Physics - Unit & Dimension Question with Solution | TestHub

PhysicsUnit & DimensionMiscellaneousMedium2 minQB

PhysicsMediumsingle choice

Taking force $F$ , length $L$ and time $T$ to be the fundamental quantities, find the dimensions of (a) density (b) pressure (c) momentum and (d) energy.

Options:

Answer:

Solution:

(a) $[$ density
$] = ([F])^{x} ([L])^{y} ([T])^{z} : [M L^{- 3}]$
$= ([M L T^{- 2}])^{x} ([L])^{y} ([T])^{z}$ gives
$x = 1, y = - 4, z = 2$
, so $[F L^{- 4} T^{2}]$ ;
the other parts follow similarly: (b) pressure
$= \frac{F}{L ^{2}} = [F L^{- 2}]$
, (c) momentum $= Ft = [FT]$
, (d) energy $= F \times L = [F L]$ .

Stream:JEESubject:PhysicsTopic:Unit & DimensionSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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