JEE Important Chapter - Units and Measurement

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 Chapter-wise Solutions 2022-23 

Important Topics of Units and Measurement

• Fundamental Forces in Nature

• Dimensional Analysis

• Accuracy, Precision and Error in Measurement

• Error Arithmetic Operations of Significant Figures

• Errors - Absolute Error, Relative Error,Percentage Error

Units and Measurement Important Concept for JEE

List of Important formulas for Units 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%)

Conclusion

In this article we have talked about the measurements, physical quantities and its various types, fundamental forces, various types of error in measurement and also get the answer about what is Units and measurement. Also provide the problem that helps us to understand the application of these concepts.

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