If force (F) , length (L) and time (T) are assumed to be fundamental units, then dimensional f

English

Gujarati

Hindi

1.Units, Dimensions and Measurement

medium

If force $(F)$, length $(L) $ and time $(T)$ are assumed to be fundamental units, then the dimensional formula of the mass will be

A

$F{L^{ - 1}}{T^2}$

B

$F{L^{ - 1}}{T^{ - 2}}$

C

$F{L^{ - 1}}{T^{ - 1}}$

D

$F{L^2}{T^2}$

Solution

(a) Let $m = K{F^a}{L^b}{T^c}$

Substituting the dimension of

$[F] = [ML{T^{ – 2}}],$ $[C] = [L]\;and\;[T] = [T]$

and comparing both sides, we get

$m = F{L^{ – 1}}{T^{ – 2}}$

Standard 11

Physics

Share

Similar Questions

The equation of state of a real gas is given by $\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}$, where $\mathrm{P}, \mathrm{V}$ and $\mathrm{T}$ are pressure. volume and temperature respectively and $R$ is the universal gas constant. The dimensions of $\frac{a}{b^2}$ is similar to that of :

hard

(JEE MAIN-2024)

Write the dimensions of $a/b$ in the relation $P = \frac{{a – {t^2}}}{{bx}}$ , where $P$ is pressure, $x$ is the distance and $t$ is the time

medium

According to Newton, the viscous force acting between liquid layers of area $A$ and velocity gradient $\Delta v/\Delta z$ is given by $F = – \eta A\frac{{\Delta v}}{{\Delta z}}$ where $\eta $ is constant called coefficient of viscosity. The dimension of $\eta $ are

medium

(AIPMT-1990)

Pressure exerted by a gas is found to be $50\, N/m^2$ , then the value of pressure in $CGS$ system is

easy

If velocity$(V)$, force$(F)$ and time$(T)$ are chosen as fundamental quantities then dimensions of energy are

hard