
The ratio of SI units to CGS units of acceleration is:
A. ${{10}^{-2}}$
B. ${{10}^{2}}$
C. $10$
D. ${{10}^{-1}}$
Answer
529.8k+ views
Hint: This problem can be solved by writing the units of acceleration as the base units of its dimensions in the SI system and the CGS system. Then upon comparing these, we can get the required ratio. Acceleration of a body is the rate of change of the velocity of the body. Velocity is the rate of change of the displacement of the body with respect to time.
Formula used:
$\text{acceleration = }\dfrac{\Delta \text{velocity}}{\Delta \text{time}}$
$\text{velocity = }\dfrac{\Delta \text{displacement}}{\Delta \text{time}}$
Complete step by step answer:
We can solve this problem by writing the units of acceleration in SI units and CGS units and then use metric conversion to get a ratio between the two.
Acceleration of a body is the rate of change of its velocity with respect to time and velocity is the rate of change of displacement with respect to time.
Hence,
$\text{acceleration = }\dfrac{\Delta \text{velocity}}{\Delta \text{time}}$ --(1)
$\text{velocity = }\dfrac{\Delta \text{displacement}}{\Delta \text{time}}$ ---(2)
The dimension of displacement is length and that of time is time. From (2), we get, the dimension of velocity is $\left[ \text{length}\text{.tim}{{\text{e}}^{-1}} \right]$ --(3)
From (1) and (3), we get that the dimension of acceleration is $\left[ \left( \text{length}\text{.tim}{{\text{e}}^{-1}} \right)\text{tim}{{\text{e}}^{-1}} \right]=\left[ \text{length}\text{.tim}{{\text{e}}^{-2}} \right]$ --(4)
Now, in the SI system, the units of length and time are metre $\left( m \right)$ and second $\left( s \right)$ respectively. Hence, from (4), the units of acceleration are metre per second squared $\left( m.{{s}^{-2}} \right)$
Also, in the CGS system, the units of length and time are centimetre $\left( cm \right)$ and second $\left( s \right)$ respectively. Hence, from (4), the units of acceleration are centimetre per second squared $\left( cm.{{s}^{-2}} \right)$
Therefore, the ratio of the units of acceleration in the SI units to the CGS units will be
$\dfrac{m.{{s}^{-2}}}{cm.{{s}^{-2}}}=\dfrac{m}{cm}=\dfrac{100cm}{cm}=100={{10}^{2}}$ $\left( \because 1m=100cm \right)$
Therefore, the required ratio is ${{10}^{2}}$.
Hence, the correct option is B) ${{10}^{2}}$.
Note: It is important to know the proper metric conversions between different units of different systems. Often students make a mistake while doing the metric conversions of different units. It is an absolute silly error to make and should be avoided. To solve such types of questions, it is essential that students know the proper definitions of physical quantities so that they can write the unit as a combination of the fundamental units. In physics, understanding and definition of the different physical quantities is key to acing the subject fully.
Formula used:
$\text{acceleration = }\dfrac{\Delta \text{velocity}}{\Delta \text{time}}$
$\text{velocity = }\dfrac{\Delta \text{displacement}}{\Delta \text{time}}$
Complete step by step answer:
We can solve this problem by writing the units of acceleration in SI units and CGS units and then use metric conversion to get a ratio between the two.
Acceleration of a body is the rate of change of its velocity with respect to time and velocity is the rate of change of displacement with respect to time.
Hence,
$\text{acceleration = }\dfrac{\Delta \text{velocity}}{\Delta \text{time}}$ --(1)
$\text{velocity = }\dfrac{\Delta \text{displacement}}{\Delta \text{time}}$ ---(2)
The dimension of displacement is length and that of time is time. From (2), we get, the dimension of velocity is $\left[ \text{length}\text{.tim}{{\text{e}}^{-1}} \right]$ --(3)
From (1) and (3), we get that the dimension of acceleration is $\left[ \left( \text{length}\text{.tim}{{\text{e}}^{-1}} \right)\text{tim}{{\text{e}}^{-1}} \right]=\left[ \text{length}\text{.tim}{{\text{e}}^{-2}} \right]$ --(4)
Now, in the SI system, the units of length and time are metre $\left( m \right)$ and second $\left( s \right)$ respectively. Hence, from (4), the units of acceleration are metre per second squared $\left( m.{{s}^{-2}} \right)$
Also, in the CGS system, the units of length and time are centimetre $\left( cm \right)$ and second $\left( s \right)$ respectively. Hence, from (4), the units of acceleration are centimetre per second squared $\left( cm.{{s}^{-2}} \right)$
Therefore, the ratio of the units of acceleration in the SI units to the CGS units will be
$\dfrac{m.{{s}^{-2}}}{cm.{{s}^{-2}}}=\dfrac{m}{cm}=\dfrac{100cm}{cm}=100={{10}^{2}}$ $\left( \because 1m=100cm \right)$
Therefore, the required ratio is ${{10}^{2}}$.
Hence, the correct option is B) ${{10}^{2}}$.
Note: It is important to know the proper metric conversions between different units of different systems. Often students make a mistake while doing the metric conversions of different units. It is an absolute silly error to make and should be avoided. To solve such types of questions, it is essential that students know the proper definitions of physical quantities so that they can write the unit as a combination of the fundamental units. In physics, understanding and definition of the different physical quantities is key to acing the subject fully.
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