If Surface tension (S), Moment of Inertia (I) | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\begin{array}{l}
P = k\,{s^a}{i^b}{h^c}\
Where\,k\,is\,{m{dimensionless}}\,{m{constant}}\
ML{T^{ - 1}} = {\left( {M{T^{ - 2}}} ight)^a}\le

Vedclass · Accepted Answer

$S^{1 /2} I^{1 /2} h^0$

Vedclass · Answer

$\begin{array}{l}
P = k\,{s^a}{i^b}{h^c}\
Where\,k\,is\,{m{dimensionless}}\,{m{constant}}\
ML{T^{ - 1}} = {\left( {M{T^{ - 2}}} ight)^a}\left( {M{L^2}} ight){\left( {M{L^2}{T^{ - 1}}} ight)^c}\
a + b + c = 1\
2b + 2c =  1\
 - 2a - c =  - 1\
a = \frac{1}{2}\,\,\,\,\,\,\,b = \frac{1}{2}\,\,\,\,\,\,c = 0\
{S^{1/2}}{i^{1/2}}{h^0}
\end{array}$

If Surface tension $(S)$, Moment of Inertia $(I)$ and Planck’s constant $(h)$, were to be taken as the fundamental units, the dimensional formula for linear momentum would be

Similar Questions

The formula $X = 5YZ^2$, $X$ and $Z$ have dimensions of capacitance and magnetic field respectively. What are the dimensions of $Y$ in $SI$ units?

The speed of light $(c)$, gravitational constant $(G)$ and planck's constant $(h)$ are taken as fundamental units in a system. The dimensions of time in this new system should be

In the equation $\left[X+\frac{a}{Y^2}\right][Y-b]= R T, X$ is pressure, $Y$ is volume, $R$ is universal gas constant and $T$ is temperature. The physical quantity equivalent to the ratio $\frac{a}{b}$ is

If we use permittivity $ \varepsilon $, resistance $R$, gravitational constant $G$ and voltage $V$ as fundamental physical quantities, then

The equation of the stationary wave is
$y = 2A\,\,\sin \,\left( {\frac{{2\pi ct}}{\lambda }} \right)\,\cos \,\,\,\left( {\frac{{2\pi x}}{\lambda }} \right)$
Which statement is not true?