Question

Consider the following equation of Bernoulli’s theorem.
\[P + \dfrac{1}{2}\rho {V^2} + \rho gh = K\text{(constant)}\]
 The dimensions of \[\dfrac{K}{P}\] are the same as that of which of the following
A. Thrust
B. Pressure
C. Angle
D. Viscosity

Answer 1

Hint: Here, we need to recall the principle of homogeneity of dimensions for an equation and also the condition for the addition of dimensions of two physical quantities. In this question determine the dimensions of constant K and check which of the given physical quantities in the options have the dimensions same as that of K.

Formula Used:
Bernoulli’s equation is,
\[P + \dfrac{1}{2}\rho {V^2} + \rho gh = K\]
Where,
P is the pressure
g is acceleration due to gravity
a is acceleration
h is height
V is volume
\[\rho \] is density

Complete step by step solution:
We have the Bernoulli’s equation,
\[P + \dfrac{1}{2}\rho {V^2} + \rho gh = K\]
According to the homogeneity principle, the dimensions on the right-hand side and left-hand side of an equation should have the same dimensions. Hence, each term on the left-hand side and the term on the right-hand side of Bernoulli's equation should have the same dimensions. So, we can say that the dimension of K is the same as that of the pressure P.

The dimensions of pressure are, \[\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]\].
The dimension of K is also\[\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]\].
Then from the above dimensional formula, we can say that the dimensions of K are equal to the dimensions of P. Therefore \[\dfrac{K}{P}\] is dimensionless which can also be written as \[\left[ {{M^0}{L^0}{T^0}} \right]\].

If we consider the angle, then,
\[\theta = \dfrac{s}{r}\]
Since the arc length dimension is \[\left[ {{M^0}L{T^0}} \right]\]and the distance r dimension is also \[\left[ {{M^0}L{T^0}} \right]\].
Then, the dimension of angle is \[\left[ {{M^0}{L^0}{T^0}} \right]\].
Therefore, the dimensions of \[\dfrac{K}{P}\] are the same as that of the dimension of the angle.

Hence, option C is the correct answer.

Note: In the given solution, we have taken the dimensions of the constant K equal to the dimensions of pressure, P. But we can also determine the dimensions of K by calculating the dimensions of any of the terms kinetic energy or potential energy in Bernoulli's equation since they all have the same dimensions.