Question

1 Newton = K dyne, what is the value of K?
A. $1$
B. ${{10}^{2}}$
C. ${{10}^{5}}$
D. $10$

Answer 1

Hint: Dyne is the CGS unit of Force, and Newton is the SI unit of Force. There are two methods to do this problem. Either you can do this in the dimension method, or you can do this by converting the constituent units.

Formula Used:
From Newton’s Laws, we get,
$F=ma$
Where,
F = force on the particle
m = mass of the particle
a = acceleration of the particle

Complete step by step solution:

Equation (1) gives us the relation of force with mass and acceleration,

Unit of mass in the SI system is Kg.

Unit of acceleration in the SI system is,
$\dfrac{m}{{{s}^{2}}}$

Hence, the unit of Force in the SI system is,
$kg\dfrac{m}{{{s}^{2}}}$
So,
$1N=kg\dfrac{m}{{{s}^{2}}}$………………(2)

We know that, the unit of mass in the CGS system is gram (gm)
And the unit of length in the CGS system is centimeter (cm).

The conversion is done as follows,
$1kg=1000gm$
$1m=100cm$

Hence, we can put these values in Equation (2), and arrive at,
$1N=(1000gm)\dfrac{(100cm)}{{{s}^{2}}}={{10}^{5}}gm\dfrac{cm}{{{s}^{2}}}$
So, 1 Dyne =
$gm\dfrac{cm}{{{s}^{2}}}$
Hence, we can write 1 Newton as,
$1N={{10}^{5}}$ Dyne

So, the value of K is,
${{10}^{5}}$
The correct choice is (C).

Note: We can also solve the problem using the dimensions of Force. We know that the dimension of Force is,
$[F]=ML{{T}^{-2}}$

Now, let us assume that there are two systems, and the units are as follows:
${{M}_{1}}{{L}_{1}}{{T}_{1}}^{-2}$ and
${{M}_{2}}{{L}_{2}}{{T}_{2}}^{-2}$

Where,
${{M}_{1}}{{L}_{1}}{{T}_{1}}^{-2}$ is the unit in the first system
${{M}_{2}}{{L}_{2}}{{T}_{2}}^{-2}$ is the unit in the second system

Now, we can write,
$\dfrac{1N}{1Dyne}=\dfrac{{{M}_{1}}{{L}_{1}}{{T}_{1}}^{-2}}{{{M}_{2}}{{L}_{2}}{{T}_{2}}^{-2}}$

Here,
$\dfrac{{{M}_{1}}}{{{M}_{2}}}=\dfrac{1kg}{1gm}=\dfrac{1000gm}{1gm}={{10}^{3}}$
$\dfrac{{{L}_{1}}}{{{L}_{2}}}=\dfrac{1m}{1cm}=\dfrac{100cm}{1cm}={{10}^{2}}$
$\dfrac{{{T}_{1}}}{{{T}_{2}}}=\dfrac{1s}{1s}=1$
Hence,
$\dfrac{1N}{1Dyne}=\left( \dfrac{{{M}_{1}}}{{{M}_{2}}} \right)\left( \dfrac{{{L}_{1}}}{{{L}_{2}}} \right){{\left( \dfrac{{{T}_{1}}}{{{T}_{2}}} \right)}^{-2}}$
$\Rightarrow \dfrac{1N}{1Dyne}=\left( {{10}^{3}} \right)\left( {{10}^{2}} \right){{\left( 1 \right)}^{-2}}$
$\Rightarrow \dfrac{1N}{1Dyne}={{10}^{5}}$

So, the value of K is
${{10}^{5}}$

The above solution is useful if a random unit system is given to you, and you need to find the unit in that particular system. Following this technique will be useful, because you can avoid mistakes easily. However, if the system is known and easy, you should stick to the first method. That is fast and accurate.