A) 1 new unit
B) 129.6 new unit
C) 125.7 new unit
D) \[{{10}^{3}}\,\]new unit
Correct Answer: B
Solution :
Key Idea: The product of the numerical value of a physical quantity and its corresponding unit is a constant. Let the numerical value of a physical quantity\[p,\] are \[{{n}_{1}}\]and \[{{n}_{2}}\]in two different systems and the corresponding units are \[{{u}_{1}}\]and \[{{u}_{2}},\] then \[{{n}_{1}}[{{u}_{1}}]={{n}_{2}}[{{u}_{2}}]\] Dimensions of force \[=[ML{{T}^{-2}}]\] \[\therefore \] \[{{n}_{1}}[{{M}_{1}}{{L}_{1}}T_{1}^{-2}]={{n}_{2}}[{{M}_{2}}{{L}_{2}}T_{2}^{-2}]\] \[\Rightarrow \] \[{{n}_{2}}={{n}_{1}}\left[ \frac{{{M}_{1}}}{{{M}_{2}}}\frac{{{L}_{1}}}{{{L}_{2}}}{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{-2}} \right]\] \[{{n}_{2}}={{n}_{1}}\left[ \frac{{{M}_{1}}}{{{M}_{2}}}\frac{{{L}_{1}}}{{{L}_{2}}}{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{-2}} \right]\] \[=1\left[ \frac{\text{1}\,\text{kg}}{1\,\text{quintal}}\times \frac{1\,m}{1\,km}\times {{\left( \frac{1s}{1\,h} \right)}^{-2}} \right]\] \[=1\left[ \frac{1}{100}\times \frac{1}{1000}\times 3600\times 3600 \right]\] \[=129.6\,\]new units
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