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)


ANSWER

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


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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