## Important Concept of Unit and Measurements for NEET

In this chapter, we will get an overall understanding of studying units and measurements. This chapter deals with the basics of physics. In this chapter, we will get to know about physical quantities and their units. Physical quantities are divided into two major categories. One is fundamental physical quantities and the second one is derived physical quantities. Similarly, units are also categorized as fundamental or standard units and derived units. This chapter deals with these physical quantities and their units. In this chapter, we will get to know how to define a unit for any physical quantity.

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The quantity measurement consists of two parts. The first part shows the frequency of the standard unit of measurement and the second part shows the name of the unit. This chapter describes the fundamental units of fundamental physical quantities such as length, time, mass, and temperature, and the derived quantities expressed in standard units of measurement. It may be a quantity. You can do a dimensional analysis of the size and then check if the formula of any size is correct. Suppose you measure something and the measurement causes an error. After learning this chapter, we can find the error in the measurement. We can learn how to use the basic quantity to find the dimension of the derived physical quantity.

Let's discuss the important concepts and formulae related to NEET physics units and measurement exams along with a few solved examples.

### Important Topics of Unit and Measurements

Physical Quantities

Fundamental and derived quantities

System of units

Practical Units

Dimensions of physical quantities

Dimensionless quantities

Applications of dimensional analysis

Significant figures

Errors of measurement

### Some Important Concept of Unit and Measurements for NEET

### List of Important Formulas of Unit and Measurements

### Solved Examples of Unit and Measurements

1. If P, Q, R are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity?

(a) $\dfrac{P-Q}{R}$

(b) $\dfrac{P \times Q}{R}$

(b) ${PQ}-{R}$

(b) $\dfrac{PR-Q}{R}$

Sol:

Given that,

P, Q, R are three different physical quantities.

To find : Invalid quantity among the options given.

Different physical quantities can not be added or subtracted. As if we add or subtract a physical quantity then units will create the conflict. So from the option we can choose the option which is having direct addition or subtraction of physical quantities.

In option (a), $P-Q$ which is not valid so option (a) is not valid.

In option (b), (c), and (d), it is possible to multiply and divide the quantities with each other. So it is not violating the rule of mathematical operation on physical quantities. So it is valid.

Hence, the correct answer is option (a).

Key Point: Validation of any derived unknown physical quantity can be done with the rules of mathematical operation on physical quantities. If it is violating the rule, it can’t be a valid physical quantity.

2. A physical quantity X is related to four measurable quantities a, b, c and d as follows: $X = a^2 b^4 c^{3/2} d^{–2}$. The percentage error in the measurement of a, b, c and d are 1%, 2%, 6% and 4%, respectively. What is the percentage error in quantity X?

Sol:

Given that,

$X = a^2 b^4 c^{3/2} d^{–2}$

Percentage error in $a = 1\%$

Percentage error in $b = 2\%$

Percentage error in $c = 6\%$

Percentage error in $d = 4\%$

To find : Percentage error in $X$

Percentage error in $X$ is given by-

$X = a^2 b^4 c^{3/2} d^{–2}$

$X = \dfrac{\Delta X}{X} \times 100$

$\dfrac{\Delta X}{X} \times 100 = 2(\dfrac{\Delta a}{a} \times 100) + 4(\dfrac{\Delta b}{b} \times 100) + \dfrac{3}{2}(\dfrac{\Delta c}{c} \times 100) + {2}(\dfrac{\Delta d}{d} \times 100) $

Here, $\dfrac{\Delta a}{a} \times 100 = 1\%$

, $\dfrac{\Delta b}{b} \times 100 = 2\%$

, $\dfrac{\Delta c}{c} \times 100 = 6\%$

, $\dfrac{\Delta d}{d} \times 100 = 4\%$

By putting all the value together in the above equation-

$\dfrac{\Delta X}{X} \times 100 = 2(1) + 4(2) + \dfrac{3}{2}(6) + {2}(4) $

$\dfrac{\Delta X}{X} \times 100 = 2 + 8 + 9 + 8 $

$\dfrac{\Delta X}{X} \times 100 = 27 \%$

Hence, the percentage error in $X$ is $27 \%$.

Key Point: Percentage error in any quantity can be calculated by using the standard formula of percentage error.

### Previous Year Questions of Unit and Measurements from NEET Paper

1. The unit of thermal conductivity is ($\text{NEET}-2019$)

a. $J~{m^{-1}}{K^{-1}}$

b. $W~m~{K^{-1}}$

c. $W~{m^{-1}}{K^{-1}}$

d. $J~m~{K^{-1}}$

Sol:

To Find: Unit of thermal conductivity

Formula Used: $\dfrac{\text{d}Q}{\text{d}t}=KA\dfrac{\Delta{T}}{L}~~J/s~\text{(or Watt)}$

To get the unit of thermal conductivity we have to go with the definition of thermal conductivity which gives us the formula in which thermal conductivity is present and using that we can get the formula for thermal conductivity.

We know that, the rate of heat flow through a conductor or length $L$ and area of cross-section $A$ is given by -

$\dfrac{\text{d}Q}{\text{d}t}=KA\dfrac{\Delta{T}}{L}~~J/s~\text{(or Watt)}$

Where, $K$ = Coefficient of thermal conductivity,

$\Delta{T}=$ Change in temperature

By rearranging the equation to get the formula for thermal conductivity, we get-

$\Rightarrow K={\dfrac{L}{A\Delta{T}}}~{\dfrac{\text{d}Q}{\text{d}t}}$

Units of quantities which are on the right side of the equation are-

$L$ = Length = $meter$= $m$,

$A$ = Area = $meter^{2}$ = $m^{2}$,

$\Delta{T}$ = Temperature Difference = $Kelvin$ = $K$,

${\dfrac{\text{d}Q}{\text{d}t}}$ = Rate of heat flow= $Js^{-1}~\text{or Watt}$

Putting all these units together in above equation-

$\Rightarrow \text{Unit of K}={\dfrac{meter}{meter^2\times {Kelvin}}}~{Watt}$

$\Rightarrow \text{Unit of K}={W~{m^{-1}}~{K^{-1}}}$

So, option (C) is the correct answer.

Key Point: To get the unit of thermal conductivity we must know the formula of rate of heat flow where thermal conductivity comes in the role. By using that formula and with a little rearrangement in it we can get the unit of thermal conductivity.

2. The unit of permitivitty of free space, $\varepsilon{_o}$ is ($\text{PMT}-2003$)

a. $\text{coulomb}/\text{newton}-\text{meter}$

b. $\text{newton}-\text{meter}^2/\text{coulomb}^2$

c. $\text{coulomb}^2/\text{newton}-\text{meter}^2$

d. $\text{coulomb}^2/{(\text{newton}-\text{meter})}^2$

Sol:

To Find: Unit of permitivitty of free space

Formula Used: $F=\dfrac{1}{4\pi\varepsilon{_o}}\times\dfrac{{q_1}{q_2}}{r^2}$

To get the unit of permitivitty of free space we have to use Coulomb's law of electrostatic force. In Coulomb’s law, when there is no medium between two charges it used permitivitty of free space for electrostatic force calculations.

By Coulomb’s law of electrostatic force-

$\Rightarrow F=\dfrac{1}{4\pi\varepsilon{_o}}\times\dfrac{{q_1}{q_2}}{r^2}$

Here, ${q_1},{q_2}$ = charges,

$r$ = distance between two charges ${q_1}\text{ and }{q_2}$,

$\varepsilon{_o}=$ permitivitty of free space.

After rearring the formula to get the permitivitty of free space, we get-

$\Rightarrow \varepsilon{_o}=\dfrac{1}{4\pi}\times\dfrac{{q_1}{q_2}}{{r^2}F}$

Here, units of

${q_1}$ and ${q_2}$ (Charges) = columb ($C$),

$r$ (distance)= meter ($m$),

and $F$ (Force)= newton ($N$)

If we put all these units together in the above equation, we get-

$\Rightarrow \text{Unit of }\varepsilon{_o}=\dfrac{{C}\times{C}}{{m^2}N}$

$\Rightarrow \text{Unit of }\varepsilon{_o}=\dfrac{C^2}{N{m^2}}$

$\Rightarrow \text{Unit of }\varepsilon{_o}=\dfrac{\text{columb}^2}{\text{newton}-{\text{meter}^2}}$

Hence the correct option is (C).

Key Point: To solve this question and get the unit or permitivitty of free space we must have the knowledge about the Coulomb’s electrostatic force.

3. A screw gauge gives the following readings when used to measure the diameter of a wire.

Main scale reading: 0mm

Circular scale reading: 52mm

Given that, 1mm on the main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is $\text{NEET}-2021$

a. $0.52$ cm

b. $0.026$ cm

c. $0.26$ cm

d. $0.052$ cm

Sol:

Given that,

Main scale reading = MSR = 0mm

Circular scale reading = CSR = 52mm

To Find: The diameter of the wire

Formula Used: $D = MSR + (CSR \times LC)$

Here, D = Diameter of the wire,

MSR = Main Scale Reading

CSR = Circular Scale Reading

LC = Least Count

First of all we have to calculate the Least Count of the scale. Least count of screw gauge is calculated as

$\Rightarrow LC = \dfrac{p}{n}$

$\Rightarrow LC = \dfrac{1}{100}\text{mm}$

$\Rightarrow LC = 0.01\text{mm}$

$\Rightarrow LC = 0.001\text{cm}$

Now using the formula we can put all the values in the above formula to get the diameter of the wire.

$\Rightarrow D= 0 + (52\text{mm} \times 0.01\text{mm})$

$\Rightarrow D= 0.52\text{mm}$

$\Rightarrow D= 0.052\text{cm}$

Hence, the option (d) is correct.

Key Point: In this type of question we need the Least count of the measurement device which is defined as the smallest value which can be measured through that device. After calculating the Least count we can use the direct formula to get the answer.

### Practice Questions

1. If the L, Q, R represents inductance, charge and resistance respectively, the unit of-

(a) $\dfrac{QR}{L}$ will be those of current

(b) $\dfrac{{Q^2}{R^3}}{L^2}$ will be those of energy

(c) $\dfrac{QL}{R}$ will be those of current

(d) $\dfrac{{Q^3}{R^2}}{L}$ will be those of power

(Ans. option a)

2. A calorie is a unit of heat (Energy in transit) and it equals around $4.2~J$ where $1~J = 1 \text{Kg}~\text{m}^2\text{s}^{-2}$. Suppose we employ a system of units in which the unit of mass equals $\alpha~\text{Kg}$, the length equals $\beta~\text{m}$, the unit of time is $\gamma~\text{s}$. Then what will be the magnitude of calories in terms of new units?

(Ans. $4.2~{\alpha}^{-1}~{\beta}^{-2}~{\gamma}^{2}$)

### Conclusion

In this chapter we discussed units and measurements. We get to know that in measurement every unit which is not fundamental, can be represented as a fundamental unit. In a measurement device, the least count is the ability of the device to measure the smallest possible value. We also discussed errors in measurement. We went through some formula based problems which are very useful while we are solving the questions in the NEET exam.

NEET Important Chapter - Unit and Measurements

## FAQs on NEET Important Chapter - Unit and Measurements

1. What is the weightage of Unit and Measurement in the NEET exam?

In the NEET exam, there are 45 questions asked in physics. Out of 45, there is 1 question on average that comes under chapter Units and Measurements, which is nearly 2% of the whole physics section.

2. How can someone secure full marks for questions from Units and Measurement in the NEET exam?

Unit and Measurement is a very simple and fundamental chapter of physics. It is the basics of physics, as in whole physics we are going to use units and dimensions. To use these we first need to know what it is and how it is derived, which comes under the chapter Units and Measurement. If we practice Previous Year Question Paper and reference books then anyone can easily get full marks in the exam.

3. What is the difficulty level of questions in the NEET Exam from chapter Units and Measurement?

As far as NEET is concerned, the difficulty level is medium for chapter Units and Measurement. According to previous year's question papers, most of the time questions are asked from derived units or from errors in measurement. If someone goes through the Previous year's question paper then it can be helpful to understand the level of difficulty of questions that come under this chapter.