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બે પદ્વિતમાં વેગ,પ્રવેગ અને બળ વચ્ચેનો સંબંધ ${v_2} = \frac{{{\alpha ^2}}}{\beta }{v_1},$ ${a_2} = \alpha \beta {a_1}$ અને ${F_2} = \frac{{{F_1}}}{{\alpha \beta }}.$ હોય,તો દળ, લંબાઇ અને સમય વચ્ચેનો સંબંધ
${M_2} = \frac{\alpha }{\beta }{M_1},{L_2} = \frac{{{\alpha ^2}}}{{{\beta ^2}}}{L_1},{T_2} = \frac{{{\alpha ^3}{T_1}}}{\beta }$
${M_2} = \frac{1}{{{\alpha ^2}{\beta ^2}}}{M_1},{L_2} = \frac{{{\alpha ^3}}}{{{\beta ^3}}}{L_1},{T_2} = {T_1}\frac{\alpha }{{{\beta ^2}}}$
${M_2} = \frac{{{\alpha ^3}}}{{{\beta ^3}}}{M_1},{L_2} = \frac{{{\alpha ^2}}}{{{\beta ^2}}}{L_1},{T_2} = \frac{\alpha }{\beta }{T_1}$
${M_2} = \frac{{{\alpha ^2}}}{{{\beta ^2}}}{M_1},{L_2} = \frac{\alpha }{{{\beta ^2}}}{L_1},{T_2} = \frac{{{\alpha ^3}}}{{{\beta ^3}}}{T_1}$
Solution
${v_2} = {v_1}\frac{{{\alpha ^2}}}{\beta }$ $ \Rightarrow \,\,[{L_2}T_2^{ – 1}]\, = \,[{L_1}T_1^{ – 1}]\,\frac{{{\alpha ^2}}}{\beta }$……$(i)$
${a_2} = {a_1}\alpha \beta $ $ \Rightarrow \,\,\,[{L_2}T_2^{ – 2}]\, = [{L_1}T_1^{ – 2}]\,\alpha \beta $……$(ii)$
${F_2} = \frac{{{F_1}}}{{\alpha \beta }}$ $ \Rightarrow \,\,\,[{M_2}{L_2}T_2^{ – 2}]\, = \,[{M_1}{L_1}T_1^{ – 2}]\, \times \frac{1}{{\alpha \beta }}$……$(iii)$
Dividing equation $(iii)$\ by equation $(ii)$ we get ${M_2} = \frac{{{M_1}}}{{(\alpha \beta )\,\alpha \beta }}$ $ = \frac{{{M_1}}}{{{\alpha ^2}{B^2}}}$
Squaring equation $(i)$ and dividing by equation $(ii)$ we get ${L_2} = {L_1}\frac{{{\alpha ^3}}}{{{\beta ^3}}}$
Dividing equation $(i)$ by equation $(ii)$ we get ${T_2} = {T_1}\frac{\alpha }{{{\beta ^2}}}$
