
Find the CGS unit of the following derived quantities: Give the relation of these CGS units with respect to their SI units.
(a). Velocity\[\left[ {{\text{Hint:Velocity = }}\dfrac{{{\text{length}}}}{{{\text{time}}}}} \right]\]
(b). Acceleration\[\left[ {{\text{Hint:Acceleration = }}\dfrac{{{\text{Velocity}}}}{{{\text{time}}}}} \right]\]
(c). Momentum \[\left[ {{\text{Hint:Momentum = Mass}} \times {\text{Velocity}}} \right]\]
Answer
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- Hint: In this question CGS stands for centimeter–gram–second system of units. It is a variant of the metric system based on the centimeter as the unit of length, the gram as the unit of mass, and the second as the unit of time.
Complete step-by-step solution -
(a). Given, \[{\text{Velocity = }}\dfrac{{{\text{length}}}}{{{\text{time}}}}\]
We have to find out CGS unit of velocity
\[\therefore \]CGS unit of velocity = \[\dfrac{{{\text{CGS unit of length}}}}{{{\text{CGS unit of time}}}} = \dfrac{{{\text{cm}}}}{{\text{s}}}\]
\[{\text{1cm }}{{\text{s}}^{{\text{ - 1}}}}\]
(b). Given, \[{\text{Acceleration = }}\dfrac{{{\text{Velocity}}}}{{{\text{time}}}}\]
We have to find out CGS unit of acceleration
\[\therefore \] CGS unit of acceleration = \[\dfrac{{{\text{CGS unit of velocity}}}}{{{\text{CGS unit of time}}}} = \dfrac{{{\text{cm }}{{\text{s}}^{ - 1}}}}{{\text{s}}}\]
\[\therefore \] CGS unit of acceleration = \[{\text{cm }}{{\text{s}}^{ - 2}}\]
1 \[{\text{cm }}{{\text{s}}^{ - 2}}\]
(c). Given, \[{\text{Momentum = Mass}} \times {\text{Velocity}}\]
We have to find out CGS unit of Momentum
\[\therefore \] CGS unit of momentum = \[{\text{gram}} \times cm{{\text{s}}^{{\text{ - 1}}}}{\text{ = g}}{\text{.cm}}{{\text{s}}^{{\text{ - 1}}}}\]
\[\therefore \] CGS unit of momentum =\[{\text{g}}{\text{.cm}}{{\text{s}}^{{\text{ - 1}}}}\]
\[{\text{1g cm}}{{\text{s}}^{ - 1}}\]
Note: In this question, the meaning of velocity of an object can be defined as the rate of change of the object’s position with respect to a frame of reference and time.
Acceleration is a vector quantity as it has both magnitude as well as direction. It is also the second derivative of position with respect to time or it is a first derivative of velocity with respect to time. Momentum is the tendency of the object to be in motion and therefore is a vector quantity.
Complete step-by-step solution -
(a). Given, \[{\text{Velocity = }}\dfrac{{{\text{length}}}}{{{\text{time}}}}\]
We have to find out CGS unit of velocity
\[\therefore \]CGS unit of velocity = \[\dfrac{{{\text{CGS unit of length}}}}{{{\text{CGS unit of time}}}} = \dfrac{{{\text{cm}}}}{{\text{s}}}\]
\[{\text{1cm }}{{\text{s}}^{{\text{ - 1}}}}\]
(b). Given, \[{\text{Acceleration = }}\dfrac{{{\text{Velocity}}}}{{{\text{time}}}}\]
We have to find out CGS unit of acceleration
\[\therefore \] CGS unit of acceleration = \[\dfrac{{{\text{CGS unit of velocity}}}}{{{\text{CGS unit of time}}}} = \dfrac{{{\text{cm }}{{\text{s}}^{ - 1}}}}{{\text{s}}}\]
\[\therefore \] CGS unit of acceleration = \[{\text{cm }}{{\text{s}}^{ - 2}}\]
1 \[{\text{cm }}{{\text{s}}^{ - 2}}\]
(c). Given, \[{\text{Momentum = Mass}} \times {\text{Velocity}}\]
We have to find out CGS unit of Momentum
\[\therefore \] CGS unit of momentum = \[{\text{gram}} \times cm{{\text{s}}^{{\text{ - 1}}}}{\text{ = g}}{\text{.cm}}{{\text{s}}^{{\text{ - 1}}}}\]
\[\therefore \] CGS unit of momentum =\[{\text{g}}{\text{.cm}}{{\text{s}}^{{\text{ - 1}}}}\]
\[{\text{1g cm}}{{\text{s}}^{ - 1}}\]
Note: In this question, the meaning of velocity of an object can be defined as the rate of change of the object’s position with respect to a frame of reference and time.
Acceleration is a vector quantity as it has both magnitude as well as direction. It is also the second derivative of position with respect to time or it is a first derivative of velocity with respect to time. Momentum is the tendency of the object to be in motion and therefore is a vector quantity.
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