Answer:
Here,
\[\text{c}=\left[
\text{L}{{\text{T}}^{-\text{1}}} \right]=\text{3}\times
\text{1}{{0}^{\text{1}0}}\text{cm }{{\text{s}}^{-\text{1}}}\]
\[\text{G}=\left[ {{\text{M}}^{-\text{1}}}{{\text{L}}^{\text{3}}}{{\text{T}}^{-\text{2}}}
\right]=\text{6}.\text{67}\times \text{1}{{0}^{-\text{8}}}\text{dyne
c}{{\text{m}}^{\text{2}}}{{\text{g}}^{-\text{2}}}\]
\[\text{h}=\left[
\text{M}{{\text{L}}^{\text{2}}}{{\text{T}}^{-\text{1}}} \right]=\text{
6}.\text{6}\times \text{1}{{0}^{-\text{27}}}\text{erg}.\text{sec}.\]
Now, \[\frac{hc}{G}=\frac{6.6\times
{{10}^{-27}}\times 3\times {{10}^{10}}}{6.67\times {{10}^{-8}}}\]
\[\frac{M{{L}^{2}}{{T}^{-1}}\times
L{{T}^{-1}}}{{{M}^{-1}}{{L}^{3}}{{T}^{-2}}}=\text{ 2}.\text{9685}\times
\text{1}{{0}^{-\text{9}}}\]
\[{{\text{M}}^{\text{2}}}=\text{
2}.\text{9685}\times \text{1}{{0}^{-\text{9}}}\]
\[\text{M
}=\sqrt{2.9685\times {{10}^{-9}}}\]
\[\text{1g
}=\frac{1}{0.5448\times {{10}^{-4}}}\]
new unit of mass
\[=\text{
1}.\text{8355}\times \text{1}{{0}^{\text{4}}}\]
new unit of mass
Again, \[\frac{h}{M{{C}^{2}}}=\frac{6.67\times
{{10}^{-27}}}{0.5448\times {{10}^{-4}}\times {{(3\times
{{10}^{10}})}^{2}}}=1.3603\times {{10}^{-43}}\]
\[\frac{M{{L}^{2}}{{T}^{-1}}}{M{{L}^{2}}{{T}^{-2}}}=\text{
1}.\text{36}0\text{3 }\times \text{ 1}{{0}^{-\text{43}}}\text{s}\]
or \[\text{T
}=\text{1}.\text{36}0\text{3}\times \text{1}{{0}^{-\text{43}}}\text{s}\]
\[\therefore\]
\[1s\,=\frac{1}{1.3603\times
{{10}^{-43}}}=0.735\times {{10}^{43}}\]
new unit of time
Again,
\[\text{c}\times \text{T}=\text{3}\times
\text{1}{{0}^{\text{1}0}}\times \left( \text{1}.\text{36}0\text{3 }\times
\text{1}{{0}^{-\text{43}}} \right)\]
\[\left( \text{L}{{\text{T}}^{-\text{1}}}
\right)\times \text{T}=\text{L}=\text{4}.0\text{8}0\text{9}\times
\text{1}{{0}^{-\text{33}}}\text{cm}\]
\[\text{1 cm }=\frac{1}{4.0809\times
{{10}^{-33}}}=\text{ }0.\text{245}0\,\times \,{{10}^{33}}\] new unit of length.
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