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1.Units, Dimensions and Measurement
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તારનો યંગ મોડયુલસ $Y = \frac{F}{A}*\frac{L}{{\Delta L}};$ જયાં $L=$ લંબાઇ, $A=$ આડછેદનું ક્ષેત્રફળ અને $\Delta L = $ લંબાઇમાં થતો ફેરફાર, તો $CGS$ માંથી $MKS$ માં જવા માટે કેટલા વડે ગુણાકાર કરવો પડે?
A$1$
B$10$
C$0.1$
D$0.01$
Solution
$Y= [M{L^{ – 1}}{T^{ – 2}}]$
$C.G.S.$ unit $: gm\ c{m^{ – 1}}{\sec ^{ – 2}}$
$M.K.S.$ unit $: kg.\ m^{-1}\ sec^{-2} .$
${n_2} = {n_1}{\left[ {\frac{{{M_1}}}{{{M_2}}}} \right]^1}{\left[ {\frac{{{L_1}}}{{{L_2}}}} \right]^{ – 1}}{\left[ {\frac{{{T_1}}}{{{T_2}}}} \right]^{ – 2}}$ $ = {\left[ {\frac{{gm}}{{kg}}} \right]^1}{\left[ {\frac{{cm}}{{meter}}} \right]^{ – 1}}{\left[ {\frac{{\sec }}{{\sec }}} \right]^{ – 2}}$
$\therefore \frac{{{n_2}}}{{{n_1}}} = {\left[ {\frac{{gm}}{{{{10}^3}gm}}} \right]^1}{\left[ {\frac{{cm}}{{{{10}^2}cm}}} \right]^{ – 1}}{\left[ {\frac{{\sec }}{{\sec }}} \right]^{ – 2}}$ $ = \frac{1}{{10}} = 0.1$
$C.G.S.$ unit $: gm\ c{m^{ – 1}}{\sec ^{ – 2}}$
$M.K.S.$ unit $: kg.\ m^{-1}\ sec^{-2} .$
${n_2} = {n_1}{\left[ {\frac{{{M_1}}}{{{M_2}}}} \right]^1}{\left[ {\frac{{{L_1}}}{{{L_2}}}} \right]^{ – 1}}{\left[ {\frac{{{T_1}}}{{{T_2}}}} \right]^{ – 2}}$ $ = {\left[ {\frac{{gm}}{{kg}}} \right]^1}{\left[ {\frac{{cm}}{{meter}}} \right]^{ – 1}}{\left[ {\frac{{\sec }}{{\sec }}} \right]^{ – 2}}$
$\therefore \frac{{{n_2}}}{{{n_1}}} = {\left[ {\frac{{gm}}{{{{10}^3}gm}}} \right]^1}{\left[ {\frac{{cm}}{{{{10}^2}cm}}} \right]^{ – 1}}{\left[ {\frac{{\sec }}{{\sec }}} \right]^{ – 2}}$ $ = \frac{1}{{10}} = 0.1$
Standard 11
Physics
