Question

1. Determine whether the following equations are dimensionally correct, if NOT, how can you make them dimensionally correct? * 12mv2=mgh

Answer 1

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies \frac{1}{2} {mv}^{2} = mgh \\ \\ \red{\underline \bold{To \: Prove :}} \\ \tt: \implies Dimensionally \: correct \: or \: not\:?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies \frac{1}{2} m {v}^{2} = mgh \\ \\ \tt \: \circ \:Mass = kg= M \\ \\ \tt \: \circ \: Length = m = L\\ \\ \tt \: \circ \: Time = s = T \\ \\ \bold{L.H.S} \\ \tt: \implies kg \times ( {ms}^{ - 1} )^{2} \\ \\ \tt: \implies kg \times {m}^{2} \times {s}^{ - 2} \\ \\ \tt: \implies M \times {L}^{2} \times {T}^{ - 2} \\ \\ \bold{R.H.S} \\ \tt: \implies kg \times {ms}^{ - 2} \times m \\ \\ \tt: \implies kg \times {m^{2} \times s}^{ - 2} \\ \\ \tt: \implies M \times L^{2} \times {T}^{ - 2} \\ \\ \green{\tt \therefore L.H.S = R.H.S} \\ \\ \green{ \huge{\boxed{ \tt Proved }}}$

Answer 2

We have to find that the dimensional formula of 1/2 mv² = mgh is correct or not.

In 1/2 mv².. 1/2 is constant, dimensional formula of m is M (in kg), v² is L²T-² (m/s)².

For mgh:

m is M (in kg), g is LT-² (m/s²) and h is L (m).

As per given condition

1/2 mv² = mgh

Substitute values from above,

m is M, v² is L²T-², g is LT-² and h is L.

So,

[ML²T-²] = [M][LT-²][L]

[ML²T-²] = [ML²T-²]

L.H.S. = R.H.S.

Hence, dimensionally the formula is correct.