Question

In C.G.S. system of units the unit of pressure is ${{{\rm{dyne}}} {\left/
 {\vphantom {{{\rm{dyne}}} {{\rm{c}}{{\rm{m}}^{\rm{2}}}}}} \right.
} {{\rm{c}}{{\rm{m}}^{\rm{2}}}}}$ in a new system of units, the unit of mass is ${\rm{1}}\;{\rm{milligram}}$, unit of length is ${\rm{1}}\;{\rm{mm}}$ and unit of time is ${\rm{1}}\;{\rm{millisecond}}$ . Let the unit of pressure in this new system be dyne. The value of ${\rm{1}}\;{\rm{dyne}}$ is equal to
A.\[{10^{ + 4}}\;{{{\rm{dyne}}} {\left/ {\vphantom {{{\rm{dyne}}} {{\rm{c}}{{\rm{m}}^{\rm{2}}}}}} \right.} {{\rm{c}}{{\rm{m}}^{\rm{2}}}}}\]
B.\[1\;{{{\rm{dyne}}} {\left/ {\vphantom {{{\rm{dyne}}} {{\rm{c}}{{\rm{m}}^{\rm{2}}}}}} \right.} {{\rm{c}}{{\rm{m}}^{\rm{2}}}}}\]
C.\[{10^{ - 2}}\;{{{\rm{dyne}}} {\left/ {\vphantom {{{\rm{dyne}}} {{\rm{c}}{{\rm{m}}^{\rm{2}}}}}} \right.} {{\rm{c}}{{\rm{m}}^{\rm{2}}}}}\]
D.\[{10^{ - 3}}\;{{{\rm{dyne}}} {\left/ {\vphantom {{{\rm{dyne}}} {{\rm{c}}{{\rm{m}}^{\rm{2}}}}}} \right.} {{\rm{c}}{{\rm{m}}^{\rm{2}}}}}\]

Answer 1

Hint: To solve this question we will first write the S.I. unit of Force which is Newton. Then we write Newton in terms of mass, length and time. Now. We will write the C.G.S. unit of force by converting the units. Now we write the expression for the new unit. We need to find the relation between the new unit and the C.G.S. unit of force.

Complete step by step solution:
Dyne is the C.G.S. unit of force. The S.I. system, unit of force is Newton. We know that Newton can be represented as
$1\;{\rm{N}} = \dfrac{{{\rm{kg}} \cdot {\rm{m}}}}{{{{\rm{s}}^{\rm{2}}}}}$
Therefore, in C.G.S. unit we can write dyne as
${\rm{1}}\;{\rm{dyne}} = \dfrac{{{\rm{g}} \cdot {\rm{cm}}}}{{{{\rm{s}}^{\rm{2}}}}}$
It is given in the question that in the new system of unit mass is given as ${\rm{1}}\;{\rm{milligram}}$, unit length is given as ${\rm{1}}\;{\rm{mm}}$ and unit time is given as ${\rm{1}}\;{\rm{millisecond}}$.
We will rewrite the expression of dyne in the new units.
\[{\rm{1}}\;{\rm{dyne}} = \dfrac{{\left( {{\rm{1}}\;{\rm{mg}}} \right) \cdot \left( {{\rm{1}}\;{\rm{mm}}} \right)}}{{{{\left( {{\rm{1}}\;{\rm{ms}}} \right)}^{\rm{2}}}}}\]
But we know that $1\;{\rm{mg}}$ can be replaced by \[{\rm{1}}{{\rm{0}}^{ - 3}}\;{\rm{g}}\] , $1\;{\rm{mm}}$ can be replaced by ${10^{ - 1}}\;{\rm{cm}}$ and $1\;{\rm{ms}}$ can be replaced by ${10^{ - 3}}\;{\rm{s}}$ . Hence dyne can be written as
$
{\rm{1}}\;{\rm{dyne}} = \dfrac{{\left( {{\rm{1}}{{\rm{0}}^{ - 3}}\;{\rm{g}}} \right) \cdot \left( {{\rm{1}}{{\rm{0}}^{ - 1}}\;c{\rm{m}}} \right)}}{{{{\left( {{\rm{1}}{{\rm{0}}^{ - 2}}\;{\rm{s}}} \right)}^{\rm{2}}}}}\\
\implies {\rm{1}}\;{\rm{dyne}} = {10^{ - 2}}\dfrac{{{\rm{g}} \cdot {\rm{cm}}}}{{{{\rm{s}}^{\rm{2}}}}}
$
It is given that pressure in the C.G.S. unit is expressed as $\dfrac{{{\rm{dyne}}}}{{{\rm{c}}{{\rm{m}}^{\rm{2}}}}}$ . But in the new unit the unit of pressure is expressed as ${\rm{dyne}}$.
From the above expression. we can write pressure as \[{\rm{Pressure}} = {10^{ - 2}}{\rm{dyne}}\]
From this expression we can conclude that
$1\;{\rm{dyne}} = {10^{ - 2}}\dfrac{{{\rm{dyne}}}}{{{\rm{c}}{{\rm{m}}^{\rm{2}}}}}$
Therefore, the value of ${\rm{1}}\;{\rm{dyne}}$ is ${10^{ - 2}}\dfrac{{{\rm{dyne}}}}{{{\rm{c}}{{\rm{m}}^{\rm{2}}}}}$

So, the correct answer is “Option C”.

Note:
In order to solve this question, we should have the prior knowledge of unit conversion. This question is totally based on the conversion of units. The units in S.I. can be converted into units in C.G.S. by converting the length, mass and time. Also, we need to check for the units given in the question, since it can vary from question to question.