Question

check the correctness of the equation f = mv2 /r using dimensional analysis, where the symbols have their usual meaning

Answer 1

Answer:

Dimensionally correct.

Explanation:

Dimension formula of LHS:

=> [F]

=> [MLT-²]

Dimension formula of RHS:

=> [mv²/r]

=> [m][v]² / [r]

=> [M][LT-¹]² / [L]

=> [ML²T-²] / [L]

=> [MLT-²]

LHS = RHS = [MLT-²], this means the given equation f = mv^2 /r is dimensionally correct.

Answer 2

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 :

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

Where,

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

Given that ,

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

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

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

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

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

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.

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

As , We can see that ,

Here ,

• L.H.S = R.H.S

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