If mass is written as mathrm{m}=mathrm{kc}^{m | vedclass

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

Question

$\mathrm{m}=\mathrm{kc}^{\mathrm{P}} \mathrm{G}^{-1 / 2} \mathrm{~h}^{1 / 2}$
$\mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0=\left[\mathrm{LT}^{-1}\right]

Vedclass · Accepted Answer

$1 / 2$

Vedclass · Answer

$\mathrm{m}=\mathrm{kc}^{\mathrm{P}} \mathrm{G}^{-1 / 2} \mathrm{~h}^{1 / 2}$
$\mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0=\left[\mathrm{LT}^{-1}ight]^{\mathrm{P}}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}ight]^{-1 / 2}\left[\mathrm{ML}^2 \mathrm{~T}^{-1}ight]^{1 / 2}$
$	ext { By comparing } P=1 / 2$

If mass is written as $\mathrm{m}=\mathrm{kc}^{\mathrm{p}} \mathrm{G}^{-1 / 2} \mathrm{~h}^{1 / 2}$ then the value of $P$ will be : (Constants have their usual meaning with $\mathrm{k}$ a dimensionless constant)

Similar Questions

The foundations of dimensional analysis were laid down by

Young's modulus of elasticity $Y$ is expressed in terms of three derived quantities, namely, the gravitational constant $G$, Planck's constant $h$ and the speed of light $c$, as $Y=c^\alpha h^\beta G^\gamma$. Which of the following is the correct option?

The time dependence of a physical quantity $P$ is given by $ P = P_0 exp^{(-\alpha t^{2})} $ where $\alpha$ is a constant and $t$ is time. The constant $\alpha$

A force defined by $F=\alpha t^2+\beta t$ acts on a particle at a given time $t$. The factor which is dimensionless, if $\alpha$ and $\beta$ are constants, is:

Let $[ {\varepsilon _0} ]$ denote the dimensional formula of the permittivity of vacuum. If $M =$ mass, $L=$ length, $T =$ time and $A=$ electric current, then: