In a new system,the unit of mass is 10 kg,th | vedclass

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(C) The given density is $\rho_1 = 20 \ g/cm^3$.In the new system,the unit of mass $M_2 = 10 \ kg = 10^4 \ g$ and the unit of length $L_2 = 5 \ m = 50

Vedclass · Accepted Answer

$2.5 	imes 10^5$ new units

Vedclass · Answer

(C) The given density is $ho_1 = 20 \ g/cm^3$.In the new system,the unit of mass $M_2 = 10 \ kg = 10^4 \ g$ and the unit of length $L_2 = 5 \ m = 500 \ cm$.The density in the new system is $ho_2 = N \frac{M_2}{L_2^3} = N \frac{10^4 \ g}{(500 \ cm)^3}$.Since the physical quantity remains the same,$ho_1 = ho_2$.$20 \ g/cm^3 = N \frac{10^4 \ g}{125 	imes 10^6 \ cm^3}$.$N = \frac{20 	imes 125 	imes 10^6}{10^4} = 20 	imes 125 	imes 10^2 = 2500 	imes 10^2 = 2.5 	imes 10^5$ new units.

List-$I$	List-$II$
$(i)$ Curie	$(A)$ $MLT^{-2}$
$(ii)$ Light year	$(B)$ $M$
$(iii)$ Dielectric strength	$(C)$ Dimensionless
$(iv)$ Atomic weight	$(D)$ $T$
$(v)$ Decibel	$(E)$ $ML^2T^{-2}$
	$(F)$ $MT^{-3}$
	$(G)$ $T^{-1}$
	$(H)$ $L$
	$(I)$ $MLT^{-3}I^{-1}$
	$(J)$ $LT^{-1}$

In a new system,the unit of mass is $10 \ kg$,the unit of length is $5 \ m$,and the unit of time is $10 \ s$. Then,in this new system,$20 \ g/cm^3$ is equal to:

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