Question

Obtain dimension of hpg where, h = height, p= density and g = gravitational acceleration ​

Answer 1

Answer:

dimensions of hpg

'' of h = L ( meter)

dimensions of p = ML^-3 (mass/volume)

dimensions of g= LT^-2 (meterper second square)

so, dimensions of hpg= (L)(ML^-3)(LT^-2)

=ML^-1T^-2 ans.

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Answer 2

The dimension of hρg is [ML⁻¹T⁻²].

• The pressure due to a column of a liquid is called gauge pressure, and it is given by the formula,

       $P = h \rho g$

• 'h' is the height of the liquid column below its free surface and is thus, measured by length. So, the dimensions of h is [L].
• 'ρ' is the density of the fluid.
• Density of a substance is given by the formula,

        $d = \frac{m}{V}$

• Here, d is the density, m is the mass and V is the volume of the fluid.
• So, the dimension would be [ML⁻³].
• 'g' is the acceleration due to gravity and has units ms⁻². So, its dimension would be [LT⁻²].
• So, the dimension of hρg will be [ML⁻¹T⁻²].