Question

On a part-time job, you are asked to bring a cylindrical iron rod of length 85.8 cm and diameter 2.85 cm from a storage room to a machinist. Will you need a cart? (To answer, calculate the weight of the rod.)

Short answer

No, you will not need a cart.

Step-by-step solution

Identification of given data

• The length of the iron rod is,$L=855\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.855\mathrm{m}$
• The diameter of the iron rod is,$D=2.85\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.0285$
• The radius of the iron rod is,$r=\frac{D}{2}=1.425\times {10}^{-2}\mathrm{m}$
• The density of the iron rod is,$p=7.8\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$

Significance of density

The density of a material is defined as the mass per unit volume of the substance.

It can be represented by the given equation,

$p=\frac{m}{V}$……………(1)

Where, $m$are$V$ the mass and volume of the object.

Determination of the mass of the iron rod

The volume of the iron rod can be evaluated as,

$$\begin{gathered} v={\mathrm{\pi r}}^2\mathrm{L} \\ =\mathrm{\pi }{\left(1.425\times {10}^{-2}\mathrm{m}\right)}^2\left(0.855\mathrm{m}\right) \\ =5.45\times {10}^{-4}{\mathrm{m}}^3 \end{gathered}$$

Re-arranging equation (1) gives,

$m=pV$…………(2)

Substitute the given values in equation (2),

$$\begin{gathered} m=\left(7.9\times {10}^{3}kg/m^3\right)\times \left(5.45\times {10}^{-4}{m}^3\right) \\ =4.25kg \end{gathered}$$

Thus, the mass of the iron rod is nearly fifteen times smaller as compared to the mass of a normal person, so there is no need for a cart.