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1.Define fundamental units and derived units.

A set of independent units of basic physical quantities which can easily be obtained and in terms of which units of all other physical quantities may be expressed is called fundamental units. Units of length, mass and time are taken as fundamental units in mechanics.

Physical quantities which can be expressed in terms of fundamental quantities are called derived quantities.

2.Write a note on the basic forces in nature.

There are four basic forces in nature. They are gravitational force, electromagnetic force, strong nuclear force and weak nuclear force.

Gravitational force

It is the force which holds the earth in its orbit round the sun. It is also the force that holds us on to the earth. It is the force of mutual attraction between any two object by virtue of their masses. It is the weakest force in nature and is always attractive.

Electromagnetic force

It is the force between electrically charged particles. The electric force can be attractive or repulsive. Like charges repel and unlike charges attract. The electrostatic force varies inversely as the square of the distance between the charges. It is very strong compared to the gravitational force. It is the combination of electrostatic and magnetic forces.

Strong Nuclear Force

It is the strongest of all the basic forces of nature. This is the force that binds protons and neutrons in a nucleus. Without this force, the nucleus of an atom would be unstable due to repulsive force between protons within the nucleus. The strong nuclear force is charge independent. Its range is very small, which is of the order of the size of a nucleus.

Weak Nuclear Force

This force appears only in certain nuclear processes such as β-decay of a nucleus. In β-decay the nucleus emits an electron and an uncharged particle called antineutrino. If the nucleus decays by emitting a positron, the emission is accompanied by a neutrino. Neutrino and anti neutrino experience only weak interactions. So any process involving them is governed by the weak nuclear force. The range of weak nuclear force is extremely small of the order of 10^{-15}m.

3.What are the limitations of dimensional analysis?

(1) In mechanics, this method is not suitable if the physical quantity depends on more than three other physical quantities. By equating dimensions of M,L and T on the two sides, we can obtain only three independent equations.

(2) This method does not tell us anything about dimensionless quantities such as number,angle,trignometrical ratio etc.

(3) This method cannot be employed if the right hand side of the equation contains more than one term.

(4) Quite often it is difficult to guess the parameters on which the physical quantity depends.

4.Explain the principle of homogeneity of dimensions.

According to this principle,a physical relation is dimensionally correct if the dimensions of fundamental quantities(mass,length and time) are the same in each and every term on either side of the equation.This principle is based on the fact that only quantities of the same kind (or dimensions) can be added or subtracted.For example,consider the equation ,A = B + C.Here the quantities A,Band C have the same dimensions.

5.What are the uses of dimensional analysis?

The method of dimensional analysis is used to

i) convert a physical quantity from one system of units to another.

Since a physical quantity is expressed in terms of appropriate units of the same nature in all systems, the dimensions should remain the same even though the number indicating its magnitude may differ. If a physical quantity of dimensions a, b and c in mass, length and time respectively has magnitude n_{1} and n_{2} in two systems having fundamental units M_{1}, L_{1}, T_{1} and M_{2}, L_{2}, T_{2} respectively then,

Example:

Convert one newton into dynes

They are the units of force in SI and CGS respectively.

We have the dimensional formula of force as MLT ^{-2}

Let n_{1} newton be equal to n_{2} dynes.

ii) Check the correctness of an equation.

An equation is correct only if the dimensions of each term on either side of the equation are equal. In this case the equation is dimensionally correct, but, may or may not be the correct equation for the physical quantity. If the homogeneity of dimensions does not hold good, the equation is definitely correct.

Example:

Check the accuracy of the equation,

Dimension of the term S = L

Dimension of the term ut = LT^{ -1} × T = L and the dimensions of the term (½)at^{2} = LT^{ -2 }× T^{2} = L.

Thus, we find that each term has the same dimensions. So the equation is dimensionally correct.

iii) derive the correct relationship between physical quantities.

When one physical quantity depends on several physical quantities, then the relationship between the quantities can be derived using dimensional method.

Example:

To derive an expression for the period of oscillation of a simple pendulum.

The period of oscillation t of a simple pendulum may depend on

(a) the length of the pendulum l_{1}

(b) the mass of the bob m and

(c) the acceleration due to gravity g

Let the form of the equation be t = kl^{x}m^{y}g^{z}

Taking dimensions of both sides, T = L^{x}M^{y}(LT ^{-2})^{z} = L^{(x+z)}M^{y}T^{ -2z} [Since k is dimensionless]

equating dimensions of L, M and T

6.The sides of a rectangular lamina are (8.5 ± 0.2)cm and (5.6 ± 0.1)cm. Calculate the perimeter of the lamina with error limits.

l = 8.5cm

Δl = 0.2cm

b = 5.6cm

Δb = 0.1cm

Perimeter, P = 2(l + b) = 2(8.5 + 5.6)

= 28.2cm

ΔP = ±2(Δl + Δb)

= ±2(0.2 + 0.1)

= ±0.6cm

Perimeter with error limits, P = (28.2 ± 0.6)cm

7.Define the following:

(a) Angstrom (Å)

(b) Astronomical unit (A.U)

(a) Angstrom (Å) : It is 10^{-10} th part of a metre. It is expressed by the symbol Å. That is

1Angstrom (Å) = 10^{-10} metre

= 10^{-8} cm = 10^{-1} nm

Therefore 1micron = 10,000 Å

and 1 nm = 10Å

Nowadays, nm is preferred over the Å. The wavelength of light, inter-atomic or inter-molecular separation etc. are now commonly expressed in nm.

(b) Astronomical Unit (A.U) : One Astronomical unit is equal to the mean distance between the earth and the sm that is;

1A.U = 1.496 × 10^{11} metre

8.Define Parsec.

One parsec is the distance from where the semi major axis of orbit of earth subtends an angle of 1″ at the centre of sun. One parsec is 3.26 times the light year. that is,

1parsec = 3.26 light year

= 3.26 (9.46 × 10^{15}m)

= 3.08 × 10^{16}m

9.Physics is an exciting subject. Comment.

The study of physics is exciting in many ways.

Example:

(i) Journey to the moon with controls from the ground.

(ii) Lasers and their ever increasing applications.

(iii) Live transmission of events, thousands of kilometers away on the T.V.

(iv) The speed and memory of the fifth generation of computers.

(v) Study of various types of forces in nature.

(vi) Technological advances in health science.

(vii) Use of robots is quite exciting

(viii) Telephone calls over long distance and so on.

Thus, physics is exciting not only to the scientist but also to a layman, children, women etc. All the musical instruments, toy guns, toy trains etc, all are constructed using simple principle of physics like collision, potential energy etc. Today, the situation is that even our thought process and social values are affected by physics. Thus, it is quite amazing.

10.A student measures the thickness of a human hair with the help of a microscope of magnification 100. He takes 20 readings and finds average field of view of microscope as 3.5mm. What is the thickness of the hair?

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