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# If force, acceleration and time are taken as fundamental qualities, then the dimension of length will be

$(a)\;FT^{2}\quad (b)\;F^{-1}A^2T^{-1}\quad (c)\;FA^2T\quad (d)\;AT^2$

Can you answer this question?

$L=F^{\large x}A^{\large y} t^{\large z}$
$M^0L^1T^0=[MLT^{-2}]^{\large x}\;[LT^{-2}]^ {\large y}\;T^ {\large z}$
$x=0\;y=1\;z=2$
$L=AT^{2}$
The dimension of length will be $L=AT^2$
Hence D is the correct answer.

answered Jun 15, 2013 by
edited Jan 9, 2014 by meena.p