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1.Units, Dimensions and Measurement
hard
In $C.G.S$. system the magnitutde of the force is $100$ dynes. In another system where the fundamental physical quantities are kilogram, metre and minute, the magnitude of the force is
A$0.036$
B$0.36$
C$3.6$
D$36$
Solution
(c) ${n_2} = {n_1}{\left( {\frac{{{M_1}}}{{{M_2}}}} \right)^1}{\left( {\frac{{{L_1}}}{{{L_2}}}} \right)^1}{\left( {\frac{T}{{{T_2}}}} \right)^{ – 2}}$
= $100{\left( {\frac{{gm}}{{kg}}} \right)^1}{\left( {\frac{{cm}}{m}} \right)^1}{\left( {\frac{{\sec }}{{min}}} \right)^{ – 2}}$
= $100{\left( {\frac{{gm}}{{{{10}^3}gm}}} \right)^1}{\left( {\frac{{cm}}{{{{10}^2}cm}}} \right)^1}{\left( {\frac{{\sec }}{{60\sec }}} \right)^{ – 2}}$
$n_2 = \frac{{3600}}{{{{10}^3}}} = 3.6$
= $100{\left( {\frac{{gm}}{{kg}}} \right)^1}{\left( {\frac{{cm}}{m}} \right)^1}{\left( {\frac{{\sec }}{{min}}} \right)^{ – 2}}$
= $100{\left( {\frac{{gm}}{{{{10}^3}gm}}} \right)^1}{\left( {\frac{{cm}}{{{{10}^2}cm}}} \right)^1}{\left( {\frac{{\sec }}{{60\sec }}} \right)^{ – 2}}$
$n_2 = \frac{{3600}}{{{{10}^3}}} = 3.6$
Standard 11
Physics
