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(B) Given the relations:$v_2 = v_1 \frac{\alpha^2}{\beta} \Rightarrow [L_2 T_2^{-1}] = [L_1 T_1^{-1}] \frac{\alpha^2}{\beta} \dots (i)$$a_2 = a_1 \alp

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$M_2 = \frac{1}{\alpha^2 \beta^2} M_1, L_2 = \frac{\alpha^3}{\beta^3} L_1, T_2 = T_1 \frac{\alpha}{\beta^2}$

Vedclass · Answer

(B) Given the relations:$v_2 = v_1 \frac{\alpha^2}{\beta} \Rightarrow [L_2 T_2^{-1}] = [L_1 T_1^{-1}] \frac{\alpha^2}{\beta} \dots (i)$$a_2 = a_1 \alpha \beta \Rightarrow [L_2 T_2^{-2}] = [L_1 T_1^{-2}] \alpha \beta \dots (ii)$$F_2 = \frac{F_1}{\alpha \beta} \Rightarrow [M_2 L_2 T_2^{-2}] = [M_1 L_1 T_1^{-2}] \frac{1}{\alpha \beta} \dots (iii)$Dividing equation $(iii)$ by equation $(ii)$:$\frac{[M_2 L_2 T_2^{-2}]}{[L_2 T_2^{-2}]} = \frac{[M_1 L_1 T_1^{-2}]}{[L_1 T_1^{-2}]} 	imes \frac{1}{(\alpha \beta)(\alpha \beta)} \Rightarrow M_2 = \frac{M_1}{\alpha^2 \beta^2}$Dividing equation $(ii)$ by equation $(i)$:$\frac{[L_2 T_2^{-2}]}{[L_2 T_2^{-1}]} = \frac{[L_1 T_1^{-2}]}{[L_1 T_1^{-1}]} 	imes \frac{\alpha \beta}{\alpha^2 / \beta} \Rightarrow T_2^{-1} = T_1^{-1} \frac{\beta^2}{\alpha} \Rightarrow T_2 = T_1 \frac{\alpha}{\beta^2}$Substituting $T_2$ into equation $(i)$:$L_2 (T_1 \frac{\alpha}{\beta^2})^{-1} = L_1 T_1^{-1} \frac{\alpha^2}{\beta} \Rightarrow L_2 = L_1 \frac{\alpha^2}{\beta} \cdot \frac{\alpha}{\beta^2} = L_1 \frac{\alpha^3}{\beta^3}$Thus,$M_2 = \frac{1}{\alpha^2 \beta^2} M_1, L_2 = \frac{\alpha^3}{\beta^3} L_1, T_2 = T_1 \frac{\alpha}{\beta^2}$.

If the relationship between velocity,acceleration,and force in two systems is given by $v_2 = \frac{\alpha^2}{\beta} v_1$,$a_2 = \alpha \beta a_1$,and $F_2 = \frac{F_1}{\alpha \beta}$,then what is the relationship between mass,length,and time?

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