If force (F), velocity (V) and time (T) are t | vedclass

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

Question

Let, mass, m,  $\propto \,{F^a}{V^b}{T^e}$
or,  m$ = k{F^a}{V^b}{T^c}$  ....................($i$)
Where k is a dimensionless constant

Vedclass · Accepted Answer

$\;\left[ {F{V^{ - 1}}T} \right]$

Vedclass · Answer

Let, mass, m,  $\propto \,{F^a}{V^b}{T^e}$
or,  m$ = k{F^a}{V^b}{T^c}$  ....................($i$)
Where k is a dimensionless constant and a, b and c are the exponents
Writing dimension on both sides we get

$\left[ {M{L^0}{T^0}} \right] = {\left[ {ML{T^{ - 2}}} \right]^a}{\left[ {L{T^{ - 1}}} \right]^b}{\left[ T \right]^c}$

$\left[ {M{L^0}{T^0}} \right] = \left[ {{M^a}{L^{a + b}}{T^{ - 2ab + c}}} \right]$
Applying the principle of homogeneity of

dimension we get

$a=1$     ..................($ii$)

$a + b = 0$         ........................($iii$)

$- 2a - b + c = 0$     ....................($iv$)
Solving eqns.,($ii$), ($iii$), and ($iv$), we get
$a = 1,b =  - 1,c = 1$
From eqn. $(i),\,\left[ m \right] = \left[ {F{V^{ - 1}}T} \right]$

If force $(F),$ velocity $(V)$ and time $(T)$ are taken as fundamental units, then the dimensions of mass are

Similar Questions

Given that $K =$ energy, $V =$ velocity, $T =$ time. If they are chosen as the fundamental units, then what is dimensional formula for surface tension?

The force of interaction between two atoms is given by $F\, = \,\alpha \beta \,\exp \,\left( { - \frac{{{x^2}}}{{\alpha kt}}} \right);$ where $x$ is the distance, $k$ is the Boltzmann constant and $T$ is temperature and $\alpha $ and $\beta $ are two constants. The dimension of $\beta $ is

Force $(F)$ and density $(d)$ are related as $F\, = \,\frac{\alpha }{{\beta \, + \,\sqrt d }}$ then dimension of $\alpha $ are