Question

Let us consider of time period 1/2 mv2=mgh where m=mass of the body.v=is the velocity,g is the acceleration due to gravity and h is the height. check whether this equation is deminsionally correct or not
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Answer 1

Solution :-

Given equation ,

• 1/2 mv² = mgh

Here substituting 1/2 mv² = K.E ( Kinetic energy )

Now ,

→ K.E = mgh

Taking RHS and proving

Now substituting the dimensions ,

• m → [M]
• g → [LT-²]
• h → [L]

→ [M] [LT-²] [L]

→ [ML²T-²] = mgh

Now , K.E substituting [ML²T-²]

And comparing with LHS

• [ML²T-²] = [ML²T-²]

LHS = RHS

Hence , proved !

Answer 2

Solution: - For checking the dimensional consistency, the essential condition is: -

Every correct equation must have the identical dimensions on

both of the side of the equation. The dimensions of LHS are

$\begin{aligned}\left[\frac{1}{2} m v 2\right]=[\mathrm{M}][\mathrm{L} 2 \mathrm{~T}-2] \\&=\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-2}\right]\end{aligned}$

The dimensions of RHS are

$\begin{array}{l}{[m g h]=[M]\left[L T^{-2}\right][L]=[M]\left[L^{2} T^{-2}\right]} \\\\{[m g h]=\left[M L^{2} T^{-2}\right]}\end{array}$

The dimension of LHS and RHS are the identical. Hence, the given equation is dimensionally correct.