Question

Ratio of SI unit of torque to its CGS unit is
$\text{A}\text{. }{{10}^{5}}:1$
$\text{B}\text{. }{{10}^{7}}:1$
$\text{C}\text{. }{{10}^{6}}:1$
$\text{D}\text{. }{{10}^{17}}:1$

Answer 1

First find the SI unit and CGS unit of torque using the formula of torque i.e. $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$. Therefore, the magnitude of torque will be $\tau =Fr\sin \theta$. SI units F and r are newton and metre. CGS units of F and r are erg and centimetre.
Formula used:
$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$

Complete answer:
When we apply a force on a body that is fixed about an axis, the body rotates about the fixed axis. This axis is called the axis of rotation. We analyse and understand the rotational motion of the body by a term called torque. It is also called a moment or moment of force.

Suppose, we apply a force (F) at distance r from the point or axis of rotation. Then torque generated about the axis or point is given as $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$.
The magnitude of the torque is given as $\tau =Fr\sin \theta$ …. (i),
where $\theta$ is the angle between the force F and r.
Let's calculate the SI and CGS units of torque.
The SI unit of torque will be the product of the SI units of force F and distance r. $\sin \theta$ is a ratio so it has no unit.
The SI unit of force is Newton (N) and SI unit of distance r is metre (m).
Hence, the SI unit of torque is Nm.

Similarly, the CGS unit of torque will be the product of the CGS units of force F and distance r.
The CGS unit of force is erg and CGS unit of distance r is centimetre (cm).
Hence, the CGS unit of torque is erg cm.
Therefore, the ratio of SI unit of torque to its CGS unit is:
$\dfrac{Nm}{erg cm}$ … (ii).
We know that $1N={{10}^{5}}erg$ and $1m={{10}^{2}}cm$.
Substitute the values of 1N and 1m in equation (ii).
$\Rightarrow \dfrac{{{10}^{5}}erg{{.10}^{2}}cm}{erg cm}={{10}^{7}}$.
This means the ratio of the SI unit of torque to its CGS unit is ${{10}^{7}}:1$.

So, the correct answer is “Option B”.

Note:
We can also use the dimensions of force and distance.
The dimensional formula of force is $\left[ ML{{T}^{-2}} \right]$.
The dimensional formula of distance is [L].

Hence, the dimensional formula of torque is $\left[ M{{L}^{2}}{{T}^{-2}} \right]$.
SI units of mass (M), length (L) and time (T) are kg, m and s.
Hence, the unit of torque is $kg{{m}^{2}}{{s}^{-2}}$.
CGS units of mass (M), length (L) and time (T) are g, cm and s.
Hence, the unit of torque is $gc{{m}^{2}}{{s}^{-2}}$.
Therefore, the ratio of SI to CGS units of torque is:
$\dfrac{kg{{m}^{2}}{{s}^{-2}}}{gc{{m}^{2}}{{s}^{-2}}}$ …. (1)
We know that $1kg={{10}^{3}}g$ and $1m={{10}^{2}}cm$.
Hence, ratio (1) can be written as:
$\dfrac{{{10}^{3}}g{{\left({{10}^{2}}cm\right)}^{2}}{{s}^{-2}}}{gc{{m}^{2}}{{s}^{-2}}}=\dfrac{{{10}^{3}}g{{10}^{4}}c{{m}^{2}}{{s}^{-2}}}{gc{{m}^{2}}{{s}^{-2}}}={{10}^{7}}$