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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Units and Measurement
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The position $x$ of a particle at time $t$ is given by $x=\large\frac{v_0}{a} $$(1-e^{-at})$ where $v_0$ is constant and $a >0$ find the dimensions of $v_0$ and $a$

\[(a)\;M^0LT^{-1}\;and\;T^{-1} \quad (b)\;M^0LT^0\;and \;T^{-1}\quad (c)\;M^0LT^{-1}\;and\;LT^{-2} \quad (d)\;M^0LT^{-1}\;and\;T\]

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1 Answer

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$a \times t$ is dimensionless quantity
$a=\large\frac{1}{t}$$=[T^{-1}]$
also $x=\large\frac{v_0}{a}$ and $v_0=x \;a$
$v_0=LT^{-1}$
$\quad=[M^0LT^{-1}]$
Hence the dimensions of $v_0=M^0LT^{-1}\;and\; a=\;T^{-1}$
Hence a is the correct answer.
answered Jun 17, 2013 by meena.p
edited May 29, 2014 by lmohan717
 

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