Physics and Measurement Chapter - Physics JEE Main

This chapter deals with the study of the physical world that surrounds us and various physical quantities that help us understand this world. It also tells us about what measurement is in physics. Basically, the process of determining the length, size, or amount of a thing is known as measurement. People have been measuring length in various ways since prehistoric times. 

A physical quantity (such as length) must be measured in relation to some fixed quantity. A unit is a set quantity against which a physical quantity is measured and it serves as a unit of measurement. There are three types of physical quantities like fundamental, derived and supplementary quantities. 

This chapter tells us about that there are four fundamental forces in nature and the name of them are:

• Gravitational forces

• Electromagnetic forces 

• Strong nuclear forces

• Weak nuclear forces

This chapter also deals with the dimensional analysis, error in measurement and tells us about the extent to which a physical measurement quantity is measured accurately through the concept of significant figures.

Now, let's move onto the important concepts and formulae related to JEE and JEE main exams along with a few solved examples.

JEE Main Physics Chapters 2024

Important Topics of Units and Measurement Chapter

Units and Measurement Important Concept for JEE Main

List of Important formulas for Units and Measurement Chapter

Dimensional Formula

The dimensional formula of a physical quantity is like a secret code that tells us which fundamental units (like mass, length, and time) are used to measure that quantity.

How to Write a Dimensional Formula:

• Start with the formula of the quantity you're interested in. Put this formula on the left side.

• Take everything on the right side of the formula and express it using the basic units like mass, length, and time.

• Replace mass with 'M,' length with 'L,' and time with 'T.'

• Note the powers (like squared or cubed) for each of these terms.

Characteristics of Dimensions:

• Dimensions are the same no matter which system of units you use.

• If two things have the same dimensions, you can add or subtract them.

• You can figure out dimensions from the units of quantities, and vice versa.

• Different things can have the same dimension.

• When you multiply or divide dimensions, you get the dimension of a third quantity.

Dimensions Physics and Measurement

Dimensional Analysis - What you can do:

• Check if an equation is correct using dimensional formulas.

• Convert units of a quantity from one system to another.

• Figure out how different physical quantities are related to each other using their dimensional formulas.

Limitations of Dimensional Analysis in Simple Terms

• Proportionality Constant Mystery: Sometimes, when using dimensional analysis to find formulas, we can't figure out the exact number (proportionality constant) that relates the quantities.

• Complex Dependencies: If a physical quantity relies on more than three independent factors, we can't deduce its equation using dimensional analysis alone.

• Too Many Parameters: If a physical quantity depends on more things than the fundamental quantities (like length, time, and mass), then this method won't work.

• Tricky Functions: Equations with trigonometric or exponential functions can't be derived through dimensional analysis. It's not suitable for those situations.

Principle of Homogeneity

The Principle of Homogeneity in physics says that when you use units to measure different things in an equation, all the units must match.

Example: Let's take the equation: s = ut + (½)at^2, where s is distance, u is initial velocity, a is acceleration, and t is time.

If you look at the units for each term:

• s is measured in meters (L).

• ut is (m/s) * s = m/s * s = m, so it's also in meters (L).

• (½)at^2 is (m/s^2) * s^2 = m/s^2 * s^2 = m, so it's also in meters (L).

So, in this equation, all the units match up, which is what the Principle of Homogeneity tells us. It ensures that the equation makes sense in terms of units, which is essential in physics to avoid errors.

Physics and Measurement Points to Remember

• Physical quantities are characteristics that help us understand how the laws of physics work. For instance, things like length, mass, and time are physical quantities.

• We can group physical quantities into two categories: fundamental quantities and derived quantities. Fundamental quantities are the basic building blocks, while derived quantities are formed from combinations of fundamental quantities.

• To measure physical quantities, we use units as reference standards. These units are divided into two types: fundamental units (the basic units for measuring fundamental quantities) and derived units (units made from combinations of fundamental units).

• The most commonly used system of units is the International System of Units (SI). SI is based on seven fundamental units and two supplementary units.

• The dimensional formula of a physical quantity is a formula that tells us which fundamental units are involved in measuring that quantity. This formula ensures that the units used are consistent with the laws of physics.

• In essence, physical quantities are what we use to understand the rules of the physical world, and units help us measure them accurately. The dimensional formula of a quantity shows us how it's measured in terms of fundamental units, and this follows the principle of homogeneity, meaning that the units on both sides of the formula match up correctly.

JEE Main Physics and Measurement Solved Examples 

1. A student produces a positive error of 2% in the length of the pendulum and a negative error of 4% in the amount of time period while measuring the acceleration due to gravity with a pendulum. His percentage error in measurement g using the ratio $g=4\pi^2( \dfrac{l}{T^2})$ will be

1. 6 %

2. 8 %

3. 10%

4. 12% 

Sol:

Given that,

Percentage error in length, $\dfrac{\Delta l}{l}\times 100=2\%$

Percentage error in time period, $\dfrac{\Delta T}{T}\times 100=4\% $

The expression of acceleration due to gravity, $g=4\pi^2( \dfrac{l}{T^2})$

The percentage error in the measurement of g is given as,

$\dfrac{\Delta g}{g} \times 100= \dfrac{\Delta l}{l} \times 100 +2\times \dfrac{\Delta T}{T}\times 100$

After putting the values, we get

$\dfrac{\Delta g}{g}\times 100= 2\%+2\times 4\%=10\%$

Hence, the option c is correct.

Key point: The formula of percentage error and combinational error is important to solve this type of problem.

2. The potential energy of the particle varies with distance x as $U=\dfrac{Ax^{2/3}+B}{x^2}$, where A and B are constants. The dimensional formula for AB is 

1. $[M^2 L^{19/3} T^{-4}]$

2. $[M^2 L^{21/3} T^{-4}]$ 

3. $[M^2 L^{20/3} T^{-4}]$

4. $[M^2 L^{22/3} T^{-4}]$

Sol:

Given,

Potential energy, $U=\dfrac{Ax^{2/3}+B}{x^2}$

$U=$[M^1 L^2 T^{-2}]$

As $x$ is a distance hence its dimension is $[L]$.

Now using the principle of homogeneity of dimensional analysis, we can write;

$U=\dfrac{Ax^{2/3}}{x^2}$ and $U=\dfrac{B}{x^2}$

Now putting the dimensional formula of each quantity in the above equations we get:

$[M^1 L^2 T^{-2}]= A\dfrac{[L^{2/3}]}{[L^2]}=A [L^{-4/3}]$

We can also write,

$A=\dfrac{[M^1L^2 T^{-2}]}{[L^{-4/3}]}=[M^1 L^{10/3} T^{-2}]$

Similarly, 

$U=\dfrac{B}{x^2}$

After putting the dimensional formula of the quantities we get;

$[M^1 L^2 T^{-2}]=\dfrac{B}{[L^{2}]}$ 

We can also write,

$B=[M^1 L^2 T^{-2}][L^{2}]=[M^1 L^4 T^{-2}]$

As the dimensional formula of A and B is known, so in order to obtain

Dimensional formula of AB we can write;

$AB=[M^1 L^{10/3} T^{-2}][M^1 L^4 T^{-2}]=[M^2 L^{22/3} T^{-4}]$

Hence option d is correct.

Key point: Always use the principle of homogeneity of dimensional analysis to find the dimension of unknown quantity.

Previous Year Questions from JEE Paper

1. If E, L, M and G denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of P in the formula $P=EL^2M^{-5}G^{-2}$ are :(JEE Main 2018)

1. $[M^0 L^1 T^0]$

2. $[M^{-1} L^{-1} T^{2}]$

3. $[M^1 L^1 T^{-2}]$

4. $[M^0 L^0 T^0]$

Sol:

Given the formula,$P=EL^2M^{-5}G^{-2}$  

So in order to find the dimension of P, we have to use the dimensional formula of quantities E, L, M and G which are energy, angular momentum, mass and constant of gravitation.

Dimensional Formula of energy, $E=[M^1 L^2 T^{-2}]$

Dimensional Formula of angular momentum,$L=[M^1 L^2 T^{-1}]$

Dimensional Formula of mass,$M=[M^1]$

Dimensional Formula for constant of gravitation,$G=[M^{-1} L^3 T^{2}]$ 

After putting the dimensional formula of all the quantities in the given formula we can obtain the dimension of P as,

$P=[M^1 L^2 T^{-2}][M^2 L^4 T^{-2}][M^{-5}][M^2 L^{-6} T^{4}]$

After adding and subtracting the powers of M,L and T, we get;

$P=[M^0 L^0 T^0]$

Hence, option d is correct.

Trick: Remembering the dimensional formula for various standard quantities like force, energy, momentum, moment of inertia etc is important to solve such a problem.

2. Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then the error in the value of resistance of the wire is (JEE Main 2017)

1. 3 %

2. 6 %

3. Zero

4. 1 %

Sol:

Given that,

Percentage error in measurement of current, $\dfrac{\Delta I}{I}\times 100=3\%$

Percentage error in measurement of potential difference, $\dfrac{\Delta V}{V}\times 100=3\%$ 

We know that the formula of between potential difference and current is given as,

$V=IR$

And in the form of percentage error we can write the above relation as,

$\dfrac{\Delta V}{V}\times 100=\dfrac{\Delta I}{I}\times 100+\dfrac{\Delta R}{R}\times 100$

From the above relation the error in the measurement of resistance is given as,

$\dfrac{\Delta R}{R}\times 100=\dfrac{\Delta V}{V}\times 100+\dfrac{\Delta I}{I}\times 100$

After putting the values of known quantities we get;

$\dfrac{\Delta R}{R}\times 100= 3\%+3\%=6\%$

Hence error in the measurement of resistance is $6\%$. Therefore, option c is correct.

Trick: To solve such problems we need to have knowledge about the percentage error formula and combinational error.

Practice Questions

1. The density of a substance in the shape of a cube is found by measuring the cube's three sides and mass. Calculate the maximum error in predicting density if the relative errors in measuring mass and length are 1.5% and 1%, respectively. (Ans: 4.5%)

2. A copper wire is stretched to lengthen it by 0.5 percent. If its volume remains constant, the percentage change in its electrical resistance is ?(Ans: 1%)

JEE Main Physics Units and Measurement Study Materials

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JEE Main Physics Study and Practice Materials

Explore an array of resources in the JEE Main Physics Study and Practice Materials section. Our practice materials offer a wide variety of questions, comprehensive solutions, and a realistic test experience to elevate your preparation for the JEE Main exam. These tools are indispensable for self-assessment, boosting confidence, and refining problem-solving abilities, guaranteeing your readiness for the test. Explore the links below to enrich your Physics preparation.

Conclusion

The chapter on "Physics and Measurement" is the foundational bedrock upon which the edifice of scientific knowledge is built. In the context of JEE Main, it serves as a gateway to understanding the intricacies of the physical world, nurturing analytical skills, and fostering precision in measurement. This chapter imparts the fundamental principles of physics and lays the groundwork for solving complex problems. It also reinforces the significance of accurate measurements, an indispensable aspect of both scientific inquiry and daily life. Thus, "Physics and Measurement" not only sets the stage for JEE Main success but also cultivates an enduring appreciation for the importance of exactitude in the realm of science.