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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Units and Measurement
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The following equation describe the motion of object where $x$ is displacement, $v,v_0$ are velocities, $m$ mass $V=Axe^{Bm}-\large\frac{c^2}{v^0}$. What must be the unit of $A$ and $B$?

\[\begin{array}{1 1}(a)\;A\;is\;\frac{1}{second}\;B\;is \;meter \\(b)\;A\;is\;\frac{1}{second}\;B\;is \;\frac{1}{kilogram} \\ (c)\;A\;is\;second\;B\;is \;meter \\ (d)\;A\;is\;second \;B\;is \frac{1}{kilogram}\end{array}\]

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1 Answer

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$meter/second=A(meters) e^{B(\large kilogram)}+c^2/(meter/second)$
$\qquad\qquad\qquad=A(meters)e^{(\large unitless)}-(meters)/second$
Therefore $A=\large\frac{1}{second};$$\quad B=\large\frac{1}{kilogram};$$\quad C=\large\frac{meter}{second}$
Hence B is the correct answer. 


answered Jun 18, 2013 by meena.p
edited Jan 9, 2014 by meena.p

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