Planck's constant h, speed of light c and gravitational constant G are used to form a unit of length L and a unit of mass M. Then the correct option(s) is(are)
- $M \propto \sqrt{c}$
- $M \propto \sqrt{G}$
- $L \propto \sqrt{h}$
- $L \propto \sqrt{G}$
The Correct Option is D
Solution and Explanation
Top Questions on Dimensional Analysis
Match List-I with List-II.
Choose the correct answer from the options given below :- JEE Main - 2025
- Physics
- Dimensional Analysis
A temperature difference can generate e.m.f. in some materials. Let $ S $ be the e.m.f. produced per unit temperature difference between the ends of a wire, $ \sigma $ the electrical conductivity and $ \kappa $ the thermal conductivity of the material of the wire. Taking $ M, L, T, I $ and $ K $ as dimensions of mass, length, time, current and temperature, respectively, the dimensional formula of the quantity $ Z = \frac{S^2 \sigma}{\kappa} $ is:
- JEE Advanced - 2025
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A quantity \( X \) is given by: \[ X = \frac{\epsilon_0 L \Delta V}{\Delta t} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( L \) is the length,
- \( \Delta V \) is the potential difference,
- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of:- WBJEE - 2025
- Physics
- Dimensional Analysis
- The dimensions of \( (\mu \epsilon)^{-1} \), where \( \epsilon \) is permittivity and \( \mu \) is permeability of a medium, are:
- CBSE CLASS XII - 2025
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- The dimensional formula of magnetic permeability \(\mu_0\) is:
- CBSE CLASS XII - 2025
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- Dimensional Analysis
Questions Asked in JEE Advanced exam
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
- JEE Advanced - 2025
- binomial expansion formula
- The total number of real solutions of the equation $$ \theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2} \sin^{-1} \left( \frac{6 \tan \theta}{9 + \tan^2 \theta} \right) $$ is
(Here, the inverse trigonometric functions $ \sin^{-1} x $ and $ \tan^{-1} x $ assume values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $(-\frac{\pi}{2}, \frac{\pi}{2})$, respectively.)- JEE Advanced - 2025
- Inverse Trigonometric Functions
- Let $$ \alpha = \frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \cdots + \frac{1}{\sin 118^\circ \sin 119^\circ}. $$ Then the value of $$ \left( \frac{\csc 1^\circ}{\alpha} \right)^2 $$ is \rule{1cm}{0.15mm}.
- JEE Advanced - 2025
- Some Applications of Trigonometry
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
- JEE Advanced - 2025
- Differential equations
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.