Question

Fill in the blanks
(A) 1kg{m^2}/{s^2} = \\_\\_\\_\\_\\_gc{m^2}/{s^2}
(B) 1m = \\_\\_\\_\\_\\_ly
(C) 3.0m{s^{ - 2}} = \\_\\_\\_\\_\\_km{h^{ - 2}}
(D) G = 6.67 \times {10^{ - 11}}N{m^2}{\left( {kg} \right)^{ - 2}} = \\_\\_\\_\\_\\_\\_\\_{\left( {cm} \right)^3}{s^{ - 2}}{g^{ - 1}}

Answer 1

Hint
To solve this problem we have to convert each unit in formula in the LHS to the units that we desire given in the RHS. Then by combining those units, we get the conversion factor for the answer we need.

Complete step by step answer
Let us consider the first option.
-We are said to convert $1kg{m^2}/{s^2}$ to $gc{m^2}/{s^2}$. Let the first convert the individual units in $1kg{m^2}/{s^2}$. So we have,
$\Rightarrow 1kg = {10^3}g$,
$\Rightarrow 1m = {10^2}cm$, so taking the square we get, $1{m^2} = {10^4}c{m^2}$
Now we can substitute the values in $1kg{m^2}/{s^2}$. Hence we get,
$\Rightarrow 1kg{m^2}/{s^2} = \dfrac{{{{10}^3}g \times {{10}^4}{m^2}}}{{{s^2}}}$
On doing calculation we get,
$\Rightarrow 1kg{m^2}/{s^2} = {10^7}g{m^2}/{s^2}$.
-Here we are asked to convert 1 meter to light year.
So according to the conversion factor we have,
$\Rightarrow 1m = 1.057 \times {10^{ - 16}}ly$
-Here we need to convert $m{s^{ - 2}}$ to $km{h^{ - 2}}$.
Let us first see the individual conversion of the units. So we get,
$\Rightarrow 1m = {10^{ - 3}}km$
and $1s = 2.77 \times {10^{ - 4}}h$, so we get, $1{s^2} = {\left( {2.77 \times {{10}^{ - 4}}} \right)^2}{h^2}$
Hence by substituting in $m{s^{ - 2}}$ we get,
$\Rightarrow 3.0m{s^{ - 2}} = 3 \times \dfrac{{{{10}^{ - 3}}km}}{{{{\left( {2.77 \times {{10}^{ - 4}}} \right)}^2}{h^2}}}$
On doing the calculation we get,
$\Rightarrow 3.0m{s^{ - 2}} = \dfrac{{3 \times {{10}^{ - 3}}km}}{{7.71 \times {{10}^{ - 8}}{h^2}}}$
Therefore, we get,
$\Rightarrow 3.0m{s^{ - 2}} = 38880km{h^{ - 2}}$
-Here we are asked to convert the $N{m^2}{\left( {kg} \right)^{ - 2}}$ to ${\left( {cm} \right)^3}{s^{ - 2}}{g^{ - 1}}$. To do so we need to first bring the units in the LHS to the form of the units in the RHS. So we have,
$\Rightarrow 1N = \dfrac{{kgm}}{{{s^2}}}$
Substituting this in the LHS we get,
$\Rightarrow 1N{m^2}{\left( {kg} \right)^{ - 2}} = 1 \times \dfrac{{kgm}}{{{s^2}}} \times \dfrac{{{m^2}}}{{k{g^2}}}$
On cancelling the similar terms from numerator and denominator we get,
$\Rightarrow 1N{m^2}{\left( {kg} \right)^{ - 2}} = 1{m^3}{s^{ - 2}}{\left( {kg} \right)^{ - 1}}$
Now the individual conversion factor for the units is,
$\Rightarrow 1m = {10^2}cm$, so on taking cube, $1{m^3} = {10^6}c{m^3}$
$\Rightarrow 1kg = {10^3}g$
So substituting we have,
$\Rightarrow 1{m^3}{s^{ - 2}}{\left( {kg} \right)^{ - 1}} = \dfrac{{{{10}^6}{{\left( {cm} \right)}^3}}}{{{s^2} \times {{10}^3}g}}$
Hence we get,
$\Rightarrow 1{m^3}{s^{ - 2}}{\left( {kg} \right)^{ - 1}} = {10^3}{\left( {cm} \right)^3}{s^{ - 2}}{g^{ - 1}}$
So in the question we need to convert $6.67 \times {10^{ - 11}}N{m^2}{\left( {kg} \right)^{ - 2}}$. So we get,
$\Rightarrow 6.67 \times {10^{ - 11}}{m^3}{s^{ - 2}}{\left( {kg} \right)^{ - 1}} = 6.67 \times {10^{ - 11}} \times {10^3}{\left( {cm} \right)^3}{s^{ - 2}}{g^{ - 1}}$
On calculating we get,
$\Rightarrow 6.67 \times {10^{ - 11}}{m^3}{s^{ - 2}}{\left( {kg} \right)^{ - 1}} = 6.67 \times {10^{ - 8}}{\left( {cm} \right)^3}{s^{ - 2}}{g^{ - 1}}$.

Note
The light year is a unit of length that is used to represent the astronomical units of length. It is defined as the distance that light travels in vacuum in 1 year. It is often used to denote the distances to stars and other astronomical bodies.