Question

$\sf{k.e = \frac{1}{6} m {v}^{2} }$
$\mathfrak{is \: it } \: \sf{ \orange{dimensionally}} \: \mathfrak{correct}$
$\star \: full \: solution \: required \\ \star \: thanks \: for \: helping$
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Answer 1

Answer:

If an equation is dimensionally correct, it does not mean that the equation must be true. On the other hand, when the equation is dimensionally correct, the equation cannot be true. Dimensional analysis is a technique used to check whether a relationship is correct a

Answer 2

Answer:

Yes! The equation is dimensionally correct.

Explanation:

METHOD I

we're all very familiar with the expression for kinetic energy of an object in terms of mass and velocity of that object.

it's :-

K.E. = 1/ 2 × m × v²

• m => Mass
• v => velocity

note that 1/ 2 is a constant and will not be included while finding out the dimensions of both the sides.

[K.E] = [m v²]

so we're left with Fundamental quantities raised to some exponent that make up their dimensional equation.

(and yeah this being a standard equation is dimensionally correct)

now, if we replace 1/ 2 with 1/ 6 in the same equation

it's obvious that 1/ 6 being a constant term is left out during the dimensional consistency test and the remaining equation we get is same as that we got before

[K.E] = [m v²]

and since the above equation was dimensionally correct this has to be correct too. :}

METHOD II

You can also do the consistency test yourself and find out if it's dimensionally correct or not.

L.H.S.

[K.E.] = [E] = [ML²T⁻²]

R.H.S.

[m v²] = [M (L T⁻¹)²]

= [ML²T⁻²]

Since, the dimensions of LHS equal the dimensions of RHS, the equation is dimensionally correct.

________________________

This is one major drawback of dimensionally analysis

as one cannot draw the exact values of dimension-less constants.

Hence, the method of dimensions can only test the dimensionally validity but not the exact relationship between physical quantities in any equation.

The given question itself proves it.