If the present units of length,time,and mass | vedclass

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(B) Let the original units be $L_1 = 1 \; m$,$T_1 = 1 \; s$,$M_1 = 1 \; kg$. The new units are $L_2 = 100 \; m$,$T_2 = 100 \; s$,$M_2 = 0.1 \; kg$.$1$

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The new unit of force is decreased $\frac{1}{1000}$ times

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(B) Let the original units be $L_1 = 1 \; m$,$T_1 = 1 \; s$,$M_1 = 1 \; kg$. The new units are $L_2 = 100 \; m$,$T_2 = 100 \; s$,$M_2 = 0.1 \; kg$.$1$. Velocity $(v = L T^{-1})$: New unit $v_2 = L_2 T_2^{-1} = 100 \; m / 100 \; s = 1 \; m/s$. Thus,velocity remains the same.$2$. Force $(F = M L T^{-2})$: New unit $F_2 = M_2 L_2 T_2^{-2} = (0.1 \; kg) 	imes (100 \; m) / (100 \; s)^2 = 0.1 	imes 100 / 10000 = 0.001 \; N = \frac{1}{1000} \; N$. The new unit of force is $\frac{1}{1000}$ times the original.$3$. Energy $(E = M L^2 T^{-2})$: New unit $E_2 = M_2 L_2^2 T_2^{-2} = (0.1 \; kg) 	imes (100 \; m)^2 / (100 \; s)^2 = 0.1 	imes 10000 / 10000 = 0.1 \; J = \frac{1}{10} \; J$.$4$. Pressure $(P = M L^{-1} T^{-2})$: New unit $P_2 = M_2 L_2^{-1} T_2^{-2} = (0.1 \; kg) / (100 \; m 	imes (100 \; s)^2) = 0.1 / (100 	imes 10000) = 0.1 / 10^6 = 10^{-7} \; Pa$.

If the present units of length,time,and mass $(m, s, kg)$ are changed to $100 \; m, 100 \; s, 0.1 \; kg$,then:

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