Physics, asked by hindilearn8545, 1 year ago

Check correct ness of equation f=mv2\r dimensionally

Answers

Answered by kvvijin77
4

[f]=[MLT^-2]


[MV^2/r]=[M][LT^-1]^2/[L]


=[ML^2T^-2]/[L]

= [MLT^-2]

Therefore LHS = RHS

Answered by llsmilingsceretll
13

Given that , F ( or Force ) = m v² / r .

Exigency To Check : It is Correct or not using dimensional analysis ?

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¤ Dimensional Formula of Force :

\begin{gathered}\qquad \star\:\:\underline {\boxed {\pmb{\sf{ F \: ( \:or \: Force \:)\:=\:\:\bigg\lgroup \sf{ M^1 \:L^1 \; T^{-2} }\bigg\rgroup}}}}\\\\\end{gathered}

Where,

  • M is Mass ,
  • L is Length &
  • T is Time.

Given that ,

\begin{gathered}\qquad \dashrightarrow \sf F \: ( \:or \: Force \:)\:=\:\:\bigg\lgroup \sf{ \dfrac{m v^2}{r} }\bigg\rgroup\\\\\end{gathered}

⠀⠀⠀⠀⠀⠀\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\end{gathered}

\begin{gathered}\qquad \dashrightarrow \sf F \: ( \:or \: Force \:)\:=\:\:\bigg\lgroup \sf{ \dfrac{m v^2}{r} }\bigg\rgroup\\\\\end{gathered}

\begin{gathered}\qquad \dashrightarrow \sf M^1 \:L^1 \; T^{-2}\:=\:\:\bigg\lgroup \sf{ \dfrac{m v^2}{r} }\bigg\rgroup\\\\\end{gathered}

\begin{gathered}\qquad \dashrightarrow \sf M^1 \:L^1 \; T^{-2}\:=\:\: \dfrac{m v^2}{r} \\\\\end{gathered}

Using Dimensional Analysis :

  • m is Mass which is M in Dimensional,
  • r is measurement which is L in Dimensional &
  • v is velocity or speed which is L¹T-¹ in Dimensional.

\begin{gathered}\qquad \dashrightarrow \sf M^1 \:L^1 \; T^{-2}\:=\:\: \dfrac{m v^2}{r} \\\\\qquad \dashrightarrow \sf M^1 \:L^1 \; T^{-2}\:=\:\: \dfrac{M ( L^1 T^{-1})^2}{L} \\\\\qquad \dashrightarrow \sf M^1 \:L^1 \; T^{-2}\:=\:\: \dfrac{M ( L^2 T^{-2})}{L} \\\\\qquad \dashrightarrow \sf M^1 \:L^1 \; T^{-2}\:=\:\: M ( L^1 T^{-2}) \\\\\qquad \dashrightarrow \sf M^1 \:L^1 \; T^{-2}\:=\:\: M L^1 T^{-2} \\\\ \qquad \dashrightarrow \underline { \boxed { \pmb { \sf{ M^1 \:L^1 \; T^{-2}\:=\:\: M L^1 T^{-2} \:}}}}\\\\\end{gathered}

As , We can see that ,

Here ,

L.H.S = R.H.S

\begin{gathered}\qquad \therefore \:\underline {\sf Hence, \: It \: is \:\pmb{\bf Correct \;} \: using \: Dimensional \: analysis \:.}\\\end{gathered}

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