
- If density \[\rho \] , acceleration due to gravity g and frequency f, are the basic quantities, find the dimensions of force
Answer
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Hint: This is a problem of dimensional analysis. we can deduce the value from the dimensional formula of these individual elements by finding the degree of dependence of a physical quantity on another. The principle of consistency of two expressions can be used to find the equation relating these two quantities.
Complete step by step answer:
Let us write the dimensional formula of the three quantities first.
Acceleration is defined as the rate of change of velocity and is given by the formula, \[a=\dfrac{v}{t}\], where v is the velocity and t are the time. Dimensions of g is \[[L{{T}^{-2}}]\]
Dimensions of\[\rho \] is \[[M{{L}^{-3}}]\] and the dimension of frequency is \[[{{T}^{-1}}]\]
Now, the dependency can be written as \[F \propto {{\rho }^{A}}{{g}^{B}}{{f}^{C}}\]
On removing the proportionality sign, we multiply by a constant, k.
\[F=k{{\rho }^{A}}{{g}^{B}}{{f}^{C}}\]------(1)
Now, equating the dimensions of the LHS and RHS, we get,
\[F=k{{[L{{T}^{-2}}]}^{A}}{{[M{{L}^{-3}}]}^{B}}{{[{{T}^{-1}}]}^{C}}\]
$\implies$ \[[ML{{T}^{-2}}]=k{{[L{{T}^{-2}}]}^{A}}{{[M{{L}^{-3}}]}^{B}}{{[{{T}^{-1}}]}^{C}}\]
On comparing we get,
A=1
-3A+B=1 and -2B-C=2
On solving we get, A=1, B=4, C=-6
Thus, \[F=k{{\rho }^{1}}{{g}^{4}}{{f}^{-6}}\]
Additional Information:
Dimensional analysis is also called Factor Label Method. Using the dimensional analysis, we can check the correctness of Physical Equation. However, it should be kept in mind that dimensional analysis cannot help us determine any dimensionless constants in the equation. This method also uses chemistry and there in place of using designated symbols we use units.
Note:
Though this method is widely used, there are few drawbacks for this method. Dimensionless quantities cannot be determined by this method. Constant proportionality cannot be determined by this method. It does not apply to trigonometric, logarithmic and exponential functions physical quantities which are dependent upon more than three physical quantities, this method will be difficult.
Complete step by step answer:
Let us write the dimensional formula of the three quantities first.
Acceleration is defined as the rate of change of velocity and is given by the formula, \[a=\dfrac{v}{t}\], where v is the velocity and t are the time. Dimensions of g is \[[L{{T}^{-2}}]\]
Dimensions of\[\rho \] is \[[M{{L}^{-3}}]\] and the dimension of frequency is \[[{{T}^{-1}}]\]
Now, the dependency can be written as \[F \propto {{\rho }^{A}}{{g}^{B}}{{f}^{C}}\]
On removing the proportionality sign, we multiply by a constant, k.
\[F=k{{\rho }^{A}}{{g}^{B}}{{f}^{C}}\]------(1)
Now, equating the dimensions of the LHS and RHS, we get,
\[F=k{{[L{{T}^{-2}}]}^{A}}{{[M{{L}^{-3}}]}^{B}}{{[{{T}^{-1}}]}^{C}}\]
$\implies$ \[[ML{{T}^{-2}}]=k{{[L{{T}^{-2}}]}^{A}}{{[M{{L}^{-3}}]}^{B}}{{[{{T}^{-1}}]}^{C}}\]
On comparing we get,
A=1
-3A+B=1 and -2B-C=2
On solving we get, A=1, B=4, C=-6
Thus, \[F=k{{\rho }^{1}}{{g}^{4}}{{f}^{-6}}\]
Additional Information:
Dimensional analysis is also called Factor Label Method. Using the dimensional analysis, we can check the correctness of Physical Equation. However, it should be kept in mind that dimensional analysis cannot help us determine any dimensionless constants in the equation. This method also uses chemistry and there in place of using designated symbols we use units.
Note:
Though this method is widely used, there are few drawbacks for this method. Dimensionless quantities cannot be determined by this method. Constant proportionality cannot be determined by this method. It does not apply to trigonometric, logarithmic and exponential functions physical quantities which are dependent upon more than three physical quantities, this method will be difficult.
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