Question

Using screw gauge radius of wire was found to be \[2.50\,mm\] . The length of wire found by mm Scale is \[50.0\,cm\] . If the mass of the wire was measured as \[25\,gm\], the density of the wire is correct S.F. will be \[\left( {use\,\pi = 3.14\,exactly} \right)\] in \[gm/c{m^3}\] .
A. \[2.5\]
B. \[2.50\]
C. \[25\]
D. $0.25$

Answer 1

Hint: As we know that density is defined as mass per unit volume. It is denoted by\[\rho \]. Density plays an important role in calculating mass and volume.
\[V = m\rho \]
Here, \[V\] is volume,\[m\]is the mass and\[\rho \]the density.
\[\rho \]= \[\dfrac{m}{V}\] ------- (1)
Volume of the wire can be given by- \[V\]=\[\pi {r^2}l\]
By substituting the volume and mass in equation (1), we can easily determine the value of density of the wire.Density is a scalar quantity because it does not show any direction.

Complete step by step answer:
Given that, radius of wire = \[2.50\]\[mm\]= \[2.50\]\[cm\]
Length of wire =\[50.0\]\[cm\]
Mass of wire = \[25\]\[gm\]
Given, \[\pi \]=\[3.14\]
As we know that
\[V = m\rho \]
\[\Rightarrow\rho \]= \[\dfrac{m}{V}\]
\[\Rightarrow\rho \]= \[\dfrac{{25}}{{\pi {r^2}l}}\]
\[\Rightarrow\rho \]= \[\dfrac{{25}}{{3.14 \times {{\left( {0.25} \right)}^2} \times 50.0}}\]
$\therefore\rho = 2.5 \,gm/c{m^3}$

Hence, option A is correct.

Additional Information:
Volume is the amount of space occupied by a substance, while mass is the amount of matter.
Mass is physical value while volume is geometrical value. The SI unit of mass is \[kg\]and the SI unit of volume is \[{m^3}\].

Note:Volume plays an important role in water conservation, and also when we fill up our vehicles at pumps. Density shows how much a substance occupies volume. The most common example of density is water and oil, when we mix oil and water. Oil is less dense than water so it floats. And a boat also floats on water because the density of the boat is less than water.