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Convert 10 Joules to erg using the dimensional method.

dimensional analysis and its application
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Convert 10 J into erg


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Solution:

Let, 10 Joules = N2 erg

Here, Joule is the SI unit of energy and erg is the CGS unit of energy. The dimensional formula of energy is [ M L2 T-2 ], comparing this with [ Ma Lb Tc ], we get

a = 1, b = 2 and c = -2

Given System (SI-Unit)

N1 = 10

M1 = 1 kg

L1 = 1m

T1 = 1 s

To be converted (CGS-System)

N2 = ?

M2 = 1g

L2 = 1cm

T2 = 1s

Now, using the dimensional method of unit conversion, we have

\(\(\therefore N_2=N_1\left[\frac{M_1}{M_2}\right]^a\left[\frac{L_1}{L_2}\right]^b\left[\frac{T_1}{T_2}\right]^c\)

= \(10\left[\frac{1kg}{1g}\right]^1\left[\frac{1m}{1cm}\right]^2\left[\frac{1s}{1s}\right]^{-2}\)

= \(10\left[\frac{1000g}{1g}\right]^1\left[\frac{100cm}{1cm}\right]^2\left[\frac{1s}{1s}\right]^{-2}\)

∴ N2 = 108

Hence, 10 Joule = 108 erg

 

Things to remember from Physical Quantities
  • The four applications of dimensional analysis are:
    • To check the correctness of physical relation.
    • To derive the relation between various physical quantities.
    • To convert the value of physical quantities from one system of units into another system of units.
    • To find the dimensions of constants in the given equation.

  • The correctness of physical relation can be determined by comparing the dimension of the left-hand side (LHS) and the dimension of the right-hand side (RHS)
  • We can use the formula, \(N_2=N_1\left[\frac{M_1}{M_2}\right]^a\left[\frac{L_1}{L_2}\right]^b\left[\frac{T_1}{T_2}\right]^c\)\) to convert unit from one to another system using dimension.

  • Despite the usefulness of dimensions, there are some limitations. They are:
    • The dimensional analysis does not give any information about dimensionless constants.
    • If the quantity depends on more than three other physical quantities having dimensions, the formula cannot be derived.
    • We cannot derive the formula containing trigonometric functions, logarithmic functions, exponential functions, etc. It is best suited for linear functions only.
    • The exact form of a relationship cannot be determined when there is more than one part in any relationship.
    • It gives no information about the physical quantity,

 

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