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Q.
In a system of units, the units of mass, length and time are 1 quintal, 1 km and 1 h respectively. In this system 1 N force will be equal to
VMMC MedicalVMMC Medical 2007
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