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:

  • A
    $2.5 \times 10^2$ new units
  • B
    $2.5 \times 10^4$ new units
  • C
    $2.5 \times 10^5$ new units
  • D
    $2.5 \times 10^3$ new units

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Similar Questions

Which physical quantities have the same dimension?

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

Match the List-$I$ with List-$II$. The correct match in the following is:
List-$I$ List-$II$
$A$. Boltzmann constant $I$. $[ML^0T^0]$
$B$. Coefficient of viscosity $II$. $[ML^{-1}T^{-1}]$
$C$. Water equivalent $III$. $[ML^2T^{-2}K^{-1}]$
$D$. Coefficient of thermal conductivity $IV$. $[MLT^{-3}K^{-1}]$

Match the following (Take the relative strength of the strongest fundamental forces in nature as $1$):
$I$ (Fundamental forces in nature)$II$ (Relative strength)
$(A)$ Strong nuclear force$(e)$ $10^{-2}$
$(B)$ Weak nuclear force$(f)$ $1$
$(C)$ Electromagnetic force$(g)$ $10^{10}$
$(D)$ Gravitational force$(h)$ $10^{-13}$
$(i)$ $10^{-39}$

The correct match is:

Some physical quantities are given in Column $I$ and some possible $SI$ units in which these quantities may be expressed are given in Column $II$. Match the physical quantities in Column $I$ with the units in Column $II$.
Column $I$ Column $II$
$(A)$ $GM_e M_s$ ($G$: universal gravitational constant,$M_e$: mass of the earth,$M_s$: mass of the Sun) $(p)$ $(\text{volt})(\text{coulomb})(\text{metre})$
$(B)$ $\frac{3RT}{M}$ ($R$: universal gas constant,$T$: absolute temperature,$M$: molar mass) $(q)$ $(\text{kg})(\text{m})^3(\text{s})^{-2}$
$(C)$ $\frac{F^2}{q^2 B^2}$ ($F$: force,$q$: charge,$B$: magnetic field) $(r)$ $(\text{m})^2(\text{s})^{-2}$
$(D)$ $\frac{GM_e}{R_e}$ ($G$: universal gravitational constant,$M_e$: mass of the earth,$R_e$: radius of the earth) $(s)$ $(\text{farad})(\text{volt})^2(\text{kg})^{-1}$

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