Dimensional analysis ? Write limitation of dimensional analysis.

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1.Units, Dimensions and Measurement

easy

What is dimensional analysis ? Write limitation of dimensional analysis.

Option A

Option B

Option C

Option D

Solution

$(1)$ In the dimensional equations containing $\mathrm{M}, \mathrm{L}$ and $\mathrm{T}$, we get at the best only three equations by equating the indices of $\mathrm{M}, \mathrm{L}$ and $\mathrm{T}$. Hence this method is of no avil in deducting the exact form of a physical relation which happens to depend upon more than three quantities.

$(2)$ The information about dimensionless constant cannot be obtained.

$(3)$ The equations containing exponential trigonometric functions lie quite outside the preview of this method.

$(4)$ This method is of no use if a constant of proportionally is not dimensionless.

Standard 11

Physics

If $\varepsilon_0$ is permittivity of free space, $e$ is charge of proton, $G$ is universal gravitational constant and $m_p$ is mass of a proton then the dimensional formula for $\frac{e^2}{4 \pi \varepsilon_0 G m_p{ }^2}$ is

medium

In the relation $y = a\cos (\omega t – kx)$, the dimensional formula for $k$ is

easy

Which of the following is dimensionally correct

medium

The entropy of any system is given by

${S}=\alpha^{2} \beta \ln \left[\frac{\mu {kR}}{J \beta^{2}}+3\right]$

Where $\alpha$ and $\beta$ are the constants. $\mu, J, K$ and $R$ are no. of moles, mechanical equivalent of heat, Boltzmann constant and gas constant repectively. [Take ${S}=\frac{{dQ}}{{T}}$ ]

Choose the incorrect option from the following:

hard

(JEE MAIN-2021)

The Bernoulli's equation is given by $p +\frac{1}{2} \rho v ^{2}+ h \rho g = k$

where $p =$ pressure, $\rho =$ density, $v =$ speed, $h =$ height of the liquid column, $g=$ acceleration due to gravity and $k$ is constant. The dimensional formula for $k$ is same as that for

medium

What is dimensional analysis ? Write limitation of dimensional analysis.

Solution

Similar Questions

If $\varepsilon_0$ is permittivity of free space, $e$ is charge of proton, $G$ is universal gravitational constant and $m_p$ is mass of a proton then the dimensional formula for $\frac{e^2}{4 \pi \varepsilon_0 G m_p{ }^2}$ is

In the relation $y = a\cos (\omega t – kx)$, the dimensional formula for $k$ is

Which of the following is dimensionally correct

The Bernoulli's equation is given by $p +\frac{1}{2} \rho v ^{2}+ h \rho g = k$ where $p =$ pressure, $\rho =$ density, $v =$ speed, $h =$ height of the liquid column, $g=$ acceleration due to gravity and $k$ is constant. The dimensional formula for $k$ is same as that for

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The Bernoulli's equation is given by $p +\frac{1}{2} \rho v ^{2}+ h \rho g = k$

where $p =$ pressure, $\rho =$ density, $v =$ speed, $h =$ height of the liquid column, $g=$ acceleration due to gravity and $k$ is constant. The dimensional formula for $k$ is same as that for