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If force $(F),$ velocity $(V)$ and time $(T)$ are taken as fundamental units, then the dimensions of mass are
$\left[ {FV{T^{ - 1}}} \right]$
$\;\left[ {FV{T^{ - 2}}} \right]$
$\;\left[ {F{V^{ - 1}}{T^{ - 1}}} \right]$
$\;\left[ {F{V^{ - 1}}T} \right]$
Solution
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. $(i),\,\left[ m \right] = \left[ {F{V^{ – 1}}T} \right]$
Similar Questions
Match the following two coloumns
| Column $-I$ | Column $-II$ |
| $(A)$ Electrical resistance | $(p)$ $M{L^3}{T^{ – 3}}{A^{ – 2}}$ |
| $(B)$ Electrical potential | $(q)$ $M{L^2}{T^{ – 3}}{A^{ – 2}}$ |
| $(C)$ Specific resistance | $(r)$ $M{L^2}{T^{ – 3}}{A^{ – 1}}$ |
| $(D)$ Specific conductance | $(s)$ None of these |
