
If length L and force F and time T is taken as fundamental quantities. What would the dimensional equation of mass and density be?
Answer
233.1k+ views
Hint: Direct mass thickness has the measurement mass per length. The SI composite unit of direct mass thickness is the kilogram per meter. Altogether, there are seven essential measurements. Essential (a few times called essential) measurements are characterized as free or central measurements, from which different measurements can be gotten. The essential measurements are mass, length, time, temperature, electric flow, a measure of light, and a measure of issue.
Formula used:
mass formula:
mass $ = \rho v$
$\rho = $ density and
$v = $the volume
Density formula
$\rho = \dfrac{m}{v}$
$\rho = $density
$m = $mass
$v = $volume
Complete step by step solution:
Let the force be $F$ and $L$be the acceleration
mass
$L$$ = \dfrac{{force}}{{acceleration}}$
$ = \dfrac{F}{{L{T^{ - 2}}}}$
$ = F{L^{ - 1}}{T^2}$($\therefore $denominator are going to numerator)
Density
Let volume be the $L$
$ = \dfrac{{mass}}{{volume}}$($\therefore $mass$ = F{L^{ - 1}}{T^2}$)
$\dfrac{{ = F{L^{ - 1}}{T^2}}}{{{L^3}}}$
$ = F{L^{ - 4}}{T^2}$$\left( {\therefore {L^ - }^1 + {L^{ - 3}}} \right)$
Note: Essential amounts are those that are characterized straightforwardly by the cycle of estimation as it were. They are not characterized regarding different amounts; their units are not characterized as far as other units. The units of all other actual amounts, which can be gotten from principal units, are called inferred units.
Formula used:
mass formula:
mass $ = \rho v$
$\rho = $ density and
$v = $the volume
Density formula
$\rho = \dfrac{m}{v}$
$\rho = $density
$m = $mass
$v = $volume
Complete step by step solution:
Let the force be $F$ and $L$be the acceleration
mass
$L$$ = \dfrac{{force}}{{acceleration}}$
$ = \dfrac{F}{{L{T^{ - 2}}}}$
$ = F{L^{ - 1}}{T^2}$($\therefore $denominator are going to numerator)
Density
Let volume be the $L$
$ = \dfrac{{mass}}{{volume}}$($\therefore $mass$ = F{L^{ - 1}}{T^2}$)
$\dfrac{{ = F{L^{ - 1}}{T^2}}}{{{L^3}}}$
$ = F{L^{ - 4}}{T^2}$$\left( {\therefore {L^ - }^1 + {L^{ - 3}}} \right)$
Note: Essential amounts are those that are characterized straightforwardly by the cycle of estimation as it were. They are not characterized regarding different amounts; their units are not characterized as far as other units. The units of all other actual amounts, which can be gotten from principal units, are called inferred units.
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