Physics

The dimensions of '$k$' in the relation $V = kavt$ is:(Where '$V$' is the volume of a liquid passing through any point in time $t$, '$a$' is area of cross section, '$v$' is the velocity of the liquid)

${ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }$

SOLUTION

$V=k a v t$

$Let\ k = M^{\alpha }L^{\beta }T^{\gamma }$

$M^{0}L^{3}T^{0}=(M^{\alpha }L^{\beta }T^{ \gamma })(L^{2})(LT^{-1})(T)$

$M^{0}L^{3}T^{0}=M^{\alpha }L^{\beta+2+1 }T^{ \gamma+0}$

$\gamma =0$

$\alpha =0$

$\beta +3=3$ $\Rightarrow \beta =0$

Hence, $k=M^{0}L^{0}T^{O}$

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Single Correct Medium Published on 18th 08, 2020
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