© All Rights reserved @ LearnWithDash
Step 1: Identify Given Data
• Density in SI units (numerical value): $128$
• SI unit for density: $kg\,m^{-3}$
• New unit of length: $25\,\text{cm}$
• New unit of mass: $50\,\text{g}$
Step 2: Express SI Density in g/cm³
1 kg = 1000 g, and 1 m = 100 cm, so
1 m³ = (100 cm)³ = $10^6$ cm³.
Thus,
$128\, kg\, m^{-3} = 128 \times \dfrac{1000\, g}{10^6\, cm^3} = 0.128\, g\, cm^{-3}.$
Step 3: Express Density in the New Units
In the new system:
• 1 new unit of mass = $50\,g.$
• 1 new unit of length = $25\,cm,$ so 1 new unit of volume = $(25\,cm)^3 = 25^3\,cm^3 = 15625\,cm^3.$
Hence, “1 new unit of density” = $\dfrac{50\,g}{(25\,cm)^3} = \dfrac{50\,g}{15625\,cm^3} = \dfrac{50}{15625}\, g\, cm^{-3}.$
Step 4: Equate to Find the Numerical Value in New Units
Let $n_2$ be the numerical value of the density in the new system. Then,
$$
0.128\, g\, cm^{-3} = n_2 \times \dfrac{50}{15625}\, g\, cm^{-3}.
$$
Solving for $n_2:$
$$
n_2 = 0.128 \times \dfrac{15625}{50} = 0.128 \times 312.5 = 40.
$$
Answer
The numerical value of the density in the new unit system is $40$.
