Question

When 1 meter, 1 kg and 1 min are taken as fundamental units the magnitude of force is 36 units. What will be its vale in Cgs system [use formula:- n1u1 = n2u2, from dimensional analysis.]

Answer 1

Answer:-

Your Answer Is $\bf 10^3$ dyne or $\bf 10^3 gcms^{-2}.$

Where,

• g = gram, cm = centimeter and s = seconds.

Explanation:-

Given:-

• Fundamental units which are 1 m, 1 kg and 1 min.

• Magnitude of force = 36 units.

To Find:-

• The value in cgs i.e. $\tt gcms^{-2}$ or in dyne as its is cgs unit of force.

Formula Used:-

$\large\orange{\longrightarrow\boxed{\pink{\bf n_1u_1 = n_2u_2.}}}$

Where,

• $\tt n_1\:\: And\:\: n_2$ = Numerical value in first and second system respectively.

• $\tt u_1\:\: And\:\: u_2$ = Unit of 1st and 2nd system respectively.

So Here,

• $\tt n_1\:\: And\:\: n_2$ = 36 and ?? respectively [?? this means to find].

We know that dimension formula of force = $\bf [MLT^{-2}]$.

And Hence,

• $\tt u_1\:\: And\:\: u_2$ = $\tt kgms^{-2}$ And $\tt gcms^{-2}$ respectively.

Now,

By Using Formula we get:-

$\\ \tt\mapsto n_1u_1 = n_2u_2.$

$\\ \tt\mapsto 36\ \ kgm s^{-2} = n_2\ \ g cm s{-2}.$

$\\ \tt\mapsto n_2 = 36\times \dfrac{kgms^{-2}}{gcms^{-2}}.$

$\\ \tt\mapsto n_2 = 36\times\dfrac{1kg}{gm}\times\dfrac{1m}{cm}\times\bigg(\dfrac{1min}{sec}\bigg)^{-2}.$

$\\ \tt\mapsto n_2 = 36\times\dfrac{1000\ \cancel{gm}}{\cancel{gm}}\times\dfrac{100\ \cancel{cm}}{\cancel{cm}}\times\bigg(\dfrac{60\ \cancel{sec}}{\cancel{sec}}\bigg)^{-2}.$

$\tt\mapsto n_2 = 36\times 1000\times 100\times (60)^{-2}.$

$\\ \tt\mapsto n_2 = 36\times 1000\times100\times\dfrac{1}{3600}.$

$\\ \tt\mapsto n_2 = \cancel{3600000}\times\dfrac{1}{\cancel{3600}}.$

$\\ \tt\mapsto n_2 = 1000.$

$\\ \large\red{\mapsto\boxed{\blue{\bf n_2 = 10^3.}}}$

Therefore 36 given system of unit = $\bf 10^3$ dyne.