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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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.

Column-$I$	Column-$II$
$(a)$ Ratio of change in time period of a simple pendulum with temperature to its original time period	$(i) \, \alpha \Delta T$
$(b)$ Ratio of the value of a length to its scale reading	$(ii) \, T$
$(c)$ Reciprocal of coefficient of volume expansion for an ideal gas of constant pressure	$(iii) \, (1 + \alpha \Delta T)$
$(d) \, \frac{F}{YA} =$	$(iv) \, \frac{1}{2} \alpha \Delta T$

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:

Similar Questions

Out of the fundamental forces in nature,which forces have the maximum and minimum range,respectively?

The unit $J \cdot s^2$ corresponds to which physical quantity?

Find the rate of change of the volume of a sphere with respect to its surface area when the radius is $2 \ cm$.

The unit of "impulse per unit area" is the same as that of

Vedclass Test Series

Exam Paper Generator

Online Exam Module