In the following dimensionally consistent equation F- X density -Y- where F is force- then dimensional formula for X and Y will be respectively equal to

The correct option is B [ M ^ 2 L ^ 2 T ^ 2 ] :[ M ^ L ^ T ^ 2 ] Given, F= X density +Y In order to maintain dimensional homogeneity X=fo ...

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Standard XII

Physics

Dimensional Analysis

In the follow...

Question

In the following dimensionally consistent equation $F = \frac{X}{d e n s i t y} + Y$ , where $F$ is force, then dimensional formula for $X$ and $Y$ will be respectively equal to

[M L^{- 2} T^{- 2}] : [M L T^{- 2}]

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[M^{2} L^{- 2} T^{2}] : [M L T^{- 2}]

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[M L T^{- 2}] : [M L^{2} T^{- 2}]

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None of these

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Solution

The correct option is B $[M^{2} L^{- 2} T^{2}] : [M L T^{- 2}]$
Given, $F = \frac{X}{d e n s i t y} + Y$
In order to maintain dimensional homogeneity
$X = f o r c e \times d e n s i t y$
$[M L T^{- 2}] \times [M L^{- 3}] = [M^{2} L^{- 2} T^{- 2}]$
also, $Y$ must have dimensions of $F$
$∴ Y = [M L T^{- 2}]$

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In the following dimensionally consistent equation F=Xdensity+Y, where F is force, then dimensional formula for X and Y will be respectively equal to

The correct option is B [M2L−2T2]:[MLT−2]Given, F=Xdensity+YIn order to maintain dimensional homogeneityX=force×density[MLT−2]×[ML−3]=[M2L−2T−2]also, Y must have dimensions of F∴Y=[MLT−2]

In the following dimensionally consistent equation $F = \frac{X}{d e n s i t y} + Y$ , where $F$ is force, then dimensional formula for $X$ and $Y$ will be respectively equal to