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Ratio of SI unit of torque to its CGS unit is……………

seo-qna
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Answer
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Hint: In this question to basic definitional formula of torque (T) which is the cross product of force (F) and length (L) to get the S.I unit and then the basic C.G.S parameters to get the C.G.S unit of torque (T), the ratio of them will get the right answer.

Complete Step-by-Step solution:
As we know that torque (T) is the cross product of force (F) and length (L).
$ \Rightarrow T = F \times L$
As we all know that the S.I unit of force is Newton (N) and the S.I unit of length is meter.
So the S.I unit of torque is N-m.
And the C.G.S unit of torque is Dyne-Centimeter (dyn.Cm).
Now as we all know that
1 Newton = 100000 Dyne.
And 1 m = 100 cm.
So 1 N.m = 100000$ \times $100 dyn.cm
$ \Rightarrow 1N.m = {10^7}dyn.cm$
$ \Rightarrow \dfrac{{1N.m}}{{1dyn.cm}} = {10^7}$
So the ratio of S.I unit of torque to the C.G.S unit of torque is ${10^7}$.
So this is the required answer.

Note – The trick point here was simply to know the difference between the S.I unit and the C.G.S unit system that is in CGS system the unit of length, mass and time are centimeter (cm), gram (g) and second (s), respectively whereas In SI system the unit of length, mass and time are meter (m), kilogram (kg) and second (s), respectively. These basics help solving problems of this kind.