If momentum $[ P ]$, area $[ A ]$ and time $[ T ]$ are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is :
If momentum $[ P ]$, area $[ A ]$ and time $[ T ]$ are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is :
- [JEE MAIN 2022]
- A
$\left[ PA ^{-1} T ^{0}\right]$
- B
$\left[ PA T ^{-1}\right]$
- C
$\left[ PA ^{-1} T \right]$
- D
$\left[ PA ^{-1} T ^{-1}\right]$
Similar Questions
Which of the following is dimensionally correct
Which of the following is dimensionally correct
Consider the following equation of Bernouilli’s theorem. $P + \frac{1}{2}\rho {V^2} + \rho gh = K$ (constant)The dimensions of $K/P$ are same as that of which of the following
Consider the following equation of Bernouilli’s theorem. $P + \frac{1}{2}\rho {V^2} + \rho gh = K$ (constant)The dimensions of $K/P$ are same as that of which of the following
Which of the following equations is dimensionally incorrect?
Where $t=$ time, $h=$ height, $s=$ surface tension, $\theta=$ angle, $\rho=$ density, $a, r=$ radius, $g=$ acceleration due to gravity, ${v}=$ volume, ${p}=$ pressure, ${W}=$ work done, $\Gamma=$ torque, $\varepsilon=$ permittivity, ${E}=$ electric field, ${J}=$ current density, ${L}=$ length.
Which of the following equations is dimensionally incorrect?
Where $t=$ time, $h=$ height, $s=$ surface tension, $\theta=$ angle, $\rho=$ density, $a, r=$ radius, $g=$ acceleration due to gravity, ${v}=$ volume, ${p}=$ pressure, ${W}=$ work done, $\Gamma=$ torque, $\varepsilon=$ permittivity, ${E}=$ electric field, ${J}=$ current density, ${L}=$ length.
- [JEE MAIN 2021]
A physcial quantity $x$ depends on quantities $y$ and $z$ as follows: $x = Ay + B\tan Cz$, where $A,\,B$ and $C$ are constants. Which of the following do not have the same dimensions
A physcial quantity $x$ depends on quantities $y$ and $z$ as follows: $x = Ay + B\tan Cz$, where $A,\,B$ and $C$ are constants. Which of the following do not have the same dimensions
In the relation : $\frac{d y}{d x}=2 \omega \sin \left(\omega t+\phi_0\right)$ the dimensional formula for $\left(\omega t+\phi_0\right)$ is :
In the relation : $\frac{d y}{d x}=2 \omega \sin \left(\omega t+\phi_0\right)$ the dimensional formula for $\left(\omega t+\phi_0\right)$ is :