Question

The density of a material in SI units is $128kg{m}^{-3}$. In certain units in which the unit of length is $25cm$ and the unit of mass is $50g$, the numerical value of density of the material is :

• A. $40$
• B. $410$
• C. $640$
• D. $16$

Answer 1

The correct option is A

$40$

Step 1: Given data :

The density of a material in SI units is $\delta =128kg{m}^{-3}$

The unit of length is $L=25cm$

The unit of mass $M=50g$

Step 2: Formula used :

Note that the formula of density,

$Density\left(\delta \right)=Mass\left(M\right)/Volume\left(V\right)$

Now,

$Density\left(\delta \right)=\frac{M}{V}=\frac{M}{L^3}$

Know that, $n_1{u}_1=n_2{u}_2$

Where, $n_1=128$

Step 3: Calculate the value of $n_2$

Now solve,

$$\begin{gathered} \Rightarrow 128\left[M_1{L_1}^{-3}\right]=n_2\left[M_2{L_2}^{-3}\right] \\ \Rightarrow n_2=128\left[M_1{L_1}^{-3}\right]/\left[M_2{L_2}^{-3}\right] \end{gathered}$$

Know that from the question, substitute the values,

$$\begin{gathered} m=\frac{50}{1000}kg \\ L=\frac{25}{100}m \end{gathered}$$

Therefore,

$$\begin{gathered} \Rightarrow n_2=128\left[M_1/M_2\right]{\left[L_2/L_1\right]}^3 \\ \Rightarrow n_2=128\times \frac{1}{{\displaystyle \frac{50}{1000}}}\times {\left(\frac{{\displaystyle \frac{25}{100}}}{1}\right)}^3 \\ \Rightarrow n_2=128\times 20\times 1/64 \\ \Rightarrow n_2=40 \end{gathered}$$

Therefore the numerical value of density of the material is $40$.

Hence, the correct option is (A).