Explain this statement clearly : “To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary : - atoms are very small objects
- a jet plane moves with great speed
- the mass of Jupiter is very large
- the air inside this room contains a large number of molecules
- a proton is much more massive than an electron
- the speed of sound is much smaller than the speed of light.
- atoms are very small objects
- a jet plane moves with great speed
- the mass of Jupiter is very large
- the air inside this room contains a large number of molecules
- a proton is much more massive than an electron
- the speed of sound is much smaller than the speed of light.
Solution and Explanation
The given statement is true because a dimensionless quantity may be large or small in comparison to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction.
a. An atom is a very small object in comparison to a soccer ball.
b. A jet plane moves with a speed greater than that of a bicycle.
c. Mass of Jupiter is very large as compared to the mass of a cricket ball.
d. The air inside this room contains a large number of molecules as compared to that present in a geometry box.
e. A proton is more massive than an electron.
f. Speed of sound is less than the speed of light.
Top Questions on Dimensional formulae and dimensional equations
- A famous relation in physics relates ‘moving mass’ \(\text m\) to the ‘rest mass’ \(\text m_0\) of a particle in terms of its speed v and the speed of light, \(\text c\) . (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant \(\text c\). He writes : \(\text m\) = \(\frac{\text m_0}{(1-v^2)^{\frac{1}{2}}}\) . Guess where to put the missing \(\text c\).
- CBSE Class XI
- Physics
- Dimensional formulae and dimensional equations
Questions Asked in CBSE Class XI exam
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Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
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\(AlCl_3+Cl^- \rightarrow AlCl_4^-\)- CBSE Class XI
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Concepts Used:
Dimensional Analysis
Dimensional Analysis is a process which helps verify any formula by the using the principle of homogeneity. Basically dimensions of each term of a dimensional equation on both sides should be the same.
Limitation of Dimensional Analysis: Dimensional analysis does not check for the correctness of value of constants in an equation.
Using Dimensional Analysis to check the correctness of the equation
Let us understand this with an example:
Suppose we don’t know the correct formula relation between speed, distance and time,
We don’t know whether
(i) Speed = Distance/Time is correct or
(ii) Speed =Time/Distance.
Now, we can use dimensional analysis to check whether this equation is correct or not.
By reducing both sides of the equation in its fundamental units form, we get
(i) [L][T]-¹ = [L] / [T] (Right)
(ii) [L][T]-¹ = [T] / [L] (Wrong)
From the above example it is evident that the dimensional formula establishes the correctness of an equation.