Check the dimensional correctness of the relation:
$mgh = \dfrac{1}{2}m{v^2}$, where letters have their usual meanings.
Answer
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Hint: The dimensions of physical quantities are the powers to which the fundamental quantities are raised to represent that quantity. All physical quantities can be expressed as a combination of seven fundamental or base quantities. Here we have to find the dimensional correctness of the given relation.
Complete step by step solution:
Here we are given the expressions of kinetic energy and potential energy.
The expression for P.E is given as, $P.E = mgh$.
Let us first find the dimensional formula for potential energy for $mgh\;$
The mass,
$m = \left[ M \right]$
Acceleration can be written as,
$a = \dfrac{d}{{{t^2}}}$
where $d$ stands for the distance and $t$ stands for the time.
The dimensional formula for acceleration can be written as,
$a = \dfrac{L}{{{T^2}}} = L{T^{ - 2}}$
The dimensional formula for the acceleration due to gravity is, therefore,
$g = \left[ {L{T^{ - 2}}} \right]$
The dimensional formula for height
$h = \left[ L \right]$
The dimensional formula for $mgh\;$ can be written as,
$mgh = \left[ {M{L^2}{T^{ - 2}}} \right]$
The dimensional formula for $\dfrac{1}{2}m{v^2}$ can be written as,
$m = \left[ M \right]$
The formula for velocity
$v = \dfrac{d}{t}$
where $d$ stands for the displacement and $t$ stands for time.
The dimensional formula for velocity will be
$v = \dfrac{d}{t} = \dfrac{L}{T} = L{T^{ - 1}}$
Therefore,
${v^2} = {L^2}{T^{ - 2}}$
The dimensional formula for $\dfrac{1}{2}m{v^2}$ can be written as,
$\dfrac{1}{2}m{v^2} = \left[ {M{L^2}{T^{ - 2}}} \right]$
The dimensions of both $\dfrac{1}{2}m{v^2}$ and $mgh\;$ are the same.
Therefore we can say that the relation $mgh = \dfrac{1}{2}m{v^2}$ is dimensionally correct.
Additional Information:
There are seven fundamental quantities are represented as mass $\left[ M \right]$, length $\left[ L \right]$, time $\left[ T \right]$, electric current $\left[ A \right]$, Thermodynamic temperature $\left[ K \right]$, luminous intensity $\left[ {cd} \right]$, and amount of substance $\left[ {mol} \right]$.
Note:
An equation connecting the physical quantity with its dimensional formula is known as the dimensional equation of that physical quantity. The dimensions of the equivalent fundamental quantity must be equal on either side of a dimensional equation.
Complete step by step solution:
Here we are given the expressions of kinetic energy and potential energy.
The expression for P.E is given as, $P.E = mgh$.
Let us first find the dimensional formula for potential energy for $mgh\;$
The mass,
$m = \left[ M \right]$
Acceleration can be written as,
$a = \dfrac{d}{{{t^2}}}$
where $d$ stands for the distance and $t$ stands for the time.
The dimensional formula for acceleration can be written as,
$a = \dfrac{L}{{{T^2}}} = L{T^{ - 2}}$
The dimensional formula for the acceleration due to gravity is, therefore,
$g = \left[ {L{T^{ - 2}}} \right]$
The dimensional formula for height
$h = \left[ L \right]$
The dimensional formula for $mgh\;$ can be written as,
$mgh = \left[ {M{L^2}{T^{ - 2}}} \right]$
The dimensional formula for $\dfrac{1}{2}m{v^2}$ can be written as,
$m = \left[ M \right]$
The formula for velocity
$v = \dfrac{d}{t}$
where $d$ stands for the displacement and $t$ stands for time.
The dimensional formula for velocity will be
$v = \dfrac{d}{t} = \dfrac{L}{T} = L{T^{ - 1}}$
Therefore,
${v^2} = {L^2}{T^{ - 2}}$
The dimensional formula for $\dfrac{1}{2}m{v^2}$ can be written as,
$\dfrac{1}{2}m{v^2} = \left[ {M{L^2}{T^{ - 2}}} \right]$
The dimensions of both $\dfrac{1}{2}m{v^2}$ and $mgh\;$ are the same.
Therefore we can say that the relation $mgh = \dfrac{1}{2}m{v^2}$ is dimensionally correct.
Additional Information:
There are seven fundamental quantities are represented as mass $\left[ M \right]$, length $\left[ L \right]$, time $\left[ T \right]$, electric current $\left[ A \right]$, Thermodynamic temperature $\left[ K \right]$, luminous intensity $\left[ {cd} \right]$, and amount of substance $\left[ {mol} \right]$.
Note:
An equation connecting the physical quantity with its dimensional formula is known as the dimensional equation of that physical quantity. The dimensions of the equivalent fundamental quantity must be equal on either side of a dimensional equation.
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