Question

Using dimensions convert
A. 1 Newton into dynes
B. 1 erg into joules

Answer 1

Hint: The above problem can be resolved using the concept and fundamentals of the dimensional analysis, along with the primary method to derive the dimensional formulas utilizing the idea of dimensional analysis. The dimensional analysis is generally used to identify the significant expressions of the fundamental physical quantities. Moreover, the basic dimensions of the mass, length, and time are utilized to determine the proper relationship for the force and the energy in the given problem.

Complete step by step solution
(A) The dimension for the newton is,
\[N = kg \cdot \left( {\dfrac{m}{{{{\sec }^2}}}} \right)\]
Here, kg is the fundamental unit of mass, metre (m) is the fundamental unit of the length and seconds (s) is the fundamental unit to represent the time.
Solve by substituting the values as unity, for each variable as,
\[\begin{array}{l}
N = kg \cdot \left( {\dfrac{m}{{{{\sec }^2}}}} \right)\\
\Rightarrow N = \left( {1\;{\rm{kg}} \times \dfrac{{1000\;{\rm{g}}}}{{1\;{\rm{kg}}}}} \right) \cdot \left( {\dfrac{{1\;{\rm{m}} \times \dfrac{{100\;{\rm{cm}}}}{{1\;{\rm{m}}}}}}{{1\;{{\sec }^2}}}} \right)\\
\Rightarrow N = 1000\;g \times 100\;cm \times 1\;{{\rm{s}}^{{\rm{ - 2}}}}\\
\Rightarrow N = {10^5}\;{\rm{dyne}}
\end{array}\]
Therefore, the one Newton is equal to \[{10^5}\;{\rm{dyne}}\].

(B) Similarly, the dimension of Joule is given as,
\[J = \dfrac{{kg \cdot {m^2}}}{{{s^2}}}\]
Here, kg is the fundamental unit of mass, metre (m) is the fundamental unit of the length and seconds (s) is the fundamental unit to represent the time.
Solve by substituting the unit values of each variables as,
 \[\begin{array}{l}
\Rightarrow J = \dfrac{{\left( {1\;{\rm{kg}} \times \dfrac{{1000\;{\rm{g}}}}{{1\;{\rm{kg}}}}} \right) \times {{\left( {1\;{\rm{m}} \times \dfrac{{100\;{\rm{cm}}}}{{1\;{\rm{m}}}}} \right)}^2}}}{{{{\left( {1\;{\rm{s}}} \right)}^2}}}\\
\Rightarrow J = 1000\;{\rm{g}} \times {\left( {100\;{\rm{cm}}} \right)^2} \times 1\;{{\rm{s}}^{{\rm{ - 2}}}}\\
\Rightarrow J = {10^7}\;{\rm{erg}}\\
\Rightarrow {\rm{1}}\;{\rm{erg}} = {10^{ - 7}}\;{\rm{J}}
\end{array}\]
Therefore, one erg equals \[{10^{ - 7}}\;{\rm{J}}\].

Note:
To solve the given problem, the basic knowledge of dimensional quantities is needed to be considered. The fundamentals and tools of dimensional analysis are major elements to derive the critical formulas used to resolve physics problems. Moreover, the dimensional analysis is also utilized to convert any fundamental quantity into the derived quantities.