Question

In the CGS system the magnitude of the force is\[100{\rm{ dynes}}\] . In another system where the fundamental physical quantities are kilogram, meter, and minute, find the magnitude of the force.

Answer 1

Hint: This question is based on the measurement system. There are various measurement systems used to get the magnitude of any physical quantity. We have to use the dimensional analysis and conversion of the CGS system to the MKS system to get the magnitude of force in another measurement system.

Complete step by step answer:
First, we have to know what the term force is. The force is the term that comes into existence when there is a variation in the shape as well as size of the body. Its standardized unit is Newton and 1 Newton is ${\rm{kg}} \cdot {{\rm{m}} \mathord{\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.
 } {{{\rm{s}}^{\rm{2}}}}}$
We know that the unit of force in CGS system is ${\rm{1}}\;{\rm{Newton}} = {\rm{100000}}\;{\rm{dynes}}$.
Now, we have to know the term dimensional quantity. The dimensional quantity is defined as the powers to which the fundamental quantities are raised to describe that quantity.
The dimensional quantity of force is $F = \left[ {{M^1}{L^1}{T^{ - 2}}} \right]$
Here, $M$ is the mass, $L$ is the length and $T$ is the time.
Now, the conversion formula for CGS system is given as,
${n_1}{\left[ {{M_1}} \right]^a}{\left[ {{L_1}} \right]^b}{\left[ {{T_1}} \right]^c}$
Now, the conversion formula for MKS system is given as,
${n_2}{\left[ {{M_2}} \right]^a}{\left[ {{L_2}} \right]^b}{\left[ {{T_2}} \right]^c}$
On combining above two expressions we get,
${n_2} = {n_1}{\left[ {\dfrac{{{M_1}}}{{{M_2}}}} \right]^a}{\left[ {\dfrac{{{L_1}}}{{{L_2}}}} \right]^b}{\left[ {\dfrac{{{T_1}}}{{{T_2}}}} \right]^c}$
Here, ${n_1} = 100$, ${M_1} = g$, ${L_1} = {\rm{cm}}$, ${M_2} = {\rm{kg}}$, ${T_1} = {\rm{s}}$, ${T_2} = {\rm{min}}$
Now, substitute the values in the above equation to get ${n_2}$
$
{n_2} = 100{\left[ {\dfrac{{\rm{g}}}{{{\rm{kg}}}}} \right]^1}{\left[ {\dfrac{{{\rm{cm}}}}{{\rm{m}}}} \right]^1}{\left[ {\dfrac{s}{{\min }}} \right]^{ - 2}}\\
\implies {n_2} = 100{\left[ {\dfrac{{\rm{g}}}{{{\rm{1}}{{\rm{0}}^3}{\rm{g}}}}} \right]^1}{\left[ {\dfrac{{{\rm{cm}}}}{{{\rm{1}}{{\rm{0}}^2}\;{\rm{cm}}}}} \right]^1}{\left[ {\dfrac{{\rm{s}}}{{60\;{\rm{s}}}}} \right]^{ - 2}}\\
\therefore {n_2} = 3.6
$

Thus, the magnitude of force is $3.6$.

Note:
In this question, students might get confused between CGS and MKS systems. To solve this problem we have to know the dimensional analysis and the dimensional formula of force. To get the dimensional quantity of force, we have to use the formula of force that is $F = ma$ . Here \[a\] is the acceleration. So the dimension of acceleration is $L{T^{ - 2}}$ and $m$ is the mass. Thus, the dimensional formula of force is $ML{T^{ - 2}}$.