If time $(t)$, velocity $(v)$, and angular momentum $(l)$ are taken as the fundamental units. Then the dimension of mass (m) in terms of $t, v$ and $l$ is :
Correct Option: , 4
$\mathrm{m} \propto \mathrm{t}^{\mathrm{a}} \mathrm{v}^{\mathrm{b}} \ell^{\mathrm{c}}$
$\mathrm{m} \propto[\mathrm{T}]^{\mathrm{a}}\left[\mathrm{LT}^{-1}\right]^{\mathrm{b}}\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]^{\mathrm{c}}$
$\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{0}=\mathrm{M}^{\mathrm{c}} \mathrm{L}^{\mathrm{b}+2 \mathrm{c}} \mathrm{T}^{\mathrm{a}-\mathrm{b}-\mathrm{c}}$
comparing powers
$\mathrm{c}=1, \mathrm{~b}=-2, \mathrm{a}=-1$
$m \propto t^{-1} v^{-2} \ell^{1}$
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