uses of dimensional equations:

A. Dimensional equations are used to check the correctness of the equation formed by relating the fundamental quantities. This is based on the principle of homogeneity which states that only those physical quantities having same dimensions can be added, subtracted or equated together.For example:

v_{f} = v_{i} + at

Expressing each quantity in dimensional form

LT^{-1 }= LT^{-1} + LT^{-2} * T

LT^{-1} = LT^{-1 }+ LT^{-1}

Hence the quantities on both sides of the equation have same dimensions of LT^{-1}, therefore the equation is dimensionally correct. One might think that right side may be written as

LT^{-1} = 2LT^{-1}

In this case the numbers will not prove the equations as invalid. In dimensional analysis the numbers are not significant and only dimensions are considered.

B. Dimensional equations can be used to derive the relation between two physical quantities. For example, the time period of a pendulum is dependent on the length of the string and acceleration due to gravity. The mathematical relation of time period with these physical quantities can be found out as:

T ∝ l^{x}g^{y}

T = kl^{x}g^{y}

Converting in to dimensional form we get,

T=k L^{x} L^{y} T^{-2ys}

Comparing dimensions on both side of equation we get,

Comparing T |
1 = -2y |
Equation – 1 |

Comparing L |
0 = x + y |
Equation – 2 |

Solving the equation we get,

${\rm{y}} = - \frac{1}{2}$

${\rm{x}} = \frac{1}{2}$

So the time period can be written as,

${\rm{T}} = {\rm{k\: }}{{\rm{l}}^{\frac{1}{2}}}{{\rm{g}}^{ - \frac{1}{2}}}$

${\rm{T}} = {\rm{k}}\sqrt {\frac{{\rm{l}}}{{\rm{g}}}} $

And that is the correct form of time period. Although we cannot find the values of dimensionless constants but we can find out the relations between physical quantities.

C. Dimensional equations are also used in the inter conversion of the units from one system to another. For example, the unit velocity in S.I system is ($\frac{{\rm{m}}}{{{{\rm{s}}^2}}})$, but in C.G.S system, velocity has a unit of ( $\frac{{{\rm{cm}}}}{{{{\rm{s}}^2}}})$. Consider the following conversion of value of gravitational constant from CGS system to MKS system. We have a value of 6.67*10^{-8} cm^{3}g^{-1}s^{-2}, converting through dimensional analysis we get:

${\rm{G}} = 6.67{\rm{*}}{10^{ - 8}}\frac{{{{\left( {{\rm{cm}}} \right)}^3}}}{{{\rm{g\: }}{{\rm{s}}^2}}}* \frac{{{{\rm{m}}^3}}}{{{{\left( {100{\rm{\: cm}}} \right)}^3}}}* \frac{{{{10}^3}{\rm{\: g}}}}{{{\rm{kg}}}}* \frac{{{{\rm{s}}^2}}}{{{{\rm{s}}^2}}}$

${\rm{G}} = 6.67{\rm{*}}{10^{ - 8}}\frac{{{{\left( {{\rm{cm}}} \right)}^3}}}{{{\rm{g}}{{\rm{s}}^2}}}* \frac{{{{\rm{m}}^3}}}{{{{10}^6}{{\left( {{\rm{cm}}} \right)}^3}}}* \frac{{{{10}^3}{\rm{g}}}}{{{\rm{kg}}}}* \frac{{{{\rm{s}}^2}}}{{{{\rm{s}}^2}}}$

And hence we will get the result as,

G= 6.67* 10^{-11} m^{3} kg^{-1} s^{-2}

4.

Show the pressure and stress have the same dimensions.

Pressure is defined as the applied external force normal to the unit area:

${\rm{P}} = \frac{{\rm{F}}}{{\rm{A}}}$

Dimensionally, force can be expressed as ${\rm{ML}}{{\rm{T}}^{ - 2}}$

And area can be expressed as ${{\rm{L}}^2}$ so,

${\rm{P}} = \frac{{{\rm{ML}}{{\rm{T}}^{ - 2}}}}{{{{\rm{L}}^2}}}$

${\rm{P}} = {\rm{M}}{{\rm{L}}^{ - 1}}{{\rm{T}}^{ - 2}}$

Now, stress is the internal force exerted by the body on a unit area:

$\sigma = \frac{{\rm{F}}}{{\rm{A}}}$

Dimensionally, force can be expressed as ${\rm{ML}}{{\rm{T}}^{ - 2}}$

And area can be expressed as ${{\rm{L}}^2}$ so,

$\sigma = \frac{{{\rm{ML}}{{\rm{T}}^{ - 2}}}}{{{{\rm{L}}^2}}}$

$\sigma = {\rm{M}}{{\rm{L}}^{ - 1}}{{\rm{T}}^{ - 2}}$

5.

Show that the following pairs of physical quantities have identical dimension (i) momentum and impulse (ii) torque and energy and (iii) energy and work.

1. Momentum and Impulse

Momentum is defined as the product of mass and velocity. Momentum is a vector quantity and has a direction along with magnitude. The SI unit for momentum is kgms^{-1} or N-s Momentum can be written as,

Momentum (p) = m v

Converting in dimensional form we get,

p=MLT^{-1}

Impulse is defined as the integral of force acting on a body in a time interval during which the force is acting on the body. The unit of impulse is N-s. As it involves force which is a vector, impulse is also a vector quantity. Impulse is expressed as,

Impulse (J) = Ft

Converting in dimensional form we get,

J=MLT^{-2}T = MLT^{-1}

Since it is proved that the momentum (p) and impulse (J) have same dimensions.

2. Torque and Energy

Torque can be defined as the force with a tendency of rotating a body about an axis. Torque causes the body to rotate. Torque is a vector quantity and defined as the cross product of force and the distance of force from the axis of rotation, also called as turning radius. The units of torque is N-m and expressed as,

Torque (τ) = F * r

Converting in dimensional form we get,

Τ = MLT^{-2}L = ML^{2} T^{-2}

Energy can be defined as the ability of a system to perform work. As it is something that makes a system able to do work, the energy is expressed in units of work i.e. Joules. Writing energy in dimensional form we get,

Energy (E) = ML^{2} T^{-2}

Hence it is proved that the torque and energy have same dimensional forms.

3. Work and Energy

Work is said to be perform when a force acts on a body and displaces it through some distance. This physical quantity has the units of N-m or Joules. Work is expressed as,

Work (W) = F d

Converting into dimensional form we get,

W = ML^{2} T^{-2}

The term energy has been already defined and can be expressed in the dimensional form as

Energy (E) = ML^{2} T^{-2}

Hence energy and work has same physical dimensional form.