Question

In the C.G.S system the magnitude of the force is \[100{\rm{ dyne}}\]. In another system where the fundamental physical quantities are kilogram, metre, and minute, and the magnitude of force is:
A. 0.036
B. 0.36
C. 3.6
D. 36

Answer 1

Hint:We will convert the Centimetre-gram-second system unit of force dyne into S.I. unit of force that is Newton. Later we will convert Newton into its fundamental physical quantities (kilogram, metre and second). At last, we will convert the fundamental quantity second into a minute.

Complete step by step answer:
We are given that the magnitude of the force is \[P = 100{\rm{ dyne}}\].
We have to find the magnitude of the force in fundamental physical quantities (kilogram, metre and minute).

We know that the unit dyne in terms of Newton can be expressed as:
\[1{\rm{ dyne}} = {10^{ - 5}}{\rm{ N}}\]…….(1)
We can write the conversion of Newton into its fundamental units (kilogram, metre and second) as below:
\[1{\rm{ N}} = {\rm{kg}}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
} {{{\rm{s}}^2}}}\]
On substituting \[{\rm{kg}}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
} {{{\rm{s}}^2}}}\] for Newton in equation (1), we get:
\[1{\rm{ dyne}} = {10^{ - 5}}{\rm{ kg}}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
} {{{\rm{s}}^2}}}\]
Let us write the expression for the given value of force
\[
P = 100{\rm{ dyne}} \times \left( {\dfrac{{{{10}^{ - 5}}{\rm{ kg}}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
} {{{\rm{s}}^2}}}}}{{{\rm{dyne}}}}} \right)\\
= {10^{ - 3}}{\rm{ kg}}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
} {{{\rm{s}}^2}}} \times {\left( {\dfrac{{{\rm{60 s}}}}{{{\rm{min}}}}} \right)^2}\\
= 3.6{\rm{ kg}}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {{{\min }^2}}}} \right.
} {{{\min }^2}}}
\]

Therefore, the magnitude of the force in terms of fundamental physical quantities are kilogram, metre, and the minute is 3.6, and option (C) is correct.

Note: Alternate method: First, we will convert the C.G.S. system unit of force given in dyne into its fundamental quantities (centimetre, gram and second). Then using the below formula, we can find the magnitude of the force in kilogram, metre and minute:
\[{n_2} = {n_1}{\left[ {\dfrac{{{{\rm{M}}_1}}}{{{{\rm{M}}_2}}}} \right]^a}{\left[ {\dfrac{{{{\rm{L}}_1}}}{{{{\rm{L}}_2}}}} \right]^b}{\left[ {\dfrac{{{{\rm{L}}_1}}}{{{{\rm{L}}_2}}}} \right]^c}\]
Here a=1, b=1 and c= -2.
\[{n_1}\] is the magnitude of the given value of force and \[{n_2}\] is the magnitude of the required force.
And, \[\left[ {{{\rm{L}}_1}} \right] = {\rm{cm}}\], \[\left[ {{{\rm{L}}_2}} \right] = {\rm{m}}\], \[\left[ {{{\rm{M}}_1}} \right] = {\rm{ g}}\], \[\left[ {{{\rm{M}}_2}} \right] = {\rm{kg}}\], \[\left[ {{{\rm{T}}_1}} \right] = {\rm{s}}\].
We also know that the dimensional formula of force is given by \[\left[ {{{\rm{M}}^1}{{\rm{L}}^1}{{\rm{T}}^{ - 2}}} \right]\].