Units & Dimensions

JEE Main level, single correct, medium

88 of 149

Question 88 of 149

If momentum [P]\left[P\right], area [A]\left[A\right] and time [T]\left[T\right] are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is

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Answer: (A) [PA−1T0]\left[{PA}^{-1}{T}^{0}\right]

We know, Viscosity==Pascal‐‐second
Let [η]=\left[\eta \right]=PxAyTz=[M1L−1T−1]{P}^{x}{A}^{y}{T}^{z}=\left[{M}^{1}{L}^{-1}{T}^{-1}\right]
⇒[M1L+1T−1]x[L2]y[T1]z=M1L−1T−1\Rightarrow {\left[{M}^{1}{L}^{+1}{T}^{-1}\right]}^{x}{\left[{L}^{2}\right]}^{y}{\left[{T}^{1}\right]}^{z}={M}^{1}{L}^{-1}{T}^{-1}
⇒MxL+x+2yT−x+z=M1L−1T−1\Rightarrow {M}^{x}{L}^{+x+2y}{T}^{-x+z}={M}^{1}{L}^{-1}{T}^{-1}
From above, we have
x=1x+2y=−1−x+z=−1x=1 x+2y=-1 -x+z=-1
Solving above three equations, we get
y=−1y=-1 and z=0z=0
So, Viscosity =[P1A−1T0]=\left[{P}^{1}{A}^{-1}{T}^{0}\right]

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