Tuesday, October 4, 2016

Units and Dimensions Problems and Solutions Two

We are first going to solve a problem using dimensional analysis. According to principle of homogeneity, we can understand that a physics equation has to be correct as per dimension. That means to say that we can add or subtract only similar physical quantity.

Problem One


Solution

The left hand side of the equation is the velocity which is defined as the rate of change of displacement. So it has the dimensions of length divided by time.

The right hand side of the equation shall have the same dimensions. In the denominator, we have a term with exponential function. As per dimensional analysis, it can not have any dimensions. Thus by equating it to zero dimensions, we can find the dimensions of B and then we can find the dimensions of A as shown below.


Problem Two

The problem is about finding the dimensions of pressure and density when it is given in the problem that the velocity depends on them as shown in the diagram below.


Solution

According to principle of homogeneity, the dimensions of each fundamental physical quantity on both  the sides of the equations has to be the same. Thus by equating the powers of fundamental quantities like length, mass and time we can get  the values as shown in the diagram below.


Problem Three

This is also a similar problem of the above but with different physical quantities. It is given in the problem that the mass of the stone depends on velocity, density and acceleration due to gravity.


Solution

By raising the dimensions of each physical quantity to a unknown power and then equating the dimensions of each fundamental quantity, we can solve the problem as shown in the diagram below.


Problem Four

This problem links with Young modulus with different physical quantity relates with mass, acceleration due to gravity, length of the wire with a unknown power and thickness and elongation. 


Solution

We need to write the dimensional formula for all the physical quantity of the given equation so that we can find the power of the length as shown in the diagram below.


Problem five

This problem links couple per unit twist with the rigidity modulus of the wire, radius of the wire and length of the wire as shown in the diagram below. We need to find the dimensions of each physical quantity.


Solution

We shall write the dimensions of each quantity like couple per unit twist and others as shown in the diagram below. Thus by applying the dimensional analysis, we can find the powers as shown in the diagram below.



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