We will discuss about the important aspects of this chapter, the kind of questions covered in JEE Advanced , JEE mains and EAMCET.

**Importance**

Units and dimensions is basically considered as numerical topic when it comes to JEE Mains and Advanced. A brief analysis of previous year papers shows atleast 1 or 2 questions are asked every year in JEE mains and JEE Advanced from this topic.

The essence of this chapter is, it can be combined with some of the formulas of other chapters also. So you need to have a thorough practice for its numerical contents along with the basic formulas of other chapters.

You can go through the points below so as to get some idea about the topic.

A brief idea about the dimensions and basic formulas from above mentioned chapters will be very helpful for this topic.

#### Introduction

**UNITS:**
Things in which quantity is measured are known as units.
Measurement of physical quantity = (Magnitude) × (Unit)

There are three types of units

1 Fundamental or base units

2 Derived units

3 Supplementary units

**DIMENSIONS :**
Dimensions of a physical quantity are the powers to which the fundamental quantities must be raised to represent the given physical quantity.

*Example:*
Force (Quantity) = mass × acceleration

= !!mass × (velocity)/(time) = massxx(distance)/(time)^2!!
= mass × length × !!(time)^2!!

So dimensions of force : 1 in mass , 1 in length , –2 in time

and Dimensional formula : !![MLT^–2]!!

**APPLICATION OF DIMENSIONAL ANALYSIS**

- To find the unit of a given physical quantity in a given system of units.

*Example:* To find the dimensions of physical constants Planks constant (h) , Coefficient of viscosity (η)

Dimensions of h : Plank’s constant

!!E=hnu!!

!![ML^2T^(–2)] =hxx1/T !!

So !!h = [ML^2T^(–1)]!!

Dimension of η : Coefficient of viscosity.

!!F = 6pietavr!!

!!=> [MLT^(–2)] = eta[LT^(–1)][L]!!

!!eta=[ML^(–1)T^(–1)]!!

2.To check the dimensional correctness of a given relation.

*Example:* Find the correct relation !!F=(mv^2)/r!!

LHS = !![MLT^(–2)]!!

RHS = !![M(LT^(-1))^2/([L])]= [MLT^(–2)]!!

LHS = RHS

Hence dimensionally second relation is correct

3.In finding the dimensions of physical constants or coefficients.

4.As a research tool to derive new relation.

**ERRORS**

Absolute error = |experimental value – standard value|

If !!Deltax!! is the error in measurement of x, then

a. Fractional error = !!(delx)/x!!

b. Percentage error = Percentage error in experimental measurement x100%

*Some important points in errors :*

When two quantities are added or subtracted the absolute error in the result is the sum of the absolute error in the quantities.

When two quantities are multiplied or divided, the fractional error in the result is the sum of the fractional error in the quantities to be multiplied or to be divided.

If the same quantity x is multiplied together n times (i.e. !!x^n!!), then the fractional error in !!x^n!! is

n times the fractional error in x , i.e !!+-n(delx)/x!!

*Example:* In an experiment to determine acceleration due to gravity by simple pendulum, a student commit positive error in the measurement of length and 3% negative error in the measurement of time period. Calculate the percentage error in the value of g.

We know that !!T=ksqrt(l/g)!!

So !!g=k'l/T^2!!

!!=> (delg)/gxx100= (dell)/lxx100 +(2xxdelT)/Txx100!!

!!=> 1% +2xx3%=7%!!