Find Numerical Value of Density in New Unit System

 

❓ Question

In the SI unit system, density is

$$d_1=128\ \text{kg/m}^3$$

In another unit system, the unit of length is $25\text{ cm}$ and the unit of mass is $50\text{ g}$.

Find the numerical value of density in the new unit system.

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✍️ Solution

Density is given by

$$d=\frac{M}{V}$$

Therefore, the dimensional formula of density is

$$[d]=[ML^{-3}]$$

For conversion of numerical values,

$$n_1u_1=n_2u_2$$

In SI system,

$$n_1=128$$
$$u_1=\frac{1\text{ kg}}{(1\text{ m})^3}$$

In the new unit system,

$$1\text{ mass unit}=50\text{ g}$$
$$1\text{ length unit}=25\text{ cm}$$

Hence,

$$u_2=\frac{50\text{ g}}{(25\text{ cm})^3}$$

Now,

$$128\left(\frac{1\text{ kg}}{1\text{ m}^3}\right) = n_2\left(\frac{50\text{ g}}{(25\text{ cm})^3}\right)$$

Using

$$1\text{ kg}=1000\text{ g}$$

and

$$1\text{ m}=100\text{ cm}$$

we get

$$128\left(\frac{1000}{100^3}\right) = n_2\left(\frac{50}{25^3}\right)$$

Therefore,

$$n_2 = \frac{128\times1000\times25^3} {100^3\times50}$$
$$n_2=40$$

✅ Final Answer

$$\boxed{40}$$